1. Show how to prepare the following products from the given starting materials:
OH
a.
i. H2CrO4, acetone (oxidation to aldehyde)
NH
ii. CH3NH3, NaBH3CN, H+ (reductive amination)
b.
c. Generate the enolate that then undergoes a Michael reaction:
d. Enolate attacks the carbonyl, that you can then dehydrate to form the double bond (and then
saturate the double bond to give the desired product).
e. hydrolysis to the carboxylic acid, followed by addition of two equivalents of ethyl lithium to give you
the ethyl ketone directly
f. Oxidize first to the ketone, then the second arrow shows one of several possible ways to introduce a
double bond into the system:
g. SN2 to give you the alcohol, oxidation, then DCC (dicyclohexylcarbodiimide) which is a coupling
reagent, to give the product amide:
h. Convert to a bromide, then create a grignard reagent to react with the aldehyde. The resulting alcohol
is then oxidized to the aldehyde selectively using PCC. Another equally valid option would be to create a
Grignard that reacts with CO2 to generate the carboxylic acid, and then selectively reduce that to the
aldehyde:
2. Give the final products of the reactions and show the intermediates as well:
a.
b.
c.
d.
e.
f.
g.
h. here you generate an enolate that then acts as a nucleophile:
i. Again, enolate. This time intramolecular attack of the nucleophile:
j. First reagent generates an acyl chloride, that undergoes an intramolecular friedel crafts acylation:
k. Enolate acts as a nucleophile, then the LiOH hydrolyzes both esters to the corresponding carboxylic
acids, heat causes the molecule to decarboxylate (lose one carboxylic acid group), and then warm acid
causes the five membered ring (called an acetal) to hydrolyze to the corresponding aldehyde product
l. Excess base generates the thermodynamic enolate preferentially, which then attacks the aldehyde to
give you the desired product (either this product or the unsaturated aldehyde are acceptable answers
here):
m. Enolate, attacks the bromide, hydrolyze to acid, decarboxylate with heat.
3. Provide a retrosynthetic pathway for the following targets from “simple, readily available” starting
materials:
a. You can disconnect back to the aldehyde and a “negatively charged” carbon, which can simply be a
stand-in for an organolithium compound. The ester in turn could come from the aldol reaction of the
diester (which forms an enolate) and benzaldehyde.
b. Here again, the carbonyl is a result of a reaction between an electrophilic carbonyl and a carbon
nucleophile.
c. Here the six membered ring (lactone) came from the nucleophilic attack of a protecting OH on the
ester carbonyl. This brings us back to an alpha,beta unsaturated carbonyl, which should be a red flag for
you that it came from an aldol reaction. The aldol reaction could have occurred between the enolate of
the diester and the carbonyl of an aldehyde (or monoester) (which as you may note, has no alpha
hydrogens and therefore cannot form an enolate – which would complicate product formation
tremendously.).
d. Again, alpha, beta unsaturated carbonyl should scream “aldol reaction” at the top of its lungs. If you
do the standard disconnections, that brings you back to the diketone. This is where things get a little
trickier. The cyclohexanone generates an enolate (that’s your nucleophile) and it needs to attack
something that will give the OH at the desired position. It turns out that the three membered ring with
the oxygen gives you the best way to get the desired product.
e. The lactone is from nucleophilic attack of an OH at the ester of a carbonyl, and the resulting
compound came from nucleophilic attack of the enolate of the diester on cyclohexanone (electrophile).