CHEM 2320-2
Dr. Bachrach
Sep. 19, 2012

1) (12 points) Provide the IUPAC name of the following compounds.





2) (16 points) Starting with chlorocyclohexane, prepare 1,2-diethoxycyclohexane. You may use any inorganic compound you wish and any alcohol with 4 or fewer carbon atoms.

3) (18 points) A compound with molecular formula C9H12O has the following IR, 13C NMR and 1H NMR. Determine the structure of the compound. Identify the information from each spectrum that you used to solve the structure. For the 1H NMR spectrum, identify the H source of each signal.

So what do we get from the molecular formula? The degrees of unsaturation is 4 – so maybe a phenyl ring?

What do we get from the IR? Well, there is the usual stretch around 3000 cm-1 for the C-H, but no broad strong signal from 3100 to 3500 cm-1, so the molecule does NOT have an OH group!

How about the 13C NMR? Well, there are seven signals, so we have seven unique carbons. Since the molecule has 9 carbons, there has to be some symmetry!

OK, on to the 1H NMR. There are two clean doublets in the aromatic region (6.8 to 7.2 ppm). This is strongly indicative of a para-substituted phenyl ring – and this would give us the symmetry to reduce the number of carbons form 9 to 7.

Since we know from the IR that there is no OH group, the singlet at 2.3ppm is NOT due to an alcohol. This is likely a methyl group with no neighboring hydrogens. So we could have either a methyl directly attached to the phenyl ring, or a methoxy group (OCH3). The latter would be shifted farther down that just to 2.2 ppm, so we have a para-substitued toluene.

The quartet at 4 and the triplet at 1.3 look like an ethoxy group that has been shifted down field – what could do that? If it were an ethox group –OCH2CH3. Combining fragments we have para-ethoxybenzene:

Hydrogens (a) and (b) are in the aromatic region (6.8 to 7.2 ppm), hydrogens (c) are at 4.0 split into a quartet by hydrogens (e)which appear at 1.3ppm, while hydrogens (d) appears as the singlet at 2.2ppm

4) (24 points) Provide the product(s) of the following reactions. If there is more than one product, indicate the major product.





5) (12 points) Provide the reagent(s) to perform the following transformations.




6) (18 points) Compound A is reacted with MCPBA.

a) Draw the possible products.

b) In fact, only one product is formed. Which one is formed and explain why.

Compound II is the only product observed. MSPBA has to attack the π-bond from the top or from the bottom. The top direction is blocked by the methyl group, so the only available pathway to the double bond is from the bottom. The oxygen is delivered from that (bottom) direction, leading to the epoxide II.