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The compound below is fexofenadine and the only chirality center, as identified by the textbook is, the one circled in red.

I don't understand why the carbonyl carbon is not a chiral center too. Isn't it attached to a tert-butyl group? So isn't it attached to four different substituent groups? Or is it that the hydroxide group is attached to the carbon by the oxygen, thereby making a 'duplicate' substituent group of the carbon already double bonded to the oxygen?

enter image description here

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Take a look at your carbon circled in red, it has a hydrogen on it giving it four different groups.

Now look at the C circled in blue. There is no hydrogen attached to that carbon, because it already has four bonds (two to the O, one to the OH, one to the rest of the molecule.) Therefore it only has three substituent groups attached and not four.

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    $\begingroup$ So an atom connected by a double bond is still considered to be a single substituent group. That was my mistake. Thanks! $\endgroup$ – imaginov May 12 '15 at 21:11
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    $\begingroup$ Even if it were counted as two substituent groups, the carbon would be achiral because the two substituent group would be identical and you're looking for four different things attached to the carbon. $\endgroup$ – Melanie Shebel May 12 '15 at 23:53
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    $\begingroup$ Since this is the accepted answer, I'll point out the 1,1'-Bi-2-naphthol is optically active but it doesn't have a carbon with 4 different attachments. The optical activity is the result of steric hindrance into two different forms. $\endgroup$ – MaxW Nov 17 '15 at 7:42
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If in doubt, simplify, examine both centres separately and draw.

stereocentre

Can you do the same for the carboxylic acid?

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    $\begingroup$ I see that the central carbon is only connected to three substituent groups, two of which are bonded by their oxygens. So this carbon cannot be a chiral center. imgur.com/oH6gX2B $\endgroup$ – imaginov May 12 '15 at 21:18
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No.

If you look at the carbon circled in blue, it only has three not four substituents. If you examine the three-dimensional structure around the carbon, you will find that there is a plane of reflectional symmetry through the plane of the C=O bond which means it cannot be a chiral centre.

4 different things attached to a carbon in 3D space means it will be chiral. 3 does not.

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  • $\begingroup$ Can you please explain what you mean by "plane of reflectional symmetry through the plane of the C=O bond"? If I understand correctly, only sp3 hybridized C atoms can be chiral centers. $\endgroup$ – imaginov May 12 '15 at 21:16
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    $\begingroup$ @imaginov Yes, you re right. Only sp3 hybridised carbons can be chiral centres but the reason is because, if they have 4 different substituents, they lack a plane of reflectional symmetry. The carbon in a C=O bond is sp2 hybridised and, because this make it flat, there is a mirror plane through corresponding to the three substituents. Therefore, no chirality as the molecule will inevitably look just like its mirror image. $\endgroup$ – matt_black May 13 '15 at 21:24

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