Hydrocarbons NEET PYQs (2005–2026) with Solutions

10 solved Hydrocarbons PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026HardStereochemistry of addition to alkenes

   Given below are two statements : Statement I : trans-But-2-ene upon treatment with $\mathrm{Br_2}$ in $\mathrm{CCl_4}$ gives the following product. Statement II : cis-But-2-ene upon treatment with alkaline $\mathrm{KMnO_4}$ gives the following product. In the light of the above statements, choose the most appropriate answer from the options given below.

   1. AOption A: Both Statement I and Statement II are correct
   2. BOption B: Both Statement I and Statement II are incorrect
   3. COption C: Statement I is correct but Statement II is incorrect
   4. DOption D: Statement I is incorrect but Statement II is correct

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   Correct answer: (D) Statement I is incorrect but Statement II is correct

   Explanation

   Statement I: $\mathrm{Br_2}$ adds to alkenes by anti addition (through a cyclic bromonium ion). Anti addition to trans-but-2-ene gives meso-2,3-dibromobutane. Anti addition to cis-but-2-ene gives the racemic ($\pm$) form. In the Newman projection shown, the front carbon has Br (up), H and Me, and the rear carbon has Br (down), H and Me. The two Br are anti but the two Me groups are gauche. If you rotate so the Me groups are anti, the Br atoms become gauche. No conformation has a centre of symmetry, so this is one enantiomer of ($\pm$)-2,3-dibromobutane, the product from the cis alkene. Statement I is incorrect. Statement II: cold dilute alkaline $\mathrm{KMnO_4}$ adds two OH groups by syn addition (via a cyclic manganate ester). Syn addition to cis-but-2-ene gives meso-butane-2,3-diol. In the Newman projection shown, rotate the rear carbon by $120^\circ$ so the two Me groups are anti. The two OH groups and the two H atoms then also become anti, so the molecule has a centre of symmetry and is meso. Statement II is correct. So the answer is option (D).

2. Question 2 (NEET 2026)

   NEET 2026MediumPartial reduction of alkynes, HBr addition

   Consider the following reaction sequences and choose the correct option. $\mathrm{L} \xleftarrow{\mathrm{Na/liq.\ NH_3}} \mathrm{Ph{-}C{\equiv}C{-}Me} \xrightarrow[\text{(Lindlar's Catalyst)}]{\mathrm{H_2,\ Pd/C}} \mathrm{K}$ $\mathrm{L} \xrightarrow{\text{HBr, benzoyl peroxide}} \mathrm{N}$ ; $\mathrm{K} \xrightarrow{\mathrm{HBr}} \mathrm{M}$

   1. AOption A: K and L are geometrical isomers
   2. BOption B: K and L are enantiomers
   3. COption C: M and N are geometrical isomers
   4. DOption D: M and N are stereoisomers

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   Correct answer: (A) K and L are geometrical isomers

   Explanation

   Na/liquid $\mathrm{NH_3}$ reduces the alkyne by anti addition, giving the trans alkene L = trans-1-phenylprop-1-ene, $\mathrm{Ph{-}CH{=}CH{-}CH_3}$ (E). $\mathrm{H_2}$ with Lindlar's catalyst gives syn addition, so K = cis-1-phenylprop-1-ene (Z). K and L have the same structure and differ only in the arrangement around the C=C, so they are geometrical (cis–trans) isomers: option (A). They are achiral, so they cannot be enantiomers (B wrong). M: HBr adds by Markovnikov's (ionic) route through the more stable benzylic carbocation, giving $\mathrm{Ph{-}CH(Br){-}CH_2{-}CH_3}$. N: with benzoyl peroxide, HBr adds by the free-radical (anti-Markovnikov) route. $\mathrm{Br^\bullet}$ adds to C-2 to give the benzylic radical, so N = $\mathrm{Ph{-}CH_2{-}CH(Br){-}CH_3}$. M and N differ in the position of Br. They are position (structural) isomers, not geometrical isomers or stereoisomers, so (C) and (D) are wrong.

3. Question 3 (NEET 2026)

   NEET 2026MediumAromatic Hydrocarbons

   The number of chlorine atoms present in the organic products X and Y of the following reactions, respectively, are:

   1. AOption A: 3 and 3
   2. BOption B: 6 and 3
   3. COption C: 6 and 6
   4. DOption D: 3 and 6

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   Correct answer: (C) 6 and 6

   Explanation

   Benzene with excess Cl₂/anhydrous AlCl₃ gives hexachlorobenzene (6 chlorine atoms). Benzene with Cl₂ under UV undergoes addition to form benzene hexachloride (BHC), containing 6 chlorine atoms. Therefore X and Y each contain 6 chlorine atoms. Hence option (C) is correct.

4. Question 4 (NEET 2025)

   NEET 2025MediumPeroxide Effect and Nucleophilic Substitution Reactions

   Predict the major product 'P' in the following sequence of reactions:

   1. AOption A: Option (1)
   2. BOption B: Option (2)
   3. COption C: Option (3)
   4. DOption D: Option (4)

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   Correct answer: (A) Option (1)

   Explanation

   In the presence of peroxide, HBr adds to the alkene via anti-Markovnikov addition. Bromine attaches to the less substituted carbon atom. Treatment with KCN substitutes Br by CN through nucleophilic substitution, forming a nitrile. Reduction with Na(Hg)/C₂H₅OH converts the nitrile group $(-CN)$ into a primary amine $(-CH₂NH₂)$. Therefore, the final major product corresponds to option (A).

5. Question 5 (NEET 2025)

   NEET 2025MediumAromatic Hydrocarbons

   Which one of the following reactions does NOT give benzene as the product?

   1. AOption A: Option (1)
   2. BOption B: Option (2)
   3. COption C: Option (3)
   4. DOption D: Option (4)

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   Correct answer: (D) Option (4)

   Explanation

   Sodium benzoate with soda lime undergoes decarboxylation to form benzene. n-Hexane undergoes aromatization in presence of catalyst to form benzene. Acetylene polymerizes in a red hot iron tube to produce benzene. However, benzene diazonium chloride with warm water forms phenol, not benzene. Therefore option D is correct.

6. Question 6 (NEET 2025)

   NEET 2025MediumAlkanes

   How many products (including stereoisomers) are expected from monochlorination of the following compound?

   1. AOption A: 2
   2. BOption B: 3
   3. COption C: 5
   4. DOption D: 6

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   Correct answer: (D) 6

   Explanation

   The compound is 2-methylbutane. Monochlorination can occur at four different types of hydrogen positions. Considering stereoisomerism, chlorination at the CH₂ carbon produces two enantiomers. Thus total products obtained are 6.

7. Question 7 (NEET 2025)

   NEET 2025EasyUnsaturation and Bromine Water Test

   Which one of the following compounds does not decolourize bromine water?

   1. AOption A: Option (1)
   2. BOption B: Option (2)
   3. COption C: Option (3)
   4. DOption D: Option (4)

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   Correct answer: (A) Option (1)

   Explanation

   Cyclohexane is a saturated hydrocarbon and does not react with bromine water under normal conditions. Phenol, styrene and aniline decolourize bromine water due to electrophilic substitution or addition reactions.

8. Question 8 (NEET 2024)

   NEET 2024MediumBoiling Point and Branching

   Given below are two statements: Statement I : The boiling point of three isomeric pentanes follows the order n-pentane > isopentane > neopentane Statement II : When branching increases, the molecule attains a shape of sphere. This results in 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:

   1. AOption A: Statement I is correct but Statement II is incorrect.
   2. BOption B: Statement I is incorrect but Statement II is correct.
   3. COption C: Both Statement I and Statement II are correct.
   4. DOption D: Both Statement I and Statement II are incorrect.

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   Correct answer: (C) Both Statement I and Statement II are correct.

   Explanation

   Among isomeric pentanes, increased branching decreases boiling point because branching reduces surface area and weakens van der Waals forces. Therefore: $$\text{n-pentane} > \text{isopentane} > \text{neopentane}$$ Statement II correctly explains this trend. Hence, option (C) is correct.

9. Question 9 (NEET 2024)

   NEET 2024HardHydroboration Oxidation and Oxidation of Alcohols

   Identify the correct reagents that would bring about the following transformation.

   1. AOption A: Option 1
   2. BOption B: Option 2
   3. COption C: Option 3
   4. DOption D: Option 4

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   Correct answer: (D) Option 4

   Explanation

   The alkene undergoes hydroboration-oxidation: $$BH_3$$ followed by $$H_2O_2/OH^-$$ to form a primary alcohol. The primary alcohol is then oxidized using PCC to form the aldehyde. Thus the correct sequence is: (i) $$BH_3$$ (ii) $$H_2O_2/OH^-$$ (iii) PCC Hence, option (D) is correct.

10. Question 10 (NEET 2024)

    NEET 2024MediumAlkenes

    For the given reaction:

    1. AOption A: Option 1
    2. BOption B: Option 2
    3. COption C: Option 3
    4. DOption D: Option 4

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    Correct answer: (D) Option 4

    Explanation

    Acidified KMnO₄ causes oxidative cleavage of alkenes. When the double-bond carbon contains hydrogen, oxidation proceeds to carboxylic acid. Thus the given alkene forms cyclohexane carboxylic acid as the major product. Hence, option (D) is correct.