Question

Bromine test differentiates Benzene and Cyclohexane by:
(A) odour change
(B) effervescence
(C) colour change
(D) evolution of hydrogen gas

Answer 1

Hint: The basic knowledge about the bromine test will help us evaluate the given illustration. Also, one of the options can be correct as it is enough for a noticeable change to take place.

Complete Solution :
Let us first know about the bromine test in detail before we decide the answer:
Bromine test-
Bromine test, also known as bromine water test is usually used to test the presence of alkenes. Alkene goes addition reaction with bromine water only in dark conditions. The double bonds in the alkenes will react with bromine water resulting in a colour change of bromine water from intense yellow to colourless solution.
- Usually, this test differentiates alkanes and alkenes. Alkanes will not react with the bromine water and the solution will remain orange in colour. This test is also used to test the presence of aldehyde groups in a compound. Additionally, enols, alkenes, aniline, glucose, phenols and acetyl groups are the common compounds which will undergo bromine water test.
General alkene reaction can be given as:
$C{{H}_{2}}=C{{H}_{2}}+B{{r}_{2}}\to C{{H}_{2}}\left( Br \right)-C{{H}_{2}}\left( Br \right)$

- Bromine water has a highly oxidising property, so it is prepared in a laboratory with higher safety.
The illustration given have exceptions:

- Benzene-
Even though benzene has unsaturated bonds, they are stable and thus will not react with bromine water solution and thus no decolourisation of bromine water.
- Cyclohexane-
In dark conditions, cyclohexane does not show any reaction with bromine water. But when exposed to sunlight or UV light, free radical substitution reaction takes place. Hydrogen atoms from the cyclohexane are substituted with bromine atoms to give hydrogen bromide and ${{C}_{6}}{{H}_{\left( 12-n \right)}}B{{r}_{\left( n \right)}}$ .

- Overall reaction can be stated as:
${{C}_{6}}{{H}_{12}}+12B{{r}_{2}}\to {{C}_{6}}B{{r}_{12}}+12HBr$
Thus, here bromine water will decolourise to show positiveness of the reaction.
Therefore, Bromine test differentiates Benzene and Cyclohexane by colour change.
So, the correct answer is “Option C”.

Note: Cyclohexane and benzene, both are the exceptions to the general bromine test; though have valid reasons further.
Also, as $B{{r}_{2}}$ is present during the reaction thus, hydrogen bromide will surely be formed as a product. So, option (D) cannot be the solution.