5
$\begingroup$

enter image description here

For the above question, here's what I'd thought: 1. H+ ion from HCl protonates the oxygen 2. Chloride ion attacks the electrophilic carbon, opening the epoxide ring 3. So the final product formed is a chlorohydrin (which it really isn't)

Also, I didn't know what to do with methanol (guessing it just acts as the solvent, HOWEVER, the product contains a methoxy group)

Could someone please explain the mechanism to get to the actual product, and also highlight the mistake here?

$\endgroup$
  • 7
    $\begingroup$ Methanol attacks, not chloride. It's present in large excess (as it's the solvent) $\endgroup$ – orthocresol Jul 25 '17 at 7:59
  • $\begingroup$ Is it that major product is formed due to attack by methanol and minor is formed as a result of attack by chloride? $\endgroup$ – arya_stark Jul 25 '17 at 14:52
2
$\begingroup$

If you think about the quantities involved in this reaction, you'll realize why does the product contain the methoxy group and not the halide. Methanol is present in large excess since it is the solvent of this reaction. Cl- ions will also be present and a very small amount of the chlorohydrin may be formed as well. The reason why methanol is a reagent and not only act as a solvent in this reaction is because its oxygen atom acts as a nucleophile, just like halides or any other molecule containing electronegative atoms (such as O, N, S etc).

I found the following mechanism on Google and it is draw as if the reaction was being performed in water, however you can replace the blue water molecule for methanol or any other alcohol since the mechanism is the same for both.

enter image description here

Whenever you look at a reaction schemes like the one you've posted, you should always address the reactivity of solvents and other reaction components, sometimes they end up reacting with your substrate rather than your desired reagent, this is a very good example of it. So if you intend to synthesize a chlorohydrin, you cannot use a nucleophilic solvent, otherwise, it will react with the epoxide and form the correspondent adduct.

I hope this answers your question.

$\endgroup$

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.