Identify geometrical isomers of 2-butene and assign E/Z

Question

Identify the geometrical isomers of but-2-ene ( $\text{CH}_3\text{CH}=\text{CHCH}_3$ ) and assign E/Z designation to each isomer.

Solution — Step by Step

For geometrical isomers (cis-trans isomers) to exist in an alkene, each carbon of the double bond must have two different groups attached to it.

But-2-ene: $\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3$

Carbon 2 (C2): attached to $\text{CH}_3$ and $\text{H}$ — two different groups ✓
Carbon 3 (C3): attached to $\text{CH}_3$ and $\text{H}$ — two different groups ✓

Both carbons of the double bond have different substituents — geometrical isomers exist.

Isomer 1 (cis / Z): Both methyl groups ( $\text{CH}_3$ ) on the same side of the double bond.

 CH₃   CH₃
   \  /
    C=C
   /  \
  H    H

Isomer 2 (trans / E): Methyl groups on opposite sides of the double bond.

 CH₃   H
   \  /
    C=C
   /  \
  H   CH₃

These two structures are non-superimposable — they are different compounds (geometrical isomers) with different physical properties.

CIP (Cahn-Ingold-Prelog) Rules:

Assign priority to each substituent on EACH double-bond carbon based on atomic number (higher atomic number = higher priority).
If the two higher priority groups are on the same side → Z (German: zusammen = together).
If the two higher priority groups are on opposite sides → E (German: entgegen = opposite).

For but-2-ene:

Each double-bond carbon has: $\text{CH}_3$ (carbon attached to 3 H) and H.

Priority: $\text{CH}_3 > \text{H}$ (carbon has higher atomic number than hydrogen).

So on each carbon, $\text{CH}_3$ is priority 1 and H is priority 2.

Isomer 1: Both $\text{CH}_3$ groups are on the same side → Z-but-2-ene (also called cis-but-2-ene)

Isomer 2: $\text{CH}_3$ groups are on opposite sides → E-but-2-ene (also called trans-but-2-ene)

Property	Z-but-2-ene (cis)	E-but-2-ene (trans)
Boiling point	3.7°C	0.9°C
Melting point	−139°C	−106°C
Dipole moment	0.33 D (non-zero)	0 D (zero, symmetric)
Stability	Less stable	More stable (groups farther apart)

Cis isomer has a small dipole moment (methyl groups both on same side, creating a slight asymmetry). Trans isomer is symmetric — zero dipole moment.

Why This Works

The C=C double bond consists of a sigma ( $\sigma$ ) bond and a pi ( $\pi$ ) bond. The pi bond restricts rotation around the double bond (unlike single bonds, which can rotate freely). This restricted rotation is what makes geometric isomers possible — the arrangement of groups is “locked.”

If free rotation were possible, cis and trans would interconvert and couldn’t be isolated as separate compounds.

E/Z notation is more general than cis/trans. Cis/trans works only when the same groups are compared. E/Z uses CIP priorities, working even for asymmetric alkenes like $\text{ClCH}=\text{CH-Br}$ where “same groups” comparison fails.

Alternative Method — Cis/Trans vs E/Z

For but-2-ene (symmetric around the double bond), cis/trans = Z/E:

cis = both identical groups on same side = Z
trans = identical groups on opposite sides = E

But for asymmetric alkenes like $\text{Cl}-\text{CH}=\text{C}(\text{Br})(\text{CH}_3)$ , you must use CIP rules:

On C2: Cl (priority 1) vs H (priority 2). On C3: Br (priority 1) vs $\text{CH}_3$ (priority 2).

Determine if the two priority-1 groups are on same side (Z) or opposite sides (E).

Common Mistake

Applying E/Z based on hydrogen positions instead of the higher priority groups. The rule is: if the high-priority groups are on the same side → Z. Some students check if H atoms are on the same side and get E/Z backwards. Also: don’t confuse “Z = cis” universally — this shortcut only works when the two substituents are the same on each carbon.

Question

Solution — Step by Step

Why This Works

Alternative Method — Cis/Trans vs E/Z

Common Mistake

Try These Next