Q.The molecule (A) is a 2-bromobutane drawn as a wedge-and-dash structure at its chiral carbon: the H atom points straight up in the plane of the paper, the ethyl group (C2H5) points to the left in the plane, the methyl group (CH3) is on a dashed bond going behind the plane, and the Br atom is on a bold wedge coming toward the viewer. Each candidate below is the same 2-bromobutane (the four groups H, CH3, C2H5 and Br on one chiral carbon) drawn in a different three-dimensional arrangement:
Structure (i): H up in the plane, CH3 to the left in the plane, C2H5 on the dashed (rear) bond, and Br on the bold wedge (front).
Structure (ii): CH3 up in the plane, Br to the left in the plane, H on the dashed (rear) bond, and C2H5 on the bold wedge (front).
Structure (iii): H up in the plane, CH3 to the left in the plane, Br on the dashed (rear) bond, and C2H5 on the bold wedge (front).
Structure (iv): Br up in the plane, C2H5 to the left in the plane, H on the dashed (rear) bond, and CH3 on the bold wedge (front).
Which structure is the enantiomer (non-superimposable mirror image) of molecule (A)?

Concept

A molecule with one asymmetric carbon exists as two mirror-image forms (enantiomers), labelled R and S by the CIP rules. Two structures are enantiomers only if they share the same connectivity but have opposite configuration. If two drawings have the same configuration, they are the identical molecule simply drawn differently.

Assigning CIP priorities

At the chiral carbon the four groups rank: Br (highest, atomic number 35) > C2H5 (ethyl — first atom C carrying C, H, H) > CH3 (first atom C carrying H, H, H) > H (lowest).

Configuration of (A)

In (A): H is in the plane (up), C2H5 in the plane (left), CH3 to the back (dash) and Br to the front (wedge). Orienting so the lowest-priority H is directed away and tracing Br -> C2H5 -> CH3 gives an anticlockwise sense, so (A) is S.

Testing each option with the same method

• Structure (i): H up (plane), CH3 left (plane), C2H5 back (dash), Br front (wedge). Tracing Br -> C2H5 -> CH3 is clockwise -> R. This is the opposite of (A), so it is the enantiomer. …

Method: Permutation Parity at a Stereocentre

Concept: At a chiral carbon, swapping any TWO of the four attached groups inverts the configuration (R becomes S or vice versa). Swapping groups in pairs an EVEN number of times (2 swaps, or any 3-cycle, which is algebraically two swaps) returns you to the SAME configuration you started with, just redrawn. This lets you check every candidate structure against molecule (A) without having to assign full CIP priorities each time -- just track where each of the four groups moved.

Step 1: Record molecule (A)'s arrangement

• up (in-plane) = H
• left (in-plane) = C₂H₅
• dashed (back) = CH₃
• wedge (front) = Br

Step 2: Compare each candidate structure position-by-position

Structure (i): up=H, left=CH₃, dashed=C₂H₅, wedge=Br

• "up" and "wedge" are unchanged (H, Br stay put).
• "left" and "dashed" have swapped their occupants: C₂H₅ and CH₃ traded places.
• This is exactly ONE pairwise swap → an ODD permutation → inverted configuration → this is the true enantiomer of (A).

Structure (ii): up=CH₃, left=Br, dashed=H, wedge=C₂H₅

• Tracking where each group moved: H (up→dashed), CH₃ (dashed→up) — one swap. Also: C₂H₅ (left→wedge), Br (wedge→left) — a second, independent swap.
• Two swaps together = an EVEN permutation → same configuration as (A), just redrawn — NOT the enantiomer.

Structure (iii): up=H, left=CH₃, dashed=Br, wedge=C₂H₅

• "up" unchanged (H stays). Tracking the other three: C₂H₅ (left→wedge), CH₃ (dashed→left), Br (wedge→dashed) — this is a 3-cycle among left/dashed/wedge. …

Here are the common mistakes students make when identifying the enantiomer of a wedge-dash structure, and how to avoid each.

Mistake 1: Assuming any structure that "looks different" from (A) is the enantiomer

The error: Students see that (ii), (iii), or (iv) place the four groups in different positions around the stereocentre than (A) does, and conclude any of them could be the mirror image.

Why it's wrong: A structure can look completely rearranged and STILL represent the exact same 3D molecule (just rotated on the page) — only a genuine single-swap (or any odd number of swaps) of the four groups actually inverts the configuration. Structures (ii), (iii), and (iv) are all just molecule (A) rotated and redrawn, not its mirror image.

How to avoid: Don't judge "different-looking" as "different molecule." Track exactly which TWO groups (if any) have traded positions between the reference structure and the candidate.

Mistake 2: Trying to assign full R/S priorities to every structure from scratch

The error: Students painstakingly redo CIP priority assignment (Br > C₂H₅ > CH₃ > H) and a full 3D rotation for all four candidates, which is slow and error-prone under exam time pressure.

Why it's inefficient: There's a shortcut — compare each candidate to (A) position-by-position. If exactly one pair of groups has swapped places (everything else unchanged), that's an inversion (the enantiomer), full stop — no need to re-derive R/S from scratch.

How to avoid: Line up the four candidate structures against (A)'s four positions (up / left / dashed / wedge) and count how many pairwise swaps separate each one from (A). One swap = enantiomer. An even number of swaps (or a 3-cycle, which counts as two swaps) = the same molecule redrawn.

Mistake 3: Confusing "some atoms are in the same place" with "same configuration" …