Amines Test

Result

Amines Test - 59

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

In the following reaction the major product is:

A

B

C

D

Solution

The reaction mechanism is shown above.
Option D is the correct answer.
Question 2
1 / -0

Which of the following is a correct decreasing order of boiling points of the given isometric compounds:
Ethyldimethyl amine (I), n-butylamine (II), Diethylamine (III)

A
$$II> III> I$$

B
$$I> II> III$$

C
$$III> II> I$$

D
$$III> I> II$$

Solution

Answer:

Primary amine has higher boiling point than secondary followed by tertiary amine, this is because higher boiling points of the primary amines is that they can form hydrogen bonds with each other as well as van der Waals dispersion forces and dipole-dipole interactions.

Tertiary amine has no hydrogen left for forming Hydrogen bonds among themselves so can not show intermolecular H-bonding. So intermolecular force will be weak and low boiling point of tertiary amines

$$I=Ethyldimethyl\ amine=3^0$$

$$II=n-butylamine=1^0$$

$$III=diethylamine=2^0$$

In primary amines only one hydrogen is substituted by an alkyl group, in secondary amines, two hydrogens are substituted by alkyl groups and in tertiary amines, all the three hydrogens are substituted by alkyl groups thereby resulting in the absence of intermolecular hydrogen bonding. Higher the hydrogen bonding difficult is to break the bonds thereby high boiling point.

Intermolecular hydrogen bonding is maximum in primary amines and absent in tertiary amines.

So order of boiling points will be $$II>III>I$$
Question 3
1 / -0

Which is the strongest acid amongst the following compounds?

A

B

C

D
Question 4
1 / -0

Decreasing order of basicity is?
(i) $$CH_3NH_2$$
(ii) $$C_6H_5CH_2NH_2$$
(iii) $$p-CH_3O-C_6H_4-NH_2$$
(iv) $$m-CH_3O-C_6H_4-NH_2$$

A
I $$>$$ II $$>$$ III $$>$$ IV

B
II $$>$$ IV $$>$$ I $$>$$ III

C
II $$>$$ IV $$>$$ III $$>$$ I

D
IV $$>$$ I $$>$$ II $$>$$ III

Solution

Basicity depends upon the availability of the lone pair on the $$N$$atom. If the lone pair on $$N$$ is more available then it makes it a better Lewis base and if the lone pair is less available it makes the amine a weak Lewis base. Tertiary amines are the most basic followed by secondary, primary and phenyl amines. Amides are still weaker than the phenyl amines and can be considered neutral.
Therefore, the correct answer will be option A) $$I > II > III > IV$$
Question 5
1 / -0

I: $$CH_2 = CHCH_2NH_2,$$ II:$$ CH_3CH_2CH_2NH_2,$$III: $$CH\equiv CCH_2NH_2$$
What is the correct order of basicity of the above compounds?

A
I $$>$$ II $$>$$ III

B
II $$>$$ I $$>$$ III

C
III $$>$$ II $$>$$ I

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

Solution

For the basicity of compounds, specially amines, the $$+I$$ Effect on them is counted. Basicity is directly proportionality to the $$+I$$ effect.
Among the given compounds, maximum $$+I$$ effect is present in (II) as it is an alkyl group. Minimum $$+I$$effect is present in (III) as $$CH\equiv C-$$ group is present.
So, (II) compound is most basic and (III) is least basic among the given compounds.
$$II>I>III$$
Question 6
1 / -0

The order of basicity in increasing order is:

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

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

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

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

Solution
Question 7
1 / -0

Which of the following compound is strongest base?

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

B
$${ NH }_{ 2 }CO{ NH }_{ 2 }$$

C
$$NH_2 - \overset{NH}{\overset{||}{C}} - NH_2$$

D
$${ CH }_{ 3 }CONHCO{ CH }_{ 3 }$$

Solution
Question 8
1 / -0

What is the final product?

A

B

C

D

Solution
Question 9
1 / -0

The order of basic strength the above compound is:

A
$$a > b > c$$

B
$$b > a > c$$

C
$$c > a > b$$

D
$$c > b > a$$
Question 10
1 / -0

The decreasing order of boiling points of ethyldimethylamine, n- butylamine and diethylamine is n - Butylamine > Diethylamine > Ethyldimethylamine. This trend of the boiling point can be explained as:

A
boiling point increases with increase in molecular mass

B
tertiary amines have highest boiling point due to highest basicity

C
intermolecular hydrogen bonding is maximum in primary amines and absent in tertiary amines

D
intramolecular hydrogen bonding is present in tertiary amines

Solution

The decreasing order of boiling points:
n-butylamine(primary amine) > diethylamine(secondary amine) > ethyldimethylamine (tertiary amine).
$$ CH_3CH_2CH_2CH_2NH_2 > (C_2H_5)_2NH > C_2H_5NH(CH_3)_2$$
In primary amines only one hydrogen is substituted by an alkyl group, in secondary amines, two hydrogens are substituted by alkyl groups and in tertiary amines, all the three hydrogens are substituted by alkyl groups thereby resulting in the absence of intermolecular hydrogen bonding. Higher the hydrogen bonding difficult is to break the bonds thereby high boiling point.
Intermolecular hydrogen bonding is maximum in primary amines and absent in tertiary amines.