NMR - Compound A: In compound A the only thing we got from...

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Compound A: In compound A the only thing we got from the IR was the sp3 C-H bonds at 2900-3000. We had zero unsaturations so this told us that there were not double bond or rings. Since that is true, there were no carbonyl groups and no alcohol groups present in the IR so our compound had to contain an ether. The HNMR signals confirmed the symmetrical either since the intensity of the peaks were 6 and 4 protons respectively. Compound B: In compound B, we also had zero unsaturations. The IR scan showed that our compound contained an –OH group as well as sp3 hybridized C-H bonds. The most helpful graph for this compound was the HNMR. The triplet at .9 ppm told us that there were 2 adjacent protons and the multiplet at 1.5 ppm confirmed the ethyl group on one side of the chain. The singlet at 4.0 ppm is the proton connected directly to the oxygen which explains the shift downfield due to deshielding. The doublet at 1.2 told us that there had to be a methyl group split by the lone proton connected to the carbon with the –OH group attached. We had 4 separate carbon signals
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