Topic 5 · page 3 of 15

Which alkenes have geometric isomers

Geometric isomers need a C=C and two different groups on each carbon of it. Test each carbon of the double bond separately, then draw the isomers with 120° bond angles.

13 min read5.2

Read first: Restricted rotation around C=C

The two conditions 5.2

Restricted rotation is not enough on its own. A molecule has geometric isomers only if both of these are true:

  1. it has a C=C double bond, so that rotation is restricted;
  2. each carbon of the C=C carries two different groups.

The second condition is about the substituents: the atoms or groups attached to the carbons of the C=C. If a carbon carries two identical groups, swapping them gives the same molecule, so nothing is changed by the swap.

Four alkenes tested for geometric isomers. But-2-ene has two different groups on each carbon of the C=C, so it has isomers. But-1-ene has two hydrogen atoms on one carbon, and 2-methylpropene has two CH3 groups on one carbon, so neither has isomers. 1,2-dichloroethene has chlorine and hydrogen on each carbon, so it has isomers.
Test the left carbon, then the right carbon. Both must pass.

MethodTesting an alkene for geometric isomers

  1. Draw the C=C and put the four groups on it at 120°.
  2. Look at the left carbon only. Are its two groups different? If not, stop: no isomers.
  3. Look at the right carbon only. Are its two groups different? If not, stop: no isomers.
  4. If both carbons pass, there are two geometric isomers.

Worked exampleDoes each alkene have geometric isomers?

  1. But-2-ene, CH₃CH=CHCH₃. Left carbon: CH₃ and H, different. Right carbon: CH₃ and H, different. Yes.
  2. But-1-ene, CH₂=CHCH₂CH₃. The left carbon has two H atoms, identical. No.
  3. 2-methylpropene, CH₂=C(CH₃)₂. Left carbon: two H. No (the right carbon has two CH₃ as well).
  4. 2-methylbut-2-ene, (CH₃)₂C=CHCH₃. The left carbon has two CH₃ groups. No.
  5. 3-methylpent-2-ene, CH₃CH=C(CH₃)CH₂CH₃. Left carbon: CH₃ and H. Right carbon: CH₃ and CH₂CH₃. Both pairs different. Yes.

Drawing and counting the isomers 5.2

To draw geometric isomers, put the C=C in the middle and draw the other four bonds at 120° to it, two on each carbon. Place the groups of the left carbon above and below the axis, then do the same on the right. The two isomers have the same group on top at the left; on the right, one has the matching group on top and the other has it on the bottom.

The two geometric isomers of pent-2-ene, CH3CH=CHCH2CH3. In (Z)-pent-2-ene the CH3 group and the C2H5 group, drawn in orange, are on the same side of the dashed line through the C=C. In (E)-pent-2-ene they are on opposite sides.
Pent-2-ene: the two isomers. The orange groups are the ones to compare.

The alkenes with formula C₄H₈ show how structural isomerism and geometric isomerism combine.

The four alkenes with molecular formula C4H8: but-1-ene, (Z)-but-2-ene, (E)-but-2-ene and 2-methylpropene. The two but-2-enes are geometric isomers of each other; the other pairs are structural isomers.
Four alkenes of formula C₄H₈: three structural isomers, one of which has two geometric forms.
Formula Structural isomers that are alkenes Alkenes counting geometric isomers too
C₄H₈ 3 (but-1-ene, but-2-ene, 2-methylpropene) 4
C₅H₁₀ 5 (pent-1-ene, pent-2-ene, 2-methylbut-1-ene, 3-methylbut-1-ene, 2-methylbut-2-ene) 6

Only the but-2-enes and pent-2-ene pass the two-condition test, so each of those adds one extra isomer.

Check yourself

  1. State the two conditions needed for a molecule to have geometric isomers.
  2. Which of these has geometric isomers? Give a reason for each: (a) propene, (b) pent-2-ene, (c) 2-methylbut-1-ene, (d) 1,2-dichloroethene, (e) 1,1-dichloroethene.
  3. Draw the two geometric isomers of hex-3-ene, CH₃CH₂CH=CHCH₂CH₃.
  4. Write down all the alkenes with the molecular formula C₅H₁₀ that do not have geometric isomers, and say how many alkenes with this formula there are in total.
  5. A student says "but-1-ene has no geometric isomers because it has no CH₃ groups on the double bond". Explain what is wrong with this reason.

Answers cover these until you have tried the questions

  1. A C=C double bond (restricted rotation) and two different groups on each carbon of the C=C.
  2. (a) No: CH₂= has two H on one carbon. (b) Yes: CH₃ and H on one carbon, C₂H₅ and H on the other. (c) No: CH₂= has two H on one carbon. (d) Yes: Cl and H on each carbon. (e) No: both Cl are on the same carbon (and both H on the other).
  3. Each carbon of the C=C has H and C₂H₅. In the Z isomer both C₂H₅ groups are on the same side of the C=C (and both H on the other side); in the E isomer the two C₂H₅ groups are on opposite sides. Each isomer is drawn with the C=C at the centre and 120° bonds to H and CH₂CH₃.
  4. Without geometric isomers: pent-1-ene, 2-methylbut-1-ene, 3-methylbut-1-ene, 2-methylbut-2-ene (four). Pent-2-ene is the one that has isomers (Z and E), so the total is 4 + 2 = 6 alkenes.
  5. The reason is not about CH₃ groups. But-1-ene has no isomers because its first carbon has two identical groups (two H atoms). The test is whether each carbon of the C=C has two different groups.