Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Synthesis of ethyl phenyl ether may be achieved by Williamson synthesis. Reason (R) : Reaction of bromobenzene with sodium ethoxide yields ethyl phenyl ether. In the light of the above statements, choose the most appropriate answer from the options given below :

(A) is correct but (R) is not correct

Both (A) and (R) are correct and (R) is the correct explanation of (A)

(A) is not correct but (R) is correct

Both (A) and (R) are correct but (R) is not the correct explanation of (A).

Alcohols Phenols and Ethers PYQ — JEE Main 2021 · 27 Aug · Shift 1 · Q10 with Solution

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Correct answer: B

Understand the assertion

Williamson ether synthesis is used to prepare ethers by reaction of an alkoxide/phenoxide ion with a primary alkyl halide via $S_N2$ mechanism.

For ethyl phenyl ether $(\mathrm{C_6H_5OC_2H_5})$ , one valid Williamson route is: $\mathrm{C_6H_5ONa + C_2H_5Br \rightarrow C_6H_5OC_2H_5 + NaBr}$

Here, sodium phenoxide reacts with bromoethane. Since ethyl halide is a primary alkyl halide, the reaction proceeds well.

Therefore, Assertion (A) is correct.
Examine the reason

The reason says:

Reaction of bromobenzene with sodium ethoxide yields ethyl phenyl ether.

This would correspond to: $\mathrm{C_6H_5Br + C_2H_5ONa \rightarrow C_6H_5OC_2H_5 + NaBr}$

But this reaction does not occur in Williamson synthesis, because bromobenzene is an aryl halide.
Why bromobenzene does not work

Williamson synthesis proceeds by $S_N2$ attack. In aryl halides like bromobenzene:
- the carbon bonded to Br is $sp^2$ hybridized,
- backside attack required for $S_N2$ is not feasible,
- the $\mathrm{C-Br}$ bond has partial double bond character due to resonance.
Hence, sodium ethoxide cannot displace bromide from bromobenzene under normal Williamson conditions.

Therefore, Reason (R) is incorrect.
Conclusion
- Assertion (A): Correct
- Reason (R): Incorrect
So the correct option is: $\boxed{\text{B}}$
Comparison with stored answer

Stored correct answer: B

My derived answer: B

Hence, they agree.

View written solutionFree

Correct answer: B

Understand the assertion

Williamson ether synthesis is used to prepare ethers by reaction of an alkoxide/phenoxide ion with a primary alkyl halide via $S_N2$ mechanism.

For ethyl phenyl ether $(\mathrm{C_6H_5OC_2H_5})$ , one valid Williamson route is: $\mathrm{C_6H_5ONa + C_2H_5Br \rightarrow C_6H_5OC_2H_5 + NaBr}$

Here, sodium phenoxide reacts with bromoethane. Since ethyl halide is a primary alkyl halide, the reaction proceeds well.

Therefore, Assertion (A) is correct.
Examine the reason

The reason says:

Reaction of bromobenzene with sodium ethoxide yields ethyl phenyl ether.

This would correspond to: $\mathrm{C_6H_5Br + C_2H_5ONa \rightarrow C_6H_5OC_2H_5 + NaBr}$

But this reaction does not occur in Williamson synthesis, because bromobenzene is an aryl halide.
Why bromobenzene does not work

Williamson synthesis proceeds by $S_N2$ attack. In aryl halides like bromobenzene:
- the carbon bonded to Br is $sp^2$ hybridized,
- backside attack required for $S_N2$ is not feasible,
- the $\mathrm{C-Br}$ bond has partial double bond character due to resonance.
Hence, sodium ethoxide cannot displace bromide from bromobenzene under normal Williamson conditions.

Therefore, Reason (R) is incorrect.
Conclusion
- Assertion (A): Correct
- Reason (R): Incorrect
So the correct option is: $\boxed{\text{B}}$
Comparison with stored answer

Stored correct answer: B

My derived answer: B

Hence, they agree.