C_{7}H_{10}O reacts with CH_{3}MgBr to give a | vedclass

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(C) The reaction of an aldehyde or ketone with $CH_{3}MgBr$ (Grignard reagent) followed by hydrolysis yields an alcohol.For the resulting alcohol to g

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(C) The reaction of an aldehyde or ketone with $CH_{3}MgBr$ (Grignard reagent) followed by hydrolysis yields an alcohol.For the resulting alcohol to give a positive iodoform test,it must contain a $CH_{3}CH(OH)-$ group.Starting with $C_{7}H_{10}O$,if $A$ is cyclohex$-3-$ene$-1-$carbaldehyde,the reaction is:$C_{6}H_{9}CHO + CH_{3}MgBr \rightarrow C_{6}H_{9}CH(OMgBr)CH_{3}$Upon hydrolysis,this gives $C_{6}H_{9}CH(OH)CH_{3}$,which is a secondary alcohol with a $CH_{3}CH(OH)-$ group.This structure gives a positive iodoform test with $I_{2}/NaOH$.Thus,the structure of $A$ is cyclohex$-3-$ene$-1-$carbaldehyde.

$C_{7}H_{10}O$ reacts with $CH_{3}MgBr$ to give a compound $C_{8}H_{12}O$ which gives a positive iodoform test. Identify the structure of $A$.

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