Key Fact
Classification of hydrocarbons
Saturated (alkanes, single C-C). Unsaturated (alkenes C=C; alkynes C≡C). Aromatic (cyclic with delocalized π electrons obeying Hückel's 4n+2 rule).
-- NCERT Class 11 Chemistry, Ch. 9, p. 296Hydrocarbon Classification
8 MCQs9-step worked example
Source: NCERT HydrocarbonsPYQ coverage: NEET 2020, 2023, 2024Official key: NTA-verifiedLast updated: 24 Sep 2026
Hydrocarbon Classification, explained for NEET
The classification of hydrocarbons is the structural foundation of organic chemistry in NEET. The high-frequency trap here: confusing the criteria that separate one class from another — particularly misclassifying cycloalkenes as aromatic, or forgetting that aromaticity requires more than just a ring.
Hydrocarbons are compounds of carbon and hydrogen only. The primary split is between open-chain (acyclic/aliphatic) and closed-chain (cyclic) hydrocarbons (NCERT Class 11 Chemistry, Chapter 9, page 295).
Open-chain hydrocarbons subdivide by bond type:
- Saturated (alkanes): all C–C single bonds; general formula CₙH₂ₙ₊₂
- Unsaturated: contain at least one C=C (alkenes, CₙH₂ₙ) or C≡C (alkynes, CₙH₂ₙ₋₂)
Cyclic hydrocarbons subdivide into:
- Alicyclic: cyclic but NOT aromatic (cyclopropane, cyclohexene)
- Aromatic: must satisfy Hückel's rule (4n+2 π electrons in a planar, conjugated ring) — benzene is the parent compound
The trap that costs marks: a cyclic compound with alternating double bonds is NOT automatically aromatic. Cyclooctatetraene (8 π electrons, non-planar) is NOT aromatic despite appearing conjugated. The solvent/conditions trap also appears at this classification level — the same hydrocarbon behaves differently in polar protic vs polar aprotic media, affecting whether substitution or elimination dominates when the hydrocarbon acts as substrate.
Watch-out for NEET: Classification questions test whether you can assign a given structure to the correct family using general formula + bond analysis + aromaticity check. Don't rush — count carbons, count hydrogens, check the formula, then verify aromaticity criteria separately.
Can you answer these Hydrocarbon Classification MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The general formula of acyclic alkenes is:
Show answer and why every option is right or wrong
Answer: C. Acyclic alkenes have one degree of unsaturation (one C=C double bond), giving the general formula CₙH₂ₙ. This is directly stated in NCERT Class 11 Chemistry, Chapter 9.
Why A is wrong: A (CₙH₂ₙ₊₂) is the general formula for saturated acyclic hydrocarbons (alkanes), not alkenes. The presence of a double bond reduces the hydrogen count by 2.
Why B is wrong: B (CₙH₂ₙ₋₂) is the general formula for alkynes (one triple bond) or alkadienes (two double bonds), not alkenes with a single double bond.
Why D is wrong: D (CₙH₂ₙ₋₆) corresponds to aromatic hydrocarbons (benzene series), not acyclic alkenes.
MCQ 2Easy RecallPractice
Which of the following is an alicyclic hydrocarbon?
Show answer and why every option is right or wrong
Answer: A. Cyclopentane is a cyclic, non-aromatic hydrocarbon — it has a ring but no (4n+2) π electron system. It is therefore alicyclic by definition (NCERT Class 11 Chemistry, Chapter 9, page 295).
Why B is wrong: B is wrong because benzene is aromatic (6 π electrons, planar, fully conjugated), not alicyclic.
Why C is wrong: C is wrong because ethene (CH₂=CH₂) is an open-chain (acyclic) hydrocarbon, not cyclic at all.
Why D is wrong: D is wrong because naphthalene is a polycyclic aromatic hydrocarbon with 10 π electrons satisfying aromaticity.
MCQ 3Easy RecallPractice
A hydrocarbon with molecular formula C₅H₈ and no ring can be classified as:
Show answer and why every option is right or wrong
Answer: B. For an acyclic hydrocarbon with formula CₙH₂ₙ₋₂: n=5 gives C₅H₈. This matches the alkyne (or alkadiene) general formula. Since the question says no ring, it is an acyclic alkyne (NCERT Class 11 Chemistry, Chapter 9).
Why A is wrong: A is wrong because alkanes follow CₙH₂ₙ₊₂; for n=5 that would be C₅H₁₂, not C₅H₈.
Why C is wrong: C is wrong because alkenes follow CₙH₂ₙ; for n=5 that would be C₅H₁₀, not C₅H₈.
Why D is wrong: D is wrong because the question specifies no ring; aromatic hydrocarbons require a cyclic conjugated system.
MCQ 4Direct ApplicationPractice
The degree of unsaturation (index of hydrogen deficiency) in cyclohexene is:
Show answer and why every option is right or wrong
Answer: D. Cyclohexene (C₆H₁₀) — using the formula: DoU = (2×6 + 2 − 10)/2 = 4/2 = 2. One degree from the ring, one from the double bond. Total = 2.
Why A is wrong: A (DoU = 1) counts only the double bond but ignores that forming a ring also contributes one degree of unsaturation.
Why B is wrong: B (DoU = 0) would imply a saturated open-chain compound (CₙH₂ₙ₊₂), but cyclohexene has both a ring and a double bond.
Why C is wrong: C (DoU = 3) overcounts — cyclohexene has only one ring and one double bond, not three elements of unsaturation.
MCQ 5Direct ApplicationPractice
A compound has molecular formula C₄H₆. It decolourises bromine water and gives a precipitate with ammoniacal silver nitrate. The compound is classified as:
Show answer and why every option is right or wrong
Answer: D. C₄H₆ has DoU = 2. Decolourising Br₂ water confirms unsaturation. Precipitate with ammoniacal AgNO₃ (silver acetylide test) is specific to terminal alkynes. But-1-yne (CH≡C–CH₂–CH₃) is a terminal alkyne fitting both tests (NCERT Class 11 Chemistry, Chapter 9).
Why A is wrong: A is wrong because cyclobutene would decolourise Br₂ water but would NOT give a silver acetylide precipitate — that test is specific to terminal C≡C–H, which cyclobutene lacks.
Why B is wrong: B is wrong because but-1,3-diene has no triple bond and no terminal acetylenic hydrogen; it cannot form a silver acetylide precipitate.
Why C is wrong: C is wrong because but-2-yne (CH₃–C≡C–CH₃) is an internal alkyne with no terminal hydrogen; it does NOT give a positive silver acetylide test.
MCQ 6Direct ApplicationPractice
Which of the following molecular formulae corresponds to an aromatic hydrocarbon?
Show answer and why every option is right or wrong
Answer: A. C₆H₆ gives DoU = (2×6 + 2 − 6)/2 = 4. Benzene (the simplest aromatic hydrocarbon) has formula C₆H₆ with 3 double bonds + 1 ring = 4 degrees of unsaturation, consistent with aromaticity.
Why B is wrong: B (C₆H₁₂) has DoU = 1, corresponding to cyclohexane (one ring, no double bonds) — this is alicyclic, not aromatic.
Why C is wrong: C (C₆H₁₀) has DoU = 2, corresponding to cyclohexene (one ring + one double bond) — insufficient unsaturation for aromaticity.
Why D is wrong: D (C₆H₁₄) has DoU = 0, corresponding to hexane — a fully saturated acyclic alkane with no possibility of aromaticity.
MCQ 7Concept TrapPractice
Cyclooctatetraene (COT, C₈H₈) has four alternating double bonds in an eight-membered ring. It is classified as:
Show answer and why every option is right or wrong
Answer: B. COT has 8 π electrons. For aromaticity: 4n+2 rule requires 2, 6, 10, 14… π electrons. 8 ≠ 4n+2 for any integer n. Additionally, COT adopts a tub-shaped (non-planar) geometry to avoid anti-aromaticity. Since it is non-planar, it is neither aromatic nor anti-aromatic — it is non-aromatic (alicyclic). (Trap: assuming all conjugated cyclic systems are aromatic.)
Why A is wrong: A is wrong because aromaticity requires 4n+2 π electrons in a planar ring. COT has 8 π electrons (= 4×2, which is 4n not 4n+2) and is non-planar — it fails both criteria.
Why C is wrong: C is wrong because anti-aromaticity requires planarity with 4n π electrons. COT avoids anti-aromaticity by adopting a non-planar tub shape — it is non-aromatic, not anti-aromatic.
Why D is wrong: D is wrong because 'aliphatic alkene' implies an open-chain compound. COT is cyclic — it is correctly termed alicyclic (non-aromatic cyclic).
MCQ 8CalculationPractice
A hydrocarbon X has molecular formula C₇H₈. It does NOT decolourise bromine water in the dark but undergoes nitration to give ortho, meta, and para isomers. X is classified as:
Show answer and why every option is right or wrong
Answer: C. Step 1: C₇H₈ has DoU = (2×7 + 2 − 8)/2 = 4. Step 2: Does not decolourise Br₂ water → no reactive C=C or C≡C exposed to electrophilic addition (aromatic C=C requires catalyst, not simple Br₂ water). Step 3: Undergoes nitration giving positional isomers → electrophilic aromatic substitution, confirming an aromatic ring. C₇H₈ with DoU = 4 is toluene (methylbenzene) — an aromatic hydrocarbon (NCERT Class 11 Chemistry, Chapter 9).
Why A is wrong: A is wrong because an alicyclic unsaturated hydrocarbon (e.g. cycloheptene) WOULD decolourise Br₂ water due to its reactive C=C double bond. The compound in question does not react with Br₂ water.
Why B is wrong: B is wrong because an open-chain alkyne with C₇H₈ doesn't exist (alkynes are CₙH₂ₙ₋₂; for n=7 that gives C₇H₁₂). Also, alkynes decolourise Br₂ water.
Why D is wrong: D is wrong because a cyclic non-aromatic diene would decolourise Br₂ water (localised double bonds react with Br₂ by addition). Also, non-aromatic dienes don't undergo electrophilic aromatic substitution.
Free NEET study resources
Get a structured 30-day study plan and a complete formula booklet — delivered to your inbox instantly.
How do you solve a Hydrocarbon Classification question? A worked example
- 1
Given
A hydrocarbon has molecular formula C₆H₆. It does not decolourise acidified KMnO₄ under mild conditions. On catalytic hydrogenation (H₂/Ni, high temperature and pressure), it absorbs 3 moles of H₂ per mole of compound.
- 2
Required
Classify the hydrocarbon and identify it.
- 3
Concept
Classification uses: (1) molecular formula to calculate degree of unsaturation, (2) chemical tests to distinguish saturated/unsaturated/aromatic behaviour, (3) hydrogenation data to confirm the number of π bonds.
- 4
Formula
Degree of Unsaturation (DoU) = (2C + 2 − H) / 2, where C = number of carbons, H = number of hydrogens.
- 5
Substitution
DoU = (2 × 6 + 2 − 6) / 2 = (14 − 6) / 2 = 8 / 2 = 4
- 6
Calculation
DoU = 4. This means 4 degrees of unsaturation. Absorbs 3 mol H₂ → 3 C=C equivalents reduced. The fourth degree of unsaturation is a ring. Does NOT decolourise KMnO₄ → resistant to mild oxidation (characteristic of aromatic C=C, not isolated alkene C=C).
Note on exact values: the integers (6 carbons, 6 hydrogens, 3 moles H₂) are exact counting numbers and do not affect any significant-figure consideration. - 7
Final answer
The compound is benzene — an aromatic hydrocarbon. Classification: cyclic, aromatic, with 3 C=C in a planar ring satisfying Hückel's rule (6 π electrons = 4(1) + 2).
- 8
Common trap
Seeing 3 double bonds and classifying as "cyclohexatriene" (non-aromatic). The key distinguishing evidence is the failure to decolourise KMnO₄ — true isolated alkenes WOULD react. Aromatic stability prevents this.
- 9
Similar NEET-style question
"A compound C₅H₆ absorbs 2 moles of H₂ and gives a positive test with ammoniacal Cu₂Cl₂. Classify the compound." (Answer: a CYCLIC terminal alkyne — for example ethynylcyclopropane. DoU = (2×5 + 2 − 6)/2 = 3, and the Cu₂Cl₂ test fixes one terminal C≡C, which accounts for 2 of them and absorbs 2 mol H₂. Since only 2 mol are absorbed, the third degree of unsaturation takes up no hydrogen at all — so it is a RING, not another π bond. An open-chain pent-1-en-4-yne has C=C plus C≡C and would absorb 3 mol, which contradicts the data given.)
---
What to remember before solving Hydrocarbon Classification questions
Where do students lose marks on Hydrocarbon Classification?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Organic Reaction Conditions
Same starting materials give different products depending on solvent. Polar protic (water, alcohols): SN1/E1 favoured. Polar aprotic (DMSO, DMF): SN2 favoured. Affects substitution vs elimination.
When it triggers
Question contrasts product when solvent is changed; or specifies solvent type.
How to avoid
Polar protic stabilises carbocation → SN1/E1 (3° preferred). Polar aprotic doesn't solvate nucleophile → strong SN2 nucleophile (1°/2° preferred). Bulky base (t-BuOK) favours E2 over SN2.
More in Hydrocarbons: 3 exam traps and mistakes · 1 formula · 2 question patterns from its other lessons.
Hydrocarbon Classification questions from past NEET papers
3 questions from NEET 2020, 2023, 2024. Answers verified against NTA official keys.
NEET 2024Revised key
1Both Statement I and Statement II are correct.
2Both Statement I and Statement II are incorrect.
3Statement I is correct but Statement II is incorrect.
4Statement I is incorrect but Statement II is correct.
NTA Answer: Option 1(revised_final)
NEET 2023
132
230
318
416
NTA Answer: Option 1(final)
NEET 2020
Which of the following alkane cannot be made in good yield by Wurtz reaction ?
1n-Butane
2n-Hexane
32,3-Dimethylbutane
4n-Heptane
NTA Answer: Option 4(final)
Sources
NCERT refs: Class 11 Chemistry Chapter 9, p.295
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Test yourself on this topic with real past-paper questions:
Practice this topic →