Amines Test

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Amines Test - 54

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following compounds has the lowest boiling point?

A
2-propanamine

B
ethylmethylamine

C
1-propanamine

D
N,N-dimethylmethanamine

Solution

N,N-dimethylmethanamine will have the lowest boiling points as it cannot form hydrogen bonds.

Other amines can form hydrogen bonds and have higher boiling points.

Option D is correct.
Question 2
1 / -0

$$C$$ and $$D$$ in the given sequence are :

A
Benzoic acid$$+$$ aniline

B
Phthalic acid $$+$$ ethylamine

C
Phthalic acid $$+$$ aniline

D
Benzoic acid $$+$$ ethylamine

Solution

C and D in the given sequence are Phthalic acid + ethylamine.

The given reaction is Gabriel phthalimide reaction.

Option B is correct.
Question 3
1 / -0

Among the following the weakest base is:

A
$$C_6H_5CH_2NH_2$$

B
$$C_6H_5CH_2NHCH_3$$

C
$$O_2N-CH_2NH_2$$

D
$$CH_3NHCHO$$

Solution

The weakest base is $$\displaystyle CH_3NHCHO$$.

The lone pair of electrons cannot be easily donated to suitable Lewis acid as the nitro group is electron-withdrawing group.

As amides are not basic in nature and given option D is amide structure and other compounds are amines.

Option D is correct.
Question 4
1 / -0

Arrange the following resonating structure according to their contribution towards resonance hybrid?
(a) $$CH_2=\overset {\oplus}{N}=\overset {\ominus}{N}$$ (b) $$\overset {\ominus}{C}H_2-N=\overset {\oplus}{N}:$$ (c) $$\overset {\oplus}{C}H_2-\underset {. .}{N}=\overset {\ominus}{N}$$ (d) $$\overset {\ominus}{C}H_2-\overset {\oplus}{N}\equiv \overset {. .}{N}$$

A
a > d > c > b

B
b > a > c > d

C
a > c > b > d

D
d > a > b > c

Solution

The correct order of the contribution of resonating structure towards resonance hybrid is
$$\displaystyle a (CH_2=\overset {\oplus}{N}=\overset {\ominus}{N}) > d (\overset {\ominus}{C}H_2-\overset {\oplus}{N}\equiv \overset {. .}{N}) > c (
\overset {\oplus}{C}H_2-\underset {. .}{N}=\overset {\ominus}{N}
) > b (\overset {\ominus}{C}H_2-N=\overset {\oplus}{N}:)$$
a and d have higher contribution as they have less charge separation.
b and c have lower contribution as they have more charge separation.
Question 5
1 / -0

The correct order of increasing basic nature of the following base is :

A
2 < 5 < 1 < 3 < 4

B
5 < 2 < 1 < 3 < 4

C
2 < 5 < 1 < 4 < 3

D
5 < 2 < 1 < 4 < 3

Solution

In structure 2, electron withdrawing group $$NO_2$$ present in para position.

Also in structure 5, electron withdrawing group $$NO_2$$ is present in meta position.

At para position $$NO_2$$ shows more withdrawing effect than at meta position due to $$-R$$ effect.

Hence, structure 2 is least basic.

Next basic structure is aniline i.e., structure 1.

Next comes the structure 3 and 4 in which methyl and methoxy groups are present which are electron-donating groups.

So the correct answer is A
Question 6
1 / -0

Arrange in decreasing order of basic strength :

A
$$I > II > III > IV$$

B
$$II > III > I > IV$$

C
$$IV > I > III > II$$

D
$$IV > I > II > III$$

Solution

Aniline itself act as lewis base as the nitrogen of $$-NH_2$$ can donate electron pairs. Its basic strength increases by presence of $$alkyl$$ group which shows $$+I$$ effect. Addition of $$-CN \ or \ -NO_2$$ decrease its basic strength as they both show $$-M$$ effect. $$-M$$ effect of nitro group is higher than nitrile group. Therfore the order of strength is $$IV>I>III>II$$
Question 7
1 / -0

The basic character of substituted anilines (I to IV) are such that :

A
I < II < IV < III

B
I < II < III < IV

C
II < I < III < IV

D
III < IV < I < II

Solution

The basic character of substituted anilines (I to IV) order is:

$$\displaystyle III < IV < I < II $$.

When electron-withdrawing groups such as nitro and chloro are present, the basic strength of aniline is reduced.

When electron releasing groups such as methyl and methoxy are present, the basic strength of aniline is increased.

Option D is correct.
Question 8
1 / -0

Iso-propyl amine is a:

A
primary amine

B
secondary amine

C
teritary amine

D
quaternary amine

Solution

The structure of isopropyl amine is $$CH_3-\underset{\underset{\displaystyle NH_2}{|}}{C}H-CH_3$$
It is primary amine.
Question 9
1 / -0

Within the list shown below, the correct pair of structures of alanine in pH ranges 2-4 and 9-11 is:
I. $$H_3N^+ - CH(CH_3)CO_2H$$
II. $$H_2N - CH(CH_3)CO_2^-$$
III. $$H_3N^+ - CH(CH_3)CO_2^-$$
IV. $$H_2N -CH(CH_3)CO_2H$$

A
I and II

B
I and III

C
II and III

D
III and IV

Solution

The correct pair of structures of alanine in pH ranges 2-4 and 9-11 is
I. $$\displaystyle H_3N^+ - CH(CH_3)CO_2H$$
II. $$\displaystyle H_2N - CH(CH_3)CO_2^- $$
In pH range 2-4, the medium is acidic and the amino group is protonated.
In pH range 9-11, the medium is basic and carboxylic group is deprotonated.
Question 10
1 / -0

Arrange the following in the increasing order of the $$\displaystyle{pK}_b$$ values:
i) $$ C_2H_5NH_2 $$
ii)$$\,{ C }_{ 6 }{ H }_{ 5 }{ NHCH }_{ 3 }$$
iii)$$\,({ C }_{ 2}{ H }_{ 5 }{ ) }_{ 2 }NH$$
iv) $$ { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$$

A
i < ii < iii < iv

B
iii < i < ii < iv

C
i < iii < ii < iv

D
iv < ii < i < iii

Solution

$$pK_b$$ value gives a measure of the basicity of a compound. Higher the value of $$K_b$$ of a base, stronger is the base. In aniline, $$C_6H_5NH_2$$, lone pair on nitrogen atom are withdrawn into the benzene ring. Benzene ring is an electron withdrawing group and hence, the lone pair is not easily available on nitrogen atom, so it cannot attract a proton easily.

N-Methylaniline, $$C_6H_5NHCH_3$$ has benzene ring attached to it as well as a methyl group attached to the nitrogen atom. The benzene ring withdraws lone pair from nitrogen atom . Methyl group attached to the nitrogen atom is an electron donating group and it donates electron to nitrogen atom increasing electron density .

Thus the $$K_b$$ value for N-Methylaniline is higher than aniline and it is a stronger base than aniline.

In $$C_2H_5NH_2$$, ethyl groups are electron donating group. It donates electrons to nitrogen atom increasing the electron density over nitrogen atom. Thus Ethylamine is a stronger base than N-Methylaniline.

In diethylamine, $$(C_2H_5)_2NH$$, two Ethyl groups attached to nitrogen atom are electron donating group. increasing the electron density on nitrogen atom. hence, diethylamine is a stronger base than ethylamine because of the presence of two electron donating groups.

Thus the correct option is D.