Synthesis Problems – Review for Final Evaluate the starting material and the product for each. Count carbons. Evaluate functional groups. Contemplate how to transform the first into the second. a.

OH

Same number of carbons. Alcohol is turning into alkene. First thought might be “dehydrate”! This problem looks straight-up, doesn’t it? Except if you try to dehydrate that alcohol, you’ll make the Zaitsev product, not the Hoffman. You need to force the Hofmann product – and you can only do that with an E2 elimination on an alkyl halide. So – convert that alcohol to a halide (use HBr) and then form the Hofmann product using KOtBu in tBuOH.

OH Br

b. O

H Nine carbons to start. Eight carbons to end. Lost a carbon – the only type of reaction you know is the oxidative cleavage. What’s the catch? You cannot form aldehydes though when you oxidatively cleave alkynes (you always get carboxylic acids). So – what do you do to the triple bond? Convert it to a double bond (use Na, NH3 or H2 with + Pd/CaCO3/Pb) and THEN use 1. O3 2. Zn, H3O to form an aldehyde (with one less C). O

H c. O

9 Carbons to start. 12 carbons (and an oxygen) to end. Starting with an alkene. Ending with an ether. Ethers (asymmetrical primary ethers) come from alcohols in the Williamson ether synthesis. The alcohol must be forming on the less substituted end of the alkene – hydroboration. So 1. BH3 2. NaOH, H2O2 and then 1. NaH 2. CH3CH2CH2Br.

OH O

d. Br

5 carbons to start. 5 carbons to end. Alkene in starting material. Bromide in product. Catch? Its on the less substituted end (and you don’t know how to make the non- Markovnikov bromide from an alkene). But you DO know how to make a bromide from alcohols and you do know how to make the non-Markovnikov alcohol. So – 1. BH3 2.

NaOH, H2O2 and then use PBr3.

OH Br

e.

8 carbons to start with. 9 carbons in the product. Adding 1 C. Find that specific carbon (see the methyl group?) and locate where it is relative to the alkene. Which end is it attached to? More substituted end. How do we add alkyl groups? (Can’t use acetylide anion – no triple bond. Can’t use Williamson – not making an ether. What else do you have in your arsenal of reactions? The Cuprate… but first you need a halide to couple it to… Start this synthesis by adding HBr to the alkene, and then add (CH3)CuLi. Br

f.

6 carbons to start with. 8 carbons to end with. Match up the carbons. Notice that there are two methyls on the right side of the starting material and two methyls on the left side of the product. Guess what? Those are the same. Flip that starting material over. Now note that you need to add the two carbons to the end of the double bond.

Couple of options now… You could add Br2 to the alkene and then 2 equiv. of KOtBU and form a terminal alkyne, then add 1. NaNH2, NH3 2. CH3CH2Br and finally H2, Pd/C:

Or you could hydroborate the alkene to form a primary alcohol, then use PBr3 to form a primary bromide and couple that bromide with diethyl copper lithium, (CH3CH2)2CuLi. Br

g.

O 5 carbon terminal alkyne starting material. 5+1 carbon ether product. Obviously adding one carbon. And forming an ether (think Williamson ether synthesis for that part). Ethers come from alcohols. Alcohols do not come from alkynes – but they do come from alkenes. Alkenes come from alkynes. Use Na, NH3 to form the alkene (or hydrogenate with Lindlar’s catalyst, Pd/CaCO3/Pb) then hydroborate with 1. BH3 2. NaOH, H2O2 and do the Williamson Ether synthesis 1. NaH 2. CH3I.

OH O

h.

OH

Cl 7 carbons with a tertiary alcohol. 7 carbons with a secondary chloride, one carbon over from where the alcohol was positioned. How do we get to that carbon?? What can we do to the alcohol and incorporate some sort of reactivity on that carbon next door? Dehydrate to form an alkene. It will want to form Zaitsev. Now that other carbon is accessible and “reactive” and you have a “less substituted” end. Hydroboration will allow you to form an alcohol on that secondary carbon and thionyl chloride and pyridine will form the chloride you want.

OH

OH Cl i.

HO

8 carbons to start with a primary alcohol. 11 carbons to end, no alcohol. Adding three carbons (isopropyl type piece – only reaction that allows you to do that is the coupling reaction via a cuprate). You will need a halide in place first. Use PBr3 first then add

((CH3)2CH)2CuLi.

HO Br

j. OH OH

Obviously forming a diol where the alkene is located. The diol is anti so you will need to use that nucleophilic epoxide opening reaction, with NaOH. First, make an epoxide with mCPBA, then add the NaOH. Anti diol!

OH O OH