Question

Distinguish between the following by chemical tests: Propan-1-ol and Propan-2-ol

Answer 1

Hint: To answer this question, you should recall the concept of differentiation of alcohols. Chemical tests to distinguish alcohols mainly involve conversion reactions. For example, on oxidation, a primary alcohol will give an aldehyde, a secondary alcohol will give a ketone while a tertiary alcohol will not react with an oxidizing agent. Thus, three compounds will be present with different functional groups which can be easily distinguished.

Complete Step by step solution:
1.Lucas test is used to distinguish between primary and secondary alcohol. The principle of this test is the difference in reactivity of the alcohols within a \[S{{N}^{1}}\] reaction:
\[ROH\text{ }+\text{ }HCl\to RCl+{{H}_{2}}O\]
This test easily distinguishes between primary and secondary alcohol because secondary carbocations are more stable and form faster than primary carbocations. The Lucas reagent is prepared by mixing an equimolar solutions of \[ZnC{{l}_{2}}~and~HCl.\]When added to the unknown alcohol sample, secondary alcohol will react within 3 min to 5 min to form the alkyl halide, which is insoluble and forms a turbid layer whereas A primary alcohol does react with Lucas reagent at room temperature.
2.Victor Meyer test is also used to distinguish between primary and secondary alcohol. The key principle of this test is that unknown alcohol is converted into the corresponding nitroalkane which is then treated with nitrous acid followed by alkalifying the solution. The primary alcohol gives blood-red colour, secondary alcohol gives blue and tertiary alcohol remains colourless.

Note: This reaction proceeds through a backside attack by the nucleophile on the substrate. The nucleophile approaches the given substrate at an angle of\[{{180}^{o}}\] to the carbon-leaving group bond. It is important to note that the product is formed with an inversion of the tetrahedral geometry at the atom in the centre. You should remember the important points regarding $S{{N}^{2}}$ reactions:
1.$S{{N}^{2}}$ reactions are bimolecular reactions in which there are simultaneous bond-making and bond-breaking steps.
2.$S{{N}^{2}}$ reactions result in inverted stereochemistry at the reaction center.
3.Steric effects are particularly important in $S{{N}^{2}}$reactions.
4.Unhindered back of the substrate makes the formation of carbon-nucleophile bonds easy. Therefore, methyl and primary substrates undergo nucleophilic substitution easily.
5.More negative charge more the nucleophilicity and a strong nucleophile can easily form the carbon-nucleophile bond.
6.Polar aprotic solvents do not hinder the nucleophile, but polar solvents form hydrogen bonds with the nucleophile. A good solvent for this reaction is acetone.
7.Stability of the anion of the leaving group and the weak bond strength of the leaving groups bond with carbon help increase the rate.