Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
Give the IUPAC name of each: (a) ; (b) ; (c) ; (d) .
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.
Have you tried it?
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. (a). The acid outranks the ketone. The acid carbon is carbon 1, and the C=O is on carbon 3, as an oxo prefix. The name is 3-oxobutanoic acid.
Step 2. (b). The aldehyde is the principal group, and its carbon is carbon 1. Chlorine is a prefix on carbon 3. The name is 3-chlorobutanal.
Step 3. (c). The alcohol outranks the alkene, so the carbon with OH is carbon 1. The double bond is at carbon 3. The name is but-3-en-1-ol.
Step 4. (d). The amine is on the middle carbon of a three-carbon chain. The name is propan-2-amine.
Answer: 3-oxobutanoic acid, 3-chlorobutanal, but-3-en-1-ol and propan-2-amine.
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