This is a question given in our exam. In the reaction, enter image description here

Find the reagents X and Y required for steps(ii) and (iii) are _______ and ________ respectively. I have tried in this way but I have stuck in second step. The given compound reacts with PhCO3H to form an epoxide. Am i correct? Then please help me continue the reaction. The answer given is CH3MgBr and H3O+.

  • $\begingroup$ This seems like a homework question. We ‎have a policy which states that you should show your thoughts and/or efforts into solving the ‎problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this ‎question may get closed. $\endgroup$ – bon Feb 4 '16 at 11:32

You're right: The reaction of the alkene with perbenzoic acid forms an oxirane.

My suggestion for the next step is:

  1. Draw the epoxide.
  2. Draw the final product.
  3. Map the atoms: Count and assign the carbon atoms. Do you spot any difference?

Supposed that the oxygen atom of $\ce{OH}$ group in the final alkanol is the same atom introduced by the action of perbenzoic acid, the oxirane was apparently opened.

Which general ways to open an oxirane exist?

Try to think in general synthons first.

One of the two steps in the sequence might just be a neutralization, quenching or workup step.


With a bit of drawing, atom counting and mapping, you will have realized (by now) that:

  1. The final alkanol has one carbon atom more than the oxirane intermediate.
  2. The additional methyl group was added to the less substituted side of the oxirane.

Now, try to think in synthons, such as "$\ce{CH3+}$" and "$\ce{CH3-}$", rather than in reagents.

Where would a "$\ce{CH3+}$", thrown at the oxirane, add? Probabaly at the centre of high(est) electron density. This is the oxygen atom, and thus the wrong position. Apparently, "$\ce{CH3+}$" is not your synthon.

What about "$\ce{CH3-}$"? It's a strong nucleophile that would preferably attack centres of low(er) electron density, such as carbon atoms with electron withdrawing substituents. You have two of these, namely the two carbon atoms that constituted the $\ce{C=C}$ double bond and which are now in your oxirane. If "$\ce{CH3-}$" attacks one of these carbon atoms, the oxirane is cleaved in such a way that one $\ce{C-O}$ bond is broken and the oxygen atom gets the negative charge. Note that sterical hendrance comes into play here too.

In summary,"$\ce{CH3-}$" seems to be the right synthon. Now, you have to translate the synthon, "$\ce{CH3-}$", into a reagent that you can either find in the lab or make in situ. In order to get the negative charge on the carbon atom, it might be a good idea to attach an electropositive to it.

Consequently, a Grignard reagent, $\ce{CH3MgBr}$ is a good choice.

As I mentioned initially, while trying not to spoil the fun of solving, the last step is just an aqueous, acidic workup to release the alkanol ;-)


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