Alkane Conformations

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 conformation question NEET loves: which ethane conformation is most stable, and why? The topic sounds simple — rotation around a C–C single bond — but aspirants routinely confuse the two standard projection methods and misidentify the torsion-angle–energy relationship.

What conformations are. When two carbon atoms are joined by a single (sigma) bond, the groups attached to them can rotate freely around that bond axis. Each distinct spatial arrangement produced by such rotation is called a conformation (or conformer, or rotamer). Conformations are NOT different compounds — they interconvert rapidly at room temperature and cannot be isolated under normal conditions.

Two projection methods. NCERT Class 11 Chemistry Chapter 10 (Hydrocarbons) introduces two ways to visualise conformations:

Sawhorse projection: a perspective drawing showing the C–C bond as a diagonal line with all six substituents visible. Front carbon is lower-left; back carbon is upper-right.
Newman projection: a head-on view along the C–C bond axis. The front carbon is shown as a dot (intersection of three bonds); the back carbon as a circle. Bonds on the back carbon radiate from the circle's edge.

Key conformations of ethane. Ethane (C₂H₆) has two extreme conformations:

Staggered — H atoms on adjacent carbons are as far apart as possible (dihedral angle 60°). This is the most stable (lowest energy) because electron-cloud repulsion is minimised (torsional strain is zero).
Eclipsed — H atoms on adjacent carbons are directly aligned (dihedral angle 0°). This is the least stable (highest energy). The energy barrier between eclipsed and staggered ethane is approximately 12.5 kJ/mol.

Watch out: NEET questions may show a Newman or Sawhorse diagram and ask you to identify the conformation type or compare stability. The common confusion is reversing which is more stable — remember: staggered = stable, eclipsed = strained.

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

Which of the following statements about conformations of ethane is correct?

MCQ 2Easy RecallPractice

In a Newman projection of ethane, the front carbon is represented by:

MCQ 3Easy RecallPractice

The energy barrier to rotation about the C–C bond in ethane is approximately:

MCQ 4Direct ApplicationPractice

In the eclipsed conformation of ethane, the dihedral angle between adjacent C–H bonds on the two carbons is:

MCQ 5Direct ApplicationPractice

In a Sawhorse projection of the staggered conformation of ethane, the spatial relationship between the H atoms on the front and back carbons is best described as:

MCQ 6Direct ApplicationPractice

Which of the following is the reason for the instability of the eclipsed conformation of ethane?

MCQ 7Concept TrapPractice

Consider the following statements about conformational isomers: (i) They can be isolated at room temperature. (ii) They arise due to rotation about C–C single bonds. (iii) They have different structural formulae. Which of the above statements is/are correct?

MCQ 8Concept TrapPractice

When ethane is converted from its staggered conformation to its eclipsed conformation, the potential energy of the molecule:

Worked Example

1
Given
- Molecule: ethane (C₂H₆) - Rotational barrier (eclipsed vs staggered): 12.5 kJ/mol - Number of eclipsed H–H interactions in fully eclipsed ethane: 3
2
Required
- Identify the most stable conformation and draw its Newman projection - Calculate torsional strain energy per single eclipsed H–H interaction
3
Concept
In ethane, the staggered conformation (60° dihedral angle between adjacent C–H bonds) is the most stable because electron-pair repulsion between adjacent C–H bonds is minimised. The eclipsed conformation (0° dihedral) has three simultaneous H–H eclipsing interactions that collectively produce 12.5 kJ/mol of torsional strain.
4
Formula
Torsional strain per eclipsed interaction = Total torsional strain ÷ Number of eclipsed interactions
5
Substitution
Strain per interaction = 12.5 kJ/mol ÷ 3
6
Calculation
Strain per interaction = 4.17 kJ/mol (to 3 significant figures) **Note on exact values:** The number 3 (count of eclipsed H–H interactions) is an exact counting integer and does not limit significant figures. The precision of the answer is determined by 12.5 kJ/mol (3 significant figures).
7
Final answer
Each eclipsed H–H interaction in ethane contributes approximately **4.2 kJ/mol** of torsional strain energy. The most stable conformation is the **staggered** conformation (Newman projection: front carbon's three H bonds at 12, 4, and 8 o'clock positions; back carbon's three H bonds at 2, 6, and 10 o'clock positions — all bonds perfectly staggered at 60° intervals).
8
Common trap
Aspirants sometimes confuse staggered and eclipsed in Newman projections, particularly misreading which conformation has bonds overlapping (eclipsed) versus offset (staggered). In the eclipsed Newman projection, the front and back bonds visually overlap — if you can "see" all six bonds clearly without overlap, it is staggered.
9
Similar NEET-style question
"The most stable conformation of n-butane around the C2–C3 bond is the anti conformation. What is the dihedral angle between the two methyl groups in this conformation?" (Answer: 180°. Extension of the same staggered-vs-eclipsed principle to a larger alkane.) ---

Before solving, remember these

Key Fact

Alkane conformations

Free rotation around C-C single bond; staggered (lower energy, dihedral 60°) vs eclipsed (higher energy, dihedral 0°). Newman/sawhorse projections show this. Ethane: 12 kJ/mol energy difference.

-- NCERT, p. 6

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