$C_{6} H_{5} {C O N H C H}_{3}$ can be converted into $C_{6} H_{5} {C H}_{2} {N H C H}_{3}$ by?

{N a B H}_{4}

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H_{2} - P d / C

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{L i A l H}_{4}

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Z n - H g / H C L

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Solution

The correct option is C $Z n - H g / H C L$
b'
$C_{6} H_{5} C O N H C H_{3} + Z n - H g / H C l \to C_{6} H_{3} C H_{2} N H C H_{3}$
This reaction is known as Clemmenson reduction.
'

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Similar questions

C_{6} H_{5} {N H}_{2}

C_{6} H_{5} {N H C H}_{3}

C_{6} H_{5} {C H}_{2} {N H}_{2}

C_{6} H_{5} {N H}_{2}

C_{6} H_{5} {N H C H}_{3}

C_{6} H_{5} {C H}_{2} {N H}_{2}

{p K}_{b}

C_{2} H_{5} N H_{2}

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C_{6} H_{5} - C H_{2} - C H = C H_{2} + H B r \to

C_{6} H_{5} - C H = C H_{2} + H B r P e r o x i d e - ---- \to

C_{6} H_{5} C O N H C H_{3}

C_{6} H_{5} C H_{2} N H C H_{3}

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