How many carbon atoms does an alkane need before it is capable of existing in enantiomeric forms?
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2 Answers
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2
If we allow our substituents to be isotopes, then 2 is all that is really necessary:
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If we exclude such compounds from consideration, then cyclopropanes will give you smallest number of carbons at 5:
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As a side note, for a non-alkane, only 1 carbon is necessary:
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2$\begingroup$ If we were not confined to saturated compounds, I'd suggest 1,3-dimethyl allene. $\endgroup$ Oct 4, 2016 at 5:50
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2$\begingroup$ Also, come to think of it, carbon is not necessary at all. $\endgroup$ Oct 4, 2016 at 5:59
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If you are only looking for alkanes, of the form $\ce{C_nH_{2n+2}}$, with no isotopic substitution, then the most logical way is to think back to the definition of a chiral carbon. The IUPAC Gold Book writes:
The traditional name for a carbon atom that is attached to four different entities...
So, start with a carbon in the middle.
The smallest substituent you can put on this carbon is hydrogen (zero carbons). The next-smallest is a methyl (one carbon), and after that you need to use an ethyl group (two carbons).
For the last group, you need to use three carbons, because there is no other two-carbon alkyl substituent apart from the ethyl group. So, you could use a n-propyl group (only one isomer depicted here)
or an isopropyl group (again only one isomer depicted)
answered Oct 4, 2016 at 10:38