Amines Test

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Amines Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

Acetonitrile on reduction gives:

A
ethanamine

B
propanamine

C
methanamine

D
None of the above

Solution

Acetonitrile on reduction gives ethanamine shown in the above reaction.
Question 2
1 / -0

The correct order of increasing basic nature for the bases $$NH_3, CH_3NH_2$$ and $$(CH_3)_2NH$$ in aqueous solutions.

A
$$NH_3 < CH_3NH_2 <(CH_3)_2NH$$

B
$$CH_3NH_2 < NH_3 < CH_3)_2NH$$

C
$$CH_3NH_2 < (CH_3)_2NH < NH_3$$

D
$$(CH_3)_2NH < NH_3 < CH_3NH_2$$

Solution

In aqueous solution, the correct order of increasing basic nature for the basicity is:

$$\displaystyle NH_3 < CH_3NH_2 <(CH_3)_2NH$$.

It is due to combination of the inductive effect, solvation effect, steric effect etc.
Question 3
1 / -0

Pick the correct statement among the following.

A
Sodium lauryl sulphate forms an insoluble scum with hard water.

B
Sodium dodecyl benzene sulphonate used in tooth paste is a cationic detergent.

C
Cetyl-trimethyl ammonium bromide is a popular cationic detergent used in hair conditioner.

D
Non-ionic detergents is formed when polyethylene glycol reacts with adipic acid.

Solution

The option (C) is the correct answer.
(A) Sodium lauryl sulphate does not forms an insoluble scum with hard water.
Builders present in detergent form soluble calcium and magnesium salts.

(B) Sodium dodecyl benzene sulphonate used in tooth paste is a anionic detergent.

(C) Cetyl-trimethyl ammonium bromide is a popular cationic detergent used in hair conditioner. It is alkyl ammonium salt.

(D) Non-ionic detergents is formed when polyethylene glycol reacts with steric acid.
Question 4
1 / -0

Among the following substituted pyridines, the most basic compound is:

A

B

C

D

Solution

4-dimethlyamino pyridine is the most basic compound. The lone pair of electrons on N atom of dimethyl amino group is in resonance with aromatic ring . This increases the electron density on ring nitrogen. Hence, the lone pair of electrons on ring nitrogen can be easily donated to a suitable acid.
Question 5
1 / -0

Between any two of the following molecules, hydrogen bonding is not possible:

A
two primary amine molecules

B
two secondary amine molecules

C
two tertiary amine molecules

D
two ammonia molecules

Solution

Between two tertiary amine molecules, hydrogen bonding is not possible.
The hydrogen bonding is possible when $$\displaystyle H$$ atom is attached to electronegative $$\displaystyle N, O$$ or $$\displaystyle F$$ atom. In tertiary amines, no $$\displaystyle H$$ atom is attached to $$\displaystyle N $$ atom. Hence, hydrogen bonding is not possible.
Question 6
1 / -0

Aniline is reacted with bromine water and the resulting product is treated with an aqueous solution of sodium nitrite in presence of dilute hydrochloric acid. The compound so formed is reacted with a tetrafluoroborate which is subsequently heated. The final product is:

A
$$p-bromoaniline$$

B
$$p-bromofluorobenzene$$

C
$$1, 3, 5-tribromobenzene$$

D
$$2, 4, 6 - tribromofluorobenzene$$

Solution

Option D is correct.
Question 7
1 / -0

Gabriel's phthalimide synthesis can be used to prepare:

A
ethanamine

B
n-methylmethanamine

C
benzeneamine

D
$$N, N-$$ dimethylmethanamine

E
p-toluidine

Solution

Gabriel phthalimide synthesis is used for the preparation of aliphatic primary amines.
Since ethanamine is the only primary amine among the given compounds, hence ethanamine is correct answer.
Question 8
1 / -0

Which one of the following is used as a test for aliphatic primary amines ?

A
Tollen's test

B
Fehling's lest

C
Isocyanide test

D
Azo dye test

E
Phthalein fusion test.

Solution

Isocyanide test (carbylamine test) is used to confirm primary amines.
$$R-NH_2 + CHCl_3 + 3KOH \rightarrow \underset{Isocyanide }{R-NC} + 3KCl + 3H_2O$$
Question 9
1 / -0

Which one of the following can be prepared by Gabriel phthalimide synthesis?

A
Aniline

B
$$o$$-toluidine

C
Benzylamine

D
N-methylethanamine

E
4-bromoaniline

Solution

Gabriel phthallmide synthesis gives primary aliphatic amines. Therefore, the product is
Question 10
1 / -0

Which of the following is an example of carbylamine reaction?

A
$$2PhNH_2+CS_2\xrightarrow[\begin{matrix}\Delta\\(ii)\,conc.HCl\end{matrix}]{(i)KOH}Ph-NCS$$

B
$$R-NH_2+HNO_2\xrightarrow{0-5^oC} R- OH+N_2+HCl$$

C
$$Ph-NH_2+CHCl_3+3KOH(alc.)\xrightarrow{\Delta} Ph-NC+3KCl+3H_2O$$

D
$$Ph_2-NH+HONO\rightarrow Ph_2-N-N=O+H_2O$$

Solution

(a) $$2Ph-NH_2+S=C=S\xrightarrow[\begin{matrix}\Delta\\(ii)\,conc.HCl\end{matrix}]{(i) 2KOH}\underset{\text{Hotmann mustard oil reaction}}{Ph-N=C=S+Ph\overset{\oplus}{N}H_3Cl^-}$$

(b) $$R-NH_2+HNO_2\xrightarrow{0-5^oC} \underset{\text{Diazotisation reaction}}{R- OH+N_2+HCl}$$

(c) $$Ph-NH_2+CHCl_3+3KOH(alc.)\xrightarrow{\Delta} \underset{\text{Carbylamine reaction}}{Ph-NC+3KCl+3H_2O}$$

(d) $$Ph_2-NH+HONO\rightarrow \underset{\text{Lieber mann's nitroso reaction}}{Ph_2-N-N=O+H_2O}$$

Option C is correct.