Animated Solution for Chemistry - Organic Chemistry: 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

Williamson Synthesis and the Aryl Halide Trap

When it comes to synthesizing ethers, the Williamson Ether Synthesis is the undisputed king of the laboratory. It is an elegant, straightforward, and highly reliable method for creating both symmetrical and asymmetrical ethers. But like any powerful tool, it comes with a strict set of rules. Let's dive into the mechanics of this reaction and uncover why certain chemical combinations work flawlessly while others fail spectacularly.

Decoding the $S_{N}2$ Mechanism

At its core, the Williamson synthesis is a classic $S_{N}2$ (Substitution Nucleophilic Bimolecular) reaction. The recipe requires two key ingredients: a strong nucleophile and a suitable electrophile.

The nucleophile is typically an alkoxide or phenoxide ion ($R-O^{-}$), generated by reacting an alcohol or phenol with a strong base like sodium metal or sodium hydride. The electrophile is an alkyl halide ($R^{\prime }-X$). The negatively charged oxygen atom hunts down the partially positive carbon atom of the alkyl halide, attacking it from the backside. In one smooth, concerted motion, the carbon-oxygen bond forms while the carbon-halogen bond breaks, kicking out the halide ion as a leaving group.

The Perfect Match

Let's evaluate the first statement of our problem: synthesizing ethyl phenyl ether. To build this specific molecule, we need a phenyl group ($C_6{H}_5-$) and an ethyl group ($-C_2{H}_5$) connected by an oxygen bridge.

If we choose sodium phenoxide ($C_6{H}_5{O}^{-}N{a}^{+}$) as our nucleophile and ethyl bromide ($C_2{H}_{5}Br$) as our electrophile, we have a match made in heaven.

Why does this work so well? Ethyl bromide is a primary alkyl halide. In the $S_{N}2$ mechanism, steric hindrance is the ultimate enemy. Because the primary carbon in ethyl bromide is relatively uncrowded, the bulky phenoxide ion can easily approach and execute the backside attack. The reaction proceeds smoothly, yielding ethyl phenyl ether. Therefore, the assertion that this ether can be achieved via Williamson synthesis is absolutely correct.

Why Bromobenzene Refuses to React

Now, let's flip the script and examine the alternative route proposed in the Reason statement. What if we try to build the same ether by reacting bromobenzene ($C_6{H}_{5}Br$) with sodium ethoxide ($C_2{H}_5{O}^{-}N{a}^{+}$)?

At first glance, it seems logical. We still have an alkoxide and a halide. However, this reaction will completely fail. The culprit lies within the structure of bromobenzene itself. Bromobenzene is an aryl halide, meaning the halogen is attached directly to an $s{p}^2$ hybridized carbon of an aromatic ring.

The bromine atom possesses lone pairs of electrons. Because it is directly attached to the conjugated $\pi$-system of the benzene ring, these lone pairs participate in resonance. They delocalize into the ring, creating a resonance hybrid where the carbon-bromine bond acquires a partial double bond character.

This partial double bond is significantly shorter and stronger than a standard single bond. When the ethoxide nucleophile attempts its $S_{N}2$ backside attack, it hits a brick wall. The carbon-bromine bond is simply too strong to be broken under normal conditions. Furthermore, the electron-rich $\pi$-cloud of the benzene ring repels the incoming negatively charged nucleophile, and the backside of the $s{p}^2$ carbon is sterically blocked by the ring itself.

Because of these insurmountable electronic and steric barriers, aryl halides (and vinyl halides) are virtually inert to $S_{N}2$ nucleophilic substitution.

The attempt to synthesize ethyl phenyl ether using bromobenzene and sodium ethoxide will result in no reaction. Therefore, the Reason statement is factually incorrect. We are left with a true Assertion and a false Reason, making option (b) the correct choice. The golden rule of Williamson synthesis remains unbroken: always choose the least sterically hindered alkyl halide, and never rely on an aryl halide as your electrophile.