[Agent-X]

It appears that the product with a 28% output is from a primary carbocation. The product with a 72% output was formed from a secondary carbocation. Am I right about this?

As a secondary carbocation is more stable than a primary carbocation, it makes sense that more 2-chlorobutane formed.




Now, in the second formula, it appears that the Cl atom is on the end of the left-hand product with a 63% output. Why is that? Wouldn't that have been a primary carbocation before it was attacked by the Cl atom? I'm under the assumption that the product on the right was tertiary, and it should have been the major product of the reaction.

This book is saying this:

"A similar study of the chlorination of 2-methylpropane established that a tertiary hydrogen is removed 5.2 times faster than each primary hydrogen."

So, shouldn't the product on the right have more output?

Am I misreading this somehow?
If the second molecular formula is right, why is it right?

[plankk]

Look that in both examples the numbers of primary, secondary and tertiary carbons aren't the same. Divide the output by the number of the same carbons as that which was substitute, for example 28% : 6 = 4,67%.

[Agent-X]

Look that in both examples the numbers of primary, secondary and tertiary carbons aren't the same. Divide the output by the number of the same carbons as that which was substitute, for example 28% : 6 = 4,67%.

I'm sorry, but I don't understand your statement.
Where are you getting the number 6 from? Why are you trying to divide 28 by 6?
In neither reactant nor product do I see 6 carbon atoms.

[plankk]

Butane has got 6 primary carbons and 4 secondary carbons. Statistically the more favourable are primary carbons than secondary (3:2 probability). But as the experiment shows, not only statistic influence on the reaction but also reactivity of species. We can calculate the relative reactivity by divide the output by the number of the same carbons as that which was substitute. In the first example we have the first product, which output is 28%. The chlor occupy primary position. To make such product were 6 posibilities (six the same atoms). So the effectiveness reactivity is: 28% (the output) : 6 (the same atoms) = 4,67%. For the second product, the output is 72%, but we have only 4 the same atoms, so the effectiveness reactivity is: 72% : 4 = 18%. The relative reactivity primary : secondary is 4,67% : 18%.

[Darwin]

Yes, experimental data reveal that 1-chloro-2-methylpropane predominates over 2-chloro-2-methylpropane even though the radical intermediate formed from the abstraction of H⁺ from a tertiary C atom is energetically favoured over that formed from abstraction of H⁺ from a primary C atom. The experimental observations can be rationalized by considering that there are nine H atoms attached to primary C atoms in 2-methylpropane and only one H atom attached to a tertiary C atom. While abstraction of H⁺ from the tertiary C centre is favoured on energetic grounds and is faster than abstraction of H⁺ from a primary C atom, the statistical change of abstracting H⁺ from a primary C atom is higher than that of removing H⁺ from a tertiary centre. Thus, there are two competing factors, and the observed product distributiuon confirms that abstraction of H⁺ from a primary C atom wins over the increased reactivity of the tertiary C atom.

[Agent-X]

Thank you everyone, but it doesn't seem logical.

I'm going to assume there is a typographical error.

I tried looking for a secondary source on the Internet, and this correlates to my understanding:

These results suggest strongly that 2º-hydrogens are inherently more reactive than 1º-hydrogens, by a factor of about 3:1. Further experiments showed that 3º-hydrogens are even more reactive toward halogen atoms. Thus, light-induced chlorination of 2-methylpropane gave predominantly (65%) 2-chloro-2-methylpropane, the substitution product of the sole 3º-hydrogen, despite the presence of nine 1º-hydrogens in the molecule.

(CH3)3CH   +   Cl2   ——>  65% (CH3)3CCl   +   35% (CH3)2CHCH2Cl

If you are uncertain about the terms primary (1º), secondary (2º) & tertiary (3º) Click Here.

- Source: http://www.cem.msu.edu/~reusch/VirtualText/funcrx1.htm

In this website, it is shown that the tertiary radical obtained 65% output.

[plankk]

Curiously I looked for it in two books (Clayden "Organic Chemistry" and Boyd-Morrison "Organic Chemistry"). Both say that the tertiary radical obtained 36-37% output.

[Agent-X]

NOTICE:

I meant radicals rather than carbocations.

Curiously I looked for it in two books (Clayden "Organic Chemistry" and Boyd-Morrison "Organic Chemistry"). Both say that the tertiary radical obtained 36-37% output.

I'm also reading that percentage in other books.
I'm also noticing a similarity in rhetoric.

If the tertiary bond is weaker, then it should require less energy. And the need for less energy, means more stability.

Am I to assume that I'm suppose to only examine the relative reactivity from such experimental data?

For example, am I not suppose to consider primary, secondary, or tertiary radical stability when determining product output? Am I only to look for past experimental data that tabulates rate relativity in order to know the percentage of products that come out?

I thought stability was suppose to be the theoretical ground for determining experimental output. In other words, the theoretical understanding provides knowledge as to what product will overpower the other products and the reasons for doing so. The reasons would be intermediate stability versus the intermediate stability of other carbon radicals.

It seems that theory and experimental data do not coorelate. And being that the case, the theory seems disproved.

[Agent-X]

I found a nice wiki article that goes into the math.
However, it seems that the radical stability theory is disproved.
I'm seeing one theory of output based on mathematical methods.
And then I'm seeing another theory of output based on radical stability.
Why do these two theories exist?

Example of mathematical method:

chlorination of 2-methylpropane:

a = (CH₃)₃
b = CH

q = 1 (relative rate of chlorination for primary hydrogens)
r = ~5  (relative rate of chrlorination for tertiary hydrogens)

a = 9(primary hydrogens) x q = 9
b = 1(tertiary hydrogen) x r = 5

9 + 5 = 14
14 = added total

9/14 = 64% = 1-chloro-2-methylpropane
5/14 = 36% = 2-chloro-2-methylpropane

[plankk]

Both theories accomplish each other. The radical stability theory says us which radical is more favourable, but it doesn't take into account the number of primary, secondary and tertriary carbons. The mathematical theory does. So if you want to predict which product will overpower the other products you should at first consider which homolysis will make the most stable radical (the radical stability theory)), and then consider probability of that process (the mathematical theory).