The best method for preparation of ${M e}_{3} C C N$ is:

to react

{M e}_{3} C O H

with

H C N

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to react

{M e}_{3} C B r

with

N a C N

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to react

{M e}_{3} C M g B r

with

C l C N

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to react

{M e}_{3} C L i

with

{N H}_{2} C N

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Solution

The correct option is C to react ${M e}_{3} C M g B r$ with $C l C N$
The reaction is a $S_{N} 2$ reaction. Here, one bond is broken and another bond is formed synchronously.

$M e_{3} C - M g B r + C l - C N \to M e_{3} C - C N + M g (C l) B r$

So, the correct option is $C$

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H C N ?

H C N

B a (C N)_{2}

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