Introduction to stereoisomerism

Isomers are compounds sharing one molecular formula but differing in properties. The syllabus splits them into two broad families that answer two different questions.

Structural isomers answer "which atoms are bonded to which?" differently — chain, position, functional-group and metameric isomers are all built by redrawing the connectivity itself. You can tell structural isomers apart just by writing different flat (two-dimensional) structural formulas; no 3D picture is needed.

Stereoisomers answer that question identically — same connectivity, same flat formula — but differ in a second, independent question: how are the atoms or groups arranged in space around a bond or a carbon centre? A flat structural formula alone cannot distinguish two stereoisomers; you need to picture the molecule in three dimensions. This distinction is the single most important idea in the chapter, and it is worth pinning down as a compact rule before going further:

where: "connectivity" means the specific sequence of atom-to-atom bonds — which atom is directly attached to which. Two structures with the same connectivity are, chemically, the same skeleton drawn in two different spatial poses; two structures with different connectivity are genuinely different skeletons. At this level, stereoisomerism has exactly two named types, and each has its own structural trigger:

• Geometrical (cis–trans) isomerism is triggered by a carbon–carbon double bond. The $\pi$ bond of a C=C locks the two carbons and their substituents in a flat, rigid plane — unlike a C–C single bond, which allows free rotation. If each doubly-bonded carbon carries two different groups, two distinct, non-interconvertible arrangements exist: the cis form (similar groups on the same side of the double bond) and the trans form (similar groups on opposite sides).
• Optical isomerism is triggered by a molecule being non-superimposable on its own mirror image — the way a left hand and a right hand are mirror images that can never be rotated to coincide. Such a non-superimposable pair is called a pair of enantiomers. The commonest structural feature that produces this at this level is a carbon atom attached to four different groups.

Both routes share the same underlying logic: something in the molecule's shape prevents two spatial arrangements from being interconverted by simple rotation, so the two arrangements survive as genuinely different, isolable compounds — even though a flat structural formula would draw them identically.

This chapter's core claim has no numeric formula to derive — it is a structural, geometric argument — so the reasoning itself is the thing to master, in two parts.

Part 1 — why a double bond can freeze cis and trans apart. A single C–C bond is cylindrically symmetric along its axis, so the groups on either carbon can rotate freely around it without changing the molecule's identity — free rotation constantly interconverts every possible orientation, so there is nothing to "freeze." A C=C double bond is different: it is built from one $\sigma$ bond plus one $\pi$ bond, and the $\pi$ bond is formed by sideways overlap of unhybridised p-orbitals above and below the plane of the two carbons. Rotating one carbon relative to the other would have to break that sideways overlap — effectively breaking the $\pi$ bond — which costs far more energy than ordinary thermal motion supplies at room temperature. So the two carbons and their four attached groups are locked into one flat, rigid plane. If both carbons happen to carry two identical substituents each, swapping "sides" changes nothing new — but the moment each carbon carries two genuinely different groups, "same side" and "opposite sides" become two distinct, non-interconvertible molecules: cis and trans.

Part 2 — why four different groups make a mirror image "un-turnable." Any object and its mirror image can always be superimposed by rotation unless the object itself has no internal symmetry to exploit — this is exactly the everyday fact that a plain ball's mirror image is identical to the ball, but a hand's mirror image (the other hand) can never be rotated to match. A carbon atom with two or more identical attached groups has an internal mirror symmetry of its own, so rotating its mirror image back onto the original is always possible. A carbon atom bonded to four different groups has no such internal symmetry — nothing to pivot around — so its mirror image is a genuinely different spatial object, non-superimposable on the original no matter how it is rotated. That pair of non-superimposable mirror images is a pair of enantiomers, and the molecule is said to show optical isomerism.

Problem. But-2-ene, $\text{C}_4\text{H}_8$, is written as the single flat formula $\text{CH}_3\text{–CH=CH–CH}_3$. Show that this one formula actually corresponds to two different compounds, name the isomerism involved, and then contrast it with 2-chlorobutane, $\text{CH}_3\text{–CHCl–CH}_2\text{–CH}_3$, to show why that molecule shows a different type of stereoisomerism instead.

Step 1. Look at each doubly-bonded carbon of but-2-ene in turn. The left carbon carries one $\text{CH}_3$ group and one H atom; the right carbon also carries one $\text{CH}_3$ group and one H atom. Each carbon has two genuinely different substituents — exactly the trigger condition for geometrical isomerism identified above.

Step 2. Because the $\pi$ bond forbids free rotation, the two $\text{CH}_3$ groups are locked either on the same side of the double bond or on opposite sides, and thermal motion at ordinary temperatures cannot flip one into the other. Same side gives cis-but-2-ene; opposite sides gives trans-but-2-ene. Both share the identical formula $\text{C}_4\text{H}_8$ and identical connectivity, so they are geometrical stereoisomers, not structural isomers — one flat formula, two real compounds with different melting and boiling points.

Step 3. Now examine 2-chlorobutane's second carbon: it is bonded to Cl, H, a $\text{CH}_3$ group, and a $\text{CH}_2\text{CH}_3$ group — four genuinely different groups, with no C=C bond anywhere nearby. This carbon has no internal symmetry, so by the Part 2 argument its mirror image cannot be rotated back onto the original.

Step 4. Conclusion: but-2-ene shows geometrical isomerism (cis/trans, from a restricted double bond), while 2-chlorobutane shows optical isomerism (a pair of enantiomers, from a four-different-groups carbon) — two structurally distinct triggers, both falling under the umbrella of stereoisomerism, neither requiring R/S labelling at this stage.

Competitive stretch

For 1-chlorobutane, $\text{CH}_3\text{–CH}_2\text{–CH}_2\text{–CH}_2\text{Cl}$, the chlorine-bearing carbon carries two hydrogens, one Cl, and one propyl chain — only three distinct groups, since the two H atoms repeat. By the Part 2 rule, this carbon retains an internal mirror symmetry (swap the two identical H atoms and nothing changes), so its mirror image can always be rotated back onto the original. 1-chlorobutane shows no optical isomerism, unlike its isomer 2-chlorobutane — a reminder that the four-different-groups test must be checked group by group, not assumed from "there's a chlorine on a carbon."

• Calling geometrical isomers "structural isomers" because their molecular formula looks the same — stereoisomers share identical connectivity; only spatial arrangement differs, so the flat formula alone will not reveal them.
• Assuming any C=C bond automatically gives cis–trans isomers — it needs each doubly-bonded carbon to carry two different substituents; $\text{CH}_2\text{=CHCl}$ has one carbon with two identical H atoms, so it shows no geometrical isomerism.
• Trying to apply R/S (CIP priority) rules, name enantiomers "D" or "L", or predict specific rotation at this stage — the syllabus asks only for awareness that optical isomerism exists and recognition of the four-different-groups condition.
• Treating any carbon in a branched molecule as automatically "chiral-type" — all four attached groups must be genuinely different; repeat groups (like two H atoms) restore the internal symmetry and rule out optical isomerism at that carbon.

Ask the tutor: want to try a few more molecules and decide together whether each one shows geometrical isomerism, optical isomerism, or neither?