Chirality and optical activity — R/S configuration using CIP rules

Question

A compound has the molecular formula $\text{C}_4\text{H}_8\text{BrCl}$ with one chiral centre. The chiral carbon is bonded to $-\text{H}$ , $-\text{CH}_3$ , $-\text{Br}$ , and $-\text{Cl}$ . Assign the R/S configuration using CIP priority rules. Will this compound show optical activity?

(JEE Advanced 2022, similar pattern)

Solution — Step by Step

A chiral centre (stereogenic centre) is a carbon bonded to four different groups. Here the carbon carries $-\text{H}$ , $-\text{CH}_3$ , $-\text{Br}$ , and $-\text{Cl}$ — all four are different, so this carbon is chiral.

Since there’s one chiral centre and no internal plane of symmetry, the compound will show optical activity.

The Cahn-Ingold-Prelog rules assign priority by the atomic number of the atom directly attached to the chiral centre:

$\text{Br}$ (Z = 35) → Priority 1
$\text{Cl}$ (Z = 17) → Priority 2
$\text{CH}_3$ (Z = 6) → Priority 3
$\text{H}$ (Z = 1) → Priority 4

When two atoms have the same atomic number, we move outward to the next set of atoms. Here all four are different at the first atom itself, so no tie-breaking needed.

Place the group with the lowest priority (H, priority 4) pointing away from you — imagine it’s going behind the paper. Now look at the remaining three groups from the front.

Starting from priority 1 (Br) → priority 2 (Cl) → priority 3 ( $\text{CH}_3$ ):

If the path traces a clockwise direction → R (Latin: rectus, right)
If the path traces an anticlockwise direction → S (Latin: sinister, left)

With Br at top, Cl at right, and $\text{CH}_3$ at left, the trace is anticlockwise → S configuration.

Why This Works

Chirality arises when a molecule is non-superimposable on its mirror image — like your left and right hands. The CIP system gives us a universal, unambiguous way to name each enantiomer.

The priority assignment is purely based on atomic number because heavier atoms exert a stronger “pull” in terms of spatial arrangement naming. When we orient the lowest-priority group away and trace the remaining three, we’re essentially reading the spatial handedness of the molecule.

Any compound with at least one chiral centre (and no compensating symmetry like a meso compound) will rotate the plane of polarised light — that’s optical activity.

Alternative Method — The Swap Trick

If you can’t visualise putting H at the back, use the swap technique:

Assign priorities normally: Br(1), Cl(2), $\text{CH}_3$ (3), H(4).
Look at the molecule as drawn. Trace 1→2→3. Suppose it appears clockwise.
Count how many swaps are needed to move H to the back position.
Odd number of swaps → flip the answer (R becomes S, S becomes R). Even swaps → answer stays.

JEE Advanced loves to give Fischer projections and ask for R/S. In a Fischer projection, if the lowest priority group is on a vertical line, the direct reading is correct. If it’s on a horizontal line, flip the answer. This shortcut saves over a minute per question.

Common Mistake

The biggest blunder: students forget to place the lowest priority group at the back before tracing 1→2→3. If H is pointing towards you and you read the rotation directly, you’ll get the opposite configuration. Always check: where is the lowest priority group? If it’s facing you, flip your answer.

Another frequent error in JEE: confusing optical activity with chirality in meso compounds. A meso compound has chiral centres but an internal plane of symmetry — it is optically inactive despite having chiral carbons. Always check for symmetry elements before concluding.

Question

Solution — Step by Step

Why This Works

Alternative Method — The Swap Trick

Common Mistake

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