Saponification of ethyl benzoate with $N a O H$ as alkali gives :

benzyl alcohol and ethanoic acid

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sodium benzoate and ethanol

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benzoic acid and sodium ethoxide

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sodium benzoate and ethanoic acid

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Solution

The correct option is B sodium benzoate and ethanol
$C_{6} H_{5} - COO - C_{2} H_{5} NaOH - -- \to C_{6} H_{5} COONa + C_{2} H_{5} OH$

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P C l_{5}

N a O H

1.75

N a O H

25^{o} C

35^{o} C

{log}_{10} (1.75) = 0.243

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