An organic compound X with molecular formula | vedclass

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Question

(A) $1$. The molecular formula of $X$ is $C_{11}H_{14}$.$2$. Ozonolysis of $X$ gives $P$ and $Q$. $P$ gives a positive iodoform test $(I_2/NaOH)$ and

Vedclass · Accepted Answer

$Ph-C(CH_3)=CH-CH_2-CH_3$

Vedclass · Answer

(A) $1$. The molecular formula of $X$ is $C_{11}H_{14}$.$2$. Ozonolysis of $X$ gives $P$ and $Q$. $P$ gives a positive iodoform test $(I_2/NaOH)$ and does not reduce Tollen's reagent,identifying $P$ as a methyl ketone (specifically acetophenone,$Ph-CO-CH_3$).$3$. $Q$ gives a positive Fehling's test but no iodoform test,identifying $Q$ as an aliphatic aldehyde (specifically propanal,$CH_3CH_2CHO$).$4$. Combining these fragments ($Ph-CO-CH_3$ and $CH_3CH_2CHO$) via ozonolysis reversal,the structure of $X$ is $Ph-C(CH_3)=CH-CH_2-CH_3$.$5$. Hydrogenation of $X$ $(Ph-C(CH_3)=CH-CH_2-CH_3 + H_2 \rightarrow Ph-CH(CH_3)-CH_2-CH_2-CH_3)$ produces a chiral center at the carbon attached to the phenyl group,making the product optically active.

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