Amines Test

Result

Amines Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

In the above reaction sequence (D) is?

A

B

C

D

Solution
Question 2
1 / -0

Find product in the following reaction is Y. What is Y?

A
benzamide

B
benzophenone

C
benzoic acid

D
benzaldehyde

Solution

In the given reaction, reduction of the cyanide group takes place followed by steam distillation to form benzaldehyde as a product.
Question 3
1 / -0

The reactivities of these ions in azo-coupling reactions will be in the order:

A
$$3<1<4<2$$

B
$$1<4<2<3$$

C
$$1<2<3<4$$

D
$$3<1<2<4$$

Solution

In azo-coupling, the diazonium ion should be more electrophilic for more reactivity. Presence of electron-withdrawing groups increase the reactivity.
Considering these factors, the order of reactivity should be:
$$1<2<3<4$$
Question 4
1 / -0

The correct order of boiling points of the following amines $${ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2 },\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH,{ C }_{ 2 }{ H }_{ 5 }N\left( { CH }_{ 3 } \right) _{ 2 }$$ is :

A
$${ C }_{ 2 }{ H }_{ 5 }N\left( { CH }_{ 3 } \right) _{ 2 }>\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH>{ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2 }$$

B
$$\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH>{ C }_{ 2 }{ H }_{ 5 }N\left( { CH }_{ 3 } \right) _{ 2 }>{ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2 }$$

C
$${ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2} >\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH>{ C }_{ 2 }{ H }_{ 5 }N\left( { CH }_{ 3 } \right) _{ 2 }$$

D
$$\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH>{ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2 }>{ C }_{ 2 }{ H }_{ 5 }N\left( { CH }_{ 3 } \right) _{ 2 }$$

Solution

Because $${ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2 }$$ is primary amine and can form hydrogen bonding more than secondary amine $$({ { C }_{ 2 }{ H }_{ 5 } })_{ 2 }NH$$ and tertiary amine $${ C }_{ 2 }{ H }_{ 5 }N{ ({ CH }_{ 3 }) }_{ 2 }$$ .More hydrogen bonding leads to strong bonding between the molecules, hence increases the boiling point.
Question 5
1 / -0

The reagents/conditions K, L and M are respectively:

A
$$Ba(OH)_{2}$$, $$KCN$$ and $$CHCl_{3} / NaOH$$ heat, $$H_{3}O^{+}$$

B
$$H_{2}O$$ (boil), $$CO_{2} / KOH$$ and $$Ac_{2}O - AcONa$$, heat

C
Steam ; $$CHCl_{3} / NaOH$$, heat, $$H_{3}O^{+}$$ and $$AC_{2}O - AcONa$$, heat

D
$$Cu_{2}Cl_{2} / HCl$$, $$NaOH$$ / high P and $$CO_{2}|NaOH$$, heat

Solution

$$ \text { Honre option } A \text { is correct } $$
Question 6
1 / -0

Which of the N atoms is least basic in the given compound ?

A
a

B
b

C
c

D
d

Solution

Electron pair of $$NH_{2}$$ attached to benzene undergoes resonance i.e lone pair on Nitrogen is delocalised and hence less basic than other amine groups.
Question 7
1 / -0

Which of the following is the least basic?

A

B

C

D
All are equally basic

Solution

Since $$NO_{2}$$ is strong withdrawing group than $$OCH_{3}$$ and $$C_{6}H_{5}$$ , hence withdraw more electron density from benzene and leads to lone pair of $$NH_{2}$$ to be dicolocalised at faster rate.
Question 8
1 / -0

Suitable explanation for the order of basic character, $$(CH_{3})_{3}N < (CH_{3})_{2}NH$$, is:

A
steric hindrance by bulky methyl group

B
higher volatility of $$3^{\circ}$$ amine

C
stability of conjugate acid due to solvation

D
all of these

Solution

$${ ({ CH }_{ 3 }) }_{ 3 }$$ N is less basic due to steric hindrance by bulky methyl group prevent any acid to take a lone pair of nitrogen.
Question 9
1 / -0

With which one of the following substrate do diazonium salt reacts to form a colourful compound?

A

B

C

D

Solution

The answer is refer to image.
Question 10
1 / -0

Basic nature order of the given compounds is
(a) $$CH_3 - CH = \ddot{N}H $$
(b) $$CH_2= CH - CH = CH -NH_2$$
(c) $$CH_3 -NH - CH_3$$

A
c>a>b

B
a>b>c

C
a>c>b

D
c=a=b

Solution

$$C H_{2}=C H-C H=C H-N H_{2}(b)$$
In this compound lone pair on $$-N H_{2}$$ group
is involued in resonance. so givening lone
Pair is difficult. So this compound
has less basicity. than other two.
$$\mathrm{CH}_{3}\mathrm{CH}=\ddot{\mathrm{N}}\mathrm{H}(\mathrm{O})$$
$$ \begin{array}{l} \text { Nitrogen in }^{\prime}=\ddot{N} H^{\prime} \text { group is more electro. } \\ \text { negitiven than }-N H-\text { is } C H_{3}-N H-C H_{3}(C) \\ \text { due to double bond. } \\ \text { More electronegitive Nitrogen in ' }=\text { NH 'grap } \\ \text { cannot tolarate positive charge of it (when } \\ \text { lone pair is given it get positive charge) } \end{array} $$
$$ \begin{array}{l} \text { Compared to Nitrogen in }-\mathrm{NH} \text { - group } \\ \text { So } \mathrm{CH}_{3}-\mathrm{NH}-\mathrm{Ct}_{3} \text { is more basic than } \\ \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{NH} \text { (a) } \\ \text { Hence Basisity order is } c>a>b \\ \text { so option } A \text { is correct. } \end{array} $$