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

JEE ADV. 2015

A B

C D

Solution:

The absolute configuration of is :

(2S, 3R)

(2S, 3S)

(2R, 3R)

(2R, 3S)

A

B

C

D

JEE MAIN 2016

Solution:

Write the product of the givenreaction :

A B

C D

Solution:

It involves nucleophilic substitution reaction (SN2 ) followed by

oxidation with oxidising agent and removal of water molecule.

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

Solution:

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




A

B

C

D

JEE MAIN 2019

Solution:

Arrange in order of increasing acidicstrength :

X > Z >Y

Z < X >Y

X >Y > Z

Z > X >Y

A

B

C

D

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

Which amongst the following is the strongest acid?

CHBr3

CHI3

CHCl3

CH(CN)3

A

B

C

D

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.

Among the following, the strongest base is :

C6H5NH2

p-NO2C6H4NH2

m-NO2—C6H4NH2

C6H5CH2NH2

A

B

C

D

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.

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

A B

C D

Solution:

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

A B

C Dgauche

partially eclipsed

staggered

eclipsed

2004

Solution:

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

1

2

3

4

A

B

C

D

2007

Solution:

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

two

three

four

five

A

B

C

D

Solution:

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

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

Solution:

3-methyl-1-pentene

(It has one chiral centre)

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

Ethane

Ethylene

Acetylene

Hexachloroethane

A

B

C

D

Solution:

Ethane has the smallest sized group (H) bonded

to carbons, hence there will be least hindered

rotation about C—C bond.

The number of bonds in o-xylene is :

6

9

12

18

A

B

C

D

Solution:

O-dimethylbenzene or O-Xylene

It has 18 sigma bonds and 3 pi bonds.

Which of the following molecules is least resonance stabilised?

A B

C D

JEE MAIN 2017

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.

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

Solution:

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

Solution:

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

attached substituents are different.

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

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 ]

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

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.

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

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

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

2011

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

Ans : 6

2011

Solution:

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

participate in hyperconjugation.

The total number of stereoisomers that can exist

for M is :

JEE ADV. 2015

The total number of stereoisomers that can exist

for M is :

Ans : 2

JEE ADV. 2015

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.

The number of resonance structuresfor N is :

JEE ADV. 2015

The number of resonance structuresfor N is :

JEE ADV. 2015

Ans : 9

Solution:

Product ‘X’ major will be :

A B

C D

Solution:

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

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

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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