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Standard XII

Chemistry

Nomenclature of Alcohols

Reaction of a...

Question

Reaction of acetamide with bromine water and $K O H$ gives:

{C H}_{3} C O O H

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{C H}_{3} {C H}_{2} {N H}_{2}

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{C H}_{3} {C O O N H}_{4}

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{C H}_{3} {N H}_{2}

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Solution

The correct option is D ${C H}_{3} {N H}_{2}$
Acetamide reacts with bromine and aqueous or alcoholic $K O H$ to form methyl amine.

$C H_{3} C O N H_{2} + B r_{2} + 4 K O H \to C H_{3} N H_{2} + K_{2} C O_{3} + 2 K B r + 2 H_{2} O$

This is Hoffmann bromamide degradation reaction. The amine formed contains one carbon atom less than amide.

Option D is correct.

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Similar questions

Q. What is the decreasing order of strength of the bases

O H^{-}, N H_{2}^{-}, H C \equiv C^{⊝}

and

C H_{3} - C H_{2}^{⊝}

Q. In the given example B,C and B,D are :

(a)

C H_{3} - C H_{2} - C H_{2} - C H_{2} N H_{2}

Butanamine

(1^{o})

(b)

C H_{3} - C H_{2} - N H_{2} | C H - C H_{3}

Butan-2-amine

(1^{o})

(c)

C H_{3} - C H_{3} | C H - C H_{2} - N H_{2}

2-Methylpropanamine

(1^{o})

(d)

C H_{3} - C H_{3} | C | - C H_{3} N H_{2}

2-methylpropan-2-amine

(1^{o})

(e)

C H_{3} - C H_{2} - C H_{2} - N H - C H_{3}

N-methylpropanamine

(2^{o})

Q. In the given compound A,C and A,D are exbhits which of the following isomerism :

(a)

C H_{3} - C H_{2} - C H_{2} - C H_{2} N H_{2}

Butanamine

(1^{o})

(b)

C H_{3} - C H_{2} - N H_{2} | C H - C H_{3}

Butan-2-amine

(1^{o})

(c)

C H_{3} - C H_{3} | C H - C H_{2} - N H_{2}

2-Methylpropanamine

(1^{o})

(d)

C H_{3} - C H_{3} | C | - C H_{3} N H_{2}

2-methylpropan-2-amine

(1^{o})

(e)

C H_{3} - C H_{2} - C H_{2} - N H - C H_{3}

N-methylpropanamine

(2^{o})

C H_{3} - N H | | C (1) - N H_{2}

C H_{3} - C H_{2} (2) - N H_{2}

(C H_{3})_{2} (3) N H

C H_{3} - O | | C (4) - N H_{2}

The correct order of the basicities of the following compounds is:

Q. Write a chemical equations for the following conversions:
i)

C H_{3} - C H_{2} - C I

into

C H_{3} - C H_{2} - C H_{2} - N H_{2}

ii)

C_{6} H_{5} - C H_{2} - C I

into

C_{6} H_{5} - C H_{2} - C H_{2} - N H_{2}

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Reaction of acetamide with bromine water and KOH gives:

The correct option is D CH3NH2Acetamide reacts with bromine and aqueous or alcoholic KOH to form methyl amine.CH3CONH2+Br2+4KOH→CH3NH2+K2CO3+2KBr+2H2OThis is Hoffmann bromamide degradation reaction. The amine formed contains one carbon atom less than amide.Option D is correct.

Reaction of acetamide with bromine water and $K O H$ gives: