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What products are formed when a benzene ring containing a formyl group (–CHO) and a chloro substituent is treated with 50% KOH?

1. A benzene ring containing –CH2OH and –Cl substituents, along with a benzene ring containing –CH2COO− and –Cl substituents
2. A benzene ring containing –CH2OH and –OH substituents, along with another benzene ring containing –CH2OH and –OH substituents
3. A benzene ring containing –CH2OH and –Cl substituents, along with a benzene ring containing –COO− and –Cl substituents
4. A benzene ring containing –CH2OH and –OH substituents, along with a benzene ring containing –COO− and –OH substituents

Correct answer: A benzene ring containing –CH2OH and –Cl substituents, along with a benzene ring containing –COO− and –Cl substituents

Solution

An aromatic aldehyde with no alpha-hydrogen undergoes the Cannizzaro reaction with concentrated (50%) KOH: one molecule is reduced to the alcohol (-CH2OH) and one is oxidised to the carboxylate (-COO-). The ring-bound chlorine is not displaced, so the products are a ring bearing -CH2OH and -Cl and a ring bearing -COO- and -Cl.

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