Amines Test

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Amines Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

The structural formula of N-methylaminomethane is:

A
$$\left( { CH }_{ 3 } \right) _{ 2 }{ CHNH }_{ 2 }$$

B
$$\left( { CH }_{ 3 } \right) _{ 3 }N$$

C
$$\left( { CH }_{ 3 } \right) _{ 2 }NH$$

D
$${ CH }_{ 3 }{ NH }_{ 2 }$$

Solution

N-methylaminomethane contains two methyl groups directly attached to nitrogen atom.

Hence, its structural formula is $$\left( { CH }_{ 3 } \right) _{ 2 }NH$$.

Option C is correct.
Question 2
1 / -0

Aniline dissolves in $$HCl$$ due to the formation of:

A
anilinium chloride

B
o-chloroaniline

C
azo dye

D
diazonium chloride

Solution

Aniline is basic in nature and $$HCl$$ is an acid. So, acid-base reaction will take place.

In aniline, the lone pair of electrons is partially delocalized into the benzene ring and is thus, available for protonation by an acid.

Hence, aniline dissolves in acid like $$HCl$$ forming anilinium chloride salt.

Therefore, the answer is option $$A$$.
Question 3
1 / -0

$$-{ NH }_{ 2 }$$ group is only:

A
ortho director

B
only meta director

C
only para director

D
both ortho, para director

Solution

$$-NH_{2}$$ group is very strongly activating.
It activated both the ortho as well as para positions.

Note that all the positions - ortho, meta and para are activated.
But ortho and para are activated too strongly (due to +M effect) while the meta position is activated weakly (due to +I effect)

So, D is correct answer.
Question 4
1 / -0

The structure given below represents:

A
quarternary ammonium Salt.

B
primary amine.

C
secondary amine.

D
tertiary amine.

Solution

only 1 hydrogen atom of ammonia $$(NH_3$$) has been replaced by the alkyl group.

So, clearly it is a primary amine.

Note that it does not matter what kind of alkyl group is present.

To classify as primary/secondary/tertiary, we only count the number of replaced hydrogens.

If 1 has been replaced -> Primary
2 replaced -> secondary
3 replaced -> tertiary

Option B is correct.
Question 5
1 / -0

The compound formed when aniline reacts with benzene diazonium chloride is:

A
p-amino azobenzene

B
anilinium chloride

C
p-hydroxy azobenzene

D
diphenyl amine

Solution

Aniline reacts with benzene diazonium chloride to form p-amino azobenzene.
Question 6
1 / -0

Which of the following is the strongest base?

A
Aniline

B
N-methylaniline

C
o-methylaniline

D
Benzylamine

Solution

The strongest base among the given compound is benzylamine. The formula of benzylamine is $$PhCH_{2}NH_{2}$$.
In all other compounds, the lone pair of nitrogen atom is dispersed in the benzene ring. But in benzylamine, the lone pair is not in conjugation with the benzene ring. So, it is freely available for donation. Therefore, benzylamine is most basic.
Question 7
1 / -0

$${ C }_{ 3 }{ H }_{ 9 }N$$ cannot represent:

A
$${ 1 }^{ 0 }$$ amine

B
$${ 2 }^{ 0 }$$ amine

C
$${ 3 }^{ 0 }$$ amine

D
quaternary salt

Solution

Quaternary salts of ammonia contain atleast 4 carbon atoms
Here we are given only 3 carbon atoms
So, it cannot represent a quaternary salt
Question 8
1 / -0

Term tertiary, secondary and primary amine represents the:

A
number of amino groups

B
nature of C atom

C
degree of substitution on nitrogen

D
degree of uunsaturation

Solution

Amines are derivatives of ammonia.
If one hydrogen of ammonia is replaced by an alkyl group, it is called a primary amine
If two hydrogens of ammonia are replaced by alkyl groups, it is called a secondary amine
If three hydrogens of ammonia are replaced by alkyl groups, it is called a tertiary amine
So, tertiary, secondary, and primary amine represents the degree of substitution on the nitrogen
Question 9
1 / -0

The colour of dye formed in the reaction between benzene diazonium chloride and phenol is:

A
yellow

B
orange

C
purple

D
blue

Solution

In alkaline medium, benzene diazonium chloride and phenol react to form p-hydroxy azobenzene which is an orange dye.

Option B is correct.
Question 10
1 / -0

Reaction at $${ 0-5 }^{ 0 }C$$ between aniline, $${ NaNO }_{ 2 }$$ and HCl is known as____________

A
Suphonation of benzene

B
Acylation of aniline

C
Alkylation of aniline

D
Diazotisation

Solution

This is the diazotization reaction.

The reaction goes as follows :
$$PhNH_{2} + NaNO_{2} + HCl \rightarrow PhN_{2}^{+} Cl^{-}$$

The benzendiazonium salt thus formed acts as a synthetic intermediate and can be used to synthesize a wide variety of substituted benzene compounds.

For example,
$$PhN_{2}^{+} Cl^{-} + CuCN \rightarrow PhCN$$ (cyanobenzene)
$$PhN_{2}^{+} Cl^{-} + CuI (KI) \rightarrow PhI$$ (iodobenzene)

Option D is correct.

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Amines Test - 13

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Amines Test - 13

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Question 1

$$\left( { CH }_{ 3 } \right) _{ 2 }{ CHNH }_{ 2 }$$

$$\left( { CH }_{ 3 } \right) _{ 3 }N$$

$$\left( { CH }_{ 3 } \right) _{ 2 }NH$$

$${ CH }_{ 3 }{ NH }_{ 2 }$$

Solution

Question 2

anilinium chloride

o-chloroaniline

azo dye

diazonium chloride

Solution

Question 3

ortho director

only meta director

only para director

both ortho, para director

Solution

Question 4

quarternary ammonium Salt.

primary amine.

secondary amine.

tertiary amine.

Solution

Question 5

p-amino azobenzene

anilinium chloride

p-hydroxy azobenzene

diphenyl amine

Solution

Question 6

Aniline

N-methylaniline

o-methylaniline

Benzylamine

Solution

Question 7

$${ 1 }^{ 0 }$$ amine

$${ 2 }^{ 0 }$$ amine

$${ 3 }^{ 0 }$$ amine

quaternary salt

Solution

Question 8

number of amino groups

nature of C atom

degree of substitution on nitrogen

degree of uunsaturation

Solution

Question 9

yellow

orange

purple

blue

Solution

Question 10

Suphonation of benzene

Acylation of aniline

Alkylation of aniline

Diazotisation

Solution

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