Question

n-propyl alcohol and isopropyl alcohol can be chemically distinguished by which reagent:
A.${\text{PC}}{{\text{l}}_{\text{5}}}$
B.Reduction
C.Oxidation with potassium dichromate
D.Ozonolysis

Answer 1

n-propyl alcohol is ${\text{C}}{{\text{H}}_{\text{3}}}{\text{--C}}{{\text{H}}_{\text{2}}}{\text{--C}}{{\text{H}}_{\text{2}}}{\text{--OH}}$. Isopropyl alcohol is ${\text{C}}{{\text{H}}_{\text{3}}}{\text{--CH(OH)--C}}{{\text{H}}_{\text{3}}}$. In the first case, OH is attached to a primary methyl group whereas, in isopropyl alcohol, OH is attached to a secondary methyl group.

Complete step by step answer:
The first option ${\text{PC}}{{\text{l}}_{\text{5}}}$ as we know is a chlorinating agent. On reacting n-propyl alcohol and iso-propyl alcohol with ${\text{PC}}{{\text{l}}_{\text{5}}}$, both the OH will be replaced by Cl atom. So ${\text{PC}}{{\text{l}}_{\text{5}}}$ cannot be used to distinguish between them.
We know that OH is not a good leaving group and thus it cannot be easily reduced to alkanes. The OH group is first converted into other easy leaving groups (maybe by treating with ${\text{PC}}{{\text{l}}_{\text{5}}}$ ) and then it is reduced to alkanes. Therefore, a reduction cannot be used to distinguish between n-propyl alcohol and isopropyl alcohol.
We know that acidified potassium chromate is an oxidizing agent.
On treating n-propyl alcohol with acidified potassium dichromate, we get an aldehyde
${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH + }}{{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}{\text{ + }}{{\text{H}}^{\text{ + }}} \to {\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CHO + }}{{\text{H}}_{\text{2}}}{\text{O + C}}{{\text{r}}^{{\text{3 + }}}}$
We already discussed that isopropyl alcohol is a secondary alcohol. On treating secondary alcohol with potassium dichromate in an acidic medium, it acts as an oxidation agent and gives propanone (ketone) as the product.
${\text{C}}{{\text{H}}_{\text{3}}}{\text{COC}}{{\text{H}}_{\text{3}}}$
Thus, n propyl alcohol and isopropyl alcohol give different products on oxidation with potassium dichromate. Therefore, this reagent can be used to distinguish between the two.
Ozonolysis is the addition of Oxygen to alkenes or alkynes and making it saturated. Therefore, it cannot be used to distinguish between them.

Thus, the correct option is C.

Note: There are various other distinguishing steps. Some of them are treated with Fehling's reagent, Tollen’s reagent, Schiff reagent, and benedict solution. These are some of the reactions that aldehydes undergo but ketone does not. On adding Schiff’s reagent, if there is a change in the pink color, it means that aldehyde is formed. This shows the presence of primary alcohol. It shows the absence of primary alcohol if there is no change in the pink color. Another test is the iodoform test which is answerable only by ketone. If it gives the iodoform test, it shows the presence of secondary alcohol.