AAMC Unscored #38

[wheatthinners]

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[theonlytycrane]

de-protonate the alpha-H on acetyl-coa and use it as the nucleophile

 

[aldol16]

OP, your mechanism isn't right. Look at it again.

 

[wheatthinners]

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[theonlytycrane]

@wheatthinners np 🙂 All of your follow-up questions are great and ones that I asked as well when I learned about cholesterol biosynthesis in biochem. I don't have great answers except for this is the observed biochemical pathway. Strictly from an orgo standpoint, both can possibly be nucleophiles and the hydrolysis of the other -S-CoA is possible too, but it just depends on the biological enzyme's specificity to target part of the substrate.

 

[aldol16]

Also, at the end, is the S-CoA just removed by a substitution (SN2?) by water molecules? If so, then why did this not occur on the right side of the molecule as well? Only one of the S-CoA's acted as a leaving group?

When you're talking about organic synthesis, this can become a huge concern. That's why organic chemists are nowhere near the level that nature is on in terms of synthesis - we have to protect functional groups, deprotect them, worry about side reactions, etc. whereas in an enzyme active site, nature can erect huge kinetic barriers to reaction that results in 100% selectivity.

 

[NextStepTutor_2]

Thank you!!!

For some reason, I didn't expect the smaller molecule to become the nucleophile. Do you know why that occurred in this case? I always thought that bigger molecules could go through more stabilization, so the acetoacetyl-coa would have gotten the alpha-H removed. I mean based on the products of this reaction, it's not possible, but in a case where they don't give you the products?

Also, at the end, is the S-CoA just removed by a substitution (SN2?) by water molecules? If so, then why did this not occur on the right side of the molecule as well? Only one of the S-CoA's acted as a leaving group?

Size doesn't guarantee much of anything as it relates to nucleophillic ability. this is actually quite a common mechanism (loss of alpha-H to make a good electrophile into a good nucleophile through resonance) the MCAT tests (e.g aldol condensation, crossed aldol condensation, Claisen condensation).

Hope this helps, good luck!

 

[BerkReviewTeach]

OP, you have posted a great question. Theonlytycrane, you have presented perfect organic chemistry reasoning.

It is true that because of steric hindrance, size of nucleophile (and electrophile) can influence the mechanism. This is seen in simple SN2 vs SN1 studies as well as reaction rate studies involving increasing alkyl amine size in an SN2 reaction.

But what is at play here is the impact of resonance on pKa. The keto-thioester has an alpha carbon that upon deprotonation will be resonance stabilized by two carbonyl moieties. The acetyl-CoA molecule only allows for resonance stabilization via one carbonyl group. So in the OP's original proposal (the right side of the picture), the doubly-conjugated alpha carbon is the most logical expectation (exactly what you'd expect from a Claisen reaction with a standard ester as opposed to a thioester).

In fatty acid biosynthesis (after the first coupling) you see chain lengthening, although that reaction occurs after the keto-thioester is reduced and the coupling steps involve acetyl-CoA and an longer-chain thioester molecule (such as butonoyl-CoA or hexanoyl-CoA). In that pathway, it is the small molecule (acetyl-CoA) that is acting as the nucleophile, as expected by steric hindrance. So your (OP's) thoughts are valid. The smaller molecule is the better nucleophile when the alpha carbons are essentially equal.

The pathway presented in the passage goes against standard organic chemistry logic in terms of nucleophilicity of deprotonated alpha carbons, but the question is not asking for you to reason the organic chemistry. You just need to track carbons. This is the synthetic process for isoprene production, so it is a great mixture of organic chemistry and biochemistry.