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How many carbon atoms does an alkane need before it is capable of existing in enantiomeric forms?

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If we allow our substituents to be isotopes, then 2 is all that is really necessary:

$\hspace{6cm}$isotopes

If we exclude such compounds from consideration, then cyclopropanes will give you smallest number of carbons at 5:

$\hspace{5.5cm}$cyclopropanes

As a side note, for a non-alkane, only 1 carbon is necessary:

$\hspace{5.7cm}$bromochlorofluoromethane

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    $\begingroup$ If we were not confined to saturated compounds, I'd suggest 1,3-dimethyl allene. $\endgroup$ – Ivan Neretin Oct 4 '16 at 5:50
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    $\begingroup$ Also, come to think of it, carbon is not necessary at all. $\endgroup$ – Ivan Neretin Oct 4 '16 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.

enter image description here

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).

enter image description here

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)

enter image description here

or an isopropyl group (again only one isomer depicted)

enter image description here

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