Amines Test

Result

Amines Test - 9

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Weekly Quiz Competition

Question 1
1 / -0

The increasing order of basicity of the above compounds is:

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

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

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

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

Solution

$$Basicity \propto \text {stability of conjugate acid}$$

Since $$C$$ is most basic due to resonance stabilization of its conjugate acid and $$b $$ is least stable as positive charge is on $$sp^2$$ $$N$$.

Out of the given options, the answer is option $$A$$.
Question 2
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Considering the basic strength of amines in aqueous solution, which one has the smallest $$ pK_{b}$$ value?

A
$$ (CH_{3})_{3}N$$

B
$$ C_{6}H_{5}NH_{2}$$

C
$$ (CH_{3})_{2}NH$$

D
$$ CH_{3}NH_{2}$$
Question 3
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The increasing order of diazotisation of the following compounds is:

A
$$(D) < (C) < (B) < (A)$$

B
$$(A) < (D) < (B) < (C)$$

C
$$(A) < (B) < (C) < (D)$$

D
$$(A) < (D) < (C) < (B)$$

Solution

Aromatic diazonium salts are more stable than aliphatic diazonium salts. The stability of aryl diazonium salts is due to resonance. Electron donating substituents increase electron density on benzene ring. They increase the stability of diazonium salts.

Electron withdrawing substituents decrease electron density on benzene ring. They decrease the stability of diazonium salts. $$\displaystyle -COCH_3$$ group is electron withdrawing and hence, (d) is less stable than (b).

Although $$\displaystyle -O-COCH_3$$ is electron donating substituent, but it is present in meta position. Hence, it will not have significant effect on stability. The increasing order of diazotisation is $$\displaystyle (A) < (D) < (B) < (C)$$.
Question 4
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Directions For Questions

Treatment of compound O with $$KMnO_4/H^+$$ gave P, which on heating with ammonia gave Q. The compound Q on treatment with $$Br_2/NaOH$$ produced R. On strong heating, Q gave S, which on further treatment with ethyl 2-bromopropanoate in the presence of KOH followed by acidification, gave a compound T.

...view full instructions

The compound R is:

A

B

C

D

Solution

Compound Q on treatment with $$Br_2/NaOH$$ is nothing but Hoffmann Bromamide reaction which gives amine from amide.

Option A is correct.
Question 5
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Directions For Questions

Treatment of compound O with $$KMnO_4/H^+$$ gave P, which on heating with ammonia gave Q. The compound Q on treatment with $$Br_2/NaOH$$ produced R. On strong heating, Q gave S, which on further treatment with ethyl 2-bromopropanoate in the presence of KOH followed by acidification, gave a compound T.

...view full instructions

The compound T is:

A
Glycine

B
Alanine

C
Valine

D
Serine

Solution

Option B is correct.
Question 6
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The order of basicity among the following compounds is:

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

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

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

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

Solution

The order of basicity among the following compounds is $$\displaystyle IV>I>II>III$$

In IV, the extent of delocalization is maximum. When a lone pair of electrons on one of the $$N$$ atom is donated to a suitable acid, still some delocalization is possible.

However, in case of I, when a lone pair of electrons on one of the $$N$$ atom is donated to a suitable acid, further delocalization is not possible.

So, option C is correct.
Question 7
1 / -0

The major product of the following reaction is:

A

B

C

D

Solution
Question 8
1 / -0

The correct order of the basic strength of methyl substitited amines in aqueous solution is :

A
$$(CH_3)_2NH> CH_3NH_2> (CH_3)_3N$$

B
$$(CH_3)_3N> CH_3NH_2>(CH_3)_2NH$$

C
$$(CH_3)_3N> (CH_3)_2NH>CH_3NH_2$$

D
$$CH_3NH_2>(CH_3)_2NH> (CH_3)_3N$$

Solution

In aqueous solution, electron donating inductive effect, solvation effect (H-bonding) and steric hindrance all together affect basic strength of substituted amines.
Basic character:
$$\underset{2^0}{(CH_3)_2NH}> \underset{1^0}{CH_3NH_2}>\underset{3^0}{(CH_3)_3N}$$
Question 9
1 / -0

The correct statement regarding the basicity of arylamines is:

A
arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring $$\pi$$ electron system.

B
arylamines are generally more basic than alkylamines because the nitrogen lone-pair electrons are not delocalized by interaction with the aromatic ring $$\pi$$ electron system.

C
arylamines are generally more basic than alkylamines because of aryl group.

D
arylamines are generally more basic than alkylamines, because the nitrogen atom in arylamines is sp-hybridized.

Solution

Arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring $$\pi$$ electron system which makes electrons less available. While in alkyl amines, the $$+I$$ electron releasing effect of the alkyl group makes it strongly basic. Thus, option A is correct.
Question 10
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Which of the following will be most stable diazonium salt $$RN_2^+X^-$$?

A
$$CH_3N_2^+X^-$$

B
$$C_6H_5N_2^+X^-$$

C
$$CH_3CH_2N_2^+X^-$$

D
$$C_6H_5CH_2N_2^+X^-$$

Solution

Among the given diazonium salts, benzene diazonium halide $$C_6H_5N_2^+X^-$$ is most stable due to conjugation of the $$N \equiv N$$ triple bond with benzene ring.