# Textbook: *Brown 6th Ed. (2011)*

## Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

## Practice Problems (No matching mendel sets were found.)

## Individual Problems

Using curved hooks, draw a mechanism for the free radical bromination reaction shown below.

Note that the radical will form on the 3° carbon only, as this is the most stable radical.

The structure of 2-methylbutane is shown below.

a) Draw the structures of all possible monochloro products resulting from the free-radical chlorination of 2-methylbutane.

b) Based on statistics alone, what do you expect the major product to be? Is this the same structure as the expected major product? Explain.

c) How would the relative yield of the products differ if bromine was used instead of chlorine?

**a)** There are five different carbons in 2-methylbutane, so hypothetically, you should get five different products, 20% A, 20% B, 20% C, 20% D, and 20% E.

But notice that A and B are the same product! So only four products are possible.

**b)** Based on statistics alone, we would expect product A (or B) to be the major product, encompassing 40% of all the products (20% A + 20% B).

But this is not what happens in real life because A is primary, and for free-radical halogenation reactions, more substituted products are favored. So the major product will be C, which is tertiary.

This happens because the intermediate is a radical, and radical stability goes 3° > 2° > 1°.

**c) **Chlorine reacts faster than bromine, and so is less selective than bromine.

For example, we know that the 3° halide will be the major product. But with chlorine, the proportion might turn out to be ~70% 3° , 20% 2°, and 10% 1°.

But since bromine reacts more slowly, you will get better selectivity. The proportion of products will might be something like ~95% 3° , 4% 2°, and 1% 1°.

While on this topic of reactivity, it's interesting to note that iodine reacts too slowly to be useful, and fluorine reacts so violently it's dangerous!