CHEM 2320-1
TEST 2 Answers
Dr. Bachrach

1) Provide the IUPAC name of the following compounds.

a) (3E)-N,N-diethyl-5,5-dimethylhept-3-enamide

b) (3S, 8R)-8-benzyl-3-chlorodecanoyl chloride

2) Predict the product and provide the mechanism for the reaction below.

3) Determine the structure of the molecule with formula C₇H₁₄O­₂ whose IR and ¹³C NMR and ¹ H NMR spectra are shown below. Clearly justify your structure.

So the IR tells us that it has a carbonyl (the strong stretch at ~1720 cm^-1) and no –OH groups due to the lack of a broad strong stretch between 3600 and 3000 cm^-1. Note that the string stretch around 3000 cm^-1 is due to C-H stretching!

Now the ¹³C NMR has three important pieces of information; (a) there are six peaks indicating 6 unique carbons (which means that two carbons must be identical since we have 7 carbons in the molecule), (b) the peak at 175 ppm is due to the carbonyl carbon and (c) a single peak around 70 ppm means that only one carbon can be connected to the second oxygen. Thus we must have an ester, not a carbonyl and an ether (since the ether would have 2 carbons attached to it).

The degree of unsaturation here is 1, which is accounted for in the carbonyl – so no other double or triple bonds.

From the ¹H NMR we have lots of information. Starting from the right, the signal around 0.9 ppm is a doublet, which means we must CH-CH₃, i.e. a methyl group with only one proton on its neighboring carbon. Next at about 1.2ppm is a triplet – this is a methyl with a neighboring methylene group: CH₂-CH₃. And this is a bit farther down field than usual – probably near to an oxygen. Next is a mess at about 1.8ppm. Then we have a quartet at 2.4pp – whenever you see a quartet in the 1.5-2.5 ppm region AND a triplet in the 1-1.5 ppm region, you should be thinking of an ethyl group! And since the methylene is a bit farther downfield than usual, perhaps it is attached to the carbonyl: C=O-CH₂-CH₃. Lastly we have a doublet way down at 3.8ppm – this must be attached to the ether oxygen and have a neighboring C with one H: O-CH_x-CH-. So putting this together we have CH-CH_x-C=O-O-CH₂-CH₃. We still have a methyl doublet left and we still need two more carbons in the molecule – but we have already used up 5 unique carbons – so the methyl doublet must really be two methyl groups that are identical. So the molecule is:

4) Provide the product(s) of the following reactions:

a)

b)

c)

d)

5) Provide the necessary reagent(s) to enable the following transformations. Multiple steps might be required.

a)

b)

c)

d)

6) Draw the structure of sec-butyl 6-methyl-5-oxo-hept-2-ynoate.

7) Indicate the most acidic hydrogen in cyclohex-2-enone (I) and explain your answer.

The complete way of answering this problem is to examine all possible hydrogrens to see which is the most acidic. This means looking at the reaction of I going to every possible anion, and finding the most stable anion. So…

So the loss of the proton from C₄ leads to the most stable anion, one that has three resonance structures!