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Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - JEE MCQ


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26 Questions MCQ Test 35 Years Chapter wise Previous Year Solved Papers for JEE - Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen

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Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 1

PASSAGE - 1

The conversion of an amide to an amine with one carbon atom less by the action of alkaline hydrohalite is known as Hofmann bromamide degradation.



In this reaction, RCONHBr is formed from which the reaction has derived its name. Hofmann reaction is accelerated if the migrating group is more electron-releasing. Hofmann degradation reaction is an intramolecular reaction.

Q. How can the conversion of (i) to (ii) be brought about?

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 1

The reagent used in Hofmann bromamide reaction is alkaline halogen (NaOH or KOH + X2)

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 2

PASSAGE - 1

The conversion of an amide to an amine with one carbon atom less by the action of alkaline hydrohalite is known as Hofmann bromamide degradation.



In this reaction, RCONHBr is formed from which the reaction has derived its name. Hofmann reaction is accelerated if the migrating group is more electron-releasing. Hofmann degradation reaction is an intramolecular reaction.

Q. Which is the rate determining step in Hofmann bromamide degradation?

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 2

Conversion of (iii) to (iv) involving rearrangement is the slowest step. Species (iii) is electron deficient (N has only 6 electrons), hence it has a tendency to get its octet completed by migration of alkyl group.

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Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 3

PASSAGE - 1

The conversion of an amide to an amine with one carbon atom less by the action of alkaline hydrohalite is known as Hofmann bromamide degradation.



In this reaction, RCONHBr is formed from which the reaction has derived its name. Hofmann reaction is accelerated if the migrating group is more electron-releasing. Hofmann degradation reaction is an intramolecular reaction.

Q. What are the constituent amines formed when the mixture of (i) and (ii) undergoes Hofmann bromamide degradation?

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 3

Since the reaction is intramolecular, no cross product will be formed.

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 4

PASSAGE - 2

Treatment of compound O with KMnO4/H+ gave P, which on heating with ammonia gave Q. The compound Q on treatment with Br2/NaOH produced R. On strong heating, Q gave S, which on further treatment with ethyl 2-bromopropanoate in the presence of KOH followed by acidification, gave a compound T.

Q. The compound R is

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 5

PASSAGE - 2

Treatment of compound O with KMnO4/H+ gave P, which on heating with ammonia gave Q. The compound Q on treatment with Br2/NaOH produced R. On strong heating, Q gave S, which on further treatment with ethyl 2-bromopropanoate in the presence of KOH followed by acidification, gave a compound T.

Q. The compound T is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 5

                                                             

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 6

Read the following Statement-1(Asseration)  and Statement -2 (Reason) and answer as per the options given below :

Q.

Statement - 1: p-Nitrophenol is a stronger acid than o-nitrophenol.

Statement - 2 : Intramolecular hydrogen bonding makes the o-isomer weaker than the p-isomer.

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 6

In o-nitrophenol intramolecular H-bonding is possible because OH and NO2 groups are close to each other.
This makes the ortho isomer less acidic as its capacity to donate a proton (H-atom) decreases. There is no such intramolecular H–bonding in the p-isomer.

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 7

Statement - 1:  Benzonitrile is prepared by the reaction of chlorobenzene with potassium cyanide.

Statement - 2 :  Cyanide (CN) is a strong nucleophile.

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 7

Chlorobenzene is resonance stabilized.
Thus aryl halides (chlorobenzene) do not undergo nucleophilic substitution. Reason is correct.

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 8

Statement - 1 :  In strongly acidic solutions, aniline becomes more reactive towards electrophilic reagents.

Statement-2 : The amino group being completely protonated in strongly acidic solution, the lone pair of electrons on the nitrogen is no longer available for resonance.

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 8

TIPS/FORMULAE : Electron donating tendency to a double bond is called +M effect and the transfer of electrons take place towards the attacking reagent due to +E effect.
In strongly acidic conditions, aniline becomes protonated with the result lone pair of electrons is not available to produce +E and +M effects. Thus here aniline becomes less reactive towards electrophilic substitution. On the other hand, the   group exerts strong –I effect causing deactivation of the

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 9

Statement - 1 :  Aniline on reaction with NaNO2 / HCl at 0oC followed by coupling with β-naphthol gives a dark blue precipitate. and

Statement - 2 : The colour of the compound formed in the reaction of aniline with NaNO2/HCl at 0oC followed by coupling with β-naphthol is due to the extended conjugation.

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 9

The colour of the azo dye formed will be orange red but not blue. However, the colour of dye can said to be due to extended conjugation due to presence of azo group.

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 10

When primary amine reacts with chloroform in ethanolic KOH then the product is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 10

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 11

The reaction of chloroform with alcoholic KOH and p-toluidine forms

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 11

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 12

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

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 12

The alkyl groups are electron releasing group (+ I), 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

NH3 < CH3NH2 < (CH3)2 N

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 13

Ethyl isocyanide on hydrolysis in acidic medium generates

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 13

Ethyl isocyanide on hydrolysis form primary amines.

Therefore it gives only one mono chloroalkane.

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 14

Which one of the following methods is neither meant for the synthesis nor for separation of amines? 

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 14

Wurtz reaction is for the preparation of hydrocarbons from alkyl halide

RX + 2Na + XR → R – R + 2NaX

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 15

Amongst the following the most basic compound is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 15

is most basic. Inothers the basic character is suppressed due to Resonance (see applications of resonance).

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 16

An organic compound having molecular mass 60 is found to contain  C = 20%, H = 6.67% and N = 46.67% while rest is oxygen. On heating it gives NH3 alongwith a solid residue.

The solid residue give violet colour with alkaline copper sulphate solution. The compound is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 16


The compound is CH4N2O
Empirical weight = 60; Mol. wt. = 60; 
Molecular formula

On heating urea loses ammonia to give Biuret

2NH2CONH2 → H2 NCO.NH.CONH2 + NH3

Biuret with alkaline CuSO4 gives violet colour. Test for –CONH group.

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 17

Which one of the following is the strongest base in aqueous solution ?

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 17

NOTE : Aromatic amines are less basic than aliphatic amines. Among aliphatic amines the order of basicity is 2° > 1° > 3°. The electron density is decreased in 3° amine due to crowding of alkyl group over N atom which makes the approach and bonding by a proton relatively difficult. Therefore the basicity decreases. Further Phenyl group show – I effect, thus decreases the electron density on nitrogen atom and hence the basicity.
∴ dimethylamine (2° aliphatic amine) is strongest base among given choices.
∴ The correct order of basic strength is Dimethylamine > Methyl amine > Trimethyl amine > Aniline.

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 18

In the chemical reaction,

CH3CH2NH2 + CHCl3 + 3KOH → (A) + (B) + 3H2O, the compounds (A) and (B) are respectively

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 18

This is carbylamine reaction.

CH3CH2NH2 + CHCl3 + 3KOH → C2H5NC + 3KCl  + 3H2O

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 19

In the chemical reactions,

the compounds ‘A’ and ‘B’ respectively are

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 19

Primary aromatic amines react with nitrous acid to yield arene diazonium salts.

The diazonium group can be replaced by fluorine by treating the diazonium salt with fluoroboric acid (HBF4). The precipitated diazonium fluoroborate is isolated, dried and heated until decomposition occurs to yield the aryl fluoride. This reaction is known as Balz-Schiemann reaction.

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 20

A compound with molecular mass 180 is acylated with CH3COCl to get a compound with molecular mass 390. The number of amino groups present per molecule of the former compound is :

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 20

Now since the molecular mass increases by 42 unit as a result of the reaction of one mole of CH3COCl with one-NH2 group and the given increase in mass is 210. hence the number of –NH2 group is = 210/42 = 5

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 21

An organic compound A upon reacting with NH3 gives B. On heating B gives C. C in presence of KOH reacts with Br2 to given CH3CH2NH2. A is :

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 21

Reaction (III) is a Hofmann bromamide reaction formation of CH3CH2NH2 is possible only from a compound CH3CH2CONH2 which can be obtained from the compound CH3CH2COO NH+4 (B) in (II) reaction further propanic acid (CH3CH2COOH) on reaction with NH3 produce CH3CH2COONH4 (reaction I) hence the reaction will be


                       

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 22

The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was :

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 22

Methyl isocyanate CH3 – N = C = O

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 23

On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the organic compound formed is:

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 23

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 24

Considering the basic strength of amines in aqueous solution, which one has the smallest pKb value?   

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 24

Arylamines are less basic than alkyl amines and even ammonia. This is due to resonance. In aryl amines the lone pair of electrons on N is partly shared with the ring and is thus less available for sharing with a proton.
In alkylamines, the electron releasing alkyl group increases the electron density on nitrogen atom and thus also increases the ability of amine for protonation.
Hence more the no. of alkyl groups higher should be the basicity of amine. But a slight discrepancy occurs in case of trimethyl amines due to steric effect. Hence the correct  order is

(CH3)2 NH > CH3NH2> (CH3)3 N> C6H5 NH2

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 25

In the reaction

the product E is :

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 25

Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 26

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

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - Question 26

4 moles of NaOH and one mole of Br2 is required during production of one mole of  amine during Hoffmann's bromamide degradation  reaction.

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