Thursday, December 5, 2013

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 :


  1. I agree that using the DOU formula helps tremendously when trying to read and decipher mass spectrometry. However, I do still struggle with reading the actual readings on the mass spec. Sometimes its not clear to me if the graph "dips" are insinuating bonds or just unnecessary markings and also when bonds are within the same wave number range. For example, if there is a marking between 3300 and 2500, it's sometimes difficult to determine whether it's a C-H, N-H, or O-H bond. Even though the intensities are different, I often struggle with deciding which is which without other clues from the mass spec.