Amines Test

Result

Amines Test - 15

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Weekly Quiz Competition

Question 1
1 / -0

The major product expected, when Phthalmide is treated with NaOH, is :

A

B

C

D

Solution
Question 2
1 / -0

Among the following isomeric $$C_{4} H_{11}N$$ amines, one having the lowest boiling point is:

A

B

C

D

Solution

Tertiary amines have no hydrogen atoms bonded to the nitrogen atom and therefore are not hydrogen bond donors. Thus, tertiary amines cannot form intermolecular hydrogen bonding so that is why tertiary ammine have lower boiling point so the answer is option c
Question 3
1 / -0

Which of the following compounds is most basic?

A

B

C

D

Solution

Basicily $$\propto + I, +m$$ group (election denating) also line pairs of nitrogen should not be delocalised.
Question 4
1 / -0

Which one of the following on reduction with
lithium aluminium hydride yield a secondary
amine:

A
Methyl Cyanide

B
Nitroethane

C
Methylisocyanide

D
Acetamide

Solution
Question 5
1 / -0

Aniline when diazotized in cold and then treated with dimethylaniline gives a coloured product. Its structure would be:

A

B

C

D

Solution

Aniline when diazotized in cold I.e 0-5$$^{0}$$ C give benzene diazonium chloride . When this compound coupled with dimethyl aniline it give coloured compound I.e p-(N,N-dimerhyle) amino azobenzene I.e azo dye
Question 6
1 / -0

Which of the following has maximum $$pK_{b}$$ value?

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

B

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

D

Solution

The $$pK_{b}$$ values of ethanamine, N-methylaniline, N-ethylethaneamine and aniline are 3.29, 9.30, 3.00 and 9.38 respectively.
Thus among the given amines, aniline is the weakest base and has maximum $$pK_{b}$$ value.
Aliphatic amines ethanamine and N-methylaniline are more basic than aniline because in aniline, the lone pair on nitrogen is in resonance with benzene.
N-methylaniline is more basic than aniline because in N-methylaniline, the methyl group (an electron releasing group) increases the electron density on nitrogen.
Question 7
1 / -0

Which of the following statements about primary amines is false?

A
Alkyl amines are stronger bases than aryl amines

B
Alkyl amines react with nitrous acid to produce alcohols

C
Aryl amines react with nitrous acid to produce phenols

D
Alkyl amines are stronger bases than ammonia

Solution

Primary amines are those amines whose one hydrogen is substituted by some other groups like $$C{ H }_{ 3 }$$ , $$C{H}_{3}C{H}_{2}$$ etc or those amines which have one $$C-N$$ bond present in them. Example:
$${ H }_{ 3 }C-N{ H }_{ 2 }\quad ,\quad { H }_{ 3 }C-C{ H }_{ 2 }-N{ H }_{ 2 }$$
Aryl amines (if it is primary) on reaction with nitrous acid gives diazonium salt.
Question 8
1 / -0

Which of the following is most basic?

A
Diphenylamine

B
Triphenylamine

C
p-Nitroaniline

D
Benzylamine

Solution

Benzylamine $$C_{6}H_{5}CH_{2}NH_{2}$$ is more basic because benzyl group is electron donating group due to +I effect. So it is able to increase electron density of N of -NH$$_{2}$$ group.
Thus due to higher electron density rate of donation of a free pair of electron is increased I.e basic character is higher While phenyl and nitro group are electron withdrawing group so they are able to decrease the electron density of N of $$-NH_{2}$$ group. Hence they are less basic
Question 9
1 / -0

The correct order of increasing basic nature for the bases $$NH_{3} ,CH_{3} NH_{2} and (CH_{3})_{2}NH$$ in gas phase is :

A
$$(CH_{3})_{2}NH < NH_{3}< CH_{3}NH_{2}$$

B
$$NH_{3} < CH_{3}NH_{2} < (CH_{3})_{2}NH$$

C
$$CH_{3}NH_{2} < (CH_{3})_{2}NH_{2} < NH_{3}$$

D
$$CH_{3}NH_{2} < NH_{3} <(CH_{3})_{2}NH $$

Solution

In general, amines are more basic than ammonia.
Reason : The alkyl groups donate electrons to the nitrogen atom by +I effect
Among the amines, secondary amines are more basic than primary amines.
Reason : In secondary amines there are more alkyl groups so there is more electron density on the nitrogen atom as compared to the primary amine.
So, B is the correct answer.
Question 10
1 / -0

What is the decreasing order of strengths of the following bases?
$$\overset { \circleddash }{ O } H, N{ H }_{ 2 }^{ \circleddash }, HC\equiv { C }^{ \circleddash }$$ and $$C{ H }_{ 3 } - C{ H }_{ 2 }^{ \circleddash }$$

A
$$C{ H }_{ 3 } -C{ H }_{ 2 }^{ \circleddash } > N{ H }_{ 2 }^{ \circleddash } > H - C\equiv { C }^{ \circleddash } > \overset { \circleddash }{ O } H$$

B
$$H -C\equiv { C }^{ \circleddash } > C{ H }_{ 3 } -C{ H }_{ 2 }^{ \circleddash } > N{ H }_{ 2 }^{ \circleddash }>\overset { \circleddash }{ O } H$$

C
$$\overset { \circleddash }{ O } H > N{ H }_{ 2 }^{ \circleddash } > H - C\equiv { C }^{ \circleddash } > C{ H }_{ 3 } -C{ H }_{ 2 }^{ \circleddash }$$

D
$$N{ H }_{ 2 }^{ \circleddash } > H -C\equiv { C }^{ \circleddash } > \overset { \circleddash }{ O } H > C{ H }_{ 3 } -C{ H }_{ 2 }^{ \circleddash }$$

Solution

The conjugate base of strong acid is weaker and conjugate base of weak acid is stronger. So, decreasing order of basic strength:$${ CH }_{ 3 }-{ CH }_{ 2 }^{ - }$$>$${ NH }_{ 2 }^{ - }$$>$${ CH }\equiv { C }^{ - }$$>$${ OH }^{ - }$$