How many isomers does $C_{4} H_{9} B r$ $C_4H_9Br$ have?

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Answer 1 · 2016-01-27T18:35:21

4

First we should consider the formula as $C_{4} X_{10}$ $C_4X_10$ .

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

$C - C - C$ $C-C-C$

The fourth one can be allocated in the ends:

$C - C - C - C$ $C-C-C-C$ (butane)

or in the middle carbon:

$C - C - C$ $C-C-C$
$color(white)(.......)|$
$color(white)(.. ....)C$
(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 $C_4H_9Br$

How many isomers does C_4H_9BrC4H9BrC_4H_9Br have?