Take ethane, CH₃–CH₃, and turn one CH₃ group around the C–C bond. Bonding showed that nothing stops this turn: the head-on overlap of a single bond looks the same from every angle around its axis. But free rotation does not mean that every arrangement is equally good. Some arrangements cost more energy than others, and a molecule spends most of its time in the low-energy ones. This page finds those arrangements, starting with the drawing that makes them easy to see.
What a conformation is
Each arrangement that ethane passes through as it turns is a conformation. A conformation is one spatial arrangement of a molecule, and you get from one conformation to another only by rotating about single bonds. No bond breaks along the way.
So the conformations of one molecule are not different compounds. At room temperature they turn into one another so fast that no one can separate them in a bottle.
Some pairs of molecules have the same atoms, joined in the same order, yet cannot turn into each other without breaking a bond. Those are genuinely different substances, and Stereochemistry covers them. Keep that apart from a change of conformation, which takes only a turn about a single bond.
How to draw a Newman projection
Hold an ethane molecule so that its C–C bond points straight at your eye. The nearer carbon now hides the one behind it. What you see is a Newman projection: the view straight down one C–C bond, the one whose rotation you want to study.
The nearer carbon is the front carbon. It is drawn as a dot, and its three other bonds are lines that meet at the dot. The farther carbon is the back carbon. It is drawn as a circle, and its three other bonds start at the edge of the circle. The C–C bond itself points at your eye, so it gets no line of its own: the dot and the circle are its two ends.
Each carbon is tetrahedral, as in Molecular geometry. Seen end-on, its three other bonds look evenly spread around the circle, so they are drawn 120° apart.
Butane, CH₃CH₂CH₂CH₃, is a good place to learn the habit that prevents most mistakes. Pick the bond first, here C2–C3. Then mark on the structure which end is front and which is back, and list what each of those two carbons carries besides the bond you are looking down. Only then draw. Most Newman errors are not drawing errors; they come from putting a group on the wrong carbon.
Staggered versus eclipsed
Look at ethane's Newman projection and pick one front H and the back H nearest to it. The angle between their two bonds, as the projection shows it, is the dihedral angle, also called the torsion angle. Turning the back carbon changes it.
When the dihedral angle is 60°, every back bond sits in the middle of the gap between two front bonds. The front and back hydrogens are then as far apart as they can get. This arrangement is the staggered conformation.
Turn the back carbon another 60° and the dihedral angle becomes 0°. Each back bond now lies directly behind a front bond. This arrangement is the eclipsed conformation.
An exact eclipse would hide the back bonds, so drawings turn them a few degrees to keep them visible. Read that small offset as 0°, not as "almost staggered". A staggered drawing puts each back bond squarely in the middle of a gap.
Why eclipsing costs energy
Eclipsed ethane is about 2.9 kcal/mol higher in energy than staggered ethane. This extra energy is torsional strain, the cost of lining up bonds on neighboring carbons. Eclipsed ethane has three H/H pairs lined up, so each pair costs about 1 kcal/mol.
Textbooks give two explanations for it. The traditional one is that the electrons in two eclipsed bonds repel each other. The one many chemists now favor is hyperconjugation, a small stabilizing overlap that is much stronger in the staggered form. Chemists still debate how much each contributes. Either way, the prediction is the same, and the cost is called torsional strain.
Hyperconjugation rests on one idea from Orbitals. Two orbitals that combine in phase make a bond. The same two orbitals combined in opposite phase make an antibonding orbital, which is higher in energy. So every C–H bond has an antibonding partner, written σ* ("sigma star"). In ethane every σ* is empty, because all the electrons sit in the bonds.
In staggered ethane, each C–H bond on one carbon points the opposite way from one C–H bond on the other carbon. That lines the filled C–H bond up with the empty σ* of the opposite C–H bond, and a little electron density spreads from one into the other. Electrons spread over more space are lower in energy, so the molecule is stabilized. Eclipsing turns the bonds out of line and loses most of that stabilization.
This 2.9 kcal/mol is ethane's rotational barrier, the energy it must gain to turn from staggered through eclipsed. It is small: ethane crosses it some tens of billions of times a second at room temperature. The barrier is still large enough to measure, and large enough that at any instant most ethane molecules are close to staggered.
Butane: four named conformations
Ethane's three staggered conformations all look the same, because every group on both carbons is an H. Butane is different. Look down its central C2–C3 bond. The front carbon carries a CH₃ and two H, and so does the back carbon. As the back carbon turns through 360°, the two methyl groups pass through four arrangements with different energies. The energies below are in kcal/mol, measured from the lowest one.
| Conformation | Dihedral | Type | Energy |
|---|---|---|---|
| anti | 180° | staggered | 0 (reference) |
| gauche | 60° | staggered | +0.9 |
| eclipsed (Me/H) | 120° | eclipsed | +3.6 |
| syn (Me/Me) | 0° | eclipsed | +4.5 to 6 |
Each name says where the two methyl groups sit. In the anti conformation they point in opposite directions, 180° apart. In a gauche conformation they are 60° apart, and the molecule is still staggered. In the methyl/hydrogen eclipsed conformation, each methyl lines up with an H. In the syn conformation the two methyls eclipse each other at 0°; some books call it the totally eclipsed conformation.
One full turn passes through two gauche conformations with the same energy, and through two methyl/hydrogen eclipsed ones. Anti is the lowest point of all, the global minimum. At room temperature butane spends roughly 70% of its time anti, and most of the rest in the two gauche forms.
Gauche butane sits 0.9 kcal/mol above anti, even though it is staggered. The cost cannot be torsional strain, because nothing is eclipsed. At 60° apart, the two methyl groups are simply close enough to bump into each other. That cost is steric strain: the energy of groups crowding into the same space. Gauche butane has steric strain with no torsional strain mixed in, so it is the cleanest place to see the difference between the two.
The names anti and gauche are not limited to methyl groups. They describe the dihedral angle between any two groups you choose to track, usually the two largest. 1,2-Dibromoethane, BrCH₂–CH₂Br, has an anti conformation, with Br 180° from Br, and two gauche ones, just as butane does. A molecule with more substituents can be described the same way, once you say which pair you are tracking.
Estimating any conformer: the per-interaction values
Butane's four energies are not four separate facts to memorize. Each one is a sum of a few interaction costs, and knowing the costs lets you estimate a conformer you have never seen. That is what an exam question asking "how much strain is in this drawing" really tests.
| Interaction | Kind | Cost (kcal/mol) |
|---|---|---|
| H/H eclipsed | torsional | 1.0 |
| H/CH₃ eclipsed | torsional + steric | 1.4 |
| CH₃/CH₃ eclipsed | torsional + steric | 2.5–3 |
| CH₃/CH₃ gauche | steric only | 0.9 |
Check the costs against butane. Its methyl/hydrogen eclipsed conformation has one CH₃/H eclipse and two H/H eclipses: 1.4 + 1.0 + 1.0 = 3.4, close to the measured 3.6. The syn conformation has one CH₃/CH₃ eclipse and two H/H eclipses: 2.5–3 + 2.0 = 4.5–5, inside the measured range of 4.5–6. Ethane has three H/H eclipses: 3 × 1.0 = 3.0, close to its measured 2.9. The scheme is only approximate, but for simple chains like these it lands within a few tenths of a kcal/mol.
The costs also show something the butane totals hide. An eclipsed CH₃/CH₃ pair costs 2.5–3 kcal/mol, far more than the 1.0 of an eclipsed H/H pair. The extra is steric strain, which a methyl adds on top of the torsional strain whenever it eclipses another group. That is why syn is the highest point of the rotation and not just one more eclipse.
Draw 2-methylbutane as a Newman projection down the C2–C3 bond, then find its best conformer.
Step 1: name the two carbons. The chain is CH₃–CH(CH₃)–CH₂–CH₃. C2 is the carbon with the branch, and C3 is the CH₂ next to it. Put C2 in front, and mark both carbons on the structure before you draw anything.
Step 2: list what each carbon carries besides the C2–C3 bond. C2, in front, carries two CH₃ groups (C1 and the branch) and one H. C3, at the back, carries one CH₃ (C4) and two H.
Step 3: draw a starting conformer. Put the front groups on the dot: CH₃ at the top, CH₃ at the lower right and H at the lower left. Start the back carbon staggered, with its CH₃ at the upper right, between the two front methyls, and its two H at the bottom and the upper left. Call this conformer A.
Step 4: turn the back carbon and name what you get. Turn it 60° clockwise. The back CH₃ lands behind the lower-right front CH₃, and every other back bond lands behind a front bond too. This eclipsed conformer has one CH₃/CH₃ eclipse, one CH₃/H eclipse and one H/H eclipse, roughly 2.5–3 + 1.4 + 1.0 ≈ 5 kcal/mol. Turn another 60° and the molecule is staggered again, with the back CH₃ at the bottom, between the lower-right front CH₃ and the front H. Call this conformer B. One more 120° turn gives conformer C, with the back CH₃ at the upper left, between the top front CH₃ and the front H.
Step 5: pick the winner by counting. Skip the eclipsed forms: each carries torsional strain as well as any steric strain, so the best conformer is always a staggered one. Count the gauche CH₃/CH₃ pairs in each. In A the back methyl is 60° from both front methyls, so A has two gauche pairs (1.8 kcal/mol). B and C each have one (0.9). So B and C tie as the best conformers, each with a single gauche interaction. Unlike butane, 2-methylbutane has no conformer that escapes gauche strain entirely.
The reverse direction is the same bookkeeping run backward. Read the front carbon's three groups and the back carbon's three, write them as two connected carbons, and you have the whole structure.
Plot butane's energy against the dihedral angle between its two methyl groups, for one full turn. Energy, measured from anti, runs up the vertical axis. The angle runs along the horizontal axis from 0° to 360°.
The curve has three low points, or minima, and three high points, or maxima. The minima sit at 60° (gauche), 180° (anti) and 300° (the other gauche). The maxima sit at 0° (syn, the highest) and at 120° and 240° (methyl/hydrogen eclipsed).
The gap between the highest maximum and the lowest minimum, about 5 kcal/mol, is butane's rotational barrier. It is small enough to be crossed constantly at room temperature, and large enough that anti is clearly preferred.
What carries forward
Conformations look like a detail, but they decide how molecules behave. Enzymes, for example, bind one particular conformation of the molecule they act on.
The next page, Cyclohexanes, adds a third kind of strain, angle strain, to the torsional and steric strain met here; together the three make up ring strain. On a cyclohexane ring the gauche clash comes back in a new setting, the 1,3-diaxial interaction that Axial/equatorial covers.
The idea that pays off furthest ahead is the anti-periplanar arrangement. Anti means 180° apart, as in anti butane, and periplanar means that the two bonds lie in one plane. A reaction called E2, which makes a C=C double bond, needs the H and the group that leave, one on each of two neighboring carbons, to be anti-periplanar. A molecule can therefore react only from conformations that allow it, and Substitution & Elimination turns this drawing exercise into a prediction about a reaction.