Topic 5 · page 4 of 15

E–Z naming

The E–Z system names geometric isomers by the priority of the groups on each carbon of the C=C. It works for every alkene with isomers, including those where cis and trans break down.

14 min read5.3

Read first: Which alkenes have geometric isomers

cis and trans 5.3

The older way to name geometric isomers uses two words:

For but-2-ene this works well. The identical groups are the two CH₃ groups (the two H atoms are identical too). cis-but-2-ene has both CH₃ groups on the same side, and trans-but-2-ene has them on opposite sides.

The E–Z system 5.3

E–Z does not look for identical groups. It ranks the two groups on each carbon and compares the higher-priority ones.

MethodNaming an isomer with E–Z

  1. Look at the left carbon of the C=C. Of its two groups, the one whose first atom has the higher atomic number has the higher priority.
  2. Do the same, separately, for the right carbon.
  3. Are the two higher-priority groups on the same side of the C=C? Then the isomer is Z (from German zusammen, "together").
  4. Are they on opposite sides? Then it is E (from entgegen, "opposite").
  5. Write the letter in brackets in front of the name: (Z)-but-2-ene, (E)-but-2-ene.

The priority comes from the atomic number of the atom joined directly to the carbon of the C=C. The ones you need here are:

Atom H C F Cl Br
Atomic number 1 6 9 17 35
The E-Z system applied to but-2-ene. On each carbon of the C=C the group with the higher priority is drawn in orange. In (Z)-but-2-ene both orange CH3 groups are on the same side of the dashed line. In (E)-but-2-ene they are on opposite sides.
Higher-priority groups in orange. Same side: Z. Opposite sides: E.

In but-2-ene each carbon has CH₃ and H. The first atom of CH₃ is C (atomic number 6), which beats H (1). So the CH₃ group is the higher priority on both carbons, and Z means the two CH₃ groups are together. Here cis = Z and trans = E.

Worked exampleNaming the isomers of pent-2-ene

CH₃CH=CHCH₂CH₃ (see the figure on the previous page).

  1. Left carbon: CH₃ (first atom C, 6) and H (1). CH₃ is higher.
  2. Right carbon: CH₂CH₃ (first atom C, 6) and H (1). CH₂CH₃ is higher.
  3. Isomer with CH₃ and CH₂CH₃ on the same side: (Z)-pent-2-ene.
  4. Isomer with them on opposite sides: (E)-pent-2-ene.

When cis and trans break down 5.3

cis and trans need identical groups to compare. Two kinds of molecule cause trouble.

1. Four different groups. In 1-bromo-1-chloro-2-fluoroethene, the carbons carry Br, Cl, F and H. No group appears on both carbons, so there is nothing for "cis" or "trans" to refer to.

2. A pair that looks cis but is E. In 2-bromobut-2-ene, the two CH₃ groups are on the same side, so a student might say "cis" and then "Z". The E–Z rule gives a different answer, because the higher-priority group on the left carbon is Br, not CH₃.

Two alkenes where cis and trans fail. Left: 1-bromo-1-chloro-2-fluoroethene has four different groups, so no group appears on both carbons. Right: 2-bromobut-2-ene has its two CH3 groups on the same side, which looks cis, but the higher-priority groups, Br and the right-hand CH3, are on opposite sides, so it is E. Higher-priority groups are orange.
Left: Br and F are together, so Z. Right: Br and CH₃ are apart, so E, even though the two CH₃ groups look cis.

Worked exampleNaming two awkward alkenes

  1. 1-bromo-1-chloro-2-fluoroethene. Left carbon: Br (35) beats Cl (17). Right carbon: F (9) beats H (1). Br and F are on the same side: (Z)-1-bromo-1-chloro-2-fluoroethene.
  2. 2-bromobut-2-ene with the two CH₃ groups on the same side. Left carbon: Br (35) beats CH₃ (C, 6). Right carbon: CH₃ (6) beats H (1). Br and the right-hand CH₃ are on opposite sides: (E)-2-bromobut-2-ene.

Check yourself

  1. State the difference between the terms cis and trans.
  2. Name each isomer of 1,2-dichloroethene: (a) the Cl atoms on the same side; (b) the Cl atoms on opposite sides.
  3. Draw and name the two geometric isomers of hex-3-ene.
  4. Give the E or Z name of the isomer of CHBr=CHCl in which Br and Cl are on the same side.
  5. Explain why the cis–trans system cannot be used to name the isomers of CHBr=CFCl.
  6. A student names the isomer of 2-bromobut-2-ene with the two CH₃ groups on the same side as (Z)-2-bromobut-2-ene. Explain why this is wrong, and give the correct name.

Answers cover these until you have tried the questions

  1. In cis the two identical groups are on the same side of the C=C; in trans they are on opposite sides.
  2. Each carbon has Cl (17) and H (1). Cl is the higher priority on both. (a) Cl atoms together: (Z)-1,2-dichloroethene. (b) Cl atoms apart: (E)-1,2-dichloroethene.
  3. Each carbon of the C=C has C₂H₅ (first atom C, 6) and H (1), so C₂H₅ is the higher priority on both carbons. (Z)-hex-3-ene: both C₂H₅ groups on the same side. (E)-hex-3-ene: the two C₂H₅ groups on opposite sides. Draw each with 120° bonds around the C=C.
  4. Left carbon: Br (35) beats H (1). Right carbon: Cl (17) beats H (1). Br and Cl are together, so it is (Z)-1-bromo-2-chloroethene.
  5. The two carbons carry four different groups: Br and H on one, F and Cl on the other. No group appears on both carbons, so there are no "identical groups" to be on the same or opposite sides. E–Z uses priorities, so it still works.
  6. The priority is decided on each carbon. On the left carbon Br (35) beats CH₃ (C, 6), so the higher-priority groups are Br and the right-hand CH₃, which are on opposite sides. The correct name is (E)-2-bromobut-2-ene. The two CH₃ groups being on the same side is irrelevant to E–Z.