Organic Compounds Containing Nitrogen

NEET Chemistry · 61 questions · Page 6 of 7 · Click an option or "Show Solution" to reveal answer

Q51

In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br2 used per mole of amine produced are:

A Four moles of NaOH and two moles of Br2.

B Two moles of NaOH and two moles of Br2

C Four moles of NaOH and one mole of Br2

D One mole of NaOH and one mole of Br2

Correct Answer

Option C

Solution

4

moles of

NaOH

and one mole of

B{r_2}

is required during production of one mole of amine during Hoffmann's bromamide degradation reaction.

Q52

Ammonolysis of Alkyl halides followed by the treatment with NaOH solution can be used to prepare primary, secondary and tertiary amines. The purpose of NaOH in the reaction is :

A to remove acidic impurities

B to remove basic impurities

C to activate NH3 used in the reaction

D to increase the reactivity of alkyl halide

Correct Answer

Option A

Solution

During the reaction HX (acid) is formed Hence, we use NaOH to remove these acidic impurities

Q53

The correct order of increasing basic nature for the bases NH3, CH3NH2 and (CH3)2 NH is

A (CH3)2NH < NH3 < CH3NH2

B NH3 < CH3NH2 < (CH3)2NH

C CH3NH2 < (CH3)2NH < NH3

D CH3NH2 < NH3 < (CH3)2NH

Correct Answer

Option B

Solution

The alkyl groups are electron releasing group

\left( { + {\rm I}} \right),

thus increases the electron density around the nitrogen thereby increasing the availability of the lone pair of electrons to proton or lewis acid and making the amine more basic.

Hence more the no. of alkyl group more basic is the amine.

Therefore the correct order is

N{H_3} < C{H_3}N{H_2} < {\left( {C{H_3}} \right)_2}N

Q54

Given below are two statements: Statement I : Pure Aniline and other arylamines are usually colourless. Statement II : Arylamines get coloured on storage due to atmospheric reduction. In the light of the above statements, choose the most appropriate answer from the options given below :

A Both Statement I and Statement II are correct

B Statement I is correct but Statement II is incorrect

C Both Statement I and Statement II are incorrect

D Statement I is incorrect but Statement I is correct

Correct Answer

Option B

Solution

The most appropriate answer is Statement-I is correct but Statement-II is incorrect.

Pure Aniline and other arylamines are usually colorless, which is in agreement with Statement-I.

However, Statement-II is incorrect as arylamines do not get colored on storage due to atmospheric reduction.

Instead, get coloured on storage due to atmospheric oxidation and become darker in color upon exposure to air.

Q55

When a concentrated solution of sulphanilic acid and 1-naphthylamine is treated with nitrous acid

(273 \mathrm{~K})

and acidified with acetic acid, the mass

(\mathrm{g})

of 0.1 mole of product formed is : (Given molar mass in

\mathrm{g} \mathrm{mol}^{-1} \mathrm{H}: 1, \mathrm{C}: 12, \mathrm{~N}: 14, \mathrm{O}: 16, \mathrm{~S}: 32

)

A 330

B 33

C 343

D 66

Correct Answer

Option B

Solution

0.1 mole of red-azo dye $($ Molar Mass $=327 \mathrm{gm} / \mathrm{mol})$ will have 32.7 gm mass. Nearly 33 gm .

Q56

Match List I with List II 1 - Bromopropane is reacted with reagents in List I to give product in List II .tg .tg LIST I - Reagent LIST II - Product A.

\mathrm{KOH}

(alc) I. Nitrile B.

\mathrm{KCN}

(alc) II. Ester C.

\mathrm{AgNO_2}

III. Alkene D.

\mathrm{H_3CCOOAg}

IV. Nitroalkane Choose the correct answer from the options given below:

A A-I, B-II, C-III, D-IV

B A-IV, B-III, C-II, D-I

C A-III, B-I, C-IV, D-II

D A-I, B-III, C-IV, D-II

Correct Answer

Option C

Solution

For the reactions of bromopropane with the reagents in List I, the products formed in List II are as follows: A.

\mathrm{KOH}

(alc) - Alcoholic KOH is a strong base and will cause elimination of HBr, forming an alkene as the product.

So, A-III.

\mathrm{KCN}

(alc) - Potassium cyanide (KCN) will replace the bromine atom in the alkyl halide with a cyanide group, forming a nitrile.

So, B-I.

\mathrm{AgNO_2}

- Silver nitrite (AgNO2) will replace the bromine atom with a nitro group, forming a nitroalkane. So, C-IV. D.

\mathrm{H_3CCOOAg}

- Silver acetate (H3CCOOAg) will replace the bromine atom with an acetate group, forming an ester.

So, D-II.

The correct answer is Option C: A-III, B-I, C-IV, D-II.

Q57

Match List I with List II .tg .tg List I (Amines) List II (

\mathrm{pK_b}

) A. Aniline I. 3.25 B. Ethanamine II. 3.00 C. N-Ethylethanamine III. 9.38 D. N, N-Diethylethanamine IV. 3.29 Choose the correct answer from the options given below :

A A-III, B-II, C-I, D-IV

B A-III, B-II, C-IV, D-I

C A-III, B-IV, C-II, D-I

D A-I, B-IV, C-II, D-III

Correct Answer

Option C

Solution

Aromatic amines are less basic than aliphatic amines.

Among given aliphatic amines, $2^{\circ}$ amine is most basic, followed by $3^{\circ}$ amine and $1^{\circ}$ amine.

Therefore the correct basic strength $\left(\mathrm{K}_{\mathrm{b}}\right)$ order of the given amines is

\underset{(C)}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NHCH}_2 \mathrm{CH}_3}>\underset{(D)}{\left(\mathrm{CH}_3 \mathrm{CH}_2\right)_3 \mathrm{~N}}>\underset{(B)}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2}>\underset{(A)}{\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2}

The $\mathrm{pK}_{\mathrm{b}}$ order of the given amines will be just the opposite of their basic strength order.

The correct matching is $A-I I I, B-I V, C-I I, D-I$

Q58

A. Phenyl methanamine B. N,N-Dimethylaniline C. N-Methyl aniline D. Benzenamine Choose the correct order of basic nature of the above amines.

A D > C > B > A

B A > C > B > D

C D > B > C > A

D A > B > C > D

Correct Answer

Option D

Solution

In phenyl methanamine, the lone pair on nitrogen of -NH2 group is localised and does not undergoes resonance.

(B), (C) and (D) are aromatic amines in which lone pair of electrons of N-atoms goes in resonance (+R effect) with the benzene ring.

So, Lewis basicity or donation of lone of electrons of these amines will be decreased in comparison to (A).

+ I effects of two -CH3 groups increases electron density on N atom while lone pair of N atom take part in resonance with the benzene ring and decreases electron density on N atom.

Both this effect try to compensate each other.

+ I effects of one -CH3 groups increases electron density on N atom while lone pair of N atom take part in resonance with the benzene ring and decreases electron density on N atom.

Both this effect try to compensate each other.

It has no + I-effect on N-atom to increase electron density on the N atom which was decreased due to lone pair of N atom take part in resonance with the benzene ring.

So, A is purely aliphatic 1°-amine.

B is aromatic. 3°-amine with more aliphatic nature (for two -CH3 groups).

C is aromatic 2°-amine with less aliphatic nature (for one -CH3 group).

D is purely aromatic 1°-amine.

Hence basicity order is A > B > C > D.

Q59

Given below are two statements : Statement I : Piciric acid is 2,4,6 - trinitrotoluene. Statement II : Phenol - 2,4 - disulphonic acid is treated with Conc.

\mathrm{HNO}_3

to get picric acid. In the light of the above statements, choose the most appropriate answer from the options given below :

A Statement I is incorrect but Statement II is correct

B Both Statement I and Statement II are incorrect

C Both Statement I and Statement II are correct

D Statement I is correct but Statement II is incorrect

Correct Answer

Option A

Solution

Statement I is incorrect, and Statement II is correct.

The correct name for picric acid is 2,4,6-trinitrophenol, not 2,4,6-trinitrotoluene.

Meanwhile, phenol-2,4-disulphonic acid, when treated with concentrated nitric acid ( $\mathrm{HNO}_3$ ), results in the formation of picric acid.

Thus, the most appropriate option is : Option A Statement I is incorrect but Statement II is correct.

Q60

Ethylamine (C2H5NH2) can be obtained from N-ethylphatalimide on treatment with :

A CaH2

B H2O

C NaBH4

D NH2NH2

Correct Answer

Option D

Solution

N-ethyl phthalimide on treatment with NH2—NH2 gives ethylamine.

It is the final step of Galbriel phatalimide synthesis reaction.

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