Question

Identify the heteroatom from the following?

(A) Carbon

(B) Hydrogen
(C) Bromine
(D) Bromide

Answer 1

Hint: Heteroatoms are those atoms other than hydrogen or carbon in a molecular structure. Among the given options there is only one atom other than carbon and hydrogen. This atom would be the heteroatom involved in the given compounds.

Complete step by step answer:
 -Let's start with the concept of heteroatoms. Heteroatoms can be defined as those atoms other than carbon or hydrogen. Even though this leaves a number of possibilities, there are only a limited number
of atoms that are encountered in organic molecules.

 -When we look at a hydrocarbon chain, we can see that one or more hydrogen atoms can be replaced by atoms other than hydrogen atoms depending on their valencies. So more specifically we can describe a heteroatom as non-carbon or non-hydrogen atoms that have replaced carbon or hydrogen from their backbone of molecular structure.

 -Every heteroatom would have a lesser or greater attraction for electrons than carbon does. Therefore, each bond between a carbon atom and a heteroatom is said to be polar in nature.
 -This degree of polarity will depend upon the difference between electron attracting properties of two atoms involved. This may contain such reactive polar groups and will also be able to generate functional groups in hydrocarbons.

 -These heteroatoms will also affect the reactivity of organic molecules.
-The most common examples of heteroatoms are Oxygen, Sulphur, nitrogen, phosphorus, bromine, chlorine, iodine and also metals such as lithium and magnesium.

- Among the four compounds, Bromine is the only heteroatom present. It has replaced hydrogen in every compound and thus it acts as a heteroatom.
So, the correct answer is “Option C”.

Note: There might be a confusion among bromine and bromide. In the given organic compounds, the bromine exists as bromide ion ($B{{r}^{-}}$). But we are asked to find the heteroatom involved in the compounds and thus the answer would be bromine not bromide.