# Monochlorination Problem, Please give a hand.

#### Intent

5+ Year Member
7+ Year Member
Here is a link the problem: (Problem # 10.4)

I understand how to get the four different products. I cannot figure out how they got the percentages. Here is what I know, maybe someone can fill in a missing link.

Here are the four products:
1-chloro-2-methylbutane
1-chloro-3-methylbutane
2-chloro-2-methylbutane
2-chloro-3methylbutane

How do I find the approximate %'s of each product? 2-methylbutane has 9 primary hydrogens, 1 secondary hydrogen and 1 tertiary hydrogen. But I don't understand how to work this out following my book example (it's mcmurry 8th ed). In the book they give the ratio's of the two products formed from chlorination of butane, (30:70) and chlorination of 2-methylpropane (35:65) as examples. It says to take the number of primary hydrocarbons and divide it by the ratio. My question is how do I get this ratios if they are not provided? Any help would be great, thanks.

#### chlrbwls

7+ Year Member
Well first think of the different hydrogen environments there are and also the idea that rearragements can occur.

But a question like this wont be asked in the DAT so i donno why you are looking into it

1st hydrogen environment (on the 2-methyl group)
- from where ever you chlorinate, they have equal probability so a total of 6 equal chlorination that results in the same product.

2nd hydrogen environment (on the tertiary carbon)
- there is only one hydrogen for this environment but you forget that this is the most stable radical and therefore would have much higher probability of forming compared to 2ndary and primary radical (i dont remember by what factor you have to multiply)

My guess is that you add the total number of hydrogens and subtract the number of hydrognes you have for that specific hydrogen environment

I.e. (Sum of hydrogens from environment other than tertiary - tertiary) = (9 + 2) - 1 = 10

3rd hydrogen environment (on the secondary carbon)
- this environment has actually 2 hydrogens i think that was a typo on your part

(Same method as 2 but this time because tertiary carbon is more stable than secondary we neglect this)

(Sum of hydrogens from environments other than secondary - secondary) = 9 - 2 = 7

4th hydrogen environment (different primary carbon)
- the prabability equals to the number of hydrogens you have over the total possibility = 3

To get a total of 26 you add all of these values (6 + 10 + 7 + 3) = 26

then you use this as the total number of possible chlorinations.

I think this is the most logical explanation.

OP

#### Intent

5+ Year Member
7+ Year Member
Thanks for the quick reply. I think finding the 7 and the 10 weren't the issue, it was where the 6 and the 3 came into play that threw me off. The book doesn't give that useful of a breakdown with the example provided. You don't think there will be a similar problem on the DAT? I have the DAT destroyer and odyssey and tried looking through for a similar example but couldn't find one. I suppose you may be right about it not being on the DAT. Thanks for the help again, much appreciated.