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The reaction between p-methylbenzaldehyde and $\ce{NaOH}$ is an example of:
A) Cannizzaro reaction
B) Aldol condensation
C) Hydrolysis
D) Haloform reaction

I selected B, aldol condensation, as the given compound is an aldehyde and it reacts with $\ce{NaOH}$. But the given answer is A. Why is it so? The $\ce{NaOH}$ is not mentioned to be concentrated.

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The aldol condensation needs $\alpha$-hydrogens next to the carbonyl group in order to proceed. Check out this reaction equation from the Wikipedia article: aldol reaction

One of these hydrogen atoms is removed to form the enolate. See this mechanism (also from the Wikipedia article).

Aldol mechanism

Now, look at the structure of p-methylbenzaldehyde. This molecule does not have any $\alpha$-enolizable hydrogen atoms. The $\alpha$-position is an $sp^2$-hybridized carbon atom in the aromatic ring. The aldol condensation is thus not possible.

4-methylbenzaldehyde

We can also rule out the haloform reaction because there are no halogen compounds given. There are also no hydrolyzable functional groups (halides, ethers, esters, anhydrides, amides, nitriles, etc.).

The only remaining possibility is the Cannizzarro reaction, which requires a benzaldehyde derivative or other aldehyde lacking $\alpha$-enolizable hydrogen atoms.

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  • $\begingroup$ Is it possible to provide Pka of hydrogen of methyl group in paramethylbenzaldehyde? $\endgroup$ – Groverkss May 16 '19 at 19:40

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