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(B) Sodium bisulphite $(NaHSO_3)$ reacts with all aldehydes and most methyl ketones to form white crystalline addition products. The reaction is sensi

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$3$

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(B) Sodium bisulphite $(NaHSO_3)$ reacts with all aldehydes and most methyl ketones to form white crystalline addition products. The reaction is sensitive to steric hindrance. $1. HCHO$ (Formaldehyde): Reacts. $2. CH_3CHO$ (Acetaldehyde): Reacts. $3. CH_3COCH_3$ (Acetone): Reacts. $4. C_6H_5COCH_3$ (Acetophenone): Does not react significantly due to steric hindrance from the phenyl group. $5. C_6H_5COC_6H_5$ (Benzophenone): Does not react due to significant steric hindrance from two phenyl groups. Thus,only $3$ compounds $(HCHO, CH_3CHO, CH_3COCH_3)$ form white crystalline products with $NaHSO_3$.

How many compounds among the following give white crystals when reacted with sodium bisulphite $(NaHSO_3)$?
$1. HCHO$
$2. CH_3CHO$
$3. CH_3COCH_3$
$4. C_6H_5COCH_3$
$5. C_6H_5COC_6H_5$

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How many compounds among the following give white crystals when reacted with sodium bisulphite $(NaHSO_3)$? $1. HCHO$ $2. CH_3CHO$ $3. CH_3COCH_3$ $4. C_6H_5COCH_3$ $5. C_6H_5COC_6H_5$