Question

Primary, secondary and tertiary alcohol cannot be differentiated by:
A. reduced $Cu$
B. acidified $KMn{O_4}$
C. Lucas reagent
D. $PC{l_3}$

Answer 1

Hint: For a compound to distinguish alcohols, it must produce different products with each type of alcohol. Reduced copper acts as a mild oxidising agent, while potassium permanganate is a strong one. Lucas reagent is concentrated hydrochloric acid and zinc chloride. Phosphorus trichloride produces the corresponding alkyl halides on reaction with alcohols.

Complete step by step answer:
 Let us examine what each of the given reagents do to the different types of alcohols:
Reduced copper acts as a mild oxidising agent. Thus, when vapours of the alcohols are passed over finely divided copper, primary alcohols produce aldehydes, secondary alcohols give ketones and tertiary alcohols undergo elimination reaction to give alkenes. Since there are three different products being formed, these can be distinguished easily and hence, reduced $Cu$ can be used to differentiate alcohols.

$C{H_3}OH\xrightarrow[{300K}]{{Cu}}HCHO + {H_2}$

${(C{H_3})_2} - CH - OH\xrightarrow[{300K}]{{Cu}}{(C{H_3})_2} - C = O + {H_2}$

${(C{H_3})_3} - C - OH\xrightarrow[{300K}]{{Cu}}{(C{H_3})_2} - C = C{H_2} + {H_2}$

Let us now check the second reagent, acidified $KMn{O_4}$. This is a strong oxidising agent and thus, oxidises primary alcohols into aldehydes which on further oxidation gives carboxylic acids; secondary alcohols into ketones which undergo hydrolysis to give a mixture of carboxylic acids, and tertiary alcohols undergo elimination to yield alkenes. Thus, in this too, since there are three different products being formed, these can be distinguished easily and hence, acidified $KMn{O_4}$ can be used to differentiate alcohols.

$C{H_3}OH\xrightarrow[{{H_3}{O^ + }}]{{KMn{O_4}}}HCHO\xrightarrow[{{H_3}{O^ + }}]{{KMn{O_4}}}HCOOH$

${(C{H_3})_2} - CH - OH\xrightarrow[{{H_3}{O^ + }}]{{KMn{O_4}}}{(C{H_3})_2} - C = O\xrightarrow[{{H_3}{O^ + }}]{{KMn{O_4}}}C{H_3}COOH + C{H_3}COOH$

${(C{H_3})_3} - C - OH\xrightarrow[{{H_3}{O^ + }}]{{KMn{O_4}}}{(C{H_3})_2} - C = C{H_2}$

Lucas reagent ($con.HCl + ZnC{l_2}$) produces the corresponding alkyl halides on treatment with alcohols. Their reactivity is used as a measure to distinguish them:
Primary alcohols do not produce any turbidity when the alkyl halides are formed.
Secondary alcohols produce turbidity in the solution within five to six minutes of the reaction, and
Tertiary alcohols produce immediate turbidity.
Thus, alcohols can be clearly distinguished using the Lucas test.

$C{H_3}OH\xrightarrow[{ZnC{l_2}}]{{con.HCl}}C{H_3}Cl$
                                              (no turbidity)

${(C{H_3})_2} - CH - OH\xrightarrow[{ZnC{l_2}}]{{con.HCl}}{(C{H_3})_2} - CH - Cl$
                                              (turbidity in five to six minutes)

${(C{H_3})_3} - C - OH\xrightarrow[{ZnC{l_2}}]{{con.HCl}}{(C{H_3})_3} - C - Cl$
                                                            (immediate turbidity)

The last reagent we have is $PC{l_3}$. This reaction with alcohols produces the corresponding alkyl halide, but there are no ways to distinguish between them since all follow the same reaction pathway and produce similar results visually too. Thus, $PC{l_3}$ is not a good distinguisher of alcohol.
$3R - OH + PC{l_3} \to 3R - Cl + {H_3}P{O_3}$

So, the correct answer is Option D .

Note: Tertiary alcohols do not undergo oxidation due to the fact that they do not contain a hydrogen atom bonded to the carbon atom having the alcohol group. Thus, they produce dehydrogenation reactions to yield alkenes. Note that the best method of distinguishing alcohols is the Victor-Meyers’s test, which gives different coloured products with each type of alcohol.