Amines Test

Result

Amines Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

Arrange these compounds 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

$$iv>i>iii>ii$$
Electron - donating substituents increase the electron density of N, thereby increasing the Lewis basicity. Amongst $$NO_2$$ and $$CN,NO_2$$ is slightly resonance stabilization.
Hence, the answer is $$iv>i>iii>ii.$$
Question 2
1 / -0

In the reaction
$$Me_{3}N \xrightarrow{BrCN} A \xrightarrow[ii)\Delta]{i)H_{2}O} B$$.
B is

A
$$MeNH_{2}$$

B
$$Me_{3}NO$$

C
$$Me_{2}NH$$

D
$$Me_{2}NCOOH$$

Solution

$$Me_3N\xrightarrow {BrCN}{}\underset{A}{Me_2N}-CN$$
$$Me_2N-Cn\xrightarrow[\triangle ]{H_2O}Me_2NCOOH$$
Question 3
1 / -0

The decreasing order of the basic character at $$a, b, c, d$$ in the compound:

A
$$a > c > b > d$$

B
$$d > c > b > a$$

C
$$c > b > d > a$$

D
$$b > c > d > a$$

Solution

$$d>c>b>a.$$
Due to ease of delocalisation of electron pair is least in $$(d)$$ Nitrogen. For $$(a), (b), (c)$$ Nitrogen atoms , basic character order is governed by electronic effects of rings and $$-CH_3$$ group
Question 4
1 / -0

Arrange the following in order of decreasing basic strength.

A
IV > I > II > III

B
IV > I > III > II

C
IV > II > I > III

D
I > II > III > IV

Solution

$$IV>I>II>III$$
Question 5
1 / -0

Which of the following compounds is an imine?

A

B

C

D

Solution

The structure of imine is the given figure.
The nitrogen atom can be attached to hydrogen or an organic group.
Hence compound $$B$$ is an imine.
Question 6
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What is strong base?

A

B

C

D
Question 7
1 / -0

Which is most basic?

A

B

C

D

Solution
Question 8
1 / -0

The correct order of basicity of the above compounds is:

A
II > I > IIl > IV

B
I > III > II > IV

C
III > I > II > IV

D
I > II > III > IV

Solution

(Refer to Image)
The order of basicity is follows
$$I >III >II >IV$$
$$\rightarrow $$ in $$IV$$ electron withdrawing is attached to amine. So, it is least basic.
Question 9
1 / -0

The relative basics strenghts of $$NH_{3},CH_{3}NH_{2}$$ and $$NF_{3}$$ are in the order:

A
$$CH_{3}NH_{2} > NH_{3}>NF_3$$

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

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

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

Solution

$$CH_3-$$ group is electron donating group.
$$F$$ is electron withdrawing group.
So $$CH_3NH_2$$ is more basic. $$NH_3$$ is more basic than $$NF_3$$ because $$H$$ is not a withdrawing group. $$NF_3$$ is less basic.
$$\therefore$$ Basic nature order= $$CH_3NH_2 > NH_3 >NF_3$$
Question 10
1 / -0

Acetic anhydride with ammonia gives the product:

A
$$ CH_{3}CONH_{2} $$

B
$$ CH_{3}CONHCH_{3} $$

C
$$ CH_{3}CN$$

D
$$ CH_{3}COONH_{4} $$

Solution