Concept used: monochlorination of alkanes

On chlorination in ultraviolet light, one hydrogen atom of the alkane is replaced by chlorine: $\text{R-H} + \text{Cl}_2 \xrightarrow{h\nu} \text{R-Cl} + \text{HCl}$

If an alkane gives only one monochloroalkane, then all hydrogen atoms must be equivalent by symmetry.

Check each option

Option A: pentane, $\mathrm{CH_3-CH_2-CH_2-CH_2-CH_3}$

Pentane has three different types of hydrogens:

• terminal $\mathrm{CH_3}$ hydrogens
• next $\mathrm{CH_2}$ hydrogens
• middle $\mathrm{CH_2}$ hydrogens

So monochlorination can occur at different positions, giving more than one monochloro product.

Hence, A is not correct.

Option B: isopentane (2-methylbutane)

Structure: $\mathrm{CH_3-CH(CH_3)-CH_2-CH_3}$

It has multiple non-equivalent hydrogens:

• tertiary hydrogen on the second carbon
• secondary hydrogens on $\mathrm{CH_2}$
• primary hydrogens on different $\mathrm{CH_3}$ groups

Therefore, it gives more than one monochloroalkane.

Hence, B is not correct.

Option C: neopentane (2,2-dimethylpropane)

Structure: $\mathrm{C(CH_3)_4}$

In neopentane, all four methyl groups are identical due to high symmetry. Therefore, all 12 hydrogens are equivalent.

Replacing any one hydrogen by chlorine gives the same product.

Hence, C is correct.

Option D: propane, $\mathrm{CH_3-CH_2-CH_3}$

Propane has two types of hydrogens:

• primary hydrogens on terminal $\mathrm{CH_3}$ groups
• secondary hydrogens on middle $\mathrm{CH_2}$ group

So it gives two monochloro products:

• 1-chloropropane
• 2-chloropropane

Hence, D is not correct.

Final conclusion

The alkane that forms only one monochloroalkane is: $\boxed{\text{neopentane}}$

So the correct option is: $\boxed{\text{C}}$