[Xenos]

Hello. I am having some difficulty in knowing where to start when naming cycloalkanes. For example I worked a problem that is a 5 sided ring with one of the carbons connected to TWO methyl groups and another carbon connected to a CH3-CH-CH2-CH3 (sec-butyl) group. My textbook states that when the numbering could being with either of the two alkyl groups, begin w/ the one that is alphabetically first. For alphabetical reasons, you use the "B" , not s in the sec-butyl convention according to my book. The prefixes di, tri, don't count for alphabetical rules either. So, this is how I named the cycloalkane:
1-sec-butyl-3,3-dimethylcyclopentane. I started the "1" with sec-butyl since "b" comes before the "m" in dimethyl. But according to my solutions manual the answer is: 3-sec-butyl-1,1-dimethylcyclopentane.  Is there some rule I am missing, where disubstituted alkanes (ex: dimethyl) are the ones you begin numbering? Why did they start with the dimethyl substituents first? Any help is appreciated.

[Alpha-Omega]

http://en.wikipedia.org/wiki/IUPAC_nomenclature

Not link at bottom:  On Line Version Of Blue Book -Organic Chemistry

[Xenos]

Thank you for the reply. I am still unsure on this question though, and I have consulted many online websites w/ downloaded pdfs. One for example statyes that if there are two or more substituents,  list them in alphabetical order and number the parent chain from the end that gives the lowest number to the substituent encountered first, the lowest number is given  to the substituent highest in alphabet. This rule doesnt apply to any of my book examples, hence I see it as a contradiction to the statement in my book.

[Kryolith]

Well I am no IUPAC expert, but perhaps the substituents are ordered by their CIP-priority, analog to stereoisomers.

http://en.wikipedia.org/wiki/Cahn-Ingold-Prelog_priority_rule

[RBF]

In this case, the substituents are numbered so that the lowest possible numbers are used.  Having both methyls at the '1' position and the sec-butyl at the '3' position results in a lower combination of numbers than having both methyls at the '3' position and the sec-butyl at the '1' position.  The substituents are then listed in alphabetical order, but the lower number does not have to be first in the name.  Hence '3-sec butyl-1,1-dimethyl....