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A chain of ten carbon atoms carries methyl groups on carbons 3, 4 and 8, counted from the left…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
A chain of ten carbon atoms carries methyl groups on carbons 3, 4 and 8, counted from the left. Number the chain from both ends. Decide which numbering to use, and write the name. Give the sum of the locants of that numbering.
Need a reminder? The key ideaShow
Alkanes , alkenes , alkynes . A homologous series differs by , 14 u. Name = prefixes (branches, in alphabetical order) + parent chain + suffix (principal group). Use the longest chain, with the principal group and the multiple bonds in it. Number to give the lowest locants at the first point of difference: principal group first, then multiple bonds, then branches. Seniority: acid, ester, amide, nitrile, aldehyde, ketone, alcohol, amine, alkene, alkyne.
The worked solution
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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. From the left. The locants are 3, 4, 8.
Step 2. From the right. The methyl groups are on , and . The locants are 3, 7, 8.
Step 3. Compare. The first numbers agree, 3 and 3. The second numbers are 4 and 7, and 4 is lower. So the numbering from the left wins.
Step 4. The name and the sum. The name is 3,4,8-trimethyldecane. The sum is .
Answer: 3,4,8-trimethyldecane, with locants summing to 15.
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