# If given a starting product and end result, how do you determine which steps are used to reach the end result? How do you determine the correct order?

##### 1 Answer
Apr 10, 2016

There is generally no single "correct order".

MULTI-STEP SYNTHESES

There is almost always more than one way to do a multi-step synthesis.

My professor had usually given synthesis problems that have more than one solution to test which reactions you knew. The only reason why your solution would be wrong is if one or more of the steps are wrong.

I simply practice using the reactions I know in syntheses that I make up (in essence, putting to muscle memory what the products of certain reactions are), and get used to envisioning which reaction accomplishes what in the end. Ultimately, if you don't practice, of course it will be hard.

Here's an example synthesis problem:

For the following, propose a reasonable synthetic route.

You should be able to find more than one way to do this. Here are the ways I found:

So, you can see there are at LEAST three ways to do this. The first path is five steps, and the other two are six steps. These other two paths are more creative, but the first path is the most efficient one that I can think of that uses first-semester reactions.

MULTI-STEP REACTION MECHANISMS

However, there are generally fewer ways, or perhaps only one way, to do a multi-step mechanism. Here's an example mechanism problem from one of my homework papers:

Draw a complete mechanism - all lone pairs, correct charges, correct arrows, etc. - for the following transformation.

It seems intimidating at first, but if you focus only on what changed, it gets easier. Once you recognize what changed, it should be easier to figure out how that happened.

The basis of this problem is carbocation rearrangements. Here's what's going on.

1. You can see the $\text{OH}$ disappear and a $\pi$ bond take its place on the first product, so it indicates an elimination reaction occurred. That is, $\text{OH}$ was protonated, ${\text{OH}}_{2}^{+}$ left, and the adjacent proton was eliminated, thus generating the new $\pi$ bond. This is the expected product in the absence of carbocation rearrangements.
2. The second one is a bit harder. You can see the ring size change, so you have to assume a ring contraction occurred, which is a type of carbocation rearrangement. From experience, this occurs to stabilize the seven-membered ring, especially when there is a positive charge on it, thus indicating that, again, ${\text{OH}}_{2}^{+}$ left. The rest is figured out generally by actually doing the problem.
3. If you pay close attention to the difference between the second and third products, it's that there are two alkyl groups attached to the second-to-the-furthest-right carbon, rather than one alkenyl like in the second product. That indicates that a 1,2-alkyl shift occurred in a carbocation intermediate AFTER a ring contraction. Specifically, it was the one that was adjacent to the $\text{OH}$ in the beginning.

So, here is how I worked it out:

It's really one of those problems that you have to do yourself in order to figure out. So practice doing your mechanisms so you can figure out the general behavior of electrons in various contexts.

That's how I manage to figure this kind of stuff out. I don't merely memorize specific steps, because then if I forget how to do that step, it becomes very difficult to improvise on the spot. I get it down to muscle memory.