Test 4
Dr. Bachrach
CHEM 2319-2
4/30/2014

1) Provide the IUPAC name for the following compounds.

a)


(3S)-3-butyl-2-methylhept-3-yn-2-ol

b)


(2S,3S)-2-methyl-2,3-dipropyloxirane

2) Provide the appropriate reagent(s) to affect the following reactions.

a)

b)

c)

3) The 1H NMR below is for a compound with molecular formula C11H16. Determine the structure of the molecule and justify it by matching each proton with its corresponding peak.

The DOS is 4. The mess down at δ 7.1-7.4 means a phenyl ring, so that takes care of the DOS.

The triplet at δ 0.7 is a methyl with an adjacent CH2 group. The quartet at δ1.6 is a –CH2- group split by three equivalent hydrogens. This looks like a classic ethyl group!

The singlet at δ 1.3 looks like a methyl with no adjacent protons.

So no the key is to consider the integral. The integral of the peak at δ 0.7 is a methyl so its 3Hs. The singlet at δ 1.3 is about twice as large, so that’s 6Hs, and the peak at δ1.6 is 2Hs, and that looks right with the integral. So that adds up to 3+6+2=11 Hs, leaving 5 left over, and that means the phenyl ring has 5 hydrogens on it, so only one substituent on that ring.

The ethyl and the two methyls account for 4 carbons, so there is one more C and it can’t have any Hs on it – we’ve assigned all the proton NMR signals. So put this all together and we get

Where (a) is triplet at δ 0.7 (b) is quartet at δ1.6 (c) is singlet at δ 1.3 and the phenyl Hs (d) are at δ 7.1-7.4

4) The IR and 1H NMR spectra below are for a molecule with molecular formula C7H16O2. Determine the structure of this compound, match each proton with its peak, and identify all important IR signals.

The DOS is 0, so there are no π-bonds and no rings.

In the IR, the broad strong stretch from 3100-3400 cm-1 indicates an alcohol. So we should be on the lookout for a singlet in the NMR!

In the NMR, the triplet at δ 0.8ppm and the quartet at δ 1.3 indicate an ethyl group. These are clearly coupled to each other, and since they are such nice triplets and quartets, they are not coupled to anything else! That means that the two peaks around δ 3.5ppm must be two different hydrogens and both are singlets. (If this was a doublet – what’s it coupled to?) One of them is the OH proton. The other is a hydrogen connected to a carbon, and it’s quite downfield, so this carbon must be connected to the OH. Further, this is a singlet so it can have no H neighbors, so we have as a fragment:

What’s meant here is that there is a carbon that has the OH group, at least one H, maybe two, and its carbon neighbor has NO hydrogens.

So we have this fragment and an ethyl fragment, and NO OTHER types of carbon! So we have to have some symmetry to get to 7 carbons and 2 oxygens. We can start to draw out 7 carbon chains and play with possibilities, or one might suggest that we have two of the above OH fragments and 2 ethyl fragments and get the final answer:

5) Predict the product(s) of the following reactions

a)

b)

c)

6) Using any organic compound with four or fewer carbon atoms, prepare 4-methylpentane-1,2-diol.