C2H4⟶HClAlCl3A→KOH(aq.)B Conc H2SO41400CCWhat is the final product?

Book Online Demo

Check Your IQ

Try Test

Courses

Offline Centres

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

What is the final product?

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

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

JEE

NEET

Foundation JEE

Foundation NEET

CBSE