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My teacher told us the middle carbon in 3‐bromopentane‐2,4‐diol is chiral:

$$\ce{CH3-CH(OH)-\overset{\star}{C}H(Br)-CH(OH)-CH3}$$

Since there are two identical groups around it, I would guess it should not be chiral.

It was a pre-recorded class and I was not able ask this question. Has my teacher made a mistake?

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Carbons #2 and #4 are chiral centres themselves. There are four possible diastereoisomers:

Diastereomers of 3-bromopentane-2,4-diol

If C2 and C4 have different configurations, which is the case in molecules A and D, then C3 is termed a pseudoasymmetric centre and labelled with a small 'r' or 's'. These diastereomers are meso compounds: although they contain chiral centres, they are not chiral, because they have an internal plane of symmetry.

On the other hand, in molecules B and C, C2 and C4 have the same configuration and C3 is not a chiral centre (and not a pseudoasymmetric centre either). These diastereomers are chiral, though: the chirality comes from C2 and C4.

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  • $\begingroup$ So it's chiral in some cases and not in others ? Like Zhe said in the comments ? $\endgroup$ – Glowingbluejuicebox Jan 20 at 17:26
  • $\begingroup$ Actually, some of the diastereoisomers are chiral molecules, i.e. without internal mirror image or rotational symmetry. I hope this helps. $\endgroup$ – z1273 Jan 20 at 18:07
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    $\begingroup$ Rotational symmetry isn't related to chirality. To be more precise, the lack of improper rotations $S_n$ is associated with chirality; but proper rotations $C_n$ have no relation to chirality. PS sorry about the edits. I need to double check this. $\endgroup$ – orthocresol♦ Jan 20 at 19:00
  • $\begingroup$ Think that should be right now.. $\endgroup$ – orthocresol♦ Jan 20 at 19:07

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