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Topic: Minimum number of enantiomers  (Read 13368 times)

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Offline Charkol

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Minimum number of enantiomers
« on: October 26, 2010, 10:03:55 PM »
I have this question that I do not understand.

"How many carbon atoms does an alkane (not a cycloalkane) need before it is capable of existing in enantiomeric forms?"

The answer to this question is in the solutions manual.  The answer is 7 carbon atoms.

I thought even a methane molecule could be enantiomeric.  For instance clorofluorobromomethane.

What am I missing in this question?

Offline karlosshughes

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Re: Minimum number of enantiomers
« Reply #1 on: October 26, 2010, 11:38:43 PM »
The question says alkane, which implies that only carbon and hydrogen atoms are present. While bromochlorofluoromethane is chiral and can exist in two enantiomers, an alkane containing only carbon and hydrogen would need seven carbons:
                                                     Et               
                                                     |                   
                                              Me - c -H
                                                     |
                                                     Pr           
Me = 1 carbon, Et = 2 carbons, Pr = 3 carbons. Therefore, the minimum number of carbons needed is 7.

(For a clearer structure, just Google 3-methyl hexane.     
Chemistry student & organic kid.
Dabbles in other chemistry too (but only because I have to XD).

Offline Charkol

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Re: Minimum number of enantiomers
« Reply #2 on: October 27, 2010, 01:43:30 PM »
I see.  Yes, I must have generalized the term alkane into any single bonded carbon chain.  This makes sense.  Thanks much.

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