Match List I with List II. .tg .tg List I List II (A) Benzenesulphonyl chloride (I) Test for primary amines (B) Hoffmann bromamide reaction (II) Anti Saytzeff (C) Carbylamine reaction (III) Hinsb

(A) Benzene sulphonyl chloride is also known as Hinsberg reagent. (B) Hoffmann bromamide reaction involves conversion of amide to amine having one $\mathrm{C}$ atom less. This reaction involves isocyanate as intermediate. (C) Carbylamine reaction is a test given by all primary amines. (D) Hoffmann orientation refers to the addition of molecules to unsymmetrical alkenes according to anti Saytzeff's rule. Correct match is A-III, B-IV, C-I, D-II

Organic Compounds Containing Nitrogen

NEET Chemistry · 61 questions · Page 4 of 7 · Click an option or "Show Solution" to reveal answer

Q31

Compound A is converted to B on reaction with CHCl3 and KOH. The compound B is toxic and can be decomposed by C. A, B and C respectively are :

A primary amine, nitrile compound, conc. HCl

B secondary amine, isonitrile compound, conc. NaOH

C primary amine, isonitrile compound, conc. HCl

D secondary amine, nitrile compound, conc. NaOH

Correct Answer

Option C

Solution

Primary amine reacts in presence of chloroform gives isonitrile which is toxic in nature which further reacts in presence of HCl/H3O+ to give primary amine and acid.

Hence, A-primary amine, B-isonitrile compound and C-conc.

HCl.

Q32

Given below are two statements, one is labelled as Assertion (A) and other is labelled as Reason (R). Assertion (A) : Gabriel phthalimide synthesis cannot be used to prepare aromatic primary amines. Reason (R) : Aryl halides do not undergo nucleophilic substitution reaction. In the light of the above statements, choose the correct answer from the options given below :

A Both (A) and (R) true but (R) is not the correct explanation of (A)

B (A) is false but (R) is true.

C Both (A) and (R) true and (R) is correct explanation of (A).

D (A) is true but (R) is false.

Correct Answer

Option C

Solution

Gabriel phthalimide synthesis

Q33

Given below are two statements : Statement I : Aniline is less basic than acetamide. Statement II : In aniline, the lone pair of electrons on nitrogen atom is delocalized over benzene ring due to resonance and hence less available to a proton. Choose the most appropriate option;

A Statement I is true but statement II is false.

B Statement I is false but statement II is true.

C Both statement I and statement II are true.

D Both statement I and statement II are false.

Correct Answer

Option B

Solution

Explanation : aniline is more basic than acetamide because in acetamide, lone pair of nitrogen is delocalized to more electronegative element oxygen.

In Aniline lone pair of nitrogen delocalized over benzene ring.

Q34

Which of the following is not a correct statement for primary aliphatic amines?

A The intermolecular association in primary amines is less than the intermolecular association in secondary amines.

B Primary amines on treating with nitrous acid solution from corresponding alcohols except mythyl amine.

C Primary amines are less basic than the secondary amines.

D Primary amines can be prepared by the Gabriel phthalimide synthesis.

Correct Answer

Option A

Solution

The intermolecular association is more prominent in case of primary amines as compared to secondary, due to the availability of two hydrogen atom.

Q35

Given below are two statements : Statement I : In Hofmann degradation reaction, the migration of only an alkyl group takes place from carbonyl carbon of the amide to the nitrogen atom. Statement II : The group is migrated in Hofmann degradation reaction to electron deficient atom. 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 Both Statement I and Statement II are incorrect.

C Statement I is correct but Statement II is incorrect.

D Statement I is incorrect but Statement II is correct.

Correct Answer

Option D

Solution

Hofmann bromamide degradation In this degradation, the migration of the alkyl/aryl group occurs to the electron-deficient nitrogen (nitrene).

Statement (I) is not absolutely correct as it mentions only the alkyl group, whereas migration of aryl groups may also occur depending on migratory aptitude.

Statement (II) is correct as the migration occurs to the electron-deficient atom.

Q36

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Experimental reaction of

\mathrm{CH}_{3} \mathrm{Cl}

with aniline and anhydrous

\mathrm{AlCl}_{3}

does not give

o

and

p

-methylaniline. Reason (R): The

-\mathrm{NH}_{2}

group of aniline becomes deactivating because of salt formation with anhydrous

\mathrm{AlCl}_{3}

and hence yields

m

-methyl aniline as the product. In the light of the above statements, choose the most appropriate answer from the options given below :

A Both A and R are true and (R) is the correct explanation of (A).

B Both A and R are true but (R) is not the correct explanation of (A).

C (A) is true, but (R) is false.

D (A) is false, but (R) is true.

Correct Answer

Option C

Solution

Aniline does not undergo Friedel Craft reaction because the reagent AlCl3 being electron deficient acts as a Lewis acid.

Q37

Reaction of propanamide with

\mathrm{Br_2/KOH(aq)}

produces :

A Propanenitrile

B Propylamine

C Ethylnitrile

D Ethylamine

Correct Answer

Option D

Solution

Propanamide

\mathrm{\overset{{B{r_2}/KOH}}\longrightarrow}

Ethylamine

Q38

Which of the following nitrogen containing compound does not give Lassaigne's test ?

A Glycene

B Phenyl hydrazine

C Urea

D Hydrazine

Correct Answer

Option D

Solution

Hydrazine

(\mathrm{N}_2 \mathrm{H}_4)

doesn't contain any carbon atom and hence doesn't give Lassaigne test.

Q39

Hinsberg's reagent is :

A C6H5COCl

B C6H5SO2Cl

C SOCl2

D (COCl)2

Correct Answer

Option B

Solution

The correct answer is Option B: C6H5SO2Cl.

Hinsberg's reagent is benzenesulfonyl chloride (C6H5SO2Cl).

It is used to differentiate primary, secondary, and tertiary amines.

Here's why the other options are incorrect : Option A: C6H5COCl is benzoyl chloride, a reagent used in Friedel-Crafts acylation reactions.

Option C: SOCl2 is thionyl chloride, a reagent used to convert alcohols into alkyl chlorides.

Option D: (COCl)2 is oxalyl chloride, a reagent used for the conversion of carboxylic acids into acid chlorides.

Q40

Match with . .tg .tg

List - I		List - II
(A)	Benzenesulphonyl chloride	(I)	Test for primary amines
(B)	Hoffmann bromamide reaction	(II)	Anti Saytzeff
(C)	Carbylamine reaction	(III)	Hinsberg reagent
(D)	Hoffmann orientation	(IV)	Known reaction of Isocyanates.

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

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

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

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

Correct Answer

Option C

Solution

(A) Benzene sulphonyl chloride is also known as Hinsberg reagent.

(B) Hoffmann bromamide reaction involves conversion of amide to amine having one $\mathrm{C}$ atom less.

This reaction involves isocyanate as intermediate.

(D) Hoffmann orientation refers to the addition of molecules to unsymmetrical alkenes according to anti Saytzeff's rule.

Correct match is A-III, B-IV, C-I, D-II

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