Identify $Z$ in the following reaction series.
$C H_{3} . C H_{2} C H_{2} B r a q . N a O H ⟶ (X) A l_{2} O_{3} ⟶ H e a t (Y) H O C l ⟶ (Z)$

mixture of

C H_{3} - C ∣ C l H - C ∣ C l H_{2}

and

C H_{3} - C ∣ O H H - C ∣ O H H_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C H_{3} - C ∣ O H H - C ∣ C l H_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C H_{3} - C ∣ C l H - C ∣ O H H_{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C H_{3} - C ∣ C l H - C ∣ C l H_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $C H_{3} - C ∣ C l H - C ∣ O H H_{2}$
When $1$ -bromopropane is treated with aq $N a O H$ , propan- $1$ -ol (compound $X$ ) is obtained. Bromide is replaced with $- O H$ group.
When propan- $1$ -ol is heated in presence of alumina, propene (compound $Y$ ) is obtained due to dehydration.
When propene is treated with $H O C l$ (chlorine water), $2$ -chloropropan- $1$ -ol (compound $Z$ ) is obtained.

Suggest Corrections

Similar questions

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)

P h - M e ∣ C ∣ H - O H S O C l_{2} ⟶ C_{5} H_{5} N m e d i u m

{[\begin{matrix} - C H_{2} - C H ∣ O H - C H_{2} - C H ∣ O H - C H_{2} - C H ∣ O H - \end{matrix}]}_{n}

H_{3} C ╲ C ╱ H = H ╱ C ╲ C H_{3} m - C P B A - -------- \to P C l_{5} - -- \to P r o d u c t (s)

C H_{3} - C | H C l - C | H B r - C | H O H - C H_{3}

Join BYJU'S Learning Program