Organic Chemistry · Structure and Bonding

sp3 Hybrid Orbitals and the Structure of Ethane

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On this page 9 sections

In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

Ethane (CH3CH3) is the second member of the alkane family and the direct test of the sp³ model built for methane in Topic 6. Two tetrahedral sp³ carbons join through a C–C σ bond formed by head-on overlap of one sp³ orbital from each carbon; every remaining sp³ orbital overlaps a hydrogen 1s orbital to make a C–H σ bond. All bond angles are close to 109.5°, the C–C bond is 1.54 Å, and the molecule contains seven σ bonds in total. Because a σ bond is , the two CH3 groups can spin relative to each other without breaking the bond — giving conformations that differ in energy. Viewing down the C–C axis (a ) shows two extreme arrangements: (C–H bonds spread apart, lower energy) and (C–H bonds lined up, higher energy). The energy cost of eclipsing is about 12 kJ/mol (≈ 3 kcal/mol), a small but measurable barrier called .

Why this matters

Conformational analysis — the study of shapes that interconvert by bond rotation without breaking any bonds — begins with ethane and never stops being useful. It reappears in cyclohexane ring flipping and chair conformations (Chapter 4), in the staggered arrangements of protein backbones and polymer chains, and whenever a drug must adopt a particular shape to fit a receptor. Ethane also establishes the pattern that single bonds rotate freely while double bonds do not (Topic 8) — a distinction that underlies stereochemistry (Chapter 5). Newman projections are a standard exam tool for representing three-dimensional shape on a two-dimensional page, so learning to read and draw them now pays off repeatedly.

The college version

Core Concepts

Building ethane from two sp³ carbons

Each carbon in ethane is sp³: it carries four electron groups (one C–C σ bond plus three C–H σ bonds) and no lone pairs. The carbon–carbon bond forms by sp³–sp³ overlap along the C–C axis, and each carbon keeps three sp³ orbitals for its C–H bonds. The result is CH3CH3 with seven σ bonds total: six C–H plus one C–C.

The C–C σ bond allows rotation

Because σ-bond electron density is cylindrically symmetric around the C–C axis, rotating one methyl group relative to the other does not change the overlap. The bond stays intact as the molecule turns — this is why chemists say rotation about single bonds is "free," meaning the bond itself never breaks.

Newman projections

To compare rotational arrangements, look straight down the C–C bond. The front carbon appears as a dot with three lines (its three C–H bonds); the back carbon appears as a circle with three lines. The angle between a front C–H bond and the nearest back C–H bond is the . A dihedral angle of 60° defines the staggered ; 0° defines the eclipsed conformation.

Staggered vs. eclipsed: torsional strain

In the staggered conformation, front and back C–H bonds are as far apart as possible. In the eclipsed conformation they line up, and their bonding electrons experience extra repulsion — torsional strain. Staggered ethane is therefore the energy minimum and eclipsed ethane the maximum, about 12 kJ/mol (≈ 2.9 kcal/mol) higher. At room temperature, ethane rotates rapidly through all angles, spending most of its time near the staggered arrangement.

Bond lengths and angles

Ethane's C–C bond is about 1.54 Å and its C–H bonds about 1.10 Å, with all angles near 109.5°. The C–C bond is longer than the C–H bonds because it joins two large sp³ lobes rather than one sp³ lobe and a small hydrogen 1s orbital.

How It Works / Step-by-Step Process

Drawing and comparing Newman projections:

Choose the view: look along the C1–C2 bond axis.
Draw the front carbon as a dot and add three lines at 120° for its C–H bonds.
Draw the back carbon as a circle and add three lines at 120° for its C–H bonds.
Measure the dihedral angle between a front bond and the nearest back bond.
If it is 60°, the conformation is staggered; if 0°, it is eclipsed.
Compare energies: staggered is lower; eclipsed is higher by about 12 kJ/mol.

Common Confusions

Do Not Confuse	With	The Difference
"Free rotation"	Costless rotation	Rotation happens, but eclipsed is ≈ 12 kJ/mol higher in energy than staggered
Staggered	Eclipsed	Staggered (60°) is the low-energy minimum; eclipsed (0°) is the maximum
Conformer	Constitutional isomer	Conformers share identical connectivity and interconvert by rotation; isomers differ in connectivity
Torsional strain	Steric strain	Torsional = repulsion between eclipsed bond electrons; steric = through-space crowding of atoms or groups (later chapters)
Single bonds rotate	Double bonds rotate	The C=C π bond (Topic 8) blocks rotation about a double bond

Eli explains

The same idea, in plain words

Explain it like I’m 10

Imagine two bundles of pencils taped end to end. You can spin one bundle while the other stays still — the tape (the σ bond) does not break. But when the pencils of one bundle line up directly behind the pencils of the other, they crowd each other (eclipsed); when they sit in the gaps between each other's pencils, everyone has room (staggered). Staggered is the comfortable position, which is why the molecule prefers it.

Worked example

Example 1: Converting the rotation barrier (dimensional analysis)

The ethane rotation barrier is about 12 kJ/mol. Express it in kcal/mol using 1 kcal = 4.184 kJ. Set up the conversion so the units cancel:

$12 kJ/mol \times 1 kcal 4.184 kJ = 2.9 kcal/mol$

The eclipsing cost is therefore ≈ 2.9 kcal/mol — far smaller than a typical C–C bond energy (≈ 83 kcal/mol), which is why rotation happens freely at room temperature even though the eclipsed form costs energy.

Example 2: Converting the C–C bond length (dimensional analysis)

Ethane's C–C bond is 1.54 Å. Convert to picometers and nanometers using 1 Å = 100 pm and 1 nm = 1000 pm:

$1.54 Å \times 100 pm 1 Å = 154 pm$

$154 pm \times 1 nm 1000 pm = 0.154 nm$

Compare with the C–H bond (≈ 1.10 Å = 110 pm): the C–C bond is longer because it links two bulky sp³ lobes instead of one sp³ lobe and a small 1s orbital.

Example 3: Hybridization and σ-bond count in ethane

Assign hybridization to each carbon and count the total σ bonds. Each carbon holds one C–C σ bond plus three C–H σ bonds = four electron groups → sp³, tetrahedral. Total σ bonds in CH3CH3: 1 (C–C) + 6 (C–H) = 7. There are no π bonds and no lone pairs, so hybridization is complete: every orbital on both carbons is sp³.

Key takeaways

Both carbons of ethane are sp³ (four electron groups each, no lone pairs).
The C–C σ bond is sp³–sp³ head-on overlap; C–H σ bonds are sp³–1s; total σ bonds = 7.
σ bonds are cylindrically symmetric → rotation about single bonds is allowed.
Newman projection: view down the C–C axis; front carbon = dot, back carbon = circle.
Dihedral angles: 60° = staggered (low energy); 0° = eclipsed (high energy).
Torsional strain: eclipsing costs ≈ 12 kJ/mol (≈ 2.9 kcal/mol).
C–C ≈ 1.54 Å; C–H ≈ 1.10 Å; angles ≈ 109.5°.

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

How many electron groups surround each carbon in ethane, and what hybridization follows?
Show answer
Four electron groups per carbon (one C–C σ + three C–H σ), so each carbon is sp³ with tetrahedral geometry.
Why can the two CH3 groups rotate relative to each other without breaking the C–C bond?
Show answer
The C–C σ bond is cylindrically symmetric around the bond axis, so rotation does not change the orbital overlap and the bond never breaks.
In a Newman projection of ethane, what dihedral angle defines the staggered conformation?
Show answer
60°.
Which conformation of ethane is lower in energy, and what is the approximate energy difference?
Show answer
Staggered is lower in energy; eclipsed is about 12 kJ/mol (≈ 2.9 kcal/mol) higher because of torsional strain.
What is torsional strain?
Show answer
The extra energy that results from repulsion between the bonding electrons of eclipsed bonds (here, eclipsed C–H bonds).
Convert ethane's C–C bond length of 1.54 Å into picometers.
Show answer
1.54 Å × (100 pm / 1 Å) = 154 pm.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

conformation: Any 3-D arrangement reached by rotating about single bonds
Newman projection: Drawing viewed straight down a C–C bond (front dot, back circle)
staggered: Conformation with 60° dihedral angles; H's maximally apart
eclipsed: Conformation with 0° dihedral angles; H's lined up
dihedral angle: Angle between a front bond and the nearest back bond
torsional strain: Extra energy caused by repulsion between eclipsed bond electrons
cylindrically symmetric: Electron density spread evenly around a bond axis

Sources & references

openstax.org — Organic Chemistry

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