Amines Test

Result

Amines Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

The order of the basicity of the following compound is
(1) is refer to image
(2) $${ CH }_{ 3 }CH_{ 2 }{ NH }_{ 2 }$$
(3) $${ ({ CH }_{ 3 }) }_{ 2 }NH$$
(4) $${ CH }_{ 3 }CO{ NH }_{ 2 }$$

A
2>1>3>4

B
1>3>2>4

C
3>1>2>4

D
1>2>3>4

Solution

Amines are basic in nature as they have lone pair of electrons on Nitrogen the stability of conjugate acids formed is
dependent on the extent of $$H-$$bonding and greater the number of $$H-$$atom on $$N$$ more stable is the conjugate acidHence, option $$A$$ is correct answer.
Question 2
1 / -0

Which of the following arylamine is most reactive towards diazotisation?

A

B

C

D

Solution

Diazotisation is an electrophilic substitution reaction. Electrophile $$(NO)^+$$ attacks the $$-NH_2$$ group to form diazo compound. Electron donating groups increase the reactivity of aryl amines towards electrophilic substitution reaction while electron withdrawing groups decrease the reactivity.
So, compounds C and D having electron withdrawing groups $$CN$$ and $$Cl$$ at a para position will be less reactive towards diazotization. Both $$CH_3$$ and $$OCH_3$$ will increase its reactivity, but $$OCH_3$$ being strong electron donating group when present at ortho and para positions will increase its reactivity more effectively. So, B will be most reactive.
Question 3
1 / -0

Number of isomeric primary amines obtained from $$C_4H_{11}N$$ are :

A
3

B
4

C
5

D
6

Solution

The possible isomers are
$$CH_{3}CH_{2}CH_{2}CH_{2}NH_{2}$$

$$(CH_{3})_{2}-CH-CH_{2}NH_{2}$$

$$(CH_{3})_{3}-NH_{2}$$

$$CH_{3}-CH(NH_{2})CH_{2}-CH_{3}$$
Question 4
1 / -0

Amongst the following the strongest base is:

A
benzylamine

B
aniline

C
m-nitroaniline

D
p-nitroaniline

Solution

Correct option is $$A$$
Benzyl amine is strongest base than aniline because
The lone pair of electron on $$N$$-atom in aniline is delocalised & not involved in resonance
Question 5
1 / -0

Which of the following will be least basic?

A

B

C

D
Question 6
1 / -0

Amongest the following the most basic compound is:

A
p-nitro aniline

B
Acetanilide

C
Aniline

D
Benzylamine

Solution

Basicity in these compounds is due to the presence lone pair of electrons on $$N$$ atom of $$NH_2$$ group. In p-nitroaniline, acetanilide and aniline the $$NH_2$$ group is attached directly to the benzene ring and the lone pair of electrons on $$N$$ atom will get involved in resonance in the benzene ring. As a result its ability to donate electrons to the incoming group will decrease, thereby decreasing its basicity. While in benzylamine, $$NH_2$$ group is not attached directly to the benzene ring and its lone pair of electron is available for donation. Therefore, it is most basic among these compounds.
Question 7
1 / -0

When methyl iodide treated with ammonia, the product obtained is:

A
methyl amine

B
dimethyl amine

C
trimethyl amine

D
all of these

Solution
Question 8
1 / -0

Which of the following aromatic rings having greater electronics density than the given compound?

A

B

C

D

Solution

The compound given in the problem contains both $$+I$$ and $$+H$$ group. Aniline contains $$-NH_2$$ as a functional group which is a $$-I$$ and $$+R$$ group. Since $$+R$$ effect is predominant when compared to $$+I$$ and $$+H$$ effect, it will have the greatest electron density
Question 9
1 / -0

Which nitrogen atom is most basic in the above given compound?

A
$$3$$

B
$$4$$

C
$$5$$

D
$$2$$
Question 10
1 / -0

Ethylamine can be prepared by all except:

A
Curtius reaction

B
Hoffman bromamide reaction

C
Gabriel Phthalimide reaction

D
Reduction of formaldoxime

Solution

In Curtius reaction, first the conversion of acyl azide $$(R-CO-N_3)$$ into isocyanate $$(R-N=C=O)$$ takes place followed by its reaction with water to give primary amine $$(RNH_2)$$. So, for the preparation of ethyl amine,
$$C_2H_5-CO-N=N^+=N^- \underset{-N_2}{\overset{\Delta}{\rightarrow}}C_2H_5-N=C=O\overset{H_2O}{\rightarrow}C_2H_5-NH_2$$
Hoffmann bromamide reaction is the conversion of primary amide into primary amine with one carbon less. Primary amide reacts with bromine $$(Br_2)$$ and a base $$(NaOH)$$ to form an intermediate isocyanate $$(R-N=C=O)$$ which upon hydrolysis gives primary amine (with one carbon less) along with the evolution of $$CO_2$$ gas.
$$C_2H_5-CO-NH_2 \underset{NaOH}{\overset{Br_2}{\rightarrow}}C_2H_5-N=C=O\overset{H_2O}{\rightarrow}C_2H_5-NH_2$$
Gabriel Phthalimide reaction involves the conversion of primary alkyl halides into primary amines. Phthalimide undergoes deprotonation and reacts with alkyl halide to form an intermediate which reacts with hydrazine to give primary amine as the final product.
Reduction of formaldoxime does not yield primary amine.