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(A) The reaction shows that compound $M$ reacts with $CH_3MgBr$ (a Grignard reagent) to release $CH_4$ gas. This indicates that $M$ contains an acidic

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$CH_3COCH_2COCH_3$

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(A) The reaction shows that compound $M$ reacts with $CH_3MgBr$ (a Grignard reagent) to release $CH_4$ gas. This indicates that $M$ contains an acidic hydrogen atom.Upon treatment with $H^+$,the intermediate $N$ yields $CH_3COCH_2COCH_3$ (acetylacetone).Acetylacetone has an active methylene group ($-CH_2-$ between two carbonyl groups) with acidic protons.Therefore,$M$ must be $CH_3COCH_2COCH_3$ itself,as the Grignard reagent acts as a base to abstract the acidic proton from the active methylene group,forming the enolate $N$,which is then protonated back to the original ketone upon treatment with $H^+$.

In the following sequence of reactions,compound $M$ is:
$M$ $\xrightarrow{CH_3MgBr} N + CH_4 \uparrow$ $\xrightarrow{H^+} CH_3COCH_2COCH_3$

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In the following sequence of reactions,compound $M$ is:$M$ $\xrightarrow{CH_3MgBr} N + CH_4 \uparrow$ $\xrightarrow{H^+} CH_3COCH_2COCH_3$