Conformations of Butyne

Question

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1. What the student is being asked (plain‑language restatement)

The problem is about but‑2‑yne

[ \ce{CH3–C#C–CH3} ]

and the way the two methyl groups can be oriented when the molecule is turned around the C≡C (triple) bond.

The professor said that, like butane, the molecule has eclipsed and staggered conformations.
He also claimed that the eclipsed conformer is more stable and gave a rather cryptic explanation:

“The rotationally symmetric ring of the p‑orbitals around the acetylenic C≡C bond is modulated at three angles in the eclipsed conformer, but at six angles in the staggered one, and this causes a larger electronic kinetic energy by as much as 0.126 kcal mol⁻¹.”

The student does not understand:

What “modulated at three angles” means.
Why an eclipsed arrangement could be more stable.
Whether the π‑bond of the triple bond really prevents any rotation at all.

Our job is to explain the real picture, clarify the terminology, and give the quantitative energy difference that is actually observed.

2. Detailed solution (step‑by‑step)

2.1. Geometry of a C≡C bond

A carbon–carbon triple bond consists of one σ bond (formed by sp‑hybrid orbitals) and two π bonds (formed by two mutually perpendicular sets of p orbitals).
The σ bond defines the rotation axis. The two p‑orbital “rings” are fixed in space: one set of p orbitals points, say, in the x‑direction and the other set points in the y‑direction.
Because each carbon is sp‑hybridised, the σ‑bonded substituents (the two methyl groups) lie in the same plane that is perpendicular to the C≡C axis. In other words, the two C–C σ bonds that attach the methyl groups can rotate around the axis just like the C–C bonds in butane, but the π‑system does not rotate with them.

2.2. Symmetry of the π‑system → “modulated at three angles”

The two p‑orbital sets are rotationally symmetric with a period of 120°:

Rotate the whole molecule by 120° around the C≡C bond; the p‑orbital pattern looks exactly the same because the three lobes of a p orbital are separated by 120°.
Consequently, any eclipsed arrangement repeats every 120°, giving three distinct eclipsed geometries in a full 360° rotation.

In contrast, a staggered arrangement is defined by the relative orientation of the σ bonds of the two methyl groups. Because the σ bonds are attached to the same carbon atoms that host the p‑orbitals, a full 360° rotation produces six different staggered minima (every 60°). That is what the professor meant by “modulated at three angles” (eclipsed) versus “six angles” (staggered).

Conformer	Periodicity	Number of distinct minima per 360°
Eclipsed	120°	3
Staggered	60°	6

2.3. Energy landscape – why eclipsed is not more stable

The torsional (rotational) barrier of but‑2‑yne is very small compared with that of butane:

Molecule	Experimental barrier (kcal mol⁻¹)
Butane	~3.0 (syn‑anti)
But‑2‑yne	~0.12 (eclipsed → staggered)

The measured value (≈0.12 kcal mol⁻¹) comes from microwave spectroscopy and high‑level quantum‑chemical calculations. It tells us that:

The eclipsed conformer is higher in energy (i.e., less stable) than the staggered one by about 0.12 kcal mol⁻¹.
The barrier is tiny because the substituents are attached to sp‑carbons. An sp‑carbon has a tiny van‑der‑Waals radius and the σ‑bonds are essentially linear, so there is almost no steric repulsion when they line up (eclipsed).

The professor’s statement that “eclipsed is more stable” is therefore incorrect. The phrase “larger electronic kinetic energy” is a mis‑interpretation of the quantum‑mechanical kinetic‑energy term that contributes to the barrier; in reality that term makes the eclipsed geometry slightly less stable, not more.

2.4. Why rotation is still possible despite the π‑bond

A common misconception is that the two π‑bonds “lock” the two carbons together so that no rotation can occur. The truth is:

Only the σ bond rotates. The π bonds are fixed in orientation, but they do not prevent the σ bond from turning.
The rotational barrier of a C≡C bond is ≈12–14 kcal mol⁻¹ for a free alkyne (i.e., rotation of the whole molecule).
In but‑2‑yne, we are not rotating the triple bond itself; we are rotating the σ bonds attached to the sp carbons (the C–C bonds to the methyl groups). Those σ bonds can rotate freely because the sp‑carbons are linear and the attached groups are small.

Thus, rotation about the C≡C axis is allowed, and the resulting conformational energy profile is the shallow sinusoid described above.

2.5. Visualising the conformers

      CH3                 CH3
       \                 /
        C≡C   ← axis →  C≡C
       /                 \
   CH3                 CH3

Eclipsed: the two C–C σ bonds are aligned (0°, 120°, 240°).
Staggered: the two C–C σ bonds are offset by 60° (60°, 180°, 300°).

Because the methyl groups are identical, all eclipsed positions are energetically equivalent, and all staggered positions are also equivalent.

2.6. Quantitative description of the torsional potential

A simple cosine Fourier series captures the potential (V(\phi)) as a function of the dihedral angle (\phi) (measured from an eclipsed reference):

[ V(\phi)=\frac{V_3}{2}\bigl[1-\cos (3\phi)\bigr] ]

The factor 3 reflects the 120° periodicity (threefold symmetry) of the eclipsed minima.
The fitted amplitude for but‑2‑yne is

[ V_3 \approx 0.12\;\text{kcal mol}^{-1}. ]

The minimum (staggered) occurs at (\phi = 60^\circ, 180^\circ, 300^\circ) and the maximum (eclipsed) at (\phi = 0^\circ, 120^\circ, 240^\circ).

3. Final answer (concise)

“Modulated at three angles” means that because the two π‑orbitals of the C≡C bond repeat every 120°, there are three distinct eclipsed conformations in a full 360° rotation.
The staggered conformations repeat every 60°, giving six minima per 360°.
The eclipsed conformer is higher in energy, not lower; the experimental torsional barrier for but‑2‑yne is ≈0.12 kcal mol⁻¹, i.e. the staggered form is more stable by that amount.
Rotation about the C≡C σ‑bond is allowed; the π‑system fixes the orientation of the p‑orbitals but does not stop the attached σ‑bonds (the C–C bonds to the methyl groups) from rotating. The barrier is small because the substituents are attached to linear sp‑carbons.

4. Common mistakes for this type of problem

Mistake	Why it’s wrong	How to avoid it
Assuming the triple bond locks the molecule completely	The π‑bonds are fixed, but the σ bond can still rotate; only the overall C≡C rotation is high‑energy.	Remember the distinction σ‑rotation vs. π‑orientation.
Thinking eclipsed is always more stable	In most alkanes eclipsed is less stable because of steric and hyperconjugation repulsion; the same holds for alkynes, though the penalty is tiny.	Compare energies from experimental barriers or quantum calculations, not just intuition.
Confusing “three angles” with “three bonds”	The phrase refers to the periodicity of the π‑orbital pattern (120°), not to the number of bonds.	Visualise the p‑orbital “ring” and count how many rotations bring it back onto itself.
Using the butane torsional barrier (≈3 kcal mol⁻¹) for but‑2‑yne	Substituents on sp‑carbons are much smaller and have less steric interaction; the barrier is an order of magnitude lower.	Look up or calculate the specific barrier for the molecule you are studying.
Neglecting symmetry when counting distinct conformers	Symmetry reduces the number of unique conformations; in but‑2‑yne the two methyl groups are identical, so many positions are equivalent.	Apply symmetry operations (rotation, reflection) before counting distinct minima.

Original question: Conformations of Butyne on Chemistry Stack Exchange, licensed CC BY-SA.