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

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

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JEE • CHEMISTRY • HYDROCARBONS

Hydrocarbons practice · Medium

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Showing 1-30 of 118 questions

1 Medium

1 of 118

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 Medium

2 of 118

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.

3 Medium

3 of 118

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.

4 Medium

4 of 118

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.

5 Medium

5 of 118

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.

6 Medium

6 of 118

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^-

7 Medium

7 of 118

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.

8 Medium

8 of 118

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.

9 Medium

9 of 118

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.

10 Medium

10 of 118

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.

11 Medium

11 of 118

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.

12 Medium

12 of 118

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.

13 Medium

13 of 118

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.

14 Medium

14 of 118

Understanding +4 | -1

Which is not deflected by a non-uniform electrostatic field?

View answer and explanation

Correct option: D

Hexane

Option D (Hexane) is correct because hexane is a nonpolar molecule and does not possess a dipole moment, hence it is not deflected by a non-uniform electrostatic field.

15 Medium

15 of 118

Remembering +4 | -1

The compound in which the C-atoms form a closed chain are termed as

View answer and explanation

Correct option: B

cycloalkanes

Option B (cycloalkanes) is correct because cycloalkanes are hydrocarbons in which the carbon atoms form a closed ring or chain.

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16 Medium

16 of 118

Understanding +4 | -1

S-I: Benzene does not decolourise alkaline

\mathrm{KMnO}_4

.
S-II: Benzene is stabilized by resonance and

\pi

-electrons are delocalized.

View answer and explanation

Correct option: A

S-I

\&

S-II are true

Option A is correct because benzene is indeed stable due to resonance and does not decolorize alkaline KMnO4 under normal conditions.

17 Medium

17 of 118

Remembering +4 | -1

The main source of aromatic compounds is:

View answer and explanation

Correct option: D

Both (2) and (3)

Option D (Both (2) and (3)) is correct because both petroleum and coal are major sources of aromatic compounds, obtained through refining and processing.

18 Medium

18 of 118

Understanding +4 | -1

1-Butyne reacts with cold alkaline

\mathrm{KMnO}_4

to produce

View answer and explanation

Correct option: C

\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}+\mathrm{CO}_2

Option C (

\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}+\mathrm{CO}_2

) is correct because cold alkaline KMnO4 oxidizes 1-butyne to propanoic acid and carbon dioxide.

19 Medium

19 of 118

Understanding +4 | -1

The presence of

\mathrm{A}{\mathrm{g}^ + }

ion increases the solubility of alkenes due to the formation of

View answer and explanation

Correct option: C

p\pi - d\pi

bonding

Option C (

p\pi - d\pi

bonding) is correct because Ag+ ion increases the solubility of alkenes by forming

p\pi - d\pi

bonds with the alkene's pi system.

20 Medium

20 of 118

Analyzing +4 | -1

A hydrocarbon gives two moles of an aldehyde of molar mass 44 on reductive ozonolysis. IUPAC name of hydrocarbon is

View answer and explanation

Correct option: C

But-2-ene

Option C is correct because but-2-ene undergoes reductive ozonolysis to produce two moles of ethanal (molar mass 44).

21 Medium

21 of 118

Analyzing +4 | -1

Which of the following will not produce ethane?

View answer and explanation

Correct option: B

Reduction of

\mathrm{C}{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_3}

with HI and red P

The correct answer is Reduction of

\mathrm{C}{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_3}

with HI and red P because it will produce propane, not ethane.

22 Medium

22 of 118

Remembering +4 | -1

The reaction,

{C_6}{H_6} + C{H_3}Cl\xrightarrow[{AlC{l_3}}]{{Anhydrous}}{C_6}{H_5}C{H_3} + HCl

is an example of:

View answer and explanation

Correct option: A

Friedel-Crafts reaction

Option A (Friedel-Crafts reaction) is correct because the reaction of benzene with methyl chloride in the presence of anhydrous aluminum chloride is a Friedel-Crafts alkylation.

23 Medium

23 of 118

Remembering +4 | -1

When two halogen atoms are attached to two adjacent carbon atoms, the dihaloalkane is called :

View answer and explanation

Correct option: C

Alkylene dihalide

Option C (Alkylene dihalide) is correct because when two halogen atoms are attached to two adjacent carbon atoms, the dihaloalkane is called an alkylene dihalide (or vicinal dihalide).

24 Medium

24 of 118

Remembering +4 | -1

n-hexadecane (cetane) has cetane number:

View answer and explanation

Correct option: A

100

Option A (100) is correct because n-hexadecane (cetane) is the reference fuel for cetane number determination and is assigned a cetane number of 100.

25 Medium

25 of 118

Understanding +4 | -1

The treatment of

R'\mathrm{Mg}X

with

R\mathrm{C} \equiv \mathrm{CH}

produces

View answer and explanation

Correct option: B

R'\mathrm{H}

Option B (

R'\mathrm{H}

) is correct because the Grignard reagent (

R'\mathrm{Mg}X

) abstracts the acidic proton from the terminal alkyne (

R\mathrm{C} \equiv \mathrm{CH}

), forming an alkynyl magnesium halide and an alkane (

R'\mathrm{H}

26 Medium

26 of 118

Application +4 | -1

The lowest possible alkane with ethyl group as substituents possesses mol. mass equal to:

View answer and explanation

Correct option: D

128

Option D (128) is correct because the lowest possible alkane with ethyl groups as substituents is tetraethylmethane, which has a molecular mass of 128.

27 Medium

27 of 118

Understanding +4 | -1

Which of the following groups deactivates the benzene ring towards electrophilic substitution reaction?

View answer and explanation

Correct option: B

-{NO}_2

Option B (

-{NO}_2

) is correct because the nitro group is an electron-withdrawing group that deactivates the benzene ring towards electrophilic substitution due to its -M effect.

28 Medium

28 of 118

Application +4 | -1

C{H_3} - C \equiv CH\xrightarrow[{HgS{O_4}}]{{{H_2}S{O_4}}}A\xrightarrow[{}]{{{\text{Isomerisation}}}}C{H_3} - \mathop C\limits_{\begin{array}{cccccccccccccccccccc} {||} \\ O \end{array}} - C{H_3}

Structure of

A

and type of isomerism in the above reactions respectively.

View answer and explanation

Correct option: C

Prop-1-en-2-ol, Tautomerism

Option C (Prop-1-en-2-ol, Tautomerism) is correct because the intermediate A is prop-1-en-2-ol, which then undergoes tautomerism to form acetone.

29 Medium

29 of 118

Remembering +4 | -1

Conformation of ethane in which hydrogen atoms attached to two carbon atoms are closest to each other is known as

View answer and explanation

Correct option: A

eclipsed

The correct answer is A (eclipsed) because in the eclipsed conformation, the hydrogen atoms on adjacent carbon atoms are as close as possible.

30 Medium

30 of 118

Understanding +4 | -1

Bromination of toluene gives

View answer and explanation

Correct option: C

Mixture of

\mathrm{o}

and

\mathrm{p}

substituted product

Option C is correct because the methyl group (

\mathrm{-CH}_3

) in toluene is an ortho- and para-directing group, leading to a mixture of ortho- and para-substituted products.

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Key Takeaways

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