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Starts at 24th May 2021,

End at 3rd July 2021

90

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Compound(s) that on hydrogenation produce(s) opticallyinactive compound(s) is (are)

JEE ADV. 2015

A B

C D

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Compound(s) that on hydrogenation produce(s) opticallyinactive compound(s) is (are)

JEE ADV. 2015

A B

C D

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Solution:

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The absolute configuration of is :

(2S, 3R)

(2S, 3S)

(2R, 3R)

(2R, 3S)

A

B

C

D

JEE MAIN 2016

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The absolute configuration of is :

(2S, 3R)

(2S, 3S)

(2R, 3R)

(2R, 3S)

A

B

C

D

JEE MAIN 2016

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Solution:

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Write the product of the givenreaction :

A B

C D

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Write the product of the givenreaction :

A B

C D

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Solution:

It involves nucleophilic substitution reaction (SN2 ) followed by

oxidation with oxidising agent and removal of water molecule.

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A gaseous hydrocarbon gives upon combustion 0.72 g of waterand 3.08 g of CO2 . The empirical formula of the hydrocarbon is :

C2H4

C3H4

C6H5

C7H8

A

B

C

D

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A gaseous hydrocarbon gives upon combustion 0.72 g of waterand 3.08 g of CO2 . The empirical formula of the hydrocarbon is :

C2H4

C3H4

C6H5

C7H8

A

B

C

D

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Solution:

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25 g of an unknown hydrocarbon upon burning produces 88 g ofCO2 and 9 g of H2O. This unknown hydrocarbon contains :

20 g of carbon and 5 g of hydrogen

22 g of carbon and 3 g of hydrogen

24 g of carbon and 1 g of hydrogen

18 g of carbon and 7 g of hydrogen

A

B

C

D

JEE MAIN 2019

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25 g of an unknown hydrocarbon upon burning produces 88 g ofCO2 and 9 g of H2O. This unknown hydrocarbon contains :

20 g of carbon and 5 g of hydrogen

22 g of carbon and 3 g of hydrogen

24 g of carbon and 1 g of hydrogen

18 g of carbon and 7 g of hydrogen

A

B

C

D

JEE MAIN 2019

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Solution:

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Arrange in order of increasing acidicstrength :

X > Z >Y

Z < X >Y

X >Y > Z

Z > X >Y

A

B

C

D

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Arrange in order of increasing acidicstrength :

X > Z >Y

Z < X >Y

X >Y > Z

Z > X >Y

A

B

C

D

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Solution:

● Carboxylic acid is stronger acid than ammonium ion,

hence —COOH(X) is most acidic.

● Z (NH3+) is more acidic than Y (NH3

+) due to – I effect of

—COOH on Z.

Hence, overall acid strength order is X > Z > Y

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Which amongst the following is the strongest acid?

CHBr3

CHI3

CHCl3

CH(CN)3

A

B

C

D

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Which amongst the following is the strongest acid?

CHBr3

CHI3

CHCl3

CH(CN)3

A

B

C

D

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Solution:

We know, a stronger acid produces its stable or weaker conjugate base. Here, CH(CN)3 produces the most stable conjugate base (NC)3C- . Stronger -R and -I effects of the CN- group, make the carbanion (conjugate base) very stable. The resonance hybrid structure of [(NC)3C]- is as follows:

However, halogen (X = Cl, Br, I) show -I effect but + R effect of halogens, destabilises the carbanion, X3C– (conjugate base of the haloform, HCX3 ).

Thus, CH(CN)3 is the strongest acid among the given options.

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Among the following, the strongest base is :

C6H5NH2

p-NO2C6H4NH2

m-NO2—C6H4NH2

C6H5CH2NH2

A

B

C

D

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Among the following, the strongest base is :

C6H5NH2

p-NO2C6H4NH2

m-NO2—C6H4NH2

C6H5CH2NH2

A

B

C

D

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Solution:

Lone pair is not taking part in resonance, most basic. In

other cases, lone pair of nitrogen is part of delocalisation

which decreases Lewis base strength.

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The compound which gives the most stable carbonium ion ondehydration is :

A B

C D

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The compound which gives the most stable carbonium ion ondehydration is :

A B

C D

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Solution:

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C2 is rotated anti-clockwise 120° about C2- C3 bond.The resulting conformer is :

A B

C Dgauche

partially eclipsed

staggered

eclipsed

2004

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C2 is rotated anti-clockwise 120° about C2- C3 bond.The resulting conformer is :

A B

C Dgauche

partially eclipsed

staggered

eclipsed

2004

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Solution:

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The number of stereoisomers obtained by bromination of trans-2-butene is :

1

2

3

4

A

B

C

D

2007

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The number of stereoisomers obtained by bromination of trans-2-butene is :

1

2

3

4

A

B

C

D

2007

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Solution:

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The maximum number of isomer for an alkene with the molecularformula C4H8 is :

two

three

four

five

A

B

C

D

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The maximum number of isomer for an alkene with the molecularformula C4H8 is :

two

three

four

five

A

B

C

D

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Solution:

The maximum number of isomer for an alkene with molecular formula C4H8 is four

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Out of the following the alkene that exhibits optical isomerism is :

3-methyl-2-pentene

4-methyl-1-pentene

3-methyl-1-pentene

2-methyl-2-pentene

A

B

C

D

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Out of the following the alkene that exhibits optical isomerism is :

3-methyl-2-pentene

4-methyl-1-pentene

3-methyl-1-pentene

2-methyl-2-pentene

A

B

C

D

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Solution:

3-methyl-1-pentene

(It has one chiral centre)

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Which of the following will have least hindered rotation aboutcarbon-carbon bond ?

Ethane

Ethylene

Acetylene

Hexachloroethane

A

B

C

D

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Which of the following will have least hindered rotation aboutcarbon-carbon bond ?

Ethane

Ethylene

Acetylene

Hexachloroethane

A

B

C

D

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Solution:

Ethane has the smallest sized group (H) bonded

to carbons, hence there will be least hindered

rotation about C—C bond.

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The number of bonds in o-xylene is :

6

9

12

18

A

B

C

D

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The number of bonds in o-xylene is :

6

9

12

18

A

B

C

D

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Solution:

O-dimethylbenzene or O-Xylene

It has 18 sigma bonds and 3 pi bonds.

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Which of the following molecules is least resonance stabilised?

A B

C D

JEE MAIN 2017

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Which of the following molecules is least resonance stabilised?

A B

C D

JEE MAIN 2017

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Solution:

are aromatic and stabilised by resonance. They follow Huckel’s rule.

Aromatic compounds are stable due to resonance while non-aromatics are not. According to Huckel’s rule (or 4n + 2 rule), “For a planar, cyclic compound to be aromatic, its π cloud must contain (4n + 2)π electrons, where, n is any whole number.” Thus,

is non-aromatic, hence, least stabilised by resonance.

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The correct stability order for the following species is :

A B

C DII > I > IV > III

II > IV > I > III

I > III > II > IV

I > II > III > IV

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The correct stability order for the following species is :

A B

C DII > I > IV > III

II > IV > I > III

I > III > II > IV

I > II > III > IV

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Solution:

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Among the following four structure I and IV

It is true that :

A B

C Donly III is a chiralcompound

all four are chiralcompounds

only II and IV arechiral compounds

only I and II are chiralcompounds

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Among the following four structure I and IV

It is true that :

A B

C Donly III is a chiralcompound

all four are chiralcompounds

only II and IV arechiral compounds

only I and II are chiralcompounds

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Solution:

Only (I) and (II) are chiral compounds as all the four

attached substituents are different.

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Taking into account of hybridization and resonance effect, rank the following bonds in order of decreasing bond length.

A B

C DI > III > II

I > II = III

II = III = I

II > III > I

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Taking into account of hybridization and resonance effect, rank the following bonds in order of decreasing bond length.

A B

C DI > III > II

I > II = III

II = III = I

II > III > I

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Solution:

Taking resonance into consideration, all the bonds have partial double bond character. As seen by resonance, all the C−O bonds are identical and have the same bond length.∴ II = III = I [ order of bond length ]

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Which of the following orders is not correct regarding −I -effect ofthe substituents?

-I < -Cl < -Br < -F

-NR3+ < -OR2

+

-NR3+ < -OR < -F

-SR < -OR < -OR2+

A

B

C

D

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Which of the following orders is not correct regarding −I -effect ofthe substituents?

-I < -Cl < -Br < -F

-NR3+ < -OR2

+

-NR3+ < -OR < -F

-SR < -OR < -OR2+

A

B

C

D

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Solution:

The correct order regarding —I effect of the

substituents is :

-I < -Cl < -Br < -F

These groups have the ability to withdraw electron

density toward themselves.

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The correct order of acidity for the following compounds is :

A B

C DIII > IV > II > I

I > II > III > IV

I > III > IV > II

III > I > II > IV

JEE ADV. 2016

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The correct order of acidity for the following compounds is :

A B

C DIII > IV > II > I

I > II > III > IV

I > III > IV > II

III > I > II > IV

JEE ADV. 2016

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Solution:

● -OH group displays both kinds of effect; an electron withdrawing acid-strengthening

inductive effect from the meta-position and an electron-releasing acid weakening

resonance effect from the para-position (at this position, resonance effect overweighs

the inductive effect). Thus, III > IV.

● o-hydroxybenzoic acid (II) is far stronger than the corresponding meta and para

isomers as the carboxylate ion is stabilised by intramolecular H-bonding.

● 2,6-dihydroxybenzoic acid (I) forms carboxylate ion which is further stabilised by

intramolecular H-bonding, Thus, correct order is I > II > III > IV

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The total number of contributing structures showinghyperconjugation (involving C — H bonds)for the following carbocation is :

2011

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The total number of contributing structures showinghyperconjugation (involving C — H bonds)for the following carbocation is :

Ans : 6

2011

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Solution:

These are total 6α-H to sp2 carbon and they all can

participate in hyperconjugation.

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The total number of stereoisomers that can exist

for M is :

JEE ADV. 2015

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The total number of stereoisomers that can exist

for M is :

Ans : 2

JEE ADV. 2015

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Solution:

Although the compound has two chiral carbons (indicated by stars), it does

not has four optically active isomers as expected. It is due to its existence in

cis-form only.

The above shown transformation does not exist due to restricted rotation

about the bridge head carbons, hence only cis-form and its mirror image

exist.

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The number of resonance structuresfor N is :

JEE ADV. 2015

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The number of resonance structuresfor N is :

JEE ADV. 2015

Ans : 9

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Solution:

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Product ‘X’ major will be :

A B

C D

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Product ‘X’ major will be :

A B

C D

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Solution:

It involves Electrophilic addition of alkenes, followed by Nucleophilic substitution mechanism.

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The correct IUPAC name of the following compound is

2-methyl-5-nitro-1-chlorobenzene

3-chloro-1-methyl-1-nitrobenzene

2-chloro-1-methyl 1-4-nitrobenzene

5-chloro-4-methyl 1-1-nitrobenzene

A

B

C

D

JEE MAIN 2019

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Solution:

The IUPAC name of the given compound is

2- chloro-1-methyl- 4-nitrobenzene

Here, the given compound contains two or more functional groups. So, the numbering is done in such a way that the sum of the locants is the lowest.

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