1) Provide the IUPAC name of the following compounds.

2) Provide the reagent(s) needed to perform the following transformations.

a)

b)

c)

d)

3) Determine the structure of the molecule with formula C₅H₉OCl whose IR and ¹H NMR spectra are shown below. Identify every major peak in the IR and indicate which protons are responsible for each signal in the NMR.

IR: signal at 1700cm^-1 indicates a carbonyl

NMR: four distinct signals – four types of protons.

Triplet at 3.6ppm: far downfield, protons on a carbon with a chlorine, adjacent to –CH₂- (protons d)
Triplet at 2.6ppm: downfield due to adjacent to carbonyl, adjacent to –CH₂- (protons b)
Singlet at 2.2ppm: methyl group adjacent to carbonyl (protons a)
multiplet at 2.0ppm: complicated splitting due to adjacent groups, farther downfield than usual –CH₂- group (protons c)

4) Starting with any alcohol containing four or fewer carbon atoms, prepare 3-methyl-4-heptanone. You may use any inorganic reagents you wish.

Note that there are other possible ways to caryy out this synthesis.

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

a)

b)

c)

d) Note that Ph is the same thing as φ

6) Compound A is reacted with MCPBA.

a) There are two possible products. They are diastereomers. Draw them.

b) Only one of these diastereomers is actually formed. Argue on mechanistic grounds for which one is produced.

MCPBA attacks the top of the π-bond, see below.

So, it must gain access to either the top or bottom face of the alkene. In A, the top face is hindered by the methyl group that sits right above the double bond, sterically blocking the approach of MCPBA to that face. The bottom face of the double bond is open and MCPBA can attack the &-pi-bond from the bottom, forming the two C-O bonds of the epoxide simultaneously, leading to product C.