Practice Start Test

Organic Compounds Containing Nitrogen

JEE Chemistry · 46 questions · Page 2 of 5 · Click an option or "Show Solution" to reveal answer

Q11
Amongst the following the most basic compound is
A benzylamine
B aniline
C acetanilide
D p-nitroaniline
Correct Answer
Option A
Solution

In others the basic character is suppressed due to Resonance (see applications of resonance).

Q12
Which one of the following methods is neither meant for the synthesis nor for separation of amines?
A Hinsberg method
B Hofmann method
C Wurtz reaction
D Curtius reaction
Correct Answer
Option C
Solution

Wurtz reaction is for the preparation of hydrocarbons from alkyl halide

RX+2Na+XR⟶ R−R+2NaXRX + 2Na + XR\overset{\,}\longrightarrow R - R + 2NaX
Q13
The test to distinguish primary, secondary and tertiary amines is :
A Carbylamine reaction
B C6H5SO2Cl
C Sandmeyer’s reaction
D Mustard oil test
Correct Answer
Option B
Solution

Hinsberg’s reagent (C6H5SO2Cl) forms monoalkyl sulphonamide with 1o amines which is soluble in KOH.

With 2o amines it gives dialkyl sulphonamide which is insoluble in KOH and with 3o amines there is no reaction.

In mustard oil test, 1o amines on action of CS2 and HgCl2 give alkyl isothiocyanate having mustard oil smell. 2o amines react with CS2 but not with HgCl2 while 3o amines give no reaction.

However, this test is not able to distinguish 2o and 3o amines.

Q14
A compound 'X' on treatment with Br2/NaOH, provided C3H9N, which gives positive carbylamine test. Compound 'X' is :
A CH3CH2CH2CONH2
B CH3CON(CH3)2
C CH3CH2COCH2NH
D CH3COCH2NHCH3
Correct Answer
Option A
Solution

Br2/NaOH(Hoffman Hypobromide reagent) converts amide into primary amine having one carbon atom less, which gives carbylamine test.

In the Amide there should be -CONH2 group to produce primary amine.

Q15
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.

Q16
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

Q17
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.

Q18
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.

Q19
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.

Q20
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Experimental reaction of CH3Cl\mathrm{CH}_{3} \mathrm{Cl} with aniline and anhydrous AlCl3\mathrm{AlCl}_{3} does not give oo and pp-methylaniline. Reason (R): The −NH2-\mathrm{NH}_{2} group of aniline becomes deactivating because of salt formation with anhydrous AlCl3\mathrm{AlCl}_{3} and hence yields mm-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.

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