Amines Test

Result

Amines Test - 56

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Weekly Quiz Competition

Question 1
1 / -0

The correct order of basicity of the following compounds is:

A
$$1 < 2 < 3 < 4$$

B
$$1 < 2 < 4 < 3$$

C
$$2 < 1 < 3 < 4$$

D
$$4 < 3 < 2 < 1$$

Solution

The correct order of basicity is:
$$2 < 1 < 3 < 4$$

The basic strength order depends on:
(i) Accumulation of negative charge on $$N$$ (double bonded) by another $$-NH_2 $$group, thus intensifying the donor ability of $$N$$.

(ii) The higher donor ability of $$sp^3$$ hybrid $$N$$ as compared to $$sp^2\ N$$.

Option C is correct.
Question 2
1 / -0

Which of the following is true?

A
Aniline is less basic than ethyl amine due to resonance effect.

B
Aniline is less basic than ethyl amine due to inductive effect.

C
Aniline is more basic than ethyl amine due to resonance effect.

D
Aniline is more basic than ethyl amine due to inductive effect.

Solution

Aniline is less basic than ethyl amine due to resonance effect.

In aniline, the $$-NH_2$$ group is directly attached to the benzene ring. The lone pair of electrons on N atom is in conjugation with benzene ring.

Hence, it is less available for protonation because of resonance.

Option A is correct.
Question 3
1 / -0

Which of the following is true?

A
Aralkylamines are more basic than arylamines. Benzylamine is more basic than aniline.

B
Aralkylamines are less basic than arylamines. Benzylamine is less basic than aniline.

C
Aralkylamines are more basic than arylamines. But, benzylamine is less basic than aniline.

D
Aralkylamines are less basic than arylamines. But, benzylamine is more basic than aniline.

Solution

Aralkylamines are more basic than arylamines. Benzylamine is more basic than aniline.

The lone pair of electrons on N atom in arylamines is delocalised as it is in conjugation with benzene ring.

However in case of aralkylamines, the lone pair of electrons on N atom in arylamines is not delocalised as it is not in conjugation with benzene ring.

Hence, the lone pair of electrons on N atom in aralkylamines is more readily available for protonation than the lone pair of electrons on N atom in arylamines.

Option A is correct.
Question 4
1 / -0

The correct basicity order of the following compound is :

A
$$I > III > II > IV$$

B
$$III > I > II > IV$$

C
$$II > I > III > IV$$

D
$$I > II > III > IV$$
Question 5
1 / -0

Which of the following is the correct order of the solubility in water?

A
propane $$<$$ ethylamine $$<$$ ethyl alcohol

B
propane $$<$$ ethyl alcohol $$<$$ ethylamine

C
ethyl alcohol $$<$$ propane $$<$$ ethylamine

D
ethyl alcohol $$<$$ ethylamine $$<$$ propane
Solution
Correct Answer: Option A
Explanation:
- We know, like dissolves like. So, the polar compounds are soluble in the polar solvents. And the non-polar solvents are soluble in non-polar solvents.
- Water is a polar solvent.
- The order of polarity of the given compounds is:
$$propane < ethylamine < ethyl\ alcohol$$
- So, the order of solubility in the water is:
$$propane < ethylamine < ethyl\ alcohol$$
Question 6
1 / -0

Benzylamine is a stronger base than aniline because:

A
the lone pair of electrons on the nitrogen atom in benzylamine is delocalised

B
the lone pair of electrons on the nitrogen atom in aniline is delocalised

C
the lone pair of electrons on the nitrogen atom in aniline is not involved in resonance

D
benzylamine has a higher molecular mass than aniline

Solution

Benzylamine is a stronger base than aniline because the lone pair of electrons on the nitrogen atom in aniline is delocalised over benzene ring.

Hence, due to resonance in aniline, the lone pair of electron on nitrogen is less available for donation to suitable acid.

However in case of benzylamine ($$C_6H_5-CH_2-NH_2$$), the N atom is attached to saturated C atom.

Hence, no resonance is possible and the lone pair of electron can be easily donated.
Question 7
1 / -0

Which of the following represents the correct order of basic strength?

A
N,N-Dimethylaniline $$>$$ N-methylaniline $$>$$ aniline

B
N,N-Dimethylaniline $$>$$ aniline $$>$$ N-methylaniline

C
aniline $$>$$ N-methylaniline $$>$$ N,N-Dimethyl aniline

D
N-methylaniline $$>$$ N,N-Dimethylaniline $$>$$ aniline

Solution

The following represents the correct order of basic strength
N,N-Dimethylaniline $$>$$ N-methylaniline $$>$$ aniline
When hydrogen atoms of the amino group of arylamines are replaced by electron donating alkyl groups, the basic character of the resulting amine increases.
Question 8
1 / -0

Gabriel synthesis is used for synthesis of:

A
primary amines

B
secondary amines

C
aldehydes

D
acids

Solution

Gabriel synthesis is a reaction that converts primary alkyl halides into primary amines. This reaction uses potassium phthalimide. This reaction is named after the German chemist Siegmund Gabriel.
Hence option A is correct.
Question 9
1 / -0

Gabriel phthalimide synthesis is used in the preparation of primary amine from phthalimide which of the following reagent is not used during the process?

A
HCl

B
NaOH

C
Alkyl Halides

D
KOH

Solution

Gabriel phthalimide synthesis is used in the preparation of primary amine from phthalimide. HCl is not used during the process.

Hence the correct option is A.
Question 10
1 / -0

Which one of the following will be most basic?

A
Aniline

B
P-methoxyaniline

C
P-nitroaniline

D
Benzylamine

Solution

Here, option A, B, and C all are aromatic amines in which the lone pair on nitrogen involved in ring resonance while option D is aliphatic amine so it is most basic.