Amines Test

Result

Amines Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

What is the product $$A$$?

A

B

C

D

Solution

In Diazotisation reactions, Diazonium ion forms which then followed by carbocation rearrangement forms this product.
Question 2
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Consider the Hofmann ammonolysis reaction:
If $$R = {C}_{2}{H}_{5}$$, the correct order of basic character of the above amines in aqueous medium is :

A
$$ 2 > 3 > 1 > N{H}_{3}$$

B
$$ 2 > 1 > 3 > N{H}_{3}$$

C
$$ 3 > 2 > 1 > N{H}_{3}$$

D
$$ N{H}_{3} > 1 > 2 > 3$$

Solution

If $$R = {C}_{2}{H}_{5}$$, the correct order of basic character of the above amines in aqueous medium is $$ 2 > 3 > 1 > N{H}_{3}$$. The factors that determine basicity are the +I effect of alkyl groups and the solvation /hydration in aqueous medium.
Question 3
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Give the decreasing order of basic strength of the following:
a. $$N{H}_{3}$$
b. $$N{H}_{2}OH$$
c. $$N{H}_{2}-N{H}_{2}$$

A
$$c > b > a$$

B
$$a > b > c$$

C
$$c > a > b$$

D
$$a > c > b$$

Solution

The decreasing order of basic strength is $$c > a > b$$ or $$N{H}_{2}-N{H}_{2} > N{H}_{3} > N{H}_{2}OH$$.
$$N{H}_{2}-N{H}_{2}$$ is most basic as it has two lone pairs of electrons.
The basic strength of ammonia is lower than the basic strength of hydrazine as ammonia contains only one lone pair.
Hydroxyl amine is least basic due to $$-I$$ effect of $$-OH$$ group which decreases the ease with which lone pair of an electron can be donated.
Question 4
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Consider the Hofmann ammonolysis reaction:
If $$R = {Me}_{3}C-$$ ($$t-butyl$$), the correct order of basic character of the above amines in aqueous medium is :

A
$$N{H}_{3} > 1 > 2 > 3$$

B
$$3 > 2 > 1 > N{H}_{3}$$

C
$$1 > 2 > N{H}_{3} > 3 $$

D
$$1 > N{H}_{3} > 2 > 3 $$

Solution

If $$R = {Me}_{3}C-$$ (t-butyl), the correct order of basic character of the above amines in aqueous medium is $$N{H}_{3} > 1 > 2 > 3$$
The factors that determine basicity are the +I effect of alkyl groups and the solvation /hydration in aqueous medium.
Question 5
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Indicate the following compound as $$1^o, 2^o$$ and $$3^o$$.

$$Et_3N$$

A
$$1^o$$

B
$$2^o$$

C
$$3^o$$

D
Non of the above

Solution

$$C_2H_5 -\underset{\underset {\displaystyle {C_2H_5}}|}{N}-C_2H_5$$
Triethylamine, is tertiary amines, as the three hydrogen atoms are replaced by ethyl substituents.

Option C is correct.
Question 6
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Consider the Hofmann ammonolysis reaction:
If $$R = {Me}_{2}CH-$$ (isopropyl), the correct order of basic character of the above amines in aqueous medium is :

A
$$ 3 > 2 > 1 > N{H}_{3}$$

B
$$N{H}_{3} > 1 > 2 > 3$$

C
$$1 > 2 > N{H}_{3} > 3$$

D
$$1 > N{H}_{3} > 2 > 3$$

Solution

If $$R = {Me}_{2}CH-$$ (isopropyl), the correct order of basic character of the above amines in aqueous medium is $$1 > N{H}_{3} > 2 > 3$$.
The factors that determine basicity are the +I effect of alkyl groups and the salvation /hydration in aqueous medium.
Question 7
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Mendius reaction involves the:

A
Reduction of aldehydes to give alcohols

B
Reduction of nitriles with sodium and ethanol

C
Oxidation of nitriles

D
Hydrolysis of cyanides

Solution

b
$$\displaystyle R-C\equiv N\xrightarrow [ Mendius\quad reduction ]{ Na+EtOH } { RCH }_{ 2 }{ NH }_{ 2 }$$
Question 8
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(A) is subjected to reduction with $$\displaystyle Zn-{ Hg }/{ HCl }$$ and the product formed is N-methylmethanamine. (A) can be:

A
Ethane nitrile

B
Nitroethane

C
Carbylaminoethane

D
Carbylaminomethane

Solution

Reduction with $$Zn-Hg/HCl$$ is Clemmensen reduction
which converts $$R-C\equiv N\rightarrow RCH_2NH_2$$ and
$$R-\overset{\oplus}{N}\equiv \overset{\ominus}{C}\rightarrow RNHCH_3$$

Carbylamino methane is $$\left ( {Me-\overset{\oplus}{N}\equiv \overset{\ominus}{C}} \right )$$

$$\therefore Me-\overset{\oplus}{N}\equiv\overset{\ominus}{C}\xrightarrow {Zn-Hg/HCl}\underset{N-Methylmethanamine}{Me-NH-CH_3}$$

Option D is correct.
Question 9
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Which of the following cannot be prepared by Gabriel phthalimide synthesis?

A
Aliphatic primary amines

B
Aromatic primary amines

C
Cyclic primary amines

D
None of the above

Solution

Gabrial phthalimide synthesis is a method for preparation of 1 amine.
therefore aliphatic primary amines can be prepared.
Question 10
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Consider the Hofmann ammonolysis reaction:
For any value of $$R$$, the correct order of basic character of the above amines in gaseous phase of nonpolar solvent is :

A
$$ 3 > 2 > N{H}_{3} > 1 $$

B
$$3 > 2 > 1 > N{H}_{3}$$

C
$$2 > 3 > N{H}_{3} > 1 $$

D
$$1 > 2 > 3 > N{H}_{3}$$

Solution

For any value of $$R$$, the correct order of basic character of the above amines in gaseous phase of nonpolar solvent is $$3 > 2 > 1 > N{H}_{3}$$
This is because in gaseous or nonpolar solvent, the basic order is $$ 3 > 2 > 1 > N{H}_{3}$$.