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I read my book and learned everything about enantiomers, racemization, $\mathrm{S_N1}$ and $\mathrm{S_N2}$ reactions, but when I'm trying to identify molecules as chiral or achiral, I get confused.

Consider the two molecules below:

1.CH3-CH-OH
      |
      Br

2.CH3-CH-Br
      |
      Br

How can I tell if a compound is chiral?

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    $\begingroup$ Look at your molecule, if you see any plane of symmetry the molecule must be achiral. If you don't see a plane of symmetry that doesn't mean the molecule is chiral. But checking for a plane is an easy first step and may often make the time consuming step of drawing and comparing structures unnecessary. $\endgroup$ – ron May 10 '15 at 18:25
  • $\begingroup$ @ron why does the lack of planes of symmetry not imply chirality? $\endgroup$ – DHMO Jan 28 '17 at 5:31
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    $\begingroup$ Compounds that possess any $S_n$ axis are achiral and a plane of symmetry is only one example of such an axis. $\endgroup$ – orthocresol Jan 28 '17 at 13:46
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Chirality can be determined by visualising the molecule in space. Once you do that, you can check to see if there exists a non-superimposable mirror image. If this is indeed the case, then your molecule is chiral.

You can look for an asymmetric carbon atom--because this often results in chirality.

An asymmetric carbon (or chiral carbon) is, very simply, a carbon atom attached to four different groups.

With this in mind, consider your examples: 1) should thus be chiral and 2) is achiral.

Hope this makes sense.

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  • $\begingroup$ I drew out the wedge and dash representations as per your steps. It turns out that 1st one is Chiral and 2nd is achiral. Thanks $\endgroup$ – Sharad Gautam May 10 '15 at 6:49
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    $\begingroup$ Good job. The same reasoning can be applied to more complicated molecules (structurally). With some practice, the process of identification becomes almost intuitive. $\endgroup$ – getafix May 10 '15 at 6:52
  • $\begingroup$ @getafix I downvoted this question because it could mislead the OP into thinking that the presence of chiral carbons implies chirality. $\endgroup$ – DHMO Jan 28 '17 at 5:35
  • $\begingroup$ @DHMO Well that's up to you. But I don't think that's the case. The answer opens with, "Chirality can be determined by visualising the molecule in space. Once you do that, you can check to see if there exists a non-superimposable mirror image. If this is indeed the case, then your molecule is chiral." Which is general. "You can look for an asymmetric carbon atom--because this often results in chirality." Which is also true. Nowhere do I say asymmetric carbons automatically imply chirality in all cases. $\endgroup$ – getafix Jan 28 '17 at 5:41
  • $\begingroup$ @getafix That is why I said "could" and "mislead". These words mean that you did not say directly that the presence of chiral carbons implies chirality. I made that judgment because right after you define chirality on the first paragraph, the second paragraph starts with "[y]ou can look for an asymmetric carbon". You did use the word "often", but I feel that the point "not always" has not been emphasized enough, albeit not the main topic of the question. $\endgroup$ – DHMO Jan 28 '17 at 5:48
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Look for carbons with four different groups attached to identify potential chiral centers.

Draw your molecule with wedges and dashes and then draw a mirror image of the molecule. If the molecule in the mirror image is the same molecule, it is achiral. If they are different molecules, then it is chiral.

Here you're particularly interested in the C in the CH. However, take a look at the C in CH3. Why isn't it a chiral carbon?

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    $\begingroup$ I don't think CH3 Groups can be chiral centers. Am I wrong? $\endgroup$ – Sharad Gautam May 10 '15 at 7:07
  • $\begingroup$ Nope, you're correct. :P It was just a hint toward why the carbon attached to the two bromines wasn't chiral. $\endgroup$ – Melanie Shebel May 10 '15 at 7:08
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  1. Draw both molecules. Don't draw them flat. Alternatively, model them using Avogadro.

  2. Exchange the positions of $\ce{Br}$ and $\ce{OH}$ in the first case. Is the new molecule identical to the original one?

  3. Do the same for the dibromo-substituted compound. How is the result here?

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    $\begingroup$ Thanks for pointing out on using avagadro. I did exchange the postitions of OH and Br and the new molecule was indeed non-superimposable to the first one and hence chiral. $\endgroup$ – Sharad Gautam May 10 '15 at 6:54
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The first method I used to understand chirality was to grab some different colored zip ties and make a couple small tetrahedral structures with them, assigning a color to each different radical (say blue for $\ce{CH3}$, white for $\ce{H}$, red for $\ce{OH}$ and yellow for $\ce{Br}$.)

Then look at it by rotating the first to see if I can get the second molecule, if you can't then that molecule is chiral.

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