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Hyperconjugation

Hyperconjugat...

Question

Hyperconjugation phenomenon is possible in:

C H_{2} = C H_{2}

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C H_{3} - C H_{2} C H = C H_{2}

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C_{6} H_{5} C H = C H_{2}

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(C H_{3})_{3} C - C H = C H_{2}

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Solution

The correct option is A $C H_{3} - C H_{2} C H = C H_{2}$
Hyperconjugation is possible only in $C H_{3} - C H_{2} C H = C H_{2}$ .
$C H_{3} - C H_{2} - C H = C H_{2} (α - H)$ [Presence of 2 $α - H$
Presence of $α - H$ is necessary condition for hyperconjugation.

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Similar questions

C H_{3} - C H_{2} - C H_{3} | C H - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{3} | C H - C H_{3}

C H_{3} - C H_{2} - C H_{2} - C H_{2} - C H | C H_{3} - C | C H_{3} - C H_{3} - C H_{2} - C H_{3} | C H - C H_{2} - C H_{2} - C H_{2} - C H_{3}

C H_{3} - C H_{3}, C H_{2} = C H_{2}

C H \equiv C H

S_{N} 2

K I

C H_{3} - C H_{2} - C H_{2} - C H_{2} - C l a n d C H_{3} - C H ∣ C H_{3} - C H_{2} - C l

(I) (I I)

C H_{3} - C H_{2} - C H_{2} - C l a n d C H_{3} - C H_{2} - C H_{2} - B r

(I) (I I)

C H_{3} - C H ∣ C H_{3} - C H_{2} - C H_{2} - C l a n d C H_{3} - C H_{3} ∣ C ∣ C H_{3} - C H_{2} C l

(I) (I I)

C H_{3} - C H_{2} - C H_{2} - B r ∣ C H - C H_{3} a n d C H_{3} - C H_{3} ∣ C H - C H_{2} - B r ∣ C H - C H_{3}

(I) (I I)

A . C H_{3} - C H = C H_{2} + H C l ⟶ C H_{3} - C | C l H - C H_{3}

B . C H_{3} - C H = C H_{2} + H B r ⟶ C H_{3} - C H_{2} - C H_{2} B r

C . C H_{3} - C H = C H_{2} + H B r p e r o x i d e - ---- \to C H_{3} - C H_{2} - C H_{2} - B r

D . C H_{3} - C H = C H_{2} + H I p e r o x i d e - ---- \to C H_{3} - C | I H - C H_{3}

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