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 9 (Hydrocarbons) introduces two ways to visualise conformations:
Key conformations of ethane. Ethane (C₂H₆) has two extreme conformations:
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.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Which of the following statements about conformations of ethane is correct?
Answer: B. B is correct. In the staggered conformation of ethane, adjacent C–H bonds are maximally separated (60° dihedral), minimising torsional strain and giving the lowest energy — hence greatest stability. NCERT Class 11 Chemistry Chapter 9, page 305.
Why A is wrong: A is wrong because the eclipsed conformation has the HIGHEST energy due to maximum torsional strain (dihedral angle 0°).
Why C is wrong: C is wrong because the staggered conformation has the LEAST torsional strain, not the highest — it is the eclipsed form that has maximum torsional strain.
Why D is wrong: D is wrong because conformations interconvert rapidly at room temperature through free rotation about the C–C sigma bond; the 12.5 kJ/mol barrier is easily overcome by thermal energy.
In a Newman projection of ethane, the front carbon is represented by:
Answer: A. A is correct. In a Newman projection, the viewer looks along the C–C bond axis; the front carbon appears as a dot (or point) at the centre, and its three bonds radiate from this point. The back carbon is represented by the circle. NCERT Class 11 Chemistry Chapter 9.
Why B is wrong: B is wrong because the large circle represents the BACK (rear) carbon in a Newman projection, not the front carbon.
Why C is wrong: C is wrong because a diagonal line is a feature of Sawhorse projections, not Newman projections.
Why D is wrong: D is wrong because wedge-dash bonds are used in Fischer or three-dimensional representations of stereochemistry, not in Newman projections.
The energy barrier to rotation about the C–C bond in ethane is approximately:
Answer: A. A is correct. The rotational barrier in ethane — the energy difference between the eclipsed and staggered conformations — is approximately 12.5 kJ/mol (about 3 kcal/mol). NCERT Class 11 Chemistry Chapter 9.
Why B is wrong: B is wrong because 3.0 kJ/mol is too low; this would correspond to roughly one eclipsed H–H interaction (~4 kJ/mol each, and ethane has three such interactions in the fully eclipsed form).
Why C is wrong: C is wrong because 30.5 kJ/mol is closer to the rotational barrier in butane about the C2–C3 bond, not ethane.
Why D is wrong: D is wrong because 63.0 kJ/mol far exceeds the ethane rotational barrier; this magnitude approaches bond-dissociation energies for weak interactions and does not correspond to a simple torsional barrier.
In the eclipsed conformation of ethane, the dihedral angle between adjacent C–H bonds on the two carbons is:
Answer: C. C is correct. In the eclipsed conformation, the hydrogen atoms on the front and back carbons are aligned directly behind one another when viewed along the C–C axis, giving a dihedral (torsion) angle of 0°. This alignment maximises torsional strain.
Why A is wrong: A is wrong because 120° is the angle between two bonds on the SAME carbon in a Newman projection (sp³ projected angles), not the dihedral angle between bonds on adjacent carbons.
Why B is wrong: B is wrong because a 60° dihedral angle corresponds to the staggered conformation, where adjacent C–H bonds are maximally separated.
Why D is wrong: D is wrong because 180° would represent the anti (fully staggered) relationship between specific substituents on adjacent carbons in butane or higher alkanes, not the eclipsed conformation.
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:
Answer: D. D is correct. In the staggered Sawhorse projection, the three H atoms on the back carbon are positioned in the gaps between the three H atoms on the front carbon, maximising separation. This is equivalent to viewing a 60° dihedral angle in Newman projection.
Why A is wrong: A is wrong because direct alignment of H atoms on adjacent carbons describes the eclipsed conformation, not the staggered conformation.
Why B is wrong: B is wrong because the H atoms in any conformation of ethane project above and below the C–C bond axis in three dimensions; they are never coplanar with the C–C bond in either Sawhorse or Newman representations.
Why C is wrong: C is wrong because in staggered ethane all six H atoms are equivalently separated — none are overlapping; partial overlap is not a recognised intermediate conformation of ethane at standard descriptions.
Which of the following is the reason for the instability of the eclipsed conformation of ethane?
Answer: C. C is correct. The eclipsed conformation is destabilised by torsional strain, which arises from the repulsion between the bonding electron clouds of the C–H bonds on the two adjacent carbons when they are aligned (0° dihedral). NCERT Class 11 Chemistry Chapter 9.
Why A is wrong: A is wrong because angle strain refers to deviation of bond angles from the ideal tetrahedral value and is associated with small rings (e.g. cyclopropane), not with eclipsed open-chain conformations where bond angles remain near 109.5°.
Why B is wrong: B is wrong because hydrogen atoms are very small — steric strain (van der Waals repulsion due to atomic size) becomes significant only with bulky groups like methyl or tert-butyl, not with hydrogen in ethane.
Why D is wrong: D is wrong because ring strain applies to cyclic molecules (e.g. cyclopropane, cyclobutane), and ethane is an open-chain molecule with no ring.
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?
Answer: B. B is correct. Statement (ii) is the definition: conformational isomers arise from rotation about C–C single bonds. Statement (i) is wrong because conformers interconvert too rapidly at room temperature to be isolated. Statement (iii) is wrong because conformers have the same structural formula — only the spatial arrangement (torsion angles) differs.
Why A is wrong: A is wrong because statement (i) is false — the low rotational barrier (~12.5 kJ/mol for ethane) means conformers interconvert millions of times per second at room temperature and cannot be separated.
Why C is wrong: C is wrong because both statements (i) and (iii) are false. Conformers cannot be isolated at room temperature, and they share the same structural formula — they differ only in torsion angles about single bonds.
Why D is wrong: D is wrong because statement (iii) is false. Conformational isomers have identical structural formulae and differ only in their three-dimensional arrangement due to bond rotation.
When ethane is converted from its staggered conformation to its eclipsed conformation, the potential energy of the molecule:
Answer: D. D is correct. The staggered conformation is the energy minimum and the eclipsed conformation is the energy maximum for ethane. Going from staggered to eclipsed requires overcoming the torsional barrier of approximately 12.5 kJ/mol, so the potential energy increases by that amount.
Why A is wrong: A is wrong because the eclipsed conformation is HIGHER in energy than staggered (not lower) — moving to a less stable form increases potential energy, not decreases it.
Why B is wrong: B is wrong because 63.0 kJ/mol greatly overestimates the ethane rotational barrier. The actual value is approximately 12.5 kJ/mol.
Why C is wrong: C is wrong because while both are conformations of the same molecule, they have different potential energies. The staggered form is lower in energy due to minimal torsional strain; the eclipsed form is higher due to maximal torsional strain.
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Given
• Molecule: ethane (C₂H₆)• Rotational barrier (eclipsed vs staggered): 12.5 kJ/mol• Number of eclipsed H–H interactions in fully eclipsed ethane: 3
Required
• Identify the most stable conformation and draw its Newman projection• Calculate torsional strain energy per single eclipsed H–H interaction
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.
Formula
Torsional strain per eclipsed interaction = Total torsional strain ÷ Number of eclipsed interactions
Substitution
Strain per interaction = 12.5 kJ/mol ÷ 3
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).
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).
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.
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.)
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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 Class 11 Chemistry, Ch. 9, p. 305More in Hydrocarbons: 4 exam traps and mistakes · 1 formula · 2 question patterns from its other lessons.
1 question from NEET 2021. Answers verified against NTA official keys.
Dihedral angle of least stable conformer of ethane is :
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.
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