00:02
Monobromination is a free radical substitution where a hydrogen is replaced with a bromine.
00:16
And so the number of products will be based on the number of inequivalent types of hydrogen that are in the structure.
00:34
So we can substitute a primary carbon on the branch.
00:44
We can substitute a secondary, or sorry, primary hydrogen or a secondary hydrogen on the branch.
00:51
We can substitute the tertiary hydrogen, and we can substitute on the ring.
01:04
And because of the symmetry of the rain, there are only two positions.
01:15
We need to consider.
01:24
So let's just call these products, a, b, c, d, and e.
01:44
And for product ratios, what we do is take the number of that type of hydrogen and multiply by the relative selectivity or reactivity for that type of hydrogen.
02:10
So for product a, we have three primary hydrogens.
02:17
And from the table, we see that the relative reactivity with bromine is one.
02:26
So three times one is three.
02:33
For product b, we have two secondary hydrogens.
02:43
And the relative selectivity is 82.
02:47
So two times 82.
02:50
Is 164.
03:00
For c, we just have one tertiary, but it's got a relative reactivity of 1600.
03:17
For d, we're back to secondary hydrogens now.
03:24
Keep in mind, we have four.
03:27
That could be replaced to give that same product so four times 82 is 328 and then for product b it's a similar situation we have four secondary hydrogens of that type and relative reactivity of 82 so now we need to add these numbers together which gives us a total of 2 ,423 and now to get the percent composition, we take each individual one divided by that sum and multiply by 100.
04:48
So we can just tell off the bat that the tertiary substitution is going to be the major product.
04:59
We still need to do the math to determine what percent that will represent.
05:11
But the 1600 is more than half of 2423.
05:28
So the product a is only going to be about 0 .1 % of the total.
05:42
Product b will be about 6 .8 % of the total.
05:52
Product c would be about 66 % of the total.
06:01
Total and products d &e would both be about 13 and a half percent of the total.
06:19
Now, if we do this for chlorine, the possible products are analogous.
06:36
We just change the bromine to a chlorine...