Question

Dehydration of alcohols take place more rapidly with $POC{{l}_{3}}$ than with ${{H}_{2}}S{{O}_{4}}$.
Select the correct statement(s) about the above dehydration reaction:

This question has multiple correct options

Answer 1

An attempt to this question can be made by determining the reaction mechanism for dehydration using both the reagents $POC{{l}_{3}}$ and${{H}_{2}}S{{O}_{4}}$. Now determine the difference in the mechanism. With this you can reason why $POC{{l}_{3}}$ is preferred over ${{H}_{2}}S{{O}_{4}}$ for dehydration.

Complete step by step answer:
So in the question it is asked that, we have to select
We will write the reaction mechanism for dehydration using $POC{{l}_{3}}$.

In the above reaction we observe,
- No carbocation formation
- Nucleophilic attack used for deprotonation
- involves E2 mechanism as pyridine base abstracts proton from the adjacent carbon as the same time at which $-\text{OPOC}{{\text{l}}_{2}}$ is leaving
- involves $\text{R}-\text{OPOC}{{\text{l}}_{2}}$ with $-\text{OPOC}{{\text{l}}_{2}}$ as a better leaving group
Based on the above statements we can conclude that the below given options stand correct.
(A) It does not involve carbocation.
(B) It involves the species $\text{R}-\text{OPOC}{{\text{l}}_{2}}$ with $-\text{OPOC}{{\text{l}}_{2}}$ as a better leaving group.
(C) It involves E2 mechanism as the pyridine base abstracts proton from the adjacent carbon as the same time at which $-\text{OPOC}{{\text{l}}_{2}}$ is leaving.
The correct answer is option “A, B and C” .

Note: The reaction with sulfuric acid on the other hand involves the formation of carbocation. This is a slow step. Along with that it uses the E1 mechanism for elimination unlike $POC{{l}_{3}}$ that uses the E2 mechanism for elimination.