Amines Test

Result

Amines Test - 78

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Weekly Quiz Competition

Question 1
1 / -0

Ethylamine $$(C_2H_5NH_2)$$ can be obtained from N-ethylphthalimide on treatment with:

A
$$NaBH_4$$

B
$$CaH_2$$

C
$$H_2O$$

D
$$NH_2NH_2$$

Solution

Solution:- (D) $$N{H}_{2} N{H}_{2}$$
Question 2
1 / -0

Coupling of benzene diazonium chloride with 1-napthol in alkaline medium will give:

A

B

C

D

Solution
Question 3
1 / -0

Benzene diazonium chloride on reaction with aniline in the presence of dilute hydrochloric acid gives:

A

B

C

D

Solution

Hence option C is correct.
Question 4
1 / -0

Aniline dissolved in dilute $$HCl$$ is reacted with sodium nitrite at $${0}^{o}C$$. This solution was added dropwise to a solution containing equimolar mixture of aniline and phenol in dil. $$HCl$$. The structure of the major product is:

A

B

C

D

Solution

Aniline undergoes diazo coupling in acidic medium with $$PhN_2^{+}$$
Question 5
1 / -0

Amino acid of the above ion exist as shown above at pH:

A
1

B
7

C
11

D
13

Solution

As COOH group remains as it is i.e. it does not get ionised. So, the pH of the solution must be as that of a acid.
So, at pH=1, amino acid will exist as shown above.

Hence, Option "A" is the correct answer.
Question 6
1 / -0

The correct order of basicity of the following compounds is:

A
$$2>1>3>4$$

B
$$4>3>1>2$$

C
$$3>1>2>4$$

D
$$1>2>3>4$$
Solution
$$\text{The correct order of basicity is:}$$
$$4>3>1>2$$

$$\text{The basic strength order depends on }$$
$$\text{Accumulation of negative charge on N (double-bonded) by another }−NH_2 \text{ group, thus intensifying the}$$$$\text{ donor ability of N.}$$
$$\text{The higher donor ability of }sp^3\text{ hybrid N as compared to }sp^2\ N.$$
Option B is correct.
Question 7
1 / -0

The correct order of basicities of the following compounds
(2) $$ CH_3CH_2NH_2 $$
(3) $$ (CH_3)_2 NH $$

A
2 > 1 > 3> 4

B
1 > 3 > 2 > 4

C
3 > 1 > 2 > 4

D
3 > 2 > 1 > 4

Solution

Basicity is the ability to donate electrons.

So, the order of basicity is:

$$\mathrm{3>2>1>4}$$

Hence, Option "D" is the correct answer.
Question 8
1 / -0

The correct order of the basic strength of following compounds is:

A
$$4>2>3>1$$

B
$$4>2>1>3$$

C
$$3>1>2>4$$

D
$$1>3>4>2$$
Solution
$$\text{Higher the electron density on electron-donating group higher is the basic strength.}$$
$$\text{In compound 3 the electron density is highest because of }+ I\text{ effect of methyl groups attached to N}$$. $$\text{Hence it has the highest basicity.}$$
$$\text{In compound 1 there are two amine groups attached to the carbon in which one of the amine group is always}$$$$\text{available for donation. Hence it has the second-highest basicity.}$$
$$\text{In compound 2 again the ethyl group increases the basic strength of amine due to its }+ I\text{ effect but lower}$$ $$\text{than that of as in compound 3.}$$
$$\text{The compound 4 has the least basic strength because the carbonyl group attached to amine decreases the }$$$$\text{basicity by withdrawing the lone pair from Nitrogen through resonance effect.}$$
$$\text{The correct order of basic strength is - }3>1>2>4$$
Question 9
1 / -0

Which of the following is the correct order of basic character ?
I. 1-Amino Propane
II. Ethanamide
III. Guanidine $$[HN=C(NH_2)_2]$$
IV. Aniline

A
I > II > III > IV

B
III > I > IV > II

C
IV > III > I > II

D
III > II > I > IV

Solution
Question 10
1 / -0

Arrange the basicity of the following compounds in decreasing order.

$$ (I) \ CH_3CH_2NH_2 $$
$$(II) \ H_2C=CH-NH_2 $$
$$(III)\ HC\equiv C-NH_2 $$
$$(IV)\ C_6H_5NH_2 $$

A
I > II > III > IV

B
IV > III > II > I

C
III > II > I > IV

D
I > III > II > IV

Solution

Order of basicity depends upon the ability to donate electrons.
Among the given options, compounds II, III & IV have their lone pair of electron of nitrogen involved in resonance. So, I is the most basic compound.
Compound II has higher tendency to donate electrons than III because electronegativity of $$sp^2$$ hybridized carbon is less than that of $$sp$$ hybridized carbon.

So, the order of basicity of given compounds is:
$$\mathrm{I>II>III>IV}$$

Hence, Option "A" is the correct answer.