🥼Organic Chemistry Unit 7 Review

Alkene stereochemistry describes the spatial arrangement of substituents around a carbon-carbon double bond. Because double bonds prevent free rotation, the groups attached to each carbon are locked in place. The E,Z naming system gives us a way to specify which arrangement we're looking at, using priority rules to determine whether groups end up on the same side or opposite sides of the double bond.

E vs. Z Configurations in Alkenes

The E,Z system applies whenever both carbons of the double bond carry two different substituents. If either carbon has two identical groups, the molecule doesn't have E/Z stereochemistry at all.

Z (zusammen): The higher-priority substituent on each double-bond carbon is on the same side of the double bond.
E (entgegen): The higher-priority substituent on each double-bond carbon is on opposite sides of the double bond.

"Zusammen" is German for "together" and "entgegen" means "opposite." Z = same side, E = opposite side.

You might be tempted to just use cis/trans here. Cis and trans work fine for simple disubstituted alkenes, but they become ambiguous with three or four substituents. The E,Z system handles every case because it's based on priority rankings, not on visual intuition about which groups "match."

Cahn-Ingold-Prelog (CIP) Priority Rules

To assign E or Z, you need to rank the two substituents on each double-bond carbon from higher to lower priority. The Cahn-Ingold-Prelog (CIP) sequence rules tell you how:

Atomic number: Look at the atom directly attached to the double-bond carbon. Higher atomic number = higher priority. For example, $Br$ (35) > $Cl$ (17) > $O$ (8) > $N$ (7) > $C$ (6) > $H$ (1).
Isotopes: If two atoms are the same element, the heavier isotope wins. $^{14}C$ > $^{13}C$ > $^{12}C$ .
Multiple bonds: Treat a double or triple bond as if it were expanded into single bonds to duplicate "phantom" atoms. A $C=O$ is treated as if the carbon is bonded to two separate oxygen atoms (and the oxygen is bonded to two separate carbon atoms). A $C \equiv N$ is treated as if the carbon is bonded to three nitrogen atoms.
Tie-breaking by moving outward: If the first atoms are identical, move to the next set of atoms along each chain and compare again. Keep going until you find a point of difference.

A common mistake is listing priority by molecular weight of the whole substituent. That's wrong. You compare atom by atom, starting at the point of attachment.

E vs Z configurations in alkenes, Alkene - Wikipedia

Naming Trisubstituted and Tetrasubstituted Alkenes

Trisubstituted alkenes have three non-hydrogen substituents on the double bond (one carbon carries two different groups, the other carries one group plus a hydrogen).

To assign configuration:

On the carbon with two non-hydrogen substituents, rank them using CIP rules.
On the other carbon, hydrogen is automatically the lower-priority group.
Compare the higher-priority group on each carbon. If they're on the same side, it's Z. Opposite sides, it's E.

Example: In (E)-1-bromo-1-chloro-2-methylprop-1-ene, C1 carries $Br$ and $Cl$ . Bromine has a higher atomic number, so $Br$ is the higher-priority group on C1. On C2, the methyl group outranks hydrogen. Since $Br$ and the methyl group are on opposite sides, the configuration is E.

Tetrasubstituted alkenes have four non-hydrogen substituents (two different groups on each carbon of the double bond).

Rank the two substituents on C1 using CIP rules. Identify the higher-priority one.
Do the same for C2.
Compare the two higher-priority groups. Same side = Z, opposite sides = E.

Example: In (E)-2-bromo-3-methylbut-2-ene, you'd rank the substituents on each carbon separately, then check whether the winners sit on the same or opposite sides.

Note: A tetrasubstituted alkene with identical substituent pairs on both carbons (like $a,b$ on C1 and $a,b$ on C2) doesn't automatically become a meso compound. Meso compounds require internal mirror planes and chiral centers, which is a separate concept from E/Z designation. Don't confuse alkene stereoisomerism with chirality.

Ranking Substituents: Worked Examples

Applying CIP rules gets easier with practice. Here are a few comparisons:

$-Br$ vs. $-Cl$ : Bromine (atomic number 35) beats chlorine (17). Simple atomic number comparison.
$-I$ vs. $-Cl$ : Iodine (53) beats chlorine (17). So in (Z)-1-chloro-1-iodoprop-1-ene, iodine is the higher-priority group on that carbon.
$-CH_2OH$ vs. $-CH_3$ : The first atoms are both carbon (tie). Move to the next atoms: oxygen (8) in $-CH_2OH$ vs. hydrogen (1) in $-CH_3$ . Oxygen wins, so $-CH_2OH$ has higher priority.
$-CH=O$ (aldehyde) vs. $-CH_2OH$ : Expand the double bond in the aldehyde: the carbon "sees" $(O, O, H)$ . For $-CH_2OH$ , the carbon sees $(O, H, H)$ . At the first point of comparison both have oxygen, but at the second position the aldehyde has another phantom oxygen while $-CH_2OH$ has hydrogen. The aldehyde wins.

Stereoisomerism and Configurational Analysis

Stereoisomers share the same molecular formula and the same bonding sequence, but differ in the 3D arrangement of their atoms. E and Z isomers of alkenes are one type of stereoisomer.

These are specifically configurational isomers, meaning you cannot interconvert them without breaking the $\pi$ bond. Rotating around a double bond requires roughly 264 kJ/mol of energy, so E and Z isomers are stable, isolable compounds with distinct physical properties (different boiling points, melting points, dipole moments, and reactivities).

This is different from conformational isomers, which interconvert freely through rotation around single bonds and generally can't be isolated separately at room temperature. Don't mix up the two: E/Z isomers are locked in place, while conformations are constantly changing.

🥼Organic Chemistry Unit 7 Review