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Explore popular questions from Organic Chemistry Some Basic Principles and Techniques for JEE Advanced. This collection covers Organic Chemistry Some Basic Principles and Techniques previous year JEE Advanced questions hand picked by experienced teachers.

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Organic Chemistry Some Basic Principles and Techniques

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Q 1. Allyl isocyanide has:

A

{tex} 9 \sigma {/tex} and {tex} 4 \pi {/tex} bonds

B

{tex} 8 \sigma {/tex} and {tex} 5 \pi {/tex} bonds

{tex} 9 \sigma , 3 \pi {/tex} and {tex}2{/tex} non-bonded electrons

D

{tex} 8 \sigma , 3 \pi {/tex} and {tex}4{/tex} non-bonded electrons

Explanation

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Q 2. In the given conformation, if {tex}\mathrm{C_2}{/tex} is rotated about {tex}\mathrm{C_2-C_3}{/tex} bond anticlockwise by an angle of {tex}120^\circ{/tex} then the conformation obtained is

A

fully eclipsed conformation

B

partially eclipsed conformation

gauche conformation

D

staggered conformation

Explanation


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Q 3. The IUPAC name of the compound {tex} \mathrm { CH } _ { 2 } = \mathrm { CH } - \mathrm { CH } \left( \mathrm { CH } _ { 3 } \right) _ { 2 } {/tex} is

A

{tex}1,1{/tex}-dimethyl-{tex}2{/tex}-propene

B

{tex}2{/tex}-vinylpropane

{tex}3{/tex}-methyl-{tex}1{/tex}-butene

D

{tex}1{/tex}-isopropylethene

Explanation

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Q 4. Which compound will not give positive chromyl chloride test?

A

{tex} \mathrm { CuCl } _ { 2 } {/tex}

{tex} \mathrm { HgCl } _ { 2 } {/tex}

C

{tex} \mathrm { ZnCl } _ { 2 } {/tex}

D

{tex} \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { NH } _ { 3 } ^ { + } \mathrm { Cl } ^ { - } {/tex}

Explanation

{tex} \mathrm { HgCl } _ { 2 } {/tex} does not give chromyl chloride test

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Q 5. {tex} \mathrm { PbCl } _ { 2 } + \mathrm { Kl } \rightarrow \underset { \text { Yellow ppt } } { [ \mathrm { A } ] } + \mathrm { KCl } {/tex}
{tex} \underset { \text { Yellow ppt } } { [ \mathrm { A } ] } + \underset { \text { excess } } { \mathrm { Kl } } \longrightarrow \underset{\mathrm {soluble}} \mathrm {[ B } ] {/tex}
Compound [A] and [B] are

A

{tex} \mathrm { PbI } _ { 4 } {/tex} and {tex} \mathrm { K } _ { 2 } \left[ \mathrm { PbI } _ { 4 } \right] {/tex} respectively

B

{tex} \mathrm { K } _ { 2 } \left[ \mathrm { PbI } _ { 4 } \right] {/tex} and {tex} \mathrm { PbI } _ { 4 } {/tex} respectively

{tex} \mathrm { PbI } _ { 2 } {/tex} and {tex} \mathrm { K } _ { 2 } \left[ \mathrm { PbI } _ { 4 } \right] {/tex} respectively

D

{tex} \mathrm { PbI } _ { 2 } {/tex} and {tex} \mathrm { K } _ { 2 } \left[ \mathrm { PbI } _ { 2 } \right] {/tex} respectively

Explanation

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Q 6. When concentrated {tex} \mathrm { H } _ { 2 } \mathrm { SO } _ { 4 } {/tex} is added to dry {tex} \mathrm { KNO } _ { 3 } , {/tex} brown fumes evolve. These brown fumes are of

A

{tex} \mathrm { SO } _ { 2 } {/tex}

B

{tex} \mathrm { SO } _ { 3 } {/tex}

C

{tex}\mathrm {NO} {/tex}

{tex} \mathrm { NO } _ { 2 } {/tex}

Explanation

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Q 7. In the fifth group {tex} (\mathrm { NH } _ { 4 }) _ { 2 } \mathrm { CO } _ { 3 } {/tex} is added to precipitate out the carbonates, we do not add {tex} \mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 } {/tex} because

A

{tex} \mathrm { MgCO } _ { 3 } {/tex} is soluble in {tex} \mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 } {/tex}

B

{tex} \mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 } {/tex} increases the solubility of fifth group carbonates

{tex} \mathrm {MgCO}_ { 3 } {/tex} will also be precipitated out in fifth group

D

{tex} \mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 } {/tex} will decrease the solubility product of {tex} \mathrm { MgCO } _ { 3 } {/tex}

Explanation

{tex} \mathrm {MgCO}_ { 3 } {/tex} will also be precipitated out in fifth group

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Q 8. Match the list type :


List-I

List-II
(P)
 As2S3 sol(1)
 Negatively charged
(Q)
 Fe2O3.xH2O sol(2)
 Positively charged
(R)
 Starch sol(3)
 Coagulation near cathode during electrophoresis
(S)
 AgI / I : K+ sol(4)
 Coagulation near anode during electrophoresis

 (5)
 May be coagulated by adding Na2SO4

 (6)
 May be coagulated by adding gold sol.
The correct option is

A

P → 1,3,5 ; Q → 2,4,5,6 ; R → 1,4,6 ; S → 1,4,5

P → 1,4,5 ; Q → 2,3,5,6 ; R → 1,4,5 ; S → 1,4,5

C

P → 1,4,5 ; Q → 2,4,6 ; R → 1,3,5 ; S → 1,4,6

D

P → 2,3,5,6 ; Q → 1,4,5 ; R → 1,4,5 ; S → 2,3,5,6

Explanation

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Q 9.


List-I
Compound

List-II
(P)
 H2O2(1)
 Bleaching agent
(Q)
 Ca(OCl)Cl(2)
 Absorbs PH3 gas
(R)
 CuSO4(3)
 Colourless/white in colour
(S)
 SiO2(4)
 Paramagnetic in nature

 (5)
 Acidic in nature
The CORRECT option is

A

P → 1,2,4; Q → 3,5; R → 1,3; S → 4,5

B

P → 2,4,5; Q → 1,2; R → 3,4; S → 1,2,3

P → 1,3,5; Q → 1,2,3; R → 2,4; S → 3,5

D

P → 4,5; Q → 1,5; R → 2,3,4; S → 1,5

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Q 10.


List-I
Radicals

List-II
Correct statement regarding them
(P)
 Br(1)
 Can be confirmed by Brown ring test
(Q)
 I(2)
 Can be confirmed by Layer test
(R)
 S2–(3)
 Can be confirmed by methylene blue test
(S)
 NO3(4)
 On reaction with conc. H2SO4, oxidation takes place

 (5)
 All salts are soluble
The CORRECT option is

A

P → 2,4; Q → 2,4; R → 3,5; S → 1,5

B

P → 1,4; Q → 2,3; R → 3,5; S → 1,5

P → 2,4; Q → 2,4; R → 3,4; S → 1,5

D

P → 2,4; Q → 1,4; R → 3,4; S → 2,5

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Q 11.

How many chiral compound are possible on mono chlorination of 2-methyl butane?

2

B

4

C

6

D

8

Explanation

On chlorination of 2-methyl butane

2-chiral compound are formed.

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Q 12.

Forty millilitre of CO was mixed with 100 ml of and the mixture was exploded. On cooling, the reaction mixture was shaken with KOH. What volume of gas is left?

A

60ml of

80 ml of

C

20 ml of CO

D

40 ml of

Explanation

40 ml 20 ml 40 ml

absorbed

Volume of left

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Q 13.

A certain compound has the molecular formula . If 10 gm of has 5.72 gm X, the atomic mass of X is:

B

C

D

Explanation

Let the atomic weight of X be

mol of = mol of

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Q 14.

The number of - and -bonds in -oxohexanoic acid, respectively, is:

18,2

B

18,1

C

17,2

D

17,1

Explanation

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Q 15.

An organic compound contains 4% sulphur. Its minimum molecular weight is:

A

200

B

400

800

D

1600

Explanation

Four gram of S is in 100 gm of compound

32 gm S is in

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Q 16.

The IUPAC name of the following compound, is

A

4-bromo-3-cynophenoal

2-bromo-5-hydroxybenzonitrile

C

2-cyano-4-hydroxybromobenzene

D

6-bromo-3-hydroxybenzonitrile

Explanation

Cyano group has the highest priority therefore, parent name must be benzonitrile. Br occurs at 2-position, and hydroxyl at 3-position, hence the IUPAC name is 2-bromo-5-hydroxy benzonitrile.

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Q 17.

An enantiomerically pure acid is treated with racemic mixture of an alcohol having one chiral carbon. The ester formed will be

Optically active mixture

B

Pure enantiomer

C

Meso compound

D

Racemic mixture

Explanation

(a) When optically active acid reacts with racemic mixture of an alcohol, it forms two types of isomeric esters. In each, the configuration of the chiral centre of acid will remain the same.

So, the mixture will be optically active.

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Q 18.

Which of the following is a soft base?

B

C

D

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Q 19.

A compound (A) with molecular formula gives one monochlorination product. Compound (A) is:

A

C

D

Explanation

D.U. in (A)

1 D.U. suggests that either it is an alkene or a cyclopentane . Only cyclopentane gives one product on monochlorination

So the answer is (b)

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Q 20.

Which of the following carbocations is least stable?

A

C

D

Explanation

Positive charge on -hybridised C atom

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Q 21.

The process of separation of racemic modifications into and enantiomers is called:

Resolution

B

Dehydration

C

Revolution

D

Dehydrohalogenation

Explanation

The separation of a racemic mixture is called resolution

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Q 22.

Give the decreasing order of hyperconjugative effect of R in , where R is:

B

C

D

Explanation

In Me– (no

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Q 23.

Two litre air formed 1915 ml of ozonised air when passed through Brodio's apparatus. The volume of ozone formed is:

A

85 ml

170 ml

C

225 ml

D

42.5 ml

Explanation

There is a reduction of 1 volume

When reduction in volume is 1, volume of is 2

Volume of air = 2000 ml

Volume of ozonised air ml

Reduction in volume

When reduction of 1 volume, volume of = 2

When reduction is 85 ml , volume of

= 170 ml

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Q 24.

The decreasing order of effect of the orbitals is:

B

C

D

Explanation

The more the character, the more is the penetration effect of orbital towards the nucleus, and hence more -withdrawing effect. So,

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Q 25.

The minimum number of carbon atoms an alkane should contain in order to exihibit optical activity is:

A

5

B

6

7

D

8