Amines Test

Result

Amines Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following compounds will not undergo azo coupling reaction with benzene diazonium chloride?

A
Aniline

B
Phenol

C
Anisole

D
Nitrobenzene

Solution

Diazonium cation is a weak electrophile and hence it reacts with electron-rich compounds containing electron-donating groups such as $$-OH$$, $$-NH_2$$ and $$OCH_3$$ groups and not with compounds containing electron-withdrawing groups such as $$-NO_2,$$ etc.

Hence, option D is correct.
Question 2
1 / -0

Among the following compounds, the most basic compounds is:

A

B

C

D

Solution

The most basic compound is [refer image] as it is ready to act as Lewis base by donating electrons and not undergoing resonance as these are no double bonds adjacent to the molecules.

Option D is correct.
Question 3
1 / -0

The correct decreasing order of basic strength of the following species is:
$$H_2O, \, NH_3, \, OH^-, \, NH_2^-$$

A
$$NH_2^- \, > \, OH^- \,>\, NH_3 \, > \, H_2O$$

B
$$OH^- \, >\, NH_2^- \,> \, H_2O \, > \, NH_3$$

C
$$NH_3 \, > \, H_2O \, > \, NH_2^- \,> \, OH^-$$

D
$$H_2O \, >\, NH_3 \, > \, OH^- \, > \, NH_2^-$$

Solution

$$NH_3$$ is more basic than $$H_2O$$, therefore $$NH^-_2$$ (Conjugate base of weak acid $$NH_3$$) is a stronger base than $$OH^-$$.Thus decreasing order of basic strength is $$NH^-_2$$ > $$OH^-$$ > $$NH_3$$ > $$H_2O$$,
Question 4
1 / -0

What is the decreasing order of relative basic strengths of following species?

A
II > IV > I > III

B
III > I > IV > II

C
III > IV > I > II

D
II > I > IV > III

Solution

The relative basic strength depend on, the electron donating nature comparison of the species.
Therefore as we all know that $$-NH_2$$ is an EDG so, it would increase basicity but $$-Cl,-NO_2(-NO_2>-Cl)$$ in electronic withdrawing nature would decrease basicity significantly.
Question 5
1 / -0

Among the isomeric amines select the one with the lowest boiling point.

A

B

C

D

Solution

The boiling point of amine $$\propto$$ tendency to form H-bonds.

No. of H-bonds - $$3^oAmine<2^oAmine<1^oAmine$$[In gas phase]

The $$1^o$$ and $$2^o$$ amine have less steric repulsion than $$3^o$$ amine.

Here in $$3rd$$ compound, it is a tertiary amine that is sterically hindered by the methyl group, that is the reason it has less tendency to form hydrogen bond and has a lowest boiling point.

The answer is C.
Question 6
1 / -0

In order to prepare a 1 amine from an alkyl halide with simultaneous addition of one $$CH_2$$ group in the carbon chain, the reagent used as source of nitrogen is:

A
Sodium amide $$NaNH_2$$

B
Sodium azide $$NaN_3$$

C
Potassium cyanide $$KCN$$

D
Potassium phthalimide $$C_6H_4(CO)_2N^-K^+$$

Solution

$$4R-X \xrightarrow{KCN} R-CN \xrightarrow {Na\ / C_2H_5OH} R-CH_2NH_2$$
Question 7
1 / -0

In the given pair identify most acidic compound in $$(A)$$ and $$(B)$$. Most basic in $$(C)$$ and $$(D)$$.

A
$$A - (I),\ B - (II),\ C - (I),\ D - (II)$$

B
$$A - (II),\ B - (I),\ C - (I),\ D - (II)$$

C
$$A - (II),\ B - (II),\ C - (II),\ D - (II)$$

D
$$A - (I),\ B - (II),\ C - (I),\ D - (I)$$

Solution

Acidity increases with electron withdrawing group. Ortho effect increases acidity.
Basicity increases with electron donating group. Ortho effect decreases basicity.
Question 8
1 / -0

Among the following amines, the strongest Bronsted base is:

A

B
$$NH_3$$

C

D

Solution

Option $$D$$ is the strongest Bronsted base as there is no delocalisation of lone pair of electron of $$N$$ atom which is not possible in aniline and in pyrrole.
Question 9
1 / -0

The correct increasing order of basic strength for the following compounds is:

A
II < III < I

B
III < I < II

C
III < II < I

D
II < I < III

Solution

Electron donating $$-CH_3$$ group increases the basicity while electron withdrawing $$NO_2$$ group decreases the basicity of amines.

Option D is correct.
Question 10
1 / -0

Which of the following cyclic amine has lowest $$\Delta G $$ for inversion?

A

B

C

D

Solution

More is the steric hindrance the N of amine, lesser is the $$\Delta G$$ for inversion because it has the least tendency to invert)
Refer to Image
Most steric hindrance is in C. Therefore it has lowest $$\Delta G$$ for inversion.