# How many isomers does C_4H_9Br have?

Jan 27, 2016

4

#### Explanation:

First we should consider the formula as ${C}_{4} {X}_{10}$.

$10 = 2 \times 4 + 2$: 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:

$C - C - C$
$\textcolor{w h i t e}{\ldots \ldots .} |$
$\textcolor{w h i t e}{. . \ldots .} C$
(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:

$B r - C - C - C - C$ (butylbromide)

or in the second which is equivalent to the third:

$C - C - C - C$
$\textcolor{w h i t e}{\ldots \ldots .} |$
$\textcolor{w h i t e}{\ldots \ldots} B r$
(sec-butyl bromide)

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

$B r - C - C - C$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots .} |$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots .} C$
(isobutyl bromide)

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

$\textcolor{w h i t e}{\ldots \ldots} B r$
$\textcolor{w h i t e}{\ldots \ldots .} |$
$C - C - C$
$\textcolor{w h i t e}{\ldots \ldots .} |$
$\textcolor{w h i t e}{\ldots \ldots .} C$
(ter-butyl bromide)

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

So there are four isomers of ${C}_{4} {H}_{9} B r$