Hyperconjugation and Aromaticity- GOC

Hyperconjugation and aromaticity- GOC

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*Vedantu Pro Subscription Link available in description Use the code: ABSRPRO Mesomeric effect

When a group releases or withdraw orbital electron in any conjugated system then it is called M effect group and effect is known as mesomeric effect.

+M group -M group Mesomeric effect

Electron releasing group

ーーーーーーーー OH, OR, NH2 NHR, NR2 SH, SR, X etc. Mesomeric effect

When a group releases or withdraw orbital electron in any conjugated system then it is called M effect group and effect is known as mesomeric effect.

+M group -M group Mesomeric effect

Electron withdrawing group

ーーーーーーー ⊕ ー ⊕ NO2, CN, CHO, COR, COOH, SO3H, NR3 , NH3 , etc. Dual behaviour

ーN＝O Both electron releasing as well as electron withdrawing Mesomeric effect

Resonance Eﬀect doesn’t depend upon distance unlike inductive eﬀect Let’s solve

Which of the following have -M eﬀect (e- withdrawing mesomeric eﬀect) ?

A C = O B ㅡOH

C ㅡOR D ㅡBr Let’s solve

Which of the following have -M eﬀect (e- withdrawing mesomeric eﬀect) ?

A C = O B ㅡOH

C ㅡOR D ㅡBr Solution

❖ The carbonyl group contain double bonds so they can accept lone pair

of electrons and thus have -M eﬀect (electron withdrawing mesomeric

eﬀect).

❖ On the other hand ㅡOH. ㅡOR and ㅡBr have lone pair of electrons and

have +M eﬀect (electron releasing mesomeric eﬀect). Aromatic Compounds

● Aromatic compounds are specialized cyclic compound which are known for their distinct smell. (aroma-fragrance). This group is called arenes

❖ Benzene was considered as parent aromatic compound Aromatic Compounds

Huckel’s rule :

Characteristics for ring being aromatic- ❖ Planarity (2-D) [sp2 carbon] ❖ Conjugated system (de-localisation of 휋 e-in ring) ❖ Cyclic ❖ Presence of (4n + 2)휋 e- in the ring. Where, (n = 0, 1, 2, 3...) Aromatic Compounds

Π e- are loosely bonded electrons, lone pair are also Π e- . Aromatic Compounds

● Cyclic, Planar ● 6 e- system (4π e-’s + 2 e- from only one lone pair ) ● Follow (4n + 2) rule and 2 e-’s from LP are in complete delocalisation or in resonance ● Hence aromatic

Only one lone pair is required for delocalisation Aromatic Compounds

❖ Cyclic ❖ Planar ❖ Conjugated System ❖ 4n휋 e- should present in the ring where (n = 1 2,3,...)

Cyclobutadiene (n = 1 ⇒ 4휋e-)

(Highly Unstable) Anti-Aromatic Compounds

Some important points about Anti-Aromatic-

❖ Dimerisation Anti-Aromatic Compounds

Some important points about Anti-Aromatic-

❖ Reaction with active metals Anti-Aromatic Compounds

Some important points about Anti-Aromatic-

❖ Reaction with active metals Non-Aromatic Compounds

Compounds which are neither aromatic nor antiaromatic.

Stability order of compounds :

Aromatic > non aromatic > Anti - Aromatic Let’s solve

Which of the following is least stable?

A B N

C D O O Let’s solve

Which of the following is least stable?

A B N

C D O O Solution

● Cyclic

● Non-conjugated

● Non-aromatic Let’s solve

Which of the following chemical system is non-aromatic ?

A B

C D

S Let’s solve

Which of the following chemical system is non-aromatic ?

A B

C D

S Solution

sp3

sp2 sp2

● Cyclopentane-1,3-diene is a non aromatic compound ● Cyclic delocalization of electron is not possible because of the presence of sp3 hybridized carbon atom. Let’s solve

Identify the following as aromatic, antiaromatic or non-aromatic:

⊕ (a) (b)

S Solution

⊕

n = 1 ⇒ 4휋e- Anti-aromatic

Cyclic, conjugated, planer, (4n+2) 휋e- n = 1 ⇒ 6휋 e-

S Aromatic 2 e- from only one lone pair Hyperconjugation

● When a sigma CH bond of sp3 hybridised carbon is in conjugation with ힹ bond at p-orbital, half ﬁlled p-orbital or vacant p-orbital, then the bond pair e of sigma CH bond overlap with adjacent p-orbital.This phenomenon is called hyperconjugation

Unshared p Orbital σ

Unsaturated System Carbocation Hyperconjugation

Free Radicals Alpha carbon and alpha hydrogen

❖ 훼-Carbon is the carbon attached to a functional group such as C＝C. ❖ The hydrogen attached to 훼-carbon is called 훼-hydrogen. ❖ For an 훼 CーH bond to be eligible for hyperconjugation, 훼 C must be sp3 hybridized. Origin of hyperconjugation: Stability of alkene Origin of hyperconjugation: Stability of alkene

ー＝ーーーー CH3 CH CH CH3 + H2 CH3 CH2 CH2 CH3

ΔH3(expected) > ΔH2(expected) > ΔH1

But ΔH3(actual) < ΔH2(actual) < ΔH1 Origin of hyperconjugation: Stability of alkene

As we increase the number of methyl groups on double bond, the stability of alkene increases

Order of electron release based on inductive eﬀect : Origin of hyperconjugation: Stability of alkene

As we increase the number of methyl groups on double bond, the stability of alkene increases

Order of electron donation based on hyperconjugation:

● 훼 H ↑ ● 휋 bond delocalisation ↑ ● Stability of alkene ↑ ● Heat of hydrogenation↓ Origin of hyperconjugation: Stability of alkene

Mechanism of Electron Donation to explain Anomaly :

No-bond resonance Baker nathan eﬀect 휎 - 휋 eﬀect Origin of hyperconjugation: Stability of alkene Let’s solve

Mark the number of 훼-C and 훼-H in the given compounds (mark

훼-H eligible for hyperconjugation only).

＝ー (a) CH2 CH CH3

ー＝ー (b) CH3 CH CH CH3

S Solution

훼 ＝ー (a) CH2 CH CH3 훼C = 1, 훼 H = 3 훼 훼 ー＝ー (b) CH3 CH CH CH3 훼C = 2, 훼H = 6 Let’s solve

Mark the number of 훼-C and 훼-H in the given compounds: ー＝ーー (a) (CH3)2CH C(CH3) CH CH2 CH3

ー (b) CH3 CHO

ー ⊕ (c) CH3 CH2

ー • (d) CH3 CH2 Solution Application of hyperconjugation

● Stability of alkenes :

흰 hydrogen ↑ hyperconjugative structure ↑ stability ↑ no-bond

resonance energy ↑

● Acidic character of alkenes :

Hyperconjugation weakens the 힪 C-H bond Application of hyperconjugation Application of hyperconjugation

● Stability of carbocation :

훅⊕ H 훅⊕ 훅⊕ H C CH Positive charge on C is 2 delocalized over 훂H to give 훅⊕ stability to the carbocation. H

Hyperconjugation Hybrid Application of hyperconjugation

● Stability of carbocation :

Number of 훂H ⬆ Stability of carbocation ⬆

3o 2o 1o Methyl 훂 훂 훂 훂 ( H = 9) ( H = 6) ( H = 3) ( H = 0) Let’s solve

Which has more acidic 훂-H? 一一 I. R CH2 CH=CH2 一一 II. R CH2 CHO

A I B II

C Both D None Let’s solve

Which has more acidic 훂-H? 一一 I. R CH2 CH=CH2 一一 II. R CH2 CHO

A I B II

C Both D None Solution

In (ll), negative charge is delocalized on O while in (l), negative charge is delocalized on C. Therefore, (ll) is more stable than (l). Hence, (ll) is more acidic than (l). Let’s solve

Which is more stable carbocation?

⊕ I. C(CH3)2H ⊕ II. ((CH3)2CH)3C

A I B II

C Both equally D None stable Let’s solve

Which is more stable carbocation?

⊕ I. C(CH3)2H ⊕ II. ((CH3)2CH)3C

A I B II

C Both equally D None stable Solution

❖ (I) has 2훂C and 6훂H.

❖ (II) has 3훂C and 3훂H.

❖ (I) is 2o carbocation while (II) is a 3o carbocation.

❖ But since (I) has greater number of 훂H, it is more stable. Let’s solve

The number of hyperconjugation structures possible in

A 3 B 2

C 6 D None of these Let’s solve

The number of hyperconjugation structures possible in

A 3 B 2

C 6 D None of these Solution JEE MAIN 2021 CRASH COURSE

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