Organic Compounds Containing Nitrogen Test

Result

Organic Compounds Containing Nitrogen Test - 11

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Weekly Quiz Competition

Question 1
1 / -0

The correct IUPAC name for CH₂=CHCH₂NHCH₃ is

A
N-methylprop-2-en-1-amine

B
4-aminopent-1-ene

C
2-amino-4-pentene

D
allylmethylamine

Solution

The correct IUPAC name is N-methylprop-2-en-1-amine.
Question 2
1 / -0

Which of the following reagents can be used to convert primary amides into primary amines containing the same number of carbon atoms?

A
Br₂ + NaOH

B
LiAlH₄

C
Sn + HCl

D
Na + C₂H₅OH

Solution

Reduction with LiAlH₄ converts CONH₂ to CH₂NH₂ giving 1^o amines with the same number of carbon atoms as the amide. In contrast, Br₂ + NaOH will bring about Hofmann bromamide reaction giving 1^o amines with one carbon atom less than the parent amide.
Question 3
1 / -0

Gabriel phthalimide synthesis can be used for the preparation of amine from

A
CH₃CH₂Br

B
(CH₃)₃CBr

C
p-CH₃OC₆H₄Br

D
p-CH₃C₆H₄Br

Solution

By the use of Gabriel phthalimide synthesis, 1^o alkyl halides can be converted into corresponding 1^o amines.
Question 4
1 / -0

The reaction by which a primary amine is formed from a primary amide is called

A
Hofmann bromamide reaction

B
Gabriel phthalimide reaction

C
Carbylamine reaction

D
Coupling reaction

Solution

Hofmann bromamide reaction can be used to prepare 1^o amine from a 1^o amide.
Question 5
1 / -0

Which of the following amines has the highest solubility in water?

A
Methylamine

B
sec-butylamine

C
Aniline

D
Isobutylamine

Solution

Solubility decreases as the molecular mass increases.

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Organic Compounds Containing Nitrogen Test - 11

Result

Organic Compounds Containing Nitrogen Test - 11

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Question 1

N-methylprop-2-en-1-amine

4-aminopent-1-ene

2-amino-4-pentene

allylmethylamine

Solution

Question 2

Br2 + NaOH

LiAlH4

Sn + HCl

Na + C2H5OH

Solution

Question 3

CH3CH2Br

(CH3)3CBr

p-CH3OC6H4Br

p-CH3C6H4Br

Solution

Question 4

Hofmann bromamide reaction

Gabriel phthalimide reaction

Carbylamine reaction

Coupling reaction

Solution

Question 5

Methylamine

sec-butylamine

Aniline

Isobutylamine

Solution

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Br₂ + NaOH

LiAlH₄

Na + C₂H₅OH

CH₃CH₂Br

(CH₃)₃CBr

p-CH₃OC₆H₄Br

p-CH₃C₆H₄Br