Hyperconjugation is not observed in:

A free radical

(- ∙ C H_{2})

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A carbanion

(- - C H_{2})

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A carbocation

(- + C H_{2})

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A double bond

>C=C<

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Solution

The correct option is B A carbanion $(- - C H_{2})$
Hyperconjugation :
It is a stabilising effect due to the reason that octet of carbon is always complete. Stability is also increased due to the delocalisation of sigma-pi electrons.

Condition for hyperconjugation :
Atleast one $α$ hydrogen has to be present.

Hyperconjugation is mainly observed in:
1. A free radical $(- ∙ C H_{2})$
2. A carbocation $(- + C H_{2})$
3. A double bond $>C=C<$
It is not observed in carbanion because of its fully filled p-orbital (or octet).

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Similar questions

(1) C H_{3} C H = C H_{2} + H B R - ---------- \to C H_{3} C H_{2} C H_{2} - B r

(2) C H_{3} C H = C H_{2} + C I_{2} - --------- \to C H_{3} C H (C l) C H_{3}

(3) C H_{3} C H = C H_{2} + H C I - ---------- \to C l - C H_{2} C H = C H_{2}

(4) C H_{3} C H_{3} + C l_{2} h r - \to C I - C H_{2} C H = C H_{2}

C H_{2} = C H_{2} (i) H^{\oplus} - --- \to (i i) D_{2} O (A)

C H_{3} - C H = C H_{2} (1) T H F : B D_{3} - ------- \to (2) C H_{3} C O_{2} T (A)

(H_{2} C = C = C = C H_{2})

{CH}_{2} = CH - {CH}_{2} - CH = {CH}_{2}

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