Chemistry Test

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Chemistry Test - 19

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Weekly Quiz Competition

Question 1
5 / -1

A primary amine on treatment with carbon disulphide and HgCI₂ produces black ppt and

A
alkyl cyanide

B
alkyl isothiocyanate

C
alkyl cyanate

D
carbylamine

Solution

Explanation:
The reaction is the Hoffman mustard oil reaction
Now Di thio carbonic acid is react with HgCl₂ to produce Alkyl isothiocyanate.
∴ The product is alkyl isothiocyanate
Question 2
5 / -1

An isocyanide on hydrolysis gives

A
N - substituted amide

B
1° amine and formic acid

C
carboxylic acid and ammonia

D
an amide

Solution

Explanation:
⇒ \(R-N\equiv C + H_2O → R-NH_2 + HCOOH\)
→ An isocyanide, on hydrolysis, the triple bond will break and form 1° amine and formic acid.
Question 3
5 / -1

Maximum basic in gas phase is

A
NH₃

B
CH₃CH₂NH₂

C
(CH₃CH₂)₂NH

D
(CH₃CH₂)₃N

Solution

Explanation:
In the Gaseous phase, Basicity depends on carbocation
the more stable the carbonation the more basic the compound is
⇒ Basic Strength ∝ + I Effect & Alkyl group (- R) show + I effect
So in (CH3CH2)3N - (3 ° Amine), it has more alkyl group
⇒ Stability of 3 °> 2° carbonation > 1° > NH₃
Given compounds,
⇒ NH3, CH3CH2NH2 (1 ° Amine), (CH3CH2)2NH (2 ° Amine), (CH3CH2)3N (3 ° Amine)
∴ (CH3CH2)3N is the most stable.
Question 4
5 / -1

Among the following amines, the strongest Bronsted base is ______

A

B
NH₃

C

D

Solution

Explanation:
The substance which accepts a proton from the other substance is called the Bronsted base.
Option 1:
→ The lone pair on N is not delocalized as it is part of the resonance and hence it is not a bronsted base.
Option 2:
→ The lone pair on N is localized and hence it is a bronsted base. when protonated forms NH₄ and not stabilized easily.
Option 3:
→ The lone pair is delocalized and part of aromaticity hence it is not a bronsted base.
Option 4:
→ The lone pair is localized easily and hence it is a strong bronsted base. and is stabilized easily.
Question 5
5 / -1

Reduction of aromatic nitro compounds using Fe and HCI gives ______

A
aromatic oxime

B
aromatic hydrocarbon

C
aromatic primary amine

D
aromatic amide

Solution

Explanation:
Reduction of nitro aryl compounds in presence of Fe and HCl gives aromatic primary amines.
Question 6
5 / -1

In the nitration of benzene using a mixture of conc. H₂SO₄ and conc. HNO_3. the spices which initiates the reaction is ______

A
NO₂

B
NO⁺

C
\(NO^+_2\)

D
\(NO^-_2\)

Solution

Explanation:
In the nitration of benzene using a mixture of concentrated H₂SO₄ & Concentrated HNO₃ the species initiates is NO₂⁺
Question 7
5 / -1

Mixture of 1°, 2°, and 3° amines can be separated by

A
Hinsberg's method

B
Hofmann's method

C
distillation

D
All of the above

Solution

Explanation:
→ Hinsberg test is employed to separate primary, secondary and tertiary amines from a mixture.
→ In this test, the mixture of amines is treated with benzene sulphonyl chloride C₆H₅SO₂Cl (Hinsberg's reagent) followed by treatment with aqueous KOH (5%) solution and then shaken with ether in a separatory funnel.
→ Hofmann's method to separate amines in a mixture uses the reagent diethyl oxalate. The mixture of the three amines is heated with diethyl oxalate. Primary amine forms solid oxamide, the secondary amine forms a liquid oxamic ester and the tertiary amine does not react at all.
→ The Mixture of 1°, 2°, and 3° amines can be separated by distillation as they differ in boiling points.
Question 8
5 / -1

The compound obtained by heating a mixture of a primary amine and chloroform ethanolic potassium hydroxide (KOH) is

A
an isocyanide

B
an halide

C
an amide

D
an amide and nitro compound

Solution

Explanation:
We know that
CH₃CH₂ NH₂ + CHCI₃ + 3KOH → CH₃ CH₂ NC + 3KCI + 3H₂O
In this reaction, the bad-smelling compound ethyl isocyanide (CH3CH2 NC)is produced This equation is known as the carbylamine reaction.
(or)
RNH₂ + CHCI₃ + CHCI₃ + 3KOH(alc.) → RNC (Alkyl isocyanide) + 3KCI + 3H₂O
This reaction is known as carbylamine test. (only 1° amine gives this reaction)
Question 9
5 / -1

Consider the following compounds.
Rank the above compounds in order of increasing basicity (weakest → strongest)

A
4 < 2 < 1 < 3

B
4 < 1 < 3 < 2

C
4 < 3 < 1 < 2

D
2 < 1 < 3 < 4

Solution

Explanation:
The order of basic strength in amine and amide is:
→ Aliphatic amine > Aromatic amine > amide
Given,
→ NO₂ group will exert its electron-withdrawing resonance effect and thus will decrease the basic strength.
→ Thus the order of basic strength in the given compound is:
→ 4 < 3 < 1 < 2
Question 10
5 / -1

Amides can be converted into amines by a reaction named after

A
Perkin

B
Claisen

C
Hofmann

D
Kekule

Solution

Explanation:
⇒ When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, degradation of amide takes place leading to the formation of primary amine.
⇒ This reaction involves the degradation of amide and is popularly known as the Hoffmann bromamide degradation reaction.
Amide to amine