MCQ Of Compounds Containing Nitrogen, Past Year Questions JEE Mains, Class 12, Chemistry


26 Questions MCQ Test Class 12 Chemistry 35 Years JEE Mains &Advance Past yr Paper | MCQ Of Compounds Containing Nitrogen, Past Year Questions JEE Mains, Class 12, Chemistry


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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?

Solution:

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

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?

Solution:

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.

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?

Solution:

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

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

Solution:
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

Solution:

                                                             

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.

Solution:

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.

QUESTION: 7

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

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

Solution:

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

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.

Solution:

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

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.

Solution:

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.

QUESTION: 10

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

Solution:

QUESTION: 11

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

Solution:

QUESTION: 12

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

Solution:

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

QUESTION: 13

Ethyl isocyanide on hydrolysis in acidic medium generates

Solution:

Ethyl isocyanide on hydrolysis form primary amines.

Therefore it gives only one mono chloroalkane.

QUESTION: 14

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

Solution:

Wurtz reaction is for the preparation of hydrocarbons from alkyl halide

RX + 2Na + XR → R – R + 2NaX

QUESTION: 15

Amongst the following the most basic compound is

Solution:

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

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

Solution:


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.

QUESTION: 17

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

Solution:

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.

QUESTION: 18

In the chemical reaction,

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

Solution:

This is carbylamine reaction.

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

QUESTION: 19

In the chemical reactions,

the compounds ‘A’ and ‘B’ respectively are

Solution:

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.

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 :

Solution:

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

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 :

Solution:

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


                       

QUESTION: 22

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

Solution:

Methyl isocyanate CH3 – N = C = O

QUESTION: 23

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

Solution:

QUESTION: 24

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

Solution:

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

QUESTION: 25

In the reaction

the product E is :

Solution:

QUESTION: 26

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

Solution:

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