Question

On mixing a certain alkane with chlorine and irradiating it with U.V light it forms only one mono chloroalkane. This alkane could be
(a) Pentane
(b) Isopentane
(c) Neopentane
(d) Propane

Answer 1

Hint: In halogenation reaction, depending on the types of hydrogen atom present, the chemical compound on photochemical reaction gives single or multiple monochlorinated products.

Complete step by step answer:
Halogenation is the reaction where one or more than one hydrogen atom present in the chemical compound is replaced by a halogen group. It is a type of substitution reaction where the C-H bond is broken and a new C-X bond is formed.
In pentane five carbon atoms are present in a straight chain. It contains three types of the hydrogen atom which on monochlorination give three products. They are 1-chloropentane, 2-chloropentane and 3-chloropentane.
In isopentane five carbon atoms are present out of which one is in branched form. It contains four types of a hydrogen atom which is present which on monochlorination gives four products. They are: 1-chloro-3-methylbutane, 2-chloro-3-methylbutane, 2-chloro-2-methylbutane and 1-chloro-2-methylbutane.
In neopentane five carbon atoms are present which is doubly branched. It contains only one type of hydrogen atom which on monochlorination gives one product which is 1-chloro-2,2-dimethylbutane.
In propane three carbon atoms are present in a straight chain. It contains two types of hydrogen atom which on monochlorination give two products. They are 1-chloropropane and 2-chloropropane.
The reaction of neopentane with chlorine is shown below.

Neopentane reacts with chlorine in presence of U.V light to form 2,2-dimethyl 1-chloropropane.

So, the correct answer is Option C.

Note: Isopentane, neopentane, and n-pentane are the isomer of pentane as they have the same molecular formula but they differ from each other by their structural representation.