On mixing certain alkane with chlorine and irradiating with ultraviolet light, it forms only one monochloroalkane. This alkane could be:

propane

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pentane

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isopentane

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neopentane

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Solution

The correct option is D neopentane
Number of monochloro products formed depends on number of different sets of hydrogen.
In propane; $C H_{3}^{a} - C H_{2}^{b} - C H_{3}^{a}$ ,there are two different types of hydrogens as labelled.
In pentane; $C H_{3}^{a} - C H_{2}^{b} - C H {_{2}}^{b} - C H {_{2}}^{b} - C H_{3}^{a}$ also there are two types of hydrogens
In isopentane; $C H_{3}^{a} - C H^{c} (C H 3^{a}) - C H {_{2}}^{b} - C H_{3}^{a}$ , there are three different types of hydrogens.
In neopentane ; $C H_{3}^{a} - (C H {_{3}}^{a}) C (C H {_{3}}^{a}) - C H_{3}^{a}$ , there is only one type of hydrogen .
Thus amongst the given options, neopentane gives only one monochloroalkane on chlorination.

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