Super simple.
This is Ibuprofen. You may be familiar with its name. It's the main ingredient in the super-effective Advil tablet which many of you are well-acquainted with.
It's structure can easily be broken down. I've attached a photo of a 4-step synthesis below (click photo to enlarge):
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Step 2: If you look above, you'll notice that the alkyl group that we want attached doesn't have an oxygen double-bonded to a carbon. We'll get rid of this by performing a Clemmensen Reduction using Zinc and hydrochloric acid. Done.
Step 3: Now, we'll add the next substituent on the opposite side using a friedel-crafts alkylation. The reactant will be a 3-chloropropanol. We have to use AlCl3 again as our catalyst to make this reaction go. Our benzene might still be a little unreactive, even after adding the alkyl group. The only problem with this step is that another group might get added onto the ortho positions, since alkylations are runaway-type reactions... but maybe the steric hindrance will prevent that from happening too fast.
Step 4: Now, all we have to do is oxidize the alcohol into an acid. We can use the king of oxidation, KMnO4 for this step.
That's all it takes to make ibuprofen. Nice and straight forward.
Note: In Step 1, I went with an acylation instead of an alkylation for 2 reasons.
#1: Alkylations are runaway reactions, and we'll get the same substituent all over the ring - we only want it in one spot.
#2: When AlCl3 pulls the Cl group off of our reactant (which happens when you alkylate), the molecule will undergo a rearrangement into a tertiary carbocation... and then our benzene ring will attack the wrong carbon.
Note #2: KNNO4 is used incorrectly in the synthesis, as it would oxidize at both benzylic positions, unwantedly.
Can u please include the full synthesis and also the amount of substances used, please?
ReplyDeleteChemistry sure is a really difficult subject. These days, I can't even concentrate on it because of other subjects. Going to join a academy for it soon and try to learn it.
ReplyDeleteWouldn't the KMNO4 attack both benzylic positions? instead of just the C at the alcohol?
ReplyDeleteYou are absolutely right, I didn't catch that in my synthesis. Thanks!
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