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Home/JEE MAINS/chemistry/haloalkanes and haloarenes
Curated PYQ Collection

Top 50 Most Repeated HALOALKANES AND HALOARENES PYQs | JEE MAINS

A curated collection of the most important questions from HALOALKANES AND HALOARENES, fully solved with step-by-step concepts to prepare for JEE MAINS.

Question #1

Practice Question

A.CH₃Cl
B.(CH₃)₂CHCl
C.(CH₃)₃CCl
D.CH₃CH₂Cl

Concept Applied

SN1 rate depends on carbocation stability. Tertiary carbocation is most stable....

Read Full Step-by-Step Solution →

Question #2

Practice Question

A.CH₃Br + OH⁻ → CH₃OH
B.(CH₃)₃CCl + H₂O → (CH₃)₃COH + HCl
C.CH₃CH₂Br + NaI → CH₃CH₂I + NaBr
D.C₂H₅Cl + KCN → C₂H₅CN + KCl

Concept Applied

SN1 reactions are favored for tertiary alkyl halides in polar protic solvents because a stable carbocation can form. Option B involves a tertiary chlo...

Read Full Step-by-Step Solution →

Question #3

Practice Question

A.(R)-2-iodobutane
B.(S)-2-iodobutane
C.racemic mixture of 2-iodobutane
D.No reaction occurs

Concept Applied

NaI in acetone promotes an SN2 substitution. SN2 proceeds with a backside attack, causing Walden inversion of configuration. Hence the (R) substrate g...

Read Full Step-by-Step Solution →

Question #4

Practice Question

A.CH₃Cl
B.CH₃CH₂Cl
C.(CH₃)₂CHCl
D.(CH₃)₃CCl

Concept Applied

SN1 rate depends on carbocation stability. Tertiary carbocation from (CH₃)₃CCl is most stable....

Read Full Step-by-Step Solution →

Question #5

Practice Question

A.0.5 mol
B.1.0 mol
C.1.5 mol
D.2.0 mol

Concept Applied

Each mole of CCl₂F₂ yields two moles of Cl atoms. Therefore, $0.5\,\text{mol}$ of CCl₂F₂ produces $0.5\times2 = 1.0\,\text{mol}$ of chlorine atoms....

Read Full Step-by-Step Solution →

Question #6

Practice Question

A.A fundamental principle of Chemistry.
B.A complex derivation in JEEMains syllabus.
C.An experimental observation.
D.A theoretical assumption.

Concept Applied

This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Haloalkanes and Halo...

Read Full Step-by-Step Solution →

Question #7

Practice Question

A.CH₃Cl
B.(CH₃)₂CHCl
C.(CH₃)₃CCl
D.CH₃CH₂Cl

Concept Applied

SN1 rate depends on carbocation stability. Tertiary carbocation is most stable....

Read Full Step-by-Step Solution →

Question #8

Practice Question

A.CH₃Br + OH⁻ → CH₃OH
B.(CH₃)₃CCl + H₂O → (CH₃)₃COH + HCl
C.CH₃CH₂Br + NaI → CH₃CH₂I + NaBr
D.C₂H₅Cl + KCN → C₂H₅CN + KCl

Concept Applied

SN1 reactions are favored for tertiary alkyl halides in polar protic solvents because a stable carbocation can form. Option B involves a tertiary chlo...

Read Full Step-by-Step Solution →

Question #9

Practice Question

A.(R)-2-iodobutane
B.(S)-2-iodobutane
C.racemic mixture of 2-iodobutane
D.No reaction occurs

Concept Applied

NaI in acetone promotes an SN2 substitution. SN2 proceeds with a backside attack, causing Walden inversion of configuration. Hence the (R) substrate g...

Read Full Step-by-Step Solution →

Question #10

Practice Question

A.CH₃Cl
B.CH₃CH₂Cl
C.(CH₃)₂CHCl
D.(CH₃)₃CCl

Concept Applied

SN1 rate depends on carbocation stability. Tertiary carbocation from (CH₃)₃CCl is most stable....

Read Full Step-by-Step Solution →

Question #11

Practice Question

A.0.5 mol
B.1.0 mol
C.1.5 mol
D.2.0 mol

Concept Applied

Each mole of CCl₂F₂ yields two moles of Cl atoms. Therefore, $0.5\,\text{mol}$ of CCl₂F₂ produces $0.5\times2 = 1.0\,\text{mol}$ of chlorine atoms....

Read Full Step-by-Step Solution →

Question #12

Practice Question

A.A fundamental principle of Chemistry.
B.A complex derivation in JEEMains syllabus.
C.An experimental observation.
D.A theoretical assumption.

Concept Applied

This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Haloalkanes and Halo...

Read Full Step-by-Step Solution →

Question #13

Practice Question

A.CH₃Cl
B.(CH₃)₂CHCl
C.(CH₃)₃CCl
D.CH₃CH₂Cl

Concept Applied

SN1 rate depends on carbocation stability. Tertiary carbocation is most stable....

Read Full Step-by-Step Solution →

Question #14

Practice Question

A.CH₃Br + OH⁻ → CH₃OH
B.(CH₃)₃CCl + H₂O → (CH₃)₃COH + HCl
C.CH₃CH₂Br + NaI → CH₃CH₂I + NaBr
D.C₂H₅Cl + KCN → C₂H₅CN + KCl

Concept Applied

SN1 reactions are favored for tertiary alkyl halides in polar protic solvents because a stable carbocation can form. Option B involves a tertiary chlo...

Read Full Step-by-Step Solution →

Question #15

Practice Question

A.(R)-2-iodobutane
B.(S)-2-iodobutane
C.racemic mixture of 2-iodobutane
D.No reaction occurs

Concept Applied

NaI in acetone promotes an SN2 substitution. SN2 proceeds with a backside attack, causing Walden inversion of configuration. Hence the (R) substrate g...

Read Full Step-by-Step Solution →

Question #16

Practice Question

A.CH₃Cl
B.CH₃CH₂Cl
C.(CH₃)₂CHCl
D.(CH₃)₃CCl

Concept Applied

SN1 rate depends on carbocation stability. Tertiary carbocation from (CH₃)₃CCl is most stable....

Read Full Step-by-Step Solution →

Question #17

Practice Question

A.0.5 mol
B.1.0 mol
C.1.5 mol
D.2.0 mol

Concept Applied

Each mole of CCl₂F₂ yields two moles of Cl atoms. Therefore, $0.5\,\text{mol}$ of CCl₂F₂ produces $0.5\times2 = 1.0\,\text{mol}$ of chlorine atoms....

Read Full Step-by-Step Solution →

Question #18

Practice Question

A.A fundamental principle of Chemistry.
B.A complex derivation in JEEMains syllabus.
C.An experimental observation.
D.A theoretical assumption.

Concept Applied

This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Haloalkanes and Halo...

Read Full Step-by-Step Solution →

Question #19

Practice Question

A.CH₃Cl
B.(CH₃)₂CHCl
C.(CH₃)₃CCl
D.CH₃CH₂Cl

Concept Applied

SN1 rate depends on carbocation stability. Tertiary carbocation is most stable....

Read Full Step-by-Step Solution →

Question #20

Practice Question

A.CH₃Br + OH⁻ → CH₃OH
B.(CH₃)₃CCl + H₂O → (CH₃)₃COH + HCl
C.CH₃CH₂Br + NaI → CH₃CH₂I + NaBr
D.C₂H₅Cl + KCN → C₂H₅CN + KCl

Concept Applied

SN1 reactions are favored for tertiary alkyl halides in polar protic solvents because a stable carbocation can form. Option B involves a tertiary chlo...

Read Full Step-by-Step Solution →

Question #21

Practice Question

A.(R)-2-iodobutane
B.(S)-2-iodobutane
C.racemic mixture of 2-iodobutane
D.No reaction occurs

Concept Applied

NaI in acetone promotes an SN2 substitution. SN2 proceeds with a backside attack, causing Walden inversion of configuration. Hence the (R) substrate g...

Read Full Step-by-Step Solution →

Question #22

Practice Question

A.CH₃Cl
B.CH₃CH₂Cl
C.(CH₃)₂CHCl
D.(CH₃)₃CCl

Concept Applied

SN1 rate depends on carbocation stability. Tertiary carbocation from (CH₃)₃CCl is most stable....

Read Full Step-by-Step Solution →

Question #23

Practice Question

A.0.5 mol
B.1.0 mol
C.1.5 mol
D.2.0 mol

Concept Applied

Each mole of CCl₂F₂ yields two moles of Cl atoms. Therefore, $0.5\,\text{mol}$ of CCl₂F₂ produces $0.5\times2 = 1.0\,\text{mol}$ of chlorine atoms....

Read Full Step-by-Step Solution →

Question #24

Practice Question

A.A fundamental principle of Chemistry.
B.A complex derivation in JEEMains syllabus.
C.An experimental observation.
D.A theoretical assumption.

Concept Applied

This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Haloalkanes and Halo...

Read Full Step-by-Step Solution →

Question #25

Practice Question

A.CH₃Cl
B.(CH₃)₂CHCl
C.(CH₃)₃CCl
D.CH₃CH₂Cl

Concept Applied

SN1 rate depends on carbocation stability. Tertiary carbocation is most stable....

Read Full Step-by-Step Solution →

Question #26

Practice Question

A.CH₃Br + OH⁻ → CH₃OH
B.(CH₃)₃CCl + H₂O → (CH₃)₃COH + HCl
C.CH₃CH₂Br + NaI → CH₃CH₂I + NaBr
D.C₂H₅Cl + KCN → C₂H₅CN + KCl

Concept Applied

SN1 reactions are favored for tertiary alkyl halides in polar protic solvents because a stable carbocation can form. Option B involves a tertiary chlo...

Read Full Step-by-Step Solution →

Question #27

Practice Question

A.(R)-2-iodobutane
B.(S)-2-iodobutane
C.racemic mixture of 2-iodobutane
D.No reaction occurs

Concept Applied

NaI in acetone promotes an SN2 substitution. SN2 proceeds with a backside attack, causing Walden inversion of configuration. Hence the (R) substrate g...

Read Full Step-by-Step Solution →

Question #28

Practice Question

A.CH₃Cl
B.CH₃CH₂Cl
C.(CH₃)₂CHCl
D.(CH₃)₃CCl

Concept Applied

SN1 rate depends on carbocation stability. Tertiary carbocation from (CH₃)₃CCl is most stable....

Read Full Step-by-Step Solution →

Question #29

Practice Question

A.0.5 mol
B.1.0 mol
C.1.5 mol
D.2.0 mol

Concept Applied

Each mole of CCl₂F₂ yields two moles of Cl atoms. Therefore, $0.5\,\text{mol}$ of CCl₂F₂ produces $0.5\times2 = 1.0\,\text{mol}$ of chlorine atoms....

Read Full Step-by-Step Solution →

Question #30

Practice Question

A.A fundamental principle of Chemistry.
B.A complex derivation in JEEMains syllabus.
C.An experimental observation.
D.A theoretical assumption.

Concept Applied

This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Haloalkanes and Halo...

Read Full Step-by-Step Solution →

Question #31

Practice Question

A.CH₃Cl
B.(CH₃)₂CHCl
C.(CH₃)₃CCl
D.CH₃CH₂Cl

Concept Applied

SN1 rate depends on carbocation stability. Tertiary carbocation is most stable....

Read Full Step-by-Step Solution →

Question #32

Practice Question

A.CH₃Br + OH⁻ → CH₃OH
B.(CH₃)₃CCl + H₂O → (CH₃)₃COH + HCl
C.CH₃CH₂Br + NaI → CH₃CH₂I + NaBr
D.C₂H₅Cl + KCN → C₂H₅CN + KCl

Concept Applied

SN1 reactions are favored for tertiary alkyl halides in polar protic solvents because a stable carbocation can form. Option B involves a tertiary chlo...

Read Full Step-by-Step Solution →

Question #33

Practice Question

A.(R)-2-iodobutane
B.(S)-2-iodobutane
C.racemic mixture of 2-iodobutane
D.No reaction occurs

Concept Applied

NaI in acetone promotes an SN2 substitution. SN2 proceeds with a backside attack, causing Walden inversion of configuration. Hence the (R) substrate g...

Read Full Step-by-Step Solution →

Question #34

Practice Question

A.CH₃Cl
B.CH₃CH₂Cl
C.(CH₃)₂CHCl
D.(CH₃)₃CCl

Concept Applied

SN1 rate depends on carbocation stability. Tertiary carbocation from (CH₃)₃CCl is most stable....

Read Full Step-by-Step Solution →

Question #35

Practice Question

A.0.5 mol
B.1.0 mol
C.1.5 mol
D.2.0 mol

Concept Applied

Each mole of CCl₂F₂ yields two moles of Cl atoms. Therefore, $0.5\,\text{mol}$ of CCl₂F₂ produces $0.5\times2 = 1.0\,\text{mol}$ of chlorine atoms....

Read Full Step-by-Step Solution →

Question #36

Practice Question

A.A fundamental principle of Chemistry.
B.A complex derivation in JEEMains syllabus.
C.An experimental observation.
D.A theoretical assumption.

Concept Applied

This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Haloalkanes and Halo...

Read Full Step-by-Step Solution →

Question #37

Practice Question

A.CH₃Cl
B.(CH₃)₂CHCl
C.(CH₃)₃CCl
D.CH₃CH₂Cl

Concept Applied

SN1 rate depends on carbocation stability. Tertiary carbocation is most stable....

Read Full Step-by-Step Solution →

Question #38

Practice Question

A.CH₃Br + OH⁻ → CH₃OH
B.(CH₃)₃CCl + H₂O → (CH₃)₃COH + HCl
C.CH₃CH₂Br + NaI → CH₃CH₂I + NaBr
D.C₂H₅Cl + KCN → C₂H₅CN + KCl

Concept Applied

SN1 reactions are favored for tertiary alkyl halides in polar protic solvents because a stable carbocation can form. Option B involves a tertiary chlo...

Read Full Step-by-Step Solution →

Question #39

Practice Question

A.(R)-2-iodobutane
B.(S)-2-iodobutane
C.racemic mixture of 2-iodobutane
D.No reaction occurs

Concept Applied

NaI in acetone promotes an SN2 substitution. SN2 proceeds with a backside attack, causing Walden inversion of configuration. Hence the (R) substrate g...

Read Full Step-by-Step Solution →

Question #40

Practice Question

A.CH₃Cl
B.CH₃CH₂Cl
C.(CH₃)₂CHCl
D.(CH₃)₃CCl

Concept Applied

SN1 rate depends on carbocation stability. Tertiary carbocation from (CH₃)₃CCl is most stable....

Read Full Step-by-Step Solution →

Question #41

Practice Question

A.0.5 mol
B.1.0 mol
C.1.5 mol
D.2.0 mol

Concept Applied

Each mole of CCl₂F₂ yields two moles of Cl atoms. Therefore, $0.5\,\text{mol}$ of CCl₂F₂ produces $0.5\times2 = 1.0\,\text{mol}$ of chlorine atoms....

Read Full Step-by-Step Solution →

Question #42

Practice Question

A.A fundamental principle of Chemistry.
B.A complex derivation in JEEMains syllabus.
C.An experimental observation.
D.A theoretical assumption.

Concept Applied

This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Haloalkanes and Halo...

Read Full Step-by-Step Solution →

Question #43

Practice Question

A.CH₃Cl
B.(CH₃)₂CHCl
C.(CH₃)₃CCl
D.CH₃CH₂Cl

Concept Applied

SN1 rate depends on carbocation stability. Tertiary carbocation is most stable....

Read Full Step-by-Step Solution →

Question #44

Practice Question

A.CH₃Br + OH⁻ → CH₃OH
B.(CH₃)₃CCl + H₂O → (CH₃)₃COH + HCl
C.CH₃CH₂Br + NaI → CH₃CH₂I + NaBr
D.C₂H₅Cl + KCN → C₂H₅CN + KCl

Concept Applied

SN1 reactions are favored for tertiary alkyl halides in polar protic solvents because a stable carbocation can form. Option B involves a tertiary chlo...

Read Full Step-by-Step Solution →

Question #45

Practice Question

A.(R)-2-iodobutane
B.(S)-2-iodobutane
C.racemic mixture of 2-iodobutane
D.No reaction occurs

Concept Applied

NaI in acetone promotes an SN2 substitution. SN2 proceeds with a backside attack, causing Walden inversion of configuration. Hence the (R) substrate g...

Read Full Step-by-Step Solution →

Question #46

Practice Question

A.CH₃Cl
B.CH₃CH₂Cl
C.(CH₃)₂CHCl
D.(CH₃)₃CCl

Concept Applied

SN1 rate depends on carbocation stability. Tertiary carbocation from (CH₃)₃CCl is most stable....

Read Full Step-by-Step Solution →

Question #47

Practice Question

A.0.5 mol
B.1.0 mol
C.1.5 mol
D.2.0 mol

Concept Applied

Each mole of CCl₂F₂ yields two moles of Cl atoms. Therefore, $0.5\,\text{mol}$ of CCl₂F₂ produces $0.5\times2 = 1.0\,\text{mol}$ of chlorine atoms....

Read Full Step-by-Step Solution →

Question #48

Practice Question

A.A fundamental principle of Chemistry.
B.A complex derivation in JEEMains syllabus.
C.An experimental observation.
D.A theoretical assumption.

Concept Applied

This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Haloalkanes and Halo...

Read Full Step-by-Step Solution →
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