Hydrocarbons NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
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Every Hydrocarbons question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Hydrocarbons previous year questions
34 questions. Tap an option to check your answer.
Q1NEET 2021
The major product of the following chemical reaction is : (CH₃)₂CH–CH=CH₂ + HBr —(C₆H₅CO)₂O₂→ ?
Not quite — the answer is A.
Benzoyl peroxide makes HBr add by a free-radical chain (peroxide or Kharasch effect), giving anti-Markovnikov addition. Br adds to the terminal CH₂ so that the more stable secondary radical forms, giving (CH₃)₂CH–CH₂–CH₂–Br. The Markovnikov product (CH₃)₂CH–CH(Br)–CH₃ would form without peroxide.
Q2NEET 2021
Dihedral angle of least stable conformer of ethane is :
Not quite — the answer is D.
The least stable conformer of ethane is the eclipsed form, where the C–H bonds on adjacent carbons line up and torsional strain is maximum; its dihedral angle is 0°. 60° corresponds to the most stable staggered form.
Q3NEET 2021
CH₃CH₂COO⁻Na⁺ —(NaOH, + ?; Heat)→ CH₃CH₃ + Na₂CO₃. Consider the above reaction and identify the missing reagent/chemical.
Not quite — the answer is C.
Heating a sodium carboxylate with soda lime (NaOH + CaO) removes CO₂ as Na₂CO₃ and gives an alkane with one carbon fewer: sodium propanoate → ethane. The missing reagent is CaO; red phosphorus is used with HI or in HVZ halogenation.
Q4NEET 2020 · Q133
An alkene on ozonolysis gives methanal as one of the product. Its structure is :
Not quite — the answer is C.
Methanal (HCHO) forms only from a terminal =CH₂ group. Allylcyclohexane, C₆H₁₁–CH₂–CH=CH₂, has a terminal double bond and gives HCHO plus cyclohexylethanal. The prop-1-enyl compound gives ethanal, and the ring alkenes open to dialdehydes with no methanal.
Q5NEET 2020 · Q134
Which of the following alkane cannot be made in good yield by Wurtz reaction ?
Not quite — the answer is C.
Wurtz coupling works well for symmetrical alkanes made from one alkyl halide: n-butane (2 ethyl), n-hexane (2 n-propyl) and 2,3-dimethylbutane (2 isopropyl). n-Heptane has an odd number of carbons and needs two different halides (C₃ + C₄), which gives a mixture of hexane, heptane and octane, so the yield is poor.
Q6NEET 2019 · Q55
The most suitable reagent for the following conversion, is: H₃C–C≡C–CH₃ → cis-2-butene
Not quite — the answer is B.
Lindlar's catalyst (H₂ over partially poisoned Pd with quinoline) adds hydrogen syn to the alkyne, giving the cis-alkene. Na in liquid NH₃ gives anti addition and hence trans-2-butene.
Q7NEET 2019 · Q74
An alkene "A" on reaction with O₃ and Zn–H₂O gives propanone and ethanal in equimolar ratio. Addition of HCl to alkene "A" gives "B" as the major product. The structure of product "B" is:
Not quite — the answer is C.
Joining propanone and ethanal at their carbonyl carbons gives A = (CH₃)₂C=CH–CH₃ (2-methylbut-2-ene). Markovnikov addition of HCl puts Cl on the more substituted carbon via the tertiary carbocation, giving 2-chloro-2-methylbutane. The secondary chloride would come from anti-Markovnikov orientation.
Q8NEET 2018 · Q62
Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to gaseous hydrocarbon containing less than four carbon atoms. (A) is
Not quite — the answer is D.
CH₄ + Br₂ → CH₃Br; Wurtz coupling 2CH₃Br + 2Na → C₂H₆, a gas with fewer than four carbons. Ethane would give butane (four carbons), and ethene/ethyne add bromine rather than substitute.
Q9NEET 2017 · Q2
Predict the correct intermediate and product in the following reaction: H₃C–C≡CH —(H₂O, H₂SO₄, HgSO₄)→ intermediate (A) → product (B)
Not quite — the answer is D.
Mercuric-ion catalysed hydration of propyne adds water by Markovnikov's rule, giving the enol CH₃–C(OH)=CH₂. The enol is unstable and tautomerises to the ketone, acetone (CH₃COCH₃). No sulphate-containing product is formed.
Q10NEET 2017 · Q17
Which one is the correct order of acidity?
Not quite — the answer is B.
Acidity of C–H rises with s-character of the carbon: sp (50%) > sp² (33%) > sp³ (25%), so alkynes > alkenes > alkanes. Between the alkynes, the electron-donating methyl group makes propyne less acidic than ethyne, which is why ethene cannot be placed above propyne.
Q11NEET 2017 (Gujarati paper) · Q7
Amongst the following compounds the one which is most easily sulphonated is:
Not quite — the answer is C.
Sulphonation is electrophilic substitution, so the ring with the most electron-donating group reacts fastest. The methyl group of toluene activates the ring, while –NO₂ and –Cl deactivate it compared with benzene.
Q12NEET 2017 (Gujarati paper) · Q44
The correct statement regarding ethane conformation is:
Not quite — the answer is C.
Ethane has only a C–C σ bond, and a σ bond is cylindrically symmetric about the bond axis, so its overlap does not change as the groups rotate; this free rotation produces the staggered and eclipsed conformations. Ethane has no π bond, so the statements saying rotation is blocked by a π bond are false.
Q13NEET 2016 Phase 2 · Q80
Which of the following compounds shall not produce propene by reaction with HBr followed by elimination or direct only elimination reaction?
Not quite — the answer is C.
Cyclopropane adds HBr to give 1-bromopropane, which eliminates HBr to give propene; propan-1-ol dehydrates and 1-bromopropane dehydrohalogenates directly to propene. Ketene (H₂C=C=O) has only two carbons, so it cannot give a three-carbon alkene.
Q14NEET 2016 Phase 2 · Q84
In the given reaction, benzene + cyclohexene —(HF, 0 °C)→ P, the product P is
Not quite — the answer is C.
HF protonates cyclohexene to give the cyclohexyl carbocation, which alkylates benzene (Friedel–Crafts type): the product is cyclohexylbenzene. Fluoride is not retained in the product, and the low temperature and 1:1 reactants do not give a diphenyl product.
Q15NEET 2016 Phase 1 · Q47
In the reaction H–C≡CH --(1) NaNH₂/liq. NH₃, (2) CH₃CH₂Br--> X --(1) NaNH₂/liq. NH₃, (2) CH₃CH₂Br--> Y X and Y are:
Not quite — the answer is A.
NaNH₂ removes the acidic terminal H of ethyne, and the acetylide is ethylated: HC≡C–CH₂CH₃ (1-butyne). Repeating on the remaining terminal H puts a second ethyl group on the other carbon: CH₃CH₂C≡CCH₂CH₃, 3-hexyne. 2-Butyne cannot form because the new group always attaches to the former terminal carbon.
Q16NEET 2016 Phase 1 · Q69
Consider the nitration of benzene using mixed conc. H₂SO₄ and HNO₃. If a large amount of KHSO₄ is added to the mixture, the rate of nitration will be:
Not quite — the answer is B.
The electrophile NO₂⁺ forms when H₂SO₄ protonates HNO₃: HNO₃ + 2H₂SO₄ ⇌ NO₂⁺ + H₃O⁺ + 2HSO₄⁻. Adding a lot of HSO₄⁻ (from KHSO₄) suppresses the ionisation of H₂SO₄ by the common ion effect, so less NO₂⁺ forms and nitration slows down.
Q17NEET 2016 Phase 1 · Q81
The correct statement regarding the comparison of staggered and eclipsed conformations of ethane, is:
Not quite — the answer is D.
In the staggered form the C–H bonds on adjacent carbons are as far apart as possible, so there is minimum repulsion and no torsional strain, making it the more stable conformation. The eclipsed form has the bonds aligned, which creates torsional strain and makes it less stable.
Q18AIPMT 2015
A single compound of the structure OHC–CH₂–CH(CH₃)–CH₂–CO–CH₃ is obtainable from ozonolysis of which of the following cyclic compounds?
Not quite — the answer is B.
In 1,4-dimethylcyclopentene the ring C1(CH₃)=C2 bond is cleaved: C1 becomes a methyl ketone and C2 an aldehyde, still joined through C5–C4(CH₃)–C3. This gives the single product CH₃CO–CH₂–CH(CH₃)–CH₂–CHO. 1,2-Dimethylcyclopentene would give a diketone.
Q19AIPMT 2015
The reaction of C₆H₅CH=CHCH₃ with HBr produces:
Not quite — the answer is B.
H⁺ adds to the carbon away from the ring so that the positive charge sits on the benzylic carbon, where it is stabilised by resonance with the ring. Br⁻ then attacks there, giving C₆H₅CH(Br)CH₂CH₃. Ring bromination does not occur with HBr.
Q20NEET 2013
Nitrobenzene on reaction with conc. HNO₃/H₂SO₄ at 80–100°C forms which one of the following products?
Not quite — the answer is B.
–NO₂ is a strong deactivating, meta-directing group, so the second nitro group enters the meta position, giving 1,3-dinitrobenzene. Ortho/para dinitro products would need an ortho/para-directing group.
Q21NEET 2013
Some meta-directing substituents in aromatic substitution are given. Which one is most deactivating?
Not quite — the answer is D.
All four groups withdraw electrons and are meta directing, but –NO₂ deactivates the ring most strongly through its large −I and −R effects. –C≡N, –SO₃H and –COOH are deactivating but less so.
Q22NEET 2013
Which of the following compounds will not undergo Friedel-Crafts reaction easily?
Not quite — the answer is C.
The –NO₂ group strongly deactivates the ring and also complexes with AlCl₃, so nitrobenzene does not undergo Friedel–Crafts reactions easily. Cumene, xylene and toluene have activating alkyl groups and react readily.
Q23AIPMT 2012 Mains
Which of the following reagents will be able to distinguish between 1-butyne and 2-butyne?
Not quite — the answer is A.
1-Butyne is a terminal alkyne with an acidic ≡C–H and reacts with NaNH₂ to form a sodium acetylide; 2-butyne has no such hydrogen. Br₂ and HCl add to both alkynes, so they cannot tell them apart.
Q24AIPMT 2012 Prelims
Among the following compounds the one that is most reactive towards electrophilic nitration is:
Not quite — the answer is D.
The –CH₃ group of toluene donates electron density to the ring (+I and hyperconjugation), activating it for electrophilic nitration, so toluene reacts faster than benzene. –COOH and –NO₂ withdraw electrons and deactivate the ring.
Q25AIPMT 2011 Prelims
Considering the state of hybridization of carbon atoms, find out the molecule among the following which is linear?
Not quite — the answer is C.
In CH₃–C≡C–CH₃ both triple-bonded carbons are sp hybridised (180°), so all four carbons lie on a straight line. Butane and but-2-ene contain sp³ or sp² carbons that bend the chain, and pent-1-en-4-yne has sp³ and sp² centres.
Q26AIPMT 2011 Prelims
In the following reactions, (a) CH₃–CH(CH₃)–CH(OH)–CH₃ —(H⁺/heat)→ A (major product) + B (minor product) (b) A —(HBr, dark, in absence of peroxide)→ C (major product) + D (minor product) The major products (A) and (C) are respectively
Not quite — the answer is C.
Protonation and loss of water from 3-methylbutan-2-ol gives a secondary carbocation, which undergoes a 1,2-hydride shift to the tertiary cation; loss of H⁺ gives the more substituted alkene 2-methylbut-2-ene (Saytzeff). HBr adds by Markovnikov's rule without peroxide, giving 2-bromo-2-methylbutane. The 2-methylbut-1-ene or anti-Markovnikov products are the minor/trap options.
Q27AIPMT 2010 Prelims · Q84
The reaction of toluene with Cl₂ in presence of FeCl₃ gives X and reaction in presence of light gives Y. Thus, X and Y are
Not quite — the answer is D.
With FeCl₃ (Lewis acid) chlorine substitutes the ring at the o- and p-positions activated by CH₃, giving o- and p-chlorotoluene (X). In light, free-radical substitution occurs on the side chain; with excess Cl₂ this goes to C₆H₅CCl₃, trichloromethylbenzene (Y). Benzyl chloride is a side-chain product, so it cannot be X.
Q28AIPMT 2010 Prelims · Q85
Liquid hydrocarbons can be converted to a mixture of gaseous hydrocarbons by
Not quite — the answer is C.
Cracking (pyrolysis) breaks large hydrocarbon molecules into smaller ones at high temperature, turning liquid hydrocarbons into gaseous ones. Distillation under reduced pressure only separates components without changing them.
Q29AIPMT 2009 Prelims · Q172
Benzene reacts with CH₃Cl in the presence of anhydrous AlCl₃ to form:
Not quite — the answer is D.
This is Friedel–Crafts alkylation: AlCl₃ generates CH₃⁺ which substitutes on the ring, giving C₆H₅CH₃ (toluene). Chlorobenzene would need Cl₂, benzyl chloride needs side-chain chlorination of toluene, and xylene needs a second methylation.
Q30AIPMT 2009 Prelims · Q195
Nitrobenzene can be prepared from benzene by using a mixture of conc. HNO₃ and conc. H₂SO₄. In the mixture, nitric acid acts as a/an:
Not quite — the answer is B.
H₂SO₄ is the stronger acid and protonates HNO₃: HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻. Accepting a proton means nitric acid acts as a base here, producing the nitronium electrophile.
Q31AIPMT 2007 Prelims · Q64
Predict the product C obtained in the following reaction of butyne-1. CH₃CH₂–C≡CH + HCl → B —HI→ C
Not quite — the answer is C.
HCl adds by Markovnikov's rule to give CH₃CH₂C(Cl)=CH₂ as product B. HI then also adds Markovnikov, with I going to the carbon already carrying Cl (stabilised carbocation), giving CH₃CH₂C(I)(Cl)CH₃. Putting I and Cl on the terminal carbon would be anti-Markovnikov.
Q32AIPMT 2007 Prelims · Q80
Which of the compounds with molecular formula C₅H₁₀ yields acetone on ozonolysis?
Not quite — the answer is D.
Acetone needs a (CH₃)₂C= unit. 2-Methyl-2-butene (CH₃)₂C=CHCH₃ gives acetone + acetaldehyde on ozonolysis. 2-Methyl-1-butene gives butanone + formaldehyde, and cyclopentane has no double bond.
Q33AIPMT 2007 Prelims · Q93
The order of decreasing reactivity towards an electrophilic reagent, for the following would be (i) benzene (ii) toluene (iii) chlorobenzene (iv) phenol
Not quite — the answer is C.
–OH strongly activates the ring by resonance, –CH₃ activates weakly by hyperconjugation, and –Cl deactivates (–I effect dominates). So phenol > toluene > benzene > chlorobenzene.
Q34AIPMT 2005 Prelims · Q52
Products of the following reaction : CH₃C≡C·CH₂CH₃ —[(1) O₃, (2) Hydrolysis]→ ...... are :
Not quite — the answer is C.
Ozonolysis of an alkyne followed by hydrolysis cleaves the C≡C and gives carboxylic acids. Pent-2-yne CH₃C≡CCH₂CH₃ gives CH₃COOH + CH₃CH₂COOH. Aldehydes would form from an alkene, not an alkyne.
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