A student writes down every aldehyde and ketone with the formula C₅H₁₀O, using open carbon chains…
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A student writes down every aldehyde and ketone with the formula , using open carbon chains only. Find the number of aldehydes, the number of ketones, and the number of compounds in all. Then give the name of the aldehyde that has no hydrogen 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
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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. Aldehydes. The alkyl group on the CHO has four carbons: butyl, isobutyl, sec-butyl and tert-butyl. These give pentanal, 3-methylbutanal, 2-methylbutanal and 2,2-dimethylpropanal. That is 4.
Step 2. Ketones. The two alkyl groups hold three carbons in all. Methyl and propyl give pentan-2-one. Methyl and isopropyl give 3-methylbutan-2-one. Ethyl and ethyl give pentan-3-one. That is 3.
Step 3. Total. .
Step 4. The special aldehyde. Carbon 2 of 2,2-dimethylpropanal carries two methyl groups and no hydrogen. This is the one with no alpha hydrogen.
Answer: 4 aldehydes, 3 ketones, 7 compounds, and 2,2-dimethylpropanal.
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