Identify the product (P) in the reaction: ${R_3}{\rm{C}}-{\rm{H}}\xrightarrow{{{\rm{Alk}}{\rm{.KMn}}{{\rm{O}}_4}}}{\rm{P}}$

${R_3}{\rm{C}}-{\rm{C}}{{\rm{R}}_3}$

${R_3}{\rm{C}}-{\rm{O}}-{\rm{C}}{R_3}$

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JEE • CHEMISTRY • HYDROCARBONS

Hydrocarbons 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 13

Application +4 | -1

\mathop {C{H_3} - C \equiv CH}\limits_{{\text{Propyne}}} \xrightarrow[{\,\,\,\,\,}]{{\,\,\,\,\,}}\mathop {C{H_3} - C \equiv C - C{H_2} - C{H_3}}\limits_{{\text{Pent}}\,\,2 - {\text{yne}}}

The most suitable reagents required for above
conversion?

View answer and explanation

Correct option: B

NaN{H_2},\,C{H_3}C{H_2}I

Option B (

NaNH_2, CH_3CH_2I

) is correct because it uses a strong base to deprotonate the alkyne, followed by alkylation with ethyl iodide.

2 Hard

2 of 13

Analyzing +4 | -1

1,3-Butadiene is treated with hot

\mathrm{KMnO}_4 / \mathrm{NaIO}_4

, the product(s) formed is(are)

View answer and explanation

Correct option: C

2 \mathrm{CO}_2+\mathrm{HOOC}-\mathrm{COOH}

The correct answer is C (

2 \mathrm{CO}_2+\mathrm{HOOC}-\mathrm{COOH}

) because hot KMnO4/NaIO4 strongly oxidizes 1,3-butadiene to carbon dioxide and oxalic acid.

3 Hard

3 of 13

Understanding +4 | -1

Polynuclear hydrocarbons are formed by an incomplete process of organic materials. This process is termed as

View answer and explanation

Correct option: B

combustion

Option B (combustion) is correct because polynuclear hydrocarbons are formed during the incomplete combustion of organic materials.

4 Hard

4 of 13

Application +4 | -1

The major product formed when a 3, 3-dimethyl butan-2-ol is heated with concentrated sulphuric acid, is

View answer and explanation

Correct option: A

2,3-dimethyl-2-butene

The correct answer is A (2,3-dimethyl-2-butene) because the major product formed is the most stable alkene according to Zaitsev's rule, which involves the elimination of water from the alcohol.

5 Hard

5 of 13

Analyzing +4 | -1

The isomer of

\mathrm{C}_6 \mathrm{H}_{14},

which will give maximum number of monochloroderivative is

View answer and explanation

Correct option: B

2-methylpentane

Option C (inorganic graphite) is correct because diborane reacts with ammonia at high temperatures to form boron nitride, which has a graphite-like structure.

6 Hard

6 of 13

Analyzing +4 | -1

Consider the following statements : A hydrocarbon of molecular formula

{C}_5 {H}_{10}

is a

I. monosubstituted alkene

II. disubstituted alkene

III. trisubstituted alkene

Which of the following statement(s) is(are) correct?

View answer and explanation

Correct option: A

I, II and III

Option A (I, II and III) is correct because a hydrocarbon with the formula

{C}_5 {H}_{10}

can be mono-, di-, or trisubstituted.

7 Hard

7 of 13

Application +4 | -1

Ca{C_2}\xrightarrow{{{H_2}O}}A\xrightarrow[{H{g^{2 + }}}]{{dil.{H_2}S{O_4}}}B\xrightarrow[{{H_2}}]{{Ni}}C

end product is

View answer and explanation

Correct option: C

{C_2}{H_5}OH

Option C (

{C_2}{H_5}OH

) is correct because the sequence of reactions involves hydration of calcium carbide to acetylene, followed by hydration of acetylene to acetaldehyde, and finally, reduction of acetaldehyde to ethanol.

8 Hard

8 of 13

Analyzing +4 | -1

The decreasing order of reactivity of hydrogen bromide on the following alkenes is

\mathop {{\text{C}}{{\text{H}}_2} = {\text{C}}{{\text{H}}_2}}\limits_{\text{I}} ,\mathop {{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{CH}} = {\text{C}}{{\text{H}}_2}}\limits_{{\text{II}}} ,\mathop {{\text{C}}{{\text{H}}_2} = {\text{CHCl}}}\limits_{{\text{III}}} {\text{,}}\mathop {{{\left( {{\text{C}}{{\text{H}}_3}} \right)}_2}{\text{C}} = {\text{C}}{{\text{H}}_2}}\limits_{{\text{IV}}}

View answer and explanation

Correct option: B

>

>

>

III

The correct answer is B (IV

>

>

>

III) because the reactivity of alkenes with HBr increases with the stability of the carbocation formed, and is decreased by electron-withdrawing groups.

9 Hard

9 of 13

Understanding +4 | -1

Identify the product (P) in the reaction:

{R_3}\mathrm{C}-\mathrm{H}\xrightarrow{{\mathrm{Alk}\mathrm{.KMn}{\mathrm{O}_4}}}\mathrm{P}

View answer and explanation

Correct option: C

{R_3}\mathrm{C}-\mathrm{OH}

Option C (

{R_3}\mathrm{C}-\mathrm{OH}

) is correct because alkaline

KMnO_4

oxidizes the C-H bond of a tertiary carbon to a hydroxyl group.

10 Hard

10 of 13

Remembering +4 | -1

Angle strain in cyclopropane is

View answer and explanation

Correct option: A

24^\circ 44'

Option A (

24^\circ 44'

) is correct because it represents the calculated angle strain in cyclopropane based on its geometry.

11 Hard

11 of 13

Application +4 | -1

At constant temperature and pressure, the percentage of butane by volume in a mixture of butane, propane and propyne is

40 \%.

What will be the volume of

\mathrm{CO}_{2(\mathrm{g})}

(in ml) under same pressure and temperature produced, when 150 ml of gaseous mixture is burnt in excess of

\mathrm{O}_2?

View answer and explanation

Correct option: C

510 ml

Option C (510 ml) is correct because it's the total volume of CO2 produced from the combustion of the mixture.

12 Hard

12 of 13

Understanding +4 | -1

S-I: Friedal Crafts Acylation of benzene with acetic anhydride in presence of anhydrous

{AlCl}_3

yields acetophenone and not poly substituted products.
S-II: It is due to stearic hindrance of bulky actyl group

C{H_3} - \mathop {{\text{ }}C}\limits^{\begin{array}{cccccccccccccccccccc} O \\ \parallel \end{array}}-

and also, it is a deactivating group.

View answer and explanation

Correct option: A

S-I

\&

S-II are true

Option A is correct because Friedel-Crafts acylation indeed yields acetophenone and not polysubstituted products due to steric hindrance and the deactivating nature of the acyl group.

13 Hard

13 of 13

Analyzing +4 | -1

S-I : Heavy metal acetylides can be used to purify alkynes
S-II: Terminal alkynes form acetylides which are soluble in acids.

View answer and explanation

Correct option: A

S-I

\&

S-II are true

Option A is correct because heavy metal acetylides can be used to purify alkynes, and terminal alkynes form acetylides soluble in acids.

Revision notes and key takeaways

Review HYDROCARBONS in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 13 single-correct-answer MCQs on HYDROCARBONS.
✓ Try each question before opening its answer. Read the explanation where one is available.
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