Consider the following reactions. What is A?| | vedclass

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(B) $1$. $B$ gives a yellow precipitate with $I_2 + NaOH$ (iodoform test) and a silver mirror with $Ag_2O$ (Tollens' test). This indicates that $B$ is

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$3-$ethylpent$-2-$ene

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(B) $1$. $B$ gives a yellow precipitate with $I_2 + NaOH$ (iodoform test) and a silver mirror with $Ag_2O$ (Tollens' test). This indicates that $B$ is an aldehyde containing a $CH_3CO-$ group. The simplest such aldehyde is acetaldehyde $(CH_3CHO)$.$2$. $C$ does not give a yellow precipitate with $I_2 + NaOH$,but upon reduction with $LiAlH_4$,it gives $D$,which forms white turbidity with Lucas reagent $(ZnCl_2 + conc. HCl)$ within $5$ minutes. This indicates $D$ is a secondary alcohol. Thus,$C$ must be a ketone,specifically diethyl ketone $(CH_3CH_2COCH_2CH_3)$.$3$. The ozonolysis of $A$ $(C_7H_{14})$ yields $B$ $(CH_3CHO)$ and $C$ $(CH_3CH_2COCH_2CH_3)$.$4$. Reconstructing $A$ from the products: $CH_3CH=O + O=C(CH_2CH_3)_2 \rightarrow CH_3CH=C(CH_2CH_3)_2$.$5$. The structure $CH_3CH=C(CH_2CH_3)_2$ is $3$-ethylpent-$2$-ene.

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