Amines Test

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Amines Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

The source of nitrogen in Gabriel synthesis of amines is:

A
sodium azide $$Na{ N }_{ 3 }$$

B
potassium phthalimide $${ C }_{ 6 }{ H }_{ 4 }{ \left( CO \right) }_{ 2 }{ N }^{ - }{ K }^{ + }$$

C
sodium cyanide

D
potassium nitrate

Solution

In the Gabriel synthesis of primary amines, N-potassiophthalimide is used as a source of the nitrogen atom
Question 2
1 / -0

Coupling of diazonium salts of the following takes place in the order:

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

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

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

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

Solution

Coupling of diazonium salts takes place in the given order due to the groups attached to the benzene nucleus. (More the donating group more is the reactivity)
Question 3
1 / -0

What is the name of red azo dye ?

A
$$\beta$$-naphthyl azo benzene

B
p-hydroxy azo benzene

C
p-amino azo benzene

D
p-N, N-dimethyl amino azo benzene

Solution
Question 4
1 / -0

Compare boiling point of isomeric alkyl amines.

A
$${ 1 }^{ o }>{ 2 }^{ o }>{ 3 }^{ o }$$

B
$${ 1 }^{ o }>{ 2 }^{ o }<{ 3 }^{ o }$$

C
$${ 1 }^{ o }<{ 2 }^{ o }<{ 3 }^{ o }$$

D
$${ 1 }^{ o }<{ 2 }^{ o }>{ 3 }^{ o }$$

Solution

Among isomeric amines, the primary and secondary amines have highest boiling point due to their large tendency to form hydrogen bonds whereas tertiary amines have a lowest boiling point because of their inability to form hydrogen bond due to the absence of hydrogen atom.

Therefore, the order of boiling point of isomeric amine is as follows $${ 1 }^{ o }>{ 2 }^{ o }>{ 3 }^{ o }$$ amines
Question 5
1 / -0

Which of the following is more basic than aniline ?

A
Diphenylamine

B
Triphenylamine

C
p-nitroaniline

D
Benzylamine

Solution

Benzyl amine, $$C_6H_5CH_2\ddot{N}H_2$$ is more basic that aniline because be group $$C_6H_5CH_2$$ electron donating group due to $$+I$$ effect. So, It is able to Increase the electron of N of $$-NH_2$$ group. Thus, due to higher electron density, rate of donation of free pair of electron is increased i.e., basic character is higher. Phenyl and nitro groups are electron attractive group. So, they are able to decrease the electron density of $$N$$ of $$\ddot{N}H_2$$ group. Hence, they are less basic with aniline.
Question 6
1 / -0

The strongest base among the following is :

A
$$ C_6 H_5 NH_2 $$

B
$$ (C_6 H_5 )_2 NH $$

C
$$ NH_3 $$

D
$$ (C_2 H_5)_2 NH $$

Solution

$$ (C_2 H_5)_2 NH(2^o $$ amine) is the strongest base. Basic nature of amines is due to the presence of lone pair of electrons on nitrogen atom which is available for the bond formation with Lewis acid. Due to the $$+I$$ effect, $$2^o$$ amine is better base than $$1^o$$ amine and $$NH_3.$$ In case of aromatic amines, the lone pair on nitrogen atom involved in resonance, therefore, not available for bond formation, so aromatic amines are less basic.
Question 7
1 / -0

When ammonium cyanate is heated for a long time, the product is:

A
Nitrous oxide

B
Ammonium cyanide

C
Nitrogen and water

D
Biuret

Solution

When ammonium cyanate is heated urea is first formed
$$\underset { Ammonium\\ Cyanate }{ { NH }_{ 4 }CNO } \xrightarrow [ ]{ Heat } \underset { urea }{ { NH }_{ 2 }CO{ NH }_{ 2 } } \quad \quad $$

Urea on further heating gives biuret
$${ NH }_{ 2 }CO{ NH }_{ 2 }\xrightarrow [ ]{ Heat } \underset { Biuret }{ { NH }_{ 2 }CONHCO{ NH }_{ 2 } } +{ NH }_{ 3 }$$
Question 8
1 / -0

What is the molarity of 0.2 N Na2 CO3 solution ?

A
0.1M

B
0M

C
0.4M

D
0.2M

Solution

Given: Normality = 0.2N
Solution: Molarity refers to the number of moles dissolved per litre of the solution.
As we know that,
$$Normality= molarity\times n$$ (1)
where n is the valency factor.
For $$Na_{2}CO_{3}$$ n=2.
From equation 1 we get,
$$molarity= \frac{Normality}{n}$$
Substituting the values we get,
$$molarity= \frac{0.2}{2}$$
molarity = 0.1M
Hence, the correct optionis (A).
Question 9
1 / -0

Half-life period of a radioactive element is 100 yr. How long will it take for its 93.75% decay ?

A
400 yrs

B
300 yrs

C
200 yrs

D
193 yrs

Solution

Given : Half-life period of a radioactive element is 100 yr. How long will it take for its 93.75% decay ?

Solution :
Since it is decayed by 93.75 % so left is 6,25%
That is $$\frac {1}{16}$$ of total element

$$\frac{1}{2^n}$$= $$\frac{1}{16}$$ where n is number of half life.
n=4
To decay 93.75% time needed is $$4\times half -life $$ = $$4 \times 100$$ = 400 yr
The Correct Opt = A
Question 10
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Which of following statements is/are the not correct?

A
$$II$$ is more soluble than $$I$$ in $${H}_{2}O$$

B
$$II$$ is more soluble than $$III$$ in $${H}_{2}O$$

C
$$II$$ is less soluble than $$III$$ in $${H}_{2}O$$

D
$$III$$ is more soluble than $$I$$ in $${H}_{2}O$$

Solution

Among the given compounds, compound (I) is very less soluble in $$H_2O$$ because of large hydrophobic part, i.e. Carbon ring presence.
Compound (II) is less soluble than compound (III) because there is more H-bonding in case of compound (III).

So, statement (B) is not correct.