C13 NMR Interpretation Homework - C,H,O 116g/mol

[gbpackfan]

Homework Statement

I'm having a bit of difficulty interpreting the spectrum. I've given it a shot but I think I might be missing something. The compound contains only C,H,O and any assistance would be appreciated. The molecular weight is 116g/mol.

Homework Equations

none

The Attempt at a Solution

Peak at 13.73= CH3
Peak at 22.32= CH2R2
Peaks at 27.09 and 33.85= CH3
Peak at 51= C-O
Peaks in the grouping at 77= C-C double bonding?
Peak at 174 is a possible carboxylic acid?

 

[TeethWhitener]

Peaks in the grouping at 77= C-C double bonding?

Remember to check for solvent peaks. In this case, the 1:1:1 triplet at 77 ppm is CDCl₃. Also, the OP has (tacitly) correctly assumed that the peak at 0 ppm is TMS.

Other than those, we have 6 peaks, meaning 6 inequivalent carbons. Now we get to work interpreting the rest of the spectrum.

Peak at 174 is a possible carboxylic acid?

A carboxyl group: either an acid or an ester. This means we have at least 2 oxygens. If we add up the molar masses of 6 carbons and 2 oxygens, we get 12*6 + 16*2 = 104 g/mol; in other words 12 g/mol shy of the given molar mass. It's pretty inconceivable that these 12 g/mol will be another carbon atom, so since the compound only has C, H, and O, the molecular formula for the compound is C₆H₁₂O₂.

Peak at 51= C-O

A single bond C-O indicates that the compound is an ester instead of a carboxylic acid.

Peak at 13.73= CH3

Only 1 alkyl CH₃ group indicates that there's probably no branching; in other words, all the alkyl chains in the molecules are straight.

The above considerations severely limit the identity of the molecule. It's a straight chain ester with a formula of C₆H₁₂O₂. The next part is a little tricky. The NMR spectrum gives only one CH₃ unfunctionalized alkyl resonance. What does that say about either side of the ester? Well, alpha carbons on the carboxyl side of the ester are generally shifted downfield from plain alkyl groups (in fact, this carbon corresponds to the peak at 33.85 ppm), and we've already identified the peak corresponding to the carbon attached to the alcohol side of the ester (51 ppm). Since there is only one CH₃ unfunctionalized alkyl peak, that means that either we have a methyl group on the alcohol side or an acetate group on the carboxyl side. This constrains us to one of two molecules: either methyl pentanoate, or butyl acetate.

Personally, if I were giving this as a homework problem, I'd accept either one of these molecules as an answer (it's extremely tough to tell the difference between them without some other information). However, in the interest of completeness, you have to compare typical shifts of a CH₃COO carbon vs. a RCH₂COO carbon. Doing this will tell you (roughly) that the peak at 33.85 ppm is a RCH₂COO carbon, instead of an acetate. This would correctly identify the molecule as methyl pentanoate. Again, though, the spectra of the two molecules are similar enough that I'm not sure I'd worry if I didn't get it exactly right.