Hydrocarbons

Class 11

20 previous-year questions from this chapter every option and the correct answer, free · where the marks are

1. Alkanes: conformations and reactions

Exam focus: Ethane's anti-staggered conformation is the most stable (zero torsional strain) and the fully eclipsed one the least — tested via potential-energy-vs-dihedral-angle graphs, not just the word "staggered." In free-radical halogenation, bromination is far more selective for 3° C–H than chlorination — "both give the same product ratio" is the trap.

Alkanes: Conformations and Reactions

Next Toppers - 11th Science

Dedicated alkanes chapter revision; not a strongly-preferred channel, but content is on-topic and thorough.

ChemistryNEET 2025show ▾

How many products (including stereoisomers) are expected from the monochlorination of the compound shown? Question 75 as printed in the NEET 2025 booklet

Question 75 as printed in the NEET 2025 booklet

A.2
B.3
C.5
D.6✓

Solution

Monochlorination of 2-methylbutane at each distinct hydrogen: C-1 (the two equivalent methyls) gives 1-chloro-2-methylbutane, chiral at C-2 (2 enantiomers); C-2 gives 2-chloro-2-methylbutane (1); C-3 gives 2-chloro-3-methylbutane, chiral (2); C-4 gives 1-chloro-3-methylbutane (1). Total $2 + 1 + 2 + 1 = 6$ .

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ChemistryNEET 2024show ▾

Given below are two statements: Statement I: The boiling points of the three isomeric pentanes follow the order n-pentane > isopentane > neopentane. Statement II: When branching increases, the molecule attains the shape of a sphere. This results in a smaller surface area for contact, due to which the intermolecular forces between the spherical molecules are weak, thereby lowering the boiling point. In the light of the above statements, choose the most appropriate answer from the options given below:

A.Both Statement I and Statement II are incorrect
B.Statement I is correct but Statement II is incorrect
C.Statement I is incorrect but Statement II is correct
D.Both Statement I and Statement II are correct✓

Solution

Branching makes a molecule more compact and spherical, reducing surface contact and hence the van der Waals forces, so boiling point falls from n-pentane to isopentane to neopentane. Both statements are correct and II explains I.

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ChemistryNEET 2023show ▾

The weight (g) of two moles of the organic compound obtained by heating sodium ethanoate with sodium hydroxide in the presence of calcium oxide is:

A.16
B.32✓
C.30
D.18

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ChemistryNEET 2021show ▾

C H_{3} C H_{2} CO O^{-} N a^{+} ? heat C H_{3} C H_{3} + N a_{2} C O_{3}

Consider the above reaction and identify the missing reagent/chemical.

A. $B_{2} H_{6}$
B.Red phosphorus
C.CaO✓
D.DIBAL-H

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ChemistryNEET 2021show ▾

The dihedral angle of the least stable conformer of ethane is:

A.120°
B.180°
C.60°
D.0°✓

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ChemistryNEET 2020show ▾

Which of the following alkanes cannot be made in good yield by the Wurtz reaction?

A.n-Hexane
B.2,3-Dimethylbutane
C.n-Heptane✓
D.n-Butane

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2. Alkenes and alkynes: addition reactions, Markovnikov

Exam focus: The peroxide (anti-Markovnikov) effect works only for HBr — not HCl or HI — because only the bromine free-radical chain steps are both energetically favourable; with HCl and HI, normal Markovnikov addition holds regardless of peroxide. That HBr-only exclusivity, not just "peroxide flips the rule," is what NEET actually tests.

Hydrocarbons — Full Chapter (incl. Alkene/Alkyne Addition and Markovnikov's Rule)

Competition Wallah

A full-chapter one-shot covering all of Class 11 Hydrocarbons (alkanes, alkenes, alkynes, aromatics). It does include electrophilic addition to alkenes/alkynes and Markovnikov's rule with the peroxide effect, but as one segment inside a much longer video.

ChemistryNEET 2025show ▾

Which one of the following compounds (shown) does not decolourise bromine water? Question 77 as printed in the NEET 2025 booklet

Question 77 as printed in the NEET 2025 booklet

A.Structure (1)✓
B.Structure (2)
C.Structure (3)
D.Structure (4)

Solution

Bromine water is decolourised by addition to C=C (styrene) and by electrophilic bromination of activated rings (phenol, aniline give tribromo products). Cyclohexane (1) is saturated and unactivated, so it does not react.

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ChemistryNEET 2024show ▾

Match List I (reactions, shown as structures) with List II (reagents/conditions): I. (reagent shown in figure) · II.

Cr O_{3}

· III.

KMn O_{4} /KOH

,

Δ \cdot

IV. (i)

O_{3}

(ii)

Zn

–

H_{2} O

Choose the correct answer from the options given below: Question 59 as printed in the NEET 2024 booklet

Question 59 as printed in the NEET 2024 booklet

A.A-III, B-I, C-II, D-IV
B.A-IV, B-I, C-II, D-III✓
C.A-I, B-IV, C-II, D-III
D.A-IV, B-I, C-III, D-II

Solution

A: cleaving a C=C to two ketones is ozonolysis, (i) $O_{3}$ (ii) Zn– $H_{2} O$ (IV). B: benzophenone from benzene is Friedel–Crafts acylation with benzoyl chloride/ $AlC l_{3}$ (I). C: cyclohexanol to cyclohexanone uses $Cr O_{3}$ (II). D: ethylbenzene to benzoate needs $KMn O_{4} /KOH$ , $Δ ($ III). A-IV, B-I, C-II, D-III.

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ChemistryNEET 2022show ▾

Compound X on reaction with

O_{3}

followed by

Zn/ H_{2} O

gives formaldehyde and 2-methylpropanal as products. The compound X is:

A.Pent-2-ene
B.3-Methylbut-1-ene✓
C.2-Methylbut-1-ene
D.2-Methylbut-2-ene

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ChemistryNEET 2021show ▾

The major product of the following chemical reaction is:

(C H_{3})_{2} CH - CH = C H_{2} + HBr (C_{6} H_{5} CO)_{2} O_{2} ?

A. $(C H_{3})_{2} CH - C H_{2} - C H_{2} - Br$ ✓
B. $(C H_{3})_{2} CH - C H_{2} - C H_{2} - O - CO C_{6} H_{5}$
C. $(C H_{3})_{2} CH - CH (Br) - C H_{3}$
D. $(C H_{3})_{2} C (Br) - C H_{2} - C H_{3}$

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ChemistryNEET 2020show ▾

An alkene on ozonolysis gives methanal as one of the products. Its structure is: Question 133 as printed in the NEET 2020 booklet

Question 133 as printed in the NEET 2020 booklet

A.1-Propylidenecyclohexane
B.3-Allylcyclohexene
C.Allylcyclohexane✓
D.3-Propylcyclohexene

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ChemistryNEET 2019show ▾

The number of sigma (

σ

) and pi (

π

) bonds in pent-2-en-4-yne (

C H_{3} - CH = CH - C \equiv CH

) is:

A.10 $σ$ bonds and 3 $π$ bonds✓
B.8 $σ$ bonds and 5 $π$ bonds
C.11 $σ$ bonds and 2 $π$ bonds
D.13 $σ$ bonds and no $π$ bonds

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ChemistryNEET 2019show ▾

An alkene A, on ozonolysis (reaction with

O_{3}

followed by Zn/

H_{2} O

), gives propanone and ethanal in equimolar amounts (A is 2-methylbut-2-ene). Addition of HCl to A gives B as the major product (Markovnikov addition). The structure of B is:

A. $Cl - C H_{2} - C H_{2} - CH (C H_{3})_{2}$
B. $H_{3} C - C H_{2} - CH (C H_{2} Cl) - C H_{3}$
C. $H_{3} C - C H_{2} - C (Cl) (C H_{3}) - C H_{3}$ ✓
D. $H_{3} C - CH (Cl) - CH (C H_{3})_{2}$

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ChemistryNEET 2019show ▾

The most suitable reagent for converting but-2-yne (

H_{3} C - C \equiv C - C H_{3}

) into cis-2-butene is:

A.Na / liquid $N H_{3}$
B. $H_{2}$ , Pd/C, quinoline (Lindlar's catalyst)✓
C.Zn/HCl
D. $H g^{2 +} / H^{+}$ , $H_{2} O$

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3. Aromatic hydrocarbons: electrophilic substitution, directing effects

Exam focus: Halogens are the one substituent that's ortho/para-directing yet still deactivating — their lone pair donates into the ring by resonance (directing) while their electronegativity withdraws inductively (deactivating), a combination no other common group shows. Friedel–Crafts reactions also simply fail on an already strongly deactivated ring, such as nitrobenzene, no matter how much catalyst is used.

Hydrocarbons — Full Chapter (incl. Aromatic Hydrocarbons and Electrophilic Substitution)

PW Class 11 Science

Bundled inside the full Hydrocarbons chapter one-shot rather than a video dedicated solely to aromatic hydrocarbons/EAS.

ChemistryNEET 2025show ▾

Which one of the following reactions (shown in the figure) does not give benzene as the product? Question 60 as printed in the NEET 2025 booklet

Question 60 as printed in the NEET 2025 booklet

A.Reaction (1)
B.Reaction (2)
C.Reaction (3)
D.Reaction (4)✓

Solution

Sodium benzoate with soda lime decarboxylates to benzene; n-hexane over $M o_{2} O_{3}$ at 773 K aromatises to benzene; ethyne passed through a red-hot iron tube trimerises to benzene. Benzenediazonium chloride warmed with water gives phenol, not benzene — reaction (4).

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ChemistryNEET 2023show ▾

Consider the compounds/species shown in the figure. The number of compounds/species which obey Hückel's rule is ______. Question 89 as printed in the NEET 2023 booklet

Question 89 as printed in the NEET 2023 booklet

A.4✓
B.6
C.2
D.5

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ChemistryNEET 2023show ▾

Identify the product in the reaction shown in the figure. Question 59 as printed in the NEET 2023 booklet

Question 59 as printed in the NEET 2023 booklet

A.Structure (1)
B.Structure (2)✓
C.Structure (3)
D.Structure (4)

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ChemistryNEET 2022show ▾

Which compound amongst the following is not an aromatic compound? Question 54 as printed in the NEET 2022 booklet

Question 54 as printed in the NEET 2022 booklet

A.Structure (1)✓
B.Structure (2)
C.Structure (3)
D.Structure (4)

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ChemistryNEET 2020show ▾

Identify compound X in the following sequence of reactions. Question 95 as printed in the NEET 2020 booklet

Question 95 as printed in the NEET 2020 booklet

A.Chlorobenzene
B.Benzyl chloride ( $C_{6} H_{5} C H_{2} Cl$ )
C.Benzal chloride ( $C_{6} H_{5} CHC l_{2}$ )✓
D.Benzotrichloride ( $C_{6} H_{5} CC l_{3}$ )

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ChemistryNEET 2019show ▾

Among the following, the reaction that proceeds through electrophilic substitution is:

A.Benzenediazonium chloride + $C u_{2} C l_{2}$ $\to$ chlorobenzene + $N_{2}$ (Sandmeyer reaction)
B.Benzene + $C l_{2}$ , $AlC l_{3}$ $\to$ chlorobenzene + HCl (Friedel–Crafts halogenation)✓
C.Benzene + $C l_{2}$ , UV light $\to$ benzene hexachloride (free-radical addition)
D.Benzyl alcohol + HCl, heat $\to$ benzyl chloride + $H_{2} O$

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