An alkene (A) C_{16}H_{16} on ozonolysis give | vedclass

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Question

(C) $1$. Ozonolysis of $(A)$ $C_{16}H_{16}$ gives only one product $(B)$ $C_8H_8O$. This implies $(A)$ is $C_6H_5-CH=CH-C_6H_5$ (stilbene derivative)

Vedclass · Accepted Answer

It gives Tollen's test.

Vedclass · Answer

(C) $1$. Ozonolysis of $(A)$ $C_{16}H_{16}$ gives only one product $(B)$ $C_8H_8O$. This implies $(A)$ is $C_6H_5-CH=CH-C_6H_5$ (stilbene derivative) or similar. Given $(B)$ is obtained from cyanobenzene $(C_6H_5CN)$ and $CH_3MgBr$,$(B)$ is acetophenone $(C_6H_5COCH_3)$.$2$. Acetophenone $(C_6H_5COCH_3)$ contains a methyl ketone group,so it gives the iodoform test (Option $A$ is correct).$3$. Reduction of acetophenone with $LiAlH_4$ gives $1$-phenylethanol $(C_6H_5CH(OH)CH_3)$,which is a secondary alcohol with a $CH_3CH(OH)-$ group,thus it gives the iodoform test (Option $B$ is correct).$4$. Acetophenone is a ketone,not an aldehyde,so it does not give Tollen's test (Option $C$ is incorrect).$5$. Wolff-Kishner reduction of acetophenone using $NH_2NH_2, EtO^-$ gives ethylbenzene (Option $D$ is correct).$6$. Therefore,the incorrect statement is $C$.

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