At the beginning of the chapter, we mentioned the fact that carbonyl compounds react with cyanide. We are now going to deduce a mechanism. This is the reaction:

We must decide what happens. NaCN is an ionic solid so the true reagent must be cyanide ion, whose structure was discussed on p. 112. As it is an anion, it must be the nucleophile and the carbonyl group must be the electrophile. Starting the arrow on the nucleophile’s negative charge and heading for the C=O group, and then using a second arrow to break the C=O bond gives us this:

This is a good mechanism but it doesn’t quite produce the product. There must be a second step in which the anionic oxygen picks up a proton from somewhere. The only source of protons is the solvent, water, so we can write the full mechanism in one sequence:

Try a more complicated example: primary alcohols can be converted into symmetrical ethers in acid solution. Suggest a mechanism for this acid-catalysed conversion of one functional group into another.

The acid must do something, so we need to start with the reaction between ethanol and H+. H+ has to be an electrophile, so the nucleophile must be ethanol, using its HOMO, one of the O lone pairs, as the source of electrons. The first intermediate we get is called an oxonium ion. The positively charged oxonium ion has to be the electrophile in the second step of the reaction, and the only possible nucleophile is another molecule of ethanol. But how do they react? It’s tempting to allow the ethanol’s lone pair to attack the positively charged oxygen atom, but that would give us an oxygen atom with ten electrons—as with H3O+ this positive charge is not an empty orbital. Attacking the H–O bond is a good alternative, but that just takes us back to where we started.

What we need is a new C–O bond, so the lone pair must attack at carbon, putting electrons into the C–O σ* and expelling a molecule of water. Here’s the full mechanism. The last step is loss of the proton to give the ether.

Now for something completely new: try drawing a mechanism for this reaction.

You might well protest that you don’t know anything about the chemistry of either of the functional groups, the thiol or the cyclic ether. Be that as it may, you can still draw a mecha nism. Ask fi rst of all: which bonds have been formed and which broken? Clearly the S–H bond has been broken and a new S–C bond formed. The three-membered ring has gone by the cleavage of one of the C–O bonds. The main chain of carbon atoms is unchanged. All this is sketched in the diagram in the margin. We suggest you now cover the rest of this page and try to work out a mechanism yourself before reading further. The hydroxide must do something, and since it is negatively charged, a reasonable starting point is going to be to use it as a nucleophile to break the S–H bond. Hydroxide is after all a base; it likes to remove protons. So, here’s the fi rst step:

Now we have a negatively charged sulfur atom, which must be the nucleophile. We want to make a bond to carbon, so the C–O bond in the three-membered ring must be the electro phile. So … just draw the arrows and see what happens. Here goes …

That is not quite the product: we need to let this anion pick up a proton from somewhere. Where can the proton come from? It must be the proton originally removed by the hydroxide. The anion attacks water and the hydroxide is regenerated.

Your mechanism possibly didn’t look as neat as the printed version, but if you got it roughly right, you should be proud. This is a three-step mechanism involving chemistry that is new to you and yet you could draw a mechanism for it.