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Diborane

C 6 H 5 CONHC...

Question

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

A

N a B H_{4}

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B

L i A l H_{4}

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C

H_{2} - P d / C

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D

Z n - H g / H C l

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Solution

The correct option is B $L i A l H_{4}$
Here we need to convert an amide to a secondary amine.

$N a B H_{4}$ is used to convert ketones and aldehydes to alcohols

Hydrogenation can reduce amides to amines, but only at extreme conditions of pressure and temperature

Clemmensen reduction reduces ketones in acidic conditions

$L i A l H_{4}$ is suitable for the reduction of amide to amine.

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Similar questions

Q. Which of the comparison regarding

Z n, C d, H g

is/are correct?

Z n {C l}_{2}

is ionic whereas

C d {C l}_{2}

H g {C l}_{2}

Z n

C d

dissolves in dilute acid

(H C l)

H_{2}

H g

Z n

C d

forms precipitate of

Z n (O H)_{2}

C d {(O H)}_{2}

H g

forms coloured precipitate of

H g (O H)_{2}

A_{2}^{2 +}

Q. A compound C

_{5}

_{10}

O does not reduce Fehling's solution, form a phenyl hydrazone, shows the haloform reaction, and can be converted into n-pentane by Zn - Hg and conc. HCl. What is this compound ?

H_{2} S

S O_{2}

can be distinguished by

H_{3} C - C O C l H_{2} / P d - B a S O_{4} - -------- \to A Z n - H g / H C l - ------- \to B

What are A and B?

A) Grignard reagent	1) $H_{2} / P d - B a S O_{4}$
B) Clemmensen reduction	2) $N_{2} H_{4} / K O H / C H_{2} O H - C H_{2} O H$
C) Rosenmunds reduction	3) $C H_{3} M g X$
D) Wolf-Kishner reduction	4) $Z n - H g / C o n c . H C l$ 5) $H_{2} / N i$

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