Chirality — chiral centres, enantiomers, diastereomers, meso compounds: Exercises

Before we start, one reminder written in plain words so nothing is assumed:

Level 1 — Recognition

Q1. Count the chiral centres

For each molecule, decide whether the central carbon is a chiral centre.

(a) $C H F C l B r$
(b) $C H_{3} - C H C l - C H_{3}$ (2-chloropropane)
(c) $C H C l_{3}$ (chloroform)
(d) $C H_{3} - C H B r - C_{2} H_{5}$ (2-bromobutane)

Recall Solution

WHAT we do: list the four groups on the carbon; if all four differ, it is a chiral centre.

(a) Groups: $H, F, C l, B r$ — all four different → chiral centre ✔
(b) Groups: $H, C l, C H_{3}, C H_{3}$ — two $C H_{3}$ are identical → not chiral ✘
(c) Groups: $H, C l, C l, C l$ — three identical $C l$ → not chiral ✘
(d) Groups: $H, B r, C H_{3}, C_{2} H_{5}$ — all four different → chiral centre ✔

WHY: swapping two identical groups changes nothing, so the molecule would equal its mirror image (achiral). Only when all four differ does a swap build a genuinely new mirror object.

Answer: (a) and (d) have a chiral centre; (b) and (c) do not. Count = 2 chiral molecules out of 4.

Q2. Name the relationship word

Fill the blank with enantiomers, diastereomers, or meso:

(a) Two mirror-image forms of $C H F C l B r$ are ________.
(b) A molecule with chiral centres yet an internal mirror plane is ________.
(c) Stereoisomers that are NOT mirror images are ________.

Recall Solution

(a) enantiomers — they are mirror images and cannot be stacked.
(b) meso — internal plane cancels the handedness.
(c) diastereomers — same connectivity, different 3-D shape, but not mirror twins.

Level 2 — Application

Q3. Maximum stereoisomer count

Using $max = 2^{n}$ (where $n$ = number of chiral centres), give the upper bound for:

(a) $n = 1$
(b) $n = 2$
(c) $n = 3$
(d) $n = 4$

Recall Solution

WHY the formula: each centre is an independent yes/no ("R or S") choice; $n$ independent binary choices multiply → $2^{n}$ . $2^{1} = 2, 2^{2} = 4, 2^{3} = 8, 2^{4} = 16$ Answers: (a) 2, (b) 4, (c) 8, (d) 16.

Q4. Racemate rotation

A pure enantiomer has specific rotation $[α] = + 1 2^{\circ}$ . You mix it 1:1 with its mirror twin. What is the net specific rotation of the mixture?

Recall Solution

WHAT: a racemate is a $1 : 1$ mix of the two enantiomers. One rotates $+ 1 2^{\circ}$ , the other rotates the exact opposite $- 1 2^{\circ}$ . $α_{net} = \frac{1}{2} (+ 1 2^{\circ}) + \frac{1}{2} (- 1 2^{\circ}) = 0^{\circ}$ WHY zero: equal amounts twist light in opposite directions → external compensation. See Optical Isomerism and Polarimetry. Answer: $0^{\circ}$ (optically inactive).

Level 3 — Analysis

Q5. Tartaric acid — sort the four labels

Tartaric acid $H O O C - C^{*} H (O H) - C^{*} H (O H) - C O O H$ has two chiral centres. The four label-combinations are $(R, R), (S, S), (R, S), (S, R)$ . Answer:

(a) Which pair are enantiomers?
(b) Which two labels are actually the same molecule, and why?
(c) How many distinct stereoisomers exist?

Recall Solution

The picture you should build in your head first (then check against the figure): draw the molecule as a vertical chain with the top carbon $C_{2}$ and the bottom carbon $C_{3}$ , connected by a bond. Now imagine a horizontal line drawn exactly halfway between $C_{2}$ and $C_{3}$ . Because both ends of the molecule are identical (each carbon carries $C O O H$ , $O H$ , $H$ ), that horizontal line is a candidate mirror plane — whatever sticks out above it, an identical group sticks out below it.

The figure makes this concrete: the amber dashed line is that halfway mirror plane, and you can read off that $C_{2}$ 's substituents ( $C O O H$ up-left, $O H$ down-left, $H$ right) are mirrored by $C_{3}$ 's substituents below.

Figure — Chirality — chiral centres, enantiomers, diastereomers, meso compounds

Figure: tartaric acid drawn vertically; the amber dashed line halfway between $C_{2}$ and $C_{3}$ is the internal mirror plane. Each cyan group above the line has an identical partner below it.

(a) $(R, R)$ and $(S, S)$ are mirror images and non-superimposable → enantiomers. (Reflecting an all- $R$ molecule turns every centre into $S$ .)
(b) $(R, S)$ and $(S, R)$ : reflect $(R, S)$ across the halfway plane and you get $(S, R)$ — but now spin the whole molecule 180° end-over-end and the top becomes the bottom, mapping $(S, R)$ right back onto $(R, S)$ . Since one can be superimposed on the other by a rotation, they are the same single molecule: the meso form. The internal mirror plane means its two halves cancel → achiral.
(c) Distinct isomers: $(R, R)$ , $(S, S)$ , and one meso $=$ 3.

$count = 2^{n} - (meso forms) = 2^{2} - 1 = 3$

Q6. Enantiomer vs diastereomer physical properties

You are told compound A and compound B are stereoisomers. A and B have different melting points and different solubilities. Are they enantiomers or diastereomers?

Recall Solution

WHY: enantiomers share identical physical properties (mp, bp, solubility) and differ only in the sign of optical rotation. Diastereomers are essentially different substances → different mp, bp, solubility. Answer: diastereomers. Different physical properties ⇒ not mirror twins. See Stereochemistry.

Level 4 — Synthesis

Q7. Meso vs enantiomer diagnosis

For 2,3-dibromobutane $C H_{3} - C^{*} H B r - C^{*} H B r - C H_{3}$ :

(a) How many chiral centres?
(b) What is the naive $2^{n}$ count?
(c) Is there a meso form? Justify with symmetry.
(d) What is the true number of stereoisomers?

Recall Solution

Build the picture: draw the two central carbons stacked vertically, each carrying $C H_{3}$ , $B r$ , $H$ . Since both ends are the same group set, imagine a horizontal mirror line halfway between them — exactly as in Q5. The figure shows this mirror line in amber, with each cyan substituent above matched by an identical one below.

Figure: 2,3-dibromobutane drawn vertically; the amber dashed line halfway between the two $C^{*}$ carbons is the internal mirror plane that produces the meso form.

(a) Two centres → $n = 2$ .
(b) Naive $2^{2} = 4$ .
(c) The two ends are identical, so the $(R, S)$ arrangement has an internal mirror plane between the central carbons: reflect the top half and it lands exactly on the bottom half. Its "left/top" $C^{*}$ twists light one way, its "right/bottom" $C^{*}$ twists it back → internal compensation, net $α = 0$ . This is one meso molecule (and reflecting-then-rotating shows $(R, S)$ and $(S, R)$ are the same, just like Q5).
(d) True count: $2^{2} - 1 = 3$ Namely: $(R, R)$ and $(S, S)$ enantiomer pair, plus one meso $(R, S)$ . Answer: 3.

Q8. Specific rotation calculation

A $2 dm$ tube holds a solution of $c = 0.10 g/mL$ of a pure chiral sugar. The polarimeter reads $α = + 5.2 4^{\circ}$ . Compute the specific rotation $[α]$ using $[α] = \frac{α}{l \cdot c} .$

Recall Solution

WHY this formula: dividing by path length $l$ and concentration $c$ removes "how much / how long," leaving a pure material constant (see Optical Isomerism and Polarimetry). Units bookkeeping (do NOT skip): $α$ is in degrees ( $^{\circ}$ ), $l$ in $dm$ , $c$ in $g/mL$ . So the ratio carries units $\frac{^{\circ}}{dm \cdot ( g/mL )} =^{\circ} \cdot mL \cdot g^{- 1} \cdot dm^{- 1} .$ These are the standard specific-rotation units; by convention we quote only the number and the degree sign, but $l$ must be in dm and $c$ in g/mL for the number to be correct. $[α] = \frac{+ 5.2 4 ^{\circ}}{( 2 dm ) ( 0.10 g/mL )} = \frac{+ 5.2 4 ^{\circ}}{0.20} = + 26.2^{\circ} mL g^{- 1} dm^{- 1}$ Answer: $[α] = + 26. 2^{\circ}$ (in the units above), dextrorotatory (the $+$ sign = clockwise).

Level 5 — Mastery

Before the last two problems, one new term — defined properly, with a picture, before we use it:

Intuition How to assign lower-case

$r$ / $s$ to a pseudo-asymmetric carbon The normal R-S Nomenclature (CIP rules) rules rank the four groups by priority, then look from opposite the lowest group: clockwise $= R$ , anticlockwise $= S$ . For a pseudo-asymmetric carbon the two arms are constitutionally tied, so ordinary CIP cannot separate them. The tie-breaker rule: an $R$ -arm outranks an $S$ -arm (a "like $R$ " reference beats a "like $S$ " reference). Apply this extra rule to rank the two arms, then read clockwise/anticlockwise exactly as usual — but write the descriptor in lower case ( $r$ or $s$ ) to signal it came from the pseudo-asymmetric tie-breaker. Result you will use below: for the middle carbon, once you fix the outer centres as one $R$ -arm and one $S$ -arm, the two possible spatial arrangements of the middle carbon's $H$ and $O H$ give the two labels $r$ and $s$ — two genuinely different molecules.

Intuition A recipe for counting meso forms when

$n > 2$ (fills the "subtract every meso" gap) "Subtract every meso" is only useful if you can find every meso. Here is the recipe:

Check the molecule has a symmetric backbone (identical halves about a centre). If not, there are no meso forms → answer is $2^{n}$ .
A configuration is meso exactly when reflecting one half maps it onto the other half — i.e. the two halves carry opposite configurations that mirror each other.
Count the meso forms = count the arrangements of the outer centres in which the top half is the mirror of the bottom half. For a symmetric molecule with an even number of centres this is $2^{n /2}$ mirror-matched arrangements collapsing appropriately; for an odd number, the extra middle carbon becomes pseudo-asymmetric and splits the meso family in two. The safe method for exams: enumerate all $2^{n}$ labels, then physically test each for an internal mirror plane — the ones that pass are meso, and mirror-image pairs among the rest are enantiomers.

Q9. The $n = 3$ molecule with a hidden symmetry (full enumeration)

Consider a chain symmetric about its middle carbon: $H O O C - C_{2} H (O H) - C_{3} H (O H) - C_{4} H (O H) - C O O H,$ where $C_{2}$ and $C_{4}$ are the two identical outer carbons and $C_{3}$ is the middle carbon. The two end-arms $(- C H (O H) C O O H)$ are constitutionally identical.

(a) Naive $2^{n}$ count?
(b) When is $C_{3}$ actually a stereocentre?
(c) Enumerate every distinct stereoisomer and classify it. How many are there?

Recall Solution

(a) Three chiral positions → naive $2^{3} = 8$ labels.

(b) $C_{3}$ is bonded to $H$ , $O H$ , and the two arms toward $C_{2}$ and $C_{4}$ .

If $C_{2}$ and $C_{4}$ have the same configuration (both $R$ or both $S$ ), the two arms are truly identical → $C_{3}$ is not a stereocentre (only $H$ / $O H$ differ, two arms match).
If $C_{2}$ and $C_{4}$ have opposite configurations (one $R$ , one $S$ ), the arms differ → $C_{3}$ becomes a pseudo-asymmetric carbon (labelled $r$ or $s$ ).

How we assign the middle label $r$ / $s$ (using the tie-breaker above): fix $C_{2} = R$ and $C_{4} = S$ . The middle carbon's four groups are $O H$ , $H$ , the $R$ -arm (toward $C_{2}$ ) and the $S$ -arm (toward $C_{4}$ ). Rank by priority: $O H$ (highest, oxygen) $>$ $R$ -arm $>$ $S$ -arm (the $R$ -beats- $S$ tie-breaker) $>$ $H$ (lowest). Point the lowest ( $H$ ) away and read the remaining three:

if $O H \to R -arm \to S -arm$ runs clockwise, the middle carbon is $r$ ;
if it runs anticlockwise, the middle carbon is $s$ . The two spatial placements of the middle $H$ / $O H$ give exactly these two outcomes → two distinct molecules, $(R, r, S)$ and $(R, s, S)$ .

(c) Full enumeration. Write each label as $(C_{2}, C_{3}, C_{4})$ . Because $C_{3}$ 's label only exists when $C_{2} \neq = C_{4}$ , we split into cases.

Case A — outer carbons SAME ( $C_{2} = C_{4}$ ): $C_{3}$ is not a stereocentre, so its geometry is fixed; only the $C_{2} = C_{4}$ choice matters.

$(R, -, R)$ and $(S, -, S)$ . These two are mirror images of each other, and neither has an internal mirror plane (both halves same handedness twist the same way, so nothing cancels) → they form a genuine enantiomer pair (2 molecules, optically active).

Case B — outer carbons OPPOSITE ( $C_{2} \neq = C_{4}$ ): now $C_{3}$ is pseudo-asymmetric, so it can be $r$ or $s$ .

$(R, r, S)$ and $(R, s, S)$ . In each, the top half is the mirror of the bottom half → internal mirror plane → each is a meso compound (achiral, $α = 0$ ). The $r$ vs $s$ label of the middle carbon distinguishes them, giving two different meso molecules.
What about $(S, r, R)$ and $(S, s, R)$ ? Rotating the molecule 180° end-over-end turns $(S, r, R)$ into $(R, r, S)$ and $(S, s, R)$ into $(R, s, S)$ — the same two meso molecules already counted. No new species.

Tally of all $2^{3} = 8$ labels: the 4 labels of Case A ( $(R, -, R)$ , $(S, -, S)$ , each written once since the middle is not a centre) collapse to the enantiomer pair (2 molecules); the 4 labels of Case B ( $(R, r, S), (R, s, S), (S, r, R), (S, s, R)$ ) collapse by reflect-then-rotate to 2 distinct meso molecules. So $8 ⟶ enantiomer pair 2 + meso r, meso s 2 = 4.$

Final answer (c): 4 distinct stereoisomers — one enantiomer pair (optically active, $R, R$ and $S, S$ ) plus two meso forms (achiral, $R, r, S$ and $R, s, S$ ). WHY fewer than 8: the four opposite-outer labels merge (reflect-then-rotate) into just two meso molecules distinguished by the pseudo-asymmetric middle, and the four same-outer labels merge into one enantiomer pair. $8 \to 4$ .

Q10. Full reasoning — a drug enantiomer

A chiral drug's active enantiomer is $(S)$ with $[α] = - 8 8^{\circ}$ . The manufacturer accidentally sells the racemate.

(a) What optical rotation would a polarimeter read for the racemate?
(b) Does "zero rotation" mean "no active drug present"? Explain.
(c) What is $[α]$ of the pure $(R)$ enantiomer?

Recall Solution

(a) Racemate = $1 : 1$ of $(S)$ at $- 8 8^{\circ}$ and $(R)$ at $+ 8 8^{\circ}$ : $α_{net} = \frac{1}{2} (- 8 8^{\circ}) + \frac{1}{2} (+ 8 8^{\circ}) = 0^{\circ} .$
(b) No! Zero net rotation is external compensation — the active $(S)$ molecules are still physically present at half concentration; their twist is merely masked by the $(R)$ twin. Biologically the $(S)$ still works (and the $(R)$ may be inactive or harmful). See Drug Chirality and Biological Activity.
(c) Enantiomers rotate light by equal magnitude, opposite sign: $[α]_{R} = - [α]_{S} = - (- 8 8^{\circ}) = + 8 8^{\circ} .$ Answers: (a) $0^{\circ}$ , (b) active drug still present, rotation just cancelled, (c) $+ 8 8^{\circ}$ .

Active recall

Q distinct stereoisomers of tartaric acid

3 (

d

l

, and one meso), since

2^{2} - 1 = 3

Q when the middle carbon of a symmetric n=3 diacid is a stereocentre

Only when the two outer carbons have opposite configuration — then it is pseudo-asymmetric (

r

s

Q net specific rotation of a racemate

0^{\circ}

(external compensation).

Q specific rotation from alpha=+5.24, l=2 dm, c=0.10

+ 26. 2^{\circ} mL g^{- 1} dm^{- 1}

Q specific rotation of the R enantiomer if S is -88

+ 8 8^{\circ}

(equal magnitude, opposite sign).