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.

We begin by recalling the general principle of the Williamson ether synthesis. The statement of the method is: An ether can be prepared by the reaction of an alkoxide ion with a primary alkyl halide under S}_\mathrm{N}2 \text{ conditions. Symbolically, $$\mathrm{R'O^-Na^+ \;+\; R''-X \;\longrightarrow\; R'O\!\!-\!R'' \;+\; NaX}$$ where $$X$$ is generally $$\mathrm{Cl,\, Br}$$ or $$\mathrm{I}$$ and the $$R''$$ group should be such that it can undergo the back-side attack required for the $$S_\mathrm{N}2$$ mechanism.

Now we look at the target molecule ethyl phenyl ether, also called phenetole, having the structure $$\mathrm{C_6H_5-O-CH_2CH_3}.$$ To obtain this by Williamson synthesis, we choose the pair of reactants so that one provides the $$\mathrm{O^-}$$ and the other provides the alkyl group. A convenient combination is:

$$\mathrm{C_6H_5O^-Na^+ \;+\; BrCH_2CH_3 \;\longrightarrow\; C_6H_5OCH_2CH_3 \;+\; NaBr}$$

Here $$\mathrm{C_6H_5O^-}$$ is the phenoxide ion, and $$\mathrm{BrCH_2CH_3}$$ (ethyl bromide) is a primary alkyl halide, perfectly suited for the $$S_\mathrm{N}2$$ reaction. Every requirement of the Williamson synthesis is satisfied, so the preparation is indeed feasible. Thus, the Assertion (A) that ethyl phenyl ether may be achieved by Williamson synthesis is correct.

Next, we examine the Reason (R): “Reaction of bromobenzene with sodium ethoxide yields ethyl phenyl ether.” Writing the suggested reaction, we would have

$$\mathrm{C_6H_5Br \;+\; Na^+O^-CH_2CH_3 \;\longrightarrow\; ?}$$

For this to give the desired ether, the ethoxide ion would have to displace the bromide ion directly from the aromatic ring of bromobenzene. However, an aryl halide such as $$\mathrm{C_6H_5Br}$$ cannot undergo the $$S_\mathrm{N}2$$ mechanism because the carbon bearing the halogen is part of a rigid, conjugated, planar $$\mathrm{sp^2}$$ system. The back-side attack required in $$S_\mathrm{N}2$$ is sterically and electronically impossible on such a substrate. Therefore the ethoxide ion does not replace bromide, and no ether is formed under ordinary Williamson conditions. Hence the Reason (R) is false.

Summarising, we have:

• Assertion (A): correct.
• Reason (R): incorrect.

Therefore the option that matches this evaluation is Option B.

Hence, the correct answer is Option B.