Amines Test

Result

Amines Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

The conjugate base of $$\displaystyle ({ CH }_{ 3 }{ ) }_{ 2 }{ NH }_{ 2 }^{ \oplus }$$ is:

A
$$\displaystyle ({ CH }_{ 3 }{ ) }_{ 3 }N$$

B
$$\displaystyle ({ CH }_{ 3 }{ ) }_{ 2 }NH$$

C
$$\displaystyle ({ CH }_{ 3 }{ ) }_{ 2 }{ N }^{ \ominus }$$

D
$$\displaystyle ({ CH }_{ 3 }{ ) }_{ 2 }{ N }^{ \oplus }$$

Solution
Question 2
1 / -0

Arrange the following in correct activating order towards EAS.
$$-\underset{I}{NHCOCH_3} \quad - O - \underset{II}{\overset{O}{\overset{||}{C}}} - CH_3 \quad - \underset{III}{\overset{\!\!\!\!\!+}{NR_3}}\quad - \underset{IV}{NH_2}$$

A
$$III<IV<I<II$$

B
$$IV<I<II<III$$

C
$$I<II<III<IV$$

D
$$II<III<I<IV$$

Solution

In Electrophilic aromatic substitution reaction, the electrons donating group is favoured over electron withdrawing group.
$$-NH_2 > \overset{+}{-NR_3} > - O - \overset{O}{\overset{||}{C}} - CH_3 > - NHCOCH_3$$
This is the required order of electrophilic aromatic substitution.
Question 3
1 / -0

Which one is more basic, $$CH_3NH_2 \,\, or \,\, CF_3NH_2$$?

A
$$CH_3NH_2$$

B
$$CF_3NH_2$$

C
Both have similar basicity

D
None of the above

Solution

Between $${ H }_{ 3 }C\ddot { N } H_{ 2 },{ F }_{ 3 }C\ddot { N } { H }_{ 2 },{ H }_{ 3 }\ddot { N } { H }_{ 2 }$$ is more basic than $$F_3C\ddot NH_2$$ because $$-CH_3$$ group is an electron donating group so shows posetive inductive effect and donates electron density to nitrogen which makes lone pair on nitrogen more destabilished and are available for reaction .but $$CF_3$$ group is an electron density from nitrogen as it shows negative inductive effect and so lone pairs on nitrogen are not available for reactions.
Question 4
1 / -0

Which of the following statements is correct?

A
Methyl amine is slightly acidic.

B
Methyl amine is less basic than ammonia.

C
Methyl amine is less basic than dimethyl amine.

D
Methyl amine is less basic than aniline.

Solution

Dimethyl amine ($$\displaystyle { Me }_{ 2 }NH$$) is more basic than $$\displaystyle { MeNH }_{ 2 }$$, due to (+I) effect of two (Me) groups.
Question 5
1 / -0

Tertiary butyl amine is a:

A
$${1}^{o}$$ amine

B
$${2}^{o}$$ amine

C
$${3}^{o}$$ amine

D
quaternary salt

Solution

Tertiary butyl amine is a primary amine as it contains $$R-NH_2$$ group.

If only one hydrogen of ammonia is replaced by one alkyl group then it is primary amine.

Option A is correct.
Question 6
1 / -0

Which one of the following is the strongest base in aqueous medium :

A
$$(C_2H_5)_3N$$

B
$$C_2H_5NH_2$$

C
$$NH_3$$

D
$$(C_2H_5)_2NH$$

Solution

$$+I$$ effect of $$C_2H_5$$ group increases electron density on $$N$$ and make it stronger base. So the order of basic strength would be $$3^o>2^o>1^o>NH_3$$ but in aqueous medium presence of hydrogen bonding and bulkier groups affect the basicity. In the $$(C_2H_5)_3N$$, alkyl groups hinder the attack of proton on $$N$$. Therefore it become less basic. And due the additive effects of steric hindrance and induction $$(C_2H_5)_2NH$$ is strongest base in aquous medium.
Question 7
1 / -0

Which of the following amines will form stable diazonium salt at 273-283 K?

A
$$\displaystyle { C }_{ 2 }{ H }_{ 5 }{ NH }_{ 2 }$$

B
$$\displaystyle { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$$

C
$$\displaystyle { C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }{ NH }_{ 2 }$$

D
$$\displaystyle { C }_{ 6 }{ H }_{ 5 }{ N\left( { CH }_{ 3 } \right) }_{ 2 }$$

Solution
Question 8
1 / -0

Pyridine is less basic than triethylamine because :

A
Pyridine has aromatic character

B
Nitrogen in pyridine is $$sp^2$$ hybridised

C
Pyridine is a cyclic system

D
In pyridine, lone pair of nitrogen is delocalised

Solution

Pyridine, is much less basic than typical aliphatic amines, but for a very different reason: the unshared pair is in an $$sp^2$$ , which is much lower in energy than the electron pair of aliphatic amines, which is in an $$sp^3$$ . Therefore, pyridine is less easily protonated than typical aliphatic amines such as triethylamine .
Question 9
1 / -0

Arrange the following in increasing order of pH value

A
II < I < III

B
III < I < II

C
III < II < I

D
II < III < I

Solution

(I) Benzene ring in aniline with single amine group is having electron density on nitrogen atom which releases the proton from it.

(II) Two benzene rings causes maximum inductive effect and maximizes stearic hindrance which is counterd by the highly electronegative nitrogen atom making it least acidic.

(III) In cyclohexane the electron withdrawing group are maximum causing inductive effect and releasing the proton very easily.

Option A is correct.
Question 10
1 / -0

Phenyl cyanide on reduction with $$Na/{C}_{2}{H}_{5}OH$$ yields :

A
$${C}_{6}{H}_{5}{CH}_{2}{NH}_{2}$$

B
$${C}_{6}{H}_{5}NH{CH}_{3}$$

C

D
$${C}_{6}{H}_{5}{NH}_{2}$$

Solution

Phenyl cyanide (benzonitrile) on reduction with sodium ethanol produces benzyl amine.
$$\displaystyle -C \equiv N $$ group is reduced to $$\displaystyle CH_2-NH_2 $$ group.
$$\displaystyle Ph-C \equiv N \xrightarrow {Na/ ethanol} Ph-CH_2-NH_2 $$