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Aldehydes form Schiff's base when they react with

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${NH}_{2} - OH$

${NH}_{2} - {NH}_{2}$

$Ph - NH - {NH}_{2}$

$R - {NH}_{2}$

answer is D.

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Detailed Solution

$- \overset{|}{C} = O + H_{2} N - R \overset{Conc . H_{2} {SO}_{4}}{\to} - \overset{|}{C} = N - R N - alkyl imine (Schiff' s base)$

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