JEE Chemistry 2023 Question with Solution

Given: The decreasing order of dehydration of alcohols $a$, $b$, $c$ and $d$ is to be identified.

Find: The correct order based on the ease of dehydration.

Dehydration of alcohols depends on the stability of the carbocation intermediate formed during the reaction. The more stable the carbocation, the easier the dehydration.

From the solution, the relative carbocation stability is described as:

• $b$ forms a tertiary carbocation.
• $d$ forms a secondary carbocation.
• $c$ forms a secondary carbocation, slightly less stable than $d$.
• $a$ forms a primary carbocation.

Therefore, the decreasing order of dehydration is:

$$b > d > c > a$$

So, the correct option is B.

There is a discrepancy on the page because one header says D, but the worked solution repeatedly concludes $b > d > c > a$, which matches option B. Hence option B is the defensible answer.

Given: Dehydration proceeds through carbocation formation.

Find: Which alcohol dehydrates fastest to slowest.

The governing principle is:

$$\text{tertiary carbocation} > \text{secondary carbocation} > \text{primary carbocation}$$

The extracted solution states:

1. $b$ gives the most stable carbocation, so it dehydrates most easily.
2. $d$ gives a less stable carbocation than $b$, so it comes next.
3. $c$ is below $d$.
4. $a$ is least favorable because it corresponds to a primary carbocation.

Hence,

$$b > d > c > a$$

Therefore, the correct option is B.