${\rm{Ca}}{{\rm{C}}_2} + {{\rm{H}}_2}{\rm{O}} \to X\xrightarrow{{{{\rm{O}}_3}/\frac{{{{\rm{H}}_2}{\rm{O}}}}{{{\rm{O}}{{\rm{H}}^ + }}}}}{\rm{HCOOH}} + {\rm{HCOOH}},X\;{\rm{is}}$

${\rm{Ca}}{\left( {{\rm{OH}}} \right)_2}$

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JEE • CHEMISTRY • HYDROCARBONS

Hydrocarbons practice · Mixed difficulty

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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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Showing 1-30 of 269 questions

1 Medium

1 of 269

Understanding +4 | -1

\mathrm{Ca}{\mathrm{C}_2} + {\mathrm{H}_2}\mathrm{O} \to X\xrightarrow{{{\mathrm{O}_3}/\frac{{{\mathrm{H}_2}\mathrm{O}}}{{\mathrm{O}{\mathrm{H}^ + }}}}}\mathrm{HCOOH} + \mathrm{HCOOH},X\;\mathrm{is}

View answer and explanation

Correct option: B

{\mathrm{C}_2}{\mathrm{H}_2}

Option B (

{\mathrm{C}_2}{\mathrm{H}_2}

) is correct because the reaction of calcium carbide with water produces ethyne (acetylene), which upon ozonolysis followed by hydrolysis yields formic acid.

2 Easy

2 of 269

Understanding +4 | -1

In benzene, there is a delocalisation of π-electrons. Hence, each π-electron is attached by….carbon nuclei.

View answer and explanation

Correct option: C

Option C (6) is correct because in benzene, the π-electrons are delocalized over all six carbon atoms in the ring.

3 Easy

3 of 269

Understanding +4 | -1

Which statement is false?

View answer and explanation

Correct option: C

Chlorination of methane follows an ionic mechanism

Option C is correct because chlorination of methane follows a free radical mechanism, not an ionic mechanism.

4 Easy

4 of 269

Analyzing +4 | -1

Assertion (A) In comparison to ethyl chloride, it is not easy to carry outnucleophilic substitution on vinyl chloride.

Reason (R) Vinyl group is an electron donating.

View answer and explanation

Correct option: C

If Assertion is true but Reason is false.

Option C is correct because nucleophilic substitution is difficult on vinyl chloride due to resonance, which gives the C-Cl bond partial double bond character, but the vinyl group is electron-withdrawing, not donating.

5 Medium

5 of 269

Analyzing +4 | -1

If 2-oxopropanal and methanal are obtained by the reductive ozonolysis of compound (A), then the number of

\sigma

-bonds presents in

View answer and explanation

Correct option: C

Option C (12) is correct because the reductive ozonolysis of isoprene yields 2-oxopropanal and methanal, and isoprene has 12 sigma bonds.

6 Easy

6 of 269

Remembering +4 | -1

Fischer-Tropsch process is used in the manufacture of:

View answer and explanation

Correct option: A

Synthetic petrol

Option A (Synthetic petrol) is correct because the Fischer-Tropsch process is a chemical reaction used to produce synthetic hydrocarbons, which can be refined into synthetic petrol (gasoline).

7 Easy

7 of 269

Understanding +4 | -1

Among the following group, which deactivates benzene ring for electrophilic substitution:

View answer and explanation

Correct option: D

Chlorine

Option D (Chlorine) is correct because halogens deactivate the benzene ring due to their strong -I effect, even though they are ortho- and para-directing.

8 Medium

8 of 269

Understanding +4 | -1

1-Alkyne and 2-alkyne can be distinguished by:

View answer and explanation

Correct option: A

Tollens reagent

Option A (Tollens reagent) is correct because terminal alkynes (1-alkynes) react with Tollens' reagent to form a precipitate, while internal alkynes (2-alkynes) do not.

9 Medium

9 of 269

Analyzing +4 | -1

An alkene on vigarous oxidation with

KMn{O_4}

gives only acetic acid. The alkene is

View answer and explanation

Correct option: B

C{H_3}CH = CHC{H_3}

Option B (

C{H_3}CH = CHC{H_3}

) is correct because upon oxidation with

KMn{O_4}

, it yields only acetic acid.

10 Easy

10 of 269

Understanding +4 | -1

Which of the following annulenes is anti-aromatic?

View answer and explanation

Correct option: B

Cyclobutadiene

Option B is correct. Cyclobutadiene is anti-aromatic because it is cyclic, planar, has 4π electrons (which is not a Hückel number), and has alternating single and double bonds.

11 Easy

11 of 269

Remembering +4 | -1

By coaltar distillation which is not obtained?

View answer and explanation

Correct option: D

Mobil oil

Option D (Mobil oil) is correct because light oil, middle oil, and heavy oil are all obtained through coal tar distillation, while mobil oil is not.

12 Medium

12 of 269

Analyzing +4 | -1

Statement-1 : All the hydrogen atoms in

{CH}_2={C}={CH}_2

lie in one plane.

Statement-2 : Carbon atoms are

s p^2

and

s p

hybridized.

View answer and explanation

Correct option: D

Assertion is incorrect, reason is correct.

Option D is correct because both the assertion and the reason are false. The standard cell potential is not always zero, and the given reaction has a non-zero standard cell potential.

13 Easy

13 of 269

Remembering +4 | -1

An anaestheticnarcylene is commercial name of:

View answer and explanation

Correct option: B

{\mathrm{C}_2}{\mathrm{H}_2}

Option B (

{\mathrm{C}_2}{\mathrm{H}_2}

) is correct because narcylene is the commercial name for acetylene, which was historically used as an anesthetic.

14 Hard

14 of 269

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.

15 Easy

15 of 269

Remembering +4 | -1

The main constituent of light oil fraction is:

View answer and explanation

Correct option: A

Benzene

Option A (Benzene) is correct because benzene (

C_6H_6

) is the main constituent of the light oil fraction obtained from coal tar distillation.

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16 Easy

16 of 269

Remembering +4 | -1

Benzene is converted to toluene by:

View answer and explanation

Correct option: A

Friedel-Crafts reaction

Option A (Friedel-Crafts reaction) is correct because the Friedel-Crafts alkylation reaction is used to add alkyl groups, such as methyl groups, to aromatic rings like benzene, converting it to toluene.

17 Easy

17 of 269

Understanding +4 | -1

Which of the following compounds can be best prepared by Wurtz reaction?

View answer and explanation

Correct option: D

n-butane

Option B (2 molecules of benzoic acid, 1 molecule of benzoic acid and 1 molecule of ethanoic acid) is correct because it shows the hydrolysis products of the given anhydrides.

18 Medium

18 of 269

Understanding +4 | -1

Assertion (A): Benzene under goes nitration to form nitro benzene
Reason (R): Benzene nitration on electrophile is

{\text{NO}}_2^\Theta

View answer and explanation

Correct option: C

A is correct, R is incorrect

Option C is correct because while benzene does undergo nitration to form nitrobenzene, the electrophile in the nitration reaction is

\mathrm{NO}_2^+

, not

\mathrm{NO}_2^-

19 Easy

19 of 269

Understanding +4 | -1

Assertion (A): Acetylene does not react with

\mathrm{NaOH}

Reason (R): Acetylene has sp hybridised carbons.

View answer and explanation

Correct option: B

Both A

\&

R are correct but R is not the correct explanation of A

Option B (Both A

\& R are correct but R is not the correct explanation of A) is correct because acetylene is a weak acid and does not react with NaOH, and it does have sp hybridized carbons, but the hybridization is not the reason for its lack of reactivity with NaOH.

20 Medium

20 of 269

Understanding +4 | -1

S-I: 2-butene can show geometrical isomerism
S-II: B.P of cis isomer is higher than trans isomer

View answer and explanation

Correct option: A

Both S-I

\&

S-II are correct

Option A is correct because 2-butene exhibits geometrical isomerism, and the cis isomer generally has a higher boiling point than the trans isomer.

21 Medium

21 of 269

Application +4 | -1

By Wurtz reaction, a mixture of methyl iodide and ethyl iodide gives

View answer and explanation

Correct option: D

A mixture of the above three

Option D (A mixture of the above three) is correct because Wurtz reaction with a mixture of alkyl halides results in a mixture of alkanes due to different coupling possibilities.

22 Medium

22 of 269

Application +4 | -1

1 - {\text{Butyne}}\xrightarrow[\begin{subarray}{l} {\text{catalyst}} \\ {\text{controlled}} \\ {\text{readuction}} \end{subarray} ]{{{\text{Lindlar's}}}}A\xrightarrow[\begin{subarray}{l} {\text{Benzoyl}} \\ {\text{peroxide}} \end{subarray} ]{{HCl\,\,{\text{in}}}}B\xrightarrow[{KOH}]{{{\text{alc}}}}C,\,\,\,C

In the above reaction

View answer and explanation

Correct option: B

2-butene

The correct answer is 2-butene because the reaction sequence involves Lindlar's catalyst for reduction to cis-alkene, followed by HCl addition and then elimination.

23 Easy

23 of 269

Understanding +4 | -1

Which of the following compound is not prepared by Kolbe’s electrolytic method

View answer and explanation

Correct option: C

C{H_4}

Option C is correct because Kolbe's electrolytic method involves the electrolysis of an aqueous solution of sodium or potassium salts of carboxylic acids, leading to the formation of alkanes, alkenes, or alkynes with an even number of carbon atoms. Methane (CH4) cannot be prepared this way.

24 Medium

24 of 269

Understanding +4 | -1

Hydrogen atoms of ethyne attached to triply bonded carbon atom are acidic in nature due to

View answer and explanation

Correct option: C

Both (1) and (2)

Option C (Both (1) and (2)) is correct because the acidity of ethyne's hydrogen atoms is due to both the high s-character of the sp hybrid orbitals and the resulting higher electronegativity.

25 Medium

25 of 269

Analyzing +4 | -1

S-I: Addition of

\mathrm{HBr}

\mathrm{HC} \equiv \mathrm{CCH}_2 \mathrm{CH}=\mathrm{CH}_2

give

\mathrm{HC}=\mathrm{CCH}_2 \mathrm{CHBr}-\mathrm{CH}_3

and not

\mathrm{H}_2 \mathrm{C}=\mathrm{CBrCH}_2 \mathrm{CH}=\mathrm{CH}_2

S-II: A triple bond is less reactive than a double bond towards electrophilic addition reaction.

View answer and explanation

Correct option: A

S-I

\&

S-II are true

Option A (S-I & S-II are true) is correct because both statements accurately describe the reactivity and product formation in the given reaction.

26 Medium

26 of 269

Understanding +4 | -1

Which of the following statements are correct?

(i) Cis form of alkene is polar whereas trans form is non-polar

(ii) Cis form of alkene is non-polar whereas trans form is polar.

(iii) In case of solid alkenes the trans isomer has higher melting point than the cis isomer.

(iv) Cis and trans both form have same properties.

View answer and explanation

Correct option: A

(i) and (iii)

Option A ((i) and (iii)) is correct because cis alkenes are generally polar, trans alkenes are non-polar, and trans isomers of solid alkenes have higher melting points due to better packing.

27 Easy

27 of 269

Understanding +4 | -1

Butyne-1 and butyne-2 can be distinguished by:

View answer and explanation

Correct option: D

Ammoniacal cuprous chloride

Option D (Ammoniacal cuprous chloride) is correct because terminal alkynes (butyne-1) react with ammoniacal cuprous chloride to form a red precipitate, while internal alkynes (butyne-2) do not.

28 Medium

28 of 269

Understanding +4 | -1

2 \mathrm{CH}_4+\mathrm{O}_2 \xrightarrow{\mathrm{Cu} / 523 \mathrm{~K} / 100 \mathrm{~atm}} \mathrm{~A}, \mathrm{~A}

is :

View answer and explanation

Correct option: C

\mathrm{CH}_3 \mathrm{OH}

Option C (

\mathrm{CH}_3 \mathrm{OH}

) is correct because the reaction of methane with oxygen under specific conditions yields methanol.

29 Hard

29 of 269

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.

30 Easy

30 of 269

Remembering +4 | -1

The number of

\pi

-electrons in benzene molecule is :

View answer and explanation

Correct option: A

3 ×2

Option A (3 ×2) is correct because benzene has 6 pi electrons, which can be represented as 3 double bonds, hence 3 x 2 = 6 pi electrons.

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Key Takeaways

✓ This set contains 269 single-correct-answer MCQs on HYDROCARBONS.
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