$C_{2} H_{4} \underset{A l {Cl}_{3}}{\overset{HCl}{⟶}} A \overset{KOH (aq .)}{\to} B \frac{Conc H_{2} {SO}_{4}}{140^{0} C} C$
What is the final product?

see full answer

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*

An Intiative by Sri Chaitanya

$C_{2} H_{5} - O - C_{2} H_{5}$

$C_{2} H_{4}$

$C_{2} H_{5} OH$

$C_{2} H_{5} - O - {SO}_{3} H$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Given:- It is given to find the final product in the given reaction.

Explanation:-

The reaction can be completed as below.

$C_{2} H_{4} \underset{A l {Cl}_{3}}{\overset{HCl}{⟶}} {CH}_{3} - {CH}_{2} - Cl \overset{aq . KOH}{\to} {CH}_{3} - {CH}_{2} - OH$

${CH}_{3} - {CH}_{2} - OH \to_{140 ° C}^{Con H_{2} {SO}_{4}} C_{2} H_{5} - O - C_{2} H_{5}$

Hence, the correct answer is option (B) $C_{2} H_{5} - O - C_{2} H_{5}$ .

Watch 3-min video & get full concept clarity