Introductory Reaction Mechanism

GOC-III(Introductory Reaction Mechanism) Introductory Mechanisms in Organic Reactions A reaction mechanism is a step by step narration of the bond-breaking and bond-making processes that occur when reagents react to form the products. The proof of the mechanism, that the reaction is really going via this route, not via the other route, has been established by many experimental corroborations. Let us study the mechanisms of several organic reactions. Substitution Reactions: Substitution reactions are of three types

(a) Free Radical Substitution (SF)

(b) Nucleophilic Substituion(SN)

Free Radical Substitution(SF)

Usually alkanes or alkenes undergo free radical substitution at the sp3 hybrid carbon, where a stable carbon free radical is formed. For introductory purpose we take halogenation of alkane in presence

of heat and light for explaining SF mechanism. light/heat R H + X2 R X + HX

A H atom in alkane is substituted by a halogen atom (X) when alkane reacts with halogen in presence of uv/visible light or heat or both. This reaction proceeds with the participation of free radicals(not any

ionic species) and hence its name SF. A free radical mechanism proceeds via the following three steps (i) Initiation (ii) Propagation (iii) Termination

Alkane reacts with halogen like Cl2 and Br2(written as X2) in the vapour phase in presence of light or heat to produce haloalkane(alkyl halide) This reaction goes via the above three steps. Let us discuss one by one. Initiation: light/heat XX 2 X

The reagent X2 undergoes homolytic cleavage of the bond in presence of light or heat or both to produce two halogen free radicals(better call them atoms). These are primary free radicals which will take part in the propagation step. Note that X–X(halogen-halogen) bond is weaker than C–C and C–H bonds and hence light/heat preferentially breaks X–X bond. Propagation:

R H ++ X R HX (i) alkyl radical

+ R X X R X + X (ii) alkyl halide The halogen atom abstracts a H atom from alkane(R–H) to produce alkyl free readical(its a carbon free radical) and HX. If there are more than one non-eqivalent –H atoms in a molecule, more than one alkyl free radicals will be formed with the order 30 > 20 > 10 (according to stability of alkyl free radical)

This radical again reacts with a diatomic halogen molecule(X2) to abstract a X atoms and produce the product R–X(alkyl halide) and a halogen atom. This halogen atom attacks another neutral alkyl halide and repeat the cyclic process again and again till the reactions stops. In each propagation step, a radical(or Dr. S. S. Tripathy 1 GOC-III(Introductory Reaction Mechanism) atom) is consumed but another radical(or atom) is produced and thus the chain propagation continues reapeating the above two sequential cyclic process. These two steps together belong to the propagation. Note that in step (i) of propagation, the halogen atom does not abstract a alkyl radical to produce the product R–X and a hydrogen atom. This is an unfavourable process and its ΔG0 is hugely positve. In stead the step that happens shown above invoves lowering of free energy.

R H + X R X + H (does not take place)

Other minor termination processes also occurs, which explain the formation of dihalo and polyhaloalkanes alongwith monohaloalkane(major prouduct) when one mole of halogen reacts with one mole of alkane. A specific case of chloriantion of methane is given below.

CH2 Cl + Cl CH2 Cl + HCl

CH2 Cl + Cl Cl CH2Cl2 + Cl

We can write two more chain propagating pairs to show the formation of CHCl3 and CCl4.

Termination: When propagation is complete, then the chain reaction is broken or terminated. Free radicals and atoms are not stable. There will be some residual radicals and atoms of every kind at the end of propagation steps. All these radicals present in the will undergo mutual combination with each other and several byeproducts are formed during.

R + R RR R + X RX

X + X XX The combination of two alkyl radicals give a higher alkane(R–R) having double the number of carbon atoms. Sometimes disproportionation reaction occurs if the carbon free radical has at least two carbon atoms to form an alkene and alkane. See this example.

CH CH + CH CH CH2 CH2 + CH3 CH3 2 2 3 2 ethane ethane

So ethane, on halogenation gives ethane and ethene, alongwith butane(C2H5–C2H5), ethyl halide(Et–X) and X2 as the small byeproducts formed in the termination steps. Is it not interesting ??!! Unless we know the mechanism of this reaction, can we rationalise the formation of these unwanted byeproducts while we halogenate an alkane ? Exampe

Br Br light/heat 2 CH3 CH2 CH3 ++ Br2 CH3 CH CH3 CH3 CH2 CH2 + 2HBr major minor Propane reacts with bromine in presence of heat and light to produce 2-bromopropane as the major product(>97%) and 1-bromopropane as the minor product(<3%). Let us see how it happens,?

2 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

h Initiation: Br Br 2 Br Propagation:

CH3 CH CH3 + Br CH3 CH CH3 + HBr 20 stable free radical

Br CH CH CH Br Br 3 3 + CH3 CH CH3 + Br In this case, there are two types of non-equivalent H atoms : one in the primary carbon atom(two terminal H atoms) and in the other seconndary carbon atom(middle carbon). The bromine atom formed in the initiation step will preferntially abstract a H atom from sec- position to produce a more stable 20 free radical. You already know from GOC-I, that the stability order of alkyl free radicals follows the order 30 > 20 > 10 which is explained by both hyperconjugation and +I effects. Hence isopropyl free radical is formed as major carbon free radical(>97%) while n-propyl radical is formed as minor(less than 3%).

CH3 CH2 CH2 ++ Br CH3 CH2 CH2 HBr n-propyl radical(minor) (unstable) Termination:

CH3 CH + CH CH3 CH3 CH CH CH3 CH CH3 CH3 3 CH3

CH3 CH Br CH3 CH Br

CH3 CH3

Br + Br Br2

CH2 CH + CH CH3 CH2 CH CH3 + CH3 CH2 CH3 propene propane CH3 CH3 (disproportionation) So four byeproducts are formed in termination, out of which the two namely isopropyl bromide is already

formed as the major product and Br2 was taken as reactant. But most interesting byeproducts are 2,3- dimethylbutane formed by mutual combination of isopropyl radicals and a mixture of propane and propene formed by the disproportionation reaction. N.B: Here we have ignored the primary free radical i.e n-propyl free radical that would form in less than 3% extent. That willl form, besides n-propyl bromide as the minor haloalkane, insignificant amount of the byeproducts from termination steps. We have not included those equations here. I suggest you to write those in your rough copy for the sake of practice. The byeproducts will be n-hexane, propane, propene, the latter two are already there. There can be also combination between isopropyl and n-propyl radicals to form 2-methylpentane. Got it? N.B: More on halogenation of alkanes, their reactivity vs. selectivities will be disucssed in the chapter

‘alkanes’. This is introduction to SF reaction mechanism. NO RADICAL REARRAGEMENT:: Unlike carbocations, which are prone to rearrangement from a less stable to more one, radicals do not undergo rearrangement of such type. This has been experimentally Dr.proved. S. S. Tripathy 3 GOC-III(Introductory Reaction Mechanism) N:B: Free radical substitution of alkane is an example of CHAIN REACTION, a reaction which involves a series of steps(initiation and propagation), each of which generates a reactive species(in this case free radicals) that brings about the next step. In chain reaction, the quantum yield(QY) is very high. One single photon of light(uv/visible) brings about the formation of several thousand product molecules. If QY for a reaction is 10,000, then 10,000 product molecules are formed from merely one photon of light. One photon dissociates one halogen molecule(X2) and each halogen atom(X) starts a chain consisting of 5000 repetitions of propagating cycles before it stops. Graph:

The above graph shows variation of potential energy in the two propagation steps in chlorination of methane. The net energy evolved is 104.6 kJ/mole(–104.6 kJ/mole). The first step of converting methane and chlorine atom to methyl radical and HCl is endothermic absorbing 8.4 kJ/mole but second step is highly exothermic evolving 112.9 kJ/mole(–112.9 kJ/mole). Inititation step is endothermic(BDE) of 241.3 kJ/mole. Since the QY is very high, for the endothermic dissociation of one mole mole of Cl2, several thousands of methane and other chlorine molecules react to form several thousands of product molecules. Hence its a hugely exothermic process in totallity.

Nucleophilic Substititution(SN) Reaction:

Since nucloephile is involved in bringing about the substitution, it is called SN reaction. Usually alkyl halides( R–X) and other compound bearing good leaving groups like X–(halide ions) undergo substitution by this mechanim. Principle: A strong base acting as a nucleophile can displace a weak base acting as leaving group. See the forllowing examples.

CH3 Cl + OH CH3 OH + Cl stronger base weaker base (nucleophile) (leaving group – – Since OH is stronger base than Cl (as HCl is a stronger acid than H2O: see BL theory given elsewhere), the above SN reaction is feasible. A stronger base OH– acting as nucelophile can displace a weaker base Cl– actiing as leaving group. Note that the reaverse reaction is not feasible, as a weaker base Cl– cannot displace a stronger base OH– .

HC C H + CH3 MgBr CH4 + HC C MgBr stronger base weaker base

– A stronger base (CH3 from CH3MgBr) can displace a weaker base HC C . Conversely, we can say that a stronger acid(acetylene) can displace a weaker acid (CH4).

4 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

OH O

++H CO (Not feasible) HCO3 2 3 weake base stronger base

Bicabonate ion cannot produce H2CO3(carbon dioxide) from phenol. Because bicarbonate ion is weaker base than phenoxide ion since carbonic acid is stronger than phenol. We can also say that a weaker acid phoenol cannot displace a stronger acid carbonic acid.

CH3COO H + HCO3 CH3COO + H2CO3 (feasible) stronger base weaker base

– – A stronger base HCO3 can displace a weaker base CH3COO , as CH3COOH is a stronger acid than

H2CO3. Conversely, a strogner acid CH3COOH can displace a weaker acid H2CO3.

Types of SN reactions in organic chemistry ′ ′ ′ (a) SN1 (b) SN2 (c) SNi (d) SN1 (e) SN 2 (f) SNi

(g) SN(acyl) (h) ArSN

SN Reactions: Nucleophilic substitution in aliphatic halides or other acyclic(aliphatic) substrates 3 having a good leaving group attached to a sp hybrid carbon come under SN category.

OSO2 CH3 Sustrates: R–X, R–OTs, ROH2 etc. ( –OTS : (p-toluene sulfonate)

Alkyl halides(except alkyl fluoride), Alkyl toluene sulphonate(R–OTs) and protonated alcohol( ROH2 )

– are good substrates for SN reactions in aliphatic compounds as the leaving groups are respectively X – (halide ion), TsO (tosylate ion) and H2O, which are all weak bases. We know before that weak bases are good leaving groups and strong bases are good nucleophiles.

So for all SN reactions, leavability of the leaving group in the substrate is important. Let us discuss about

it first, before going to the specific SN mechanism one by one. Leavability or Fugicity: Leavability of leaving group is inversely related to its base strength. Weaker the base, better is

the leavability of the group. Greater is the leavability the SN rate is greater. – – – – The leaving group can be anionic like Cl , Br , I , TsO or it can be neutral like H2O, ROH, NH3 etc.

(a) OH– is a strong base and hence is a poor leaving group. Hence neutral alcohol R–OH cannot

undergo S reaction. However, protonated alochol ( ) does undergo substitution reaction as the N ROH2

leaving group is neutral H2O which is weak base and a good leaving group.

R OH + Nu R Nu + OH (NOT FAVOURABLE) strong base

H RO ++Nu R Nu H2O (FAVOURABLE H weak base

So alochol in presence of mineral acids act as good substrate for SN reactions. (b) Cl–, Br–, I– are weaker bases being the conjugates of strong acids, and hence are good leaving

Dr. S. S. Tripathy 5 GOC-III(Introductory Reaction Mechanism) groups. Leavability order: I– > Br– > Cl– >>>>> F– Basicity order I– < Br– < Cl– <<<<< F– Since HF is really a weak acid, its conjugate base F– is very strong and hence is not a good leaving group.

So alkyl fluorides(R–F) are not used as substrates for SN reactions. Other alkyl halides are good substrates.

Relative order of SN reaction rate RI > RBr > RCl

– – Since I is the best leaving group R–I reacts fastest in a SN reaction and Cl is least powerful leaving group among the three, hence R–Cl reacts slowest. R–Br reacts with intermediate rate as the leavability of Br– is intermediate among the three. (c) Triflate, mesylate, tosylate groups:

O O O

CF S O CH3 S O 3 H3CSO O > O > O trifluoromethanesulfonate (p-toluenesulfonate: methanesulfonate: triflate ion) tosylate ion) mesylate ion

All the three ions namely TsO–, TfO– and MsO–, the negative charge is highly delocalised due to electronic effects(M-effect) and hence are poor bases and poor nucleophiles. Moreover they are the conjugates of strong sulfonic acids. They are hence very good leaving groups in the above order. Halide ions come next to them. Moreover these ions are very bulky. To relieve the steric strain in the molecule, its leavability is further enhenced. Hence when such groups are there in the substrate the SN rate is must faster.

R OTs + Nu R Nu + OTs ( FAVOURABLE) weak and bulky base In fact, alochol(R–OH) is first converted to R–X or R-OTs or R-OTf etc by by suitable reaction and then taken as substrate for any SN reaction. Now let us take individual SN mechanisms.

SN1 Mechanism(Unimolecular SN) (i) This is the predominant mechanism for tert-(30) halides(substrates) as 30 carbocations are extremely stable (ii) The mechanism involves two steps. (a) In the first step, heterolytic cleavage of the C–X bond occurs to form a very stable carbocation. In this step the leaving group (X–) is expelled out. This step is slow and hence rate determining. We shall know later in our ‘Kinetics’ chapter that in multi-step reactions, the slowest step is rate determining. (b) In the second step, the carbocation reacts with the nucleophile(Nu–) to form the product. This step is fast. step-I step-II

R' R' R' R' slow fast RC X RC + X RC+ Nu RCNu R'' R'' R'' R'' most stable 30 carbocation Rate = k [substrate]1

6 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

The rate of the reaction is only dependent on the concentration of the substrate and not on nucleophile as rate is determined from the slow step, which does not involve nucleophile. The order of the reaction

is 1(power of the substrate concentraion) and hence it is called SN1. If the concenration of nucleophile is increased or decreased, the rate won’t change while if the concentration of the substrate is increased or decreased, accordingly rate will be increased or decreased as shown in the above rate law.

Example:

+ (CH3)3C OH + HBr (CH3)3C Br H2O ( S 1 mechanism) tert- halide nucleophile product N

Tert-butyl bromide reacts with water to form tert-butyl alochol in SN1 mechanism. Here H2O acts as neutral nucleophile. Let us see the steps.

CH3 CH3 CH3 H slow OH2 fast CH3 C Br CH3 C CH3 C O (CH3)3COH fast -H -Br CH H CH3 3 CH3

When the nucleophile is neutral, one more step is involved i.e deprotonation. H2O, as a nucleophile reacts with the 30 carbocation in the 2nd step to form protonated alcohol first which readily deprotonates to form neutral alcohol. Had the nucleophile been anionic like OH–, CN–, then the neutral product would form in 2nd step, as shown before.

Stereochemical Consequence:

If the substrate is chiral, then racemisation would be the consequence of SN1 mechanism. Since the carbocation is planar(sp2 hybridised), the nucleophile(Nu–) would attack from either side of the plane with equal probabilities hence racemic product will be formed. d- reactant will give d/l product, also the l-reactant will result d/l product.

p 50% p p p Nu q C X q C q C Nu + Nu C q r r r d or l r 50% d/l pair racemic mixture planar carbocation

Effect of Solvent:

Dr. S. S. Tripathy 7 GOC-III(Introductory Reaction Mechanism) A more polar solvent facilitates polarisation of the C–X bond hence favours the heterolytic cleavage at a faster rate. So greater the polarity, greater is the SN1 rate. H2O and EtOH are often used as the SN1 solvent. Since many organic compounds have poor solubility in water, a mixed solvent of H2O and EtOH is used with minimum amount of EtOH to dissolve the compound. Greater the % of H2O in the mixed solvent greater is the rate, as H2O has greater polarity than EtOH.

SAQ: Arrange in the decreasing order SN1 rate with the following mixed solvents.

(a) 80% H2O + 20% EtOH (b) 50 % H2O + 50% EtOH (c) 20% H2O + 80% EtOH Solution : a > b > c

Between acetone, EtOH, H2O and CCl4, arrange in the decreasing order of SN1 rate.

Solution: H2O > EtOH > acetone > CCl4 0 N.B: For 3 subtrates, acetone or CCl4 are never recommended as solvents, as the mechanism will be surely SN1. Such solvents will merely slow down its rate.

Graph:

Two transition states(TS) are involved in SN1 mechanism. Carbocation formation is represented by a minimum point between two TS maxima points. Although carbocations are highly reactive, they are intermediates formed and have existence for a short period and hence have lower energy than the TS. The first TS is when the leaving group(LG) is further polarized from the carbon atom and the second TS when carbocation forms a partial bond with the nucleophle withh reduced polarity of each.

SN2 Mechanism(Bimolecular SN)

0 (i) 1 (primary) i.e R–CH2–X reacts by this mechanismj. (ii) It involves only one step and there is no intermediate species are formed. (iii) The nucleophile(Nu–) attacks from the rear side of the leaving group and this attack and the departure of the leaving group take place simultaneously via a high energy TS.

(iv) Since the rear side of a primary substrate (RCH2–X) is not sterically hindered, rear side 0 attack can take place. Moreover, since 1 carbocation is unstable, primary substrate cannot react by SN1 0 mechanim. Conversely, due to huge steric hindrance in 3 substrate, rear attack of nucleophile for a SN2 0 mechanism is also not possible in case of a 3 substrate, which has to react by SN1 mechanism.

8 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

R R R - - H C Nu C H Nu + HXC Nu X H H H

As Nu– approaches closer to the substrate, the three bonds flatten more and more and in the TS, all the four atoms namly R–, two H atoms and the carbon atom lie in one plane. while the nucleophile makes a partial bond with the central carbon atoms and the C–X bond is partially broken. Thus a fractional negative charge(–δ) is created at both Nu and X. This is the pentavalent TS. Then the nucleophile enters closer to the carbon atoms and the leaving group departs further and finally when LG is expelled out and Nu gets attached to the carbon atom, the the three other bonds have undergone a permanent inversion from left side to right side. Rate law: This is a one-step reaction. Do not mistake from the above figure that these are two steps. TS is not a step as it is simply a passing stage. Rate = k [Substrate]1 [Nucleophile]1

The SN2 rate depends on both substrate and nucleophile concentration to equal extents. Increase in both

substrate concentration and nucleophile concentration increase the rate. In SN1, nucleophile concentration

does not influence the rate. The order of the reactions 2 (1+1), hence the name SN2. Stereochemical Consequence:

In SN2, there is inversion of configuration occurs. This is often called Walden Inversion. d changes to l while l changes to d. However, in 10 substrate, since chirality is not present, the inversion cannot be verified. However, as we 0 shall see a little later, when we have 2 substrate reacting on SN2 conditions, we can observe complete inversion of configuration. Here we presume that the priority of nucleophile(CIP rule) remains the same as the leaving group vis-a-vis the other three groups. However, if the nucleophile acquires a lower priority as against the leaving group, the configuration can remain the same, despite the inversion of configuration(flipping of the umbrella from left to right).

NUCLEOPHILICITY:

(1) For SN2, nucleophilicity term is as important as leavability. In SN1, it is leavability of LG that

Dr. S. S. Tripathy 9 GOC-III(Introductory Reaction Mechanism) was the sole determinant of the rate. But in SN2, both leavability of LG and the nucleophilicity of the Nu– combinedly determine the rate. Hence the discussion on nucleophilicity is made here in details.

(2) For general purpose, basicity and nucleophilicity go parallel i.e directly related. But there is a fundamental difference between their actions. While basicity pertains to the power of accepting a H+ ion from an acid, nucleophilicity refers to the power of attacking any electrophilic centre. In organic chemistry, the usual electrophilic centre is a carbon atom having a partial positive charge(+δ). Usually the species which can accept a H+ ion efficiently can also bond with an electrophilic carbon centre efficiently. But it is not always the case. Sometimes, a more basic species(H+ wise) is so hard that it has limited tendency to overap with electron deficient orbital of the electrophilic centre. We shall look to this issue now in greater details. (3) In a period, the nucleophilicity parallels basicity among similar kinds of species i.e either all anionic or all neutrals. – – – – – (a) CH3 (R ) > NH2 > OH > F From BL theory you know that the above order is opposite to the acid strength of their conjugate acids i.e CH4, NH3, H2O, HF respectively.

(b) NH3 > H2O > HF In these cases, both bascity and nucleophilicity decrease across a period. (4) A negative ion is more basic and hence more nucleophilic than its conjugate acid acting as nucleophile, because it is a weaker base. – OH >> H2O – NH2 >> NH3 RO– >> ROH RCOO– >> RCOOH

– OH is much stronger base than H2O, hence the former is a strong nucleophile while the latter is a weak nucleophile. Same is the case with other pairs. (5) Down a group, the nucelophilicity runs opposite to their basicity for similar type of species i.e either all anionic or all neutral. This is because of the size factor. Down the group size increases drastically which is not the case in a period, as size decreases across a period smoothly(not drastically). More the size of the species, more is the softness of the base i.e it is a softer base, though in BL sense it is a weaker base. More the softness of the base, more is its polarizability and hence its attacking power is greater. (a) F– < Cl– < Br– < I – (nucleophilicity order) F– is most basic(BL wise) but is the hardest base and hence is the weakest nucelophile. I– is the weakes base(BL wise) but is the softest base and hence it is the strongest nucleophile among the halide ions. N.B: Iodide ion enjoys double benefit being the best leaving group and best nucleophile among halide ions. – – (b) OH < SH (c) NH3 < PH3 – – (d) H2O < H2S (e) RO < RS and so on. Note that while comparing, anion is compared with another anion and a neutral molecule is compared with a neutral molecule. An anion is not compared with a neutral molecule, as the former invariably is a stronger base and hence a stronger nucleophile. Effect of Solvent: (a) A more polar solvent solvates the nucleophile by polar interaction and reduce its mobility i.e nucleophilicity. Hence less polar solvents are suitable for enhencing the SN2 rate. It is just the opposite of SN1 mechanim. Usually acetone is the best solvent for SN2 mechansm, as it is not only weakly polar but also dissoves most of the organic substrates and reagents.Non-polar solvents like CCl4, benzene are avoided as most inorganic reagents are not soluble in them.

(b) Addition of small amount of polar aprotic solvent like DMF or DMSO enchances the SN2 rate many fold. Such solvent molecule has a strong –ve pole but a diffiused or weak +ve pole. Hence the positive ions of the reagent say K+ from KCN get solvated to a large extent and lose the bad tendency of attracting the nucleophile(CN–). Since the nucleophile cannot be solvated strongly by these type of solvents, and their counter ions are jailed by the solvent, the mobilty of Nu– is largely enhanced and hence the SN2 rate. 10 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

O CH3 CH3 S O HC N (DMSO) CH3 CH3 N,N-dimethylformamide(DMF) Dimethylsulfoxide(DM SO

20 Substrate:

0 0 We know why 3 substrate cannot react by SN2 mechanism and why 1 substrate react by SN1 mechanism. The answers respectively are the huge steric crowding will not allow rear nucleophilic attack for 30 substrate and 10 carbocation is unstable. But what about 20 substrate for which the steric crowding is not severe like 30 substrate and moreover 20 carbocation is moderately stable.

0 So 2 substrate react by a mixed mechanism, i.e both SN1 ande SN2. Some react by SN1 because 0 2 carbocation is stable and some other molecules react by SN2 because there is less severe steric hindrance for the rear attack of the nucleophile. However, for 20 substrate, we can manipulate the nature of nucleophile and solvent in such a manner that

the mechanism can be exclusively SN1 or exclusively SN2. (1) Using a polar protic solvent like water or alcohol and carrying out solvolysis reaction with

them in which the H2O/ROH etc acting both as solvent and nucleophile, the mechansim will tilt in

complete favour of SN1. Polar solvent will greatly favour carbocation formation and at the same time

such solvent molecule like H2O, ROH are very weak bases; are weak nucleophiles, which will not favour

SN2.

R' R' H2O RCH X RCHOH + HX (SN1 : 100%) (the steps not shown)

The above hydrolysis reaction takes place by 100% SN1 pathway. In this case, a pure chiral substrate

shows gives a racemic product, as SN1 mechanism involves racemisation.

Me Me Me

H2O Et I Et OH HO Et

H H H

R-2-iodobutane d/l pair (R/S-butan-2ol)

(2) Using a weakly polar solvent like acetone(a SN2 solvent) and using a strong nucleophile, we

can tilt the mechanism in complete favour of SN2.

R' R' acetone RCH X + OH RCHOH + HX (SN2 : 100%) OH– is a strong base and strong nucleophile. In presence of acetone the formation of carbocation will

be disfavoured and the strong nucleophile will hijack the mechanism in complete favour of SN2. It is in this condition, a chiral substrate will show complete inversion of configuration.

H H

Me Br + OH HO Me + Br (solvent : acetone) Et Et R-2-bromobutane S-butan-2-ol N.B: In actual practice, 100% racemisation or 100% inversion of configuration is never found, that is because as against our presumption, all molecules do not react by a single mechanism. We shall see later that in SN i mechanism there will be retention of configuration. Dr. S. S. Tripathy 11 GOC-III(Introductory Reaction Mechanism)

SNi : (Substitution nucelophilic internal)

This is a rare kind of SN reaction in which there is retention of configuration (i.e neither racemisation of SN1 nor inversion of SN2).

This happens when alcohol reacts with SOCl2(thionyl chloride) to give alkyl chloride.

The intermediate of this reaction is a chlorosulphite. When freshly prepared chlorosulphite is decomposed, also retention of configuration occurs. The famous scientist Paul Walden(on whom inversion of configuration is being named as Walden inversion) for the first time observed the retention of configuration when he compared the reaction of S(+)malic acid with PCl5 and SOCl2 separately. In case of the former, he found inversion of configuration [the product was R(–)chlorosuccinic acid], as expected from SN2 mechanism and in case of the latter, surprisingly, got retention of configuration [S(+) chlorosuccinic acid]. Make sure that for observing all these, the molecule should be chiral. So a 20 alcohol is required to hope for all these

Cl H H H HO SOCl Cl PCl5 2 SNi SN2 HOOC CH COOH HOOC CH2COOH 2 HOOC CH2COOH R(-)chlorosuccinic acidS(+) malic acid S(+)chlorosuccinic acid

Mechanism of SN i :

First the –OH group of alcohol makes a nucleophilic attack onto SOCl2 to form a chlorosulphite intermediate. –OSOCl being a good leaving group then is delinked from the C-atom but cannot go away farther from it and makes an intimate ion-pair. So a free carbocation is not formed like SN1 in this case and the strongly attracted ion-pair (R+ and –OSOCl) does not allow racemisation to happen. Interestingly, while the chlorosulphite group starts to move out from there, the –Cl group present in the –OSOCl group makes a nucleophilic attack on the chiral carbon from the same side and finally give retention chloro product with the expulsion of SO2. Is it not interesting !!!!

O O H H H H O - Cl H OH + S Cl O S Cl - H R R R O S R' Cl R' Cl R' Cl

O H O H H R O S C OS + R R Cl SO2 R' Cl R' Cl R' alkyl chlorosulphite intimate ion pair (intermediate)

In presence of Pyridine: Note that if the same reaction is carried out in presence of an aromatic base pyridine, inversion of configuration was observed. Pyridine as a nucleophile attacks the -OSOCl group and releases Cl– free while blocking the front as before. This freed Cl– has no alternative than to make a rear attack to give inverted product like SN2.

12 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

O O O H N H H -Cl O S O S N O S N R pyridine R R R' R' Cl R' Cl H - pyridine Cl R + SO2 R' inverted product

– N.B: In case of PCl5, there is a stable R–O–PCl4 intermediate and the expelled out Cl has no alternative than to make a rear attack to give the inverted product.

Cl H Cl Cl H Cl H Cl H Cl Cl OH + PCl- Cl O P -H R R R O P Cl R' R' Cl Cl Cl Cl R' Cl

H + POCl + Cl Cl R 3 R' inverted product

SN reaction with rearrangment / Conjugate SN reaction: (Allylic Rearrangement) ′ ′ (SN1 and SN2 Mechanism)

Allyl halides undergo SN reaction always with rearragement. Often more than one product is obtained; 3 2 one normal SN at the sp carbon and the other at sp carbon. SN reaction at the latter site is called ′ ′ conjugate SN or SN1 or SN2 depending on conditions. See these examples.

Cl NaOH(aq) OH + 1-chlorobut-2-ene OH (Normal SN1 product (SN1' product) Major : but-3-en-2-ol Minor : but-2-en-1-o

OH OH Cl NaOH(aq) +

(S 1' product) (Normal SN1 product) 1-chloro-3-methylbut-2-ene N 25% : 3-methylbut-2-en-1-o 85% : 2-methylbut-3-en-2-ol

2 3 If both sp and sp carbon atoms to be attacked are unhindered(unsubstituted) and SN2 conditions of ′ solvent and strong nucleophile are used, then the mechanism can be SN2 .

Dr. S. S. Tripathy 13 GOC-III(Introductory Reaction Mechanism) ′ SN1 Mechanism:

Cl CH3 CH CH CH2 CH3 CH CH CH2 CH3 CH CH CH2 more stable carbocation OH SN1 OH (SN1') OH OH

CH3 CH CH CH 2 CH3 CH CH CH2 (major product)

Note that carbocation in this case will definitely be formed as allyl carbocation is stable due to resonance effect. So the mechanism will be SN1. So which product will be major and which minor will be determined from the stability of two carbocations. In the 2nd example given before, the more stable carbocation is 0 ′ a 3 one(not shown in the mechanism). Hence the % of the rearranged product(SN1 ) is 85%. Note that the reaction will be kinetically controlled at room temperature conditions(see later to know what is the meaning of ‘kinetically controlled and ‘thermodynamically controlled). SN2 will have limited possibility here for the normal product as water is taken as solvent which favours carbocation formation. Make sure that we have said limited possibilty, not no possibility. If we use a strong nucleophile like OH–, there is ′ 0 0 always some chance of both SN2 and SN2 for both 1 and 2 substrates.

Allyl chloride will also react by SN1(both direct and conjugate) mechanism with equal probabilities if SNI conditions are employed and by SN2(both direct and conjugate) mechanism with equal probabilities if SN2 conditions are employed.

C2H5OH CH2 CH CH2 Cl EtO CH2 CH CH2 + CH2 CH CH2 OEt (-HCl) SN1' product SN1 product (50%) (50%)

This solvolysis reaction has to go by SN1 mechanism. If we isotopically label the carbon bonded to –Cl then we get equal abundance of allyl ethyl ether with labelled carbon in sp3 and sp2 carbon with equal extents. ′ SN2 Mechanism: 0 0 When a strong nucleophile is used and the substrate is less hindered(1 or 2 ), then an SN2 and SN2’ mechanism can occur in place of SN1 and SN1’ to give the same products.

Simultaneous attack of Nu– to the sp2 carbon of allyl chloride, shifting of pi-bond and removal of leaving – ′ ′ group(Cl ) leads to SN2 mechanim. Both SN2 and SN2 occur simultaneously with the latter predominating 0 α ′ for unhindered substrates. For 2 substrates with bulky substituents at -carbon, the SN2 product predomintes as direct attack is less favoured.

SH NaSH CH2 CH CH Cl HSCH2 CH CH R + CH2 CH CH R acetone achiral R SH chiral SN2' product chiral SN2 product ′ In this case SN2 mechanism predominates and the achiral product is major. SN2 will be less favourable 0 as it is a 2 substrate offering steric hindrance to direct attack. SN2 conditions will also not favour – carbocation formation, hence SN1 is rurled out. Moreover SH is a strong nucleophile(soft base) favouring 0 SN2. Had the mechanism been SN1, then the the chiral product would have been major as 2 carbocation is more stable than 10. If the substrate disfavours carbocation formation due to the presence of –I/–M group then also the ′ mechanism will be in favour of SN2 .

14 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

EtO EtO CH CH CH Cl EtOCH CH CH Cl + CH CH CH Cl 2 acetone 2 2 SN2' product EtO chiral Cl major SN2 product The presence of 2nd Cl atom does not favour carbocation formation. EtO– being a strong base favours ′ the SN2 mechamism, in which SN2 product (rearranged product) predominates. Steric crwoding at the sp3 carbon disfavaours direct attack. (READ ‘achiral’ for ‘chiral’ written under substrate. I shall change it later)

SN i′ Mechanism:

Allyl alcohol on treatment with SOCl2 gives both normal SNi product as well as rearranged ′ product by the mechanism called SN i.

OH Cl Cl SOCl2 CH3 CH CH CH2 CH3 CH CH CH2 + CH3 CH CH CH2 SN i product SN i' product

O OS Cl Cl SN i ' CH CH CH CH + SO CH CH 3 2 2 3 rearranged product CH CH2 Allyl chlorosulphite

We have already discussed in SN i the intermediate is chlorosulphite. Here the ion-pair is not shown. The –Cl attacks the sp2 carbon with shifting of pi bond to give rearranged product. For rearranged product, ‘retention of configuration’ carries little menaing as, even if a new chiral centre(not in this case) might be formed(if the sp2 carbon is substittuted), we cannot compare the configuration of the substrate with the configuration of the newly formed chiral centre(if any). ′ But the fact is rearrangement taking place in SN i mechanism and hence called SN i .

Nucleophilic Acyl Substitution:

This type of SN reaction is observed in carboxylic acids and their derivatives. These compounds substitute in a different style, quite different from alkyl halides and related substrates in which the leaving group is hanging on a sp3 carbon. Here the leaving group is attached to sp2 carbon i,e acyl carbon. Nucleophile O O - RC L + Nu RCNu + L- acyl group Leaving group Relative reactivity of carboxylic acids and their derivatives depend on the leavability of the leaving group, analogous the the SN1 and SN2 cases discussed before. The same principle will be applicable here, less the base strength better is its leavability. Reactivity Order:

Dr. S. S. Tripathy 15 GOC-III(Introductory Reaction Mechanism)

O O O O O O

RC Cl > R C OCR' > R C OH > R C OR' > R C NH2 acid chloride acid anhydride c. acid ester acid amide This is because the leavability of the leaving groups follow in the same order.

O (increasing base strength) Leavability :Cl >> O C R' OH > OR' > NH2

The stability order of these compounds follows the reverse order.

O O O O O O

RC Cl << R C OCR' R C OH << R C OR' R C NH2 acid chloride acid anhydride c. acid ester acid amide Acid chloride is the best substrate for substitution and acid anhydride is next to it. Others are – ′– – too bad substrates because OH , OR and NH2 are strong bases. However under acidic conditions, ′ protonation of –OH, –OR and -NH2 do occur and the leaving groups become neutral i.e H2O, ′ R OH and NH3 respectively and are good leaving groups. We shall see all these details later. But as such, carboxylic acids, ester and acid amides are stable to acyl substitution.

Just like we did for normal SN1, SN2 in tetrahedral substrates earlier, here too, the principle is - a less reactive(more stable) substrate cannot displace a more reactive(less stable) substrate. See these examples.

O O RCOR' + Cl RC Cl + OR' (NOT POSSIBLE) less reactive more reactive (more stable) (less stable)

O O O

RC OCR + NH3 RC NH + RCOOH (POSSIBLE) 2 more reactive less reactive (less stable) (more stable)

O O

RCNH2 + OH RCOH + NH (NOT POSSIBLE) more stable less stable 2

O O RCCl + R'OH RCOR' + HCl (POSSIBLE) more reactive less reactive

General Mechanism of Acyl Substitution:

16 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

O O O slow fast RC L + Nu RCL RCNu + L

Nucleophile attacks the sp2 - carbon which is the electrophilic centre(more electrophilic than a sp3 carbon in R–X). The pi bond of C=O gets polarized simultaneously to form an alkoxide ion intermediate as shown. This is often the rate determining step and is second order(first order with respect to each of substrate and nucleophile), hence it is called Ac-2 mechanism(Acyl bimolecular). In the second step the pi bond is again formed and the leaving group departs in a fast step. See one specific example. Alkaline hydrolysis of ester:

O O O slow fast RC OR' + OH RCOR' RCO H + OR' OH O RCO + R'OH

An ester is hydrolysed by OH– to form c. acid is an axample of Ac-2 mechanism. Since this hydrolysis is achieved in the presence of base, this mechanism is also called BAc2. Here there is an additional step of proton exhange beween c.acid and alkoxide ion to form alkanoate ion(salt of c.acid) and neutral alcohol. This is because a strong acid RCOOH displaces a weak acid ROH in this last step. Also note that ester and c.acid have similar stabilities(though ester is little more stable), the above substitution looks improbable. However it happens by taking the help of Le

Chatelier’s principle. Its Keq is less than 1. The product alochol is continuously removed with some vacuum technique to drive the equilibrium to right. Formation of ester from acid chloride:

H O O O slow fast RC Cl + R'OH RCCl RCO - Cl R' H O R' - H O RCOR'

Formation of acid amide from acid anhydride:

O O O O O O

RCOCR + NH3 RCOCR RCOCR acid anhydride N H H N H H H H O O

RCNH2 + R C OH

In the above two examples in which acid chloride and acid anhyride act as substrates, Keq >1 and Dr. S. S. Tripathy 17 GOC-III(Introductory Reaction Mechanism) equilibrium lies to the right. In the first case, the there is an additional step of proton-loss at the end as the nucleophile is neutral i.e R′OH(alcohol). In the 2nd case, there is a proton exchange so as to produce two neutral products.

Hydrolysis of Ester: We shall discuss this in details for having a better idea on acyl substitution. Interestingly, in the name of acyl substitution at the sp2 carbon we shall also study the alkyl substitution at sp3 carbon in this reaction mechanism. This is unique to ester as there is a sp3 carbon atom bonded to - O(alcohol part of ester) along with sp2 carbon(acyl) which is not the case with other substrates. Hydrolysis of ester can be either acid actalysed or base catalysed. Let us study one by one.

O O

RC O R' + H2O RC OH + R'OH ester c.acid alcohol There are two types of mechanism in this case. (i) Acyl Substitution: Ac- mechansim

O Nu RC OR' Acyl substitution (Ac) (Acyl cleavage) acyl carbon Here nucleophile attacks the sp2- acyl carbon and C–O single bond adjacent to acyl group breaks. This mechanism is the same we discussed before(Ac 2) Evidence in favour of Acyl Substitution: 18 18 (a) If we use labelled H2O , then all O will be found in the c. acid and not in alcohol. Again if we take labelled ester RCO–O18R’, then all O18 will be found in alcohol and not in c.acid. (b) If R′ (alcohol part) is chiral, then there will be retention of configuration, as this carbon is not attacked by nucleophile. (c) Allyl ester of c.acids do not give any rearranged alcohol. We shall see later, that in the alkyl substution, a carbocation is formed which can undergo rearrangement to form more than one alcohol. (d) Neopentyl ester of c.acid also do not give rearranged(tert-pentyl) alcohol. (ii) Alkyl Substituion: AL- mechanism

O Nu R C O R' Alkyl substitution (A (Al) (Alkyl cleavage) alkyl carbon Here nucleophile attacks the sp3 carbon of R′(alcohol part) and that happens only when the carbocation formed is very stable like tert-butyl, benzyl, allyl etc. The mechanism is same as SN1. But it is called AL- mechanism in hydrolysis of ester. Evidence in favour of Alkyl Substitution: 18 18 18 (a) If labelled H2O is used, all O is found in alcohol and if labelled RCOO R is used, then all O18 is found in c.acid. (b) substituted allyl esters give rearrangeed alcohols. (c) if R′ is chiral, then a racemic product is obtained in agreement with carbocation mechanism.

18 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism) Now we shall study acid catalysed and base catalysed hydrolysis separately. ACID CATALYSED HYDROLYSIS:

The most common mechanism in this catetory is AAC2 and AAL1. The uncommon one is

AAC1. AAL2 is not observed. The first letter ‘A’ stands for ACIDIC. All the steps in this case are reversible.

ACYL CLEAVAGE: (AC mechanism)

AAC2 Mechanism (Acid catalysed bimolecular acyl substitution)

H H H O O O O + H H2O RCOR' RCOR' RCOR' RCOR' H O OH bimolecular H H

O H O - R'OH - H RCOH RCOH

The presence of acid (H+) increases the electrophilicity of acyl carbon atom by protonating =O atom. The protonated ester has two more RSs (not shown). H2O (neutral nucleophile) then attacks the protonated ester(oxonium ion) in the rate determining step to form a tetrahedral intermediate also another oxonium ion. This step is bimolecular i.e protonated substrate and H2O molecule take part in this step. Hence the title AAC2. Kinetically it is a first order reaction, but bimolecular. Hence it is called a pseudounimolecular reaction. In the 3rd step, proton exchange occurs to protonate the alkoxy group. In the 4th step, lone pair on one -OH group is transferred to form a pi bond with the expulsion of neutral alcohol molecule(good leaving group). In the last step deprotonation occurs from the oxonium ion to form c. acid. Thus H+ acts as real catalyst.

Fischer Esterfication: The opposite of this reaction is formation of ester from carboxylic acid and alcohol in presence of acid. A mixture of the two is refluxed and continuously the product H2O is removed to drive the equilibium to right. This is also a AAC2 mechanism.

H H H O O O O H+ HO R' RCOH RCOH RCOH RCOH H O OR' bimolecular R' H

O H O -H O - H 2 RCOR' RCOR'

The reactivity order is primarilty decided from the nature of alcohol. 0 0 0 CH3OH > RCH2OH(1 ) > R2CHOH (2 ) > R3COH (3 ) (purely on steric hindrace ground)

Dr. S. S. Tripathy 19 GOC-III(Introductory Reaction Mechanism)

AAC1 Mechanism (Acid catalysed unimolecular acyl substitution) It happens very rarely, as the C–OR single bond has partial double bond character(due to M- effect) and does not cleave in the rate determining step, even though the acyl carbocation is very stable. However, in case where the RCO- group is sterically hindered, then formation of acyl carbocation takes place first in unimolecular mechanism. See the hydrolysis of methyl mesitoate i.e ester of mesitoic acid(2,4,6-trimethylbenzoic acid).

CH CH CH3 3 O 3 O O H H O - R'OH 2 H3C C OR' H3C C H3C C OH slow - H

CH3 CH3 CH3 alkyl mesitoate(protonated) acyl carbocation The first protonation step has not been shown. In this case unimolecular mechanism occurs because the attack of nucleophile H2O is hindered by the bulky substituted phenyl group.

ALKYL CLEAVAGE (AL mechanism):

AAL1 Mechanism (Acid catalysed Unimolecular alkyl substitution): If the alkyl part(R′) can produce stable carbocation, then alkyl substitution will occur in stead of acyl. Tert-butyl, benzyl and allyl esters and similar type of esters react by this mechanism.

O R O H R1 1 R2 H slow RC CH3 C OCR2 CH3 C OCR2 1 + CH3COOH stable 30 R R R 3 3 3 carbocation

H2O H R2 R2 R2 H R - H 2 RO+ ROH1 + HO R 1 OR1 1 H R H R R R 3 3 3 3 d/l pair

In this case, cleavage at the –O–R′ bond i.e at tetrahedral carbon – O bond occurs in the rate determining step to form a very stable 30 carbocation. Hence the alcohol product is bound to be racemic if there is a chiral centre. Other esters which react by this mechanism are benzyl and allyl esters.

O O

CH3 C OCH2 CH3 C OCH2 CH CH CH3 benzy acetate but-2-enyl acetate 18 18 Interestingly if labelled H2O is used , all of O will be found in the alcohol, not acid and if labelled ester RCO–O18R′ is used, then all O18 is found in the c. acid. This mechanism is further corroboarated by the formation of rearranged alcohols. From but-2-enyl acetate shown above, we get two alcohols namely but-2-en-1-ol and but-3-en-1-ol(due to rearrangement of allyl carbocation).

20 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism) Exceptional case: ′ 0 Even when the R group is 2 , and we use conc. H2SO4 instead of dilute, AAL1 will operate to give racemic alochol products. But with dil acid, AAC2 will operate to form alcohol with retention of configuration. See this example.

OH O Et conc. H2SO4 CH3 C OCH CH3COOH + Et CH Me Me AAL1 d/l pair

O OH Et dil H2SO4 + CH3 C OCH CH3COOH Et CH Me Me AAC 2 retention of config

N.B: AAL2 mechanism is not observed. Bimolecular (SN2 type) attack to alkyl carbon is impropable in the face of greater electrophlicity of the acyl carbon.

BASIC HYDROLYSIS OF ESTER: Basic hydrolysis of ester is faster than acidic hydrolysis as in case of the former, the steps are irreversible while of the latter, steps are reversible. The salt RCOO– formed is soluble in water and can easilty removed from the reaction mixture, thus driving the equilbrium till completion.

ACYL CLEAVAGE: (AC mechansim) K

BAC 2 Mechanism: K This is most common mechanism encountered in basic hydrolysis. Its bimolecular and also 2nd order, first order w.r.t each of substrate and OH–.

O O O slow fast RC OR' + OH RCOR' RCO H + O R'

OH O RCO + H OR' salt of c.acid

The last step i.e proton exchange step is irreversible, and hence the whole reaction is considered to be irreversible. Though R′O– is a poor leaving group, it takes place in the fast step. The irreversibility of the last step and continuous removal of the salt (RCOO–), drives the reaction forward. ′ – N.B BAC 1 Mechanism is not observed. As R O cannot be a leaving group in the rate determining step to form acyl carbocation. Sterically hindered acyl group acyl group also has to react by bimolecular mechansim, though slowly, cannot accept BAC1.

Dr. S. S. Tripathy 21 GOC-III(Introductory Reaction Mechanism) ALKYL CLEAVAGE (AL mechanism):

BAL1 Mechanism:

When the alkyl group can form very stable carbocation like tert-butyl, allyl or benzyl, the alkyl cleavage can also occur under basic conditions, but it should be diluted basic solution (dil

NaOH) or weak base like Na2CO3 or even neutral solution. Here too, like acidic alkyl cleavage, unimolecular mechanism occurs showing all evidence for racemisation and rearrangement. Under concentrated basic condition, acyl cleavage takes over.

RC OCH2 RCOO + CH2

HOCH2

BAL2 Mechanism: This mechanism also is rare. When the R′ part is primary, and there is huge steric hindrance in the acyl part, then bimolecular SN2 type of subsitution can occur in the alkyl part. See this case.

O O OEt C OCH 3 C O + CH3 OEt

Methyl 2,4,6-tritert-butylbenzoate was found to react by bimolecular mechanism with strong base like OEt–. Though, its not a strict case of hydrolysis as one of the products is an ether, the mechanism has been found to be in fact BAL 2. Here the only reason for this mechanism is to – relieve the huge steric strain in the molecule. You can argue in favour of BAC1, but CH3O is a poor leaving group for that to happen. So smoothly the work is done by bimolecular alkyl substitution. Note that if the R′ part would be chiral, there will be perfect inversion of configuration will be found the product alcohol or ether(not shown here).

(Acyl cleavage(BAC2 and AAC2) is often called Addition - Elimnation type of substituion. This means, addition of nucleophile occurs in the first step and then the leaving group is eliminated in the 2nd step.)

22 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

Aromatic Electrophilic Substution: (ArSE Mehanism) (Addition-Elimination Mechanism) Benzene ring is a better nucleophile than an electrophile. Hence it undergoes easy electrophilic substitution. See the steps in the mechansim.

E E E E slow fast E H H H + H

resonance stabilised arenonium ion(benzenonium ion) In the first rate determining step, the electrophile reacts(addition step) with the benzene ring to form a resonance stabilised benzenonium(in general arenonium) ion(a carbocation). This was earlier called as σ-complex. In the second step which is fast, the carbocation loses a proton(the + leaving group). This is the elimination step. H is the most common leaving group in ArSE reaction. Note that this stabilised carbocation is not aromatic, it should be better called as cyclohexadienylium ion(rather than benzenonium ion). Kinetic Isotope Effect: When isotopically labelled substrate is taken in place of a normal substrate, and if there is a change of rate of reaction, then we say, there is a kinetic isotope effect. For isotopic effect to exist, the following holds good K K n  1 H 1 Ki K D

Kn= rate constant for normal substrate, Ki = normal constant for isotopically labelled substrate) If there is deuterium(D) substitution for H, then the ratio will be greater than 1, if there is isotope effect, becasue C–D bond energy is greater than C–H bond energy.

Usually in ArSE reaction, no kinetic isotope effect is observed. i.e K H 1 (ArSE) K D This is because, C–H or C–D bond breaking does not take place in the rate determining step, rather takes place in the 2nd step which is fast. So C6H6(benzene) and deuterated benzene(C6D6) undergo any particular ArSE reaction at the same rate.

Exampe: Nitration of benzene:

H NO2 HNO (conc.) + NO 3 2 + H H2SO4(conc)

+ + Here the electrophile (E ) is nitronium ion(NO2 ) which is produced in the mixture of conc.

HNO3 and conc. H2SO4 by the following reaction.

HNO3 + H2SO4 NO2 ++ HSO4 H2O Similarly we have halogenation, alkylation, acylation and sulfonation of benzene, which we shall take up later in the chapter ‘aromatic hydrocarbons’. Sulfonation of benzene only shows small kinetic isotope effect as the above ratio has been found to be 2.

Dr. S. S. Tripathy 23 GOC-III(Introductory Reaction Mechanism)

Aromatic Nucleophilic Substitution(ArSN): 1. Addition-Elimination Mechanism: Nucleophilic substitution in benzene is not easy, as benzene itself is a better nucleophile than an electrophile. Hence SN reaction of halobenzene takes place with much difficulty only under drastic conditions and hence is not commercially viable. See this example.

Cl OH Cl NH2

conc.NaOH NH3 + NaCl + HCl 3500C/150 atm Cu2O/2000C

See the conditions for converting chlorobenzene to phenol. 150 atm and 3000C along with conc. NaOH solution are drastic conditions and even then the yield is poor. Some diphenyl ether(Ph– O–Ph) is also formed along with phenol in the above reaction. Similar is the case with preparation of aniline from chlorobenzene and ammonia. Reason: Carbon in benzenen sp2 hybridised and there is partial double bond character in the C–Cl bond due to some +M effect(though the – I effect is dominant) of halo group. Hence the electrophilicity of the ipso carbon is reduced and also the leavability of Cl– group is also poor. Hence drastic conditions are required to undertake this reaction.

Mechanism of ArSN Reaction(Addition-Elimination Mechanism)

This is most common ArSN mechanism in organic reactions. Other ArSN reactions are uncommon, which will be discussed just a bit later. It involves addition of nucleophile first and then elimination of leaving group, hence called addition-elimination mehanism. It is analogous to ArSE.

Nu Nu Nu Nu Cl Nu Cl Cl Cl + Cl resonance stabilised cyclohexadienide ion (chlorophenide ion) In the first rate determining state the nucleophile attacks to the benzene ring to form a resonance stabilised cyclohexadienide ion( a carbanion) which loses the leaving grou (Cl–) in the second step which is fast. The carbanion is also called a phenide ion and earlier called as σ-complex, and also called Meisenheimer complex when –M groups are present. Stable Meisenheimer complexes(salts) can be isolated when the ring is activated strongly(see later). Aromatic substutution is more or less similar to acyl substutition which involves the addition step first and then elimination of leaving group occurs, in contrast to the aliphatic SN reaction at the sp3 hybrid carbon. – – Note that chlorobenzene in presence of strong base like OH , NH2 react in a different mechanism which we will discuss later. The above mechanism works when the ring is activated for ArSN(see – later) or in presence of a weak nucleophile like NH3. The first example of OH is given for the sake of introduction.

ACTIVATION IN ArSN Reaction: (Addition-Elimination Mechanism)

We knew before that a EDG(+M/+I) group activates the benzene ring for ArSE reaction by increasing the electron density in the ring. Also EWG(–M/–I) group deactivates the ArSE reaction by decreasing the electron density.

24 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

On the contrary, a EWG activates ArSN reaction by increasing the electrophilicity of the benzener ring. Particularly when a –M group is located o/p positions w.r.t the leaving group (–X), then the activation is pronounced. Presence of –M groups like –NO2, –CN etc at the o/p positions enhances the magnitude of +δ charge on the carbon bearing the the leaving group(ipso carbon). Hence the nucleophilc attack becomes favourable.

Cl Cl Cl Cl Cl Nu +

+ + N N N N NO OO O O O O O O 2

δ Due to –M-effect groups like –NO2 will produce + charge o/p w.r.t to it. If the laving group – X is located on the carbon o/p to the –M group, then the the nucleophilic attack will be more favourable on that carbon and the ArSN reaction becomes relatively easier. More the number of such groups in the o/p positions more is the ease of ArSN. See the following.

OH Cl Cl OH NO2 NO NaOH(1M) NaOH(0.001M) 2 0 + NaCl NaCl 150 C 1000C +

NO NO2 NO2 2 NO2

Cl OH NO O2N 2 O2N NO2 H2O 400C + HCl

NO2 NO2

Presence of one –NO2 group at the o/p position has reduced the conditions to diluted NaOH(1M) and at lower temperature. Introduction of two –NO2 groups decreased the conditions still further and for 2,4,,6-trinitrocompound, the substution occurs with only water, no base is required and that too at 400C. Note that such high activation would not have happened if the –M groups would occupy the meta postion w.r.t to the leaving group, though there would be some activation due to –I effect at that position. The activation can also be explained by the greater stabilistion of the σ-complex by the –M group. This is an alternative explanation. See below.

O Nu Nu O Cl Nu O Cl O O Cl Cl N N Cl Nu N O N N Nu O O O O Nu

NO2 + Cl

Dr. S. S. Tripathy 25 GOC-III(Introductory Reaction Mechanism)

Presence of one –NO2 group gives additional delocalisation(one more RS) of the –ve charge as

–NO2 group participates in the process. More the number of –NO2 group in the o/p positions, more is the extent of delocalisation(more RSs) and hence the σ-complex becomes more and more stable. That is why some Meisenheimer complex from trinitroaromatic compounds become stable and reversibley converted back to the starting compound.

Na+ (stable Meisenheimer salts)

When 2,4,6-trinitroanisole is treated with NaOMe, a red coloured substance is produced. This is a Meisenhemier salt(1st example) There can be many more by changing -OMe at the ipso position with other leaving groups including H(i.e trinitrobenzene only) or –Cl (2nd example).. This salt on acidification gives back the original compound. On heating complex further the – – leaving groups ie CH3O and Cl respectively departs to give the substituted product. In the first example, we get back the reactant as the nucleophile and the leaving groups are same. However, isotopic labelling can be done with one of them to corroborate the mechanism.

2. Aromatic SN1 reaction:

Aromatic SN2 reaction is not possible as the bulky benzene ring will not allow for a rear attack.

SN1 reaction is possible in some cases. The most important among them is reaction of benzene – – diazonium salt with a nucleophile like H2O, CN , Cl etc. We shall study the reactions in details in the chemistry of aniline and benzene diazonium chloride.

N N Nu Nu

- N2 benzene diazonium ion phenyl carbocation (unstable)

3. Cine Substitution(Elimination-Addtion Mechanism) :

In this case, elimination takes place first followed by addition which is opposite to what we have studied earlier in ArSN. This takes place via a very unstable benzyne intermediate and hence nucleophile is bonded to the carbon bearing the leaving group as well as its ortho position. So more than one products are formed. The latter unusual product is called Cine substituion product. Sometimes the Cine substitution product becomes major while the normal product minor.

CH CH CH3 3 3 Cl NH2 KNH2 + liq NH3 NH2 normal product Cine substn. product

When o-chlorotoluene is treated with KNH2 in liquid NH3, we get two products namely o- toluidine(normal product) and m-toluidine(Cine product). 26 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

Cl NH2 KNH2 + liq NH3 NH2 When only chlorobenzene is treated with the same reagents by labelling the ipso carbon with C-

14, the product contains 50% aniline in which the labelled carbon bears the –NH2 and the other

50% the adjacent non-labelled carbon(C-12) bears the –NH2 group. Is it not interesting ?

Mechanism:

CH CH3 H 3 NH2 NH2

CH3 CH3 NH3 Cl more stable carbanion normal product NH2 CH3 CH3 - NH3 H - Cl NH benzyne 3 H NH2 NH2 Cine product

β – Benzyne intermediate is formed by the -elimination of HCl by the strong base NH2 . In the next step, the nucleophile NH3 has the option to attack on either carbon atom of the triple bond. Both the attacks take place, though with unequal probabilities. This is dependent on the relative stability of the carbanion formed after the attack. In the present case, the carbanion formed from the ortho attack(ortho w.r.t –CH3) is more stable than formed from the meta attack as the –ve charge is farther away from the +I group in case of the former. Thus the final product formed after the proton transfer contains the normal product as major and the Cine product as minor. But the presence of appreciable quantity of Cine product corroborates the formation of benzyne. There is always some steric hindrance factor for the ortho attack which will decrease the yield due to the that despite the stability of the carbanion. The formation of benznye intermediate has also been confirmed by other experimental study(not discussed here). Structure of Benzyne: The triple bond in benzyne is different from triple bond in alkyne. In benzyne, the carbon atoms are sp2 hybridised like other 4 carbon atoms, unlike in alkyne the hybridisation is sp. Out of the two pi-bonds one is between two unhybridised p-orbitals and the other between two hybrid orbitals. Have you ever heard that two hybrid oribtals go for side wise overlapping to form a pi bond ? Yes, it is the case here in benzyne. This pi bond is very weak which breaks down by the attack of the nucleophile easily.

In case of chlorobenzene, 50% of the product of each type(–NH2 bonded to labelled carbon and bonded to non-labelled carbon) will be found as there is no other group to influence the composition.

It was further found that chlorobenzene with drastic conditions reacts with NaOH(not only by

NaNH2 in liq. NH3) by this mechanism to give two products, which could be be revealed by labelling the ipso carbon. Eariler it was thought to be addition-elimination mechanism(ArSN), but was found to be actually elimination-addition mechanism involving benzyne intermediate.

Dr. S. S. Tripathy 27 GOC-III(Introductory Reaction Mechanism)

Cl OH OCH OH OCH3 3 NaOH(aq.) Cl + NaNH 3500C 2 liq. NH3 NH2 CH CH3 CH3 3 (nearly 100%) p-chlorotoluene reacts with NaOH(aq) at high temperature to give a equimolar mixture of p- cresol and m-cresol. The presence of methyl group at a far away postion does not influence the stability of the intermediate carbanion and hence the result. The steric factor is not present here.

But o-chloroanisole reacts with NaNH2 in liquid NH3 to give almost 100% of Cine product. This is because of two reasons; (i) steric hidrance for ortho attack onto benzyne which does not favour normal product and (ii) the intermediate carbanion is more stablised in the meta attack by the

–I effect of -OCH3 as the negative charge is nearer to –OCH3(at the ortho position). Note that here, +M-effect of –OCH3 is not considered as there is another +M group(–NH2) to counter its effect. Only I(–I here) effect is considered to assess the relative stability of carbanions.

CH CH CH3 3 3 CH3 NH2 NaNH 2 + + liq. NH3 Cl NH2

NH2

Similarly m-chlorotoluene with same reagents give three products with measurable yields. This is because in this case two benzyene intermediates are fomed by the β-elimination of HCl no either side of the –Cl atom at the meta position. So all the three products due to ortho, meta and para attacks are formed.

CH3 CH3

and

N.B: In many cases, the % yield of the products cannot be predicted with logical considerations. So i suggest you to not to harp on their relative yields, as the literature does not give concrete yield data. That three products are formed, one normal and two abnormal(cine products) itself is great evidence in favour of this elimination-addition mechanism.

4. SRN1 (Radical nucleophilic aromatic substituion unimolecular)

This is a modification in the cine substitution and is very uncommon type. When –

I(iodine) replaces –Cl in the benzene ring and we use the same reagents like KNH2/liq. NH3, we get predominantly the ipso product, not the cine product. While chlorobenzene gives 1 : 1 ratio of both ipso and ortho products. Of course, due to electronic and steric factor, in substituted chlorobenzene, this ratio is bit biased. In case of iodobenzene, irrespective of the presence of substituents, the ipso product is major, the ortho product is minor. We are avoiding to show you the detailed mechanism involving radicals here. Sandmeyer reaction, that we shall study in benzene diazononium chloride later also is truley SRN1 type not ArSN1. We shall return to this later. I am now impatient to switich over to elimination reactions.

28 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism) β-Elimination Reactions: You know that when two leaving groups go away from adjacent carbon atoms a new pi- bond is formed and a single bond (C–C) is converted to a double bond (C=C) or double bond is converted to a triple bond. This is β-elimination. There are different mechanisms operating in different reactions. Let us discuss in some details about each. (a) E2 (b) E1 (b) E1CB (d) SYN Ei (e) Hoffmann

E2 Mechanism:

(1) It is the predominant mechanism in alkyl halides when elimination is the major product. (2) One leaving group -X and the other -H depart simultaneously in a one step process via a high energy TS. (3) The reaction is bimolecular and 2nd order. Rate law :

Rate = k [Substrate)[Base]

This mechansim shows primary kinetic isotope effect. KH/KD >1, since C–H bond breaking is a part of the rate determining step. – – – β (4) A base, usually a strong base like OH , NH2 , RO etc. abstracts the proton from the - position w.r.t to –X group. While H atom is gradually broken from the β-carbon atom, a new pi- bond is formed betwee α−β carbon atoms gradually while the C–X group breaks gradually simultaneously. In the TS, a fractional negative charge develops on the –X(halogen atom) while the magnitude of –ve charge is decreases from the the base(B–). (5) Antiperiplanarity is the stereoelectronic requirement for the E2 elimination. This means that the two leaving groups should remain in anti conformation(with a dihedral angle of 1800) in the TS for the elimination to occur. The breaking of the C–H sigma bond and formation of a new pi bond between C-atoms and breaking of the C–X bond can occur most favourably if the electron approaches from the rear side of the the leaving group(X), rather than approaching from the same side. The latter will generate unfavourable repulsion factor which makes the TS less attainable than the anti approach. Reactivity order in E2 30 halide >> 20 halide > 10 halide and R–I > R–Br > R–Cl >> R–F (6) Types: (a) Saytzeff Elimination: If the β-H is removed from the more hindered position to form a more substituted alkene, it is called Saytzeff elimination. This is thermodynamically stable product and we say it is thermodynamically controlled reaction. The free energy change(ΔG0) and hence Keq are the key points to drive the reaction in favour of a stable product. The reason for the stability of more substututed alkene, you already know. It is stabilisation by hyperconjugation effect(Refer GOC-I if you have not studied that). In β-elimination, we get a mixture of Sayzeff product and anti-Saytzeff product(see below), the former being the major one. This is called regioselectivity. E2 elimination is thus regioselective. This means, both the products are formed with unequal yields, one of them is formed to a greater extent selectively.

- B B less hindered position

H H H H RC C CH R'' R CCCH2 R'' 2 RC CH CH2R'' - R' X R' X R' more substituted alkene

E2 Saytzeff Elimination + BH + X Dr. S. S. Tripathy 29 GOC-III(Introductory Reaction Mechanism) (b) Hoffmann (Anti-Saytzeff) Elimination: If steric hindrance factor makes the TS unattainable for Saytzeff elimination due to very high energy requirement, then β-H is more favourably abstracted from less hindered position to form less substituted alkene. This is called kinetically controlled reaction. The activation energy(EA) is the key point to determine the course of reaction. Thermodynamics fails to win over kinetics in this case. It means, even though the more substutitued alkene is more stable, the less substituted alkene is the major product, as the reaction is pushed forward in the direction involving less EA. Conversely, the reaction with very high EA is pushed back and is not favoured.

less hindered position - B B

H H H H H

R CCCH R'' R CH C CH R'' RCH CH CH R

R' X R' X- R' less substituted alkene

E2 Anti-Saytzeff(Hoffmann) Elimination + BH + X

In real practice, we get a mixture of both the products and most often the Saytzeff product predominating.

H H OH(alcoholic) CH CH CH CH CH CH CH CH CH3 CH CH CH2 3 3 3 2 2 but-2-ene but-1-ene Br (major) (minor)

That is why we said, the reaction is regioselective. Unequal yields of both are always formed with one being the major product.

0 – – – – (8) For 3 halides, any strong base like OH , CN , RO , NH2 , CH C etc, bring about nearly 100% elimination by E2 mechanism. Tert-butyl chloride reacts with alcoholic KOH to give only one product i.e 2-methylpropene(isobutene).

CH CH3 H 3 OH(alcoholic) CH C CH CH3 C CH2 3 2 E2 100% Br We would have got same product with same yield had we taken KCN(alc.) or NaOEt(alc.), CH CNa (alc.), NaNH2 or any strong base in place of alcoholic KOH. Substitution reaction always competes with elimination, as the base is also a nucleophile. In stead of abstracting a β- proton, it could as well undertake SN reaction at the a-carbon. But in this case, it does not. We shall review both elimination and substitution together at the end. You have to wait a bit for that. Since tert-butyl bromide is a symmetrical compound, we get only one product. In unsymmetrical product we could get two products with Saytzeff being major.

CH H 3 H CH3 CH3 OH(alcoholic) CH C CH CH CH C CH CH CH C CH CH 2 3 E2 3 3 + 2 2 3 major minor Br

30 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism) In 20 halide, though behaves similar to 30 halide in presence of strong base, some substitution products are also formed as minor product, though elimination being the major.

OH H OH KOH(alc.) CH CH CH CH CH3 CH CH CH3 3 3 + CH3 CH2 CH CH3 CH CH CH CH S 2 product 0 Br 3 2 2 N 2 halide E2 products (inversion) (major) minor

In case of 10 halide, the substitution reaction takes the lead and major product becomes substituion.

OEt H KOEt(alc.) CH3 CH2 CH CH2 Br CH3 CH2 CH2 CH2 OEt + CH3 CH2 CH CH2 10 halide SN2 product E2 product (major)90% (minor)10% Conclusion: You found that elimination and substitution go hand in hand. In some cases, the elimination product dominates and in some other cases the substitution product dominates. We shall return to a more logical discussion on this at the end.

Hoffmann Product : Major

If we use a very bulky but strong base like t-BuO–, (t-BuOK), then it prefers to abstract a proton from the less hindered position and the kinetically controlled product i.e less substituted alkene becomes major. The thermodynamically favoured product will require a high energy TS due to steric hindrance, as t-BuO– is a very bulky group.

CH3

OCCH3 CH H H 3 (t-BuOK) CH3 CH CH CH2 CH3 CH2 CH CH2 + CH3 CH CH CH3 but-1-ene but-2-ene Br (major) (minor)

Hofmann Exhaustive Methylation(HEM):

less hindered carbon

CH 3 CH2 H CH2 heat CH3 N CH CH CH3 OH CH CH CH3 + (CH3)3N CH CH3 3 CH3 Hoffmann product more hindered carbon (3-methylbut-1-ene) trimethyl 3-methylbutan-2-ylammonium hydroxide

Dr. S. S. Tripathy 31 GOC-III(Introductory Reaction Mechanism) Tetraalkyl ammonium hydroxide on heating undergoes dehydration(elimination of water) to give less substituted alkene(anti-Saytzeff product) as the major product. The β-H from the less hindered position is preferably removed by the OH– present in the molecule in this case. Steric hindrance of the bulky leaving group( a 40 ammonium salt) makes the TS for the Saytzeff elimination unfavourable and hence the anti-Saytzeff(Hofmann) product becomes the major alkene. Later, we shall study how we prepare the tetraalkyl ammonium hydroxide by repeated (exhaustive) methylation of a primary amine (3-methylbutan-2-amine in this case) by excess CH3I and then treating the iodide salt with AgOH. This is just the introduction. If it becomes hard to understand now, leave it for a later study.

Regioselectivity in cyclic compounds for bimolecular Elimination:

SYN

anti CH CH3 Br 3 1 H CH3 CH3 OEt H 6 2 3 5 anti-Saytzeff product 4 (2,3-dimethylcyclohexene) E-1-bromo-1,2-dimethylcyclohexane on hehydrobromination(–HBr) by the strong base OEt– (NaOEt), gives the less substituted(anti-Saytzeff product) alkene as the only product. The more substituted alkene(Saytzeff product) is not formed at all. This is because, H- on C-2 lies on the same side as -Br on C-1 (SYN), while the E2 elimination demands the groups to remain ANTI. An anti H- atom is available in C-6 (Br on C-1 and H on C-6 lie on opposite sides), thus E2 elimination occurs on that side to give less substituted anti-Saytzeff alkene. To make further confirmation, when one of H on C-6 is replaced by deuterium(D) lying on the same side of -Br on C-1, the compound gives the same product while retaining the D atom on C-6.

SYN

anti CH CH3 Br 3 1 D H D CH3 CH3 OEt H 6 2

5 3 4 This proves beyound any doubt that E2 elimination requires the two leaving groups to remain anti w.r.t each other. This can be explained more logically by taking the chair conformation of cyclohexane. Chair Conformation of Cyclohexane: The aniperiplanarity in cyclohexane requires the two leaving groups to remain DIAXIAL, not diequatorial.

e H Me H OEt tBu Me a sl o w e a Cl 2-menthene Cl menthyl chloride Unfovarable TS Me Menthyl chloride on on dehydrohalogenation (–HCl) by using strong base(NaOEt) give 2-

32 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism) menthene(anti-Saytzeff product) at a very slow rate. This is because, the other chair form in which –Cl occupies axial position, the bulky t-butyl broup occupies also axial position. Due to huge 1-3 diaxial interaction, this is very unstable(high energy TS). Moreover, the β-axial H atom is available on the other side, as the H-atom on the carbon bearing t-Bu group occupies equatorial position. Thus less substituted alkene is formed with extremely slow rate. Some texts say, there is no reaction in this case.

Cl Me OEt a tBu Me fast

3-menthene a (Saytzeff product H neomenthyl chloride On the other hand, neomenthyl chloride undergoes E2 elimination in a much faster rate to give 3-menthene, a more substituted alkene. The only difference between menthyl and neomenthyl chlroide is the location of Cl- atom, occupying the axial position in neomenthyl chloride and equatorial position in menthyl chloride. That is why flipping to the other chair form is not required in neomenthyl chloride. Axial H atom is available on the side of t-Bu and hence Saytzeff alkene is formed. Since the bulky t-Bu group occupies in the equatorial position, the TS is very stable and hence the rate of elimination is very high. Even, in the presence of weak base like EtOH, this elimination takes place at a moderately faster rate giving the same product as mentioned before. Stereospecificity in E2 Elimination: (1)

CH H 3 CH3 H Br Zn (ethanol) Br heat H H CH3 CH meso-2,3-dibromobutane 3 E-but-2-ene

Meso(R,S)-2,3-dibromobutane on debromination(–Br2) with Zn/alcohol gives E-but-2-ene only. This is because, the two Br atoms (leaving groups) lie anti w.r.t each other in the TS of meso, as shown above.

H CH3 CH3 H Br Zn (ethanol) Br heat CH 3 H CH3 H active isomer Z-but-2-ene Any active(d or l) i,e (R,R) or (S,S)2,3-dibromobutane on debromination gives Z-but-2-ene only. Again this is possible by meeting the antiperiplanarity condition of the two leaving groups. These are called stereospecific reaction. A particular stereoisomeric reactant gives a particular stereoisomeric product in stereospecific reaction. (2)

CH Et CH Et 3 CH3 Et 3 CH3 Et Br KOH(alc.) Br KOH(alc.) H H - HBr - HBr H H CH CH3 CH CH3 H 3 H 3 E-3-methylpent-2-ene threo-2-bromo-3-methylpentane Z-3-methylpent-2-ene erythro-2-bromo-3-methylpentane Dr. S. S. Tripathy 33 GOC-III(Introductory Reaction Mechanism) Another such reaction is given above. Dehydrobromination (–HBr) of any threo-2-bromo-3- methylpentane gives Z-alkene while any erythro isomer of the same compound gives E-alkene. This proves that for E2 elimination the two leaving groups should remain anti w.r.t each other in the TS.

E1 Mechanism:

(1) Like SN1, E1 mechanism occurs when there is formation of carbocation in the rate determining step(1st order). So this step is common in both the two mechanisms. This mechanism does not show kinetic isotope effect as C–H bond breaking is not included in the rate determining step. ≈ KH/KD 1. (2) 30 and 20 alkyl halides eliminate by E1 mechanism when elimination is the minor product.That happens when weak bases like H2O(hydrolysis), ROH(alcoholysis) are used with these halides. 30 halide > 20 halide >>> 10 halide (3) 30 and 20 alcohols often react by this mechanism in presence of acid and heat. Often the carbocation undergoes rearrangement from a less stable to more stable state by hydride or alkyl shift before the elimination of β-H atom. (3) The major product in all these cases is due to Saytzeff elimination(more substituted alkene).

(4) SN1 reaction always competes with E1. At room temperature, often SN1 product is major. Action of heat favours E1 product. (5) Antiperiplanarity of the leaving groups is not a requirement in the E1 mechanism. Solvolysis of 30 halides:

CH3 H CH3 fast CH sl o w 3 H2O CH C Br CH C CH C CH 3 2 - + 2 3 - Br H3O CH3 CH3 30 stable carbocation Note that this is hydrolysis of tert-butyl bromide, which gives tert-butyl alcohol as the major product by SN1 mechanism. However, there will some elimination product(minor) and that is β formed by E1 mechanism. H2O is a weak base and cannot abstract a -H atom efficiently, hence – – favours SN1 product. Note that if we take strong base like OH , OEt etc. then the major, in fact the exclusive product will be elimination and that too by E2 mechanism. We shall return to a 0 comparative study between SN and E mechanisms soon. 2 halides will behave similarly but there will be some SN2 and E2 that will contaminate the mechanism. But still the elimination will be the minor product. Acid Catalysed Dehydration of Alcohols:

CH3 CH3 H CH3 fast CH3 H H sl o w H2O CH C CH C CH C CH 3 OH CH3 C O 2 - + 2 3 fast H3O H heat CH3 CH3 CH3 30 stable carbocation

Here in the first step, the –OH group is protonated, so that it is lost as H2O(a good leaving group) in the second step, which is slow and rate determining. Then on heating a β-proton is eliminated to give alkene. In case of unsymmetrical alcohols, we get a mixture of at least two alkenes, more substututed(major) and less substituted(minor). Rearrangement of carbocations:

34 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

CH3 CH3 CH3 CH3 CH3 CH3 conc.H SO CH C CH CH 2 4 3 3 heat CH3 C C CH3 + CH3 C CH CH2 + CH2 C CH CH3 major CH3 OH CH3

minor Mechanism:

CH3 CH3 CH3 H - H2O CH3 C CH CH3 CH3 C CH CH3 CH3 C CH CH3 20 carbocation CH3 CH CH3 OH 3 OH2 (I)

CH3 shift rearrangement

CH3 CH3 - H CH3 CCCH3 CH3 CCCH3

CH3 CH H Major product(rearranged) 3 (II) more stable 30 carbocatio (Saytzeff Product)

CH3 CH3

CH3 C CH CH2 CH3 C CH CH2 CH H 3 CH3 (I) minor product (anti-Saytzeff product)

CH H 3 CH3

CH CCHCH 2 3 CH2 CCHCH3 CH (II) 3 CH3 minor product (anti-Saytzeff product) First a less stable 20 carbocation(I) is formed at the site of –OH group after dehydrataion which quickly rearranges by a methyl carbanion shift from the adjacent position to form a more stable 30 carbocation(II). This on deprotonation from the more hindered position gives the major product(more substituted alkene). However, carbocation (I) independently gives another minor product(anti-Saytzeff) by deprotonating from terminal position. Similarly carbocation (II) also give the anti-Saytzeff elimination product which is also minor. So out of the three products formed from dehydration of the above reactant, the major product is 2,3-dimethylbut-2-ene, which is the Saytzeff product after carbocation rearrangement. Fortunately none among the three exhibits stereoisomerism(geometrical isomerism), otherwise the number of products would have been more. N.B: Once carbocation is formed, it undergoes all types of possible rearrangments from less stable to more stable state and then gives different products. We shall discuss more about carbocation rearragement later.

Dr. S. S. Tripathy 35 GOC-III(Introductory Reaction Mechanism) Pyrolysis of sulfonate salts of Menthol: You remember that in E2, we took the example of menthyl chloride and neomenthyl chloride for elimination under antiperiplanarity(diaxial) requirements. If we convert menthol to a sulfonate ester(p-acetylaminobenzenylsulfonate) and heat, it we get a mixture of both 3-menthene and 2- menthene with former being the major product(Saytzeff). This happens via E1 mechanism. Here carbocation is formed due to heavy size of the leaving group whose expulsion relieves large steric strain in the molecule. Once carbocation is formed, then antiperiplanarity is not a requirement. β-proton is abstracted from either side to form two alkenes, Saytzeff product being the dominant one.

H CH3 CH ClSO2 NHCOCH3 H 3 H tBu tBu H H H OH O p-acetylaminobenzene sulfonate ester SO2 of menthol Menthol NHCOCH3 E1 heat

H H CH CH3 3 tBu + tBu

major minor 3-menthene 2-menthene

E1CB Mechanism( Elimination Unimolecular Conjugate Base) :

(1) This is rare case of elimination mechanism, which is opposite of E1. Here thε β-proton is lost first to the base in the rate determining step to form a carbanion(conjugate base of the carbon acid). This step is first order as it is only dependent on substrate concentration(not on base concentration). (2) In the 2nd step the lone pair of carbanion forms the pi bond and the other leaving group at the α-carbon is lost.

(3) This mechanism operates particularly when there is –M(EWG) group like –NO2, –CN, –COR etc bonded to β-carbon which stabilises the carbanion. In other words, the β-H of the substrate should be acidic in nature. Moreover, the other requirement is that the leaving group at α- position. should be poor(unlike halide ions) like RO–, CN–.

H B- RC CH2 L RC CH2 L RC CH2 - BH - L EWG sl o w EWG EWG

Where EWG = an electron withdrawing group i.e a –R/–M group like –NO2, –COR etc. and L = leaving group which is usually poor in nature –OR, –CN etc.(will leave as RO–, CN–).

36 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism) Example 1:

H Me Me OH- HC CH OMe HC CH OMe HC CH Me - H O - OMe C O 2 sl o w C O CO Ph Ph Ph 2-methoxy-3-benzoylpropane 1-benzonylpropene You can see here that the carbanion is resonance stabilised by the Ph–CO– group. This has enabled a poor leaving group like MeO– to leave in the 2nd step. This is because, a stronger base(carbanio) can displace a weaker base RO–. Is it not interesting ???? Only under acidic conditions –OR group leaves because it is protonated and the actual leaving group is neutral ROH which is a good leaving group. But under basic conditions, –OR does not usually leave as RO–. It happens in rare cases like this. Example 2:

H Ph Ph N(CH3)3 NC C CCN NC CCCN NC CCPh - CN CN CN CN CN CN CN 1-phenylethane-1,1,2,2-tetracarbonitrile 1-phenylethene-1,2,2 tricarbonitrile

In this case, a weak base like (CH3)3N is able to facilitate the formation of carbanion as it is highly stablised by so many EWGs. Here the the other leaving group is CN–(incredible)!!!). Example 3:

OR O OR - OH - OR C - H2O O OH carbonyl compoun conjugate base hemiacetal Hemiacetals of carbonyl compounds eliminate under basic conditions by E1CB mechanism. First the conjugate base of the hemiacetal is formed. In this case the oxide ion(not carbanion) and then RO– group leaves in the second step. ______SYN ELIMINATION It can be divided into two types. (i) SYN not involving cyclic TS (Use of Reagents) (ii) SYN involving cycic TS called Ei (No use of Reagent) OR Thermal SYN

(i) SYN not involving cyclic TS: When E2 TS(antiperiplanar) is unfavourable due to large steric hindrance, syn elimination occurs.

Both Threo and Erythro- 3-bromo-2,2,5,5-tetramethyl-3,4-diphenylhexane with alcoholic KOH gives the same E-alkene. Erthro isomer elminates by E2 mechanism as the antiperiplnanar TS is stable, while Threo isomer eliminates by SYN mechanism as the antiperiplanar TS is unstable(unattainable) due unfavourable steric hindrance.

Dr. S. S. Tripathy 37 GOC-III(Introductory Reaction Mechanism)

Br Ph tBu Ph OH- tBu ++H O Br- H 2 SYN tBu tBu Ph Ph Threo compound trans-alkene

Ph tBu Ph tBu

- - H Br OH ++H2O Br E2 tBu tBu Ph Ph Erythro compound trans-alkene Antiperiplanar state of threo compound will bring the bulky tert-butyl groups close and so also the Ph- groups, which would make the TS highly unstable. Hence it prefers to eliminate by SYN mechanism. On the other hand the erythro isomer has a stable antiperiplanar TS and hence eliminate by E2 mechanism.

(ii) Ei (Elimination intramolecular/internal): This is also calle Thermal SYN or Pericyclic SYN mechanism which goes via a cyclic TS. Cyclic TS is energetically more favourable than the E2 elimination in this case.

Pyrolysis of xanthate esters(Chugaev reaction), 30-amine oxides(Cope reaction) and acetates produce alkenes by this Ei mechanism. Interestingly, this mechanism additionally makes the Hoffmann elimination the major product. Chugaev Reaction: Alcohol is first converted to xanthate ester by treating with NaH followed by CS2 and finaly with CH3I.

S Et t Bu Et t Bu NaH C CH3 CH CH OH CH3 CH CH O S -(NaOH +H2)

Et t Bu S Et t Bu S

CH3 I CH3 CH CH OCS CH3 CH CH OCSMe - I xanthate ester

(ii) Heating of the xanthate ester to 120 - 2000C in vapour phase brings out syn elimination of β α the two leaving groups namely H-from the -carbon and –OCS2Me from the -carbon.

S C MeS O tBu  H tBu H H + COS + MeSH 2000C  Et Me Me E-alkene Et Parti cul ar di astereoi somer (xanthate ester) 38 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism) This occurs by a one step(concerted or pericyclic mechanism) via a six-membered cyclic TS. The stability of the other products namely carbonyl sulfide(COS) and MeSH(methyl mercaptan) make the mechanism in favour of SYN. It is a first order reaction. If a particular diastereomeric reactant is taken, we get a particular diastereomeric product. In the above example, if we take the other diastereoisomer(by making one switch at any one chiral centre), we would get Z-alkene in stead of E-alkene. I have taken the above example to explain the mechanism. I am not sure whether such a molecule has actually taken for actual reaction. I could have simplified it for 10 alcohol, where i could not have shown you the stereospecificity of the reaction. I could have taken a 20 alcohol with alkyl group like i-Pr or Me instead of t-Bu, in which the Hofmann product would have been the major product, which would actually have been better. But in no other source, you will get such Hofmann product, though there is mention in some rare literature, that Ei mechanism is not only SYN, but also Hofmann. To avoid all these confusion i took a 20 alcohol which bonded to t-Bu group such that β-H atom is available on one side.

Cope Reaction: N-Oxides of 30 amine is first prepared by oxidising the amine with peroxyacids like mCPBA(m- chloroperoxybenzoic acid). Sometimes H2O2/MeOH also brings about its formation.

R' R' mCPBA (R )(R )CH C(R )(R )N O (R1)(R2)CH C(R3)(R4)N 1 2 3 4 R' 30 amine R' Amine N-oxide Then the N-oxide is merely heated 80 - 1600C to get an alkene from the alkyl part containing β-H atom. If more of such alkyl groups are available, then a mixture of alkenes are formed with Hofmann product being the major one. The mechanism is SYN elimination with 5- membered cyclic TS.

NR'2 R3  R4 R H 3 R4 80-1600C + R'2NOH  dialkyl hydroxyl R amine R2 1 R1 R2 (particular diastereoisomer) (particular diastereoisomeric 30-amine N-Oxide alkene) (N:B: Don’t confuse this Cope reaction with Cope rearrangement. The latter one is completely different. Of course that is also a concerted pericyclic reaction and is also thermally initiated, but is different one. We shall talk about it later. Example:

N N O H H2O2/MeOH heat + (CH3)2NOH

Pyrolysis of Acetates:

Alkyl acetates in which there is at least one β-H in the alkyl part udergoes pyrolytic elimination of acetic acid at a higher temperature of 400 - 5000C to form alkene by SYN elimination with six membered cycli TS. Hence it is also a stereospecific reaction. Dr. S. S. Tripathy 39 GOC-III(Introductory Reaction Mechanism)

CH3

C R3 O R O 4

CH3COOH R + H 3 heat R4 R1 R R1 R2 2 Diastereomeric acetate Diastereomeric alkene

SUBSTITUTION versus ELIMINATION Since Substitution and Elimination go hand in hand, we need to discuss in greater details. 30-Substrate(Alkyl halides or the like)

(I) USE OF STRONG BASE:

– – – Eg. OH , RO , CN , CH C etc Major Product : Elimination 100% : Mechanism : E2 Minor Product : Substitution : NIL In the presence of a strong base whichis also a strong nucleophile and in the presence of aprotic weakly polar solvent eg EtOH, acetone etc, the product is 100% elimination, that too by E2 0 mechanism. SN2 attack is not possible in a 3 substrate. Here you might be having some apprehension about the E2 mechanism. If it is a 30 substrate, then why not E1 mechanism ??

Sorry. This is not true. If carbocation will be formed, the reaction will be hijacked by SN1 i.e substitution will be major. In a 30 substrate eg. tert-butyl chloride, there are 9 β-atoms and the probability of H-abstraction from periphery by a strong base is nine times greater. So E2 opertes, not E1.

CH H 3 CH OH 3 - + H2O + Cl CH2 C CH3 CH2 C E2 CH3 Cl methylpropene(100%)

(II) USE OF WEAK BASE: (Solvolysis Reactions)

0 3 substrates react with weak bases like H2O(hydrolysis reaction), ROH(alcoholysis) etc. to give substitution as the major product by SN1 mechanism. The minor product is elimination by E1 mechanism. Since the nucleophile itsself is a polar solvent, it facilitates the formation of carbocations. Since these are weak bases, abstraction of β- H atoms is disfavoured.

Major Product: Subsitution : Mechanism: SN1 Minor Product : Elimination Mechanism: E1

40 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

CH3

CH3 C OH H2O CH CH3 3 - H CH3 H2O SN1 (about 80%) Majo CH3 C Br CH3 C - Br CH3 CH3 E1 - H CH3

CH2 C CH3 (about 20%) Minor

Tert-butyl bromide react with water to give about 80% tert-butyl alcohol(substitution product) by

SN1 mechanism and nearly 20% methylpropene(elimination product) by E1 mechanism. 10- Substrate ( 10- alkyl halide or the like)

(I) Use of both Strong and Weak Base:

Major Product : Substituion - Mechanism : SN2 Minor Product : Elimination: Mechanisml : E2

Here in no case, primary carbocation can form, as it is too unstable. So SN1 and E1 are ruled out. Using a strong base hijacks reaction in favour of substitution because strong bases are also in general strong nucleophiles. Hence they like to play better roles as nucleophiles to go for rear 0 β attack in a 1 halide(SN2) rather than act as base to undertake abstraction of -H atom(elimination).

EtO- Na+(alc.) CH3CH2CH2CH2 Br CH3CH2CH2CH2 OEt + CH3CH2CH CH2 ethyl n-butyl ether but-1-ene (SN2) 90% (E2) 10%

(II) Use of strong but bulky base like t-BuO– K+ :

Major Product : Elimination : Mechansim: E2

Minor Product : Substitution : Mechanism: SN2

Since the bulky base t-BuO– cannot favour rear attack even to a 10 halide, it will prefer abstraction of β-H atom from the periphery. So elimination becomes the major product.

H t-BuO K CH3CH2CH CH2 CH3CH2CH CH2 + CH 3CH2CH2CH2 Ot Bu n-butyl t-butyl ether Br but-1-ene (85%)E2 (15%) SN2

N.B: We have already discussed that by using a bulky strong base like t-BuOK, we get elimination to give major product. 2-bromobutane also give but-1-ene as major product, we already know such bulky bases prefer Hoffmann elimination.

Dr. S. S. Tripathy 41 GOC-III(Introductory Reaction Mechanism) 20- Substrates (20- halides or the like):

All SN1, E1, SN2 and E2 are possible in this case which depend on the nucleophilicity and/or base strength, solvent polarity, temperature.

(A) Substitution Favoured: (i) Strong Nucleophile but NOT a strong base eg. RS–, HS– etc.

Mechanism: SN2 In this case the elimination will be almost absent. (ii) Weak base :

eg. H2O, ROH (solvolysis) 0 Mechanism: SN1 (similar to 3 halides) In this case too, the minor elimination product is formed by E1 as the base strength is low.

(B) Elimination Favoured: Use of strong base (Similar to 30 halides) Mechanism: E2

In this case, there will be minor substituion product by SN2, unlike total extinction of the substitution product for 30 halides.

SEt - + CH3CH2S Na (SN2) strong nucleophile CH3 CH CH3 Br OEt EtOH CH3 CH CH3 Weak base/nucleophile CH3 CH CH3 + CH3 CH CH2 (major) SN1 (minor) E1 OEt EtO-K+ CH CH CH strong base 3 2 + CH3 CH CH3 (major) E2 (minor) SN2

– EtS is a strong nucleophile(soft base down the group), but a weaker base. Hence Here SN2 product is favoured. There will almost no elimination product. 0 EtOH is weak base but a polar solvent, so like 3 halide, the SN1 product will be major, E1 product will be minor. EtO– is a strong base,hence like 30 halide the major product is elimination(E2) and the minor is 0 SN2, as it is also strong nucleophile. In 3 halide, there was no substitution product, but here there will be appreciable substitution product, though minor.

42 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism) ADDITION REACTIONS: There are the following mechanisms for addition reactions.

(a) Electrophilic addition (AdE)

(b) Nucleophilic addition(AdN)

Electrophilic Addition (AdE) Reaction Mechansim:

(1) Alkenes and alkynes most often undergo addition by this mechanism. This is because they mostly act as nucleophile and hence they seek electrophiles. (2) The slow step often involves the formation of carbocation by the addition of the electrophile from the reagent(addendum), hence the name AdE. Since this is bimolecular(2nd order: i.e first order w.r.t substrate and first order w.r.t electrophile) it is best designated as AdE2 (3) In the second step, which is fast, the carbocation reacts with the nucleophile part of the addendum to form the product. (4) Stability of carbocation governs the direction of reaction. A more stable carbocation always is formed leading to the formation of the corresponding product.

(5) For simple alkenes and alkynes, which are unsymmetrical in structure(C=C bonded to different number of H atoms), only one product is formed depending on the stability of carbocation.

Hence AdE is usually regiospecific. For unsymmetrical alkenes/alkynes and the use of polar addendum the product is predicted from Markonikov rule. Markonikov Rule: The negative part of the polar addendum adds onto that carbon of the C=C, which bears less number of H atoms or more hindered carbon. Later we shall know that addition going via carbocations cannot be stereospecifc as we always get R/S mixture from either Z or E- alkene. However, those addition reactions which involve a cyclic intermediate or TS(no free carbocation) however, are mostly stereospecific. Z and E alkene gives different diastereoisomers. Wait a moment. (Note the difference between elimination and addition, while the former is regioselective, the latter is regiospecific especially for unsymmetrical alkenes)

(6) For alkenes, in which each carbon of C=C bears one H atom and the alkene is unsymmetrical structure-wise, then the reaction becomes regioselective. Two products are formed with appreciable amounts. But the major product can be predicted from the relative stabilities of the carbocations formed at the two positions. For that we have to use the hyperconjugation effect.

(7) Halogenation of alkenes/alkynes, though belong to AdE mechanism, takes place in a different style. In stead of formation of carbocation, a cyclic halonium ion intermediate is formed. We shall discuss about it later. (8) The reaction is exothermic in nature, as the breaking of a pi-bond requires less energy than formation of two sigma bonds.

(9) For addtion of HX, non-nucleophilic solvents such as nitromethane, acetonitrile, CCl4, hexane etc are used. Nucleophilc solvents like H2O, ROH etc are aovided as it completes with the nucleophile (X–) in the second step to form alcohol and ether respectively as co-products.

Greater the polarity of the solvent, greater is the rate of AdE reaction, as the formation of carbocation is facilitated with a polar solvent. Nitromethane(CH3NO2) and acetonitrile(CH3CN) solvents give much higher rates than nonpolar solvents like hexane and CCl4. (10) Since carbocations are very prone to rearrangements from a less stable to more stable state by either hydride shift or alkyl shift, often rearranged addition products are found in AdE2.

Dr. S. S. Tripathy 43 GOC-III(Introductory Reaction Mechanism)

Addition of Polar Addendum(HX) with Unsymmetrical Alkene( R–CH=CH2 type) + HX : eg HCl, HBr, HI, H2O(H ), H2SO4(H–OS3H) etc. (Note that weak acids like HCN and most organic acids do not add onto alkenes as they do not produce good amount of H+)

CH3 CH CH2 + HX ?

X slow X CH CH CH + H CH CH CH CH CH CH 3 2 3 3 fast 3 3 Unsymmetrical alkene stable 20 carbocation only product Propene reacts first with H+ to form a more stable 20 carbocation. Note that if H+ would have added on to C-2(middle one), the a 10 carbocation would have formed at the terminal position(1), which is extremely unstable. So only the more stable carbocation is formed. In the second step, the carbocation reacts with the nucleophile(X–) to form the product; 2-halopropane. Note that 1-halopropane is not formed at all. Example:

CH3

CH3 C CH2 + HBr ?

CH3 CH3 CH3 Br CH3 C CH2 + H CH3 C CH3 C Br

CH3 CH3 most stable 30 carbocation CH CH 3 H 3 H2O - H CH3 C O CH3 C OH H CH3 CH3

Isobutene reacts with H+ to form the most stable 30 tert-butyl carbocation. Then it reacts with Br– to form tert-butyl bromide as the ONLY product. If water is present in the reaction mixture, then the competing hydrolysis reaction will occur. H2O will react with the carbocation to form the protonated alochol first, which on deprotonation gives the alcohol product. That is why addition with HX is preferred in pesence of a non-nucleophilic solvent like acetonitrile or nitromethane.

Note that only H2O does not react with alkenes/alkynes, as it is non-electrolyte and cannot + + produce electrophile(H ). Only in presence of acids(H ), H2O does react with alkenes to form alcohol.

44 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

Like, SN1 mechanism, the carbocation is shown by a minimum point in the middle. The difference is that in SN1, heterolytic cleavage of C–X bond is responsible for its formation, while in addition, the addition of the electrophile does the job. Since TS-1 is much greater than TS-2, the first step is slow and rate determining. The second step is common to both the mechanisms. The nucleophile attacks the carbocation to form the product. N.B: Later we shall see that carbocations which are not simple and contains branches are prone to rearrangement. In such case a 1,2-hydride shift and alkyl shift occurs to result more stable carbocations. Hence AdE products will be more than one, in such case. Addition of HX with Structurally symmetrical and unsymmetrical alkenes

For symmetrical alkenes like ethene, but-2-ene etc. only one product is obtained, and for unsymmtrical alkenes with unequal number of H atoms bonded to sp2 carbon(C=C), we get also one product according to Markonikoff’s rule, just explained before. But for alkene having same number of H atoms bonded to sp2 carbons but structurally unsymmtrical will give two different products, with one being major. In some other cases, where stable carbocation is formed on one direction, one product is obtained almost exclusively. In still other cases, more than 2 two products are formed with rearranged skeleton, because of rearragement in carbocations. See the following cases.

CH2 CH2 + HCl CH3 CH2 ethylene ethyl chloride

CH3 CH CH CH3 + HBr CH3 CH2 CH CH3 but-2-ene 2-bromobutane In the two examples given above, only one product is obtained as the alkenes are structurally symmerical. So Markonikoff’s rule has no role to play in this case.

CH3 CH CH CH2 CH2 CH3 + HI CH3 CHCH2 CH2 CH2 CH3 hexa--2-ene I 2-iodohexae (major)

+ CH3 CH2 CH CH2 CH2 CH3 I 3-iodohexane In this case, both 2-iodo and 3-iodohexanes are produced from hex-2-ene, the former being major. Can you tell its reason ? It is due to relative stability of the 20 carbocation intermediate form during the reaction. More the number of α-H atoms, more is the number of hyperconjugation strcutures and hence more is the stablity.

CH CHCH CH CH CH CH3 CH CH CH2 CH2 CH3 + H 3 2 2 2 3 5 -H atoms(more stable)

+ CH3 CH2 CH CH2 CH2 CH3 4 -H atoms (less stable) I

I CH3 CHCH2 CH2 CH2 CH3 major I CH3 CH2 CH CH2 CH2 CH3 minor I

Dr. S. S. Tripathy 45 GOC-III(Introductory Reaction Mechanism) Out of the two sec- carbocations, the first one is relatively more stable as it has 5 α-H toms(6 hyperconjugation structures) and hence form the major product. The second one has 4 α-H atoms. Rearranged product:

CH3 CH3 Cl HCl CH3 CH CH CH CH3 CH3 CH CH2 CH CH3 +

CH3 CH3

CH3 CH CHCH2 CH3 + CH3 C CH2 CH2 CH3 Cl Cl rearranged product (major)

In this case, we get three products, two normal products (addition ocurring at the C=C) and the third one is the rearranged product in which(-Cl bonds to the adjacent carbon atom of C=C. There is a hydride ion transfer taking place for converting a less stable 20 carbocation to a more stable 30 carbocation. This forms the major product as the it is resulted from the most stable carbocation. You can predict which among the two normal products is in higher quantities by comparing the relative stabilities of the 20 carbocations, as we did before. See the mechanism for rearranged product.

CH3 H CH3 C CHCH CH3

CH3 CH3 H shift CH3 C CHCH2 CH3 CH3 C CH2 CH2 CH3 30 carbocation H 20 carbocation Cl

CH3

CH3 C CH2 CH2 CH3

The mechanism of the formation of the two normal products is not shown above. This is exactly the same as given in the previous example. N.B: More about rearrangement of carbocations will be discussed later.

More examples:

(1) CH3 CH CH OCH3 + HCl CH3 CH2 CH OCH3 (only product)

46 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism) Mechansim:

Cl H + CH3 CH CH OCH3 CH3 CH2 CH O CH3

CH3 CH2 CH O CH3 resonance stabilised carbocatio Cl

CH3 CH2 CH O CH3

Presence of a +M group(–OCH3) adjacent to the carbocation stabilised it. Hence the corresponding product is the only product formed.

Br (2) CF3 CH CH2 + HBr CF3 CH2 CH2 Mechansim:

+ CF3 CH CH2 + H CF3 CH CH3 (so does not form) unstable carbocation

Br Br- H + CF3 CH CH2 CF3 CH2 CH2 CF3 CH2 CH2 Here anti-Markonikoff addition product is the only product formed. This is because the 20 carbocation is adjacent to a very high EWG(–I Group) which will destablise it. Hence it will prefer to go via less stable 10 carbocation.

(3) HCl

In this case, ring expansion occurs from very strained four membered ring to a 5-membered ring having small strain. The high angle strain in the reactant (methylene cyclobutane) always seeks an opportuntity to relieve the strain by expanding the ring. That happens here through the formation of a 10 carbocation, in stead of a 30 carbocation at the vertex. Nearly 100 % product is ring expansion product i.e chlorocyclopentane. Mechanism:

1 H 1 3 2 CH2 Cl 3 2 Cl H 4 5 4 5

Cl HCl (4) CH CH CH 3 CHCH2 CH3

In this case, the carbocation adjacent to benzene ring will be formed. This is the benzyl carbocation which is more stable by M-effect than the other 20 alkyl carbocation in the other position. Hence the major product is due to that.

Dr. S. S. Tripathy 47 GOC-III(Introductory Reaction Mechanism) Mechanism:

H CH CH CH3 CHCH2 CH3

more stable benzyl carbocation Cl Cl

CHCH2 CH3

Conclusion: So you now know that addition of HX or in fact any polar addendum like this will not always follow Markonikoff’s rule. You have analyse through mechanism, as to acertain which carbocation is more stable or if there is a possibility of relief of angle strain due to ring expansion, then predict the product. Halogenation of Alkenes:

X CCl4 R CH CH R' X2 RCH CH R' alkene X Vicinial dihalide Salient Features:

1. Chroine(Cl2) and bromine(Br2) react with alkene to form vicinal dichloride and dibromide. Flourination is dangerously explosive and iodination is extremely slow. So diflouride and diiodide cannot be prepared by this method.

2. The mechanism is electrophilic addition(AdE2), but no carbocation is formed here. 3. Alkene being a nucleophile induces polarity in Cl–Cl or Br–Br bond and finally attacks onto it to capture a Cl+ and form a 3-membered cyclic halonium(chloronium or bromonium) ion intermediate, which is unstable. In this step, a halide ion (X–) is expelled out. So no carbocation is formed in this case, still it is an electrophilic addition reaction. Hence no rearranged product + is expected in this case. X2 provides the electrophile i.e X in the rate determining step. We can also justify this to be electrophilic addition by saying that the more hindered carbon possesses partial +ve charge(carbocationic character) in the resonance hybrid. 4. In the second step, the expelled halide ion (X–) attacks from the opposite side of the bromonium ion as a nucleophile. Since the side of the halonium ion is sterically hindered, X– ion has to attack from the opposite side. There is equal probabilities of its attack onto each of the two carbon atoms. The two carbon atoms develop appreciable electrophilicity due to their connection with bromine atom with +ve charge. While X– attacks, the haloinium ion ring breaks and the halogen atom of haloinium ion is opened up on the opposite side. Thus halogenation process becomes a ANTI addition process, i.e ultimately the two X atoms add onto the C=C on the two opposite sides. 5. Thus halogenation to C=C is a stereospecific reaction. A particular diastereoisomer(E or Z) gives a particular diastereoisomeric product(RR/SS or RS/SR). We shall see this case in details very shortly.

6. Aprotic solvents like CCl4, CHCl3 etc are used for halogenation. Protic solvents like – H2O, ROH etc. competes with X as nucleophile in the second step to join as –OH group(alcoholic) and –OR(ether) group. So alongwith dihalide, we get halohydrin and alkoxy halide respectively.

48 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

-  + Br Br Br C C Br2 + Br C C

bromonium ion cyclic intermediate

C C Br + Br C C

d/l pair OR MESO

Br Br Br

R CH CH 2 RCHCH 2 R CH CH2 most stable

More about Halonium Ion: In the above example, bromonium ion cyclic intermediate is formed, and if we take Cl2, then chloronium ion intermediate is formed. In this species, there are only two lone pairs on Br atom. One lone pair has been used to form bond with a carbon atom. Br is divalent here with a +ve charge. It has two more RSs from which 20 carbocation contributes more. Hence there is a partial +ve charge in the 20-carbon atom(more hindered carbon). Therefore we say this to be electrophilic addition reaction. Examples:

Br Br CH3 H CCl4 CC + Br2 H3C CCCH3 1. H CH3 H H E-but-2-ene meso-2,3-dibromobutane

H Br Br H H H CCl4 CC + Br2 H3C CCCH3 + H3C CCCH3 H3C CH3 Br H H Br Z-but-2-ene d/l-2,3-dibromobutane(rac. mix.)

Anti Addition: (Only the addition is shown, not the detailed mechanism)

-  + Br Br H3C Br H C H H3C C C Br2 H3C H Br H CH H C 3 Br C C CH3 H CH Br 3 E-but-2-ene MESO- product

Dr. S. S. Tripathy 49 GOC-III(Introductory Reaction Mechanism)

-  + Br Br H3C Br H C H H3C C C Br2 H3C H Br H3C H + H3C C Br C C H H C H Br 3 Z-but-2-ene d/l pair

Halogenation in presence of protic solvent(Cl2/H2O):

OH Cl

H2O CC+ Cl2 CC + CC Cl Cl chlorohydrin vic. dichloride major minor

If alkene reacts with Cl2 or Br2 in presence of H2O or ROH, then the mjaor product becomes halohydrine or alkoxy halide. As mentioned earlier, the solvent molecules(present in abundant quantity) will compete as a nucleophile with X– ion in the second step to give the halohydrin or alkoxyhalide as the major product, alongwith vicinal dihalide as the other minor product. Below, the formation of product from the competing reaction by the solvent molecule is shown.

-  + Cl Cl Cl C OH2 C + Cl2 C

C chloronium ion cyclic intermediate - H - H Cl

C C Cl OH C C

OH d/l- chlorohydrin

Halogenation in presence of halide ion: – When ethene reacts with Br2 in presence of Cl (NaCl), then the product contains 1- chloro-2-bromoethane(minor) alongwith dibromoethane(major). This indicates that Cl– competes with Br– in the second step(like H2O in the previous example) to form chlorobromo product. Br Br Br Cl Br2/NaCl CH2 CH2 CH2 CH2 + CH2 CH2 major minor 50 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism) Stereospecificity will also be observed if there would be chiral centres, thus confirming the formation of bromonium ion intermediate(rather than carbocation).

Exception: 1. Not always the halogenation is ANTI in terms of stereospecificity. There are cases where, stable carbocations are fomed and the stereospecificity is lost. This happens particularly when there is phenyl(Ph–) group in conjugation with C=C or conjugate dienes(discussed later). The stability of benzylic carbocation drives the reaction in favour of carbocation formation and hence the loss of ANTI addition.

Addtion of Br2 to cis and trans 1-phenylpropenein CCl4 is nonstereospecific. Both the diastereoisomers are fomed from each of cis and trans reactant.

Br Br Br Ph CH CH CH Br- Ph CH CH CH3 3 + benzylic carbocation E or Z Br-

Br Br Br Ph CH CH CH + Ph CH CH CH3 + 3 (I) Br (II)

Br Ph CH CH CH Ph CH CH CH3 + 3 Br (III) Br Br (IV)

Each of E and Z alkene gives the same mixture with varying composition. There will be one erythro pair ( I and IV) and one threo pair( II and III). Had the reaction gone via bromonium ion cyclic intermediate, the E-alkene would have given erythro pair and Z-alkene would have given the threo pair. But in this case the stereospecificity is lost due to formation of carbocation. This happens due to high stability of the benzylic carbocation which drives the mechanism in its favour and against the bromonium ion intermediate. 2. Change of solvent also in some cases removes stereospecificity. For example, Stilbene

Ph-CH=CH-Ph, on reaction with Br2 in CCl4 (dielectric constant = 2.24), the product is 90 – 100% ANTI stereospecific, i.e trans-stilbene gives meso product and cis-stilbene gives rac. mixture. Using solvents with increasing dielecric constants, the there is a gradual loss in stereospecificity. When dielectric constant becomes 35 or more ( eg. acetonitrile solvent having d.c = 36.64) the stereospecificity is completely lost, as the the mechanism goes via carbocation formation, in stead of bromonium ion intermediate.

3. Where there is scope for carbocation rearrangement, it happens. For example addition of Cl2 with trans- 1,2-ditert-butylethene(not IUPAC name), we get a rearranged dichloride(not vicinal dichloride) via carbocation formation.

Me Me Me3C H CC + Cl2 Me3CCHCH C Me H CMe3 Cl Cl (all the four optical isomers) (You are advised to work out the mechanism of the above reaction yourself in a sheet of paper)

Dr. S. S. Tripathy 51 GOC-III(Introductory Reaction Mechanism) Other Stereospecific Additions to C=C involving cyclic intermediate or TS :

Note that the following addition reactions do not come strictly under AdE type. We feel like discussing here as each of them involves cyclic intermidate or TS similar to halogenation.

(I) Hydroxylation by Baeyer’s reagent or OsO4/NaHSO3 :

1. Alkenes react with cold dilute alkaline KMnO4 called the Baeyer’s reagent to form vicinal diols. On heating or using conc. alkaline KMnO4, the C=C breaks down to form two compounds namely ketones/carboxylic acids(to be discussed later). 2. The addition mechanism is SYN (in contrast to halogention of alkene which was an ‘anti’ addition) – 3. In the first step, MnO4 ion forms a five membered intermediate(manganate ester) only – by attacking onto the same face of the C=C, hence the addition process is SYN. Here MnO4 acts as electrophile which makes use of vacant d-orbital(extending beyond -O- atoms) to facilitate electron donation from the nucleophilic alkene C=C. There is also back bonding between oxygen lone pair to antibonding π* of C=C before cyclisation. These two interactions result in cyclization to form the manganate ester. Note that there is no metal–carbon bond in the cyclic ester. 4. In the second step, the manganate cyclic ester is broken down by alkali to form the vic. diol and green manganate ion. Thus alkene is oxidised while permanganate is reduced to manganate. 5. Often this reaction is used in the laboratory as a test for unsaturation. The purple colour of permanganate is discharged in this test. Usually the green colour is not detected by eye. It is the discharge of purple colour which is usually observed.

OH O O O O OH2 pH > 8 2- Mn Mn + MnO4 OH2 O O O with 2 proton transfer O OH

Vic. Syn diol In this case, E-but-2-ene will give d/l(rac.) butane-2,3-diol and Z-but-2-ene will give the meso product, just the opposite of halogenation of alkene.

OH H H3C H3C H OH H C Baeyer's reagent 3 H OH OH OH H H3C H3C H C H 3 H OH OH MESO-butane-2,2-diol Z-butene

OH H H3C H3C H OH H C Baeyer's reagent 3 H OH OH + OH CH H H 3 H CH3 CH 3 OH OH E-butene (d/l) butane-2,3-diol (ACTIVE) 52 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

Use of OsO4: 1. This is a substitute for Baeyer’s reagent to form SYN vicinal hydroxylation product.

2. Alkene reacts with OsO4 in presence of pyridine(solvent) to form a 5-membered cyclic osmate ester like the manganate ester with Baeyer’s reagent.

3. In the second step the cyclic intermediate is broken down by NaHSO3 to form vicinal diol along with H2OsO4. The ON of Os was +8 in OsO4 which changes to +6 in osmate ester intermediate and is retained in H2OsO4. Alkene is oxidised and OsO4 is reduced. Note that to – regenerate OsO4 for reuse, sometimes t-BuO–O-H/OH (tert-butyl hydroperoxide) is used in stead of NaHSO3 in the 2nd step. NaHSO3 provides the alkaline environment and catalyses the breaking down of the cyclic ester.

O OH O O NaHSO3 HO O O Pyridine Os Os + Os O O 2 H2O O O (proton transfer) HO O OH SYN Addition product(vicinal diol)

The stereospecificity of this reaction is exactly the same as in case with Baeyer’s reagent.

(II) Hydroxylation of Alkenes through Epoxidation: 1. Alkenes are first converted to epoxides (alkene oxides/oxiranes) by reacting with a peroxy acid like m-CPBA(m-chloropeoxybenzoic acid) or the like or even heating(4000C) with

O2 in presence of Ag. – + 2. In the 2nd step, the epoxide is hydrolysed either by a alkali(OH ) or H2O(H ) to form the vicinal diol. 3. The addition mechanism is ANTI. The two OH group join on the opposite side of C=C. This is similar to halogenation of alkenes. Just like the cyclic bromonium ion intermediate can only be attacked by the nuncleophile from the opposite side of the ring bromonium bridge, in – epoxide too, the nucleophile (H2O) or OH will attack from the opposite side of oxide bridge in two possible sites of C=C, thus resulting ANTI diol. 4. Stereospecificity: It is analogous to halogenation of alkenes. Z-alkene (say Z-but-2-ene) d/l vicinal diol E-alkene(say E-but-2-ene): meso vicinal diol (or diastereomeric product) Base Catalysed Hydrolysis :

CH3 H CH3 CH3 - OH H H OH OH m-CPBA O CH3 H d/l pair

CH3 OH- OH H CH3 H CH3 H Z-but-2-ene

CH3 H

OH (d/l) butane-2,3-diol

Dr. S. S. Tripathy 53 GOC-III(Introductory Reaction Mechanism)

H CH3 H H - OH CH3 OH CH3 OH m-CPBA O CH3 H MESO diol CH3 OH- OH CH3 H H H CH E-but-2-ene 3

CH3 H

OH meso- butane-2,3-diol

Acid Catalysed Hydrolysis: The epoxide(oxirane) is protonated in presence of acid, thus the leavability of –OH becomes greater here(in contrast to –O– in basic hydrolysis). In last step, similar nucleophilic attack by water takes below the –O– bridge on both the carbon atoms like basic medium. This is followed by an additional step i.e deprotonation to form the vicinal diol.

HCH3 H H OH CH3 H2O H CH3 OH (i) m-CPBA O - H CH3 H (ii) H MESO diol CH3 OH H2O H CH3 - H H H CH E-but-2-ene 3

CH3 H

OH meso- butane-2,3-diol

Thus E-but-2-ene will give meso diol(shown above) and Z-isomer will give d/l pair(shown before). Unsymmetrical Alkene:

R-CH=CH2 Note that for unsymmetrical alkene, the nucleophile attacks onto the less hindered position i.e the carbon atom of C=C bearing more number of H atoms for basic hydrolysis and onto the more hindered position in acidic hydrolysis. Since after protonation of epoxide, the leavability of the leaving group is enhanced, the breaking of C–O bond forming a more stable carbocation is held first before the nucleophile attacks the carbon. We shall discuss more about it in the chapter – – ‘Alkenes”. The nucleophile may be OH , H2O, RO , ROH etc, but the preference of attack will be different in acidic and basic medium.

O OH O Na OEt CH3 CH3 HOEt CH3 C CH2 C CH2 CH3 C CH2 - OEt CH3 OEt CH3 OEt

– In the first case, EtO is basic, hence it makes SN2 attack onto less hindered position and forms the terminal alkoxy alcohol( 1-ethoxy-2-methylpropan-2-ol). We take a alcoholic solution of NeOEt in this case.

54 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

H O CH3 OH CH3 CH3 O H EtOH CH3 C CH2 CH3 C CH2 CH3 C CH2

OEt H - H

CH3 OH

CH3 C CH2

OEt

When same compound reacts with EtOH in presence of acid, then a different product is formed. The -O- atoms is protonated first to form an oxonium ion. By that the electrophilicity of the C- 2 is greatly enhanced. If we draw RSs, we find the stability of 30 carbocation RS makes C-2 much more electrophilic than C-1. Hence the nucleophile(weak) EtOH attacks onto the more hindered position i.e C-2 to form 2-ethoxy-2-methylpropan-1-ol. You can alternatively say, that the most stable 30 carbocation(though not free as being pulled by oxide ion forming an intimate ion pair) is formed first and then the nucleophile attacks the carbocation. I preferred to explain through the RSs concept. In any case, the product is different if we use a weak base like EtOH in acidic condition from the product when we take strong base NaOEt with strong nucleophile OEt–.

(III) Catalytic hydrogenation of Alkenes: (1) Finely divided Ni, Pd, Pt catalyses the hydrogenation of alkenes at 200 - 3000C.

(2) Both H2 and alkene are adsrobed by the metal surface and the two H atoms of H2 get added onto the C=C in two steps. (3) The addition mechanism is SYN, as both the reactants are simultaenously adsorbed on the surface of the metal and hence H atoms cannot approach from the opposite sides of the C=C.

CC HH H H H H CC

CC CC H H H H

In the first step, H2 is adsorbed and bonded to metal surface as atoms, H–H bond thus breaks. In the 2nd step, alkene forms a π-complex with the metal (vacant d-orbital of metal is used for the purpose). In the 3rd step, one H atom joins one carbon atom from below, as it cannot go above the C=C plane. In the final step the second H atom joins with the other carbon from the same side, thus resulting SYN addition product.

Dr. S. S. Tripathy 55 GOC-III(Introductory Reaction Mechanism) Stereospecificity:

H H Me Et Me Et H Me Et H H H Me Et Me Me Et Et H MESO H Z-3,4-dimethylhex-3-ene meso-3,4-dimethylhexane

H H Et Me Et Me H Et Me H H H + Me Et Me Me Et Et H rac. pair H E-3,4-dimethylhex-3-ene (d/l) 3,4-dimethylhexane So where there will be development of chiral centres, SYN additions to E and Z alkene produces different diastereoisomeric pair (in symmetrical compound like above, one product is MESO from Z-alkene). Reactivity order: ′ ′′ ′ ′′ ′′′ ′ CH2=CH2 > R–CH=CH2 > R–CH=CH–R > R(R )C=CHR > R(R )C=C(R )R More the degree of substitution, more is the inaccessability of the alkene to the surface of the catalyst due to increased steric hindrance.

(IV) Hydroboration-Oxidation(HBO) of Alkenes:

(1) Alkene first reacts with BH3 (B2H6/THF) to form alkyl borane. Successive addtion of two more alkene molecules finally gives trialkyl borane. (THF = tetrahydrofuran; a solvent)

These addition steps are Markonikoff in nature as in BH3, H is the –ve part and BH2 is the +ve part, as H is more electronegative than B. This takes place via a four memeberd cyclic TS in a single step(concerted) mechanism. So no carbocation is formed. Hence there is no scope of any rearranged products. – (2) In the second step, the trialkyl borane is oxidised by H2O2 in presence of OH to form an anti-Markonikoff alcohol.

(3) Essentially, HBO is equivalent to anti-Markonikoff addition of H2O to C=C, though ironically, the addition of BH3 to C=C is Markonikoff in nature. (4) The mechanism of addition is SYN, i.e the two groups (H and OH) add onto the C=C on the same side, both from TOP or both from BOTTOM, which occur to the same extent. Hence HBO is a stereospcecific reaction. The following Z-compound gives erythro rac(d/l) products while the E-isomer gives the threo d/l pair. Note that OH group(-ve part of H2O is being added onto less hindered carbon(bearing more number of H atoms) in the scheme shown below. Actually in the mechansim, it is really not addition of H2O onto C=C. This is a handy scheme for the students to work out the product formation, without writing the mechanistic details.

H H Et Me Et Me OH Et Me (i)BH3/THF OH H H - + (ii) H2O2/OH Me H Me Me H H OH erythro d/l pair OH (2S,3S) (2R,3R) Z-3-methylpent-2-ene 3-methylpentan-2-ol

56 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

H H Et Me Et Me OH Et Me (i)BH3/THF OH H H - + (ii) H2O2/OH H Me H H Me Me OH threo d/l pair OH (2R,3S) (2S,3R) E-3-methylpent-2-ene 3-methylpentan-2-ol

The other isomer E, gives the threo d/l pair alochols as products. Detailed Mechansim:

- +  + HBH2 - HBH2 H BH R H R H 2 BH3/THF + -

R' R'' R' R'' alkyl borane cyclic TS SYN addition Since the TS involves the carbocationic character of the more hindered carbon(though not a real – + carbocation is formed), hence H part of BH3 joins on it and BH2 part joins on the less hindered carbon. Thus addition is purely Markonkoff in nature. This monoalkyl borane will successively react with two molecules of alkene to form trialkyl borane in the same way as shown before, each time SYN addtion.

H H H THF 3 H2O2/ CC+ BH3 CH C B 3 CH C OH + B(OH) 3 - 3 (B2H6) OH 3 trialkyl borane

Trialkyl borane is formed by the reaction of 3 molecules of alkene and one molecule of BH3.

Note that BH3 comes from its stable dimer B2H6(diborane) during the reaction. The oxidation mechansim of one C–B bond is shown below. The other two bonds are oxidised similarly. The two alkyl units are written as BX2 below, where X stands for an alkyl group bonded to B.

OH HO OH HOO+ H2O

– Base removes a proton from H2O2 to form the hydroperoxide nucleophile(HOO )

OOH OOH

H BX2 H BX2 H OBX2 - 1,2-shift OH - OH - HOBX2

H O HOH H OH

- OH

Alcohol

Dr. S. S. Tripathy 57 GOC-III(Introductory Reaction Mechanism) HOO– ion attacks B atom of alkyl borane(Boron is electron deficient sextet, hence electrophilic) to form a B–O bond. Then alkyl group undergoes a 1,2-shift from B atom to O atom with the – – removal of OH . Then OH again attacks the electrophilic B and removes it as HOBX2 and the alkoxide ion is thus formed. On protonation of alkoxide ion from water, we get the neutral alcohol in the last step. Note that since –OH group is finally bonded to the less hindered carbon, we conclude that a anti- Markonikoff alcohol is obtained in HBO. But actually this carbon was bonded to B(which was electropositive) and later on due to the migration or 1,2-shift changed it to O atom. That is why it appears to be anti-markonikoff addition, which is not true.

Since –X2 parts are two alkyl groups of the same type as shown in the previous figure, same steps – with H2O2/OH are repeated two more times to finally give 3 molecules of alcohols(including the one shown) and H3BO3(orthoboric acid).

ADDTION TO ALKYNES:

(A) Electrophilic Addtion Reactions of ALKYNES:

Addition of HX to Alkynes:

1. Additon of HX to alkyne triple bond is Markonikoff regioselective and ANTI stereoselective. Note that we have used the word selective in both the terms ‘regio’ and ‘stereo’. So that we always get more than one products out of which one them is the major. 2. It takes place in two steps with the formation of vinyl halide in the first step with one mole of HX and then with the second mole of HX forms gem. dihalide. 3. This addition is slower than the addtion of HX with alkenes, although more exothermic. 4. The addition is more influenced by solvent type and the presence of catalyst, as these are slow reactions.

X X HX R C CH + HX RCCH2 RC CH3 X Markonikoff Regioselectivity

Cl CH3 H HCl CH3 CCCH3 + HCl CC CH3 C CH2 CH3

Cl CH3 Cl Anti- stereoselectivity Markonikoff Regioselectivity Z-alkene

In the first general example, you find that in both the steps, Markonikoff addition takes place. In the second specific example, which is a symmetrical alkyne(but-2-yne), there is no question of regioselectivity, however, there is stereoselectivity. The two groups H+ and Cl– are added onto the triple bond on opposite sides(Anti addition) analogous to halogenation of alkenes. Thus but- 2-yne gives Z-2-chlorobut-2-ene. Of course, in the second step, i.e with the addition of second mole of HCl, Markonikoff regioselectivity is observed and we get geminal dichloride.

58 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism) Mechanism: Since the addition of one mole of HX to a symmetrical alkyne; but-2-yne is stereoselective and that too ANTI additon is the norm, then certainly we can think of the kind of cyclic intermidiate formed for bromination of alkenes. Interestingly such a process occurs here also. 1. First the electrophile part of the addendum(H+) forms a π-complex with the C-C triple bond, which transforms to a three membered cyclic ring. A free carbocation is not formed like the addition of HX with alkenes. Though carbocationic character will be found more prominent at one carbon in an unsymmetrical alkyne eg RC CH ; stabilised more by the R-group on one side. But until the nucleophile attacks from the rear side of the ring, the C-C triple bond is tied up with the H+ ion on one side, from which the attack of nucleophile Cl– is forbidden. Hence ultimately the ANTI addition product is observed. In many literature, the mechanism suggests the formation of a free vinyl carbocation which is bit less stable than a 10- alkyl carbocation but more stable than methyl carbocation. Stability wise it is very much likely, but looking to the absence of carbocation rearrangement products and also the stereoselective ANTI additon sugguests, free vinylic carbocation is not formed. In stead, the +ve charge is delocalised in the three centres(two carbon atoms and one H atom). The RSs of the intermediate is shown below. The resonance hybrid will have more +ve character on the more hindered position to which the nucleophile attacks(valid for unsymmetrical alkynes like RC CH , unfortunately we have drawn for symmetrical alkyne). For symmetrical alkynes, this discussion carries no meaning.

- Cl

H+

H CH3 H CC CH3 C CCH3 CH3 CCCH3 -Cl CH -complex cyclic intermediate Cl 3 only Z alkene

H H H

CH3 CC CH3 CH3 C CCH3 CH3 C C CH3

Alkyne first forms a π-complex with the HCl, as shown. Then Cl– departs to form cyclic hydronium ion as shown. This step is slow and rate determining hence it is called electrophilic addition. The cyclic intermeidate has three RSs, in which carbocationic caracter is present in both the carbon atoms of C=C. Since this is a symmetrical molecule, it will give only one ANTI product. But actually there are two possible ways of nucleophilic attack (Cl–) to form two different structural isomers, if the alkyne is unsymmetrical. For propyne, which is unsymmetrical, we get regioselectivity as only consideration. Stereoselectiveity is not a question here.

+ H - Cl H+ CH C CH CH C CH CH3 CCH 3 2 3 3 + + regioselectivity Cl Cl stable carbocation

Cl-

regioselectivity Cl

CH3 C CH3

Cl Dr. S. S. Tripathy 59 GOC-III(Introductory Reaction Mechanism) Out of three RSs for cyclic hydronium intermediate, like we wrote before(not drawn here) the one in which +ve charge on the more hindered carbon will be more stable (30 vinyl carbocation) and contributes more to the hybrid. So the electrophilicity of that carbon(2nd carbon) will be more for which the nucleophile attacks that position preferentially, hence the regioselectivity. For the 2nd addition to C=C, the 30 carbocation is not only stablised by hyperconjugation of the 6 α-H atoms but also by +M -effect by the -Cl atom. That is why the nucleophile Cl– attacks that position to form geminal dihalide. The second addition is same as the HX addition to C=C, which we have done before. Here there is only regioselectivity and no stereoselectivity as a stable 30 carbocation is formed. Moreover, this carbocation also will not undergo further rearrangement. (N.B: Guys, i could have given the vinyl carbocation intermediate in all these mechanism, as you find in many texts. But i chose to go in favour cyclic interemidate as it is more convincing to rationalise the products of first phase addition as regards to its stereoselectivity. I still believe, there is no harm in going in the vinyl carbocation way. We can explain everything except, stereoselectivity in addition. Thats it)

Few examples of controlled AdE reactions of ethyne:

Cl HgCl CH CH HCl 2 + (on Carbon) CH2 CH vinyl chloride

OCOCH3 HgSO4 CH CH + CH3COOH CH2 CH vinyl acetate

CN Cu2Cl2 CH CH + HCN CH2 CH acrylonitrile These three products namely vinyl chloride, vinyl acetate and acrylonitrile(vinyl cyanide) are the monomers for their respective polymers namely PVC, PVAc, PAN which we shall study later in 2+ 2+ the chapter alkenes. Since Hg and Cu2 catalyse these reactions, they are AdE . We shall later know that there can be AdN alternative to these reactions, but with different catalyst.

Addtion of X2 to Alkynes: (Halogenation of alkynes):

1. In this case regioselectivity is no issue. Yes, there is stereoselectivity issue here and like halogenation of alkenes, ANTI selectivity is shown. Hence exactly like alkene halogenation, there is formation of halonium cyclic intermediate in the rate determining step, which leads to anti addition product.

2. Always organic solvents such as CH2Cl2 or CCl4 is used. Aquoues halogen leaves room for a competitive nucleophilic attack by H2O.

CH3 X X X CH2Cl2 X2 CH3 C CH + X2 CC CH3 CCH X H anti -stereoselectivity X X vicinal dihaloalkene tetrahaloalkane

60 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism) In the first addtioon, we get anti addition product(here Z-alkene) while the 2nd stereoselective anti addition has no observable effect, as both the carbon atoms are achiral. Just for fun, i have made the two halogen atoms coloured, only to highlight anti addition. Mechanism:

Br Br Br CH3 Br - Br Br CH3 CCH CH3 CCH C C Br E-alkene H

+ Alkyne abstracts a Br (electrophile) from the Br2 and forms the unsaturated cyclic bromonium ion intermediate. The nucleophile Br– then attacks from the opposite side of the bromonium ion to form the anti addition product. Here i have not shown the attack of Br– to the other carbon, as it would give same E-alkene.

The addition of 2nd mole of Br2 to the vicinal dibromo alkene goes in the same way via the bromonium ion intermediate. But in this case, a tetrabromo alkane is formed, which has no chiral centre. So why to waste our time by discussing about the stereoselectivity of this step.

Even the stereoselectivity is found in the addition of one mole of Br2 with acetylene.

Br Br Br H Br - Br Br CH CH CH CH C C Br E-alkene H Note that we use the term stereoselectivy here, which tells you that both the stereoisomeric products are obtained from the reactant, out of which one is major.

HOOC Br HOOC C C COOH + Br2 CC COOH Br (70 % trans product) In this example, we again see the first stage addition of bromine to but-2-ynedioic acid results E-alkene as the major product. In this reaction rest 30% product is the Z-alkene. E-product is due to bromonium ion intermediate and the Z-product is due to the formation of vinyl carbocation in the r.d.s.

Br Br Br HOOC C C COOH HOOC C C COOH - Br Br

Br Br Br HOOC C C COOH HOOC C C COOH cis- alkene Br trans-alkene

Since trans isomer is more stable than cis-, we can also argue that the the mechanism can be predominantly via vinyl carbocation(may not be at all via bromonium ion intermediate) in this case. However, the formation of bromonium ion intermediate in this reaction has been proved by its isolation and hence is the positive evidence for the mechanism. Howerver for cis-alkene formation, we have to accept the minor mechanism via vinyl carbocations.

Dr. S. S. Tripathy 61 GOC-III(Introductory Reaction Mechanism) Hydration of Alkynes:

1. Markonikoff addtion of H2O to a triple bond gives first an enol, which immediately tautomerizes to its stable keto form. Acetylene only gives aldehyde(acetaldehyde) and other alkynes give ketones. 2+ 2. This reaction is catalysed by dil. H2SO4 along with Hg salt(HgSO4) on mild warming conditions( 60–800C)

OH O

HgSO4/H2SO4 R C CH H OH CH3 C CH2 RC CH3 800C enol form keto form

OH O

HgSO4/H2SO4 H C CH H OH HC CH2 HC CH3 800C enol form keto form (acetaldehyde

It comes under electrophilic addition reaction as a +ve ion (carbocation or mercuric cyclic ion) is formed in the rate determining state.

Often i find in several texts, the mechanism is simplified to the extent that the role of Hg2+ is ignored. That is why i feel like giving two mechanisms for this reaction. One, without involving Hg2+ , as is found in many texts and the other involving Hg2+( which i believe to to be more credible). Mechanism without involving Hg2+:

H H O - H2O HSO4 R C C H H RC CH2 RC CH2 20 vinyl carbocation - H2SO4

H H O O - O HSO4 RC CH RCCH3 RC CH3 2 + H - H2SO4 enol

Mechanism involving Hg2+: Since the reaction does not take place in the absence of Hg2+, it is therefore certain that it must be taking part in the reaction. So the following is most plausible mechanism for the reaction.

62 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

The first step is formation of mercuric cyclic intermediate. Here Hg2+ is the electrophile which attacks the alkyne. This is the slow step(revesible sign is not shown unfortunately as the diagram is burrowed from Google on copy-paste basis. Thanks to Google). Here the preference of the nucleophile H2O is to attack the more hindered carbon due to its greater electrophilicity. We can draw three RSs for the cyclic intermediate (one cyclic and the other two acyclic) as we were writing for hydronium intermediate before. The RS which is a 20 carbocation is more stable and contributes more to the hybrid. Hence the hindered carbon has greater electrophilicity to favour nucleophilic attack. Alternatively we can say that the breaking of the cyclic ring takes place first 0 forming a more stable 2 carbocation to which H2O attacks. In this case, the mechanism becomes similar to the first approach that vinyl carbocation is formed. The only difference, is Hg remains on the less hindered carbon and making a intimate ion pair with carbocation and is removed by acid.

(Note that some authors believe this to be nucleophilic addition i.e AdN of H2O the triple bond, which is not acceptable to me. They say addition of H2O to the mercuric cyclic ring is the rate determining step. There can be a controversy on this issue. But majority of credible texts suggest this to be electrophilic mechansim, which i strongly lend my support). Nucleophilic Addition to Alkynes: Alkyne triple bond is relatively more electrophilic than alkenes. The ionisation energy of alkynes is greater than alkenes as the former is sp hybridised and the electrons are more tightly bound to the the carbon nucleus. Although both are predominantly nucleophiles, under the influence of strong outside nucleophiles, it is alkynes which can go for AdN, not alkenes. Some of the AdN reactions of alkynes are given below. CN NaCN CH CH + HCN CH2 CH acrylonitrile

OEt KOEt CH CH + HOEt CH CH 0 2 150 C ethyl vinyl ether

Dr. S. S. Tripathy 63 GOC-III(Introductory Reaction Mechanism) In this case, catalytic amounts of strong base CN–(first example) and OEt–(2nd example) are taken to initiate nucleophilic attack onto the triple bond in the rate determining step. Mechanism:

OEt OEt slow HOEt CH CH + EtO CH CH CH CH2 + OEt fast Since the nucleophile is regenerated in the 2nd step, we have to take small amount of the strong base KOEt. Similar is the case with the other example. HCN adds onto acetylene in presence of

Cu2Cl2 by AdE mechanism, we have seen before. But the same reaction takes place by AdN mechanism in presence of NaCN which carries a strong nucleophile CN–.

Hydrogenation of Alkynes:

* Addition of H2 with alkyne in presence of catalysts like Ni/Pd/Pt at higher 0 temperature(300 C) gives first alkene with first mole of H2 and then to alkyne with second mole of H2. Although the mechanism of addition is SYN it is not possible to control or restrict the addition to the first stage i.e to form alkene even with one mole of H2, as the reaction is very fast. In other words, it is not possible to isolate alkene with these catalysts. Pd-C(palladised carbon) can also be used at a lower temperature for the purpose. * Controlled Hydrogenation: (A) SYN ADDITION: Lindlar’s Catalyst Specially poisoned catalysts like Lindlar’s catalyst is used for the purpose of getting alkene by the hydration of alkyne. Further hydrogenation is not possible in this case.

* Pd-CaCO3 or Pd-BaSO4 poisoned by lead acetate and quinoline is called Lindlar’s catalyst. * This poisoned catalyst has adsorbing capacity to only alkynes and no adsorping capaicity for alkenes. Once alkene is formed, it is desorbed from the surface of catalyst and no further hydrogenation can occur, even with excess H2(more than one mole). Note that unpoisoned Pd-

CaCO3 or Pd-BaSO4 will not restrict the hydrogenation reaction. Poisoning is a must to control hydrogenation to the stage of alkene. Mechanism of Syn Addition:

Similar to hydrogenation of alkenes. both H2 and alkynes are adsorbed on the surface of the catalyst. Hydrogen is absored bas H- atoms and alkyne forms a π-complex with the vacant d-orbital of Pd. Then hydrogen atoms are successively added onto the same side of the alkyne, thus breaking one π- bond to produce Z-(Cis) alkene. Refer the details in the “hydrogenation of alkenes” before few pages. The only difference is the presence an extra pi-bond here in product.

CH3 CH3 Pd - CaCO3 CH3 CCCH3 + H2 + 36.7 Kcals quinoline/ lead acetate H H Z(cis)-but-2-ene

CH3 CH3 Pd/C + H2 n-butane + 28.3 Kcals H H

64 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism) The first step is more exothermic than the 2nd step, though the first step is relatively slower than the 2nd step. In presence of unpoisoned catalyst, the alkene formed very quickly further hydrogenates to alkane and hence it is not possible to isolate alkene. But in presence of poisoned catalyst(Lindlar’s catalyst), the alkene formed is expelled out of the catalyst surface and further hydrogenation is thus prevented.

H2/Pd/C CH3 CCCH3 CH3CH2CH2CH3

(Note that even with one mole of H2, in presence of unpoisoned catalyst, we get complete hydrogenation of half mole of the reactant leaving behing half mole unreacted.

Ni2B (Nickel Boride - P-2) catalyst:

Ni2B can replace Lindlar’s catalyst for the controlled hydrogenation of alkynes to the stage of alkenes. The mechansim of addition is SYN too in this case.

CH3 CH3 H2/Ni2B CH3 CCCH3 H H (B) ANTI ADDITION:

* Alkyne can also be hydrogenated to alkene in a controlled manner with the reducing agent Na/liq. NH3. * The mehchanism of addition is ANTI. The two H atoms add on the opposite side of the triple bond, thus but-2-yne gives E(trrans)-but-2-ene.

CH3 H Na/liq.NH3 CH3 CCCH3 [2H] H CH3 E(trans)-but-2-ene Mechanism: There are four steps involved in this reaction. (i) In the first step one Na atom gives one electron to the alkyne to form radical anion. In this step the two alkyl groups bonded C=C have to remain on opposite side, so that the two orbitals carrying the electrons (one carbon carries one electron, ie a radical, and the other carbon carries a lone pair i.e an anion) minimise repulsion between them. The secret of ANTI addition is thus fulfilled in this step. (Fish hook arrow shown indicates one electron shift)

CH3 + CH3 CCCH3 + Na CC + Na

radical anionCH3

(ii) In the second step, the carbanion abstracts a proton from NH3 to form a radical.

CH3 CH3 H

C C ++HNH2 C C NH2 CH CH radical anion 3 radical 3 (iii) The radical then accepts one electron from another Na atom to form an anion, in the third step.

Dr. S. S. Tripathy 65 GOC-III(Introductory Reaction Mechanism)

CH3 H CH3 H + C C ++Na C C Na

CH3 CH3

(iv) In the last step, the carbanion abstracts a proton from NH3 to form the E(trans)- alkene.

CH3 H CH3 H C C + + HNH2 C C NH2

CH3 H CH3 E(trans)but-2-ene

Thus for one mole of alkyne, two moles of Na and two moles of NH3 are required, so that in the products we get two moles of NaNH2 alongwith one mole of alkene product. N.B: (1)Alkyne is able to accept an electron from Na atom give a stable vinyl radical anion. An alkene does not accept an electron from Na to form alkyl radical anion,which is unstable. Hence alkene does not undergo forther reduction to alkane in this method. (2) Alkynes like ethyne, propyne or any other terminal alkyne, controlled hydrogenation does gives alkene(not alkane), but the there is no stereospecificity aspect in this case. The SYN adddition or the ANTI addition carries no practical significance in such cases.

Addition to Conjugated Dienes, Unconugated Dienes and Enyes:

1. Both 1,2-(direct addition) and 1,4-(conjugate addition) additions take place for conjugated dienes with one mole of addendum. At very low temperature, the 1,2-addition product prevails while at higher temperatures the 1,4-addition product prevails. 2. At low temperature, the reaction is kinetically controlled and so the major product is the kinetically stable product(1,2-addition product) i.e the product which is formed at the faster rate becomes the major product. At high tempeature, the reaction is thermodyanmically controlled and the thermodynamically controlled product(1,4-addition product) is the major product i.e the product formed at a slower rate forms the major product.

Br 2 4 HBr(one mole) + 1 3 - 800C (1,2-addition)Br (1,4-addtion) 81% 19%

Br 2 4 HBr(one mole) + 1 3 200C (1,2-addition)Br (1,4-addtion) 44% 56% Mechanism:

+ CH2 CH CH CH2 + H CH2 CH CH CH3 CH2 CH CH CH3 st a b l e 20 cabocation

Br Br Br Br

CH2 CH CH CH3 CH2 CH CH CH3 (1,2-addition product) (1,4-addition product) kinetically stable thermodyanically stable

66 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism) The diene first reacts with the electrophile in the rate determining step to form stable 20 allyl carbocation. The two RSs of the allyl carbocation have been shown. The hybrid (not shown before) will have +δ charge on both 1st and 3rd positions as shown below. Hence the activation energy in this step is same for both the products. + + CH2 CH CH CH3 In the 2nd step, which is faster, the nucleophile can attack both 1st position and 3rd position. If it attacks the 1st position, 1,4-addition product is obtained by the C=C bond migrating to the internal position. If the nucleophile attacks the 3rd carbon, then 1,2-addition product is obtained with the retention of C=C at the terminal position. Both the direct and conugate addition products are obtained at all temparatures. However at low temperature, the 1,2-product predominates and at higher temperature 1,4- product predominates.

Kinetic Vs. Thermodyamic Control of Reaction: In a two step process like this, the 2nd step is crucial to decide which becomes the major product. Since the activation energy of the first step is same for both 1,2- and 1,4- products as the carbocation is unique i.e the hybrid of two RSs, it is the 2nd step which distinguishes between the reaction path ways. In the 2nd step, there are two competing reactions :

(i) one involving higher EA which is the case for the 1,4 product as it will pass through less stable 10 allyl carbocation. But the 1,4-product is thermodynamically more stable and hence Δ 0 G will be more negative and hence Keq is more for it. So kinetically this is a slow reaction but is a product favoured reaction thermodynamically.

(ii) the other involving lower EA which is the case with 1,2- product as it will pass through more 0 stable 2 allyl carbocation. But this product is thermodynamically less stable and has a lower Keq. So kinetically this is a faster reaction but is not product-favoured reaction thermodynamically. Effect of Temp:

At very low temperature, the intermediate carbocation passes through 1,2 route involving less EA and proceed fast. Once it reaches the other side of the energy hill as product, it stops there and cannot return back to reactant side(here carbocation) as at low temperature, there is limited energy available to favour the backward journey. Hence it becomes irreversible. Thus kinetically controlled path(faster reaction) gives the major product. 1,2- product is the major product at low temperature.

At a higher temperature, the 1,2- path becomes reversible. So 1,2-product molecules traverse back the hill to the starting material(carbocation) and can take higher EA (energy is available now) to proceed in 1,4- path to give the 1,4-product. This path is still irrerversible, as the temperature is not very high. So the % of thermodynamically controlled product(1,4-product) rises as temperature increases. At still higher temperature, both the processes become reversible, Δ 0 then Keq ( G ) factor will decide which one will be the major product. Since 1,4-product product has greater Keq, this becomes the major product at higher temperature. Thus relatively slower reaction(higher EA i.e kinetically disfavoured) gives the major product.

Dr. S. S. Tripathy 67 GOC-III(Introductory Reaction Mechanism) Halogenation of Conjugated Dienes: Unlike halogenation of alkenes, which go via cyclic halonium intermediate to give ANTI addition product, in diene this does not happen. A stable 20 allyl carboncation is formed here to give the kinetically controlled 1,2-product and rearranges to give 1,4 product which is thermodynamically controlled. Low temperature favours 1,2- product while high temperature favours 1,4- product. This has been explained before.

Br Br Br Br Br2 CH2 CH CH CH2 CH2 CH CH CH2 + CH2 CH CH CH2 1,2-product 1,4-product (kinetically controlled) (thermodynamically controlle Mechanism:

Br Br Br Br CH2 CH CH CH2 CH2 CH CH CH2 CH2 CH CH CH2 - Br stable 20 allyl carbocation l e ss sta b l e 10 allyl carbocation Br Br Br Br Br Br

CH2 CH CH CH2 + CH2 CH CH CH2 (1,2-product) (1,4- product)cis and trans favoured at lower temp. favoured at higher temp

Addition to Conjugated and Unconjugated Enynes:

Br HBr(1 mole) CH C CH CH2 CH2 C CH CH2 but-1-en-3-yne (conjugaged diene)

C=C is more reactive than CCto electrophilic addition reactions, as the latter will form less stable vinyl carbocation. But for conjugated enyne(but-1-en-3-yne), one mole of addendum adds onto the CC to form thermodynamically stable product, a conjugated diene. However, in unconjugated enyne, the C=C reacts with one mole of the addendum.

Br HBr(1 mole) CH C CH2 CH CH2 CH C CH2 CH CH3

In this case, C=C undergoes addition as it is more reactive than CC.

REARRANGEMENT OF CARBOCATIONS(At a Glace) : We already know that carbocations are very prone to rearrangement from a less stable state to a more stable state. Let us treat this topic in a formal way applied to all the three types of organic reactions.

SUBSTITUTION (SN1) An alkyl carbocation is formed in only in SN1 mechanism. Reaction of HX with Alcohols:

68 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

CH3 CH3 HCl CH CH CH OH (1) 3 2 CH3 C CH3 Cl Mechanism:

CH CH CH3 3 3 + H - H2O CH C CH CH3 CH CH2 OH CH3 C CH2 OH 3 2 H H H 10 carbocation

hydride CH3 CH3 shift Cl- CH3 C CH3 CH3 C CH3 Cl sta b l e 30 carbocation

Actutally 10 carbocation is never formed. We have shown here for the purpose of easy understanding – by a beginner. In fact H shift and removal of H2O(leaving group) take place simultaneously resulting the stable 30 carbocation.

CH 3 CH3 - CH C CH OH H2O 3 2 CH3 C CH3 H H

CH 3 CH3 (2) CH C CH OH HBr 3 2 CH3 C CH2 CH3

CH3 Br

– 0 Here, a methyl shift(CH3 ) shift occurs in stead of hydride shift to rearrange to the stable 3 carbocation. Can you not show the mechanism of your own ?

I HI CH CH CH CH OH CH CCHCH (3) 3 2 2 3 2 3 CH3 CH3 Here hydride shift occurs two times, once for converting 10 to 20 and then 20 to 30 carbocation. This results the final product. However the 20 product is also obtained as minor product.

ADDITION(AdE) In presence of an electrophile, espcially H+, alkenes form carbocations which undergo rearrangement. See the following cases. (1)

CH3 CH3 HCl CH3 CH2 CH CH CH2 CH3 CH2 C CH2 CH3 + Cl major CH3

CH3 CH2 CH CH CH3 minior Cl

Dr.Mechansim: S. S. Tripathy 69 GOC-III(Introductory Reaction Mechanism)

CH3 Cl

CH3 CH2 C CH CH3 H minor product

Cl-

CH3 CH3 + CH3 CH2 CH CH CH2 + H CH3 CH2 C CH CH3 20 carbocation H

CH 3 CH3 - CH CH C CH CH Cl 3 2 2 3 CH3 CH2 C CH2 CH3

Cl major product 30 carbocation

First a less stable 20 carbocation is formed which rearranges to the most stable 30 carbocation by hydride shift. Thus the major product comes from the 30 carbocation and minor product from teh 20 carbocation. The latter prouduct is fortunately chiral having one chiral centre and hence there are two enantiomers(R and S). In total there will be three products formed(including stereosimers). Note that there will be no product from 10 carbocation, as it is highly unstable. (2)

CH(CH3)2 HBr CH3 C CH CH CH3 ? (How many products are formed)

CH3

Solution: In all 8 products are formed including stereoisomers. Let me explain. Each of the two 20 carbocations produce two products each as each one has a chiral centre, hence total 4. Out of the two, the 2-bromo product is more abundant as the corresponding carbocation is more stable due to greater number hyperconjugation structures. Then the internal 20 carbocation is rearranged to 30 carbocation by alkyl shift. There are two possibilities. If i-Pr– shifts then the – carbon to which it shifts becomes a chiral centre, hence 2 products. The other possibility is CH3 shift, in that case, the carbon from which it shifts becomes chiral, hence 2 more. Thus there will be in all 8 alkyl bromide products. Migratory Aptitude among Alkyl(R–) groups: 0 0 3 alkyl > 2 alkyl > 10 alkyl > CH3 So the product involving the migration of i-Pr group forms the major product.

70 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

iPr Br iPr Br

Me C CHCH2 Me Me C CH2 CH Me Me (II) d/l Me (I) d/l + - H Br Br- iPr iPr iPr

Me C CH CH Me Me C CHCH2 Me Me C CH2 CH Me 20 carbocation Me Me Me 20 carbocation iPr- shift - CH3 shift

iPr Me iPr

Me C CHCH2 Me Me C CHCH2 Me 0 30 carbocation Me 3 carbocation Br- Br-

iPr Me Br iPr

Me C CHCH2 Me Me C CHCH2 Me Br (IV) d/l Me (III) d/l (major)

The order of yields of the four products is : III > IV > I > II III and IV are formed from 30 carbocation, while I is formed from a more stable 20 carbocation and II from a less stable 20 carbocation. Since the migratory aptitude of iPr– is greater than Me– product III is major among all. Since each exists as a d/l pair, there are 8 products in all.

ELIMINATION(E1): When alcohols are dehydrated in presence of acids, often rearranged products are formed. See these examples.

CH3 CH3 CH3 H+/heat CH3 C CH2 OH CH3 C CH CH3 + CH2 C CH2 CH3 neopentyl alcohol major minor CH3

Neopentyl alcohol, as such, does not have a β-H atom and at first sight, seems not to undergo dehydration. However it does due to carbocation rearrangement. See the mechanism. Mechanism:

CH3 CH3 - H2O + CH3 C CH2 OH + H CH3 C CH2 OH2

CH3 CH3 CH3 H CH 3 CH3 C CH CH3

CH2 C CH CH3 (major) + H CH3

CH2 C CH2 CH3 (minor) Dr. S. S. Tripathy 71 GOC-III(Introductory Reaction Mechanism)

SAQ: On dehydration with conc. H2SO4/heat, 4-methylhexan-3-ol gives how many alkene products including stereoisomers ? Solution:

OH CH3 H+/Heat CH3 CH2 CH CH CH2 CH3 Alkenes (how many) 4-methylhexan-3-ol Mechanism: A total of 7 alkenes(including stereoisomers) will be formed.

CH3 CH3

CH3 CH CH CH CH2 CH3 + CH3 CH2 CH C CH2 CH3 (II) E & Z (I) E & Z

OH CH3 H CH3 H+ CH3 CH2 CH CH CH2 CH3 CH3 CH CH C CH2 CH3 - H2O 20 carbocation H hydride shift - H+

H CH2

CH3 CH2 CH C CH CH3 30 carbocation H H - H+ CH2 CH3 CH3 CH2 CH2 C CH2 CH3 CH3 CH2 CH C CH2 CH3 CH3 (I) E &Z (Repeat) (IV) CH3 CH2 CH C CH CH3 (III) E &Z

Usually hydride/alkyl shift does not occur if there there is no increase in stability. So we have not taken the rearrangment of the 20 carbocation to another 20 carbocation by hydride shift from the left side. That 20 carbocation is slightly more stable than the first formed 20 carbocation as the number of hyperconjugation structures would have been more. If we take that into account then one more alkene would add to the list. So a total of 8 alkenes. Can you not draw that ?

Nucleophilic addition to Carbonyl Function: (AdN)

1. Carbonyl compounds i.e aldehydes and ketones add onto polar addendum by this mechanism. 2. The nucleophile(Nu–) attacks the substrate first in the rate determining step to form an intermediate. 3. When this addition takes place under basic conditions using strong bases as nucleophiles, then the intermediate is an alkoxide ion. The 2nd step will require the addition of acid to protonate the alkoxide ion to form the respecitve alcohol. The strong bases(nucleophiles) under

this category can be HCN/NaOH, RMgX, RLi, LiAlH4, NaBH4, RC C Na etcc.

4. When this addition takes place with a weak nucleophile like H2O, ROH, NH3, RNH2 and other nitrogen nucleophiles such as NH2OH(hydroxyl amine), H2NNH2(hydrazine),

H2NNHPh(phenyl hydrazine), H2NNHCONH2(semicarbazide) etc. it should be done in the presence

72 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism) of acid to first protonate the carbonyl oxygen to increase the electrophilicity of carbonyl carbon. In this case too, the addition of nucleophile is the rate determining step.

O O O 5. Reactivity order: + + HCH >>RC+ H RC R Formaldehyde is most reactive as the +δ charge on carbonyl carbon is maximum. Due the presence of one and two R– groups which produce +I effect in any other aldehyde and ketone respectively, the magnitude of +δ charge decreases, hence the electrophilicity decreses at the carbonyl carbon in the above order. Thus ketones are least reactive for AdN reactions.

Case-I: (AdN under basic condition or using Strong bases/nuclophiles):

- O O OH sl o w + + H C + Nu C C (E+) Nu fast Nu Here the nucleophile is a strong base and hence a strong nucleophile. The small electrophilicity at the carbonyl carbon(+δ charge) is good enough for the former’s attack onto the latter. Thus the alkoxide ion is formed in the first rate determining step. In the second step the alkoxide ion invariably picks up a H+ (which is an electrophile) from the added acid to form the neutral product. See the example. (i) When an aldehyde or ketone reacts with HCN, it forms the respective cyanohydrin. But the reaction is very slow, as HCN is a very weak acid and produces the nucleophile CN– to a very small extent. If one is happy with the pace, then no acid is required to be added at the end. However, this reaction is often catalysed by small amount of base(NaOH), which produces greater amount of CN– and makes the reaction faster.

- - HCN + OH CN + H2O

O O OH sl o w HCN CH3 C H + CN CH3 C H CH3 C H - CN CN Nu The alkoxide ion intermediate can easily pluck out a H+ from the excess HCN and produce the nucleophile for fresh attack. Hence a small amount of NaOH is enough for catalysing the reaction.

(ii) When aldehyde or ketone reacts with a Grignard reagent(RMgX) and the intermediate hydrolysed, we get an alcohol. Aldehyde gives 20 alcohol while ketone gives 30 alcohol. Of course, formaldehyde gives 10 alcohol.

O O MgBr OH OH HOH CH C CH + CH3 MgBr CH3 C CH3 + CH3 C CH3 + Mg 3 3 (H ) Br CH3 CH3 RMgX carries the nucleophile R–, which is a strong nucleophile and attacks onto the substrate to form the alkoxide ion having (MgBr)+ as counter ion. This is hydrolysed by water to form alcohol product. Note that in the second step, addition of acid is not required, as H2O being a

Dr. S. S. Tripathy 73 GOC-III(Introductory Reaction Mechanism) stronger acid than alcohol, can bring about the acid-base reaction. However, most often a little strong acid is added to catalyse the second step. Note that acid is never added in the first step, as it would protonate the R– of GR and produce alkane.

Case-II: (AdN under acidic condition or using weak bases/nuclophiles):

Since the nucleophilicity is poor in case of weak base, the electrophilicity of the substrate is improved by protonating the carbonyl oxygen in the fist step. In the second step the nucleophile attacks to form the protonated product, which on deprotonation forms the neutral product.

H OH O O OH OH fast ROH -H C C C + H C C sl o w OR O H R The protonation of the oxygen atom of the carbonyl group increases the electrophilicity of the carbonyl carbon, as there will be one more resonance structure in which carbon carries the +ve charge. Hence in the 2nd step, the nucleophile, though weak, attacks to the strong electrophilic centre easily. Note that this step is the slow step and hence the name AdN also in this case. In the last step, deprotonation leads to the product(called hemiacetal or hemiketal, to be discussed later).

H OH O O OH fast EtOH -H CH3 C H CH3 C H CH3 C H + H CH3 C H sl o w OEt O H Et

Acetaldehyde reacts with ethyl alcohol in presence of acid(dil H2SO4) to form hemiacetal i.e 1-ethoxyethanol by the above mechanism.

AdN followed by Dehydration:

In many AdN reactions, the addition product undergoes dehydration easily to form an unsaturated product. Let me give one example now.

H OH O OH

CH3 C H + H2NOH CH3 C H CH3 CH NOH hydroxyl amine protonated aldehyde NH OH H H - H2O

CH3 CH N OH oxime

The carbonyl compound(acetaldehyde in the above example) reacts with hydroxyl amine(H2NOH) to form the addition product, which eliminates a molecule of H2O easily to form C=N compound called oxime, which exists as SYN and ANTI geometrical isomers.

Same thing happens with other nitrogen nucleophiles such as hadrazine(H2N-NH2), phenyl hydrazine(Ph-NH-NH2) and semicarbazide(NH2-NH-CO-NH2) and a few more, which we shall discuss later in the chapter ‘aldehydes and ketones’.

74 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism) SAQ: Why carboxylic acid (RCOOH) and its derivatives such as RCOCl, RCOOCOR, RCOOR,

RCONH2 do not undergo AdN reaction. In stead they undergo Nucleophilic Acyl substituion ? Answer: In all the above conpounds the carbonyl group is bonded to a hetetro atom i.e another O atom or N atom. There is an in-built M-effect in the molecule itself due to the conjugation of the lone pair on the heteroatom(O, N) with the C=O. Thus the electrophilicity of the C=O carbon is almost completely lost for the AdN reaction. However, due the greater leavability of the leaving – – groups like Cl (RCOCl), RCOO (in acid anhydride), H2O(from protonated RCOOH), ROH (from protonated ester) etc. the nucleophilic acyl substution takes place. This has been discussed before.

O O

RC Y RC Y

Nucleophilic Addition to C=C (Conjugate addtion)

1. Usually simple alkenes and alkynes prefer to be attacked by an electrophile first in the rate determining step and hence their AdE reactions are most common. However, when the C=C is in conjugation with a –M group like –CO–, –CN, –COOR, –CONH2, –NO2, –SO2R etc., nucleophilic addition often takes place at the C=C. 2. Although it appears to be a 1,2- addition to C=C, in reality it is a 1,4-addition(conjugate addtion) to conjugated system. The unstable intermediate quickly tautomerises to give the the final product. 3. Because of conjugation of C=C with the –M group, the terminal carbon of C=C acquires electrophilicity for the Nu– to attack in the rate determining step.

4. Mostly weak bases like ROH, Gillman reagent(R2CuLi), resonance stabilised O O O carbanions(enolate ions) eg. , or enolate ions from RCCH2 CH3 C CH C CH3 acetoacetic ester, maloic ester, cyanoacetic ester etc. favour 1,4- (conjugate) addition, while – strong bases such as alkyl carbanions (RLi, RMgX), H from LiAlH4 ec. favour 1,2-addition to the –M group. See below. 1,4-Addition:

O O

CH3 CH CH C CH3 CH3 CH CH C CH3 O- + CH3 CH CH C CH3

electrophilic centre

Dr. S. S. Tripathy 75 GOC-III(Introductory Reaction Mechanism)

H O OEt O EtOH CH3 CH CH C CH3 CH3 CH CH C CH3

OEt OH OEt O

CH3 CH CH C CH3 CH3 CH CH2 C CH3 unstable enol form stable keto form

The nucleophile EtOH attacks first to the terminal carbon of C=C, which acquires +δ charge due to conjugation and forms a enolate ion. On proton transfer, we get an unstable enol which quickly tautomerises to the stable keto form, which is the final product. Although the addition is 1,4- type (one part of EtOH joins on some atom, and the other part of it joins in the fourth atom. However, after tautomerisation, the product appears to be a 1,2- addition product at C=C and it seems as if C=O has not taken any part in this game. N.B: Note that if the nucleophile is a strong base, then 1,2-addition at the C=O gives the major product. See this case.

1,2-addition to C=O:

HOH

O O MgBr Et MgBr CH3 CH CH C CH3 CH3 CH CH C CH3 Et

+ H2O/H

CH3 CH CH C CH3 + Mg(OH)Br Et

In this case, the Grignard reagent, EtMgBr which carries a strong base Et–, favours the 1,2- addition at the carbonyl carbon(C=O) and after acidic hydrolysis gives the alcohol product. The conjugate product, though formed is the minor product.

4 O Et O MgBr 1 2 3 + CH3 CH CH C CH3 Et MgBr H2O/H CH3 CH CH C CH3 1,4-addtion

Et OH Et O

CH3 CH CH2 C CH3 CH3 CH CH C CH3 (minor product)

76 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

FREE RADICAL ADDITION (AdF) 1. Alkenes and alkynes add on to some addenda with intermediate polarity like HBr, RSH,

BrCCl3, HCCl3, ICF3 etc. in presence of free radical initiators like peroxides eg H2O2, benzoyl peroxide(BPO), di-tert-butyl peroxide(DBPO) opposite to Markonikoff’s rule. This is called Anti-Marknokoff Addition. The negative part of the polar addendum adds on the the carbon of C=C which bear more number of hydrogen atoms(less hindered carbon). This is often called the “Peroxide Effect” or the “Kharasch Effect”.

Br H2O2 RCHCH2 + HBr RCH2 CH2

Br Br H2O2 HBr R C CH HBr RCHCH RCH2 CH H2O2 anti-Markonikoff addition Br 2. Unsymmetrical alkenes and alkynes having unequal number of H-atoms bonded to the multiple bond give anti-Markonikoff addition product under the above conditions. 3. The reaction is regiospecificific i.e the anti-Markonikoff product is only formed. 4. However, the the reaction is not stereospecific.

3. The mechanism for this addition is FREE RADICAL(AdF).No ions are involved in this mechanism although the students often say that the negative part of HBr joins on the less hindered carbon and +ve part to the more hindered carbon. This is just handy memory tool. 4. Homolytic cleavage of a sigma bond in the addendum induced by other primary free radicals produces secondary free radicals like ( Br) in the inititation step. Polar molecules like HCl and molecules having polariazable anions like HI, do not unergo homolytic cleavage. But HBr polarity-wise and polarizability-wise is ideal for homolytic cleavage which is induced by other primary free radicals. It has been thermodynamically proved also that the ΔG0 for the process involving HBr and the like is –ve, while that for HCl and HI is +ve. Hence the latter never go for anti-Marknoikoff(free radical mechansim) addition even in presence of free radical inititators like peroxides.

5. Like Free Radical Substitution reaction, AdF also three steps. (A) INITIATION:

The peroxide like H2O2, BPO, DBPO or AIBN(azo-bis isobutyronitrie) when heated or subjected to radiation by visible or near UV light. The O–O bond in peroxide is weak (BE = 35 kcals/mole as against 100 kcals/mole for C–H bond) and can easility undergo homolytic cleavage by heat or light to form primary free radicals like hydroxy, alkoxy etc. This is the first step of inititation. A trace quantity of this free radical intiator is used as in the subsequent propagation step, free radicals are continuously generated. AIBN easily loses N2 gas to form free radicals. In the second step of initiation, the primary free radicals abstract a hydrogen atom from the addendum like HBr to produce bromine free radicals( Br).

heat OR RO OR 2 RO light

RO + HBr ROH + Br

Dr. S. S. Tripathy 77 GOC-III(Introductory Reaction Mechanism) The bromine free radicals are now ready to involve in propagation step by reacting with the alkene. (B) PROPAGATION: The bromine free radical reacts with the C=C in such a manner that a more stable carbon free radical is formed. You arleady know that the stability order among alkyl free radicals is 30 > 20 > 10 > methyl The carbon free radical then reacts with another neutral HBr molecule(addendum) to abstract a hydrogen atom to form the anti-Markonikoff product and a bromine free radical set free to propagate the process by attacking onto a fresh alkene and repeating the cyclic process again and again till one of the reactants is completely exhausted. This step is very fast.

RCHCH2 + Br RCHCH2 (more stable 20 free radical) If bromine would add onto the other carbon(more hindered) then a primary free radical would have formed, which is very less stable. Hence the regiospecificity.

Br Br

RCHCH2 + HBr RCH2 CH2 + Br Hince H atom can join from either side of the free radical(almost planar) i.e in the same side of Br as well as opposite side of Br, so that the product is a mixture due to syn and anti addition(if there is chirality). If there are two different chiral centres, then we get the two diastereoisomers each existing as enantiomeric pair. Hence the reaction is not stereoselective or stereospecific. (C) TERMINATION: All the free radicals left out in trace quantities at the end undergo mutual combinations to form varieties of bye-products in trace quantities.

Br + Br Br2 RO + Br ROBr

RO + RO ROOR

Br Br Br

RCHCH2 + Br RCHCH2 vic. dibromide

Br OR Br

RCHCH2 + RO RCHCH2 alkoxy bromide

RCH+ HC R RCHCHR CH Br CH2Br 2 CH2Br CH2Br The formation of these bye products, though in trace quantitites, is not only surprising but also a proof for the radical mechanism.

SAQ: With mechanism, show the products and trace bye products formed in the following reaction.

CH3 H O CH C CH HBr 2 2 ? 3 2 + heat 78 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism) Solution: Initiation:

HO OH 2 HO

HO + HBr H2O + Br Propagation:

Br + CH3 C CH2 Br CH3 C CH2 CH 3 CH3 m o st st a b l e 30 radical

Br Br CH CH CH + Br CH3 C CH2 + HBr 3 2 CH CH3 3

Termination:

HO + OH H2O2 HO + Br HOBr

Br Br Br

CH3 C CH2 + Br CH3 C CH2

CH3 CH3 vi c. di bromi de

Br OH Br

CH3 C CH2 + HO CH3 C CH2

CH3 CH3 bromohydrin

Br Br CH2Br CH2Br

CH3 C CH2 + CH3 C CH2 CH3 C C CH3 CH CH 3 3 CH3 CH3 Note that besides the expected addition product, we get so many surprising products like vicinal dibromide, bromoalkanol(bromhydrin) and the higher dibromoalkane i.e 1,4-dibromo-2,2,3,3- tetramethylbutane, though in very small amounts. Is it not interesnting !!!!!

SAQ: Show what products are obtained from the addition of HBr with 1-methylcyclohex-1-ene in presence of peroxide. Comment on the stereochemistry of the products. Solution:

H H H Br Br HBr + H2O2 CH3 heat/light CH3 H H CH3 d/l pair d/l pair An ti Syn We get four stereoisomers in this case as the chiral centres are different. One pair, in which H

Dr. S. S. Tripathy 79 GOC-III(Introductory Reaction Mechanism) and Br are anti, and the other pair in which H and Br are syn. We have aleady discussed that H can join onto the nearly planar free radical on either side w.r.t Br atom, hence we get one diastereoisomeric pair for anti addition and the other pair for syn addition. So no stereoselectivity or stereospecificity is found in this case.

N.B: Other addenda like BrCCl3, RSH also add in a smiliar fashion.

CCl3 BPO CH3 CH CH2 BrCCl3 CH3 CH CH2 + - heat (Br)(CCl3) Br

SCH3 peroxide CH3 CH CH2 + CH3SH CH3 CH2 CH2 heat

80 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

Neighbouring Group Participation (Anchimeric Assistance)

A neighbouring group(G) in an orgnaic molecule w.r.t the reactive site, takes part in the reaction with the help of electron pair(lone pair or pi/sigma bond pair) to usually form an intermediate and thus enhances the rate of reaction as compared to a molecule in which it is absent. The most common among the NGP is due to the presence of a heteroatom with a lone β pair at the -position w.r.t to leaving group(L) in a SN2 reaction. This heteroatom, as helper nucleophile, first interacts with the electrophilic centre and expels the leaving group by forming a 3-membered cyclic intermediate. Then the reactant nucelophile(Nu–) attacks the intermediate to form the product. Thus the net result is the enhancement of rate manyfold. In addition to this, there are two bonuses. First - if the Nu– attacks the other carbon which did not bear the leaving group, then a different product can be formed, provided the molecular structure is unsymmetrical. Second- if the Nu– attacks the carbon bearing the leaving group, then this carbon will suffer two times inversion and hence there will be retention of configuration in the final product.

Intermolecular S 2 G G Intramolecular G N S 2 - N Nu C C (fast) C C C C - L Nu L

* G = the neigbouring group can be : Halogen atom(-X), -OH, -NH2, -OR, -SR, -COOH, -COOR, -Ph, C=C bonds at the b-position w.r.t the leaving group have been found to produce NGP.

Examples:

Intermolecular SEt SEt Intramolecular SEt SN2 SN2 H2O (fast) CH2 CH2 CH2 CH2 CH2 CH2 - H OH Cl 2-ethylsulfanylethanol 1-chloro-2-ethylsulfanylethane

H2O EtCH2 CH2CH2Cl EtCH2 CH2CH2OH (very slow) 2-chloropentane The thio compound is solvolysed at a much faster rate due to anchimeric assistance, while 1- chloropentane (having no NGP) reacts at a much slower rate due to absence of the assistance.

Stereochemistry: If chirality is present at the α- carbon, then NGP will gives only one product with retention of configuration. Had the reaction been SN1, then we would have got rac. mixture and again had it been simply intermolecular SN2, then would have got the inverted product. But in presence of NGP, we get only the retention product. This speaks volumes of the mechanism of NGP and its effect both on rate and stereochemistry. Dr. S. S. Tripathy 81 GOC-III(Introductory Reaction Mechanism)

H OH OH H AgOH CCH3 CCH3 O O Br OH R-2-bromopropanoic acid R-2-hydroxypropanoic acid

Mechanism:

H H O H O OH- O CCH 3 CCH O - CCH3 3 H2O O O Br H Br

OH-

HO H H HO H O

CCH3 CCH3 O - OH O OH OH

The carboxylate ion formed makes intramolecular SN2 attack onto the a-carbon to bring about the first inversion in the cyclic lactone intermediate formed. Then the nucleophile OH- makes an intermolecular attack onto the same carbon from the opposite side of the lacone bridge, so as to effect another inversion. Thus the final result is retention of configuration. R- reactant remains as R- product.

NGP by Aromatic Ring: When a Ph- group occupies β-position w.r.t the leaving group, NGP by the aromatic ring occurs to entirely change the course of reaction. Acetolysis of the tosylate(2S,3R) gives a equimolecular mixture of entantiomeric acetates.

Me H Me Me H H

OAc + OTs AcOH OAc

Me H H Me H Me (2S,3R)-3-phenylbutan-2-yltosylate (2S,3R) acetate (2R,3S) acetate d/l pair

We get almost equimolar amounts of the enantiomeric products when a particular stereroisomeric reactant is subjected to solvolysis by AcOH. Had it been a simple SN2, then we would have got the diastereosomeric product(2R, 3R) due to inversion at C-2. But interestingly, due to NGP, two enantiomeric products are obtained. The other two enantiomeric pairs, namely (2R,3R) and

82 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism) (2S,3S) are almost absent in the product mixture. This speaks about the NGP with cyclic interemediate formation and attack by the nucleophile at both the chiral carbons with equal propbabilities from the opposite side of the ring.

Me H

RSs not show Me H OTs resonance st a b i l i se d arenonium ion

Me H Me H (2S,3R) AcOH

Me H

OAc Me H

Me H Me H

(2S, 3R) OAc (2R,3S)

d/l pair

Due to the chiral nature of the β-carbon we get two products(enantiomeric pair). Since the cyclopropane intermediate is tied up in the upper part, the nuclophile has to attack from the opposite side. That is why the other enantiomeric products are almost absent(In reality that is formed by 4% extent, indicatinve a small % of SN1 mechanism). NGP by C=C: The unsaturated tosylate given beblow for the bicycic compound reacts 1011 times faster than the saturated analogue. This is because in SN1 mechanism, the carbocation is stabilised by forming other cyclcic carbocation by resonance.

The resonance stabilisation of the first formed carbocation is shown below.

Dr. S. S. Tripathy 83 GOC-III(Introductory Reaction Mechanism) NGP in 1,2-disubstituted cyclcohexane: Trans-2-iodocyclohexylbrosylate undergoes solvolysis(acetolysis) 106 times faster than Cis-isomer. Additionally, the trans-isomer gives the retention product i.e trans- acetate by NGP(double SN2) while Cis-isomer gives a mixture of cis and trans acetates by SN1 mechanism, as NGP is not stereoelecronically not possible.

OBs OAc I flip NGP AcOH OBs

trans-brosylate I I I flip

I OAc

trans-acetate

OAc flip I SN1 AcOH

BsO OBs I cis-brosylate I trans I NGP not possible +

AcO

cis I

For cis-isomer, even we flip the ring to the other chair form, the SN2 attack from the rear side – fo the leaving group (BsO ) is not possible. Since NGP is not possible, it reacts by SN1 mechanism to give a mixture of cis and trans acetates.

84 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism) Stereospecific and Stereoselective Reactions:

(A) Stereospecific Reactopms The reaction in which particular stereoisomeric reactant gives a particular stereoisomeric product, it is called a stereospecific reactions. Examples: 1. Halogenation of alkenes is ANTI stereospecific. Trans(E) but-2-ene on bromination gives meso-2,3-dibromobutane while Cis(Z)but-2-ene gives the d/l pair. 2. Hydroxylation by Baeyer’s reagent or by OsO4 is SYN stereospecific. Trans-but-2-ene gives d/l-butane-2,3-diol while the cis-isomer gives the meso compound. 3. Hydroboration-Oxidation is SYN stereospecific. 4. Hydrogenation of alkenes is also SYN stereospeicific.

5. SN1 reactions of a suitable substrate is sterospecific, as either R or S reactant gives a rac. mixture.

6. SN2 reactions of a suitable substrate is also stereospeicific as inverted product is only formed. 7. Epoxide formation from alkene is stereospecific, as cis-alkene gives cis-epoxide and trans alkene give trans epoxide. 8. Reaction of alkene to singlet carbene to form cyclopropane in liquid phase is stereospecific 9. E2 elimnation is ANTI stereospecific i.e the two leaving groups should have ANTIPERIPLANARITY state in the TS. 10. Addition to a bicyclicalkene is always stereospecific. Exo product is formed exclusively.

O C HCOOH O H reflux H bicyclo[2.2.1]hept-2-ene Exo-2-formyloxybicylco [2.2.1]heptane Even the same compound with m-CPBA forms exclusively the EXO-oxide. (Many more examples you have come across in your journey in organic reactions in substitution, elimination and addition reactions other name reactions)

(B) Stereoselective Reactions: A single stereoisomeric reactant forms more than one stereoisomeric products, out of which one predominates in the mixture. * We can coin two terms (a) enantioselective; if the two products are enantiomers or (b) diastereoselective: if the products are diasteroisomers (a) When a new chiral centre is formed during a synthesis, usual result is a racemic mixture. Howeve, if suitable chiral catalysts or reagents are used the reaction is convincingly tilted in favour of one enantiomer. Enzymes in biological systems always favour one eantaiomer over the other.

H H chiral reagent H S H + S or Enzyme X Y X Y X Y achiral One major and other minor enantiomers

Dr. S. S. Tripathy 85 GOC-III(Introductory Reaction Mechanism) (b) If there is an existing chiral centre and a new chiral centre is being formed in the synthesis, we usually get the two diasteroisomers in equal amounts. But if suitable manipulation, like the previous one, is made, one of the diastereosimers can be formed as major product.

R H H H H R chiral reagent a a a + Y or Enzyme Y Y b b b c c c chiral(one chiral centre) One of them is major diastereoisomers

H H H Me Me Me H2, Pt + acetic acid Me H CH2 H Me cis (68%) trans(32%) diastereoselective

N.B: Stereoselectivity is dominated structural - steric and electronic features of the reactants. Some Examples 1.

I t-BuOK DMSO + + but-1-ene Trans Cis (20%) 60% 20% diastereoisomers

A WORD OF CAUTION: The term ‘stereoselectivity’ is used approximately for ‘stereospecificity” by many authors. What we say, bromination of alkene is stereospecific, others might say, it is stereoselective. Such authors consider that there is always some amount(however small it may be) of the other stereoisomer being formed in the mixture(may be 99% and 1%). Hence, they refrain from using the term ‘stereospecific’. But i strongly feel that we should use both the terms with proper logic. There are many reactions which are truly stereospecific and some more 99% stereospecific. For such ones, we SHOULD use the term stereospecific. For those, which are 60:40 or 70 :30 or 80:20 and the like, we should use the term stereoselective.

86 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism)

E-CONCEPT IN CHEMISTRY

For Class XI( 1st Year +2)

General Organic Chemistry-Part-III

Introductory Reaction Mechanism

Dr. S. S. Tripathy President, The Uranium (Formerly Senior Reader in Chemistry Ravenshaw College, Cuttack) Dr. S. S. Tripathy 87