The total number of monobromo derivatives formed by the alkanes with molecular formula $$C_5H_{12}$$ is ______ (excluding stereo isomers).

We need to find the total number of monobromo derivatives formed by all isomers of the alkane with molecular formula $$C_5H_{12}$$, excluding stereoisomers.

Identify all structural isomers of $$C_5H_{12}$$

There are three structural isomers:

(i) n-Pentane: $$CH_3-CH_2-CH_2-CH_2-CH_3$$

(ii) Isopentane (2-methylbutane): $$(CH_3)_2CH-CH_2-CH_3$$

(iii) Neopentane (2,2-dimethylpropane): $$(CH_3)_4C$$

n-Pentane — types of hydrogen

$$\underset{a}{CH_3}-\underset{b}{CH_2}-\underset{c}{CH_2}-\underset{b}{CH_2}-\underset{a}{CH_3}$$

Due to the plane of symmetry at C-3, there are 3 types of hydrogens (a, b, c), giving 3 monobromo products:

1-bromopentane, 2-bromopentane, and 3-bromopentane.

Isopentane — types of hydrogen

The structure $$(CH_3)_2CH-CH_2-CH_3$$ has 4 types of hydrogens:

(a) H on the two equivalent methyl groups at C-1 and C-1' (6H)

(b) H on the methine (CH) at C-2 (1H)

(c) H on the methylene ($$CH_2$$) at C-3 (2H)

(d) H on the terminal methyl ($$CH_3$$) at C-4 (3H)

This gives 4 monobromo products: 1-bromo-2-methylbutane, 2-bromo-2-methylbutane, 2-bromo-3-methylbutane (by substitution at C-3), and 1-bromo-3-methylbutane.

Neopentane — types of hydrogen

All four $$CH_3$$ groups are equivalent, so there is only 1 type of hydrogen (12H), giving 1 monobromo product: 1-bromo-2,2-dimethylpropane.

Total count

$$\text{Total monobromo derivatives} = 3 + 4 + 1 = 8$$

The answer is 8.