Arrange the following in decreasing order of stability. ( a ) C ⊝ H 2 - 1 NO 2 , C ⊝ H 2 - 2 CHO CH ≡ 3 C ⊝ ( b ) CH 3 - CH 2 - 1 CH 2 - C ⊝ H 2 CH 3 - C ⊝ H 2 - CH 2 CH 3 CH 3 2 C ⊝ H 3 - CH 2 CH 3

Part - a 1 . The stability of carbanion directly proportional to number of electron withdrawing species (electrophile) or their electron withdrawing capacity. 2 . Stability of carbanion directly proportional to - I effect. 3 . The order of electron withdrawing capacity ( - I effect) of various functional groups. So correct order of stability of carbanion in the given question- 1 > 2 > 3 (least stable) Part - b:- According to question, 1 is 1 0 Alkyl carbanion, 2 is 2 0 Alkyl carbanion, & 3 is 3 0 Alkyl carbanion The stability order of various Alkyl carbanion is - 1 0 > 2 0 > 3 0 . It is due to that R -group show + I Effect & stability of carbanion is indirectly proportional to + I Effect (number of nucleophile of electron donor species or their electron donor capacity).

Arrange the following in decreasing order of stability: ( a ) ( b ) ( c ) C . ¯ . Cl 3 1 C . ¯ . F 3 2

To solve this question, following concept are required- (a) stability of carbanion directly proportional to electron withdrawing group and their electron withdrawing capacity. (b) due to the conjugation, delocalisation of negative charge directly proportional to the stability of carbanion. In option (a): The stability of a-II, is maximum due to the participation of negative charge to make compound aromatic. a-III, has comparative less stability because here is the participation of negative charge to make compound aromatic but + I Effect of - alkyl group. a-I is less stable than a -III due to only; π & negative charge conjugation. a-IV is the least stable, it due to that one carbanion is present in conjugation with double bond of cycle, but there is no conjugation of - CH 2 - Group. So decreasing order of stability is a-II > a- III> a- I > a -IV. In option (b): The stability of an aromatic carbanion (b-I- participation of negative charge conjugated with double bond to make compound aromatic) is higher than benzyl carbanion (due to complete aromaticity of benzene ring and + I Effect of - CH 2 - Group). Phenyl carbanion (b-III -complete aromaticity of the compound so l.p do not take part). So decreasing order of stability is -b-I >b-II> b-III. In option (C): Tri-chloro-methyl carbanion is more stable than tri-fluoro methyl-carbanion. It is due to that chlorine has vacant d -orbital due to this possibility of negative charge distribution increases.

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Alpha Question Bank for Engineering: Chemistry>Chapter 23 - Organic Chemistry- Some Basic Principles and Techniques>Exercise-1>Q 210

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In hyperconjugation there is overlap between:

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Important Questions on Organic Chemistry- Some Basic Principles and Techniques

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Alpha Question Bank for Engineering: Chemistry>Chapter 23 - Organic Chemistry- Some Basic Principles and Techniques>Exercise-1>Q 211

Which of the following cannot exhibit hyperconjugation?

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Alpha Question Bank for Engineering: Chemistry>Chapter 23 - Organic Chemistry- Some Basic Principles and Techniques>Exercise-1>Q 212

Which of the following alkenes will show maximum number of hyperconjugation forms?

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Alpha Question Bank for Engineering: Chemistry>Chapter 23 - Organic Chemistry- Some Basic Principles and Techniques>Exercise-1>Q 215

The decreasing order of electron density on the ring is

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Alpha Question Bank for Engineering: Chemistry>Chapter 23 - Organic Chemistry- Some Basic Principles and Techniques>Exercise-1>Q 216

The stability order of alkene in following compounds is

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Alpha Question Bank for Engineering: Chemistry>Chapter 23 - Organic Chemistry- Some Basic Principles and Techniques>Exercise-1>Q 217

Select the correct statement about this compound

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Alpha Question Bank for Engineering: Chemistry>Chapter 23 - Organic Chemistry- Some Basic Principles and Techniques>Exercise-1>Q 218

Arrạnge the stability of following

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Alpha Question Bank for Engineering: Chemistry>Chapter 23 - Organic Chemistry- Some Basic Principles and Techniques>Exercise-1>Q 220

Arrange the following in decreasing order of stability.

$(a)$	$\overset{⊝}{C} H_{2} \underset{1}{-} {NO}_{2},$	$\overset{⊝}{C} H_{2} \underset{2}{-} CHO$	$CH \underset{3}{\equiv} \overset{⊝}{C}$
$(b)$	${CH}_{3} - {CH}_{2} \underset{1}{-} {CH}_{2} - \overset{⊝}{C} H_{2}$	${CH}_{3} - \overset{⊝}{C} \underset{2}{H} - {CH}_{2} {CH}_{3}$	${({CH}_{3})}_{2} \overset{⊝}{C} \underset{3}{H} - {CH}_{2} {CH}_{3}$