A (C_3H_6O) xrightarrow{NaCN/H_2SO_4} cyanohy | vedclass

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Question

(A) The molecular formula $C_3H_6O$ corresponds to either an aldehyde or a ketone. The reaction with $NaCN/H_2SO_4$ is the formation of a cyanohydrin

Vedclass · Accepted Answer

Propanal $(CH_3CH_2CHO)$

Vedclass · Answer

(A) The molecular formula $C_3H_6O$ corresponds to either an aldehyde or a ketone. The reaction with $NaCN/H_2SO_4$ is the formation of a cyanohydrin from a carbonyl compound.$1$. Propanal $(CH_3CH_2CHO)$ reacts with $HCN$ to form $CH_3CH_2CH(OH)CN$. This molecule has a chiral carbon atom (the carbon attached to $-OH$,$-CN$,$-H$,and $-CH_2CH_3$),making it optically active.$2$. Propanone $(CH_3COCH_3)$ reacts with $HCN$ to form $CH_3C(OH)(CN)CH_3$. This molecule is symmetric and achiral,so it is optically inactive.$3$. Cyclopropanol and oxetane are not carbonyl compounds and do not form cyanohydrins under these conditions.Therefore,the correct structure for $A$ is propanal.

Column-$I$ (Example)	Column-$II$ (Reaction)
$(A)$ $CH_3COCl + H_2 \xrightarrow{Pd-C/BaSO_4} CH_3CHO$	$(1)$ Friedel-Crafts acylation
$(B)$ $C_6H_5CHO \xrightarrow{NaOH} C_6H_5CH_2OH + C_6H_5COONa$	$(2)$ $HVZ$ reaction
$(C)$ $C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_6H_5COCH_3$	$(3)$ Aldol condensation
$(D)$ $R-CH_2COOH \xrightarrow{Br_2/Red P} R-CH(Br)COOH$	$(4)$ Cannizzaro's reaction
$(E)$ $CH_3-CN \xrightarrow{SnCl_2/HCl, H_2O/H^+} CH_3-CHO$	$(5)$ Rosenmund's reduction
$(F)$ $2CH_3CHO \xrightarrow{NaOH} CH_3-CH=CH-CHO$	$(6)$ Stephen's reaction

$A$ $(C_3H_6O)$ $\xrightarrow{NaCN/H_2SO_4}$ cyanohydrin of $A$. If the cyanohydrin is optically active,then the possible structure of $A$ is:

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