For the following two reactants, I want to predict the major product.

enter image description here

I would have thought the second reactant $\ce{CH3CH(OH)CH3}$ splits into $\ce{CH3CHCH+}$ and $\ce{OH-}$ , and the $\ce{OH-}$ would replace the $\ce{Cl-}$ in the first reactant $\ce{Phenyl-COCl}$, giving B. My main reason for thinking this would occur is that I always see e.g. water form $\ce{H+ and OH-}$.

However, the answer is C. The $\ce{CH3CH(OH)CH3}$ gives up only its $\ce{H+}$, and the rest of it joins $\ce{Phenyl-COCl}$ to form C.

How do I predict which atom(s) will leave/join? And for this particular question, how do I know the bond in the 2nd reactant cleaves at O-H not at C-OH?

  • 1
    $\begingroup$ I don't understand what you're asking. The answer is C. Why or what are you asking about D? $\endgroup$ – Zhe Feb 15 '17 at 3:18
  • $\begingroup$ Sorry! Typo. I am asking why the H+ leaves the CH3CH(OH)CH3, leaving the CH3CH(O)CH3- to swap with the Cl- (giving C). Rather than CH3CH(OH)CH splitting into CH3CHCH+ and OH-, and the OH- replacing the Cl- in the first reactant. $\endgroup$ – K-Feldspar Feb 15 '17 at 3:20
  • $\begingroup$ In other words, why does the 2nd reactant split at the O-H bond and not the C-OH bond. Thank you. Let me know if my question still doesn't make sense and I will try and re-word it. $\endgroup$ – K-Feldspar Feb 15 '17 at 3:23

It's a bit tricky because there's probably a reagent that's missing. You might add DMAP or triethylamine to this reaction which would further activate the acyl chloride. But the alcohol oxygen is the most nucleophilic moiety in this reaction. The most electrophilic moiety is the carbonyl group, especially since it's activated by the electron withdrawing chlorine.

Thus we would predict that a standard acylation (a carbonyl reaction) will take place where we add a nucleophile to the carbon of a carbonyl group. After which, reformation of the carbonyl group will result in kicking out a chloride anion. Finally, a very fast proton transfer will take place that breaks the O-H bond you are referring to. Notice that this is really an after thought and not really the key point of this reaction.

Take a look at the mechanism for the reaction in your textbook (or here).

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.