Amines Test

Result

Amines Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following will not undergo diazotisation ?

A
m-toluidine

B
Aniline

C
p-aminophenol

D
Benzyl amine

Solution

Benzylamine reacts with nitrous acid to form unstable diazonium salt, which in turn gives alcohol with the evolution of nitrogen gas.
Question 2
1 / -0

The molecule which does not exhibit strong hydrogen bonding is$$:$$

A
Methyl amine

B
Diethyl ether

C
Acetic acid

D
Glucose

Solution

The molecule which does not exhibit strong hydrogen bonding is $$\displaystyle diethyl \: ether$$ $$\displaystyle CH_3-O-CH_3$$.
Hydrogen bonding is possible when $$\displaystyle H$$ atom is attached to an electronegative $$\displaystyle N, O$$ or $$\displaystyle F$$ atom.
Hence, the option (B) is the correct answer.
Question 3
1 / -0

Which one of the following is strongly basic?

A
Dimethyl amine

B
Methyl amine

C
Ammonia

D
Aniline

Solution

Dimethyl amine $$\displaystyle (CH_3)_2-NH$$ is strongly basic due to electron releasing +I effect of two methyl groups that increase the electron density on N atom so that N can easily donate lone pair of electrons. Also, the dimethyl ammonium ion is stabilized due to the dispersal of positive charge by the +I effect of two methyl groups.

Aniline is least basic as the lone pair of electrons on N atom is in resonance with benzene ring and is less available for donation.
Question 4
1 / -0

Correct gradation of basic character is:

A
$$NH_3 > CH_3NH_2 > NF_3$$

B
$$CH_3NH_2 > NH_3 > NF_3$$

C
$$NF_3 > CH_3NH_2 > NH_3$$

D
$$CH_3NH_2 > NF_3 > NH_3$$

Solution

Correct gradation of basic character is $$\displaystyle CH_3NH_2 > NH_3 > NF_3$$. Due to $$\displaystyle +I$$ effect of methyl group, the electron density on $$\displaystyle N $$ increases.

Hence, $$\displaystyle CH_3-NH_2$$ is most basic as it can easily donate lone pair of electrons on N.

Due to $$\displaystyle -I$$ effect of F atom, the electron density on $$\displaystyle N $$ decreases. Hence, $$\displaystyle NF_3$$ is most basic as it can easily donate lone pair of electrons on N.
Question 5
1 / -0

Which one of the following on reduction with lithiium aluminium hydride yield a secondary amine?

A
Nitroethane

B
Methyl isocyanide

C
Acetamide

D
Methyl cyanide

Solution
Question 6
1 / -0

Which of the following compound is amphoteric in nature?

A

B
$$CH_3COOH$$

C

D
$$CH_3CONH_2$$

Solution

Due to presence of lp of electrons over $$N-$$atom of amide and having resonance structure, amides are amphoteric in nature.
Question 7
1 / -0

The product X and Y in the following reaction sequence are:

A

B

C

D

Solution

$$\displaystyle Nitrobenzene$$ on reduction with $$\displaystyle Sn/HCl$$ gives $$\displaystyle aniline$$.

Diazotization of $$\displaystyle aniline$$ with $$\displaystyle NaNO_2 / HCl$$ followed by heating with $$\displaystyle CuBr$$ gives $$\displaystyle bromobenzene$$.

Option B is correct.
Question 8
1 / -0

$$C_3H_9N$$ reacts with Hinsberg reagent and the product is insoluble in alkali but soluble in ether. This nitrogen containing compound is:

A
primary amine

B
secondary amine

C
tertiary compound

D
nitrite compound

Solution

The compound with molecular formula $$C_3H_6N$$ is a secondary amina. When it reacts with Hinsberg reagent, the product obtained is soluble in ether.
Question 9
1 / -0

In case of substituted aniline the group which decreases the basic strength is :

A
$$- OCH_3$$

B
$$-CH_3$$

C
$$-NH_2$$

D
$$-C_6H_5$$

Solution

In case of substituted aniline the group which decreases the basic strength is phenyl group $$\displaystyle (C_6H_5)$$. It withdraws the electron density from the ring through resonance.
Question 10
1 / -0

Select the basic strength order of following molecule.

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

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

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

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

Solution

Electron donating group increases basicity. Ortho-effect decreases basicity (least). So the order is given in the image.