Reaction-Mechanisms-GOC-Book.Pdf

158 Essential Notes on General Organic Chemistry (G.O.C.)

CHAPTER

Reaction Mechanisms

 Topics Coverage :-

• Substitution Reactions • Free Radical Reactions • SN1 Mechanism • SN2 Mechanism • SN1 v/s SN2 Mechanisms • E1 Mechanism • E2 Mechanism • E1 v/s E2 Mechanisms • Substitution v/s Elimination Reactions • Aromatic Substitution Electrophilic Mechanism • Aromatic Substitution Nucleophilic Mechanism • Addition Nucleophilic Mechanism • Addition Electrophilic Mechanism • Addition Free Radical Mechanism • RARE MECHANISMS : (SNi, E1cb, SN', NGP, Benzyne, Ipso) • Rearrangements

Introduction :

How exactly does an organic reaction take place is the subject of study in this chapter of Reaction Mechanisms. Just as detectives investigate a crime after it has taken place and arrive at a reasonable prediction about how the crime may have occurred, similarly organic chemists predict by a number of laboratory techniques using sophisticated instrumentation how an organic reaction may have taken place. Many of these techniques are based on the Physical Chemistry principles of Redox, Equilibria, Chemical Kinetics and Thermodynamics. The instruments used include Polarimeter, NMR Spectroscope, Electric Dipole measurement device, isotopic tracer, pH meter etc.

CLASSIFICATION OF ORGANIC REACTIONS

Organic Reactions in JEE syllabi can be classified as under :

1. Substitution reactions - (i) Free radical (ii) Nucleophilic-SN1, SN2,SNi (iii) Electrophilic 2. Elimination reactions-E1, E2 & E1cb 3. Addition reactions - (i) Electrophilic (ii) Nucleophilic (iii) Free Radical 4. Rearrangements

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1. SUBSTITUTION REACTIONS

A reaction in which one group or atom is replaced by another is called a substitution reaction. The incoming group is bonded to the same carbon to which the leaving group was bonded. Substitution reaction has been classified according to the nature of the substituents involved. Thus- • • A. Free radical substitution A: B + Q → A: Q + B Θ Θ + → + B. Nucleophilic substitution A: B : Q A: Q : B ⊕ ⊕ C. Electrophilic substitution A: B + Q → A: Q + B It will be seen that in all types of substitutions, the displaced species belong to the same class as the attacking species.

A. Free Radical Substitution Radical substitution reactions are initiated by radicals in the gas phase or in non-polar solvents. Thus, methane and chlorine react in the presence of sunlight or heat to give methyl chloride. hv or CH 4 + Cl2 →+ CH 3Cl + HCl 300°C Light energy or heat causes homolytic fission of chlorine producing chlorine radicals which attack methane to form methyl chloride. • • Cl : Cl hvor300°C →Cl+ Cl …….(i) • •

H 3C : H + Cl → H 3C+ HCl ……..(ii) • •

H 3C+ Cl : Cl → H 3C : Cl + Cl ……(iii)

B. Nucleophilic Substitution (Common in aliphatic compounds):

 Nucleophilic substitution is one of the two main reactions of alkyl halides. A nucleophile replaces a leaving group on an sp3 hybridized carbon. • − − R – X + • Nu R – Nu + X :

nucleophile leaving group The electron pair in the C – Nu bond comes from the nucleophile.

 One σ bond is broken and one σ bond is formed.  There are two possible mechanisms: SN1 and SN2

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SN1 and SN2 Mechanisms Compared

SN2 Mechanism SN1 Mechanism [1] Mechanism One step Two steps [2] Alkyl halide Order of reactivity : Order of reactivity :

CH 3 X > RCH 2 X > R2CHX > R2CX R3CX > R2CHX > RCH 2 X > CH 3 X [3] Rate equation Rate = k[RX][: Nu] Rate = k[RX] Second – order kinetics First – order kinetics [4] Stereochemistry Backside attack of the nucleophile Trigonal planar carbocation Inversion of configuration at a intermediate stereogenic center Racemization at a single stereogenic center [5] Nucleophile Favored by stronger nucleophiles Favored by weaker nucleophiles [6] Leaving group Better leaving group → faster reaction Better leaving group faster reaction [7] Solvent Favored by polar aprotic solvents Favored by polar protic solvents [8] Reactivity 3° > 2° > 1° > CH3 CH3 > 1° > 2° > 3° structure of Stability of R + Steric hindrance in R group R-determining RI > RBr > RCl > RF RI > RBr > RCl > RF factor. Nature of X

[9] Effect of Rate depends on nucleophilicity No effect on rate nucleophile I − > Br − > Cl − ; RS > RO − + [10] Catalysis Lewis acid, e.g. Ag ,AlC3, ZnCl2 None of these [11]Competition Elimination, rearrangement Elimination reactions

[12] TS of slow step − − − δ δ δ + δ Nu − C − X C X

• Factors to decide whether SN1 or SN2 : Any factors that affects the energy of activation of a given type of reaction will affect the rate and / or mechanism. These factors are useful to decide whether SN1 or SN2 occurs : 1. Electronic effect 2. Steric effect 3. The nature of the leaving group. 4. The nature of the nucleophile. 5. Participation of neighbouring group. 6. Effect of solvent.

1. ELECTRONIC EFFECT :- I-effect, R-effect, hyperconjugation etc. which can affect the stability of carbocation are collectively termed as electronic effect. Clearly if electronic effects stabilise the carbocation,

SN 1 will more probable than SN 2 .

C − X(1°) C − X(2°) C − X(3°)

S 2 S 2 or S 1 S 1 N N N N

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2. STERIC EFFECT :-

More steric restriction opposes SN 2 mechanism. Simple alkyl halides show the following

general order of reactivity in SN 2 reactions: methyl > primary > secondary >>(tertiary) Methyl halides react most react most rapidly and tertiary halides react so slowly as to be unreactive by the SN2 mechanism.

3. THE NATURE OF LEAVING GROUP :- The ease of formation of carbocation supports SN1 and this formation depend upon nature of leaving group. The best leaving group is the weakest base. Leaving group ability increases left-to-right across a row and down a column of the periodic table. The ease of leaving is as follows :

The leaving power of some nucleophilic groups is given below in decreasing order : O O O O

Θ Θ Θ Θ CF3 – S – O > Br S – O > CH 3 S – O > C 6H5 – S – O

O O O O

O O O

Θ Θ Θ Θ ⊕ Θ Θ > CH 3 – S – O > I > Br > CF3 – C – O > H 2O > Cl > F > CH 3 – C – O

4. THE NATURE OF THE NUCLEOPHILE :-

Stronger nucleophile supports SN 2 mechanism. Strength of nucleophile is as follows : − − − − − − − − PhS > CN > I > Et − O > OH > Br > PhO > Cl > Me3 N :

5. Participation of Neighbouring Groups : Some neighbouring groups as : O || COO− , X,OH,−O −C− Ph − etc. can also participate and determine the nucleophilic substitution.

6. Effect of solvent : As the dielectric constant (i.e., polarity) of solvent increases,

the rate of SN 1 reaction increases.

Illustration : Indicate the effect on the rate of SNl and SN2 reactions of the following: (a) Doubling − the concentration of substrate (RL) or Nu . (b) Using a mixture of ethanol and H2O or only acetone as solvent. (c) Increasing the number of R groups on the C bonded to the leaving group, L. (d) Using a strong Nu− .

− Ans: (a) Doubling either [RL] or [ Nu ] doubles the rate of the SN2 reaction. For SN1 reactions, the rate is doubled only by doubling [RL] and is not affected by any change in [ Nu− ].

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162 Essential Notes on General Organic Chemistry (G.O.C.)

(b) A mixture of ethanol and H2O has a high dielectric constant and therefore enhances the rate of SN1 reactions. This usually has little effect on SN2 reactions. Acetone has a low dielectric constant and is aprotic and favors SN2 reactions. (c) Increasing the number of R’s on the reaction site enhances SN1 reactivity through + electron release and stabilization of R . The effect is opposite in SN2 reactions + because bulky R’s sterically hinder formation of, and raise ∆H + for, the transition state. (d) Strong nucleophiles favor SN2 reactions and do not affect SN1 reactions

Illustration : List the following alkyl bromides in order of decreasing reactivity in indicated reactions. (I) (II) (III) CH3

CH3 C CH2CH3 CH3CH2CH2CH2CH2Br CH3CH2CHCH2CH3 Br Br

(a) SN1 reactivity, (b) SN2 reactivity, (c) reactivity with alcoholic AgNO3.

Ans. (a) Reactivity for the SN 1 mechanism is 3°(I) > 2°(III) > 1° (I). (b) The reverse reactivity for SN2 reactions gives 1° (II) > 2°(III) > 3°(1). + (c) Ag catalyzes SN1 reactions and the reactivities are 3°(I) > 2°(III) > 1° (II).

Aryl halides are less reactive towards SN reaction due to the donation of lone pair electrons to the benzene ring via resonance. Aryl halides undergo SN reaction when a strong electron-withdrawing group (or activating group) is present at the ortho and para positions. ArSN reactions are of two types viz. (i) Addition-Elimination reaction (ii) Elimination-Addition reaction via benzyne mechanism

(i) Addition-Elimination reaction (also called Bimolecular displacement mechanism)

•• • • Nu Cl • Cl • Nu Cl Nu Cl Nu Cl H H Θ Θ ≡ Θ

Θ H Fast

+ Cl −

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Any factor that stabilizes the carbanion will increase the rate of nucleophilic substitution reaction by dispersion the charge presence on resonating structures. An electron withdrawing group present at meta position does not activate the ring as much as it does from ortho and para position. This can be known by looking at following resonance structures. Z Cl Z Cl Z Cl

NO2 NO2 NO2 (meta) Θ

Z Cl Z Cl Z Cl

(para)

NO2 NO2 NO2

(ii) Elimination-Addition Mechanism : (Please refer to RARE MECHANISMS at the end of this chapter)

(D) Electrophilic Substitution (Common in Aromatic compounds) : Electrophilic Substitution involves the attack by an electrophile. It is represented as SE. If the order of the reaction is 1, it is abbreviated as SE1 (unimolecular) and if the order is 2, it is SE2 (bimolecular). (a) SE1 Mechanism : Electrophilic substitution in aliphatic compounds (SE1) are very rare. Some of the important examples are : (i) Replacement of metal atom in an organo-metallic compound by hydrogen R---M + H+  R—H + M+ (ii) Isotopic exchange of hydrogen for deuterium or tritium R---H + D+  R—D + H+ R---H + T+  R—T + H+

(b) SE2 Mechanism : Electrophilic substitution (SE2) is every common in benzene nucleus (aromatic compounds) in which π electrons are highly delocalized and an electrophile can attack the region of high electron density. Aromatic electrophilic substitution reactions involve the following 3 steps mechanism : Step-1 : The formation of an electrophile. Step-2 : The electrophile attacks the aromatic ring to form carbocation (or arenium ion) which is stabilized by resonance. Step-3 : Carbocation loses the proton to form the substitution product. e.g. Bromination of Benzene, Nitration, Sulphonation & Friedel-Crafts reactions of benzene are classic examples of electrophilic substitution reactions.

NB: (i) The second step is found to be rate – determining step by isotopic studies. (ii) The intermediate carbocation is stabilized by resonance; a minimum of three resonance structures can be drawn. The positive charge is always located ortho or para to the new C – W bond. Refer to Chapter 5 or lecture notes for more clarity on the SE2 mechanism.

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164 Essential Notes on General Organic Chemistry (G.O.C.)

2. ELIMINATION REACTIONS

Elimination reactions of alkyl halides are important reactions that compete with substitution reactions. In an elimination reaction, the fragments of some molecule (YZ) are removed (eliminated) from adjacent atoms of the reactant. This elimination leads to the introduction of a multiple bond: elimination C C C = C (-YZ) Y Z

The reaction in which two atoms or groups are removed from a compound is referred to as an elimination reaction. The elimination reaction can be classified as shown below: Elimination Reaction

1,1 1,2 1, 3 1, 4 Elimination Elimination Elimination Elimination or or α - elimination β - elimination

Syn Anti Elimination Elimination (Ei ) Elimination Intra-molecular

E1 E2 E1CB

E1 & E2 Mechanisms Compared

E2 Mechanism E1 Mechanism Mechanism One step Two step

Alkyl halide Rate : R3CX > R2CHX > RCH 2 X Rate : R3CX > R2CHX > RCH 2 X Rate equation Rate = k[RX][B :] Rate = k[RX] Second – order kinetics First – order kinetics Stereochemistry Anti periplanar arrangement of H and X Trigonal planar carbocation intermediate Base Favored by strong bases Favored by weak base Leaving group Better leaving group → faster reaction Better leaving group → faster reaction Solvent Favored by polar aprotic solvents Favored by polar protic solvents Product More substituted alkene favored More substituted alkene favored (Zaitsev rule (Satyzeff’s) When will the reaction be E1 and when will it be E2 ?

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The strength of the base determines the mechanism of elimination.  Strong bases favor E2 reactions  Weak bases favor E1 reactions Illustration : CH3

C = CH2 + H2O + Br-

Strong base CH3

- OH

CH3 E2 Same Product,

CH3– C – CH3. Different Mechanism

Br H2O

CH3 Weak base C = CH2 + H2O+ + Br-

E1 CH3

Regioselectivity : ORENTATION IN ELIMINATION REACTIONS :-

If substrate is unsymmetrical, then this will give more than one product. Major product of the reaction can be known by two empirical rules.

(1) Saytzeff’s Rule (also spelled as Zaitsev’s Rule by some books) :- According to this rule, major product is the most substituted alkene, i.e., major product is obtained ⊕ by elimination of H from that β - carbon which has the least number of hydrogens. Product of the reaction in this case is known as Saytzeff product. Cl β2 | alc. KOH / ∆ CH3 − CH − CH − CH3 →CH3 − C = CH − CH3 | α | β1 CH3 CH3 Saytzeff product (2) Hofmann Rule :- According to this rule, major product is always least substituted alkene, i.e., major product is formed from β − carbon which has maximum number of hydrogens. Product of the reaction in this case is known as Hofmann product. CH3 Br CH3 | | | alc.KOH /∆ CH 3 − C −CH 2 −CH−CH 3 →CH 3 − C −CH 2 −CH = CH 2 | α | β2 β1 CH3 CH3 Hofmann Product Note :- (i) In E1 reactions, product formation always takes place by Saytzeff rule.

(ii) In E1 cb reactions, product formation always takes place by Hofmann rule.

(iii) In E2 reaction, product formation takes place by Saytzeff as well as Hofmann rule. In almost all E2 reactions, product formation takes place by Saytzeff rule. Only in four cases, product formation takes place by Hofmann rule.

4 CASES OF HOFFMANN ELIMINATION REACTIONS :-

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(1) Dehydrohalogenation of alkyl halides when leaving group is very poor, i.e., alkyl fluoride (primary or secondary). F | alc.KOH /∆ CH 3 −CH 2 −CH 2 −CH−CH 3 →CH 3 −CH 2 −CH 2 −CH = CH 2 + CH 3 −CH 2 −CH = CH −CH 3 2 α β β1 Hofmann product 70% Saytzeff product 30% (2) Primary and sec – alkyl halides give Hoffmann elimination when the size of the base is bulky, i.e., sodium or potassium salt of tert butoxide. CH3 | Θ⊕ CH 3 − C − OK/∆ Br | | CH3 CH 3 −CH 2 −CH−CH 3 CH 3 −CH 2 −CH = CH 2 + CH 3 −CH = CH −CH 3 β α β 2 1 (Hofmann product) (Saytzeff product) 80% 20% (3) Primary and sec – alkyl halides having quarternary γ − carbon gives Hofmann elimination. CH 3 Br CH3 CH3 | | | | γ alc.KOH / ∆ CH 3 C − CH 2 − CH − CH 3 →CH 3 C − CH 2 − CH = CH 2 + CH 3 − C − CH = CH − CH 3 | α | | β2 β1 CH3 CH3 CH3 Major product Minor product (Hofmann product) (Saytzeff product)

(4) If leaving group is bulky, then compound gives Hofmann elimination reaction. The most common type of lager, bulky leaving group which leads to Hofmann products has positively ⊕ ⊕ charged nitrogen (NR3 )or positively charged sulphur (SR2 )

Illustration :

Q. Which of the following will give Hofmann product in E2 reaction? ⊕ Θ (a) CH 3 −CH 2 −CH 2 −CH− N(CH 3 )3 OH | CH3 ⊕ Θ (b) CH 3 −CH 2 −CH 2 −CH− S(CH 3 )2 OH | CH3 F | (c) CH 3 −CH 2 −CH−CH 3 (d) All of these Ans. (d)

Illustration : − + Explain the fact that whereas 2-bromopentane undergoes dehydrohalogenation with C2 H 5O K − + to give mainly 2-pentene (the Saytzeff product), with Me3CO K it gives mainly l-pentene (the anti-Saytzeff, Hofmann, product).

Ans :

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− Since Me3CO is a bulky base, its attack is more sterically hindered at the 2° H than at the − 1° H. With Me3CO , −H 1Br CH 2 = CHCH 2CH 2CH 3 ← CH 2 − CH − CH CH 2CH 3 major product less hindered | | | Br 2 H 1 H

−H 2Br → CH 3CH = CHCH 2CH 3 more hindered min or product Elimination v/s Substitution :

All nucleophiles are potential bases and all bases are potential nucleophiles. This is because the reactive part of both nucleophiles and bases is an unshared electron pair. It should not be surprising, then, that nucleophilic substitution reactions and elimination reactions often compete with each other.

Thus, elimination reactions are usually accompanied by substitution reactions. When the reagent is a good base, it accepts protons to yield elimination products (alkenes) and if it is a good nucleophile, then it attacks the carbon to give substitution products. The proportion of elimination and substitution depends upon the following:

(i) Structure of the Substrate (ii) Nature of the Base (iii) Nature of Solvent (iv) Effect of Temperature

In general, the proportion of elimination increases on using a strong base of high concentration and a solvent of low polarity. On the other hand, the proportion of substitution increases by using a weak base of low concentration and a solvent of high polarity

A Comparison Between Nucleophilic Substitution and β Elimination :

Nucleophilic substitution – A nucleophile attacks a carbon atom : Nu − H H Nu

- C – C - - C – C - + X :−

X X good substitution product leaving group β Elimination – A base attacks a proton B :

H − - C – C - C = C + H – B+ + X : good

X elimination leaving group

Similarities Differences

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In both reaction RX acts as an electrophile, In substitution, a nucleophile attacks a single reacting with an electron – rich reagent. carbon atom. − Both reaction require a good leaving group X : In elimination, a Bronsted – Lowry base that can accept the electron density in the C – X removes a proton to form a π bond, and two bond. carbons are involved in the reaction .

Summary Chart on the Four Mechanisms : SN 1,SN 2,E1 , or E2

Alkyl halide type Conditions Mechanism

strong nucleophile → SN 2 1°RCH X 2 strong bulky base → E2

strong base and nucleophile → SN 2 + E2 2°R CHX 2 strong bulky base → E2

weak base and nucleophile → SN 1 + E1

weak base and nucleophile → SN 1 + E1 3°R3CX strong base → E2

Mechanism Review : Substitution versus Elimination

SN2 SN1 & E1 Primary substrate Tertiary substrate. Carbocation intermediate Back-side attack of Nu: with respect to LG Weak nucleophile/base (e.g., solvent) Strong/polarizable unhindered nucleophile

Bimolecular in rate-determining step concerted Unimolecular in rate-determining step bond forming/bond breaking. Inversion of Racemization if SN1 stereochemistry. Favored by polar aprotic Removal of β -hydrogen if E1 solvent. Protic solvent assists ionization of LG Low temperature (SN1) / high temperature (E1)

SN2 and E2 E2

Secondary or primary substrate Tertiary or secondary substrate. Strong unhindered base/nucleophile leads to Concerted anti-coplanar TS SN2 Bimolecular in rate-determining step. Strong hindered base/nucleophile leads to E2 Strong hindered base. High temperature Low temperature (SN2) / high temperature (E2) δ − LG

δ − Nu /B H

Detailed Discussion on SN2 v/s E2 :

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SN2 and E2 reactions are both favored by a high concentration of a strong nucleophile or base. When the nucleophile (base) attacks a β hydrogen atom, elimination occurs. When the nucleophile attacks the carbon atom bearing the leaving group, substitution results: C (a) − || + Nu – H + : X Elimination (a) H – C C − E2 Nu : – C (b) X H – C (b) − + X : substitution Nu – C SN 2

• Consider the following examples with small (unhindered) nucleophiles and alkyl halides of different classes.

Primary Substrate :- When the substrate is a primary halide and the base is unhindered, like ethoxide ion, substitution is highly favored because the base can easily approach the carbon bearing the leaving group: − + C2 H 5OH CH 3CH 2O Na +CH 3CH 2 Br →CH 3CH 2OCH 2CH 3 +CH 2 = CH 2 55°C (−NaBr)

SN 2 E2 (90%) (10%) Secondary Substrate :- With secondary halides, however, a strong base favors elimination because steric hindrance in the substrate makes substitution more difficult:

− + C 2 H 5OH + = CH 3CH 2O Na + CH 3CHCH 3 CH 3CHCH 3 CH 2 CHCH 3 55°C

(−NaBr) Br OCH2CH2 SN 2 E2

(21%) (79%)

Tertiary Substrate :- With tertiary halides, steric hindrance in the substrate is severe and an SN2 reaction cannot take place. Elimination is highly favored, especially when the reaction is carried out at higher temperatures. Any substitution that occurs must take place through an SN1 mechanism: CH3 CH3 CH3

− + C2H5OH CH CH O Na + CH CCH CH CCH + CH = CCH 3 2 3 3 55°C 3 3 2 3

(-NaBr) Br OCH 2CH 3

SN 1 Mainly E2 (9%) (91%)

CH3 CH3

− + C2H5OH CH CH O Na + CH CCH CH = CCH + CH CH OH 3 2 3 3 55°C 3 3 3 2

(-NaBr) Br E2 + E1 (100%)

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• Size of the Base/Nucleophile :- Increasing the reaction temperature is one way of favorably influencing an elimination reaction of an alkyl halide. Another way is to use a strong sterically hindered base such as the tert-butoxide ion. The bulky methyl groups of the tert-butoxide ion inhibit its reacting by substitution, allowing elimination reactions to take precedence. We can see an example of this effect in the following two reactions. The relatively unhindered methoxide ion reacts with octadecyl bromide primarily by substitution; the bulky tert-butoxide ion gives mainly elimination.

Unhindered (small) Base / Nucleophile − CH3OH CH O + CH (CH ) CH CH − Br 3 3 2 15 2 2 65°C

CH 3 (CH 2 )15 CH = CH 2 + CH 3 (CH 2 )15 CH 2CH 2OCH 3

E2 SN 2 (1%) (99%)

Hindered Base / Nucleophile CH3

− (CH 3 )3 COH CH3 C O + CH 3 (CH 2 )15 CH 2CH 2 − Br 40°C

CH3 CH3

CH 3 (CH 2 )15 CH = CH 2 + CH 3 (CH 2 )15 CH 2CH 2 O − C − CH 3

E2 SN 2 CH3 (85%) (15%)

Basicity and Polarizability Another factor that affects the relative rates of E2 and SN2 reactions is the relative basicity and polarizability of the base/nucleophile. Use of a strong, − slightly polarizable base such as amide ion ( NH 2 ) or alkoxide ion (especially a hindered one) tends to increase the likelihood of elimination (E2). Use of a weakly basic ion such as − − a chloride ion (Cl ) or an acetate ion (CH 3CO2 ) or a weakly basic and highly polarizable one such as Br-, 1-, or RS- increases the likelihood of substitution (SN2). Acetate ion, for example, reacts with isopropyl bromide almost exclusively by the SN2 path: O CH 3 O CH 3 || | || | − − CH 3 C− O + CH 3 CH− Br →CH 3 C− O − CH CH 3 + Br SN 2 (~100%) The more strongly basic ethoxide ion reacts with the same compound mainly by an E2 mechanism.

• Temperature :- Increasing the reaction temperature favors elimination (El and E2) over substitution. Elimination reactions have greater free energies of activation than substitution reactions because more bonding changes occur during elimination. When higher temperature is used, the proportion of molecules able to surmount the energy of activation barrier for elimination increases more than the proportion of molecules able to undergo substitution, although the rate of both substitution and elimination will be increased.

Furthermore, elimination reactions are entropically favored over substitution because the products of an elimination reaction are greater in number than the reactants. Additionally, because temperature. is the coefficient of the entropy term in the Gibbs free-energy equation ∆G° = ∆H° −T∆S°, an increase in temperature further enhances the entropy effect.

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Detailed Discussion on Tertiary Halides  SN1 v/s E1

Because E1 and SN1 reactions proceed through the formation of a common intermediate, the two types respond in similar ways to factors affecting reactivities. El reactions are favored with substrates that can form stable carbocations (i.e., tertiary halides); they are also favored by the use of poor nucleophiles (weak bases) and they are generally favored by the use of polar solvents.

It is usually difficult to influence the relative partition between SN1 and E1 products because the free energy of activation for either reaction proceeding from the carbocation (loss of a proton or combination with a molecule of the solvent) is very small.

In most unimolecular reactions the SN1 reaction is favored over the E1 reaction, especially at lower temperatures. In general, however, substitution reactions of tertiary halides do not find wide use as synthetic methods. Such halides undergo eliminations much too easily.

Increasing the temperature of the reaction favors reaction by the E1 mechanism at the expense of the SN1 mechanism. If the elimination product is desired, however, it is more convenient to add a strong base and force an E2 reaction to take place instead.

Illustration : Given the product (or products) that you would expect to be formed in each of the following reactions. In each case give the mechanism (SN 1,SN 2,E1, or E2) by which the product is formed and predict the relative amount of each (i.e., would the product be the only product, the major product, or a minor product?).

− 50°C (a) CH 3CH 2CH 2 Br + CH 3O → CH 3OH

− 50°C (b) CH 3CH 2CH 2 Br + (CH 3 )3 CO → (CH 3 )3 COH

CH3 (c) C − Br + HS − 50°C → CH 3OH CH3CH2 H − 50°C (d) (CH 3CH 2 )3 CBr + OH → CH 3OH 25°C (e) (CH 3CH 2 )3 CBr → CH OH 3 Answer :

− (a) The substrate is a 1°halide. The base/nucleophile is CH 3O , a strong (but not a hindered one) and a good nucleophile. According to Table, we should

expect an SN 2 reaction mainly, and the major product should be CH 3CH 2CH 2OCH 3 .

A minor product might be CH3CH = CH2 by an E2 pathway.

− (b) Again the substrate is a 1° halide, but the base/nucleophile, (CH 3 )3 CO , is a strong hindered base. We should expect, therefore, the major product to be CH3CH = CH2 by an E2 pathway and a minor product to be CH3CH2CH2OC(CH3)3 by an SN2 pathway.

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(c) The reactant is (S)-2-bromobutane, a 2° halide and one in which the leaving group is attached to a stereogenic center. The base/nucleophile is HS − , a strong nucleophile but a weak base. We should expect mainly an SN2 reaction, causing an inversion of configuration at the stereogenic center and producing the (R) stereoisomer: CH3

HS C

CH3CH2 H (d) The base/nucleophile is OH-, a strong base and a strong nucleophile. However, the substrate is a 3° halide; therefore, we should not expect an SN2 reaction. The major product should be CH3CH = C(CH2CH3)2 via an E2 reaction. At this higher temperature and in the presence of a strong base; we should not expect an appreciable amount of the SN1 product, CH3OC(CH2CH3)3. (e) This is solvolysis; the only base/nucleophile is the solvent, CH3OH, which is a weak base (therefore, no E2 reaction) and a poor nucleophile. The substrate is tertiary (therefore, no SN2 reaction). At this lower temperature we should expect mainly an SN1 pathway leading to CH3OC(CH2CH3)3. A minor product, by an E1 pathway, would be CH3CH = C(CH2CH3)2.

Illustration : Indicate the products of the following reactions and point out the mechanism as :

S N 1, S N 2, E1or E2. − (a) CH 3CH 2CH 2 Br + LiAlH 4 (source of : H )

(b) (CH 3 )3 CBr + C2 H 5OH, heat at 60°C

(d) BrCH 2CH 2 Br + Mg (ether)

(e) BrCH 2CH 2CH 2 Br + Mg (ether)

(f) CH 3CHBrCH 3 + NaOCH 3 in CH 3OH Answer : − − − (a) CH 3CH 2CH 3 ; an S N 2 reaction, : H of AlH 4 replaces Br .

CH 3 (b) − + CH 3CH 2OH (CH ) CBr Br + (CH 3 )3 C (CH 3 )3 COCH 2CH 3 + CH 3C = CH 2 3 3 − H + 3°RX major ; SN1 very minor ; E1

Vinyl halides are quite inert toward S N 2 reactions.

(d) BrCH 2CH 2 Br + Mg → H 2C = CH 2 + MgBr2 This is an E2 type of β - elimination via an alkyl magnesium iodide. + Mg + BrCH2CH2Br − Br Mg CH2 – CH2 – Br MgBr2 +H2C = CH2

(e) This reaction resembles that in (d) and is an internal S N 2 reaction

CH2Br − + CH 2 MgBr CH2 H3C + Mg H3C H3C + MgBr2 S N 2 CH2Br CH2 – Br CH2

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(f) This 2°RBr undergoes both E2 and S N 2 reactions to form propylene and isopropyl methyl ether. +

CH 3CHBrCH 3 + NaOCH 3 (CH 3OH ) →CH 3CH = CH 2 + CH 3 CHCH 3 | OCH3

Illustration: Will the following reactions be primarily substitution or elimination? − (a) CH 3CH 2CH 2Cl + I → − (b) (CH 3 )3 CBr + CN (ethanol) → − (c) CH 3CHBrCH 3 + OH (H 3O) → − (d) CH 3CHBrCH 3 + OH (ethanol) →

(e) (CH 3 )3 CBr + H 2O → Answer : − (a) SN 2 substitution. I is a good nucleophile, and a poor base. (b) E2 elimination. A 3° halide and a a fairly strong base.

(e) SN 1 substitution. H 2O is not basic enough to remove a proton to give elimination.

3. ADDITION REACTIONS :

(1) Addition reactions are given by those compound which have at least one π − bond, i.e., O || ( > C = C <,−C ≡ C−,−C −,C ≡ N). (2) In this reaction, there is loss of one pi bond and gain of two sigma bonds. Thus product of the reaction is generally more stable than reactant. (3) The Addition reaction is a spontaneous reaction.

Types of Addition reactions : Addition reaction can be classified into three categories on the basis of the nature of initiating species : (1) Electrophilic Addition (2) Nucleophilic Addition, and (3) Free radical Addition

• Electrophilic Addition Reactions

(1) This reaction is mainly given by alkenes and alkynes

(2) Electrophilic addition reactions of alkenes and alkynes are generally two – step reactions.

(3) Alkenes and alkynes give electrophilic addition with those reagents which, on dissociation, give electrophile as well as nucleophile.

(4) If the reagent is a weak acid then electrophilic adition is catalysed by strong acids (generally H2SO4)

(5) Unsymmetrical alkenes and alkynes give addition reaction with unsymmetrical regents according to Markovnikov’s rule.

REACTION MECHANISMS.

174 Essential Notes on General Organic Chemistry (G.O.C.)

Regio-selective Study : (Markovnikov’s & Anti-Markovnikov’s Rules)

Addition Reaction of Alkenes : Markovnikov’s Rule According to Markovnikov’s rule, the negative part of the reagent adds on that doubly bonded carbon if the alkene which has least number of hydrogen (s).

Conditions for the use of Markovnikov’s Rule : This rule can be used only in those alkenes which fulfill the following conditions: (a) Alkenes should be unsymmetrical (b) Substituent/substituents present on doubly bonded carbon /(s) should only be + I group. (c) If phenyl group is present on doubly bonded carbon, then both doubly bonded carbons should be substituted by phenyl groups. For example, the following alkenes will give addition according to Markovnikov’s rule: H3C H3C H3C

CH3 – CH = CH2, C = CH2, C = CH – CH3, C = CH – C6H5, C6H5 – CH = CH2

H3C H3C H3C

Following alkenes will not give addition reaction according to Markovnikov’s rule.

R R C6H5 C6H5 CH2 = CH2, R – CH = CH – R C = C , C = C

R R C6H5 C6H5

CH2 = CH – CF3 ; CH2 = CH – NO2

Illustration : Q. Markovnikov’s rule can be used in which of these? (a) CH3 – CH = CH2 (b) C6H5 – CH = CH2

H3C

Ans. (d) All alkenes are unsymmetrical.

Substituted Alkenes which give Addition Reaction according to Anti – Markovnikov’s Rule :

Unsymmetrical alkenes having the following general structure give addition reaction according to anti – Markovnikov’s rule.

H 2C = CH − G where, G is a strong – I group such as : O || −CX3 ,−NO2 ,−CN,−CHO,−COR,−COOH,−C− Z(Z = Cl,OH,OR,NH 2 ) Some examples are : H 2C = CH − NO2 , H 2C = CH −CHO, H 2C = CH −CN, O || H 2C = CH −C− R, H 2C = CH −CH 3 etc.

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Illustration :

Q. Which of these will give addition reaction according to anti – Markovnikov’s rule?

(a) H 2C = C −CF3 (b) H 2C = CH − NO2

Ans. (d)

Stereo-selective Study (Syn & Anti Addition)

An addition that places the parts of the adding reagent on the same side (or face) of the reactant is called syn addition. We have just seen that the platinum-catalyzed addition of hydrogen (X = Y = H) is a syn addition

A C C + X – Y C – C syn addition X Y

The opposite of a syn addition is an anti addition. An anti addition places the parts of the adding reagent on opposite faces of the reactant.

Y An C C + X – Y C – C anti addition X

• Nucleophilic Addition Reactions

Addition of nucleophile in the first step and electrophile in the second step is called nucleophilic addition reaction. Thus this reaction is two – step reaction. In this reaction, addition of nucleophile is rate determining step. Nucleophilic addition reaction is given by :

(1) Alkenes (2) Alkynes (3) Carbonyl compounds and (4) Nitriles.

A reaction in which the substrate and the reagent add up to form a product is called addition reaction. The reaction occurs at the site of unsaturation in a molecule. Thus, compounds having multiple bonds such as :

C = C ,−C ≡ C − C = O, − C ≡ N, etc.

undergo addition reactions. The reactivity of these compounds is due to the more exposed and easily available π electrons to the electron-seeking (electrophilic) reagent.

REACTION MECHANISMS.

176 Essential Notes on General Organic Chemistry (G.O.C.)

Free Radical Addition

Radicals, like cations are electron-deficient species. Thus radicals will attack the pi-system of a double bond of an alkene because pi-electrons of an alkene can provide the electron needed to complete the outer shell of a radical. The net result of this usually is the addition of some species across the double bond. These reactions involve radical intermediate. Free radical addition reactions occur either in the gas phase or in inert non-polar solvents in the presence of UV light or sunlight, heat or catalytic amount of radical initiators such as organic peroxides, labile azo compounds like azoisobutyronitrile (AIBN), etc. The mode of the addition reaction involves the' general steps for the radical reactions, i.e., initiation, chain propagation and termination. The most important reaction of this category is hydrobromination.

Hydrobromination :- The radical addition of HBr to a carbon-carbon double bond occurs in the presence of UV light or a small amount of initiators such as dibenzoyl peroxide. HF, HCl and HI do not, give this addition. When propene reacts with hydrogen bromide in the presence of dibenzoyl peroxide, the product is 1-bromopropane. dibenzoyl peroxide CH 3 − CH = CH 2 + HBr →CH 3 − CH 2 − CH 2 − Br In this addition the hydrogen that is being added ends up on the carbon bearing the smaller number of hydrogens. Therefore, this is an anti-Markonikov addition. In general, unsymmetrical alkenes add, HBr, in the presence of a radical initiator, in an anti-Markonikov manner. radical R − CH = CH 2 + HBr → R − CH 2 − CH 2 − Br source The addition of HBr to alkenes in the presence of light or peroxide is a free radical reaction and is trans-addition at low temperature but at room temperature two isomers give the same mixture of diastereomers, e.g.,

I. Me Me HBr + Peroxide HBr + Peroxide + hv Meso + DL C = C at room temp. at - 78°C (Br) (trans – and cis - addition ) H Br trans – Addition cis – 2 – Bromo – but – 2 – ene (or E)

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II. Me H At room Meso + DL C C (trans or Z) temp. − ° Me Me at 78 C ( ± ) Product

At room temperature, the less stable radical (from cis butene) equilibrates with the more stable radical (from trans butene) by internal rotation which is now faster than the hydrogen abstraction. The result is that both butenes give the same mixture of diastereomers. In both cases, the addition of hydrogen is trans with respect to the bromine atom. Free radical addition to alkenes is trans; no rotation is now possible, e.g., H

Br Br Br Br H + Br H Br Br

NB: Free Radical Addition Reactions are exhibited by electron-rich alkenes when they react with electron-deficient radicals. Radicals react with alkenes via a radical chain mechanism that consists of initiation, propagation and termination steps analogous to Free Radical substitution.

(4) REARRANGEMENT REACTIONS

Rearrangement refers to structural changes within a species. Rearrangements, in general are of two types: (a) Shifting of a group from one atom to the other (b) Rearrangement because of resonance

(A) SHIFTING OF A GROUP FROM ONE ATOM TO THE OTHER ATOM

Although shifting of a group can occur from an atom to any other atom yet most of the shifts occur to adjacent atoms only. The main reason for this shifting is presence of six valence electrons and a vacant orbital on an atom. Such an atom has tendency to complete its octet for which the group from the adjacent atom migrates with the bonding electrons. There are two different conditions around an atom with six valence electrons.

(a) When atom with six valence electrons and a vacant orbital:

- A – B - - A – B - Θ ⊕

G G Six valence electrons Six valence electrons around A around B

REACTION MECHANISMS.

178 Essential Notes on General Organic Chemistry (G.O.C.)

In this case when a group is shifted from atom S, the octet of S becomes incomplete which is not favourble condition. So we can say that this shift will occur only if it increases stability. H /Shift CH 3 − CH 2 − CH 2 →CH 3 − CH− CH 3 ⊕

CH3

⊕ CH3 shift ⊕ CH3 – C – CH2 CH3 – C – CH2 - CH3

CH3 CH3

Ring expansion and ring contraction can also result because of adjacent shift. CH3

⊕ H shift ⊕ CH2

alkyl shift ⊕ Ring expansion CH3

CH3 shift ⊕

⊕ CH3 CH3 CH3

C CH3 ⊕ CH3 Ring Contraction

If migration of two (or more) different groups from adjacent atom can take place, major migration depends on the migrating tendency of the migrating groups (migration of group With higher migrating tendency is favoured) and the stability of resultant species (more stable resultant species is more favoured). Migrating tendency of a group is generally higher if it is more electron rich i.e. more electron donating. The general order of migrating tendency (or migratory aptitude) is as given below: Hydride < Alkyl < Aryl < Hydride

0 0 0 (If not involved in 1 R < 2 R < 3 R (If involved in hyperconjugation hyperconjugation with atom having 6 < < with atom having 6 valence electrons) valence electron )

E.W.G E.D.G

In several cases migrating tendency and the stability of resultant species favours the same migration. In such cases it will be easier to predict the major migration product.

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Hydride shift + CH3 – C - CH2 CH3 – C - CH2 + (higher migrating tendency ) CH3 CH3 more stable (Major)

H H

Hydride shift CH3 – C - CH2 CH3 – C - CH2 - CH3 + (higher migrating + tendency ) less stable CH3

+ 3 2 CH3 – C – CH2 CH – C – CH +

hydride shift most stable Highest migrating (Major) tendency)

CH3 – C – CH2 + H

phenyl shift CH3 – C – CH3 (higher migrating + tendency) H least stable

CH3 – C – CH2 + H

methyl shift (least migrating C – CH2 – CH3 tendency) + more stable

However, if migrating tendency and stability of resultant species oppose each other, following logics can be applied to predict the major product.

(i) If both resultant species are very highly stabilized (stabilized by resonance of/one pair or stabilized by formation of additional bond), small amount of stability does not matter a lot. Therefore, major migration occurs according to migrating aptitude.

REACTION MECHANISMS.

180 Essential Notes on General Organic Chemistry (G.O.C.)

CH3 – C – CH2 + OH

phenyl shift CH3 – C – CH2 (higher migrating + tendency) least stable (Major) (Both stabilized by resonance of lone pair) O H

CH3 – C – CH2 + OH

methyl shift (least migrating C – CH2 – CH3 tendency) + more stable

(ii) If both resultant species are not very highly stabilized (not stabilized by resonance of lone pair or not stabilized by formation of additional bond), small amount of stability also matters a lot. Therefore, major migration occurs according to stability of resultant species. CH3

CH3 – C – CH2 + CH3

phenyl shift CH3 – C – CH2 (higher migrating + tendency) least stable (Both stabilized by resonance of lone pair) CH3

CH3 – C – CH2 + CH3

methyl shift (least migrating C – CH2 – CH3 tendency) + more stable (Major)

(iii) If the difference of stability is large, stability will dominate over the migrating tendency. Therefore, groups containing lone pairs migrate rarely.

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(b) When atom with six valence electrons and a vacant also has lone pair

- A = B - - A = B -

G G

In this case when a group from atom B is shifted, octet of B becomes incomplete which can be completed (stabilized) by lone pairs present on A atom. Hence under these conditions stability increases always and these rearrangements are, in general, bound to take place in the given species. Moreover, in such cases formation of additional bond or stabilization by resonance of lone pair will be generally observed, therefore migrating tendency generally dominates. H Shift CH3 – C – H CH2 = CH2

CH3 CH3

H Shift CH3 – C – C – H CH3 - C = C – H

CH3 CH3

R – C – N O = C = N – R

(B)REARRANGEMENT BECAUSE OF RESONANCE (ALLYLIC REARRANGEMENT)

In these cases, actually rearrangements do not occur but more-than one product is formed because of the resonance in the intermediate. These rearrangements are also called pseudo rearrangements. These rearrangements are possible when the intermediates like carbocation, carbanion or carbon free radical etc. are resonance stabilized and more than one different resonating structure are possible. e.g. ⊕ ⊕

CH 3 − CH− CH = CH 2 ←→CH 3 − CH = CH − CH 2 ⊕ ⊕

CH 3 − CH− CH = CH 2 ←→CH 3 − CH = CH − CH 2

REACTION MECHANISMS.

182 Essential Notes on General Organic Chemistry (G.O.C.)

Following reactions illustrate allylic rearrangements: CH3 – CH – CH = CH2 CH3 – CH – CH = CH2 + CH3 – CH = CH – CH2

OH Cl Cl/hv CH3 – CH2 – CH = CH2 CH3 – CH – CH = CH2 + CH3 – CH = CH – CH2 High temp

Cl Cl

(i) Mg/dry ether CH3 – CH – CH = CH2 CH3 – CH2 – CH = CH2 + CH3 – CH = CH – CH3 (ii) H2O

Cl H ⊕ CH3 – CH – CH = CH2 CH3 – CH = CH – CH2

OH OH

In IIT-JEE syllabi, we will have to study few Rearrangement reactions :

 Benzilic Acid Rearrangement  Curtius Rearrangement  Schmidt Rearrangement  Claisen Rearrangement  Pinacol – Pinacolone Rearrangement  Fries Rearrangement  Hoffmann’s Brommamide Reaction / Degradation  Beckmann’s Rearrangement

NB: Refer Lecture Notes for the same.

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RARE MECHANISMS (NOT mentioned in JEE ADVANCED syllabi) (But a few questions asked in JEE in the past 5 years needed a background of these mechanisms)

1. E1 cb Mechanism It may be argued that a second-order elimination reaction may as well proceed in two steps as in E1 reaction. The first step involves a fast and reversible removal of a proton from the β -carbon with the formation of a carbanion which then loses the leaving group in the second slow rate-determining step. H Θ Θ Fast – C – C - + EtO – C – C - + EtOH …………(i)

Br Br

Θ Slow Θ – C – C - – C= C - + Br …………(ii)

Br The overall rate of this reaction is thus dependent on the concentration of the conjugate base of the substrate (carbanion). Hence, this mechanism has been designated as E1cB (Elimination, Unimolecular from conjugate base).

2. SNi Mechanism : Most common example of this type of mechanism is the mechanism of the reaction between alcohol and SOCl2 It is actually substitution nucleophilic internal. O O

R – O H + Cl – S – Cl −HCl→ R − O − S − Cl

R – O R O S = O S = O R – Cl + SO2 Cl Cl

In this type of mechanism the configuration is retained. i.e., R form of alcohol gives R type of halide and 5 form gives 5 type of alkyl halide. One reaction in which this has been shown to occur is in the replacement of OH by Cl through the use of thionyl chloride, SOCl2: Me Me

SOCl2 C OH C – Cl + SO2 + HCl Ph Ph H H

The reaction has been shown to follow a second order rate equation,

Rate = k2 [ROH][SOCl2 ], but clearly cannot proceed by the simple SN2 mode for this would lead to inversion of configuration in the product, which is not observed.

REACTION MECHANISMS.

184 Essential Notes on General Organic Chemistry (G.O.C.)

3. Neighbouring Group Participation (N.G.P.) A number of nucleophilic substitution reactions are known which occur with complete retention (not inversion or racemisation) of configuration and with unexpectedly greater rate of reaction. In these cases usually there is atom or group with an unshared electron pair β to the leaving group (or sometimes farther away). The mechanism operating in such cases is called neighbouring group mechanism or neighbouring participation. It consists of two consecutive SN2 substitutions with inversion of configuration, thus, the net result is retention of configuration. In the first step of this reaction the neighbouring group acts as a nucleophile pushing out the leaving group. In the second step the external nucleophile pushes out the neighbouring group. A common feature of all neighbouring group mechanisms is the formation of a cyclic intermediate. Z: Step.1 Slow ⊕ Intermolecular SN2 Z Z Intramolecular SN2 Step 2. Fast (inversion) (inversion) C – C C C C – C Θ Θ Nu − L L Nu Net result is retention of configuration

(a). Neighbouring carboxylate anion Me Me Θ Θ Slow O Br Br + O – C OH i. C H C H C

O O Me Protection Θ O C OH

C H

O Retention

C Θ COO COO Θ HNO2 HNO2 H2O ii. H NH2 H OH O

Me Me CH D(-) alanine COOH Me H OH

Me D(-) Lactic acid (Retention)

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Br Me

H2O iii. Me – CH – CH2 Θ − Br O O C = O Θ O O

H2O iv. Me – CH – CH2 – CH2 − Br Θ O Me Br C = O Θ O

(b). Neighbouring halogen atoms : Reaction of HBr on (-) threo – 3 – bromo – butan – 2- ol + O H OH Me H Me H H Me

C1 C1 C1 ⊕ + H − H 2O Br Inversion at C1 C2 C2 C2

H H Me H Me Br Me Br (-) form

Br Br Θ H Me H Me + Br (from HBr) C1 C1 + H

C2 C2

H H Me Br Me Br (+) form (-) form

Reacemate

If no N.G.P. of Br occurred and the reaction was SN2, complete inversion would have occurred l only at C1. If the reaction was SN , C1 would have been a classical carbonium ion (flat), so inversion and retention (racemisation) would have occurred only at C1. Since either retention or inversion occurs at both C1 and C2, the results are explained by the N.G.P. of bromine atom. Similarly, optically active erythro -3-bromobutane 2 - ol with fuming HBr gives meso-2-3-dibromobutane.

REACTION MECHANISMS.

186 Essential Notes on General Organic Chemistry (G.O.C.)

(4) Elimination - Addition Mechanism (Benzyne Mechanism)

In the absence of an electron withdrawing group, nucleophilic substitution takes place in presence of very strong bases, but the mechanism is entirely different from that what we have seen in bimolecular nucleophilic substitution reactions. These reactions proceed by a mechanism called benzyne mechanism the positions. . •• •• • • X • X • •• •• + NH2 • +NH3 • Θ X + • NH3 • Θ Benzyne

NH2 NH2 + + NH2 • NH2 • Θ H – NH2 Benzyne is a symmetrical intermediate and can be attacked by nucleophile at both.

14 Cl 14 NH2 14 NH2 →

NH2

Isotopic labelling confirmed that there is an equal chance of abstraction from both carbons. An equal halide which does not contain alpha hydrogen with respect to halogen does not undergo this reaction. In the reactions involving/ benzyne intermediates, two factors affect the position of incoming group, the first one is direction of aryne formation. When there are groups ortho or para to the leaving group, then, the following intermediates should be formed. Z Z Z Z X NH 2 → NH2 →

Intermediate X When a meta group is present, aryne can form in two ways, in such cases. Z Z Z

NH2 → or X More acidic hydrogen is removed, i.e. an electron attacking ‘Z’ favours removal of para hydrogen.

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(5) SN’ MECHANISM : Allylic substrates undergo nucleophilic substitution reactions rapidly, and are usually accompanied by a rearrangement known as an allylic rearrangement or an allylic shift. When allylic substrates are treated with nuc1eophiles under SN1 conditions, two products are usually formed the normal product and a rearranged product: Y Θ R – CH = CH – CH2 X R – CH = CH – CH2Y + R – CH – CH = CH2

Normal product Y Rearranged product

The formation of two products can be easily explained because the allyl cation is resonance hybrid of two structures: ⊕ ⊕

R − CH = CH − CH 3 ←→ R − CH− CH = CH 2 Thus, C - 1 and C - 2 each carry a partial positive charge, and both are attacked by Y Θ (nucleophile) resulting in the formation of two products. This mechanism is called SN1' mechanism. In SN1' reactions at equilibrium more stable product (thermodynamically controlled product) is formed in greater amount, whereas if equilibrium is not reached, less stable product (kinetically controlled) is formed in greater amount. If the double bond in one isomer is in conjugation with an aromatic ring, a triple bond, another double bond or a carbonyl group, then that isomer is more stable and predominates at equilibrium, e.g., Θ OH Ph – CH = CH – CH2 – X Ph – CH = CH – CH2 – OH + Ph – CH – CH = CH2 I 100% at equilibrium OH II major product if equilibrium not reached

100% of the I is formed at equilibrium because it is more stable due to conjugation of its double bond with the phenyl ring. If equilibrium is not reached, then II is the major product. Nucleophilic substitution at an allylic carbon may also take place by SN2 mechanism without allylic rearrangement. However, allylic rearrangement can also take place under SN2 conditions through the following mechanism in which the nuc1eophile attacks at the γ -carbon instead of at the usual position. This mechanism is called SN2' mechanism and is an allylic rearrangement:

SN2' mechanism takes place under SN2 conditions where a substitution sterically retards the normal SN2 mechanism. R R R R

R – C C – C – X R – C = C

Y R R Θ R R Y

REACTION MECHANISMS.

188 Essential Notes on General Organic Chemistry (G.O.C.)

(6) IPSO reaction : A position which is already occupied by a non-hydrogen substituent in an aromatic ring is called ipso position (Latin: ipso, on itself), the attack on this position is called ipso attack (or ipso addition), and the aromatic substitution in which a substituent already present is replaced is called ipso substitution. For example, protodealkylation of an alkylbenzene (reverse of Friedel-Crafts alkylation). In this reaction, tertiary alkyl groups are most easily removed, since they depart as stabler carbocations. Thus, t-butyl group is used to protect the most reactive position in a compound to effect reaction elsewhere. The mechanism is as follows :

C(CH3)3 C(CH3)3 H

⊕ ⊕ H + + C(CH 3 )3 ipso attack Electrophilic ipso substitution

⊕ ⊕ (CH 3 )3 C →(CH 3 )2 C = CH 2 + H

CONCEPT TESTING QUESTIONS :

1 2 Details :- SN ,SN ,SNE,SNAr,SEAr,E1 ,E2 ,SN /SE, free radical mechanism. AdN, AdE, Ad free radical, rare Mechanism etc. Note :- 1. In all the questions below, ‘product’ means major product only 2. Both students and teachers should write the detailed mechanism where ever possible to understand the answers.

1. Substitution Reactions

Q.1 How many transition states are involved inSN 2 reaction ? (a) 1 (b) 2 (c) 3 (d) 0 Ans.1 (a) It is one step second-order reaction & hence only one T.S.

Q.2 How many intermediates are involved in SN 1 mechanism? (a) 0 (b) 1 (c) 2 (d) 3

Ans.2 (b) It is two-step mechanism with one intermediate as carbocation.

Q.3 Among the following nucleophiles, which of the following order is correct for their nucleophilicity in water ?

Θ Θ Θ Θ Θ Θ Θ Θ (a) CH3 > NH2 > OH > F (b) F > OH > NH 2 > CH 3 Θ Θ Θ Θ Θ Θ Θ Θ (c) OH > NH 2 > F > CH 3 (d) CH 3 > F > NH 2 > OH

Ans.3 (a) From left to right in periodic table, nucleophilicity decreases.

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Q.4 Give correct order of leave ability of the anions : − − − − MeO ,CN , ROSO2 , MeCOO

− − − − (a) MeO < CN < ROSO2 < MeCOO − − 2 − (b) MeO < CN < ROSO2 < MeCOO − − − − (c) MeO < CN < MeCOO < ROSO2 − − − − (d) MeCOO < ROSO2 < CN < MeO

Ans.4 (c) Good L – G are conjugate bases of strong acids.

Q.5 There are two reactions (i) ROH + NaBr (ii) ROH + HBr. Which is not possible? (a) (ii) (b) (i) (c) both possible (d) both not possible

Ans.5 (b) − − Br V – weak base, OH bad L – G – reaction (i) not possible – In reaction (ii) ROH is converted to ROH2 & hence RBr possible

Q.6 List the following nucleophiles in order of decreasing order of SN 2 : − − − (i) Me3CO ,(ii) MeO ,(iii)MeCH 2O and (iv) − O (a) iv > iii, > ii, > i (b) iv > ii, iii > i (b) i > iii > iv > ii (d) ii > iii > iv > i

Ans.6 (d) Θ Bulkier Nu , steric hindrance and hence SN 2 reactivity decreases

Θ Q.7 Arrange the following Nu in correct order of nucleophilic strength : − − − (a) OH > CH 3COO > OCH 3 > C 6 H 5O − − − − (b) CH 3COO < C6 H 5O < OCH 3 < OH

(d) CH 3COO < C 6 H 5O < OH < CH 3O Ans.7 (d) Acetic acid CH3COOH & Phenol C6H5OH are acidic. Anions are stabilized by resonance. − OCH 3 and OH are not having resonance.

Q.8 The order of decreasing nucleophilicity of the following species is : Θ Θ (a) CH 3S > CH 3O > CH 3COO > CH 3OH Θ Θ Θ (b) CH 3COO > CH 2S > CH 3 O > CH 3OH Θ Θ Θ (c) CH 3OH > CH 3S > CH 3COO > CH 3O Θ Θ Θ (d) CH 3O > CH 3OH > CH 3COO > CH 3S

REACTION MECHANISMS.

190 Essential Notes on General Organic Chemistry (G.O.C.)

Θ Ans.8 (a) Anion better Nu than neutral molecule. Nucleophilicity increases as electronegativity of donor atom decreases.

Q.9 ⊕

rearranges to

⊕ ⊕ ⊕ (a) (b) (c) (d) ⊕ Ans.9 (a)

Q.10 Arrange the following as increasing order of their ability as L – G (leaving group) − − − − CH 3S (P),CH 3O (Q)CF3 (R) & F (S) (a) S < Q < R < P (b) P < Q < R < S (c) R < Q < P < S (d) R < S < Q < P

Ans.10(d) Conjugate base of strong acid is L – G (d) CHF3 will be very strong acid & hence Conjugate base will be very weak.

Q.11 Which of the following will react with EtONa to give only substitution & no elimination product?

Br (a) (b) Br

Br Ans.11 (d) no β − H for elimination

Q.12 Which of the following halide is almost inert in SN or E reaction?

Br Br

(a) (b)

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Ans.12 (c) Positive charge at bridge-head carbon is least likely & hence inert. It is 3°C also

Θ − − Q.13 Give correct order of nucleophilicity of F (P),Cl (Q) and Br (R)in DMSO. (a) P > R > Q (b) P > Q > R (c) Q > P > R (d) Q > R > P

Ans.13 (b) No solvation by polar aprotic solvent for anions. Basicity parallels nucleophilicity.

Q.14 For stability of carbocation, which is the correct order out of the following? (a) 1,2 − Ph shift > 1, 2 – Me shift > 1, 2 hydride shift Θ (b) 1, 2 Ph shift > 1, 2 H shift > 1, 2 - 1, 2 Me shift > 1,2 Ph Shift − (d) 1, 2 Me Shift > 1, 2 Ph shift > 1, 2 H shift Ans.14 (a)

Q.15 Get correct order of nucleophilicity

(a) SH > OH > Br > N 3 (b) SH > OH > N 3 > Br

(c) SH > N 3 > Br > OH (d) N 3 > SH > Br > OH Ans.15 (b) Less EN, higher nucleophilicity and larger size, higher nucleophilicity.

Q.16 Arrange the following in decreasing order of nucleophilicity in ethanol − − − − F(P) ,Cl(Q) ,Br(R) ,I(S)

(a) P > Q > R > S (b) R > Q > P > S (c) S > R > Q > P (d) Q > R > S > P Ans.16 (c) Ethanol is polar protic solvent and hence nucleophilicity is proportional to size of atom. In polar aprotic solvents, larger size ions are better solvated & hence nucleophilicity decreases with increase of size.

Q.17 Which of the following is most likely to undergo a favourable hydride shift ?

(a) (b) (c) (d) ⊕ ⊕ ⊕ ⊕ Ans.17 (a) 3°carbocation produced

Q.18 In decarboxylation of acid, intermediate is ____ (a) carbocation (b) carbanion (c) carbonium ion (d) carbene

Ans.18 (b) O R −C CO2 Θ O

REACTION MECHANISMS.

192 Essential Notes on General Organic Chemistry (G.O.C.)

− − Θ Θ Q.19 What is the decreasing order of strength of bases OH ,NH 2 ,CH ≡ C ,CH 3CH 2 ? Θ Θ Θ Θ (a) CH 3CH 2 > NH 2 > CH ≡ C > OH Θ Θ Θ Θ (b) CH ≡ C > CH 3CH 2 > NH 2 > OH − Θ Θ Θ (c) OH > NH 2 > CH ≡ C > CH 3CH 2 Θ Θ Θ Θ (d) NH 2 > CH ≡ C > OH > CH 3CH 2 Ans.19 (a) Conjugate base of strong acid is weak. Acidic strength HOH > CH ≡ CH > NH3 > C2H6

Θ Θ Q.20 CH 3Br + Nu →CH 3Nu + Br . The decreasing order of rate of above reaction Θ Θ − − − − with Nu A to D is ____ [ Nu A = PhO ,B = AcO ,C = OH D = MeO ]

(a) A > B > C > D (b) B > D > C > A (c) D > C > A > B (d) D > C > B > A

Ans.20 (c) Stronger acid, weaker conjugate base.

Q.21 In the compound given below, the correct order of acidity of positions x, y and z is + N H H N 3 3 y ⊕ z

COOH X (a) z > x > y (b) x > y > z (c) x > z > y (d) y > x > z

Ans.21 (b) Electron withdrawing group (EWG) will increase positive charge of nitrogen.

2. Free Radical Reactions

Q.22 The no. of possible enantiomers (pairs) can be produced during monochlorination of 2 – methyl butane is ___

(a) 3 (b) 4 (c) 2 (d) 1

Ans.22 (c)

* & *

Cl Cl

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Q.23

Br2 → Products h r

How many monobrominated products will be obtained by above reaction?

(a) 6 (b) 4 (c) 5 (d) 3

Ans.23 (d) 1°H → 1,2°H → 1,3°H → 1 and compound is symmetrical.

Q.24

Br2 → Products h r How many products (all isomers) are possible? (a) 1 (b) 6 (c) 4 (d) 3

Ans.24(b) 2 products optically active (d, l) 2 products non active – Total products = 2 + 2 + 2 = 6.

Q.25 H CH3 CH3

Br2 → Product. It can be h r

H H H

CH3 CH3

(a) CH3 C CH2Br (b) CH3 C CH3

H Br Br Br (c) (d)

Ans.25 (b) 3°carbon free radical is more stable than 1°

REACTION MECHANISMS.

194 Essential Notes on General Organic Chemistry (G.O.C.)

Q.26 How many total products will be obtained by monobromination of 2 – methyl butane and how many can be separated by fractional distillation ?

(a) 6, 4 (b) 5, 4 (c) 6, 2 (d) 4, 2

Ans.26 (a) Br2 Br + * hλ Br Br Br * + * + & hv Br

Q.27 Consider the following reaction : D

+ Br  X + HBr .

Identify X among the following D D • CH 2 (a) (b) •

D D

Ans.27 (b) 3°carbon free radical is most stable.

Q.28 The relative reactivity of 1°,2° & 3°H bromination has been found to be 1 : 82 : 1600 respectively.

hλ Br + Br2 + B A Br

% of A & B can be ____

(a) 99% 1% (b) 50%, 50% (c) 1%, 99% (d) 80 %, 20%

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Ans.28(a) Reactivity of 3° = 1600,1°H ⇒ 1× 9 = 9 There are 9H as 1° and 1H as 3°. 1600 ∴% of A = × 100 = 99.4% 1609

Q.29 The relative reactivity of 1°,2° and 3°H in chlorination has been found to be 1: 3.8 : 5 Cl + + Cl Cl + Cl2 h v (D) Cl (C) (A) (B)

The ratio of amount of products A, B, C & D will be ____

(a) 1 : 3.8 : 5 : 1 (b) 3 : 7 – 6 : 5 : 6 (c) 3 : 7. 6 : 5 : 3 (d) 1 : 7.6 : 5 : 1

3× 100 Ans.29(b) Reactivity 1°H = 9, 2°H = 2 × 3.8 3°H ≡ 5 % of A = and hence ans is (b) 9 + 7.6 + 5 5× 100 6 × 100 Because for % C = and % D = 9 + 7.6 + 5 9 + 7.6 + 5

Q.30 How many isomers (including stereo) can be obtained by chlorination of 2, 2, 4 – trimethyl pentane?

(a) 3 (b) 4 (c) 5 (d) 6

Ans.30 (d)

2 optically active (4) 2 optically inactive Total 6

Q.31 How many dichlorocyclohexanes would be obtained on chlorination of chlorocyclohexane (including stereo)?

(a) 4 (b) 6 (c) 8 (d) 9 Ans.31 (d) Cl Cl

Cl Cl Cl Cl + Cl 1 2 Cl 3 +1 (2) (geo + opt) (2 geo + 1 opt) (geo) = total 9 (d)

REACTION MECHANISMS.

196 Essential Notes on General Organic Chemistry (G.O.C.)

Q.32 H3C

X2 CH CH2 – CH3 h λ H3C

How many monohalogenated products are possible? (a) 3 (b) 4 (c) 5 (d) 7

Ans.32 (c) a H3C

b c d CH CH2 – CH3 H3C

at a, b & d possible 3 at c optically active 2 total 5

Q.33 How many monochlorinated product may be obtained when alkane shown below reacts in presence of Cl2 and uv ?

(a) 1 (b) 2 (c) 0 (d) 3

Ans.33 (a) All ‘H’ are 1° only.

Q.34 1 NBS, hγ 5 2 CTC 4 3

Bromination will not occur at

(a) 1 (b) 4 (c) 3 (d) 2 and 5

Ans.34 (d) Benzylic carbon undergoes free radical bromination with NBS.

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Q.35 NBS ? h λ

(a) (b)

Br Br Br

Br Ans.35 (a) Allylic free radical bromination

Q.36 4 – methyl – 2 – pentene is brominated with NBS. The major product is

(a) Br CH2CH = CH – CH – Me2 (b) H 2C = CHCHBrCHMe2

Q.37 3 – methyl methyl enecyclohexane is brominated by NBS. Total products will be ___ (a) 6 (b) 4 (c) 5 (d) 3

Ans.37 (a)

CH2

β At α and β position rearrangement of π bond is possible along with bromination. Cis and trans products are also possible. Total products = 3 + 3 = 6

Q.38 Arrange bonding energy of C – H bonds in increasing order for following compound: H b H

a d H c H

(a) b < d < c < a (b) b < c < d < a (c) b < a < c < d (d) c < a < b < d

REACTION MECHANISMS.

198 Essential Notes on General Organic Chemistry (G.O.C.)

Ans.38 (b) b < c < d < a 3° 2° 2° Vinyllic allylic allylic

Q.39 c d

a H CH3 NBS→ A major product. Bromination takes place at (a) a (b) b (c) c (d) d

Ans.39 (a) Allyic bromination (free radical mechism). And 3°C also

H2O2 Q.40 CH 3 − CH = CH 2 + HBr →CH 3CH 2CH 2 Br The mechanism is ____ (a) electrophilic addition (b) free radical substitution (c) free radical addition (d) electrophilic substitution

Ans.40 (c) Anti – Markownikov’s addition.

Q.41 When R −CH 2CH = CH 2 reacts with NBS, the mechanism is ____ (a) AdE (b) free radical substitution (c) AdN (d) Elimination

Ans.41 (b) O

NBr

O NBS gives Br and it shows allylic substitution.

Q.42 The compound is

CH 3 −CH = CH −CH 2 −CH 2 −CH(CH 3 )2

A B C D E F Arrange them in decreasing order of reactivity towards radical substitution :

(a) C > A > E > D > F > B (b) F > B > A > C > D > E (c) B > C > A > F > D > E (d) A > B > C < D > E > F

Ans.42 (a) Allylic substitution

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3. SN1 Mechanism

Q.43 Br

is subjected to solvolysis. In which solvent, the racemization will be maximum? (a) 80 % acetone & rest water (b) 100 % water (c) 100 % acetone (d) 50 % acetone 50% water

Ans.43 (b) Solvolysis is SN 1 . Polar protic solvent is required. Acetone is polar aprotic

Θ Θ Q.44 Nu + GX →NuG + X . This mechanism can be elimination-addition. In organic displacement reaction, the mechanism is called as

(a) SN 1 (b) SN 2 (c) E1 (d) E2

Ans.44 (a) It is SN 1 because bond breaking is RDS (energy of activation larger). In rate equation only GX will be involved

Q.45 SN 1 reaction mechanism is favoured by

(a) high temperature (b) Powerful base (c) solvent with higher dielectric constant (d) substrate with bad L – G

Ans.45 (c)SN 1 required ionization of covalent bond which is encouraged by solvent of higher dielectric constant.

Q.46 Correct order of reactivity towards SN 1 in case of Ph − Br (P),

Ph − CH 2 − Br(Q), Ph − CH 2CH 2 Br (R) Ph − CH(Br)CH 3 (S)is (a) S > Q > R > P (b) S > R > P > Q (c) P > Q > R > S (d) S > R > Q > P

Ans.46 (a) 2°, benzyllic (s) where as P can’t have stable carbocation. Q has 1° benzyllic carbocation

Q.47 Arrange the following in decreasing order of SN 1 I I I I

P Q R S

(a) P > Q > R > S (b) S > P > Q > R (c) R > Q > P > S (d) R > P > Q > S Ans.47 (c)

REACTION MECHANISMS.

200 Essential Notes on General Organic Chemistry (G.O.C.)

Q.48 Arrange the following in decreasing order of SN 1 : Br Br

Br Br Q P R S (a) P > Q > R > S (b) Q > R > S > P (c) Q > P > R > S (d) Q > R > P > S

Ans.48 (d) Q is 2°allylic where as S is vinyllic carbocation which is highly stable R is 1° allylic

Q.49 Arrange the following in decreasing order of 1 SN CH3OCH 2Cl (P),

CH3 N CH 2Cl(Q), | H

Ph − CH 2Cl(R),

(S) Cl (a) P > Q > R > S (b) R > Q > P > S (c) Q > P > R > S (d) Q > P > S > R Ans.49 (c) Electron-donating groups increase SN 1 . Lone pairs can be donated to stabilized by carbocation.

Q.50 In the following compound, arrange the reactivity of different Br atoms towards NaSH in decreasing order : Br (R)

Br (S) Br (Q)

Br (P)

(a) P > Q > R > S (b) S > Q > P > R (c) Q > S > P > R (d) P > S > Q > R Ans.50 (c) No SN 2 at bridge head. Less sterically-hindered & 2°will show SN 2 .

Q.51 SN 1 reaction as hydrolysis is catalysed by (a) NaOH (b) H2O (c) AgNO3 (d) Pt

Ans.51 (c) Precipitation of AgCl facilitates formation of carbocation

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Q.52 SN 1 reaction occurs most rapidly in which of the following cases? I Br

(a) Br (b) I (c) (d)

CH3 CH3

− Ans.52 (c) I is better LG and 3° Carbocation is formed.

Q.53 Which of the following is most reactive alkyl halide for SN 1 ? Cl Cl

(a) (b) Cl (c) (d) Cl

Ans.53 (d) 3° resonance stabilized carbocation

Q.54 Arrange the following in increasing order of their reactivity in SN 1 I Cl I Br

P) O Q) O R) S) O

(a) R < Q < S < P (b) R < S < Q < P (c) Q < S < P < R (d) P < Q < S < R Ans.54 (a) − I better LG & carbocation is stabilised by lone pair of oxygen.

Q.55 Cl

HOH Major product will be

OH OH

(a) (b) (c) (d) OH

Ans.55 (a) Nu is not powerful - 2°C + can be stabilised by H − shift

REACTION MECHANISMS.

202 Essential Notes on General Organic Chemistry (G.O.C.)

Q.56 Which is correct? H Et − EtO (a) Racemisation

D Br

(b) Me C2H5OH Walden inversion

Et Br EtOH (c) Me Br Racemisation

(d) All are correct

Ans.56 (c) Actually (a) will show inversion and (b) will show racemisation & for (c) SN 1

Q.57 Give correct order of rate of SN 1 for Cl Ph

Cl CHCl Me2CHCl S Q Ph R

(a) P > Q > R > S (b) R > Q > P > S (c) P > R > Q > S (d) R > P > Q > S Ans.57 (a) Stability of carbocation.

Q.58 HCl P. Product P can be

OH Cl

(a) (b)

Ans.58 (b) Benzyllic carbocation by rearrangement

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Q.59

CH3OH Product will be 25°C Br

OMe CH3

(a) (b) C OMe

Ans.59 (c) SN 1 mechanism

4. SN2 Mechanism

Θ Θ Q.60 RX + Nu →NuR + X . For this reaction following data is available.

Θ [RX] [ Nu ] Rate − 0.1 0.1 1.2 × 10 4 − 0.2 0.1 2.4 × 10 4 − 0.1 0.2 2.4 × 10 4 − 0.2 0.2 40 × 10 4

In this case, we can have (i) stability of carbocation (ii) polar aprotic solvent (iii) complete inversion (iv) rearrangement Select correct options.

(a) i, ii, iiii, iv (b) ii, iii (c) i, ii, iv (d) ii, iv

Ans.60 (b) The kinetics indicates SN 2

Q.61 SN 2 rate is – (a) directly proportional to LG Θ (b) directly proportional to NU (c) inversely proportional to steric hindrance (d) directly proportional to dielectric constant

Ans.61 (b) Θ rα [ substrate] [Nu ]

REACTION MECHANISMS.

204 Essential Notes on General Organic Chemistry (G.O.C.)

Q.62 Arrange the following in decreasing order of SN 2 . Br

(Q), CH3Br (R), CH2 = CH - Br Br (P), (s)

(a) S > P > Q > R (b) Q > S > R > P (c) Q > R > P > S (d) R > Q > P > S

Ans.62 (d) 1° prefers SN 2 . Steric hindrance will decrease SN 2 .

Q.63 Arrange the following in decreasing order of SN 2 :

(P), Cl Cl Cl (Q) (R) Cl (S)

(a) S > R > P > Q (b) S > R > Q > P (c) R > S > Q > P (d) S > P > R > Q Ans.63 (a) 3°will not prefer SN 2 ,1° will always prefer SN 2 due to steric problems

Q.64

NaOH ∆ Br OH O

This can be described as - (a) SN 1 with racemisation (b) SN 2 intermolecular with inversion (c) SN 2 intramolecular with retention (d) SN 2 Intramolecular with inversion

Ans.64 (d) Br as good L- G. Lone pair of O atom makes cyclic compound of 5 atoms by removal of proton due to heating. It is Walden inversion.

Q.65 3 – Bromo – 2 – methoxy butane undergoes SN 2 reaction withOMe. Can we have product ___ ?

(a) optically active (b) meso (c) enantiomer (d) homomer

Ans.65 (b)

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OMe OMe

H3C H H3C H

H CH3 CH3 H

Br OMe

∴meso form (optically inactive)

Q.66 Which of the two iodines will be more reactive in SN 1 and SN 2 reaction?

I I (A) (B)

(a) A will be faster in SN 1 but slower in SN 2 (b) A will be faster. Both in SN 1 and SN 2 (c) A & B will be equally reactive. (d) B will be faster in SN 1 and SN 2

Ans.66 (b) Allylic halide has greater reactivity in both SN 1 and SN 2

Q.67 Which of these compounds could be used to establish stereo-specificity of SN 2 ?

(a) (b) Cl Cl Cl

Ans.67 (c) It has a chiral carbon Q.68 Give correct decreasing order of SN 2 for following Cl

(A) Cl (B) Cl (C) Br (D) Cl (E)

(a) C > E > B > D > A (b) C > B > E > D > A (c) C > E > D > A > B (d) C > A > B > E > D

Ans.68 (b) Good L – G, 1° & steric factor to be counted.

REACTION MECHANISMS.

206 Essential Notes on General Organic Chemistry (G.O.C.)

Q.69 Which of the following is most reactive towards SN 2 ? Cl Cl Cl

Cl NO2 (a) (b) (c) (d)

Ans.69 (d) (a), (b), (c) can show SN 1 due to stability of carbocation (benzylic) but there is no chance of stability of carbocation in (d) due to (–R) effect.

Q.70 (s) – 2 – bromobutane is obtained from (s) – 2 – chlorobutane when (a) SN 1 and SN 2 are used (b) 2SN 1 steps are used 1 2 (c) SN and SN i are used (d) 2SN steps are used

Ans.70 (d) Cl can not be replaced by Br in one step

Q.71 Consider the following for reaction with HBr : CH2OH CH2OH CH2OH CH2OH

(P)

NO2 OMe Br (Q) (R) (S)

The order of decreasing reactivity will be ____ (a) R > P > S > Q (b) R > P > Q > S (c) P > R > S > Q (d) P > R > Q > S

Ans.71 (a) Stabilising of carbocation due to +R effect.

Q.72 Dehydration of excess alcohol at high temperature in presence of acid gives the product by which mechanism ? 2 1 (a) SN (b) E (c) E2 (d) AdE

Ans.72 (a) Ether product by substitution.

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Q.73

HBr O CH3 OH + CH3Br

The reaction is ____

(a) AdN (b) AdE (c) SN (d) E

Ans.73 (c)

Q.74 Which of the following is more reactive towards AgNO3? Br Br Br

(a) (b) (c) (d)

O Br O

− Ans.74 (b) Carbocation stability after Br is leaving

Q.75 Which of the following will not react with Na metal? OH

(a) (b) (c) (d)

Ans.75 (c) Stability of anion is not possible after reaction due to anti-aromatic nature.

Q.76 SN 2 reaction at an asymmetric carbon of a compound always gives (a) an enantiomer of substrate (b) a product with opposite optical rotation (c) a mixture of diastereomers (d) a single stereoisomer

Ans.76 (d) Θ Θ In SN 2 , there is inversion. LG is replaced by different Nu .

Q.77 The product produced will have Me Br F PhSΘ DMF

NO2

2 (a) SN mechanism (b) SNAr mechanism due to activator (c) SEAr mechanism due to activator (d) SN1 mechanism

Ans.77 (a) Nucleophile is strong and solvent is polar aprotic.

REACTION MECHANISMS.

208 Essential Notes on General Organic Chemistry (G.O.C.)

5. SN1 v/s SN2 Mechanisms

Q.78 CH3 CH3 CH3

H3CO NO2 1 2 3

H Cl CH3

where, hydrolysed in aq. acetone will give

− (i) substitution by OH at position 2 − (ii) substitution by OH at position 1 − (iii) substitution by OH at position 3

(a) mixture of (i) and (ii) (b) mixture of (i) and (iii) (c) mixture of (ii) and (iii) (d) only (i)

Ans.78 (a) SN mechanism, having OCH3 as activator to stabilize carbocation

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6. E1 Mechanism

Q.79 Find number of possible E1 products from following reaction :

(a) 3 (b) 4 (c) 5 (d) 2

Ans.79 (b)

CH3OH ∆ Br

, (2) (2)

Q.80 Which of the following compounds will give E1 reaction? Br Br

(a) C Et (b) Ph – CH – CH3

(e) All

Ans.80 (e) All can form stable carbocation

Q.81 Br

EtONa → EtOH How many total alkenes can be made?

(a) 4 (b) 5 (c) 3 (d) 6

REACTION MECHANISMS.

210 Essential Notes on General Organic Chemistry (G.O.C.)

Ans.81 (b)

(1) (2)

(2)

Q.82 Which product is formed in E1 reaction of 1 – bromo -2, 3 – dimethyl cyclohexane?

(a) 3, 4 – dimethyl Cyclohexane (b) 2, 3 – dimethyl Cyclohexane (c) 4, 5 – dimethyl cyclohexene (d) 1, 2 – dimethyl cylohexene

Ans.82 (d) H − E1 Shift

⊕ Br

Q.83 How many products can be formed by E1 reaction of 1 – chloro – 2, 2 – diphenyl propane ?

(a) 2 (b) 1 (c) 3 (d) 4

Ans.83 (d) Ph (1) Ph (2) (1)

Cl Ph Ph Ph Ph Ph Ph

1 Q.84 At higher temperature, E1 chances are more thanSN . This is due to (a) T. S. are increased (b) Endothermic reaction favourable at high temperature. (c) Nucleophile is converted to base (d) Carbocation is stabilised

Ans.84 (b) Elimination is bond breaking & hence endothermic.

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Q.85 For the reaction,

C – H2SO4 CH2OH the major product is __ ∆

(a) (b) (c) (d)

Ans.85 (c) + + ⊕ C H C CH 2 O H 2 2

Q.86 The rate of hydrolysis of following compounds by SN 1 is : OMe

CH2Cl Cl CH2Cl

CH2Cl OMe (I) (II) (II) CH2Cl (IV)

(a) 2 > 1 > 3 > 4 (b) 3 > 4 > 1 > 2 (c) 4 > 2 > 1 > 3 (d) 2 > 1 > 4 > 3 Ans.86 (c)

Q.87

Cl cannot undergo the reaction :

1 2 (a) SN (b) SN (c) E1 (d) SE

Ans.87 (c) There is no β - H on ring for elimination.

Q.88 Which alkyl halide would you expect to be least reactive in E1 reaction ?

(a) CH 2 = CH − CH− CH 2 (b) Ph − CH− CH 3 | | Br Br

(c) CH 2 = C − CH 2 − CH3 (d) Ph − CH 2 − CH 2 Br | Br Ans.88 (c) Vinylic carbocation is most unstable and hence no chance of E1.

REACTION MECHANISMS.

212 Essential Notes on General Organic Chemistry (G.O.C.)

7. E2 Mechanism Q.89 Which of the following is expected to give more than 2 products in E2 reaction?

(a) (b) Cl Cl

Cl Cl

Ans.89 (d) & (a) (a) → 3 (b) pair of enantiomer (d) 4 (Count cis & trans isomers also)

Q.90 Which of the following cannot undergo E2 reaction?

Br (a) Br (b) (c) (d) None

Ans.90 (c) There is no H atom at β position

− Q.91 2 – bromopentane + EtO ∆→ The total products are

(a) 2 (b) 4 (c) 3 (d) 1

Ans.91 (c) Trans & cis – 2 – pentene & 1 – pentene

Q.92 E2 dehydrobromination of (R, R) − 2,3 − dibromobutane can make

(a) E product (b) Z product (c) both E & Z (d) none of these

Ans.92 (b) Θ H B : CH3 H CH3 Br H CH3 Br

≡ ≡ Br Br CH3 Br Br CH3 H H anti CH3

CH3 H Me H

Br (z)

2 – bromo – 2 – butane

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Q.93 Meso – (R, S) – 2, 3 – dibromobutane is subjected to E2 dehydrobromination. The product can be (a) E (b) trans (c) Z (d) none of these

Ans.93 (a) H H H CH3

CH3 Br Br Me

H ≡

CH3 Br Br

H Br CH3 (E)

meso (R,S)

Q.94

E2 + Saytzeff Hofmann X product product

In the above reaction, maximum Saytzeff product will be obtained when x is ____

(a) – I (b) – Cl (c) – Br (d) – F − Ans.94(a) I is powerful base 2°carbon, anti-elimination product possible since X is simple. L > G.

Q.95 CH3

Ph H alc. Product can be KOH Ph Br

CH3

Ph Ph Ph CH3

(a) (b) C = C

3 H C H3C CH3 Ph

Ph CH3 Ph CH3

Ans.95 (b) trans-product as major

REACTION MECHANISMS.

214 Essential Notes on General Organic Chemistry (G.O.C.)

Q.96

The product obtained when it is subjected to E2 reaction will be

(a) (b) (c) (d) All

Ans.96 (a) No carbocation, no rearrangement.

Q.97 CH3

CH3

is subjected to E2. The product will be

(a) (b) (c) (d)

Ans.97 (b) Anti-elimination (- HBr)

Q.98 CH3

Br CH3

is subjected to E2 reaction Major product will be

(a) (b) (c) None (d)

Ans.98 (c) No anti-elimination @ 1, 2 position.

Q.99 Which reaction is steroselective and regioselective too ?

(a) SN 2 (b) E1 (c) E2 (d) AdN

Ans.99 (c) More substituted alkene and trans-product predominantly.

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Q.100 Br

NO2 PhSNa→The reaction mechanism is DMF

(a) SNAr (b) SN 2 by inversion 1 (c) SN (d) E1

Ans.100 (b) Polar aprotic solvent & Br as LG

Q.101 CH3

NaOEt→ Major product will be

(a) (b) (c) (d)

Ans.101 (a) E2 shows anti elimination. Apply wedge & dashed line structure concept.

Q.102 Major product obtained in reaction of 1 – phenyl – 2- bromobutane with NaOMe is : (a) (E) – 1 – phenyl but - 1 – ene (b) (E) – 1 – phenyl but - 2 – ene (c) 1 – phenyl – 2 – ethoxybutane (d) (z) – 1 – phenyl but – 2- ene

REACTION MECHANISMS.

216 Essential Notes on General Organic Chemistry (G.O.C.)

Ans.102 (a)

Ph E Ph

Q.103 Rank the following in order of decreasing rate of an E2 reaction CH3 CH3 CH3

Br Br Br

(P) (Q) (R)

(a) P > Q > R (b) R > P > Q (c) R > Q > P (d) Q > P > R

Ans.103 (b) Anti elimination

Q.104

E2 H H The major product is

Br CH3

(a) (b) (c) (d)

Ans.104 (b) The given compound is cis & hence Br H H

CH3 H Anti coplanar elimination will take place and hence product (b). If the compound is trans, then the answer will be (a)

Q.105 A highly hindered base gives (a) E1 (b) E2 (c) SN1 (d) SN2 at high temperature irrespective of nature of substrate Ans.105 (b) E2 only

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Q.106 Which of the following can be decarboxylated most readily on being heated?

O (a) (b)

OH OH O O HO O

O O (c) HO (d)

OH O O HO O

Ans.106 (d) β − keto acid. Hence, easily decarboxylated by heating

8. E1 v/s E2 Mechanisms

Q.107 Which one of the following will most readily be dehydrated in acidic medium? O OH OH

(a) (b)

O O

OH OH

Ans.107 (a) Conjugation

Q.108

CH3I CH3I y. Here Y can be ____ AgOH, ∆ AgOH, ∆ N | CH3

(a) (b) (c) (d)

N Me

⊕N(Me)3 Me

Ans.108 (b) Hoffmann’s elimination

REACTION MECHANISMS.

218 Essential Notes on General Organic Chemistry (G.O.C.)

Q.109 Hofmann’s elimination is possible when _____ (a) base is bulky (b) L.G. is very good (c) L.G. is very bad (d) alkyl halide contains one or more double bonds. Find incorrect one

Ans.110 (b)

Q.111 CH3 + H Br2 C 4 H 8Br2 (P) (H 2O) CTC

CH3 OH

How many isomers of P are possible?

(a) 2 (b) 5 (c) 6 (d)

Ans.111 (b) Dehydration gives 3 products. AdE of these products will give 4 optically active products and one meso product.

9. Substitution v/s Elimination Reactions

Q.112 Br

DMSO + CH 3O → Which of following pathways is predominantly followed? (a) SN 1 (b) SN 2 (c) SN 1 /E1 (d) SN 2 /E2

Ans.112 (d) Polar aprotic solvent is used

Q.113 t – butyl chloride can react with alcoholic CH 3ONa . Select the correct statement : (a) It must be SN 2 reaction to make ether

(b) It must be E2 reaction to form isobutene

Ans.113 (b) No steric hindrance in E2 .

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Q.114

Br EtOH → Give total products (including stereoisomers) of SN 1 & E1

(a) 5 (b) 6 (c) 7 (d) 4

Ans.114 (b)

(1)

OEt OEt (1) (2)

(1)

Q.115 The order of reactivity of alkyl halide towards elimination is ____ (a) 3° > 2° > 1° (b) 2° > 1° > 3° (c) 3° > 1° > 2° (d) 1° > 2° > 3°

Ans.115 (a) No steric hindrance in E & more substituted alkene as major product.

Q.116 The correct stability order of the following species is :

⊕ ⊕ ⊕ O O ⊕ I II III IV

(a) II > IV > I > III (b) I > II > III > IV (c) II > I > IV > III (d) I > III > II > IV

Ans.116 (d) Resonance stabilization due to lone pair of oxygen.

REACTION MECHANISMS.

220 Essential Notes on General Organic Chemistry (G.O.C.)

10. Aromatic Substitution Electrophilic Mechanism

Q.117

C – HNO3 + H2SO4 Major product can be

NO2

(a) (b) (c) (d)

O2N

NO2

Ans.117 (b) More activating group (– CH3) & hence ortho product para position is already blocked.

Q.118 OCH3

AlCl 3 → A. A can be ____ ∆

OMe OCH3

(a) (b)

OCH3

Ans.118 (a) It is a Friedal-Crafts alkylation (intramolecular at ortho position) (a)

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Q.119

CH3CoCl/ AlCl3 → P. HOH

P can be ______O

(a) (b)

C O C

Sol.119 (c) ⊕ F/C acylation as EAS. The group (–CH2) will be activator & hence E attack at ortho position. Para position blocked.

Q.120 A B C D O OCH3

Nitration is possible at

(a) A (b) B (c) C and (d) D

Ans.120 (d) ⊕ − OCH 3 activator for EAS & attack of E possible at ortho position

Q.121 Which of the following cannot be electrophilic substitution? (a) nitration (b) sulphonation (c) bromination (d) ammonolysis

Ans.121 (d) Electrophile is not possible from NH3

REACTION MECHANISMS.

222 Essential Notes on General Organic Chemistry (G.O.C.)

Q.122 CF3

⊕ E Product

(i) CF3 is activator (ii) CF3 is deactivator (iii) CF3 is m – directing (iv) CF3 is o/p directing

Select the correct options : (a) (i) and (iv) (b) (ii) and (iii) (c) (i) and (iii) (d) (ii) and (iv)

Ans.122 (b) Electron withdrawing effect

Q.123 In the following reaction, 4 O 1

C – HNO3 3 NH X. C – H2SO4 2

+ The major product X will be produced by NO2 when substituted at

(a) position 1 (b) position 2 (c) position 3 (d) position 4

Ans.123 (d) p – position for activator

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Essential Notes on General Organic Chemistry (G.O.C.) 223

11. Aromatic Substitution Nucleophilic Mechanism

Q.124 F

KSH→ P. ∆

NO2

P can be F SH Cl

SH SH SH SH

(a) (b) (c) (d)

NO2 Cl

NO2 NO2 NO2

Ans.124 (a) NO2 is activator for SNAr and para position only suitable here because L – G is only at para.

12. Addition Nucleophilic Mechanism

Q.125 Addition reaction is not shown by (a) alkene (b) alkyne (c) alcohol (d) aldehyde

Ans.125 (c) Alcohol does not have any π bond

Q.126 Which of the following will react with HCN/NaCN? (a) Alkane (b) alkene (c) Alkyne (d) benzene

Ans.126 (c) − Alkyne shows AdN due to CN .

Q.127 Alkynes are more reactive than alkenes towards AdN because of stability of (a) vinyl carbanion (b) 2°alkyl free radical (c) vinyl carbocation (d) 2°carbocation

(a) C1 ,C 5 (b) C 1 ,C7 (c) C1 ,C 3 (d) C 2 ,C 4

REACTION MECHANISMS.

224 Essential Notes on General Organic Chemistry (G.O.C.)

Ans.128 (b) conjugated triene will be produced

Q.129 Grignard reagent when reacted with carbonyl compound works as Θ ⊕ (a) Nu (b) E (c) free radical (d) cation

Ans.129 (a) Θ RM9 X makes R . The reaction mechanism will be AdN.

Q.130 OH O

CH3 – C - CN + HCN K1

CH3 OH O

CH3 – C - CN + HCN K2 H H

The relation between K1 and K 2 is

(a) K1 = K 2 (b) K1 > K 2 (c) K1 < K 2 (d) K1 = K 2 = 1

Ans.130 (b) Reverse reactivity for AdN. Aldehydes are more reactive than ketones for AdN.

Q.131 Aldol condensation reaction of aldehyde or ketone can be explained as ____ (a) AdN (b) AdE (c) SN2 (d) E1

Ans.131 (a) Θ Carbanion produced by NaOH works as NU for other molecule of carbonyl compound.

Q.132 Tautomerism is not shown by

(a) Ph – CH = CHOH (b) O O

O O

Ans.132 (b) No α − H on SP 3C

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Q.133

CN − O Product can be H2O

OH OH

(a) CN (b) (c) OH (d) OH CN CN

Ans.133 (c) Θ Nu attacks at less substituted carbon atom (basic).

Θ Θ Q.134 In case of epoxide reaction with Nu , in presence of acid the attack of Nu is on (a) less substituted ‘C’ (b) any ‘C’ (c) more substituted ‘C’ (d) α - H atom

Ans.134 (c) Stability of carbocation

Q.135 Arrange the following compounds in decreasing order of : O

- C – CH3

HCHO CH 3CHO CH 3COCH 3 IV

I II III (a) II > IV > III > I (b) I > II > III > IV (c) IV > III > II > I (d) II > III > IV > I

Ans.135 (b) Aldehydes are more reactive than ketones for AdN

Q.136 Arrange the following compounds in decreasing order of AdN reaction :

CH 3COCH 3 CH 3COCl CH 3CONH 2 CH 3COOH CH 3 −CO − O −CO −CH 3

I II III IV V

(a) II > V > I > IV > III (b) III > IV > I > V > III (c) II > I > V > III > IV (d) IV > III > V > I > II

Ans.136 (a) Greater unstability of carbonyl carbocation

REACTION MECHANISMS.

226 Essential Notes on General Organic Chemistry (G.O.C.)

Q.137 Which carbonyl group of the given compound is most reactive for nucleophilic addition reaction? O 1 2 O

3 O

(a) 1 (b) 2 (c) 3 (d) All have rqual reactivity

Ans.137 (b) At 1 and 3 carbonyl carbocation are stabilized and hence less reactive

Q.138 AdN reaction rate is increased by – (a) EWG at α − position (b) ERG at α - position (c) EWG at β − position (d) EWG on carbonyl carbon

Ans.138 (a) Positive nature of carbonyl carbon is increased

Q.139 When CH 3 MgBr reacts with CH 3C ≡ CH , the product will be ____

(a) CH 3 −CH = CH 2 (b) CH 3 −C ≡ C −CH 3

+ Ans.139 (d) Terminal alkyne is acidic ( H ) and hence produce will be CH 4 .

+ H3 O Q.140 C6 H 5CH(OH)C ≡ CH →C6 H 5CH = CHCHO In above reaction, the steps involved are _____ (a) 4 (b) 5 (c) 6 (d) 7

Ans.140 (c) + H −H2O C6 H 5CHOH −C ≡ CH →C6 H 5 − CH −C ≡ CH → | ⊕OH2 ⊕ ⊕ H2O C6 H 5 CH−C ≡ CH →C6 H 5CH = C = CH → −H + C6 H 5CH = C = CH →C6 H 5CH = C = CH →C6 H 5CH = CH −CHO | | ⊕OH2 OH

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Essential Notes on General Organic Chemistry (G.O.C.) 227

⊕ Q.141 The electrophile, E attacks the benzene ring to generate intermediate complex. Of the following, which σ - complex has lower energy?

NO2 NO2 NO2 H H (b) (c) (d) H (a) + + E + E + E H E Ans.141 (b) ⊕ NO2 is deactivator for E attack. Stability more and hence energy lower compared to complex b.

13. Addition Electrophilic Mechanism Q.142

H2O + H OH

How many transition states and intermediates will be formed during this reaction?

(a) 3 T. S. and 3 intermediates (b) 4 T. S and 3 intermediates (c) 3 T. S. and 2 intermediates (d) 5 T. S. and 4 intermediates

Ans.142 (b) AdE

H2O

⊕ ⊕ OH2

⊕

⊕ − H 4 steps in reaction

REACTION MECHANISMS.

228 Essential Notes on General Organic Chemistry (G.O.C.)

Q.143

HCl Major product is

(a) (b) Cl Cl

Ans.143 (c) Ring expansion for stability of carbocation (AdE)

Q.144 cis – isomer of 2- butene when reacted with Br2 in presence if CCl4 we get (a) d l pair (b) meso form (c) only d - (d) only l product.

Ans.144 (a) Br

H CH3 CH3 Me

Br2 H C = C Me

Br H H C is H CH3

d l pair as products Br

CH3 H Q.145 Which of the following compound is most reactive towards electrophilic addition reaction?

(a) CH 2 = CH 2

(b) (F3C)2 C = CH 2 (c) NC − CH = CH − CHO NO2

(d) NO2 CH = CH2

Nl2 Ans.145 (a) Θ + Electrophilic addition as RDS occus at C where C should not be reactive. Stability of

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Essential Notes on General Organic Chemistry (G.O.C.) 229

C + is essential in such a step.

Q.146 When 1 – methyl cyclohexene is converted to alcohol, which reagent can make syn addition only?

(a) acid catalysed hydration (b) hydroboration – oxidation (c) oxymercuration – demercuration (d) base catalysed hydration

Ans.146 (b) Anti – Markovnikov’s addition

Q.147 Two moles of aldehydes are produced by using alkene. The reaction is (a) reductive ozonolysis (b) oxidative ozonolysis (c) reductive oxidation (d) oxidative hydration

Ans.147 (a) Zn/HOAc is used along with O3

Q.148 In addition reaction for C = C, T.S. in energy profile diagram are / is (a) 1 (b) 2 (c) 0 (d) 3

Ans.148 (b) One intermediate, 2 steps in reaction

Q.149 Peroxy benzoic acid reacts with 2 – butene to form oxirane. This mechanism can be (a) reduction (b) redox reaction (c) oxidation (d) electrophilic addition

− Ans.149 (c) & (d). OH of peroxyacid can be considered as E+. However RCOO removes H form OH as O bonds to C = C, giving epoxide

Q.150 When 1 – pentene – 4 – yne reacts with one equivalent of HBr, the product will be

(a) CH 3CHBr −CH 2C ≡ CH (b) CH 2 = CH −CH 2CBrCH 2

Ans.150 (a) More stable carbocation to be taken. Alkenes react faster than alkynes for AdE

Q.151 Propyne reacts with water in presence of H 2SO4 and HgSO4. The intermediate is (a) carbocation (b) carbanion (c) ketol (d) enol

Ans.151 (d) Enol. It is Markovnikov’s addition of H2O.

REACTION MECHANISMS.

230 Essential Notes on General Organic Chemistry (G.O.C.)

Q.152 The compound C10 H 10 gives only one organic compound after oxidative cleavage. The product is acid. It must be ____ (a) monobasic (b) aibasic (c) tribasic (d) tetrabasic acid

Ans.152 (c)

HC ≡ CCH 2 CHCH 2C ≡ CH | CH2C≡CH.

Q.153

HBr P. Product P is CCl4

Br Br Br

(a) (b) (c) (d)

Ans.153 (c) AdE with ring expansion

Q.154

+ HBr P. ‘P’ can be -

Br (a) Br (b) (c) Br (d)

Ans.154 (a)

⊕ ⊕ ⊕ P

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REACTION MECHANISMS.

232 Essential Notes on General Organic Chemistry (G.O.C.)

Q.155

+ H P. ‘P’ can be ∆

(a) (b)

Ans.155 (b) Acid-catalysed dehydration, followed by carbocation stabilization to get 3°electrophile. Finally, addition-elimination product.

Q.156 How many alkenes on catalytic hydrogenation can produce isopentane? (a) 2 (b) 3 (c) 4 (d) 5

Ans.156 (b)

Q.157

+ H Product can be ∆

(a) (b) (c) (d)

Ans.157 (a) Mechanism can be given as

⊕ ⊕ Product ⊕

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Q.158 OH + H Product can be O

O O O

Ans.158 (a) OH O OH ⊕ ⊕ ⊕ − H O O O

Q.159 In a reaction HBr Br

How many transition states and intermediates are involved?

(a) 4 & 3 (b) 3 & 3 (c) 2 & 2 (d) 3 & 4

Ans.159 (a)

Q.160 Acid-catalysed hydration of alkene except ethene leads to formation of (a) mixture of 2°and 3°alcohol (b) mixture of 1° and 2°alcohol (c) mixture of 1° amd 3°alcohol (d) primary alcohol only

Ans.160 (a) R |

R −CH = CH 2 and R − C will produce mixture of 2°and 3°alcohol. || CH2

REACTION MECHANISMS.

234 Essential Notes on General Organic Chemistry (G.O.C.)

14. Addition : Free Radical Mechanism

Q.161 Both HCl and HI can’t show peroxide effect in case of alkenes. The reason may be (a) both are strong acids (b) both radicals Cl and I are unstable (c) both have propagation reactions energetically unfavorable (d) both can’t make electrophiles

Ans.161 (c) There are 2 steps of propagation and in both cases, ∆ H is – ve for HBr only. So, addition is possible to C = C by free radical mechanism in presence of peroxides.

Q.162 Anti – Markovnikov’s rule is not observed in ____ (a) propene (b) 1 – butane (c) 2 – butane (d) 2 – pentene

Ans.162 (c) Symmetrical alkene

15. RARE MECHANISMS : SN i, E1cb, SN', NGP, Benzyne, Ipso

Q.163 For E1 cb mechanism, select the incorrect statement from following : (a) poor LG required (b) substrate mostly have acidic hydrogen (c) conjugate base is resonance stabilized (d) less substituted product as major product

Ans.163 (d)

− Q.164 R – 2 – bromopropanoate is treated with OH in presence of Ag2O. The product will be

(a) R – 2 – hydroxyl propanoate (b) S – 2 – hydroxypropanoate (c) R – 2 – hydroxypropanol (d) S – 2 – hydroxyl propanol.

Ans.164 (a) It is double inversion due to NGP and hence retention − C = O is NGP working as | O− internal nucleophile.

Q.165 H + OH Θ HOH,Br The mechanism is _____

(a) SN 2 (b) SN 1 (c) SN 1 (d) E1

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Essential Notes on General Organic Chemistry (G.O.C.) 235

Ans.165 (c)

It is SN 1 with rearrangement. The product is Br

Br or Nucleophilic substitution at γ − carbon.

Q.166 Give decreasing order of heat evolved due to catalytic hydrogenation for

P Q R S

(a) Q > R > P > S (b) S > P > R > Q (c) S > R > P > Q (d) R > Q > S > P

Ans.166 (b) All are is isomers. Greater the stability, lesser the heat of hydrogenation

Q.167 HBr O Product can be ∆

(a) OH (b) Br

Ans.167 (d) Bridge head carbocation (unstable)

Q.168 The products will be O

CH3 cone. H I

(a) ketone and alkyl halide (b) alcohol and alkyl halide (c) alcohol and vinyl halide (d) alcohol and alkene

Ans.168 (a) Enol produced and then it is converts to ketone.

REACTION MECHANISMS.

236 Essential Notes on General Organic Chemistry (G.O.C.)

Q.169 HI Product can be

(a) (b)

I OH OH I

I OH (c) (b) O I

Ans.169 (b) Θ Acidic medium and Nu attack, accordingly

Q.170 Carboxylic acid when reduced by LAH the product is ___ (a) 2°alcohol (b) aldehyde (c) 1° alcohol (d) alkene

Ans.170 (c)

Q.171 Which of the following is strongest base? NH2

(a) (b) (c) (d) N N N | | H H

Ans.171 (a) 2°amine. All others are aromatic

Q.172 Find the correct order for energy required for heterolytic cleavage of C – Br bond forming carbocation. Z X Y

CH2 - Br

CH2 - Br CH2 - Br N O O | H (a) Z > Y > X (b) Z > X > Y (c) X > Y > Z (d) Y > X > Z

Ans.172 (a) More stability of C+, lesser energy required for cleavage.

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Q.173 Get the correct order for energy required for heterolytic cleavage of C – Cl bond forming carbocation. y z

Cl Cl Cl

X NH2 NH2 NH2

(a) x > y > z (b) z > y > x (c) z > x > y (d) y > x > z

Ans.173 (b) After cleavage, carbocation produced is much stabilised by resonance. Vinyllic carbocation is least stable.

Q.174 Consider the following carbanions Θ Θ CH CH (I) H3CO 2 (II) O2 N 2

Θ Θ

(III) CH 2 (IV) H 3C CH 2

Correct decreasing order of stability is

(a) II > III > IV > I (b) III > IV > I > II (c) IV > I > II > III (d) I > II > III > IV

Ans.174 (a) Resonance effect -R will stabilize carbanion and +R will destabilize.

Q.175 In which of the following compounds the hydroxylic proton is most acidic? O F O (a) (b) F H H

O O (c) (d) H H F F

Ans.175 (d) F is EWG and charge on O atom is stabilized

REACTION MECHANISMS.

238 Essential Notes on General Organic Chemistry (G.O.C.)

Q.176 The abstraction of proton will be fastest in which carbon of the following compound? O Y

P CH3 Z CH3 X

(a) x (b) y (c) z (d) p

Ans.176 (a) + Conjugation and hence stabilization after loss of H .

Q.177 Arrange the following in increasing order of their heat of hydrogenation

R P Q

(a) P < Q < R < S (b) S < R < Q < P (c) S < R < P < Q (d) P < Q < S < R

Ans.177 (a) Highly stabilized compound will have lesser heat of hydrogenation. Hydrogenation is exothermic. P is most stable and S is least stable.

Q.178 Which of the following can give E1cb? O F O CN

(a) (b) NO2

Br CN

Ans.178 (a) & (b).

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Essential Notes on General Organic Chemistry (G.O.C.) 239

Q.179 Carbanion is produced by heterolytic cleavage of a covalent bond from (a) terminal alkyne (b) carbonyl compound (c) active methylene compound (d) all the above compounds Ans.179 (d) Θ O Θ R −C ≡ C , CH 2 −C H

and – CH2 (methylene) group connected with EWG groups

Q.180 Br 1

2 3 O Θ when reacted with A ___ (a) substitution occurs at 1, when working as nucleophile (b) degree of unsaturation increased when working as base at high temperature (c) carbocation is stabilized and hence both the reaction will be first order (d) substitution can occur at position 3. Select the incorrect statement.

Ans.180 (d)

Q.181 Among the following, which one is not soluble in aq. NaOH ? NO2 OH OH COOH (a) (b) (c) (d)

NO2

Ans.181 (c) All others are acidic

Q.182 COOH OH

2 moles of This reaction is ____ NaNH2

NO2 CH

(a) SN 1 (b) Redox (c) Acid – base neutralization (d) E1 + Θ

Ans.182 (c) All are acidic. Two of them will react with Na NH 2

REACTION MECHANISMS.

240 Essential Notes on General Organic Chemistry (G.O.C.)

Q.183 Select the correct statement from following : (a) In Tollen’s reagent, silver is oxidised (b) In LAH, oxidation number of hydrogen is + 1 (c) In Fehling’s solution test, copper is oxidised. (d) In nitration of benzene, HNO3 works like base

Ans.183 (d) + Θ HO – NO2 + H2 SO4 →NO 2 + H 2O + HSO4

Q.184 When OCH3 Br

reacts with NaNH2/NH3, the mechanism is _____ (a) addition (b) elimination – addition (c) elimination – substitution (d) elimination

Ans.184 (b) It is benzyne mechanism giving OCH3 OCH3

and Products

NH2 NH2

OCH3 The intermediate is called benzyne. It is

Q.185 In benzyne mechanism, entering group occupies a position other than vacated by leaving group. The reaction is _____ (a) syn – substitution (b) cine substitution (c) syn – addition (d) anti – substitution

Ans.185 (b) The entering group does not occupy the position vacated by leaving group. The reaction of this type is cine – substitution.

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Essential Notes on General Organic Chemistry (G.O.C.) 241

Q.186 SO3H Br

Br2

OMe OMe

This substitution is called as (a) Ipso (b) pseudo (c) syn (d) Nucleophilic

Ans.186 (a) Ipso (on itself). Ipso substitution would ease formation of potential leaving group ⊕ ⊕ ⊕ ⊕ like R ,SO3 H , Me3Si ,Br etc. It is a rare reaction mechanism

16. Rearrangements

Q.187 The number of stereoisomers obtained by bromination of trans – 2 – butane is

(a) 1 (b) 2 (c) 3 (d) 4

Ans.187 (a) meso

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REACTION MECHANISMS.