Amines Test

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Amines Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

Correct gradation of basic character :

A
$$NH_{3} > CH_{3}NH_{2} >NF_{3}$$

B
$$ CH_{3}NH_{2} >NH_{3} >NF_{3}$$

C
$$ NF_{3} >CH_{3}NH_{2} >NH_{3}$$

D
$$ CH_{3}NH_{2} >NF_{3} >NH_{3}$$

Solution

An $$\ e^{-}$$ donating group (EDa) increase the basic character

An $$\ e^{-}$$ withdrawing group (EWa) decrease the basic character.
Question 2
1 / -0

Aniline when treated with $$NaNO_{ 2 }$$ and $$HCl$$ at $$0- { 5 }^{ 0 }C$$ gives ____________ as a product.

A
chloroaniline

B
benzene diazonium chloride

C
chlorobenzene

D
dichlorobenzene

Solution

Aniline reacts with sodium nitrite and HCl in cold condition to form benzene diazonium chloride.
$$\underset {aniline} {{ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }}\xrightarrow [ { NaNO }_{ 2 } ]{ HCl } \underset {benzene diazonium chloride} {[C_6H_5-N_2^+Cl^-]}$$
Question 3
1 / -0

The structural formula of methylamino methane is:

A
$$(CH_{3})_{2}CHNH_{2}$$

B
$$(CH_{3})_{3}N$$

C
$$(CH_{3})_{2}NH$$

D
$$CH_{3}NH_{2}$$

Solution

Methylamino methane has 2 methane groups attached to $$N$$ directly.
Hence option $$C$$ is correct
Question 4
1 / -0

The correct order of basic strength of given aniline is:

A
o - nitroaniline > p - nitroaniline > m - nitroaniline

B
m - nitroaniline > p - nitroaniline > o - nitroaniline

C
m - nitroaniline > o - nitroaniline > p - nitroaniline

D
o - nitroaniline > m - nitroaniline > p - nitroaniline

Solution

The pKa values of m - nitroaniline, p - nitroaniline and o - nitroaniline isomers are 2.5, 1.0 and -0.3 respectively.

Due to the presence of the electron-withdrawing nitro group, the delocalization of the lone pair of electron is improved. This effect is observed in para and ortho position.

In the ortho position, a very close inductive effect of the nitro group is also observed. (ortho effect) Hence, o - nitroaniline is less basic than p - nitroaniline.

In m - nitroaniline, only the inductive effect is present.

The decreasing order of basicity is m - nitroaniline > p - nitroaniline > o - nitroaniline.

Option B is correct.
Question 5
1 / -0

In the diazotization of aryl amines with sodium nitrite and hydrochloric acid, an excess of hydrochloric acid is used primarily to :

A
supress the concentration of free aniline available for coupling

B
supress hydrolysis of phenol

C
ensure a stoichiometric amount of nitrous acid

D
neutralise the base liberated

Solution

If excess $$HCl$$ is not used, the arene diazonium salt will react with free aniline to form azo compound.

Excess $$HCl$$ forms hydrochloride salt of aniline. This prevents azo coupling.

Option A is correct.
Question 6
1 / -0

Butanamine-1 structure is shown as:

A

B

C

D

Solution

$$CH_3CH_2CH_2CH_2NH_2$$
It is a four carbon compound with an amino group on a terminal carbon atom. Hence, its name is 1 - butanamine
Question 7
1 / -0

When aniline in hydrochloric acid is treated with ice cold solution of sodium nitrite, it gives:

A
benzene diazonium chloride

B
chlorobenzene

C
nitro Aniline

D
chloro Aniline

Solution

Aniline reacts with sodium nitrite and HCl in cold condition to form benzene diazonium chloride.
$$\underset {\text {aniline}} {{ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }}\xrightarrow [ { NaNO }_{ 2 } ]{ HCl } \underset {\text {benzene diazonium chloride}} {[C_6H_5-N_2^+Cl^-]}$$
Question 8
1 / -0

Which of the following would not react with benzene sulphonyl chloride in aq. $$NaOH$$ ?

A
Aniline

B
Methylamine

C
$$N,N$$-dimethyl aniline

D
$$N-$$methyl aniline

Solution

N,N-dimethyl aniline does not contain $$H$$ attached to nitrogen.
Hence, it would not react with benzene sulphonyl chloride in aqueous $$NaOH$$.
Question 9
1 / -0

The correct decreasing order of strength of the bases $$OH^{-}$$,$${NH_{2}}^{-}$$, $$HC\equiv C^{-}$$ and $${CH_{3}-CH_{2}}^{-}$$ is:

A
$${CH_{3}-CH_{2}}^{-} >{NH_{2}}^{-} > H-C \equiv C^{-} >OH^{-}$$

B
$$H-C\equiv C^{-} >{CH_{3}-CH_{2}}^{-} >{NH_{2}}^{-} >OH^{-}$$

C
$$OH^{-} > {NH_{2}}^{-} >H-C\equiv C^{-} >{CH_{3}-CH_{2}}^{-}$$

D
$${NH_{2}}^{-} > H-C\equiv C^{-} >OH^{-} >{CH_{3}-CH_{2}}^{-}$$

Solution

Making conjugate acids of all:
$$i)CH_{3}-CH_{3}\rightarrow CH_{3}-\bar{CH_{2}}+H^{+}$$
         Acid (Conjugate Base)
        strong weak
ii) $$NH_{3}\rightarrow NH_{2}^{-}+H^{+}$$
       Acid (CB)
strong weak
$$iii)H-C=C-H\rightarrow H-C=C+H^{+}$$
    (Acid) (CB)
    strong weak Base
iv)$$H_{2}O\rightarrow H^{+} + OH^{-}$$
       Acid (CB)
$$\therefore$$ order of Basicity : $$CH_3-\bar{CH}_{2}>\bar{NH}_{2}>H-C\equiv C^{-}>OH^{-}$$
Question 10
1 / -0

Amongst the following, the most basic compound is :

A
benzylamine

B
aniline

C
acetanilide

D
p-nitroaniline

Solution

Aliphatic amines are stronger bases as compared to aromatic amines. This is because in aromatic amines, the lone pair on nitrogen is involved in delocalization with aromatic ring.
Benzyl amine is an aliphatic amine, whereas other given amines are aromatic amines.
Hence, among the given amines, benzyl amine $$({ C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }{ NH }_{ 2 })$$ is the strongest base.