Chemistry · General chemistry II · Worked example

Compare the stereoisomers of 2,3-dibromobutane

${CH}_{3} {CHBrCHBrCH}_{3}$ has two stereocenters. How many different stereoisomers does it have, and how are they related?

Find the stereocenters and the upper limit

Carbons 2 and 3 each carry $H$ , $Br$ , ${CH}_{3}$ and the other $CHBr$ group. Two stereocenters allow at most $2^{2} = 4$ stereoisomers: (2R,3R), (2S,3S), (2R,3S) and (2S,3R).

Rank the groups at each center

At carbon 2, $Br$ is first. The other $CHBr$ group holds $(Br, C, H)$ and ${CH}_{3}$ holds $(H, H, H)$ , so the $CHBr$ group is second and ${CH}_{3}$ third. Hydrogen is last. Carbon 3 ranks the same way.

Take the mirror image of (2R,3R)

A mirror image reverses every center, so the mirror image of (2R,3R) is (2S,3S). The two centers of (2R,3R) have the same label, so no mirror plane maps one half onto the other, and (2S,3S) is not the same molecule. They are enantiomers.

Look at (2R,3S)

Its mirror image would be (2S,3R). But turn one end about the C2–C3 bond until the two $Br$ atoms are on the same side, and the molecule has a mirror plane through the middle of that bond. The carbon 2 half reflects into the carbon 3 half, so (2R,3S) is its own mirror image, and (2S,3R) is the same compound. It is meso.

Relate the three

The meso compound is not the mirror image of either (2R,3R) or (2S,3S), so it is a diastereomer of each. Four possible arrangements collapse to three compounds.

(2R,3R) (2S,3S) (2R,3S), meso

\begin{array}{c} (2R,3R) \\ (2S,3S) \\ (2R,3S), meso \end{array}

Result

Three. (2R,3R) and (2S,3S) are enantiomers. (2R,3S), the same compound as (2S,3R), is meso and is a diastereomer of both.

Your turn

1. One like the example

Tartaric acid, $HOOC - CH (OH) - CH (OH) - COOH$ , has two stereocenters. How many stereoisomers does it have, and how are they related?

Show the answer and explanation

Three. (2R,3R) and (2S,3S) are enantiomers, and the meso form is a diastereomer of both.

Two centers allow at most $2^{2} = 4$ . The two centers carry the same set of groups, so (2R,3S) and (2S,3R) have an internal mirror plane and are one meso compound, leaving $4 - 1 = 3$ .

\begin{array}{l} 2^{2} = 4 \\ 4 - 1 = 3 \end{array}

2. A variation

3-Bromobutan-2-ol, ${CH}_{3} CH (OH) {CHBrCH}_{3}$ , has two stereocenters. How many stereoisomers does it have?

Show the answer and explanation

Four, in two pairs of enantiomers.

Two centers allow $2^{2} = 4$ . One center carries $OH$ and the other $Br$ , so no combination can have an internal mirror plane, and none is meso. (2R,3R) and (2S,3S) are one enantiomer pair, (2R,3S) and (2S,3R) the other, and any two from different pairs are diastereomers.

2^{2} = 4

3. Somewhere new

Pent-3-en-2-ol, ${CH}_{3} CH = CHCH (OH) {CH}_{3}$ , has one stereocenter and one double bond that can be E or Z. How many stereoisomers does it have?

Show the answer and explanation

Four.

The stereocenter and the double bond each give two arrangements, so $n = 1 + 1 = 2$ and $2^{2} = 4$ . The two ends of the molecule differ, so none is meso. (2R,3E) and (2S,3E) are enantiomers, and (2R,3E) and (2R,3Z) are diastereomers.

2^{1 + 1} = 4

Keep exploring

The three Draw buttons put each isomer in the Skeletal tool, with wedges at carbons 2 and 3 and its name as the target. On one structure choose Analyze → Compare and pick another from the list: (2R,3R) against (2S,3S) reads “Enantiomers”, and the meso structure against either reads “Diastereomers”.

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Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 7 key terms Draw (2R,3R)-2,3-dibromobutane Draw (2S,3S)-2,3-dibromobutane Draw meso-2,3-dibromobutane Open worked example on a board Stereochemistry rules in Chemistry Reference

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