Question

The compound that will react most readily with gaseous bromine has the formula is:
A. ${\text{A }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$
B. ${\text{B }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}$
C. ${\text{C }}{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{6}}}$
D. ${\text{D }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}$

Answer 1

As we know that bromine is a chemical element with the symbol ${\text{Br}}$ and atomic number is ${\text{35}}$
It is a fuming red-brown liquid at room temperature. Moreover, its appearance is of reddish-brown. It belongs to p-block and of period 4.

Complete step by step answer: The compound that will react most readily with gaseous bromine has the formula${{\text{C}}_{\text{3}}}{{\text{H}}_{\text{6}}}$. Unsymmetrical alkenes in which the pair of ligands on one doubly bonded carbon and this is different from that on the other. Moreover, symmetrical alkenes are the alkenes in which the molecule having doubly bonded carbons bears the same ligands. Moving further, unsymmetrical alkenes generally are more reactive than the symmetrical alkenes, alkynes and alkanes. That is why; propene is more reactive than the other given compounds. Talking more about bromine, it is reactive non-metal and the electronic configuration is$\left[ {{\text{Ar}}} \right]$ ${\text{3}}{{\text{d}}^{{\text{10}}}}{\text{4}}{{\text{s}}^{\text{2}}}{\text{4}}{{\text{p}}^{\text{5}}}$. If we divide electrons according to per shell that is ${\text{2,8,18,7}}$. Its crystal structure is orthorhombic.
Hence, $\left( {\text{C}} \right)$ is the correct option.

Note: In this question addition of bromine is there where bromine is a halogen. So, basically addition of the halogen group is an electrophilic addition reaction. In this most reactive is alkenes because it has $\sigma$ and $\pi$ bond. $\pi$ Bond is the weakest bond so it breaks easily. Alkenes are more reactive towards electrophilic addition reaction towards halogens because $\pi$ bond breaks easily .