Hyperconjugation and aromaticity- GOC Why Vedantu Pro is the Best? 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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 O ● 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 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 ● 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 σ 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 ❖ Cover entire JEE Main Syllabus (as per the new pattern) with India’s Best Teachers in 90 Sessions ❖ Solve unlimited doubts with Doubt experts on our Doubt App from till Exam 8 AM to 11 PM ● 1 Batch Classes will be over in 10 Weeks ● 10 full and 10 Part syllabus test to make you exam ready ● ● Amazing tricks & tips to crack JEE Main 2021 Questions in power-packed 90 Min sessions ● ● Learn on a 2-way Interactive Platform where the teacher is always with you JEE MAIN 2021 CRASH COURSE Lightning Deal: ₹10000 ₹8000/- Batch Starts From : Every Monday *Crash Course Link Available in Description Apply Coupon Code: ABSRCC Join Vedantu JEE Telegram channel NOW! Assignments Notes Daily Update https://vdnt.in/jeepro .