Organic Compounds Containing Nitrogen

JEE Chemistry · 46 questions · Page 5 of 5 · Click an option or "Show Solution" to reveal answer

Q41

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.

Q42

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 $A-I I I, B-I V, C-I I, D-I$

Q43

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.

Q44

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.

Q45

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.

Q46

The most appropriate reagent for conversion of C2H5CN into CH3CH2CH2NH2 is

A NaBH4

B CaH2

C Na(CN)BH3

D LiAlH4

Correct Answer

Option D

Solution

CH3–CH2–C $\equiv$ N

\overset{{LiAl{H_4}}}\longrightarrow

CH3–CH2–CH2–NH2

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