Chapter 14, Question 20

Initially, I got this problem wrong. I chose D rather than F, the correct answer. The most prominent peak on the compound was the 1685 cm-1 peak. This is indicative of a carbonyl (1800-1600cm-1) The other peaks on this compound are relatively sparse. I noticed that the carbonyl was fairly close to 1600cm-1  , This means that it is not an aldehyde but a ketone. This makes only D, E, and F viable. To narrow down the results I also noticed that there was no N-H or N-C peak, making E not a valid choice. Between D and F, there was a difference in the spectrum around 3000cm-1.  For sp3 carbons, peaks would be found to be much stronger than sp2. In this IR, we see very small peaks, indicating sp2 carbons. The only compound that fits that description is F.