EasyChemistryHigh School (11–12)Organic Compounds with Functional Groups › Aldehydes, Ketones and Carboxylic AcidsConceptual
Give the IUPAC name of each compound. CH₃CH₂CH₂CHO; CH₃CH₂COCH₂CH₃; (CH₃)₂CHCOCH₃…
Contributed by StemPy LabTry 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 compound.
- cyclopentanone with a methyl group on carbon 2
Need a reminder? The key ideaShow
The carbonyl carbon is and flat. Oxygen pulls the pi electrons, so carbon is and oxygen . An aldehyde is with the carbonyl carbon numbered 1. A ketone is and its carbonyl carbon gets the lowest number. Carbonyl compounds have dipole-dipole forces but no hydrogen bonds between their own molecules.
The worked solution
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. Four carbons, CHO at the end: butanal.
Step 2. Five carbons, C=O in the middle: pentan-3-one.
Step 3. The longest chain with C=O has four carbons, with C=O on carbon 2 and a methyl on carbon 3: 3-methylbutan-2-one.
Step 4. Four carbons from CHO, with a methyl on carbon 3: 3-methylbutanal.
Step 5. A phenyl group on a two-carbon ketone: 1-phenylethanone, called acetophenone.
Step 6. Four carbons, CHO at the end and a double bond starting on carbon 2: but-2-enal.
Step 7. A ring ketone with the methyl group next to C=O: 2-methylcyclopentanone.
Step 8. Two CHO groups on a four-carbon chain: butanedial.
Check. Each name has the right number of carbons, and the aldehydes have no locant for the CHO group.
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