Question

How many isomers does `C_4H_9Br` have?

Answer 1

4

Explanation:

First we should consider the formula as `C_4X_10`.

`10=2 xx 4 +2`: the number of X is the maximum value possible, which means there are no multiple bonds or cycles.

Then, how can we arrange the carbon atoms?

The first three, only can be arranged in a linear mode:

`C-C-C`

The fourth one can be allocated in the ends:

`C-C-C-C` (butane)

or in the middle carbon:

`C-C-C`
`color(white)(.......)|`
`color(white)(.. ....)C`
(isobutane)

Now, where can we put the Br atoms?

In butane we can put the Br in the first carbon (equivalent to the last)
giving bromobutane:

`Br-C-C-C-C` (butylbromide)

or in the second which is equivalent to the third:

`C-C-C-C`
`color(white)(.......)|`
`color(white)(......)Br`
(sec-butyl bromide)

In the case of isobutane, the external carbons are equivalent, giving:

`Br-C-C-C`
`color(white)(................)|`
`color(white)(................)C`
(isobutyl bromide)

but we can also substitute in the carbon of the middle:

`color(white)(......)Br`
`color(white)(.......)|`
`C-C-C`
`color(white)(.......)|`
`color(white)(.......)C`
(ter-butyl bromide)

The remaining positions we add hydrogen atoms, making each carbon atom four bonds.

So there are four isomers of `C_4H_9Br`