The given alkene undergoes ozonolysis: 2text{ | vedclass

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(D) The starting material is $2	ext{-methylbut-2-ene}$ $(CH_3-C(CH_3)=CH-CH_3)$.Ozonolysis of this alkene yields $X$ ($CH_3CHO$,acetaldehyde) and $Y$

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$X$ shows Cannizzaro's reaction but $Y$ does not

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(D) The starting material is $2	ext{-methylbut-2-ene}$ $(CH_3-C(CH_3)=CH-CH_3)$.Ozonolysis of this alkene yields $X$ ($CH_3CHO$,acetaldehyde) and $Y$ ($CH_3COCH_3$,acetone).Reduction of $X$ $(CH_3CHO)$ with $LiAlH_4$ gives $P$ ($CH_3CH_2OH$,ethanol,a primary alcohol).Reduction of $Y$ $(CH_3COCH_3)$ with $LiAlH_4$ gives $Q$ ($CH_3CH(OH)CH_3$,propan$-2-$ol,a secondary alcohol).Evaluating the options:$(A)$ Both $CH_3CHO$ and $CH_3COCH_3$ contain the $CH_3CO-$ group,so both give a positive iodoform test. (Correct)$(B)$ $P$ is a primary alcohol and $Q$ is a secondary alcohol. They can be distinguished by the Victor Meyer test. (Correct)$(C)$ $P$ $(C_2H_6O)$ and $Q$ $(C_3H_8O)$ differ by a $-CH_2-$ group,so they are homologues. (Correct)$(D)$ $X$ $(CH_3CHO)$ does not have an $\alpha$-hydrogen,so it does not show the Cannizzaro reaction. Wait,$CH_3CHO$ has $\alpha$-hydrogens,so it undergoes aldol condensation,not Cannizzaro. $Y$ $(CH_3COCH_3)$ also has $\alpha$-hydrogens. The statement that $X$ shows Cannizzaro is incorrect.

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