Amines Test

Result

Amines Test - 33

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Question 1
1 / -0

This reaction is example of:

A
Intermolecular $$C-N$$ coupling

B
Intramolecular $$C-N$$ coupling

C
Intermolecular $$N-N$$ coupling

D
Intramolecular $$N-N$$ coupling

Solution

So this is example of $$N-N$$ intra molecular coupling.
Question 2
1 / -0

The increasing order of basicity of the given compounds is:

A
(b)< (a)< (c)< (d)

B
(b)< (a)< (d)< (c)

C
(d)< (b)< (a)< (c)

D
(a)< (b)< (c)< (d)

Solution

Order of base nature depends on electron donation tendency.
In compound (b) nitrogen is $$sp^{ 2 }$$ hybridized so least basic among all given compound
Compound (C) is a very strong nitrogeneous organic base as a lone pair of one nitrogen delocalize in resonance and make another nitrogen negatively charged and conjugate acid have two equivalent resonating structure.
Thus it is most basic in given compound.
Compound (d) is secondary amine more basic than compound (a) which is a primary amine.
Therefore, increasing order of basicity is
(b) < (a) < (d) < (c)
Question 3
1 / -0

The end product (Z) of the following reaction is?

A
A cyanide

B
A carboxylic acis

C
An amine

D
An arene

Solution
Question 4
1 / -0

The correct order of basic strength of the compounds?

A
$$I > II > III$$

B
$$II > I > III$$

C
$$III > II > I$$

D
$$II > III > I$$

Solution

the strength of basic nature of compounds depends on the ability to easily donate the lone pair of electrons.
IN $$(i)$$ lone pair is easily approachable so it has more basicity . in $$(ii)$$ the lone pair is in different plane. while in $$(iii)$$ the lone pair is in conjugation . hence order of basicity is $$I>II>III$$
Question 5
1 / -0

By heating ammonium chloride with two equivalents of formaldehyde it forms:

A
dimethylamine

B
ethylamine

C
methylamine

D
ammonium formate

Solution

$$6HCHO + 4N{H}_{4}Cl \rightarrow ({C{H}_{2}}_{6}){N}_{4} + 4HCl + 6{H}_{2}O$$
So Methylamine is formed
Question 6
1 / -0

$$sp^3-$$hybridised nitrogen is present in:

A

B

C
$$CH_2 = CH - \overset{\oplus}{N}H_3$$

D

Solution

A) $$Sp^{2}$$ hybridized Nitrogen
B) $$S p^{2}$$ hybridized Nitrogen due to participation
of lone pair into ring.
c) $$C H_{2}=C H-NH_3+ \quad\left(S p^{3}\right.$$ hybridised nitrogen $$)$$
D) (sp $$^{2}$$ hybridised) as nitrogen is in conjugation with double bond.
So, option C) is correct.
Question 7
1 / -0

$$CH_3Cl \xrightarrow{AgCN} A$$ (major) $$\xrightarrow{H-2O^+} C + D$$ here '$$C$$' is formic acid '$$D$$ is:'

A
$$2^o$$ amine

B
Carboxylic acid

C
Alkyl Cyanide

D
$$1^o$$ amine

Solution

$$\text { Hence correct Answer - D }$$
Question 8
1 / -0

Reductive amination of $$A$$ forms:

A

B

C

D

Solution
Question 9
1 / -0

The product $$B$$ of the following sequence, would be?

A
$$I$$

B
$$II$$

C
$$III$$

D
$$IV$$ and $$V$$
Question 10
1 / -0

The correct order of increasing basic nature for the bases $$NH_3,CH_3NH_2$$ and $$(CH_3)_2NH $$ is?

A
$$(CH_3)_2NH < NH_3 < CH_3NH_2$$

B
$$NH_3 < CH_3 NH_2 <(CH3)_2NH $$

C
$$CH_3NH_2 <(CH_3)_2NH < NH_3$$

D
$$CH_3NH_2 < NH_3 < (CH_3)_2NH$$

Solution

$$(H_3C)_2NH$$ , $$CH_3NH_2$$ , $$NH_3$$
Among the above molecules $$(CH_3)_2 NH$$ is most basic as two electron donating group are attached to it which increase its basicity is
$$NH_3 < CH_3NH_2 < (CH_3)_2NH$$