AROMATIC HYDROCARBONS: Part I

DR SUNIL K. SINGH Assistant Professor in Chemistry

SK SINGH, KMC Concept of Aromaticity

 Aromatic compounds apparently contain alternate double and single bonds in a cyclic structure and resemble benzene in chemical behaviour.

 They undergo substitution rather than addition reactions. This characteristic behaviour is known as aromaticity. Addition reaction: Substitution reaction: Br Br 2 H E CCl + 4 Br E + H+

Br2 No Reaction SK SINGH, KMC CCl4 Following are the main criteria of aromaticity:  Chemical behaviour: electrophilic aromatic substitution.

 Structural: bond length equalization due to cyclic delocalization of electrons.

C-C bond length in benzene id 139 pm.

SK SINGH, KMC Contd………………..  Enhanced stability (large resonance energy).

 Heat of hydrogenation of Benzene is 49.8 Kcal/mole.  Heat of hydrogenation of 1,3,5- hexatriene is 85.8 Kcal/ mole.  Smaller the heat of hydrogenation  Magnetic: "ring current" effects. more is the stability. Benzene is ~36 Kcal/mole more stable than the 1,3,5- • anomalous chemical shifts in NMR. hexatriene. • large magnetic anisotropies • high diamagnetic susceptibility.

SK SINGH, KMC Huckel’s Rule and aromaticity A molecule is aromatic if all the following conditions are fulfilled:

 It is cyclic, planar and it has continuous delocalization of p electrons (electrons in p orbitals) with or without the participation of lone pair(s)/ -ve charge/ +ve charge.

 Huckel’s rule: The delocalised p-electron system must contain a total of (4n+2)p electrons, where n is a whole number (i.e., n = 0,1,2,3,………).

Note that the p-orbitals in which the electrons are delocalizing must be parallel to each other so that a continuous overlap of electrons is possible along the ring. SK SINGH, KMC Contd…………………………….. So whenever a cyclic, planar and (4n+2)p electrons are continuously delocalized, it leads to the extra stability of the molecule referred to as aromatic compound. Whenever a cyclic, planar and (4n)p electrons are continuously delocalized, it leads to the extra destability of the molecule referred to as anti-aromatic compound. When either the molecule is not cyclic or is not planar or the continuous delocalization of p electrons is not there, the molecule is aliphatic or non- aromatic compound.

Stability Order: Aromatic > NonaromaticSK SINGH, KMC / Aliphatic > Antiaromatic In other words Condition for Aromatic Anti- Non- Aromaticity Compounds aromatic aromatic Compounds Compounds 1. Cyclic, planar Fulfill the first Fulfill the first Not fulfilled and continuous condition condition delocalization of p electrons 2. (4n+2)p (4n+2)p (4n)p No need to electrons electrons electrons look for this condition if 1st (Huckel’s rule) [(4n+2)p electrons condition is condition not not fulfilled fulfilled] SK SINGH, KMC Construction of MO diagram for conjugated cyclic polyenes Frost Diagram (Polygon Method):

 A circle is inscribed with a polygon with one vertex down;  The Vertex represent the energy levels (p molecular orbital levels);  The semicircle line shows the nonbonding level. p * * * 4 p3 p2 n n ……………………………………..p2 ……………………………………..p3 …………………………………….. p 1 p1

…………………………………….. …………………………………….. ……………………………………..

SK SINGH, KMC p-Molecular orbital level diagram of benzene:

Points to remember:  Aromaticity is observed when all bonding MO’s are filled and nonbonding MOn’s, if present, are empty of completely filled.

 Antiaromaticity is observed if it has a electron in antibonding molecular orbital or it has half-filled bonding molecular orbital(s) / nonbonding molecular orbital(s), provided it satisfies the 1st condition of aromaticity.

SK SINGH, KMC Examples: Three membered cyclic species

Practice Problems

Nonaromatic Aromatic Antiaromatic Antiaromatic (Unstable) Cyclopropene, Cyclopropenyl cation, Cyclopropenyl anion

Nonaromatic

More acidic proton

Pracice Problems: O O O Cyclopropenone is relatively more stable than anticipated. Explain. Cyclopropenone has larger dipole moment than cyclopropanone. Why ? I II Resonance Hybrid (II contribute more to the resonance hybrid) as cyclopropenyl cation is aromatic in SK SINGH,nat uKMCre, which provide it extra stability. Explanation:

Aromatic Antiaromatic

SK SINGH, KMC Four membered cyclic species

Antiaromatic Aromatic Aromatic Cyclobutadiene, Cyclobutadiene dication, Cyclobutadiene dianion

2+2+ 2-2+

2+ 2- C4H4 C4H4 SK SINGH, KMC Explanation:

SK SINGH, KMC AROMATIC HYDROCARBONS: Part II

DR SUNIL K. SINGH Assistant Professor in Chemistry

SK SINGH, KMC Revision………………………………………. Condition for Aromatic Anti- Non- Aromaticity Compounds aromatic aromatic Compounds Compounds 1. Cyclic, planar Fulfill the first Fulfill the first Not fulfilled and continuous condition condition delocalization of p electrons 2. (4n+2)p (4n+2)p (4n)p No need to electrons electrons electrons look for this condition if 1st (Hückel’s rule) [(4n+2)p electrons condition is condition not not fulfilled fulfilled]

SK SINGH, KMC Revision……………………………………….

Points to remember: Aromaticity is observed when all bonding MO’s are filled and nonbonding MOn’s, if present, are empty of completely filled.

Antiaromaticity is observed if it has a electron in antibonding molecular orbital or it has half-filled bonding molecular orbital(s) / nonbonding molecular orbital(s), provided it satisfies the 1st condition of aromaticity.

SK SINGH, KMC Five membered cyclic compounds / species

Cyclopentadiene Cyclopentadienyl cation Cyclopentadienyl anion (Nonaromatic) (Antiaromatic) (Aromatic)

Practice Problems: 1. Cyclopentadienyl anion is much more stable than allyl anion. 2. (a). Use (i) resonance theory and (ii) MO theory to predict

whether 5-bromo-1,3-cyclopentadiene readily undergo CH2 solvolysis in water. (b). Show that cyclopentadienyl cation is a diradical. Aromatic Compouns Nonaromatic Compound 3. Write the structure of the smallest aromatic compound. 1. Cyclic, Planar, delocalized p-electrons 2. Follow's Huckel's Rule

SK SINGH, KMC MO Diagram:

Practice problem: Cyclopentadiene is more acidic than + H+ ethane . H H pKa = 15 1. Cyclic, Planar, delocalized p-electrons 2. Follow's Huckel's Rule (Aromatic Compound) + H3C CH2 + H H3C CH3 Unstable pKa = 50 SK SINGH, KMC Six membered ring compound: Benzene or ([6]-annulene)

1. It is cyclic, planar and p-electons are deloclised. 2. It follow’s Hückel’s rule.

4n+2 = 6; n = 1 (an integer)

SK SINGH, KMC Seven membered ring compounds / species

Cycloheptatriene Cycloheptatrienyl cation Cycloheptatrienyl anion (Tropylium cation) (Nonaromatic) (Aromatic) (Antiaromatic)

+

fast Practice Problem: Solution: Br + Ag+ + AgBr (ppt) Tropylium bromide (7-bromocycloheptatriene) completely dissociates in water and give ppt of Tropylium cation Aromatic in nature (More Stable) AgBr instantaneously with AgNO3, unlike the open chain analog, 3-bromo-1,4-pentadiene. + slow Explain. Br + Ag + AgBr (ppt)

SK SINGH, KMC Nonaromatic (Less stable) MO Diagram:

Tropone Tropolone

O O

OH OH Tropylium cation

Aromatic SK SINGH, KMC Aromatic Solution: Practice Problem: Which of the following compounds has the greater O O dipole moment ?

O

O C

I II I More Stable Charge is separated in this resonating structure, three membered ring becomes aromatic.

O O C C

II Charge is separated in this resonating structure, but does not have any additional aromaticity. SK SINGH, KMC Heterocyclic compounds

N H O S N Pyrrole Furan Thiophene Pyridine Aromatic Aromatic Aromatic Aromatic lone-pair of electron is in p- orbital, involved to complete the aromaticity sextet.

- 6p e ’s Sp2-Hybridised 6p e-’s Not the part of p-cloud

4n+2 = 6; n = 1 SK SINGH, KMC Lone pair of e-’s is in 3p-orbital, part of the p-electron cloud.

6p e-’s

SK SINGH, KMC SK SINGH, KMC Other Heterocycles N 1. 1. It is cyclic, planar and have an fully conjugated Resonating structures of Imidazole cyclic π electron cloud. N 2. It follows Hückel‘s rule. H Contain 6 p e’s. Imidazole 4n + 2 = 6; n =1 (integer) Aromatic

Lone pair of electron is part of p-electron cloud

Perpendicular to p-orbital SK SINGH, KMC Other Heterocycles

1. 1. It is cyclic, planar and have an fully conjugated cyclic π electron cloud. 2. It follows Hückels rule. Contain 6 p e’s.

4n+2 =6; n=1

Oxazole Aromatic Compound

Practice Problem: Isoxazole (I) is aromatic or not ? Explain.

I SK SINGH, KMC Cyclooctatetraene or [8]-Annulene: Predicted to be Antiaromatic compound.

[8]-Annulene

“Tub” shaped structure of Cyclooctatetraene

Cyclooctatetraene is not planar, (Nonaromatic compound)

(Not planar)

SK SINGH, KMC Polycyclic aromatic compounds Hückel’s (4n+2 )p electron rule is strictly applicable to monocyclic compounds. However, it applies to a number of polycyclic compounds as well.

14 p-electrons: aromatic

SK SINGH, KMC Azulene: An isomer of naphthalene

1. It is cyclic, planar have an fully conjugated cyclic π electron cloud. 2. It follows Hückels rule. Contain 10 p e’s. 4n+ 2 = 10; n = 2

I Cyclopentadiene II III Cycloheptatriene Tropylium cation Cyclopentadienyl anion Aromatic Aromatic

1. Contribution of the resonating structure III to the resonance hybrid is more. In this charged resonating structure there is seven membered aromatic ring tropylium cation and five membered aromatic ring cyclopentadienyl anion is present. 2. The main contributing structure to the hybrid show charge separation, which is responsible for the higher dipole moment (1.0 D).

SK SINGH, KMC Cyclodecapentaene or ([10]-Annulene):  Cyclodecapentaene seems to be aromatic as it is cyclic, seems to be planar, conjugated system and has 10p electrons. But the fact is that it is not aromatic.  The reason is that if the molecule acquires planar geometry (I), there will be angular strain in the molecule as each angle will try to be 120o (due to sp2 hybridization) but cannot do so, also it will lead to five cis double bonds. In order to become stable, it tries to exist as (II) where there are three cis and two trans double bonds.  Looking closely we find that the two H atoms on the two trans double bonds come very close to each other and thus the molecule actually exists in a manner where the ring becomes puckered or non-planar so that the two half-cyclic rings are tilted in order to make these two hydrogen atoms slightly away from each other (III). trans

H H

II 1. Cyclic 2. fully conjugated cyclic π electron cloud I 3. It follow Hückels rule. Contain 10 p e’s. 4n+ 2 = 10; n = 2 III but SKNot SINGH, Planar KMC  Some Molecules are aromatic after breaking a bond: like azulene it has an unusual large dipole moment. Direction of dipole moment Cyclopropenyl anion 4p-electron: Antiaromatic Cyclopropenyl cation 2p-electron: Aromatic

Cyclopentadienyl cation Cyclopentadienyl anion 4p-electron: Antiaromatic 6p-electron: Aromatic Unstable Calicene More Stable

CH2 CH2

Fulvene SK SINGH, KMC Practice Problems: 1. Which of the following compound (s) have very high dipole moment ? Explain.

Fulvalene Sesquifulvalene Traifulvalene

2. Which of the following compound (s) will react with n-BuLi ? Explain.

SK SINGH, KMC Annulenes:  Annulenes are the completely conjugated monocyclic hydrocarbons containing an even number of carbon

atoms. They have the general formula CnHn (when n is an even number), they are represented by (CH=CH)n.  Annulenes with 7 or more number of carbon atoms are named as [n]-Annulene.

Benzene Cyclooctatetraene [6]-Annulene [10]-Annulene

[26]-Annulene SK SINGH, KMC Aromaticity of Annulenes  Annulenes could be aromatic, anti-aromatic or non-aromatic.  For example, [4]-Annulene which is cyclobutadiene is anti-aromatic.  [6]-Annulene (Benzene) is aromatic and  [8]-Annulene i.e., cyclooctatetraene is non-aromatic (Not Planar, “tub” shaped).

[10]-Annulene trans

H H

II 1. Cyclic 2. fully conjugated cyclic π electron cloud I 3. It follows Hückels rule. Contain 10 p e’s. 4n+ 2 = 10; n = 2 III but Not Planar SK SINGH, KMC bridgehead [10]-Annulene if two internal ‘H’ of [10] annulene are replaced by a methylene bridge above the molecule, the strain can be overcome, and it can acquire a flat geometry.

9,10 methano-[10]-annulene 1. It is cyclic. Planar (All the carbon atoms are in same plane), fully 9,10-aza-[10]-annulene H conjugated cyclic π electron N cloud 2. It follows Hückels rule. Contain 10 p e’s. 4n+ 2 = 10; n = 2

9,10-oxa-[10]-annulene 1. It is cyclic. Planar (All the carbon O 1. It is cyclic. Planar (All the carbon atoms are in same plane), fully atoms are in same plane), fully conjugated cyclic π electron conjugated cyclic π electron cloud cloud 2. It follows Hückels rule. Contain 10 p 2. Follow Hückels rule. Contain 10 e’s. p e’s. 4n+ 2 = 10; n = 2 4n+ 2 = 10; n = 2 SK SINGH, KMC [12]-Annulene 1. It is cyclic. Planar (The three H in-between the ring are far enough and do not create any strain for the planar arrangement), fully conjugated cyclic π electron cloud.  p H H 2. Do not Follow Hückel s rule. Contain 12 e’s. H 3. 4n p electron system. 4n = 12, n = 3

Antiaromatic Compound [14]-Annulene

1. It is cyclic. Nonplanar (As can be seen from the figure that ‘H’ H H present at the interior of the ring interfere with each other). H H 2. X ray analysis shows that the molecule is not planar.

SK SINGH, KMC Nonaromatic compound Dehydro – [14]-annulene:

 It is cyclic.  The removal of two interfering hydrogens of [14]annulene leads to formation of a triple bond, and a planar molecule.  The two electron from one of p bond of CC are delocalized into aromatic p system and the molecule becomes aromatic.  The other pair of p electron does not interact with the delocalized system as it is at right angles to the conjugated system of p electrons. Aromatic  It follows Hückels rule. Contain 14 p e’s. 4n + 2 = 14; n = 3 (an integer) trans-15,16-Dimethyldihydropyrene:

 It is cyclic.  planar molecule.  fully conjugated cyclic π electron cloud  It Follows Hückels rule. Contain 14 p e’s. 4n + 2 = 14; n = 3 (an integer)

Aromatic SK SINGH, KMC [16]-Annulene: [16]-Annulene is a 4np electron system so it is antiaromatic in nature.

 It is cyclic.  planar molecule.  fully conjugated cyclic π electron cloud.  4n p electron system. 4n = 16, n = 4

Antiaromatic

[18]-Annulene: [18]-Annulene is aromatic in nature.

 It is cyclic.  planar molecule.  fully conjugated cyclic π electron cloud.  It follows Hückels rule. Contain 18 p e’s.

4n + 2 = 18; n = 4

SK SINGH, KMC Aromatic Some More Examples [20] annulene: (4n) p electron system; where n= 5; Hence, anti-aromatic. [22] annulene: (4n + 2) p electron system; where n= 5; Hence, aromatic. [24] annulene: (4n) p electron system; where n= 6; Hence, anti-aromatic. [26] annulene: (4n + 2) p electron system; where n= 6; Hence, aromatic.

Practice Problem: 1. Write the structure of the smallest ring size annulene,SK SINGH, which KMC is aromatic in nature. Explain. AROMATIC HYDROCARBONS: Part III

DR SUNIL K. SINGH Assistant Professor in Chemistry

SK SINGH, KMC Electrophilic aromatic substitution reaction

 Aromatic compounds undergo substitution rather than addition reactions. This characteristic behavior is known as aromaticity.

Electrophilic addition reaction Electrophilic aromatic substitution reaction

Br Br2 CCl H E 4 Br E+ + H+

Br2 No Reaction CCl4

SK SINGH, KMC Types of electrophilic aromatic substitution reaction It includes vide variety of reactions:-

H NO2 H2SO4 Nitration + HONO2 + H2O A Nitro Compound

H Cl Fe or AlCl3 Halogenation + Cl2 + HCl

An Aryl chloride H Br Fe or AlBr3 + Br2 + HBr

An Aryl bromide

SK SINGH, KMC H SO3H SO3 Sulfonation + HOSO3H + H2O

A Sulfonic acid

H R AlCl Friedel-Crafts reaction + RCl 3 + HCl

An alkylbenzene

H COR AlCl + RCOCl 3 + HCl Acyl chloride A Ketone

SK SINGH, KMC H NO2 H2SO4 Mechanism of Nitration + HONO2 + H2O A Nitro Compound

+ Step 1: Generation of nitronium ion ( NO2) + - + HONO2 + 2H2SO4 H3O + 2HSO4 + NO2

(Acid) (Base) (An electrophile)

+ Step 2: Attack of electrophile ( NO2)

H H H Slow

BenzenonuimSK SINGH, KMC ion Step 3: Loss of proton H+ (Aromatisation)

- HSO4

H2SO4 Fast

Practice Problem: Nitration by nitric acid alone is believed to proceed by essentially the same mechanism as nitration in the presence of sulfuric acid. Write the equation for the generation of nitronium ion from nitric acid alone.

SK SINGH, KMC H SO3H SO3 Mechanism of Sulfonation + HOSO3H + H2O

A Sulfonic acid Step 1: Generation of SO3 (An electrophile): Practice Problem: Write an equation for the formation from + - H SO of each of the following sulfonating electrophiles: (a) 2H2SO4 H3O + 2HSO4 + SO3 2 4 + + H3SO4 ; (b)HSO3 ; (c)H2S2O7. (An electrophile)

O Slow H O Step 2: Attack of electrophile (SO3) + S O O S O- O

Fast H - O Step 3: Loss of proton H+ (Aromatisation) O + HSO4 S S - O- O O O

Step 4: Protonation of anion of benzene sulfonic acid O + O + H3O + H2O S S O- OH SK SINGH, KMCO O H R AlCl Mechanism of Friedel-Crafts alkylation + RCl 3 + HCl

Step 1: Generation of R (Carbocation: An electrophile):

- R-Cl + AlCl3 AlCl4 + R

+ Step 2: Attack of electrophile (R ) Slow H + R R

Step 3: Loss of proton H+ (Aromatisation) Fast - + AlCl + HCl H + AlCl4 3 R R

In certain cases, there is no free carbocation Cl involved. Instead the alkyl group is transferred- R-Cl + AlCl3 Cl Al Cl R without pair of electrons-directly to the aromatic Cl I ring from the polar complex, (I) , between AlCl3 Cl Slow and the alkyl halide: H - + R Cl Al Cl + AlCl4 SK SINGH, KMC Cl R Practice Problems:

1. How do you account for the fact that benzene in the presence of AlCl3 reacts with: (a) With n-propyl chloride to give isopropyl benzene; (b) with isobutyl bromide to yield tert-butylbenzene; (c) with neopentyl bromide to yield tert-pentylbenzene (d) By which of the alternative mechanisms for the Friedel-Crafts reaction are these products probably formed ? Solution: 1 (a) only Step 1: Generation of more stable Carbocation: An electrophile): H - Rearrangement CH CH CH Cl + AlCl AlCl + 3 2 2 3 4 CH3CHCH2 CH3CHCH3 o 1,2-Hydride ion shift 1 -carbocation 2o-carbocation Less stable More stable An Electrophile

Slow Step 2: Attack of electrophile H + CH CHCH CH3 3 3 CH

CH3

Fast + H - + AlCl + HCl Step 3: Loss of proton H (Aromatisation) + AlCl4 3 CH CH3 CH 3 CH CH3 CHSK 3SINGH, KMC Isopropylbenzene 2. Write all steps in the most likely mechanism for the reaction of benzene: Hint for 2 (b): (a) With tert-butyl alcohol in presence of H SO to yield tert-butylbenzene; 2 4 CH3CHCH3 2o-Carbocation (More Stable) (b) With propylene or (propene) in presence of H3PO4 to form isopropylbenzene. H+ CH3CH=CH2 OR

CH3CH2CH2 Solution 2 (a) only: 1o-Carbocation (Less Stable) Step 1: Generation of more stable Carbocation: An electrophile): Not Formed

Me H+ Me Me Me C OH Me C OH2 C Me + H2O Me Me Me 3o-Carbocation

Step 2: Attack of electrophile Me Slow H + C Me Me Me C Me Me

Step 3: Loss of proton H+ (Aromatisation) Fast H + H O+ + H2O Me 3 Me C C Me Me Me Me SK SINGH, KMC H COR AlCl3 Mechanism of Friedel-Crafts acyation + RCOCl + HCl Acyl chloride A Ketone Step 1: Generation of Acylium ion (Carbocation: An electrophile):

- RCO-Cl + AlCl3 AlCl4 + R C O R C O Octet of every Acylium ion (Resonance stabilized) atom is complete

Step 2: Attack of electrophile (R+) Slow H + R C O C R O Step 3: Loss of proton H+ (Aromatization) Fast - + AlCl + HC H + AlCl4 3 C R Practice Problems: C R 1. Nitrobenzene is used as a solvent in Friedel- O Crafts reaction. Why? O 2. Aniline do not undergo Friedel-Crafts HCOCl (Formyl chloride) is unstable. SK SINGH, KMC alkylation/acylation reactions. Why ? HCOCl = CO + HCl H Cl Fe or AlCl3 + Cl2 + HCl Mechanism of Halogenation or FeCl3 An Aryl chloride Step 1: Generation of an electrophile (Chloronium ion: Cl+) / Iron Complex Cl Cl+ + FeCl - Cl2 + FeCl3 Cl-Cl-Fe Cl 4 Cl Step 2: Reaction of benzene with the iron complex Step 2 can also be written by taking Cl+ as an electrophile.

Cl Slow H - + Cl-Cl-Fe Cl + FeCl4 Cl Cl

Step 3: Loss of proton H+ (Aromatization)

Practice Problem: Some activated Fast aromatic rings can be chlorinated - + FeCl + HCl H + FeCl4 3 using HOCl and this reaction is Cl catalyzed by H+. Suggest the possible Cl SK SINGH, KMC mechanism. THANKS Will be Continued….

SK SINGH, KMC AROMATIC HYDROCARBONS: Part IV

DR SUNIL K. SINGH Assistant Professor in Chemistry

SK SINGH, KMC Activating and deactivating groups:

 A group that make the ring more reactive  A group that make the ring less reactive than benzene is called an activating than benzene is called an deactivating group (they are o-, p-directors). group (they are m-directors but there are exceptions too).

Activating Groups: Ortho, para directors Deactivating: Meta-directors + Strongly activating --NO2, -N(CH3)2 , -CN, -COOH, -COOR, -NH2, -NHR, -NR2, -OH -SO3H, -CHO, -COR

Moderately activating: Deactivating: Ortho, para directors

-OCH3, -OC2H5, -NHCOCH3 -F, -Cl, -Br, -I

Weakly activating:

-C6H5, -CH3, -C2H5; etc

SK SINGH, KMC Determination of Orientation

Orientation of nitration of C6H5Y

SK SINGH, KMC Ref: Morrison Boyd Determination of relative reactivity

Relative reactivities are compared in following ways: Time required Severity of the condition

CH3 SO3H CH3 -CH3 is activating group Conc H2SO4 Under identical condition SO3H Take 1/10th time required by benzene

NO2 NO2 NO2 -NO2 ia Fuming HNO deactivating 3 o Conc H SO group 90 C 2 4 60oC O2N

SK SINGH, KMC For exact, quantitative comparison under identical conditions competitive reactions can be carried out

CH3 Small amount HNO + 3 Nitrotoluene + Nitrobenzene nitromethane (~ 25:1) 1:1

Cl Small amount HNO + 3 Nitrobenzene + Nitrochlorobenzene nitromethane (~ 30:1) 1:1

SK SINGH, KMC Theory of Reactivity

In electrophilic aromatic substitution reaction the intermediate carbocation is a hybrid of structures I, II and III, in which positive charge is distributed about the ring, being strongest at the ortho and para to the carbon atom being attacked by the electrophile.

H H H H E E+ E E E

I II III

Benzenonum ion or sigma complex

A group already attached to the benzene ring should affect the stability of the carbocation by dispersing or intensifying the positive charge, depend upon its electron-releasing or electron- withdrawing nature.

From structure I-III it is clear that the stabilizing or destabilizing effect should be especially important when the group is attached ortho or para to the carbon being attacked by an electrophile. SK SINGH, KMC Rates of substitution in benzene, toluene and nitrobenzene Compare the stability of benzenonium ion or sigma complex

H E H E H E

CH3 NO2 I II III

 By releasing the electrons methyl group (+I effect) in II tend to neutralize the positive charge of the ring; this dispersal of charge stabilizes the benzenonium ion, and thus leads to a faster reaction.

 The –NO2 group, on the other hand has an electron withdrawing inductive effect (-I effect); this tend to intensify the positive charge; destabilizes the benzenonium ion, and thus causes a slower reaction.

SK SINGH, KMC Theory of orientation

Electrophilic aromatic substitution reaction in the case of benzene:

H H H E+ -H+ E E E E

I II III

Benzenonum ion or sigma complex

Electrophilic aromatic substitution reaction in the case of toluene:

Carry out the attack of the electrophile at ortho, meta or at para-positions, then compare the stabilities of the resultant benzenonium ion (sigma complex). + I Effect of –CH3 CH3 CH CH CH 3 3 3 CH3 o H H H E+ -H+ E (Attack at ortho-position) E E E I II III Benzenonum ion or sigma complex More Stable III is More stable as the +ve chagre on carbon carrying the + I Effect methyl group

CH3 CH CH CH3 3 3 CH3 E+ -H+ H H H (Attack at meta-position) m E E E E Benzenonum ion or sigma complex IV V VI

CH3 CH CH3+ I Effect of CH3 3 CH3 E+ –CH3 group -H+

(Attack at para-position) p H E H E H E E VII VIII IX More Stable Benzenonum ion or sigma complex VIII is More stable as the +ve chagre on carbon carrying the + I Effect methyl group  The structure III is more stable. Its contribution to the resonance hybrid will be more, hence the hybrid carbocation resulted from the attack at ortho-position will be more stable than the hybrid carbocation resulting from the attack at the meta-position. Ortho-substitution therefore occurs faster than the meta-substitution.  The structure VIII is more stable. Its contribution to the resonance hybrid will be more, hence the hybrid carbocation resulted from the attack at para-position will be more stable than the hybrid carbocation resulting from the attack at the meta-position. para-substitution therefore occurs faster than the meta-substitution.  In toluene, ortho, para substitution is therefore faster than the meta substitution because

electron released by the –CH3 group is more effective during attack at the position ortho and para to it.

 -CH3 group due to hyperconjugation effect increases the electron density at ortho and para

positions (not at meta position). SK SINGH, KMC Carry out the attack of the electrophile at ortho, meta or at para-positions, Electrophilic aromatic substitution reaction in case of then compare the stabilities of the resultant benzenonium ion (sigma complex). nitrobenzene: -I Effect of –NO2 NO2 NO NO 2 2 NO2 NO2 o H H H E+ -H+ E E E (Attack at ortho-position) E I II III Benzenonum ion or sigma complex Unstable III is Unstable as the carbon carrying +ve chagre is directly linked with -I Effect nitro group

NO2 NO NO NO2 2 2 NO2 (Attack at meta-position) E+ -H+ H H H m E E E E Benzenonum ion or sigma complex IV V VI

-I Effect of –NO2 NO2 NO NO2 NO2 2 NO2 E+ -H+ (Attack at para-position) p H E H E H E E VII VIII IX Unstable Benzenonum ion or sigma complex VIII is Unstable as the carbon carrying +ve chagre is directly linked with -I Effect Nitro group  The structure III is unstable. The carbocation formed by the attack at ortho position is less stable than the carbocation formed by the attack at meta position. The hybrid carbocation (contributed by I and II only) resulted from the attack at ortho-position will be less stable than the hybrid carbocation (contributed by IV, V and VI) resulting from the attack at the meta-position. Ortho-substitution therefore occurs slower than the meta- substitution.  The structure VIII is unstable. The hybrid carbocation resulted from the attack at para- position will be less stable than the hybrid carbocation resulting from the attack at the meta-position. para-substitution therefore occurs slower than the meta-substitution.  In nitrobenzene, ortho, para substitution is therefore slower than the meta substitution

because of electron withdrawing nature of –NO2 group, which deactivates the ortho and

para-positions through –R effect. SK SINGH, KMC Carry out the attack of the electrophile at ortho, meta or at para-positions, Electrophilic aromatic substitution reaction in the case of phenol: then compare the stabilities of the resultant benzenonium ion (sigma

complex).

: OH : OH OH :OH OH OH H o H H H E E+ -H+ E (Attack at ortho-position) E E E I II III IV Benzenonum ion or sigma complex Stable

IV is Stable as every atom has a complete octet of electrons

:

: OH : OH OH OH OH (Attack at meta-position) E+ -H+ H H H m E E E E Benzenonum ion or sigma complex V VI VII

:

: OH : OH :OH OH OH OH E+ -H+ (Attack at para-position) p H E H E H E H E E VIII IX X XI more stable Benzenonum ion or sigma complex X is more stable octet of every atom is complete  The structure IV is more stable. Its contribution to the resonance hybrid is more. The carbocation formed by the attack at ortho position is more stable than the carbocation formed by the attack at meta position. The hybrid carbocation resulted from the attack at ortho-position will be more stable than the hybrid carbocation resulting from the attack at the meta-position. Ortho-substitution therefore occurs faster than the meta-substitution.  The structure X is stable. Its contribution to the resonance hybrid is more.The hybrid carbocation resulted from the attack at para-position will be more stable than the hybrid carbocation resulting from the attack at the meta-position. para-substitution therefore occurs faster than the meta-substitution.  In phenol, ortho, para substitution is therefore faster than the meta substitution because of electron donating nature of –OH group, which activates the ortho and para-positions through +R effect. SK SINGH, KMC Effect of halogens on electrophilic aromatic substitution reaction: 1. Reactivity:

H E H E -Cl group through –I effect withdraws electrons; destabilizes the benzenonium ion (deactivates the ring). Cl -I Effect I II • Inductive effect of the –Cl group intensifies the +ve charge on the carbocation II, makes the carbocation less stable, and causes a slower reaction than benzene.

2. Orientation: Carry out the attack of the electrophile at ortho, meta or at para-positions, then compare the stabilities of the resultant benzenonium ion (sigma complex).

Practice Problem: in chlorobenzene –Cl have electron withdrawing effect even though it directs the coming electrophile at ortho and para-positions. Explain with suitable mechanism. SK SINGH, KMC

: : : : : Cl : Cl Cl Cl Cl Cl H o H H H E (Attack at ortho-position) E+ -H+ E E E E

I II III IV Unstable Stable Benzenonum ion or sigma complex

IV is Stable as every atom has a complete octet of electrons

: : : : :

Cl Cl Cl : Cl Cl E+ -H+ (Attack at meta-position) H H H m E E E E Benzenonum ion or sigma complex V VI VII

: : :

: Cl : Cl :Cl: Cl Cl Cl E+ -H+ (Attack at para-position) p H E H E H E H E E VIII IX X XI Unstable more stable Benzenonum ion or sigma complex X is more stable as the octet of every atom is complete SK SINGH, KMC  The structure IV is more stable. Its contribution to the resonance hybrid is more. The carbocation formed by the attack at ortho position is more stable than the carbocation formed by the attack at meta position. The hybrid carbocation resulted from the attack at ortho-position will be more stable than the hybrid carbocation resulting from the attack at the meta-position. Ortho-substitution therefore occurs faster than the meta-substitution.  The structure X is stable. Its contribution to the resonance hybrid is more. The hybrid carbocation resulted from the attack at para-position will be more stable than the hybrid carbocation resulting from the attack at the meta-position. para-substitution therefore occurs faster than the meta-substitution.  In chlorobenzene,–Cl group activates the ortho and para-positions through +R effect.

SK SINGH, KMC Orientation in disubstituted benzenes:  The directive effect of one reinforces that of the other:

CH3 o,p-directing NHCOCH3 o, p-directing SO3H m-directing NC m-directing

NO2 m-directing NO2 m-directing

 The directive effect of one group opposes that of the other: Strongly activating group generally win out deactivating or weakly activating groups: -NH2, -OH, -OR, -NHCOCH3 > -C6H5, -CH3 > meta-directors

o, p-directing NHCOCH o, p-directing OH 3 CHO m-directing

OH CH3 o, p-directing CH3 o, p-directing o, p-directing SK SINGH, KMC CH o, p-directing  If fairly large difference in 3 CH3 CH3 the directive effect is not NO2 HNO3 + H2SO4 there then mixture of + products are obtained: NO2 Cl o, p-directing Cl Cl 58 % 42 %

Cl o, p-directing Cl Cl Cl O N HNO3 + H2SO4 NO2 2 + + Br  There is often little substitution Br Br Br NO between the two groups that o, p-directing 2 are meta to each other: 1 % 62 % 37 %

o, p-directing CH3 CH3 CH CH3 3 O N HNO3 + H2SO4 NO2 2 + + Cl Cl Cl Cl NO o, p-directing 2 9 % 32 % 59 %

SK SINGH, KMC Friedel-Crafts acylation reaction using anhydrides: O (RCO) O 2 C O Acid anhydride R C AlCl (CH3CO)2O 3 CH3 AlCl3

R = -CH3; acetic anhydride Acetophenone R = -CH2CH3; Propanoic anhydride Write its mechanism?

Intramolecular Friedel-Crafts alkylation Reaction: Cl

AlCl3 Intramolecluar FCR

SK SINGH, KMC SK SINGH, KMC Ring Closure to six membered ring is faster than rearrangement:

Major minor Intramolecular Friedel-Crafts acylation Reaction and its application:

AlCl3 Cl Intramolecluar FCR O O Application: Synthesis of Tetralone O O O O

succinic anhydride Zn-Hg SOCl2 AlCl3 Intramolecluar AlCl Conc HCl HOOC Cl 3 HOOC FCR O O

SK SINGH, KMC Fries Rearrangement:  Lewis acid catalyzed intramolecular rearrangement of phenolic esters (Phenyl esters) into acyl derivatives of phenol is known as Fries rearrangement.  Mechanism: Intramolecular Friedel-Crafts acylation reaction. OCOR OH OH COR Anhy. AlCl 3 + Work up

Phenyl ester COR For work up dil. HCl is taken  If the p-position is blocked ortho-acyl derivative is formed. O OH O O CH3 C CH3 Anhy. AlCl3 Work up

CH CH3 3 SK SINGH, KMC Mechanism

Aromatization

SK SINGH, KMC HCOCl is unstable so generated insitu. Gattermann-Koch reaction It is a formylation reaction of benzene. In this reaction benzene is treated with CO and HCl in presence of AlCl₃ to form benzaldehyde.

Mechanism

SK SINGH, KMC Practice Problems: 1. Write the mechanism of Friedel-Crafts acylation reaction using acetic anhydride as an acylating reagent. 2. Write the mechanism of Friedel-Crafts acylation reaction using succinic anhydride as an acylating reagent.

3. What happen when 4-phenyl-2-chlorobutane is treated with anhydrous AlCl3. 4. Write the mechanism for the following conversion.

5. Write the mechanism of following conversion:

AlCl3 Cl Intramolecluar FCR O O 6. Write the mechanism of following conversion and assign the major and minor products.

7. in chlorobenzene –Cl have electron withdrawing effect even though it directs the coming electrophile at ortho and para-positions. Explain with suitable mechanism. SK SINGH, KMC Practice Problems: 8. In electrophilic aromatic substitution reaction, nitrobenzene reacts very slowly while phenol reacts faster than benzene. Explain with mechanism. 9. In electrophilic aromatic substitution reaction, nitrobenzene reacts very slowly while toluene reacts faster than benzene. Explain with mechanism. 10. Some activated aromatic rings can be chlorinated using HOCl and this reaction is catalyzed by H+. Suggest the possible mechanism. 11. Nitrobenzene is used as a solvent in Friedel-Crafts reaction. Why? 12. Aniline do not undergo Friedel-Crafts alkylation/acylation reactions. Why ? 13. Write all steps in the most likely mechanism for the reaction of benzene:

(a) With tert-butyl alcohol in presence of H2SO4 to yield tert-butylbenzene; (b) With propylene or (propene) in presence of H3PO4 to form isopropylbenzene. 14. How do you account for the fact that benzene in the presence of AlCl3 reacts with: (a) With n-propyl chloride to give isopropyl benzene; (b) with isobutyl bromide to yield tert-butylbenzene; (c) with neopentyl bromide to yield tert-pentylbenzene (d) By which of the alternative mechanisms for the Friedel-Crafts reaction are these products probably formed ? 15. How will you convert benzene into triphenylmethane? Write the steps and mechanism of the reaction involved.

SK SINGH, KMC THANKS

SK SINGH, KMC