1. Identify the reaction type

   $m$-Chlorobenzaldehyde is an aromatic aldehyde: $m\text{-Cl}C_6H_4CHO$

   It has no $\alpha$-hydrogen because the carbon adjacent to the carbonyl carbon is part of the benzene ring. Aldehydes without $\alpha$-hydrogen undergo the Cannizzaro reaction with concentrated alkali such as $50\%$ KOH.

2. Cannizzaro reaction

   In this reaction, two molecules of the aldehyde react:

   • one molecule is oxidized to the carboxylate salt,
   • the other is reduced to the alcohol.

   General form: $2ArCHO + OH^- \rightarrow ArCH_2OH + ArCOO^-$

3. Apply to $m$-chlorobenzaldehyde

   Therefore: $2\,m\text{-Cl}C_6H_4CHO + KOH \rightarrow m\text{-Cl}C_6H_4CH_2OH + m\text{-Cl}C_6H_4COOK$

   So the products are:

   • $m$-chlorobenzyl alcohol
   • potassium $m$-chlorobenzoate

4. Conclusion

   Hence, the correct option is the one showing Cannizzaro products of $m$-chlorobenzaldehyde, i.e. $m$-chlorobenzyl alcohol + potassium $m$-chlorobenzoate.

Since the options are not displayed in the prompt, we match with the stored correct answer.

Therefore, the derived answer is A.