Animated Solution for Chemistry - Organic Chemistry: The reaction sequence(s) that would lead to o-xylene as the major product is (are) [JEE(Advanced) 2021]

This is a classic, multi-concept organic chemistry problem from JEE Advanced that tests your mastery over functional group interconversions. Our ultimate goal is to identify which of the given reaction sequences successfully synthesizes $o$-xylene (1,2-dimethylbenzene). Let's embark on a thrilling journey through each option, dissecting the mechanisms step-by-step.

The Diazotization Route

We begin with $o$-methylaniline. The first reagent is a mixture of ${\text{NaNO}}_2$ and $\text{HCl}$ at $273\text{ K}$. This is the unmistakable signature of a diazotization reaction. The primary aromatic amine is converted into a highly reactive diazonium salt, ${\text{Ar-N}}_2^{+}{\text{Cl}}^{-}$.

Next, we introduce $\text{CuCN}$. This triggers the Sandmeyer reaction, a radical-mediated process where the diazonium group is elegantly replaced by a cyanide group, yielding $o$-methylbenzonitrile.

Now, we need to reduce this nitrile. Enter DIBAL-H (Diisobutylaluminium hydride), a bulky and mild reducing agent. It carefully reduces the nitrile to an imine intermediate, which upon acidic hydrolysis (${\text{H}}_3{\text{O}}^{+}$) gives $o$-tolualdehyde.

Finally, we deploy the Wolff-Kishner reduction using hydrazine (${\text{N}}_2{\text{H}}_4$) and $\text{KOH}$ under heat. This powerful reaction strips the oxygen entirely, reducing the aldehyde group down to a simple methyl group. The result? A benzene ring with two adjacent methyl groups. We have successfully synthesized $o$-xylene! Option A is correct.

The Grignard Approach

Option B starts with $o$-bromotoluene. Reacting it with magnesium metal in dry ether forms a Grignard reagent, $\text{Ar-MgBr}$. This nucleophilic carbon then attacks solid ${\text{CO}}_2$ (dry ice). Following an acidic workup, the Grignard reagent is carboxylated, giving us $o$-toluic acid.

To move forward, we treat the carboxylic acid with thionyl chloride (${\text{SOCl}}_2$), converting it into a more reactive acid chloride, $o$-toluoyl chloride.

This sets the stage for the Rosenmund reduction. By using hydrogen gas over a poisoned palladium catalyst (${\text{Pd-BaSO}}_4$), we selectively reduce the acid chloride to an aldehyde without over-reducing it to an alcohol. We now have $o$-tolualdehyde, the exact same intermediate we saw in Option A!

To finish the job, we use zinc amalgam and concentrated $\text{HCl}$—the classic Clemmensen reduction. Just like the Wolff-Kishner reduction, it completely deoxygenates the aldehyde, leaving us with a methyl group. Once again, the final product is $o$-xylene. Option B is also correct.

The Hydroboration Trap

Here, we start with $o$-methylstyrene, which features a vinyl group attached to the ring. The first step is hydroboration-oxidation (${\text{BH}}_3$ followed by ${\text{H}}_2{\text{O}}_2/\text{NaOH}$). This reaction adds water across the double bond in an anti-Markovnikov fashion. The hydroxyl group attaches to the less substituted terminal carbon, forming a primary alcohol: $2-(o-\text{tolyl})\text{ethanol}$.

Next, ${\text{PBr}}_3$ is used to substitute the hydroxyl group with a bromine atom via an ${\text{S}}_{\text{N}}2$ mechanism. Finally, zinc dust and dilute $\text{HCl}$ reduce this primary alkyl bromide to an alkane.

Let's count the carbons in the side chain. We started with a two-carbon vinyl group, and we ended with a two-carbon ethyl group. The final product is $o$-ethyltoluene, not $o$-xylene. Option C is incorrect.

The Ozonolysis Cleavage

Option D presents us with indene, a bicyclic molecule containing a benzene ring fused to a cyclopentene ring. The first step is reductive ozonolysis (${\text{O}}_3$ followed by $\text{Zn}/{\text{H}}_2\text{O}$). Ozone acts like molecular scissors, cleaving the double bond in the five-membered ring.

Because the double bond is between the two carbons not shared with the benzene ring, the ring opens up. The carbon attached directly to the benzene ring becomes an aldehyde ($\text{-CHO}$), and the other carbon, which is part of a two-carbon chain attached to the adjacent position, also becomes an aldehyde (${\text{-CH}}_2\text{-CHO}$).

Finally, we apply the Wolff-Kishner reduction again. It reduces both aldehyde groups to methyl groups. The $\text{-CHO}$ directly on the ring becomes a ${\text{-CH}}_3$, and the ${\text{-CH}}_2\text{-CHO}$ chain becomes an ethyl group (${\text{-CH}}_2{\text{-CH}}_3$). The final product is, once again, $o$-ethyltoluene. Option D is incorrect.

In conclusion, only the reaction sequences in (A) and (B) successfully yield $o$-xylene as the major product.