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First of all, thanks a lot for taking the time. I'm homeschooling and don't currently have any proper teachers, so I thought this would be a good place to ask. Its an easy question I think, but it has me very confused. In my chemistry book, they showed this hydrocarbonenter image description here

Translated from Spanish, it would be something like 4-ethyl -1,2 dimethylcyclohexane. Now, my question is, according to the rules, when there is more than one different type of ramification, you number the carbons giving the radical with alphabetical preference the lowest value, in this case, ethyl is first in the alphabet than methyl, so shouldn't it be 1-ethyl- 3,4 dimethylcyclohexane? They would all still be between the first 4 carbons, so the rule of lowest value wouldn't really change. Unless, the di, indicating the 2 methyl radicals counts and gives it alphabetical preference, but to my understanding, those prefixes aren't considered when giving alphabetical preference. Thanks for your time and have a great day!

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The first name has the correct alphabetical order of groups: ethyl before dimethyl. The "d" of dimethyl is not involved in alphabetizing. The numbering is correct because the one-to-one comparisons of locants is {1,2,4 < 1,3,4}. The correct name is 4-ethyl-1,2-dimethylcyclohexane. For a brief set of IUPAC rules applicable to this question see, https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/nomen1.htm

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