Amines Test

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Amines Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following amines can be resolved into enantiomers?

A
Trimethylamine

B
2 – pentanamine

C
3 – pentanamine

D
4 – (dimethylamino) pyridine

Solution

2-pentamine will have a chiral center, therefore it can be resolved into enantiomers.
CH₃CH₂CH₂C*HCH₃NH₂
The C* in 2-pentan amine is the chiral center.
Question 2
1 / -0

Which of the following is an example of secondary amine?

A
3 – pentanamine

B
N,N – dimethylaniline

C
methylamine

D
N – ethyl – 1 – propanamine

Solution

N of this amine is bonded to two carbon atoms, therfore it is a secondary amine.
Question 3
1 / -0

When pentanal reacts with ethylamine under conditions of acid catalysis, the major organic product is __________.

A
a nitrile

B
a hydrazone

C
an imine

D
a ketone

Solution

This will yield imine. Imines are typically prepared by the condensation of primary amines and aldehydes and less commonly ketones:
RNH₂ + R'C(O)R → RN=C(R')(R) + H₂O
Question 4
1 / -0

Amines are soluble in

A
water

B
only slightly soluble in organic solvents

C
only slightly soluble in water

D
organic solvents

Solution

Amines are soluble in water due to hydrogen bonding with water.
Question 5
1 / -0

Arrange the following compounds in order of increasing boiling point: CH₃NHCH₂CH₃; CH₃OCH₂CH₃; (CH₃)₃N and CH₃CH₂CH₂OH.

A
Low to high; CH₃OCH₂CH₃.; (CH₃)₃N; CH₃NHCH₂CH₃; CH₃CH₂CH₂OH.

B
Low to high: (CH₃)₃N ; CH₃OCH₂CH₃; CH₃NHCH₂CH₃; CH₃CH₂CH₂OH.

C
Low to high: CH₃CH₂CH₂OH; CH₃NHCH₂CH₃; (CH₃)₃N; CH₃OCH₂CH₃.

D
Low to high: CH₃NHCH₂CH₃; CH₃CH₂CH₂OH.; (CH₃)₃N; CH₃OCH₂CH₃.

Solution

This is on the basis of inter molecular interactions.
Question 6
1 / -0

When a 1° amine reacts with an alkyl sulfonyl chloride, the major organic product is __________.

A
a sulfoxide

B
a sulphonamide

C
a nitrile

D
None of these

Solution

Sulphonyl chloride reacts with the primary amine to form sulphonamide.
RNH2+RI SO2Cl → RI SO2NHR + HCl
Question 7
1 / -0

Sec – Butylamine is the common name of what compound?

A
1 – butanamine

B
2 – butanamine

C
N – methyl – 1 – propanamine

D
N – ethylethanamine

Solution

IUPAC name is 2-butanamine. CH₃CH₂CHNH₂CH₃
Question 8
1 / -0

Arrange the following in order of increasing basicity: aniline, p – nitroaniline, p – toluidine,and p – methoxyaniline

A
p – nitroaniline < aniline< p – methoxyaniline < p – toluidine

B
aniline < p – methoxyaniline p – nitroaniline < p – toluidine

C
p – nitroaniline < aniline < p – toluidine < p – methoxyaniline

D
p – methoxyaniline p – nitroaniline < aniline < p – toluidine

Solution

-OMe group at a para position will increase the basicity more than-CH₃ group at the para position. While the presence of –NO₂ at a para position will decrease the basicity.

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Amines Test - 6

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Amines Test - 6

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SHARING IS CARING

Question 1

Trimethylamine

2 – pentanamine

3 – pentanamine

4 – (dimethylamino) pyridine

Solution

Question 2

3 – pentanamine

N,N – dimethylaniline

methylamine

N – ethyl – 1 – propanamine

Solution

Question 3

a nitrile

a hydrazone

an imine

a ketone

Solution

Question 4

water

only slightly soluble in organic solvents

only slightly soluble in water

organic solvents

Solution

Question 5

Low to high; CH3OCH2CH3.​​​​​​; (CH3)3N; CH3NHCH2CH3; CH3CH2CH2OH.

Low to high: (CH3)3N ; CH3OCH2CH3; CH3NHCH2CH3; CH3CH2CH2OH.

Low to high: CH3CH2CH2OH​​​​​​; CH3NHCH2CH3; (CH3)3N; CH3OCH2CH3.​​​​​​​​​​​​​​

Low to high: CH3NHCH2CH3; CH3CH2CH2OH.​​​​​​; (CH3)3N; CH3OCH2CH3.

Solution

Question 6

a sulfoxide

a sulphonamide

a nitrile

None of these

Solution

Question 7

1 – butanamine

2 – butanamine

N – methyl – 1 – propanamine

N – ethylethanamine

Solution

Question 8

p – nitroaniline < aniline< p – methoxyaniline < p – toluidine

aniline < p – methoxyaniline p – nitroaniline < p – toluidine

p – nitroaniline < aniline < p – toluidine < p – methoxyaniline

p – methoxyaniline p – nitroaniline < aniline < p – toluidine

Solution

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Low to high; CH₃OCH₂CH₃.; (CH₃)₃N; CH₃NHCH₂CH₃; CH₃CH₂CH₂OH.

Low to high: (CH₃)₃N ; CH₃OCH₂CH₃; CH₃NHCH₂CH₃; CH₃CH₂CH₂OH.

Low to high: CH₃CH₂CH₂OH; CH₃NHCH₂CH₃; (CH₃)₃N; CH₃OCH₂CH₃.

Low to high: CH₃NHCH₂CH₃; CH₃CH₂CH₂OH.; (CH₃)₃N; CH₃OCH₂CH₃.