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 $C$.

Reaction	Nature of reaction
$A. RBr \xrightarrow{Na}$	$1. \text{Partial reduction}$
$B. RCOO^-Na^+ \xrightarrow{NaOH/CaO, \Delta}$	$2. \text{Electrophilic addition}$
$C. R-C \equiv C-R' \xrightarrow{\text{Lindlar's catalyst } (H_2)}$	$3. \text{Wurtz reaction}$
$D. H_2C=CH_2 \xrightarrow{Br_2/CCl_4}$	$4. \text{Decarboxylation}$
	$5. \text{Decarboxylative dimerisation}$

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

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