
Two aromatic compounds having formula \[{C_7}{H_8}O\] which are easily identifiable by $FeC{l_3}$ solution test (violet colouration) are:
(a) ${\text{o - cresol}}$, and benzyl alcohol
(b) ${\text{m - cresol}}$, and ${\text{p - cresol}}$
(c) ${\text{o - cresol}}$, and ${\text{p - cresol}}$
(d) methyl phenyl ether, and benzyl alcohol
Answer
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Hint: Phenols give coloured compounds upon reaction with Ferric Chloride, $FeC{l_3}$. The formation of coloured complexes involves the mechanism of keto – enol tautomerism. Compounds undergoing this mechanism will give a positive $FeC{l_3}$ test.
Complete step by step answer:
The sample which is to be tested is dissolved in water (in case of water-soluble substance) or in a mixture of ethanol, and water (in case of water insoluble substance). Phenol, pyrogallol, and sulfosalicylic acid are soluble in water, but some phenols like $2 - naphthol$, Gallic acid, salicylic acid acids are insoluble in water.
Step (2):
The mixture is then added to a solution containing a few drops of Ferric Chloride, $FeC{l_3}$. Depending upon the type of phenol, it gives violet, red, green or blue colouration compounds.
Step (3):
The reaction undergoes keto – enol tautomerism, which is a process of converting an aldehyde or ketone containing $\alpha {\text{ - Hydrogen}}$ (${\text{C - H}}$ bond adjacent to the Carbonyl compound) to its corresponding enol, when the condition is acidic in nature, and to enolates, when the condition is basic in nature. Like enols, the phenolate can also form coloured complexes with $FeC{l_3}$. The reaction involved here is:
$6PhOH + F{e^{3 + }} \to {\left[ {Fe{{(OPh)}_6}} \right]^{3 - }}$,
where, ${\text{PhOH}}$ is the sample phenol,
$F{e^{3 + }}$ is the Ferric ion, comes from the dissociation of $FeC{l_3}$,
and, ${\left[ {Fe{{(OPh)}_6}} \right]^{3 - }}$ is the violet coloured complex.
In option (a) ${\text{o - cresol}}$will give positive phenol test as it contains $\alpha {\text{ - Hydrogen}}$ to form keto whereas Benzyl alcohol will not give the positive test as it does not have a $\alpha {\text{ - Hydrogen}}$. So, both can be distinguished from each other easily.
For option (b), and (c), all three ${\text{o - cresol}}$, ${\text{m - cresol}}$, and ${\text{p - cresol}}$ give positive $FeC{l_3}$ test. So, they cannot be distinguished easily.
In option (d), both the compounds will not give the $FeC{l_3}$ test. So, they also cannot be distinguished easily.
So, the correct answer is Option A .
Note:
Ferric Chloride, $FeC{l_3}$ test is not a conformational test for phenols as some phenols does not react with it, and also it is very sensitive test, it has to be done in specified ${\text{pH}}$ of the reaction medium, and concentration. Instead, the Dye test is one of the conformational tests for phenols.
Complete step by step answer:
The sample which is to be tested is dissolved in water (in case of water-soluble substance) or in a mixture of ethanol, and water (in case of water insoluble substance). Phenol, pyrogallol, and sulfosalicylic acid are soluble in water, but some phenols like $2 - naphthol$, Gallic acid, salicylic acid acids are insoluble in water.
Step (2):
The mixture is then added to a solution containing a few drops of Ferric Chloride, $FeC{l_3}$. Depending upon the type of phenol, it gives violet, red, green or blue colouration compounds.
Step (3):
The reaction undergoes keto – enol tautomerism, which is a process of converting an aldehyde or ketone containing $\alpha {\text{ - Hydrogen}}$ (${\text{C - H}}$ bond adjacent to the Carbonyl compound) to its corresponding enol, when the condition is acidic in nature, and to enolates, when the condition is basic in nature. Like enols, the phenolate can also form coloured complexes with $FeC{l_3}$. The reaction involved here is:
$6PhOH + F{e^{3 + }} \to {\left[ {Fe{{(OPh)}_6}} \right]^{3 - }}$,
where, ${\text{PhOH}}$ is the sample phenol,
$F{e^{3 + }}$ is the Ferric ion, comes from the dissociation of $FeC{l_3}$,
and, ${\left[ {Fe{{(OPh)}_6}} \right]^{3 - }}$ is the violet coloured complex.
In option (a) ${\text{o - cresol}}$will give positive phenol test as it contains $\alpha {\text{ - Hydrogen}}$ to form keto whereas Benzyl alcohol will not give the positive test as it does not have a $\alpha {\text{ - Hydrogen}}$. So, both can be distinguished from each other easily.
For option (b), and (c), all three ${\text{o - cresol}}$, ${\text{m - cresol}}$, and ${\text{p - cresol}}$ give positive $FeC{l_3}$ test. So, they cannot be distinguished easily.
In option (d), both the compounds will not give the $FeC{l_3}$ test. So, they also cannot be distinguished easily.
So, the correct answer is Option A .
Note:
Ferric Chloride, $FeC{l_3}$ test is not a conformational test for phenols as some phenols does not react with it, and also it is very sensitive test, it has to be done in specified ${\text{pH}}$ of the reaction medium, and concentration. Instead, the Dye test is one of the conformational tests for phenols.
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