Identify $^{'} Z^{'}$ in the following reaction series. $C H_{3} C H_{2} C H_{2} B r a q . N a O H - ------- \to (X) A l_{2} O_{3} - ---- \to H e a t (Y) H O C l - ---- \to (Z)$

Mixture of

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

and

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

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

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

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

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Solution

The correct option is C $C H_{3} - C H | O H - C H_{2} | C l$
Option B represents the compound Z.

$C H_{3} C H_{2} C H_{2} B r a q . N a O H - ---------- \to H y d r o l y s i s, - H B r C H_{3} C H_{2} C H_{2} O H (X) A l_{2} O_{3} - ---------- \to H e a t, d e h y d r a t i o n C H_{3} C H = C H_{2} (Y) H O C l - ---- \to C H_{3} - C H | O H - C H_{2} | C l (Z)$

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Similar questions

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

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

S_{N} 2

c o n c . H C l :

(a) C H_{3} - C H_{2} - O H A n h y Z n C l_{2} / △ - -------- \to C H_{3} - C H_{2} - C l

(b) C H_{3} - C H_{3} | C | C H_{3} - C H_{2} O H A n h y Z n C l_{2} / △ - -------- \to C H_{3} - C H_{1} | C | C H_{3} - C H_{2} - C H_{3}

(c) C H_{3} - O H | C | C H_{3} - C H_{3} A n h y Z n C l_{2} / △ - -------- \to C H_{3} - C l | C | C H_{3} - C H_{3}

(d) C H_{2} = C H - C H_{2} - O H A n h y Z n C l_{2} / △ - -------- \to C H_{2} = C H - C H_{2} - C l

C H_{3} - C H_{2} - C l + K C N Δ - \to

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

C H_{3} - C H_{2} - O H + N a B r + H_{2} S O_{4} \to

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