Two moles of $H C H O$ and 1 mol of $P h C H O$ react with conc. NaOH. What are the products quantitatively?

H C O O N a

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P h C H_{2} O H

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C H_{3} O H

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C H_{3} N a (O H)_{2}

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Solution

The correct options are
B $P h C H_{2} O H$ .
C $C H_{3} O H$ .
D $H C O O N a$

As shown in the reaction above, initially there will be crossed Cannizzaro reaction between 1 mole each of $H C H O$ and $P h C H O$ . Remaining 1 mol of $H C H O$ will give Cannizzaro reaction. So $1 / 2$ mol of $H C H O$ will react with $1 / 2$ mol of $H C H O$ to give $1 / 2$ mol of $H C O O N a$ and $1 / 2$ mol of $C H_{3} O H$ . Hence, 1.5 mol of $H C O O N a$ , 1 mol of $P h C H_{2} O H$ , and $1 / 2$ mol of $C H_{3} O H$ are formed.

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