[Big-Daddy]

I would like to ascertain the "Markovnikov rule" of free radical substitution, if such a thing exists. Take the below molecule:



If I add Cl₂ to this molecule in the presence of UV light, what is the major product?

[opsomath]

Chlorine is famously nonselective because it is a high-energy starting material. However, radical halogenation generally follows a Markovnikov-type pattern, because more substituted radicals are stabilized by hyperconjugation.

For instance, in methylcyclohexane, the major product is 1-methyl-1-chlorocyclohexane.

http://books.google.com/books?id=HyuogOtzoaYC&lpg=PA283&ots=1pjbx4E_zE&dq=radical%20chlorination%20of%20methylcyclohexane&pg=PA283#v=onepage&q=radical%20chlorination%20of%20methylcyclohexane&f=false

[Big-Daddy]

Chlorine is famously nonselective because it is a high-energy starting material.

I see, so we would expect a major product determined by the number of ways of forming each. From the book example, can you explain to me what is meant by "there are 4 ways to make each one"? I don't understand how the counting is being done.

However, radical halogenation generally follows a Markovnikov-type pattern, because more substituted radicals are stabilized by hyperconjugation.

For instance, in methylcyclohexane, the major product is 1-methyl-1-chlorocyclohexane.

http://books.google.com/books?id=HyuogOtzoaYC&lpg=PA283&ots=1pjbx4E_zE&dq=radical%20chlorination%20of%20methylcyclohexane&pg=PA283#v=onepage&q=radical%20chlorination%20of%20methylcyclohexane&f=false

Thanks. So in the case of my above molecule, we would get free radicals forming on both of the two C atoms with 3 carbon substituents. What would the product look like in the termination step where the organic radicals combine into one molecule? I'm guessing we'll have some kind of polymer but what would it look like?

[opsomath]

The short answer is that in the case of homocoupling (two identical radicals terminating in bond formation) the result would be a crazy molecule with two of those decalin molecules joined at their tertiary carbons. But I doubt that will happen much, the steric hindrance would be enormous. Coupling would be more likely at one of the exposed methylene sites on the opposite side from the bond which joins the two rings. Also, there would be a big mishmash of coupling products.

With reference to "four ways to make each one" they mean four chemically identical hydrogens, any one of which will lead to the product (the 1,2 and 1,3 - chlorinated methylcyclohexane derivatives). Does that make sense? By the same token, there are "two ways" (two hydrogens) to make the 1,4 product and only one way (one H) that makes the 1,1 product, yet the 1,1 product still predominates because of energetics of the different radicals.

[Big-Daddy]

The short answer is that in the case of homocoupling (two identical radicals terminating in bond formation) the result would be a crazy molecule with two of those decalin molecules joined at their tertiary carbons. But I doubt that will happen much, the steric hindrance would be enormous. Coupling would be more likely at one of the exposed methylene sites on the opposite side from the bond which joins the two rings. Also, there would be a big mishmash of coupling products.

With reference to "four ways to make each one" they mean four chemically identical hydrogens, any one of which will lead to the product (the 1,2 and 1,3 - chlorinated methylcyclohexane derivatives). Does that make sense? By the same token, there are "two ways" (two hydrogens) to make the 1,4 product and only one way (one H) that makes the 1,1 product, yet the 1,1 product still predominates because of energetics of the different radicals.

So because the C in the 1,1 product is more substituted (3 C substituents), this effect dominates the effect of there being more ways to make each product.

[opsomath]

Yup, basically. If there was no energy difference, the ratio of the different products would be identical to the ratios of the different equivalent groups of hydrogens in the molecule.