Amines Test

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Amines Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

Write the basicity order of the following.

A
(II) > (I) > (III) > (IV)

B
(I) > (III) > (II) > (IV)

C
(III) > (I) > (II) > (IV)

D
(I) > (II) > (III) > (IV)

Solution

1,2,3,4 is more basic as it contains two electron donating $$CH_3$$ groups which destabilise the lone pair on nitrogen by donating more electron density due to + I effect and so lone pair are available easily 2 is having only one electron donating $$H_3C$$ group so is less basic than 3 and 4 is least basic as it contains an electron .co group which stabilizes lone pair on nitrogen and therefore are not available for donation
Question 2
1 / -0

How many primary amines are possible for the formula $$C_4H_{11}N$$ ?

A
One

B
Two

C
Three

D
Four

Solution

Four primary amines are possible for this structure.

Option D is correct.
Question 3
1 / -0

Which of the following is more basic than aniline?

A
p-nitroaniline

B
Benzyl amine

C
Diphenyl amine

D
Triphenyl amine

Solution

Although both are primary amines, the nitrogen lone pair of electrons are involved in resonance with the aromatic benzene and are less available to accept a proton whereas benzylamine is an aliphatic amine with a larger $$K_b$$ than aniline.

So, Option B is correct.
Question 4
1 / -0

How many primary amines can be formulated by $${C}_{3}{H}_{9}N$$ and how many $${1}^{o}$$ hydrogen are associated with carbon atoms of each compound?

A
Two primary amines $$[3,6]$$

B
One primary amine $$[3]$$

C
Three primary amines $$[3,6,6]$$

D
Two primary amines $$[5,6]$$

Solution

Two primary amines, n propyl amine and iso propyl amine can be formulated with molecular formula $$C_3H_9N$$
In n propyl amine, five primary hydrogens are attached with carbon atoms. In isopropyl amine, six primary hydrogen atoms are attached with carbon atoms.
Question 5
1 / -0

Methyl cyanide on reaction with sodium and $${C}_{2}{H}_{5}OH$$ forms:

A
ethyl amine

B
ethanoic acid

C
ethanimine

D
none of these

Solution

Methyl cyanide on reaction with sodium and ethanol forms ethylamine.
$$CH_3-C \equiv N +4[H] \xrightarrow {Na/ ethanol} CH_3-CH_2-NH_2$$
In this reaction, the nitrile group is hydrogenated to form a primary amine. This is a useful method to increase the length of a carbon chain in preparing a primary amine which contains one more carbon atom than the alkyl halide from which alkyl cyanide is obtained.
Question 6
1 / -0

Among the following substituted pyridines, the most basic compound is:

A

B

C

D

Solution

Ring activating groups increases the basisity of the compound as the option B has two ring activation groups so it has more basisity than the others.
Question 7
1 / -0

Which of the following method(s) does work for the preparation of secondary amine?

A
$$R_3N \xrightarrow[\begin{matrix} (ii) H_3O^+\\(iii) \Delta\end{matrix}]{(i) Br-CN}?$$

B
$$R-\overset{O}{\overset{||}{C}}-NH - R' \xrightarrow{LiAlH_2}?$$

C
$$\displaystyle RNC\xrightarrow[\Delta ]{Nl/H_{2}}?$$

D
All of these

Solution

(A) Tertiary amines react with cyanogen bromide to form a dialkylcyanamide which, on hydrolysis, gives a secondary amine.

$$R_3N + BrCN \rightarrow [R_3NCN]^+Br^- \rightarrow RBr + R_2NCN \overset {H^+}{\rightarrow} R_2NH$$

(B) Secondary amides $$(RCONHR)$$ react with lithium aluminium hydride (LiAlH4) to form secondary amines $$(RCH_2NHR')$$

$$R\overset{\overset{O}{||}}{C}NHR + H-Al^-H_3 \rightarrow R\overset {\overset{O^-}{|}}{C}=NR \rightarrow RCH_2NHR$$

(C) Secondary amines are produced by the reduction of isocynaide.

$$RNC \overset {Ni / H_2}{\rightarrow} RNHCH_3$$
Question 8
1 / -0

Arrange the following in the decreasing order of boiling point:
i) $${ C }_{ 2 }{ H }_{ 5}OH$$
ii) $$ (CH_3)_2NH $$
iii) $$C_2H_5NH_2$$

A
i > ii > iii

B
ii > i > iii

C
iii > i > ii

D
iii > ii > i

Solution

All the given compounds have intermolecular hydrogen bonding due to which their boiling points are higher. $$C_2H_5OH$$ possesses the strongest hydrogen bonding because the $$O$$ is more electron negative than $$N$$ in other compounds. Therefore it has the highest boiling point.

In $$(CH_3)_2NH$$ and $$C_2H_5NH_2$$, the latter has a longer alkyl chain which makes hydrogen bonding weaker.

Hence the order of boiling points is $$i>ii>iii$$
Question 9
1 / -0

Arrange the following in the increasing order of boiling point:
(i)$$C_2H_5OH$$
(ii)$${(CH_3)_2}NH$$
(iii)$$C_2H_5NH_2$$

A
i < ii < iii

B
ii < i < iii

C
iii < i < ii

D
iii < ii < i

Solution

All the given compounds have intermolecular hydrogen bonding due to which their boiling points are higher. $$C_2H_5OH$$ possesses strongest hydrogen bonding because the $$O$$ is more electron negative than $$N$$ in other compounds. Therefore it has the highest boiling point. In $$(CH_3)_2NH$$ and $$C_2H_5NH_2$$, the latter has longer alkyl chain which makes hydrogen bonding weaker. Hence the order of boiling points is $$iii<ii<i$$
Question 10
1 / -0

Arrange the following in the decreasing order of solubility in water :

i) $$ \displaystyle { C}_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$$

ii) $$ ({ C }_{ 2 }{ H }_{5 }{ ) }_{ 2 }NH $$

iii) $$ { C }_{ 2 }{ H }_{ 5}{ NH }_{ 2 } $$

A
i > ii > iii

B
ii > i > iii

C
iii > i > ii

D
iii > ii > i

Solution

Ethylamine when added to water forms intermolecular H−bonds with water. And therefore it is most soluble in water.
Diethylamine having a bigger alkyl chain forms weaker intermolecular hydrogen bonds.
But aniline does not form H−bond with water to a very large extent due to the presence of a large hydrophobic $$−C_6H_5$$ group. Hence, aniline has the least solubility in water.
Hence the order of solubility is $$iii>ii>i$$