Posted by: Mark Foreman | May 3, 2016

Cyclopropane chemistry

Dear Reader,

There is a deep problem with some chemistry text books, this is the writers of some books create questions (problems) for teaching using their own opinion about what the answer should be. The problem is often they seek to teach a given point in chemistry.

But they unwittingly ignore some other aspect of chemistry which causes the answer to be something other than the one which they expected. What I would suggest is that anyone who is writing a question for a chemistry book should make a quick literature search of the topic to make sure that they have got the right answer.

For example I have encountered one text book which claimed that the hydrides of the oxygen group increase in Bronsted acid strength as the central atom (O, S, Se, Te and maybe Po) get more and more heavy. Their reasoning was that the bond strength between hydrogen and the central atom would become weaker, thus the acid would get stronger.

In case you do not know HF and water are both much weaker acids than you might expect based on the electronegativity of the central atoms. This is because the bond strength is so high.

However in the case of the chalcogen hydrides while the bond strength decreases the electronegativity of the central atom also decreases. As a result hydrogen selenide and hydrogen telluride have similar pKa values when dissolved in water. Now that was quite painless, lets move onto a more complex case.

Now if we ignore that fact that interconversion of hydrogen atoms between the axial and equitorial sites in cyclohexane is quite slow we can reason that only one hydrogen environment exists in this molecule. If you cool it down you can get the proton NMR spectrum of this molecule to change from a sharp line to a broad line which then changes further to two sharp lines.

If we replace one of the hydrogen atoms with a chlorine atom then only one product (chlorocyclohexane) is possible. I am quite happy with this idea.

We could extend this idea for a moment to cyclopropane, this is a molecule where we have an even better case for arguing that all the hydrogens are the same. In this case the molecule can only adopt one conformation. It is stuck in single conformation.

We could reason that if we were to replace one of the hydrogens with a chlorine atom then we would have chlorocyclopropane.

The only problem was that the question suggested using chlorine gas under free radical chain reaction conditions to form this alkyl chloride. When we look in the literature we find that cyclopropane and chlorine gas forms a large amount of allyl chloride under these conditions.

What will happen is that a chlorine atom will abstract a hydrogen from a cyclopropane to form a free radical which will then rearrange in a ring opening reaction to form an allyl radical. The allyl radical will then react with chlorine to form allyl chloride which will then be likely to react with the chlroine to form 1,2,3-trichloropropane and a host of other products.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Categories

%d bloggers like this: