Amines Test

Result

Amines Test - 60

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The compunds X and Y are respectively

A
$$C{ H }_{ 3 }C{ H }_{ 2 }N{ H }_{ 2 },C{ H }_{ 2 }COCl$$

B
$$C{ H }_{ 3 }CON{ H }_{ 2 },C{ H }_{ 3 }C{ H }_{ 2 }Cl$$

C
$$C{ H }_{ 3 }C{ OOH }_{ 2 },C{ H }_{ 3 }C{ ONH }_{ 2 }$$

D
$$C{ H }_{ 3 }CON{ H }_{ 2 },C{ H }_{ 3 }C{ OOH }$$

Solution

The solution is given to the image$$.$$
Question 2
1 / -0

Which of the following amino acid has pH greater than $$7$$?

A
Glutamic acid

B
Lysine

C
Glycine

D
Alanine

Solution

Lysine is a basic amino acid. It has pH greater than $$7$$. Note: Basic solutions have pH greater than $$7$$.
Question 3
1 / -0

Reduction of $$CH_3CH_2NC$$ with hydrogen in presence of $$Ni$$ or $$Pt$$ as catalyst gives:

A
$$CH_3CH_2NH_2$$

B
$$CH_3CH_2NHCH_3$$

C
$$CH_3CH_2NHCH_2CH_3$$

D
$$(CH_3)_3N$$

Solution

Reduction of isonitriles $$(NC)$$ in the presence of $$H_2/Ni$$ or $$Pt$$ gives a secondary amine ($$R - NH- R^{'}$$).

$$ CH_3CH_2NC\xrightarrow{H_2/Ni\ or \ Pt} CH_3CH_2NHCH_3$$

Option B is correct.
Question 4
1 / -0

What is the order of basic strength of the following?

A
1>2>3>4

B
4>1>2>3

C
4>3>1>2

D
3>4>1>2

Solution

In case of $$(1)$$ and $$(2),(1)$$ is more basic than $$(2),$$ because $$(1)$$ has lone pair on the $$N$$-atom in the non aromatic ring and ready to donate same as pyridine$$.$$ Although pyride act as a base and will be protonated in presence of a strong acid $$,$$ it is a weak base than it's non aromatic counterpast$$.$$
On the other hand methyl cynide $$+I$$ effect of methyl group pushed the electrons towards cyanide and it increasesthe density of electron on $$N$$-atom$$.$$ Hence the order is $$1>2>3>4.$$
The reaction is given to the image$$.$$
Question 5
1 / -0

The correct basic strength order of following anions is :

A
$${CH_3 -{C}{H_2}}^- >{{N}{H_2}}^- > {CH_2 = {C}{H}}^- >{ CH \equiv {C}}^- > H{O}^- > {F}^-$$

B
$${N}{H_2}^- > {C}{H_3} - {C}{H_2}^- > CH_2 = {C}{H}^- > CH \equiv {C}^- > {F}^- > H{O}^-$$

C
$$CH_3 -{C}{H_2}^- > CH_2 = {C}{H}^- >{N}{H_2}^- > CH \equiv {C}^- > H{O}^- > {F}^-$$

D
$$ {F}^- > H{O}^- > CH \equiv{C}^- > {C}{H_2} = {C}{H}^- >{N}{H_2} ^-> CH_3 -{C}{H_2}^-$$
Question 6
1 / -0

Which order for amines is incorrect?

A
Order of basicity in aq. medium.

B
Order of boiling point.

C
Order of basicity in gaseous state.

D
The order of aqueous solubility.

Solution

Order of basicity in aq. medium. for methyl amines.

$$2>1>3$$

In gaseous medium, the order of basicity is $$3>2>1$$

B.P order is $$1>2>3$$. Hence option B is incorrect.

Increase in the size of hydrophobic alkyl part increases the molar mass of amines. This usually results in a decrease in its solubility in water. Hence the given order for solubility is correct.
Question 7
1 / -0

Aniline when diazotised in cold and then treated with N, N - dimethylaniline gives a coloured product. The structure of this product is:

A

B

C

D

Solution

Hence, option $$B$$ is correct.
Question 8
1 / -0

The best method for preparing primary amines from alkyl halides without changing the number of carbon atoms in the chain is:

A
Hofmann bromamide reaction

B
Gabriel phthalimide synthesis

C
Sandmeyer reaction

D
Reaction with $$NH_3$$

Solution

Best method for preparing primary aminos form alkyl halides without changing the number of carbon atoms in the chain is Gabriel synthesis. Because this process is used to get primary amine without changing number of carbon atom.
Question 9
1 / -0

The source of nitrogen in Gabriel synthesis of amines is:

A
Sodium azide $$NaN_3$$

B
Sodium nitrite $$NaNO_2$$

C
Potassium cyanide $$KCN$$

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

Solution

Gabriel synthesis $$:$$
The reaction is given to the image$$.$$
Source of nitrogen atom is Gabriel synthesis is Potassium phthalamide $${C}_{6}{H}_{4}{(CO)}_{2}{N}^{-}{k}^{+}.$$
Question 10
1 / -0

Cyclohexylamine is stronger base than aniline because:

A
in aniline electron pair is involved in conjugation

B
in cyclohexylamine electron pair is involved in canjugation

C
in aniline - $$NH_2$$ group is protonated

D
in cyclohexylamine nitrogen has a negative charge.

Solution

Cyclohexylamine is a stronger base than aniline because in aniline electron pair is involved in conjugation, which makes the electron pair unavailable, where as in cyclohexyl amine, the $$-NH_2$$ group is out side the ring and there is no resonance in the ring also, thereby the lone pair present on the nitrogen is freely available, resulting in a stronger base.