Ph-CH_2-Cequiv N xrightarrow[THF]{LDA} xright | vedclass

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Question

(C) The reaction involves the use of $LDA$ (Lithium Di-isopropylamide),which is a strong,sterically hindered base. $LDA$ abstracts the acidic $\alpha$

Vedclass · Accepted Answer

$Ph-CH(CH_3)-C\equiv N$

Vedclass · Answer

(C) The reaction involves the use of $LDA$ (Lithium Di-isopropylamide),which is a strong,sterically hindered base. $LDA$ abstracts the acidic $\alpha$-hydrogen from the phenylacetonitrile $(Ph-CH_2-C\equiv N)$ to form a resonance-stabilized carbanion: $Ph-CH^--C\equiv N \leftrightarrow Ph-CH=C=N^-$. This carbanion then acts as a nucleophile and attacks the methyl iodide $(CH_3I)$ in an $S_N2$ reaction. The resulting product is $Ph-CH(CH_3)-C\equiv N$.

$Ph-CH_2-C\equiv N$ $\xrightarrow[THF]{LDA}$ $\xrightarrow{CH_3I} 71\%;$ The end product of this reaction will be:

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