Hydrocarbon Classification

8 MCQs9-step worked example

Source: NCERT Aldehydes, Ketones and Carboxylic AcidsPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

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 13, page 2).

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.

Practice 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:

MCQ 2Easy RecallPractice

Which of the following is an alicyclic hydrocarbon?

MCQ 3Easy RecallPractice

A hydrocarbon with molecular formula C₅H₈ and no ring can be classified as:

MCQ 4Direct ApplicationPractice

The degree of unsaturation (index of hydrogen deficiency) in cyclohexene is:

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:

MCQ 6Direct ApplicationPractice

Which of the following molecular formulae corresponds to an aromatic hydrocarbon?

MCQ 7Concept TrapPractice

Cyclooctatetraene (COT, C₈H₈) has four alternating double bonds in an eight-membered ring. It is classified as:

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:

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: open-chain terminal alkyne with one additional degree of unsaturation — pent-1-en-4-yne or similar; DoU = 3, terminal alkyne confirmed by Cu₂Cl₂.) ---

Before solving, remember these

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, p. 2

Formulas

Markovnikov's rule (and anti-Markovnikov)

H^{+} to C with more H

Without peroxide: ionic mechanism — H goes to carbon with MORE hydrogens (carbocation stability rule). With peroxide (HBr only, Kharasch): radical mechanism — anti-Markovnikov.

Symbol	Quantity	SI Unit
H,X	added atoms	-

Valid when

Asymmetric alkene
H-X with X = Cl, Br, I
Without peroxide for Markovnikov

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Markovnikov reverses with peroxide

Category: Organic Reaction Conditions

HBr addition to alkene: WITHOUT peroxide → Markovnikov (H to C with more H). WITH peroxide (Kharasch effect) → anti-Markovnikov (radical mechanism). Specific to HBr only — not HCl, HI.

When it triggers

Question gives HX addition to alkene with explicit peroxide condition or hints (e.g. ROOR, light).

How to avoid

Without peroxide: ionic mechanism, carbocation stability → Markovnikov. With peroxide: radical mechanism, radical stability → anti-Markovnikov. Effect ONLY for HBr (HCl too strong, HI too weak).

Trap

Solvent changes product distribution

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.

Mistake

Confuses ortho/para directors with meta directors.

Root cause: concept gap

Correction

o,p-directors (activators except halogens): -OH, -OR, -NH₂, -NHR, alkyl. m-directors (deactivators): -NO₂, -CN, -COOH, -CHO. Halogens: o,p-directors but DEACTIVATORS.

Mistake