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All questions of Compounds Containing Nitrogen for JEE Exam
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.- a)Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
- b)Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
- c)Statement-1 is True, Statement-2 is False
- d)Statement-1 is False, Statement-2 is True
Correct answer is option 'D'. Can you explain this answer?
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.
a)
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
b)
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
c)
Statement-1 is True, Statement-2 is False
d)
Statement-1 is False, Statement-2 is True
|
Krishna Iyer answered |
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
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
p-Chloroaniline and anilinium hydrochloride can be distinguished by- a)Sandmeyer reaction
- b)NaHCO3
- c)AgNO3
- d)Carbylamine test
Correct answer is option 'C'. Can you explain this answer?
p-Chloroaniline and anilinium hydrochloride can be distinguished by
a)
Sandmeyer reaction
b)
NaHCO3
c)
AgNO3
d)
Carbylamine test
|
Chirag Verma answered |
TIPS/Formulae : Anilinium hydrochloride has ionisable chlorine whereas chlorobenze has non ionizable chlorine. Thus anilium hydrochloride gives white precipitate of AgCl with AgNO3.
In chloroaniline, –Cl is directly attached to benzene ring, hence it is non-reactive.
Which one of the following methods is neither meant for the synthesis nor for separation of amines? - a)Curtius reaction
- b)Wurtz reaction
- c)Hofmann method
- d)Hinsberg method
Correct answer is option 'B'. Can you explain this answer?
Which one of the following methods is neither meant for the synthesis nor for separation of amines?
a)
Curtius reaction
b)
Wurtz reaction
c)
Hofmann method
d)
Hinsberg method
|
Rohit Jain answered |
Wurtz reaction is for the preparation of hydrocarbons from alkyl halide
RX + 2Na + XR → R – R + 2NaX
Examine the following two structures for the anilinium ion and choose the correct statement from the ones given below:
- a)II is not an acceptable canonical structure because carbonium ions are less stable than ammonium ions.
- b)II is not an acceptable canonical structure because it is non-aromatic.
- c)II is not an acceptable canonical structure because the nitrogen has 10 valence electrons.
- d)II is an acceptable canonical structure
Correct answer is option 'C'. Can you explain this answer?
Examine the following two structures for the anilinium ion and choose the correct statement from the ones given below:
a)
II is not an acceptable canonical structure because carbonium ions are less stable than ammonium ions.
b)
II is not an acceptable canonical structure because it is non-aromatic.
c)
II is not an acceptable canonical structure because the nitrogen has 10 valence electrons.
d)
II is an acceptable canonical structure
|
|
Rohit Jain answered |
In Ist structure N has complete octet , whereas in IInd structure N has 10e in its valence shell.
Because of 1 double and 3 single bonds of N.
Because of 1 double and 3 single bonds of N.
If two compounds have the same empirical formula but different molecular fomulae they must have- a)different percentage composition
- b)different molecular weight
- c)same viscosity
- d)same vapour density
Correct answer is option 'B'. Can you explain this answer?
If two compounds have the same empirical formula but different molecular fomulae they must have
a)
different percentage composition
b)
different molecular weight
c)
same viscosity
d)
same vapour density
|
Gaurav Kumar answered |
TIPS/Formulae : Empirical Formula = n × Molecular formula
Solution : Since the molecular formula is n times the empirical formula, therefore, different compounds having the same empirical formula must have different molecular weights.
Solution : Since the molecular formula is n times the empirical formula, therefore, different compounds having the same empirical formula must have different molecular weights.
In the compound given below the correct order of the acidity of the positions X, Y and Z is
- a)Z > X > Y
- b)X > Y > Z
- c)X > Z > Y
- d)Y > X > Z
Correct answer is option 'B'. Can you explain this answer?
In the compound given below the correct order of the acidity of the positions X, Y and Z is
a)
Z > X > Y
b)
X > Y > Z
c)
X > Z > Y
d)
Y > X > Z
|
Tarun Kaushik answered |
(i) Position (X) is most acidic due to – COOH group.
(ii) –NH3+ group at position Y is more acidic than at Z because of presence of electron withdrawing – COOH group in close proximity. Hence – NH3+ group at position Z is least acidic.
(ii) –NH3+ group at position Y is more acidic than at Z because of presence of electron withdrawing – COOH group in close proximity. Hence – NH3+ group at position Z is least acidic.
The major product obtained when Br2/Fe is treated with
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
The major product obtained when Br2/Fe is treated with
a)
b)
c)
d)
|
Ananya Das answered |
TIPS/Formulae :– NH is an activating group where as 
group is a deactivating group.
Hence electrophilic substitution will be governed by the ring having
group is a deactivating group.Hence electrophilic substitution will be governed by the ring having
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- a)CH3CH2CONH2
- b)(NH2)2 CO
- c)CH3CONH2
- d)CH3NCO
Correct answer is option 'B'. Can you explain this answer?
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
a)
CH3CH2CONH2
b)
(NH2)2 CO
c)
CH3CONH2
d)
CH3NCO
|
|
Utkarsh Pandey answered |
Given: C=20% , H=6.67% and N=46.67%
Therefore O= 100-(20 + 6.67 + 46.67) = 26.66%
Mol mass is given at 60
Wt. of each element = C= 20% = 12 i.e 1 atom of C as at mass of C=12
H = 6.67% = 4 i.e. 4 atoms of H
N = 46.67% = 28 i.e. 2 atoms of N
O = 26.66% = 15.96 = 1 atom of O
Mol formula = CH4N2O =
As the copper sulphate changes color the test implies presence of peptide bond. -CO-NH-
Also, the loss of NH3 on heating also justifies this fact
Hence the structure should be NH2-CO-NH2
Compound 'A' (molecular formula C3H8O) is treated with acidified potassium dichromate to form a product 'B' (molecular formula C3H6O). 'B' forms a shining silver mirror on warming with ammonical silver nitrate. 'B' when treated with an aqueous solution of H2NCONHNH2.HCl and sodium acetate gives a product 'C'. Identify the structure of 'C'.- a)CH3CH2CH = NNHCONH2
- b)
- c)
- d)CH3CH2CH = NCONHNH2
Correct answer is option 'A'. Can you explain this answer?
Compound 'A' (molecular formula C3H8O) is treated with acidified potassium dichromate to form a product 'B' (molecular formula C3H6O). 'B' forms a shining silver mirror on warming with ammonical silver nitrate. 'B' when treated with an aqueous solution of H2NCONHNH2.HCl and sodium acetate gives a product 'C'. Identify the structure of 'C'.
a)
CH3CH2CH = NNHCONH2
b)
c)
d)
CH3CH2CH = NCONHNH2
|
Tejas Verma answered |
NOTE : Reaction of (B) indicates that it is an aldehyde which thus should be C2H5CHO or CH3CH2CHO, hence C should be CH3CH2CH = NNHCONH2
Statement - 1: Benzonitrile is prepared by the reaction of chlorobenzene with potassium cyanide.Statement - 2 : Cyanide (CN–) is a strong nucleophile.- a)Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
- b)Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
- c)Statement-1 is True, Statement-2 is False
- d)Statement-1 is False, Statement-2 is True
Correct answer is option 'D'. Can you explain this answer?
Statement - 1: Benzonitrile is prepared by the reaction of chlorobenzene with potassium cyanide.
Statement - 2 : Cyanide (CN–) is a strong nucleophile.
a)
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
b)
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
c)
Statement-1 is True, Statement-2 is False
d)
Statement-1 is False, Statement-2 is True
|
|
Chirag Verma answered |
Chlorobenzene is resonance stabilized.
Thus aryl halides (chlorobenzene) do not undergo nucleophilic substitution. Reason is correct.
Thus aryl halides (chlorobenzene) do not undergo nucleophilic substitution. Reason is correct.
Considering the basic strength of amines in aqueous solution, which one has the smallest pKb value? - a)(CH3)2NH
- b)CH3NH2
- c)(CH3)3N
- d)C6H5NH2
Correct answer is option 'A'. Can you explain this answer?
Considering the basic strength of amines in aqueous solution, which one has the smallest pKb value?
a)
(CH3)2NH
b)
CH3NH2
c)
(CH3)3N
d)
C6H5NH2
|
|
Anaya Patel answered |
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
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
The correct stability order of the following resonance structures is
- a)(I) > (II) > (IV) > (III)
- b)(I) > (III) > (II) > (IV)
- c)(II) > (I) > (III) > (IV)
- d)(III) > (I) > (IV) > (II)
Correct answer is option 'B'. Can you explain this answer?
The correct stability order of the following resonance structures is
a)
(I) > (II) > (IV) > (III)
b)
(I) > (III) > (II) > (IV)
c)
(II) > (I) > (III) > (IV)
d)
(III) > (I) > (IV) > (II)
|
|
Sonu Kumar answered |
Greater the no. of bonds greater the stability and neg. charge on more electronegative atom are more stable
PASSAGE - 1The 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?- a)KBr
- b)KBr + CH3ONa
- c)KBr + KOH
- d)Br2 + KOH
Correct answer is option 'D'. Can you explain this answer?
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?
a)
KBr
b)
KBr + CH3ONa
c)
KBr + KOH
d)
Br2 + KOH
|
|
Geetika Shah answered |
The reagent used in Hofmann bromamide reaction is alkaline halogen (NaOH or KOH + X2)
The correct order of increasing basic nature for the bases NH3, CH3NH2 and (CH3)2 NH is- a)(CH3)2NH < NH3 < CH3NH2
- b)NH3 < CH3NH2 < (CH3)2NH
- c)CH3NH2 < (CH3)2NH < NH3
- d)CH3NH2 < NH3 < (CH3)2NH
Correct answer is option 'B'. Can you explain this answer?
The correct order of increasing basic nature for the bases NH3, CH3NH2 and (CH3)2 NH is
a)
(CH3)2NH < NH3 < CH3NH2
b)
NH3 < CH3NH2 < (CH3)2NH
c)
CH3NH2 < (CH3)2NH < NH3
d)
CH3NH2 < NH3 < (CH3)2NH
|
|
Utkarsh Pandey answered |
Except the amines containing tertiary butyl group, all lower aliphatic amines are stronger bases than ammonia because of +I (inductive) effect. The alkyl groups, which are electron releasing groups, increase the electron density around the nitrogen thereby, increasing the availability of the lone pair of electrons to proton or Lewis acids and making the amine more basic.
The observed order in the case of lower members is found to be as secondary > primary > tertiary. This anomalous behavior of tertiary amines is due to steric factors i.e. crowding of alkyl groups cover nitrogen atom from all sides and thus makes it unable for protonation. Thus the relative strength is in order (CH3)2NH > CH3NH2 > NH3.
Amongst the following the most basic compound is- a)p -nitroaniline
- b)acetanilide
- c)aniline
- d)benzylamine
Correct answer is option 'D'. Can you explain this answer?
Amongst the following the most basic compound is
a)
p -nitroaniline
b)
acetanilide
c)
aniline
d)
benzylamine
|
Nerella Dange Naveen answered |
Benzylamine is a stronger base than aniline because
The lone pair of electrons on the nitrogen atom in benzylamine is delocalized.
The lone pair of electrons on the nitrogen atom in aniline is delocalized.
The lone pair of electrons on the nitrogen atom in aniline is not involved in resonance.
Benzylamine has a higher molecular mass than aniline.
The lone pair of electrons on the nitrogen atom in benzylamine is delocalized.
The lone pair of electrons on the nitrogen atom in aniline is delocalized.
The lone pair of electrons on the nitrogen atom in aniline is not involved in resonance.
Benzylamine has a higher molecular mass than aniline.
When nitrobenzene is treated with Br2 in presence of FeBr 3, the major product formed is m-bromonitrobenzene.
Statements which are related to obtain the m-isomer are- a)The electron density on meta carbon is more than that on ortho and para positions
- b)The intermediate carbonium ion formed after initial attack of Br+ at the meta position is least destabilised
- c)Loss of aromaticity when Br+ attacks at the ortho and para positions and not at meta position
- d)Easier loss of H+ to regain aromaticity from the meta position than from ortho and para positions.
Correct answer is option 'A,D'. Can you explain this answer?
When nitrobenzene is treated with Br2 in presence of FeBr 3, the major product formed is m-bromonitrobenzene.
Statements which are related to obtain the m-isomer are
Statements which are related to obtain the m-isomer are
a)
The electron density on meta carbon is more than that on ortho and para positions
b)
The intermediate carbonium ion formed after initial attack of Br+ at the meta position is least destabilised
c)
Loss of aromaticity when Br+ attacks at the ortho and para positions and not at meta position
d)
Easier loss of H+ to regain aromaticity from the meta position than from ortho and para positions.
|
Pranavi Ahuja answered |
TIPS/Formulae : Nitro group decreases the electron density at the meta-position in comparison to ortho and para position due to –I and –M effects.
The above intermediate is a resonance hybrid of three structures, hence is more stable than the corresponding intermediate from ortho and paraattack
In the chemical reaction,CH3CH2NH2 + CHCl3 + 3KOH → (A) + (B) + 3H2O, the compounds (A) and (B) are respectively- a)C2H5NC and 3KCl
- b)C2H5CN and 3KCl
- c)CH3CH2CONH2 and 3KCl
- d)C2H5NC and K2CO3.
Correct answer is option 'A'. Can you explain this answer?
In the chemical reaction,
CH3CH2NH2 + CHCl3 + 3KOH → (A) + (B) + 3H2O, the compounds (A) and (B) are respectively
a)
C2H5NC and 3KCl
b)
C2H5CN and 3KCl
c)
CH3CH2CONH2 and 3KCl
d)
C2H5NC and K2CO3.
|
Sagar Mukherjee answered |
Correct answer is option 'A': C2H5NC and 3KCl
It is example of carbylamine reaction. so, the product will be C2H5NC and 3KCl
It is example of carbylamine reaction. so, the product will be C2H5NC and 3KCl
Among the following, the strongest base is- a)C6H5NH2
- b)p-NO2.C6H4NH2
- c)m-NO2.C6H4.NH2
- d)C6H5CH2NH2
Correct answer is option 'D'. Can you explain this answer?
Among the following, the strongest base is
a)
C6H5NH2
b)
p-NO2.C6H4NH2
c)
m-NO2.C6H4.NH2
d)
C6H5CH2NH2
|
|
Krishna Iyer answered |
Aliphatic amines are more basic than aromatic amines because in aliphatic amines electron pair on nitrogen is not involved in resonance.
In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br2 used per mole of amine produced are :- a)Two moles of NaOH and two moles of Br2.
- b)Four moles of NaOH and one mole of Br2.
- c)One mole of NaOH and one mole of Br2.
- d)Four moles of NaOH and two moles of Br2.
Correct answer is option 'B'. Can you explain this answer?
In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br2 used per mole of amine produced are :
a)
Two moles of NaOH and two moles of Br2.
b)
Four moles of NaOH and one mole of Br2.
c)
One mole of NaOH and one mole of Br2.
d)
Four moles of NaOH and two moles of Br2.
|
Vasu Kumar answered |
RCONH2 + Br2 + 4NaOH > RNH2 + 2NaBr + 2H2O + Na2CO3
Amongst the following, the most basic compound is :- a)Benzylamine
- b)Aniline
- c)Acetanilide
- d)p-Nitroaniline
Correct answer is option 'A'. Can you explain this answer?
Amongst the following, the most basic compound is :
a)
Benzylamine
b)
Aniline
c)
Acetanilide
d)
p-Nitroaniline
|
|
Muskaan Roy answered |
Most basic compound amongst the given options: Benzylamine (option A)
Explanation:
Introduction:
In organic chemistry, basicity refers to the ability of a compound or molecule to accept a proton (H+) or donate a pair of electrons. The basicity of a compound depends on the availability of lone pairs of electrons on the atom or group of atoms.
Benzylamine (option A):
- Benzylamine is an organic compound with the formula C6H5CH2NH2.
- It consists of a benzene ring (C6H5) attached to an amino group (-NH2) through a methylene group (CH2).
- The lone pair of electrons on the nitrogen atom in the amino group can act as a Lewis base and readily accept a proton (H+).
- The lone pair of electrons is available for donation due to the electron-withdrawing nature of the benzene ring, which increases the basicity.
Aniline (option B):
- Aniline is an organic compound with the formula C6H5NH2.
- It consists of a benzene ring (C6H5) attached to an amino group (-NH2).
- The lone pair of electrons on the nitrogen atom in the amino group can also act as a Lewis base and accept a proton (H+).
- However, compared to benzylamine, aniline is less basic due to the electron-donating nature of the benzene ring. The lone pair of electrons is less available for donation.
Acetanilide (option C):
- Acetanilide is an organic compound with the formula C6H5NH(C2H3O).
- It consists of a benzene ring (C6H5) attached to an acetyl group (-C2H3O) through an amino group (-NH).
- The lone pair of electrons on the nitrogen atom in the amino group can also act as a Lewis base and accept a proton (H+).
- However, the presence of the acetyl group decreases the basicity compared to aniline. The electron-withdrawing nature of the acetyl group decreases the availability of the lone pair of electrons.
p-Nitroaniline (option D):
- p-Nitroaniline is an organic compound with the formula C6H6N2O2.
- It consists of a benzene ring (C6H6) attached to an amino group (-NH2) and a nitro group (-NO2) at the para position (opposite to the amino group).
- The lone pair of electrons on the nitrogen atom in the amino group can also act as a Lewis base and accept a proton (H+).
- However, the presence of the nitro group decreases the basicity compared to aniline. The electron-withdrawing nature of the nitro group decreases the availability of the lone pair of electrons.
Conclusion:
Amongst the given options, benzylamine (option A) is the most basic compound. The electron-withdrawing nature of the benzene ring increases the availability of the lone pair of electrons on the nitrogen atom, making benzylamine more basic compared to aniline, acetanilide, and p-nitroaniline.
Explanation:
Introduction:
In organic chemistry, basicity refers to the ability of a compound or molecule to accept a proton (H+) or donate a pair of electrons. The basicity of a compound depends on the availability of lone pairs of electrons on the atom or group of atoms.
Benzylamine (option A):
- Benzylamine is an organic compound with the formula C6H5CH2NH2.
- It consists of a benzene ring (C6H5) attached to an amino group (-NH2) through a methylene group (CH2).
- The lone pair of electrons on the nitrogen atom in the amino group can act as a Lewis base and readily accept a proton (H+).
- The lone pair of electrons is available for donation due to the electron-withdrawing nature of the benzene ring, which increases the basicity.
Aniline (option B):
- Aniline is an organic compound with the formula C6H5NH2.
- It consists of a benzene ring (C6H5) attached to an amino group (-NH2).
- The lone pair of electrons on the nitrogen atom in the amino group can also act as a Lewis base and accept a proton (H+).
- However, compared to benzylamine, aniline is less basic due to the electron-donating nature of the benzene ring. The lone pair of electrons is less available for donation.
Acetanilide (option C):
- Acetanilide is an organic compound with the formula C6H5NH(C2H3O).
- It consists of a benzene ring (C6H5) attached to an acetyl group (-C2H3O) through an amino group (-NH).
- The lone pair of electrons on the nitrogen atom in the amino group can also act as a Lewis base and accept a proton (H+).
- However, the presence of the acetyl group decreases the basicity compared to aniline. The electron-withdrawing nature of the acetyl group decreases the availability of the lone pair of electrons.
p-Nitroaniline (option D):
- p-Nitroaniline is an organic compound with the formula C6H6N2O2.
- It consists of a benzene ring (C6H6) attached to an amino group (-NH2) and a nitro group (-NO2) at the para position (opposite to the amino group).
- The lone pair of electrons on the nitrogen atom in the amino group can also act as a Lewis base and accept a proton (H+).
- However, the presence of the nitro group decreases the basicity compared to aniline. The electron-withdrawing nature of the nitro group decreases the availability of the lone pair of electrons.
Conclusion:
Amongst the given options, benzylamine (option A) is the most basic compound. The electron-withdrawing nature of the benzene ring increases the availability of the lone pair of electrons on the nitrogen atom, making benzylamine more basic compared to aniline, acetanilide, and p-nitroaniline.
Hydrogen bonding plays a central role in the following phenomena- a)Ice floats in water
- b)Higher Lewis basicity of primary amines than tertiary amines in aqueous solutions
- c)Formic acid is more acidic than acetic acid
- d)Dimerisation of acetic acid in benzene
Correct answer is option 'A,B,D'. Can you explain this answer?
Hydrogen bonding plays a central role in the following phenomena
a)
Ice floats in water
b)
Higher Lewis basicity of primary amines than tertiary amines in aqueous solutions
c)
Formic acid is more acidic than acetic acid
d)
Dimerisation of acetic acid in benzene
|
Jyoti Saha answered |
In ice, water molecules are excessively H-bonded giving a cage-like structure which is lighter than water.
Primary amines are more basic than tertiary amine, because the protonated 1° amines are extensively H-bonded and hence more stable than the corresponding protonated 3° amines.
Primary amines are more basic than tertiary amine, because the protonated 1° amines are extensively H-bonded and hence more stable than the corresponding protonated 3° amines.
Acetic acid undergoes dimerisation in benzene.
A positive carbylamine test is given by- a)N, N—dimethylaniline
- b)2, 4–dimethylaniline
- c)N–methyl-o-methylaniline
- d)p-methylbenzylamine
Correct answer is option 'B,D'. Can you explain this answer?
A positive carbylamine test is given by
a)
N, N—dimethylaniline
b)
2, 4–dimethylaniline
c)
N–methyl-o-methylaniline
d)
p-methylbenzylamine
|
|
Sakshi Shah answered |
-dimethylbenzylamineb)N, N-dimethylanilinec)N, N-dimethylbenzenesulfonamided)N, N-dimethylformamide
In the reaction
the product E is :- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
In the reaction
the product E is :
a)
b)
c)
d)
|
|
Yash Modi answered |
In D, the NH2 Will be replaced by N2Cl, and in E the N2Cl will be replaced by CN.( Sandmeyer Reaction). So ans is A
Carbylamine test is performed in alcoholic KOH by heating a mixture of :- a)chloroform and silver powder
- b)trihalogenatedmethane and a primary amine
- c)an alkyl halide and a primary amine
- d)an alkyl cyanide and a primary amine
Correct answer is option 'B'. Can you explain this answer?
Carbylamine test is performed in alcoholic KOH by heating a mixture of :
a)
chloroform and silver powder
b)
trihalogenatedmethane and a primary amine
c)
an alkyl halide and a primary amine
d)
an alkyl cyanide and a primary amine
|
|
Rishika Roy answered |
NOTE : Only primary aliphatic and aromatic amines give this test.
CHX3 + RNH2 + 3KOH → RNC + 3 KX + 3H2O
The compound which on reaction with aqueous nitrous acid at low temperature produces an oily nitrosamine is- a)methylamine
- b)ethylamine
- c)diethylamine
- d)triethylamine
Correct answer is option 'C'. Can you explain this answer?
The compound which on reaction with aqueous nitrous acid at low temperature produces an oily nitrosamine is
a)
methylamine
b)
ethylamine
c)
diethylamine
d)
triethylamine
|
Akshara Deshpande answered |
NOTE : Secondary amines (aliphatic as well as aromatic) react with nitrous acid to form Nnitrosoamines.
Butanonitrile may be prepared by heating :- a)Propyl alcohol with KCN
- b)Butyl alcohol with KCN
- c)Butyl chloride with KCN
- d)Propyl chloride with KCN
Correct answer is option 'D'. Can you explain this answer?
Butanonitrile may be prepared by heating :
a)
Propyl alcohol with KCN
b)
Butyl alcohol with KCN
c)
Butyl chloride with KCN
d)
Propyl chloride with KCN
|
Rishika Mishra answered |
Butanonitrile may be prepared by heating propyl chloride with KCN.
In this reaction, chloride ion is replaced by cyanide ion in reaction.
In this reaction, chloride ion is replaced by cyanide ion in reaction.
Amongst the compounds given, the one that would form a brilliant colored dye on treatment with NaNO2 in dil. HCl followed by addition to an alkaline solution of β-naphthol is- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Amongst the compounds given, the one that would form a brilliant colored dye on treatment with NaNO2 in dil. HCl followed by addition to an alkaline solution of β-naphthol is
a)
b)
c)
d)
|
Lavanya Menon answered |
Only primary aromatic amines undergo diazotisation followed by coupling.
Which one of the following is the strongest base in aqueous solution ?- a)Methylamine
- b)Trimethylamine
- c)Aniline
- d)Dimethylamine.
Correct answer is option 'D'. Can you explain this answer?
Which one of the following is the strongest base in aqueous solution ?
a)
Methylamine
b)
Trimethylamine
c)
Aniline
d)
Dimethylamine.
|
|
Baby Ghosh answered |
Due to steric hindrance,solvanation of ions and the combined effect of the pushing effect of alkyl grp(+I) ,strric hindrance and the salvation of amines cause the basicity order should be ....NH3
The most unlikely representation of resonance structures of p-nitrophenoxide ion is- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
The most unlikely representation of resonance structures of p-nitrophenoxide ion is
a)
b)
c)
d)
|
Hansa Sharma answered |
N can't have more than 8 electrons in its valence shell as it does not have any d orbital. In (c), N has 10 electrons.
In the reaction shown below, the major product(s) formed is/ are
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
In the reaction shown below, the major product(s) formed is/ are
a)
b)
c)
d)
|
Manisha Kumar answered |
group is acetylated by acetic anhydride in methylene chloride (solvent). Note that —CONH2 group does not undergo acetylation because here lone pair of electrons is delocalised.
The correct order of basicities of the following compounds is
2. CH3 - CH2 - NH23. (CH3)2NH
- a)2 > 1 > 3 > 4
- b)1 > 3 > 2 > 4
- c)3 > 1 > 2 > 4
- d)1 > 2 > 3 > 4
Correct answer is option 'B'. Can you explain this answer?
The correct order of basicities of the following compounds is
2. CH3 - CH2 - NH2
3. (CH3)2NH
a)
2 > 1 > 3 > 4
b)
1 > 3 > 2 > 4
c)
3 > 1 > 2 > 4
d)
1 > 2 > 3 > 4
|
|
Anand Kumar answered |
4 has lowest basic character because lone pair present at N is involve in delocalization through +R of O.
In 2 NH2 is 1degree while in 3 N is 3degree and we know 3degree has high basic character than 1degree.
Finally 1, it has two spots to donate e and has highest tendency to donate e.
So, option B is correct.
In 2 NH2 is 1degree while in 3 N is 3degree and we know 3degree has high basic character than 1degree.
Finally 1, it has two spots to donate e and has highest tendency to donate e.
So, option B is correct.
with a mixture of Br2 and KOH gives R-NH2 as the main product. The intermediates involved in this reaction are :
The formation of cyanohydrin from a ketone is an example of :- a)Electrophilic addition
- b)Nucleophilic addition
- c)Nucleophilic substitution
- d)Electrophilic substiution
Correct answer is option 'B'. Can you explain this answer?
The formation of cyanohydrin from a ketone is an example of :
a)
Electrophilic addition
b)
Nucleophilic addition
c)
Nucleophilic substitution
d)
Electrophilic substiution
|
Om Roy answered |
TIPS/Formulae : The addition is initiated by the attack of CN– group which is a nucleophile.
PASSAGE - 2Treatment 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- a)glycine
- b)alanine
- c)valine
- d)serine
Correct answer is option 'B'. Can you explain this answer?
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
a)
glycine
b)
alanine
c)
valine
d)
serine
|
Siddharth Nair answered |
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 :- a)CH3COOH
- b)CH3CH2CH2COOH
- c)
- d)CH3CH2COOH
Correct answer is option 'D'. Can you explain this answer?
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 :
a)
CH3COOH
b)
CH3CH2CH2COOH
c)
d)
CH3CH2COOH
|
Devendra Singh answered |
Amines (only primary )can also be prepared by buy hofmann degradation. in this method the amines will one carbon atom less than .
PASSAGE - 1The 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?
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
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?
a)
b)
c)
d)
|
|
Jithin Roy answered |
Since the reaction is intramolecular, no cross product will be formed.
The compound that is most reactive towards electrophilic nitration is :- a)toluene
- b)benzene
- c)benzoic acid
- d)nitrobenzene
Correct answer is option 'A'. Can you explain this answer?
The compound that is most reactive towards electrophilic nitration is :
a)
toluene
b)
benzene
c)
benzoic acid
d)
nitrobenzene
|
|
Ritik Raj answered |
Becoz of the tendency to donate electrons !
Ethyl isocyanide on hydrolysis in acidic medium generates- a)propanoic acid and ammonium salt
- b)ethanoic acid and ammonium salt
- c)methylamine salt and ethanoic acid
- d)ethylamine salt and methanoic acid
Correct answer is option 'D'. Can you explain this answer?
Ethyl isocyanide on hydrolysis in acidic medium generates
a)
propanoic acid and ammonium salt
b)
ethanoic acid and ammonium salt
c)
methylamine salt and ethanoic acid
d)
ethylamine salt and methanoic acid
|
|
Nandita Chopra answered |
Ethyl isocyanide on hydrolysis form primary amines.
Therefore it gives only one mono chloroalkane.
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