The end product (P) of the following reaction | vedclass

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(C) The reaction involves the treatment of acetylacetone $(CH_3COCH_2COCH_3)$ with $2$ equivalents of $KNH_2$ in liquid ammonia $(NH_3(l))$. $KNH_2$ i

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(C) The reaction involves the treatment of acetylacetone $(CH_3COCH_2COCH_3)$ with $2$ equivalents of $KNH_2$ in liquid ammonia $(NH_3(l))$. $KNH_2$ is a strong base that removes both acidic protons from the central methylene group,forming a dianion: $[CH_3COCHCOCH_3]^{2-}$. This dianion is highly nucleophilic. When treated with $n-C_4H_9Br$,the more nucleophilic carbon (the terminal carbon of the enolate,which is less sterically hindered) attacks the alkyl halide via an $S_N2$ mechanism. After acidic workup with $H_3O^+$,the product formed is $CH_3COCH_2COCH_2CH_2CH_2CH_2CH_3$ (octane$-2,4-$dione substituted at the terminal position). Looking at the options,the structure corresponding to the alkylation at the terminal methyl group is $C$.

The end product $(P)$ of the following reaction is:

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