Identify $Z$ in the following reaction sequence.

$C H_{3} C H_{2} C H_{2} O H C o n c . H_{2} S O_{4} - ------- \to 160 - 180^{o} C X B r_{2} + C C l_{4} - ------ \to Y N a N H_{2} - ---- \to Z$ :

C H_{3} - C H (N H_{2}) - C H_{2} N H_{2}

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C H_{3} - C H (O H) - C H_{2} O H

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

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C H_{3} - C \equiv C H

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Solution

The correct option is B $C H_{3} - C \equiv C H$
$C H_{3} C H_{2} C H_{2} O H C o n c . H_{2} S O_{4} - ------- \to 160^{o} - 180^{o} C C H_{3} C H = C H_{2} B r_{2} - - \to C H_{3} C H B r (Y) - C H_{2} B r$

$[C H_{3} C (B r) (A) (B) = C H_{2} + C H_{3} C H = C H B r] N a N H_{2} - ---- \to - H B r C H_{3} C \equiv C H (Z)$

Alcoholic $K O H$ brings about dehydrobromination of $Y$ and give a mixture of vinyl bromide ( $A$ and $B$ ) while $N a N H_{2}$ being a strong base than $a l c . K O H$ readily brings about dehydrobromination of less reactive vinyl bromide to give propyne $C H_{3} C \equiv C H$ i.e. $(Z)$ .

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Similar questions

Which of the following compounds gives fastest reaction with HBr and why

(i) (CH3)3 COH

(ii) CH3 CH2 CH2OH

(iii) CH3 CHOH CH3

(iv) CH3 (CH3) CH CH2OH

C H_{3} - C H_{2} - C H (O H) - C H_{3} [O] \to A C H_{3} M g - I - ----- \to H_{2} O / H^{+} B

B

H O C l (C l_{2} + H_{2} O)

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