A2 Chemistry 9701: E/Z Isomerism and CIP Priority Rules

Compound Z is 2-bromobut-2-ene, $\text{CH}_3\text{C(Br)}=\text{CHCH}_3$ , numbering the four carbons C1 to C4 along the chain so that the double bond lies between C2 and C3. This compound exists as a pair of E/Z stereoisomers.

(a) State the Cahn-Ingold-Prelog (CIP) priority rule used to rank the two substituents attached to each carbon of a stereogenic C=C double bond, and use it to identify the higher-priority group on C2 and the higher-priority group on C3 of compound Z. [3]

(b) In one particular sample of compound Z, the bromine atom on C2 and the methyl group attached to C3 lie on opposite sides of the double bond. Using your answer to part (a), determine whether this sample is the E isomer or the Z isomer, explaining your reasoning. [2]

(c) But-2-ene, $\text{CH}_3\text{CH}=\text{CHCH}_3$ , can also be described using either cis/trans notation or E/Z notation. State, with a reason, whether the two notations give equivalent information for but-2-ene specifically. [2]

Part (a): The CIP priority rule, applied to C2 and C3 of compound Z

For a double-bond carbon bearing two different substituents, the Cahn-Ingold-Prelog (CIP) priority rule ranks the two substituents by comparing the atomic number of the atom directly bonded to that double-bond carbon: whichever substituent’s directly-attached atom has the higher atomic number is assigned the higher priority.

$\text{CH}_3-\underset{\text{C2}}{\text{C}}(\text{Br})=\underset{\text{C3}}{\text{C}}\text{H}-\text{CH}_3$

C2 carries a bromine atom and a methyl group (attached via its carbon atom). Comparing the directly-attached atoms: bromine has atomic number 35, carbon has atomic number 6. Since $35 > 6$ , the bromine atom is the higher-priority group on C2.
C3 carries a hydrogen atom and a methyl group (attached via its carbon atom). Comparing the directly-attached atoms: carbon has atomic number 6, hydrogen has atomic number 1. Since $6 > 1$ , the methyl group is the higher-priority group on C3.

Part (b): Assigning E or Z to the described sample

The two higher-priority groups identified in part (a) are:

on C2: the bromine atom
on C3: the methyl group

The question states that the bromine atom (on C2) and the methyl group attached to C3 lie on opposite sides of the double bond. Since these are precisely the two higher-priority groups identified above, this means the higher-priority group on C2 and the higher-priority group on C3 are on opposite sides of the double bond.

By definition, when the two higher-priority groups (one on each double-bond carbon) are on opposite sides, the isomer is labelled E (from the German entgegen, meaning “opposite”). If they were on the same side instead, the isomer would be labelled Z (zusammen, “together”).

This sample is therefore the E isomer, correctly named (E)-2-bromobut-2-ene.

Part (c): Comparing E/Z and cis/trans notation for but-2-ene

$\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3$

In but-2-ene, both double-bond carbons carry exactly the same pair of substituent types: one hydrogen atom and one methyl group. Applying the CIP rule to each carbon, carbon ( $Z=6$ ) always outranks hydrogen ( $Z=1$ ), so the higher-priority group at every double-bond carbon in this molecule is simply “the group that is not hydrogen” (the methyl group. This is exactly the same reference group that cis/trans notation uses (cis = the two named, larger groups) here, the two methyl groups, on the same side; trans = on opposite sides).

Because the CIP higher-priority group and the “cis/trans reference group” coincide for every carbon in this particular molecule, the two notations give identical (equivalent) information for but-2-ene: cis-but-2-ene is the same compound as Z-but-2-ene, and trans-but-2-ene is the same compound as E-but-2-ene. This equivalence is not guaranteed in general, it relies on each double-bond carbon carrying only one hydrogen and one other substituent, which is why E/Z notation, based on CIP priority, is the more generally applicable system overall.

Final answers

(a) Rank by atomic number of the directly-attached atom; higher priority on C2 is Br, higher priority on C3 is the methyl group
(b) E isomer. The two higher-priority groups (Br on C2, methyl on C3) are on opposite sides
(c) Equivalent for but-2-ene: cis = Z, trans = E, because each carbon carries only one H and one non-H group