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JEE • CHEMISTRY • HALOALKANES AND HALO ARENES

Haloalkanes and halo arenes practice · Hard

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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1 Hard

1 of 8

Understanding +4 | -1

Which of the followings statements is/are incorrect?

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Correct option: B

The compound which rotate light to left is called dextrorotatory and that towards right is called laevorotatory

The correct answer is 'The compound which rotate light to left is called dextrorotatory and that towards right is called laevorotatory' because the definitions are reversed.

2 Hard

2 of 8

Analyzing +4 | -1

Which of the following is not correct?

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Correct option: A

{PhCH}_2 {Br}>{PhChBrCH}_3>{PhCBr}\left({CH}_3\right)_2\left({S}_{{N}} 1\right)

Option A (

{PhCH}_2 {Br}>{PhChBrCH}_3>{PhCBr}\left({CH}_3\right)_2\left({S}_{{N}} 1\right)

) is not correct because the stability of the carbocation intermediate increases with more alkyl substituents, thus increasing SN1 reactivity.

3 Hard

3 of 8

Analyzing +4 | -1

Assertion : - 2-Bromo butane on reaction with alcoholic

\mathrm{KOH}

mainly gives 2-butene.

Reason : - The hydrogen on

C_3

carbon is more acidic than hydrogen on

C_1

carbon.

View answer and explanation

Correct option: C

If Assertion is True but the Reason is False.

Option C is correct because the assertion is true, but the reason is false. 2-butene is the major product due to Zaitsev's rule, but the acidity of hydrogens is not the reason.

4 Hard

4 of 8

Application +4 | -1

A solution of

Ni{(N{O_3})_2}

is electrolyzed between platinum electrodes using 0.1 Faraday electricity. How many mole of Ni will be deposited at the cathode?

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Correct option: D

0.05

Option D (0.05) is correct because 0.1 Faraday of electricity will deposit 0.05 moles of Ni, given that Ni is in the +2 oxidation state.

5 Hard

5 of 8

Analyzing +4 | -1

Arrange the following carbonyl compounds in increasing order of their reactivity in nucleophilic addition reaction

i) Benaldehyde

\quad

ii) P-Tolualdehyde
iii) P-Nitro benzaldehyde

\quad

iv) Acetophenone

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Correct option: C

iv < ii < i < iii

The correct answer is C (iv < ii < i < iii) because electron-withdrawing groups increase reactivity in nucleophilic addition, and electron-donating groups decrease it.

6 Hard

6 of 8

Analyzing +4 | -1

For the reaction

\mathrm{RCH}_{2} \mathrm{Br}+\mathrm{I}^{-} \stackrel{\text { Acetone }}{\longrightarrow} \underset{\text { major }}{\mathrm{RCH}_{2}} \mathrm{I}+\mathrm{Br}^{-}

The correct statement is

View answer and explanation

Correct option: C

The transition state formed in the above reaction is less polar than the localised anion

The correct answer is C because in SN2 reactions, the transition state is less polar than the localized anion due to charge dispersal.

7 Hard

7 of 8

Remembering +4 | -1

Reaction of red phosphorus with bromine and iodine generate

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Correct option: C

Both (1) and (2)

Option C (Both (1) and (2)) is correct because red phosphorus reacts with bromine to form phosphorus tribromide and with iodine to form phosphorus triiodide.

8 Hard

8 of 8

Analyzing +4 | -1

Among the following, the most reactive towards alcoholic

{KOH}

is :

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Correct option: B

{CH}_3 {COCH}_2 {CH}_2 {Br}

Option B (

{CH}_3 {COCH}_2 {CH}_2 {Br}

) is correct because the presence of the electron-withdrawing carbonyl group makes the beta-hydrogens more acidic, facilitating elimination by alcoholic KOH.

Revision notes and key takeaways

Review HALOALKANES AND HALO ARENES in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 8 single-correct-answer MCQs on HALOALKANES AND HALO ARENES.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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