Conformations of Ethane: Newman Projections

Grab a model of ethane, hold the two carbons between your fingertips, and twist. Nothing breaks, nothing forms — no bond is made, no bond is severed. Yet the molecule's energy rises and falls in a smooth wave. This happens as you rotate one methyl group relative to the other. This single fact — that free rotation about a σ\sigma bond is not actually free — is one of the great surprises waiting inside the simplest possible C–C single bond. It is the foundation of all conformational analysis in organic chemistry. Ethane, CH3–CH3\text{CH}_3\text{–CH}_3, looks almost too simple to be interesting. It has two sp³ carbons, six equivalent C–H bonds, and one C–C bond. But the relative orientation of the two methyl groups around that bond is a continuous variable. This is the dihedral (torsion) angle θ\theta. The energy of the molecule depends on θ\theta in a periodic way. Rotating through 360°360° costs and returns energy six times. This traces out alternating high-energy eclipsed conformations and low-energy staggered conformations. Understanding why this happens is the entry point to everything that follows in conformational chemistry. So is having a clean way to draw and communicate these three-dimensional arrangements. This includes butane's gauche interactions, cyclohexane's chair flip, the anomeric effect in sugars, and the folding of proteins around backbone dihedral angles. The essential tool of this article is the Newman projection. It is a formalism invented explicitly to make torsional geometry legible on paper. Viewed correctly, a Newman projection turns an abstract three-dimensional rotation into a simple two-dimensional picture. A front atom is drawn as a point with three bonds radiating from it. A back atom is drawn as a circle with three bonds emerging from behind it. The angle between a front bond and the nearest back bond is the dihedral angle. It can be read directly off the page. This article builds the Newman projection from first principles. It explains what exactly a dihedral angle measures. It shows how sight-line-down-the-bond geometry maps onto the drawing. It works through the staggered and eclipsed conformers of ethane. Orbital-level reasoning explains why they differ in energy. It gives the quantitative energy profile V(θ)V(\theta). It closes with worked examples. These range from basic reading of a Newman diagram to the population ratio of conformers at a given temperature. The deeper origin of the rotational barrier — hyperconjugation versus classical steric repulsion — is developed fully elsewhere. Here the focus is on getting the geometry, the vocabulary, and the energy landscape exactly right.

Key takeaways
  • A dihedral (torsion) angle θ\theta specifies the twist of one C–H (or C–X) bond relative to another across a central σ\sigma bond, measured as the angle between the two bonds when viewed along the bond axis.
  • A Newman projection looks straight down a C–C bond: the front carbon is a dot with three bonds at 120°120°; the back carbon is a circle with three bonds also at 120°120°, offset by whatever the true dihedral angle is.
  • Ethane has two special conformations as it rotates: staggered (θ=60°\theta = 60° between adjacent H's) and eclipsed (θ=0°\theta = 0°), each recurring every 120°120° due to the three-fold symmetry of each methyl group.
  • The staggered conformer is the global energy minimum; the eclipsed conformer is a local maximum, roughly 12.0 kJ/mol12.0\ \text{kJ/mol} (2.9 kcal/mol2.9\ \text{kcal/mol}) higher for ethane.
  • Rotation about the C–C bond is essentially free at room temperature (barrier ≪RT\ll RT-scale populations fully interconvert on a picosecond timescale), but the energy is never flat — it is a genuine periodic potential, not a constant.
  • The threefold periodicity, V(θ)=V(θ+120°)V(\theta) = V(\theta+120°), reflects the C3vC_{3v} local symmetry of each methyl rotor and dictates that ethane has exactly three equivalent staggered minima and three equivalent eclipsed maxima per full turn.
  • Key idea: rotation about a C–C single bond is a continuous, low-barrier, periodic motion whose staggered minima and eclipsed maxima are best visualized and quantified using the Newman projection.

Where this lives today

Conformational analysis built on the Newman projection is not a historical curiosity confined to introductory courses — it is the working vocabulary of modern structural and mechanistic organic chemistry. Every discussion of a reaction's stereochemical outcome that depends on the preferred rotamer of an open-chain intermediate — nucleophilic additions to acyclic ketones analyzed via Felkin–Anh models, radical cyclizations, or SN2 transition-state geometries — is drawn and reasoned about using Newman-style projections down the reacting bond. Pharmaceutical chemists routinely examine Newman projections (or their modern computational analog, a dihedral-angle energy scan from molecular mechanics or DFT) when assessing whether a candidate drug molecule's freely rotatable bonds allow it to adopt the bioactive conformation needed to fit a protein binding pocket; "conformational entropy penalty" upon binding is precisely the cost of freezing a torsional coordinate like the one described here. In structural biology, the same torsional logic scales up: the backbone dihedral angles ϕ\phi and ψ\psi of a protein (rotations about the N–Cα and Cα–C(=O) bonds) are graphed on a Ramachandran plot whose allowed and forbidden regions are, at heart, a more elaborate version of ethane's V(θ)V(\theta) curve, now shaped by steric clash between much bulkier substituents and by competing intramolecular hydrogen bonds. Polymer scientists reasoning about the flexibility and persistence length of a polyethylene chain are, chain segment by chain segment, applying exactly the staggered-versus-eclipsed logic developed here, since each C–C bond along the backbone samples the same three-fold torsional landscape (perturbed by neighboring substituents, as foreshadowed by Conformations of Butane: Gauche and Anti). Even NMR spectroscopy leans on this framework directly: the Karplus equation, relating a vicinal 3JHH^3J_{HH} coupling constant to the H–C–C–H dihedral angle, is an empirical fit built entirely on the geometric definition of θ\theta established by the Newman projection, and is used daily to determine unknown conformations and relative stereochemistry of natural products and synthetic intermediates from routine NMR spectra.

Ask the tutor: Why does hyperconjugation into a σ* orbital require the specific anti-periplanar geometry found in the staggered conformer, and how would you show this with an orbital diagram?

The physical picture — sighting down the C–C bond

Take ethane and orient it so the C–C bond points directly at your eye. The near carbon (call it C1) hides the far carbon (C2) completely — that is the entire point of "looking down the bond." C1's three C–H bonds fan out toward you in a tripod, each separated from its neighbors by 120°120° in the plane perpendicular to your line of sight (this is simply the projection of sp3sp^3 tetrahedral geometry onto that plane). Behind C1, invisible in a real ball-and-stick model but drawn explicitly in a Newman projection, C2's three C–H bonds also fan out in their own tripod, also separated by 120°120°. The Newman projection converts this line-of-sight view into ink: the front atom becomes a point (or a small "Y" of three lines meeting at a dot) and the back atom becomes a circle, with three lines drawn from the edge of the circle outward (never crossing through the center, since that would look like they belong to the front atom). The dihedral angle between a specified front bond and a specified back bond is simply the angular separation between those two lines as drawn on the page. Because each methyl group is itself C3vC_{3v}-symmetric (three equivalent H's, 120°120° apart), the appearance of the Newman projection repeats every 120°120° of rotation. Two special angles matter most:

  • Staggered (θ=60°,180°,300°\theta = 60°, 180°, 300° for the reference pair): back bonds sit exactly between front bonds. Looking down the bond, you see six evenly spaced lines, alternating front/back, each separated by 60°60°.
  • Eclipsed (θ=0°,120°,240°\theta = 0°, 120°, 240°): each back bond hides directly behind a front bond. Looking down the bond, you see only three lines (though chemists draw the back bond slightly offset, at a small angle, purely so both bonds remain visible on paper).
Figure
Newman projections of staggered and eclipsed ethane, viewed down the C1–C2 bond
H H H H H H staggered, θ = 60° H H H H H H eclipsed, θ = 0°

Note the geometric logic in the drawing: in the staggered projection, the six C–H lines are equally spaced at 60°60° around the full circle — front, back, front, back, alternating — while in the eclipsed drawing the back lines are drawn nearly (not exactly, for clarity) on top of the front lines, at θ≈0°\theta \approx 0°. Now consider what physically differs between the two arrangements. In the staggered form, every front C–H bond points between two back C–H bonds — the hydrogens are as far apart as three-fold symmetry allows, minimizing steric crowding of the electron clouds on adjacent atoms. In the eclipsed form, every front C–H bond points directly at a back C–H bond, forcing the C–H bonding electron pairs into closer proximity. There is also a subtler and, it turns out, energetically dominant effect: in the staggered conformation each C–H σ\sigma bond on the back carbon is anti-periplanar to a C–H σ∗\sigma^* antibonding orbital on the front carbon, allowing a small stabilizing donation of electron density (hyperconjugation) that is geometrically impossible in the eclipsed form, where the relevant orbitals are orthogonal rather than aligned. The full quantum mechanical accounting of why the staggered form wins is developed under Torsional Strain and the Origin of the Ethane Barrier; here we take the empirical energy difference as given and build the quantitative torsional potential around it.

Figure
The torsional potential energy of ethane as a function of dihedral angle θ
dihedral angle θ (degrees) V(θ), kJ/mol 0 60 120 180 240 300 360 0 12 eclipsed staggered eclipsed staggered

The curve has period 120°120°, maxima of 12.0 kJ/mol12.0\ \text{kJ/mol} at the eclipsed angles 0°,120°,240°,360°0°, 120°, 240°, 360°, and minima of 00 at the staggered angles 60°,180°,300°60°, 180°, 300°.

The core ideas — the torsional potential and its symmetry

The torsional energy of ethane, as a function of the dihedral angle θ\theta between a reference front C–H bond and a reference back C–H bond, is well described by a single cosine term:

V(θ)=V02[1+cos⁡(3θ)]V(\theta) = \frac{V_0}{2}\left[1 + \cos(3\theta)\right]

where:

  • V(θ)V(\theta) is the torsional potential energy at dihedral angle θ\theta,
  • V0V_0 is the full barrier height (the energy difference between eclipsed and staggered conformers),
  • the factor 33 inside the cosine encodes the three-fold periodicity forced by the C3vC_{3v} symmetry of each methyl rotor,
  • θ=0°\theta = 0° is chosen, by convention, to correspond to an eclipsed conformation, so that cos⁡(0)=1\cos(0) = 1 gives V(0)=V0V(0) = V_0, the maximum.

Reading this equation: at θ=0°,120°,240°\theta = 0°, 120°, 240° the cosine equals +1+1 and V=V0V = V_0 (eclipsed maxima); at θ=60°,180°,300°\theta = 60°,180°,300° the cosine equals −1-1 and V=0V=0 (staggered minima). The period of cos⁡(3θ)\cos(3\theta) is 120°120°, matching exactly the fact that rotating one methyl group by 120°120° relative to the other simply relabels which hydrogen is which — the molecule (and hence its energy) is unchanged. A second essential relationship converts an energy difference between conformers into a population ratio at thermal equilibrium, via the Boltzmann distribution:

NeclipsedNstaggered=exp⁡ ⁣(−ΔVRT)\frac{N_{\text{eclipsed}}}{N_{\text{staggered}}} = \exp\!\left(-\frac{\Delta V}{RT}\right)

where:

  • Neclipsed/NstaggeredN_{\text{eclipsed}}/N_{\text{staggered}} is the ratio of molecules instantaneously found near the eclipsed versus staggered geometry,
  • ΔV=V0\Delta V = V_0 is the barrier height (energy of eclipsed minus energy of staggered),
  • R=8.314 J mol−1K−1R = 8.314\ \text{J mol}^{-1}\text{K}^{-1} is the gas constant,
  • TT is the absolute temperature in kelvin.

This equation makes explicit something conceptually crucial: even though the eclipsed conformation is a strained, high-energy geometry, it is not "forbidden" — it is simply thermally disfavored, and it is transiently populated as every ethane molecule rotates through it on the way from one staggered well to the next.

Derivation ABuilding V(θ) from the symmetry of the methyl rotor

Goal: show, from symmetry alone (not curve-fitting), why the torsional potential of ethane must take the form V(θ)=V02[1+cos⁡(3θ)]V(\theta) = \tfrac{V_0}{2}[1+\cos(3\theta)] and not some other periodic function.

Step 1 — Identify the symmetry operation that must leave VV invariant. Ethane's two methyl groups are each locally C3vC_{3v}: rotating the back methyl group by exactly 120°120° about the C–C axis, while holding the front methyl fixed, permutes its three hydrogens among themselves and produces a geometry indistinguishable from the starting one. Because the three hydrogens on a methyl carbon are chemically identical, this operation cannot change the energy:

V(θ)=V(θ+120°)for all θ.V(\theta) = V(\theta + 120°) \quad \text{for all } \theta.

Step 2 — Translate periodicity into a Fourier constraint. Any smooth periodic function of θ\theta with period 120°120° can be expanded as a Fourier cosine series in multiples of 3θ3\theta (since 3θ3\theta has period 120°120° exactly when θ\theta has period 120°120°):

V(θ)=a0+a1cos⁡(3θ)+a2cos⁡(6θ)+a3cos⁡(9θ)+⋯V(\theta) = a_0 + a_1\cos(3\theta) + a_2\cos(6\theta) + a_3\cos(9\theta) + \cdots

No terms in cos⁡(θ)\cos(\theta), cos⁡(2θ)\cos(2\theta), cos⁡(4θ)\cos(4\theta), etc. are allowed, because those functions do not repeat every 120°120° and would make V(θ)≠V(θ+120°)V(\theta)\neq V(\theta+120°), violating Step 1.

Step 3 — Truncate using the physical size of the interactions. Experimentally (and computationally), the ethane torsional profile is extremely well approximated by keeping only the leading term a1cos⁡(3θ)a_1\cos(3\theta); higher harmonics (cos⁡6θ\cos 6\theta, cos⁡9θ\cos 9\theta, …) contribute negligibly because the dominant physical interactions — pairwise H···H steric/exchange repulsion and σC–H→σC–H∗\sigma_{\text{C–H}} \to \sigma^*_{\text{C–H}} hyperconjugative stabilization — are themselves each three-fold periodic functions of θ\theta to good approximation. Keeping one term:

V(θ)=a0+a1cos⁡(3θ).V(\theta) = a_0 + a_1\cos(3\theta).

Step 4 — Fix the two constants using the two known extrema. Choose θ=0°\theta=0° as eclipsed (maximum) and θ=60°\theta=60° as staggered (minimum), matching convention:

V(0°)=a0+a1=V0,V(60°)=a0−a1=0,V(0°) = a_0 + a_1 = V_0, \qquad V(60°) = a_0 - a_1 = 0,

since cos⁡(180°)=−1\cos(180°) = -1. Solving this pair of linear equations:

a0=a1=V02.a_0 = a_1 = \frac{V_0}{2}.

Step 5 — Assemble the final form. Substituting back:

V(θ)=V02+V02cos⁡(3θ)=V02[1+cos⁡(3θ)].V(\theta) = \frac{V_0}{2} + \frac{V_0}{2}\cos(3\theta) = \frac{V_0}{2}\left[1+\cos(3\theta)\right].

This is exactly the potential quoted above, now derived purely from the three-fold symmetry of the methyl rotor plus the empirical fact that one harmonic dominates. The single remaining unknown, V0V_0, is an experimental/computational number (about 12.0 kJ/mol12.0\ \text{kJ/mol} for ethane), not something symmetry alone can supply — symmetry fixes the shape of the curve, not its amplitude.

Go deeper · Derivation B · 2 minFrom bond vectors to the Newman-projection angle: what θ actually measures
Derivation BFrom bond vectors to the Newman-projection angle: what θ actually measures

Goal: derive, from the tetrahedral geometry of sp³ carbon, the explicit vector definition of the dihedral angle θ\theta that the Newman projection displays, and confirm that it reduces to 60°60° apart for adjacent staggered C–H bonds.

Step 1 — Set up the C–C axis and place the front methyl group. Let the C1–C2 bond lie along the zz-axis, with C1 at the origin and C2 at (0,0,d)(0,0,d), where d≈1.54 A˚d\approx 1.54\ \text{Å} is the C–C bond length. Each C–H bond on C1 makes the tetrahedral angle (109.5°109.5°) with the C1–C2 bond; projected into the xyxy-plane (the plane perpendicular to the viewing axis, which is exactly what a Newman projection draws), the three C–H bonds on C1 appear at angles ϕ1=0°,120°,240°\phi_1 = 0°, 120°, 240° around the circle.

Step 2 — Define a dihedral angle formally via bond vectors. For four atoms HaH_a–C1C1–C2C2–HbH_b, the dihedral angle θ\theta is the angle between the plane containing Ha,C1,C2H_a, C1, C2 and the plane containing C1,C2,HbC1, C2, H_b. Equivalently, project both the C1 ⁣− ⁣HaC1\!-\!H_a bond and the C2 ⁣− ⁣HbC2\!-\!H_b bond onto the plane perpendicular to the C1C1–C2C2 axis; θ\theta is the angle between these two projected vectors. This is precisely the angle read directly off a Newman projection, because the Newman drawing is that perpendicular-plane projection.

Step 3 — Place the back methyl group at general twist angle. Let the three C–H bonds on C2 project, in that same xyxy-plane, to angles ϕ2=θ,θ+120°,θ+240°\phi_2 = \theta, \theta+120°, \theta+240°, where θ\theta is the twist of the back methyl group relative to the front one (this θ\theta is exactly the rotation coordinate in V(θ)V(\theta)).

Step 4 — Impose the staggered condition and solve for θ. "Staggered" means each back C–H bond bisects the angle between two front C–H bonds. With front bonds fixed at 0°,120°,240°0°, 120°, 240°, the bisecting positions are 60°,180°,300°60°, 180°, 300°. Matching the back-bond set {θ,θ+120°,θ+240°}\{\theta, \theta+120°, \theta+240°\} to {60°,180°,300°}\{60°,180°,300°\} gives:

θ=60°\theta = 60°

as the reference dihedral angle for staggered ethane (equivalently 180°180° or 300°300°, since these describe the same physical arrangement — just a different choice of which specific back hydrogen is used as reference).

Step 5 — Impose the eclipsed condition. "Eclipsed" means each back C–H bond lies exactly on top of (same projected angle as) a front C–H bond: ϕ2=ϕ1\phi_2 = \phi_1, i.e. θ=0°\theta = 0° (or 120°120°, 240°240°).

Step 6 — Confirm consistency with the energy profile. Substituting θ=60°\theta = 60° into V(θ)=V02[1+cos⁡(3θ)]V(\theta) = \tfrac{V_0}{2}[1+\cos(3\theta)] gives cos⁡(180°)=−1\cos(180°) = -1, so V=0V = 0: the minimum, exactly at the geometrically staggered angle. Substituting θ=0°\theta = 0° gives cos⁡(0)=1\cos(0)=1, V=V0V=V_0: the maximum, exactly at the geometrically eclipsed angle. The geometric definition of θ\theta (Steps 1–5) and the energetic definition used in V(θ)V(\theta) (core equation) are thus the same variable, confirming the Newman projection is not just a qualitative cartoon — its drawn angle literally is the argument of the torsional potential.

Worked examples

Example 1 (foundational) — Reading and constructing a Newman projection. Draw the Newman projection of ethane looking from C1 toward C2, in the staggered conformation, and label the dihedral angle between a specific front hydrogen H1aH_{1a} and its nearest back hydrogen H2aH_{2a}. Reasoning. Place the front carbon's three C–H bonds, by convention, pointing straight up and to the two lower corners: ϕ1(H1a)=90°\phi_1(H_{1a}) = 90°, ϕ1(H1b)=210°\phi_1(H_{1b}) = 210°, ϕ1(H1c)=330°\phi_1(H_{1c}) = 330° (measuring counterclockwise from the positive xx-axis, a common drawing convention). For the staggered conformer, each back C–H bond must bisect a pair of front bonds. The gaps between front bonds are centered at 90°+60°=150°90°+60°=150°, 210°+60°=270°210°+60°=270°, and 330°+60°=30°330°+60°=30° (mod 360°360°). So the back bonds sit at ϕ2=30°,150°,270°\phi_2 = 30°, 150°, 270°. The nearest back hydrogen to H1aH_{1a} (at 90°90°) is the one at 150°150° (a separation of 60°60°) rather than the one at 30°30° (also 60°60° away, by symmetry — both are equally "nearest," which is the hallmark of a perfectly staggered arrangement). Either choice gives:

θ(H1a–C1–C2–H2a)=60°.\theta(H_{1a}\text{–}C1\text{–}C2\text{–}H_{2a}) = 60°.

This matches Step 4 of the second derivation exactly, confirming the drawing is self-consistent. Limiting case check. If instead we had drawn the eclipsed conformer, the back bonds would sit at ϕ2=90°,210°,330°\phi_2 = 90°, 210°, 330° (directly behind the front ones), giving θ=0°\theta = 0° for the aligned pair — recovering the eclipsed maximum of V(θ)V(\theta), exactly as expected, with no arbitrary choice needed since the eclipsing bond is unambiguous.

Example 2 (applied) — Population ratio of eclipsed versus staggered ethane at room temperature. Using V0=12.0 kJ/molV_0 = 12.0\ \text{kJ/mol} for the ethane rotational barrier, estimate the fraction of molecules found instantaneously in the eclipsed geometry relative to the staggered geometry at T=298 KT = 298\ \text{K}. Reasoning. Apply the Boltzmann ratio derived above:

NeclipsedNstaggered=exp⁡ ⁣(−V0RT).\frac{N_{\text{eclipsed}}}{N_{\text{staggered}}} = \exp\!\left(-\frac{V_0}{RT}\right).

Compute the exponent first:

V0RT=12,000 J/mol(8.314 J mol−1K−1)(298 K).\frac{V_0}{RT} = \frac{12{,}000\ \text{J/mol}}{(8.314\ \text{J mol}^{-1}\text{K}^{-1})(298\ \text{K})}.

The denominator is:

8.314×298=2478.6 J/mol.8.314 \times 298 = 2478.6\ \text{J/mol}.

So:

V0RT=12,0002478.6=4.842.\frac{V_0}{RT} = \frac{12{,}000}{2478.6} = 4.842.

Then:

NeclipsedNstaggered=e−4.842≈7.9×10−3.\frac{N_{\text{eclipsed}}}{N_{\text{staggered}}} = e^{-4.842} \approx 7.9\times10^{-3}.

Interpretation: for roughly every 126 molecules found at (or extremely near) a staggered geometry at any instant, only about 1 sits at the exact eclipsed maximum — a strong but not absolute preference. This is why ethane's rotation is described as "essentially free": with a barrier this low, thermal energy at room temperature (RT≈2.5 kJ/molRT \approx 2.5\ \text{kJ/mol}) is enough for the molecule to cross the barrier roughly 101010^{10}–101110^{11} times per second (an estimate obtainable from transition-state theory, using V0/RT≈4.8V_0/RT\approx4.8 in an Arrhenius-type prefactor of ∼1013 s−1\sim 10^{13}\,\text{s}^{-1}), so essentially no population is ever trapped in a single well. Limiting-case check. As T→∞T \to \infty, V0/RT→0V_0/RT \to 0 and the ratio →exp⁡(0)=1\to \exp(0) = 1: at very high temperature the populations of eclipsed and staggered geometries become equal, exactly as expected since thermal energy overwhelms the modest torsional preference and the molecule samples all θ\theta nearly uniformly. As T→0T\to 0, the ratio →0\to 0: the molecule freezes into the staggered well, consistent with quantum-mechanical localization at absolute zero.

Example 3 (exam / competition level) — Distinguishing ethane's barrier from a substituted analog and predicting a Newman diagram for a general dihedral. 1,2-Difluoroethane, FCH2–CH2F\text{FCH}_2\text{–CH}_2\text{F}, is not ethane, but consider a hypothetical thought experiment: suppose a symmetric, ethane-like rotor had a torsional potential with two significant Fourier terms,

V(θ)=V02[1+cos⁡(3θ)]+V22cos⁡(6θ),V(\theta) = \frac{V_0}{2}\left[1+\cos(3\theta)\right] + \frac{V_2}{2}\cos(6\theta),

with V0=12.0 kJ/molV_0 = 12.0\ \text{kJ/mol} and V2=1.50 kJ/molV_2 = 1.50\ \text{kJ/mol} (a small second-harmonic correction). Find the exact angle nearest θ=60°\theta=60° at which the true minimum occurs, to first order. Reasoning. Differentiate and set the derivative to zero:

dVdθ=−3V02sin⁡(3θ)−3V2sin⁡(6θ)=0.\frac{dV}{d\theta} = -\frac{3V_0}{2}\sin(3\theta) - 3V_2\sin(6\theta) = 0.

Near θ=60°\theta = 60°, write θ=60°+δ\theta = 60° + \delta with small δ\delta. Then 3θ=180°+3δ3\theta = 180° + 3\delta, so sin⁡(3θ)=sin⁡(180°+3δ)=−sin⁡(3δ)≈−3δ\sin(3\theta) = \sin(180°+3\delta) = -\sin(3\delta) \approx -3\delta (in radians, for small δ\delta). Also 6θ=360°+6δ6\theta = 360°+6\delta, so sin⁡(6θ)=sin⁡(6δ)≈6δ\sin(6\theta) = \sin(6\delta) \approx 6\delta. Substituting:

−3V02(−3δ)−3V2(6δ)=0-\frac{3V_0}{2}(-3\delta) - 3V_2(6\delta) = 0
9V02δ−18V2δ=0\frac{9V_0}{2}\delta - 18V_2\delta = 0
δ(9V02−18V2)=0.\delta\left(\frac{9V_0}{2} - 18V_2\right) = 0.

Since 9(12.0)2−18(1.50)=54.0−27.0=27.0≠0\frac{9(12.0)}{2} - 18(1.50) = 54.0 - 27.0 = 27.0 \neq 0, the only solution is δ=0\delta = 0: to first order in the small-second-harmonic limit, the minimum stays exactly at θ=60°\theta = 60°, undisplaced. This makes physical sense — a pure cos⁡(6θ)\cos(6\theta) correction shares the same symmetry axis as cos⁡(3θ)\cos(3\theta) at θ=60°\theta=60° (both have zero slope there by symmetry, since 60°60° is a symmetry point for both harmonics separately), so a small admixture cannot shift the extremum's location, only its curvature (and hence the vibrational frequency of the torsional oscillation about that minimum). This is exactly the kind of reasoning used, in more advanced treatments, to explain why substituted ethanes (e.g., 1,2-dihaloethanes, where gauche/anti populations shift measurably) require genuinely new physics — electrostatics, dipole–dipole terms, larger steric bulk — layered on top of the basic three-fold pattern rather than a simple change in the coefficient of the same harmonic; that richer story is the subject of Conformations of Butane: Gauche and Anti. Limiting case check. Setting V2=0V_2 = 0 recovers pure ethane: δ=0\delta = 0 trivially, minimum exactly at θ=60°\theta=60°, matching the geometric derivation above term-for-term.

Limiting cases, scope & common traps

SituationWhat happensWhy
θ→60°,180°,300°\theta \to 60°, 180°, 300°V(θ)→0V(\theta) \to 0 (global minima)back C–H bonds bisect front C–H bonds; minimal steric clash, maximal σCH→σCH∗\sigma_{CH}\to\sigma^*_{CH} hyperconjugative overlap
θ→0°,120°,240°\theta \to 0°, 120°, 240°V(θ)→V0V(\theta) \to V_0 (local maxima, all equal by symmetry)eclipsing C–H bonds are aligned; steric repulsion maximized, hyperconjugative alignment lost
T→∞T \to \inftyNeclipsed/Nstaggered→1N_{\text{eclipsed}}/N_{\text{staggered}} \to 1thermal energy RT≫V0RT \gg V_0; barrier becomes negligible, torsion approaches free rotation
T→0 KT \to 0\ \text{K}Neclipsed/Nstaggered→0N_{\text{eclipsed}}/N_{\text{staggered}} \to 0molecule localizes in one staggered well (quantized torsional ground state)
Replacing one H by a bulky group (e.g., going toward butane)Three-fold symmetry is broken; V(θ)V(\theta) acquires distinct gauche and anti minima of different energyloses the exact C3vC_{3v} local symmetry that forced all three staggered wells (and all three eclipsed maxima) to be identical
Drawing a Newman projection with back bonds through the circle's centerFormally wrong / ambiguousback bonds must emerge from the edge of the circle, never its center, or the projection cannot be distinguished from the front atom's bonds

The chemist behind it

The Newman projection is named for Melvin Spencer Newman (1908–1993), an American organic chemist at Ohio State University who introduced the "front atom/back atom" projection formalism in a short 1955 paper aimed squarely at making conformational analysis teachable. Newman's broader career centered on physical organic chemistry and synthetic methodology — steric effects in aromatic systems, the synthesis of polycyclic aromatic hydrocarbons, and reactive intermediates — but it is this single pedagogical device, born from the practical need to represent three-dimensional torsional geometry unambiguously on a flat page, for which his name is now attached to nearly every organic chemistry textbook diagram of a rotating C–C bond. The conceptual foundation Newman's projection formalized — that rotational conformers of ethane have measurably different energies — rests on much earlier physical chemistry. Rotational barriers in ethane-like molecules were inferred indirectly from heat capacity and spectroscopic anomalies starting in the 1930s, well before quantum chemical calculations could assign a clean orbital-level cause. The size of the ethane barrier itself, close to 12 kJ/mol12\ \text{kJ/mol}, was pinned down by microwave and infrared spectroscopy in the mid-20th century, and its microscopic origin — steric repulsion versus hyperconjugative stabilization — remained (and to a real extent still remains) an active topic of computational and conceptual debate, addressed in depth under Torsional Strain and the Origin of the Ethane Barrier.

Origins · 1 minMelvin Spencer Newman and how this was discovered
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Melvin Spencer Newman
1908–1993 · American organic chemist, Ohio State University, who introduced the Newman-projection formalism in 1955 to represent conformations about a single bond

History
- 1930s — Heat capacity and early spectroscopic anomalies in ethane hint that internal rotation is hindered, not free.
- 1936 — Kemp and Pitzer analyze ethane's thermodynamic data and estimate a rotational barrier on the order of a few kcal/mol.
- 1950s — Microwave and infrared spectroscopy refine the ethane barrier to close to 2.9 kcal/mol2.9\ \text{kcal/mol} (≈12 kJ/mol\approx 12\ \text{kJ/mol}).
- 1955 — Melvin S. Newman publishes the front-atom/back-atom projection now universally called the Newman projection.
- Late 20th–21st century — High-level ab initio and natural bond orbital (NBO) analyses reexamine whether hyperconjugation or steric repulsion dominates the ethane barrier, refining but not overturning the qualitative staggered-favored picture.
Why is ethane's rotational barrier considered "low" if it clearly is not zero?

"Low" is a comparison to the thermal energy scale RTRT, not a claim of zero. At room temperature, RT≈2.5 kJ/molRT \approx 2.5\ \text{kJ/mol}, while ethane's barrier is about 12.0 kJ/mol12.0\ \text{kJ/mol} — roughly 4.8 RT4.8\,RT. This is high enough to create a real, measurable preference for staggered geometries (about 126:1 in favor, as computed via the Boltzmann ratio) but far too low to trap the molecule in any single well: collisions and vibrational energy easily supply enough energy to hop over the barrier roughly 101010^{10} times per second. Contrast this with, say, the barrier to rotation about a C=C double bond (over 250 kJ/mol250\ \text{kJ/mol}), which genuinely is high enough to prevent interconversion at room temperature and gives rise to isolable cis/trans isomers. Ethane's barrier sits at the opposite extreme: real, structurally meaningful, but kinetically inconsequential at ordinary temperatures.

Does every C–C single bond have the same torsional barrier as ethane?

No. Ethane's 12.0 kJ/mol12.0\ \text{kJ/mol} barrier is specific to two unsubstituted methyl groups. Replacing hydrogens with larger substituents (as in butane, where one hydrogen on each carbon becomes a methyl group) breaks the perfect three-fold degeneracy: the three staggered conformers are no longer equal in energy (anti versus gauche), and the eclipsed conformers likewise split into distinct energies depending on which groups eclipse which. Larger substituents generally raise barriers substantially — the barrier to rotation in more hindered systems (e.g., around the central bond of 2,3-dimethylbutane) can be several times larger than in ethane, since steric repulsion between bulky eclipsing groups adds on top of the base hydrogen–hydrogen interactions.

Is the ethane torsional barrier caused by steric repulsion between hydrogens, or by something more subtle?

This is a genuinely debated question at the level of quantum-chemical analysis, not fully settled by intuition alone. The classical explanation attributes the barrier to steric (Pauli/exchange) repulsion between eclipsing C–H bonding electron pairs, which are forced unusually close together at θ=0°\theta=0°. A newer, competing explanation emphasizes hyperconjugative stabilization: in the staggered form, each filled σC−H\sigma_{C-H} orbital on one carbon is anti-periplanar to an empty σC−H∗\sigma^*_{C-H} orbital on the other, donating electron density and lowering energy, an alignment unavailable in the eclipsed form. Natural bond orbital (NBO) analyses tend to favor the hyperconjugative explanation as dominant for ethane specifically, though the steric contribution is not zero. This is precisely the mechanistic detail developed under Torsional Strain and the Origin of the Ethane Barrier.

How do you decide which hydrogen to draw first when constructing a Newman projection from scratch?

There is no unique "correct" starting hydrogen — the choice is arbitrary, exactly analogous to choosing where to start numbering carbons in an IUPAC name, provided the final structure is geometrically consistent. A common convention is to place one front C–H bond pointing straight up (at 90°90° in standard mathematical angle convention) and then space the other two front bonds at 120°120° intervals (210°210° and 330°330°). Once the front atom's bonds are fixed, the back atom's bonds are placed either directly behind the front bonds (eclipsed, θ=0°\theta=0°) or bisecting the front bonds' gaps (staggered, θ=60°\theta=60°), depending on which conformer is being illustrated.

Can Newman projections represent molecules other than ethane, such as substituted or larger systems?

Yes — the Newman projection is a general tool for any C–C (or C–N, C–O, etc.) single bond, not just the parent ethane. For substituted systems like butane, 2-chlorobutane, or the backbone bonds of a sugar or amino acid, the front and back "circle" positions are simply populated with whichever substituents are actually present (methyl, halogen, hydroxyl, and so on), and the torsional analysis becomes richer because the three-fold degeneracy of plain ethane is broken. The formalism itself — point for front atom, circle for back atom, radiating bonds at 120°120° — remains identical; only the substituent labels and the resulting energetics change, which is exactly the extension explored in Conformations of Butane: Gauche and Anti.

Why does the dihedral angle in a Newman projection matter for reactions, not just for describing a static structure?

Many reaction mechanisms require specific dihedral alignments to proceed efficiently. Anti-periplanar geometry (dihedral angle near 180°180°) is required for E2 eliminations, because the breaking C–H and C–leaving-group bonds must align with the forming π system for effective orbital overlap. Similarly, SN2 backside attack requires the nucleophile to approach opposite the leaving group along a well-defined trajectory that is naturally described in Newman-projection terms. Even without any bond-breaking, the ground-state torsional preferences described here (staggered over eclipsed) set the baseline conformational population that any reacting molecule must first access or distort away from, making the Newman projection a working tool for mechanism, not merely a static structural cartoon.

Is the ethane barrier the same in the gas phase, in solution, and in the solid state?

The intrinsic 12.0 kJ/mol12.0\ \text{kJ/mol} figure is a gas-phase (or isolated-molecule, computational) value, reflecting the molecule's own internal electronic structure. In solution, solvent packing and weak intermolecular interactions can perturb this slightly, though for a small, nonpolar molecule like ethane, the effect is modest. In the solid state, at very low temperature, ethane crystallizes and molecular rotation becomes further restricted by the crystal lattice, an entirely separate (intermolecular, packing-driven) phenomenon layered on top of the intrinsic intramolecular torsional potential discussed in this article.

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