Nomenclature problem

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iso-, sec-, tetr- are common names. I am not sure to what extent they are acknowledged in IUPAC but they are certainly less formal than the full IUPAC name.

That makes B correct answer but not the most correct answer.
 
iso-, sec-, tetr- are common names. I am not sure to what extent they are acknowledged in IUPAC but they are certainly less formal than the full IUPAC name.

That makes B correct answer but not the most correct answer.

Sometimes textbooks mix IUPAC nomenclature with common nomenclature (e.g. cis-1-ethyl-4-isopropylccyclodecane just pulled this out of an old textbook), but another rule to keep in mind is that you want to maximize the number of branches if you can number the parent chain in 2 different ways (both having the same number of carbons...in this case both A and B are pentanoic acids, but A has more branching, therefore is the best answer) in B both substituents would be on a 3 carbon, but in A you have "2 branches" on carbons 3 and 4. Thus, A is preferred to B.

Helpful links:

http://www.acdlabs.com/iupac/nomenclature/79/r79_933.htm

http://www.stolaf.edu/depts/chemistry/courses/toolkits/247/js/names/anotes.htm
 
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But OP is referring to the fact that you could also theoretically go down that chain on carbon 3 and form a pentanoic acid with an isopropyl group and a methyl group at carbon 3, this is wrong, however, b/c one must maximize the number of branches when doing the parent nomenclature.
 
Another way to say this is that Branches on branches (or complex branches) must be minimized, if possible. The isopropyl would be named 3-methyl-3-(methylethyl) pentanoic acid. The 3-(methylethyl) part should be avoided.

dsoz
 
But OP is referring to the fact that you could also theoretically go down that chain on carbon 3 and form a pentanoic acid with an isopropyl group and a methyl group at carbon 3, this is wrong, however, b/c one must maximize the number of branches when doing the parent nomenclature.

I saw it as having to label the carboxylic acid carbon as carbon 1, thus starting from the other end.
 
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