How many isomers does #C_4H_9Br# have?

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#