tons of unsaturation: Sapling Ch.14 (practice) #14

Using information from NMR and IR spectra to determine the structure of a compound was something I actually had fun doing, its like solving a jigsaw puzzle, except you won't make a mess of your kitchen table only to realize at the end that you're missing pieces. I`ll walkthrough my thought process in figuring out # 14 in sapling Ch. 14 practice.
Here`s a screenshot of the information given in the problem:

It may not seem like a whole lot of information, and I would have preferred to actually have the spectra, But It`s enough to deduce the structure of the compound.

The first thing I do when faced with one of these problems is determine the degree of unsaturation using the formula: (2C+2+N-halogens-H)/2.  For our molecule: ((2*10)+2-10)/2, which comes out to 6. This means that our structure has a total of 6 pi bonds and/or rings.

Having to place 6 pi bonds/ rings in a structure seems daunting, so lets take a look at the IR and NMR for structures with unsaturation present in the molecule. When a compound has a lot of unsaturation, I look for a benzene ring, because that would take care of 4 of our DOUs. Also, benzene is a very stable structure that is appears often in organic compounds. Sure enough, the NMR signal at 8.13 is typical of benzyllic protons. Great! Thats 4 of our DOUs and 6 of our carbon atoms!

We still have two more DOUs, as well as 2 O atoms… carbonyl groups? Indeed, an IR peak at 1681 is typical of a double bond between a C and an O. We also know that there is no OH stretch, and therefore no carboxylic acids. That means 2 C=O`s, and we`ll need to look at the NMR to decide wether they're ketones or aldehydes.

First, lets draw out our known components:

Ok, so we have to build our final structure out of those and 2 more C atoms. This is where the NMR really comes into play. We have only 2 peaks, so I expect a molecule with symmetry. Lets start with benzene, since there`s only one and it must therefore be dissected by any present plane(s) of symmetry. We have 4 equivalent benzyllic protons, so the benzene must have identical groups at carbons 1 and 4. The peak is a singlet, so no protons chemically inequivalent to these benzyllic protons are on adjacent C atoms.

Our other NMR peak has 6 protons. With 2 more C atoms needing to be added to our structure, it should be safe to say we have 2 identical methyl groups. these must be at the end of our 1,4 groups to preserve symmetry. That leaves our carbonyl Carbons, which must each have a bond to C 1 or 4 of the benzene, and to one of the methyl C`s. This means that there are no H atoms bonded to our carbonyl carbons, and therefore no splitting of the methyl protons, giving rise to the singlet peak in our NMR.

Thats it! heres our final structure, a benzene ring with 1,4 ethyl ketones :