Properties of Phenols Questions in English

(B) The acidity of phenols is increased by the presence of electron-withdrawing groups $(EWG)$ like $-NO_2$ on the benzene ring due to their $-I$ and $-M$ effects.
In $o$-nitrophenol,the $-NO_2$ group is at the ortho position,which exerts both $-I$ and $-M$ effects. However,it also forms an intramolecular hydrogen bond with the $-OH$ group,which slightly stabilizes the molecule but makes the release of the proton $(H^+)$ slightly less favorable compared to the para isomer.
In $p$-nitrophenol,the $-NO_2$ group is at the para position,exerting a strong $-M$ effect (resonance) and $-I$ effect,which significantly stabilizes the phenoxide ion through resonance. $p$-Nitrophenol is generally considered more acidic than $o$-nitrophenol due to the lack of intramolecular hydrogen bonding that hinders proton release.
In $m$-nitrophenol,the $-NO_2$ group only exerts an $-I$ effect (no resonance effect at the meta position),making it less acidic than both ortho and para isomers.
Therefore,$p$-nitrophenol is the most acidic among the given options.

(B) Aspirin (acetylsalicylic acid) is prepared by the acetylation of salicylic acid ($o$-hydroxybenzoic acid) with acetic anhydride in the presence of an acid catalyst.
The reaction is: $C_6H_4(OH)COOH + (CH_3CO)_2O \rightarrow C_6H_4(OCOCH_3)COOH + CH_3COOH$.

(A) Aspirin (acetylsalicylic acid) is prepared by the acetylation of salicylic acid ($o-$hydroxybenzoic acid) using acetic anhydride in the presence of an acid catalyst.
The reaction is: $C_7H_6O_3 + (CH_3CO)_2O \rightarrow C_9H_8O_4 + CH_3COOH$.

(C) Phenol is a versatile chemical compound.
At lower concentrations (e.g.,$0.2 \%$ solution),phenol acts as an antiseptic,which can be applied to living tissues.
At higher concentrations (e.g.,$1 \%$ solution),it acts as a disinfectant,which is used for inanimate objects like floors and instruments.
Therefore,it is used as both an antiseptic and a disinfectant.

(A) Diazonium salts undergo coupling reactions with phenols or aromatic amines in a basic or acidic medium to form azo dyes. This is a characteristic reaction used for the identification of aromatic amines and phenols.

(B) The $-OH$ group present in phenol is a strong activating group. It increases the electron density in the benzene ring through resonance,making it much more reactive towards electrophilic substitution $(S_E)$ reactions compared to benzene.

(C) Phenol $(C_6H_5OH)$ is a weaker acid than carbonic acid $(H_2CO_3)$. The $pK_a$ value of phenol is approximately $10$,while the $pK_a$ value of carbonic acid is approximately $6.35$. Since a lower $pK_a$ indicates a stronger acid,carbonic acid is stronger than phenol.

(D) When phenol reacts with dilute $HNO_3$ at low temperature,it undergoes electrophilic aromatic substitution to form a mixture of ortho-nitrophenol and para-nitrophenol. The reaction is as follows:
$C_6H_5OH + dil. HNO_3 \rightarrow o-Nitrophenol + p-Nitrophenol$

(B) $p$-Nitrophenol is a stronger acid than salicylaldehyde due to the strong electron-withdrawing effect of the $-NO_2$ group. Since it is a stronger acid,$p$-nitrophenol is more soluble in a base.

(A) The Reimer-Tiemann reaction involves the treatment of phenol with chloroform $(CHCl_3)$ in the presence of an aqueous base such as sodium hydroxide $(NaOH)$ or potassium hydroxide $(KOH)$,which results in the formation of ortho-hydroxybenzaldehyde (salicylaldehyde). The reaction shown in option $A$ represents this transformation.

(B) When phenol reacts with bromine $(Br_2)$ in a non-polar solvent like carbon tetrachloride $(CCl_4)$ at low temperature,the reaction is mono-bromination.
This results in the formation of a mixture of ortho-bromophenol ($o$-bromophenol) and para-bromophenol ($p$-bromophenol) due to the activating effect of the $-OH$ group.
In contrast,if the reaction were carried out in an aqueous medium (like bromine water),it would lead to the formation of $2, 4, 6$-tribromophenol.

(C) The oxidation of phenol with $KClO_3$ and $HCl$ yields chloranil ($2,3,5,6$-tetrachlorocyclohexa-$2,5$-diene-$1,4$-dione).
This is a specific reaction where phenol undergoes both oxidation and chlorination.

(D) The given reaction is the Kolbe-Schmitt reaction.
In this reaction,sodium phenoxide reacts with carbon dioxide $(CO_2)$ at $390 \ K$ and high pressure to form sodium salicylate as an intermediate product $X$.
This intermediate $X$ (sodium salicylate) upon acidification with $HCl$ gives salicylic acid.
The structure of $X$ is the ortho-substituted sodium salt of phenol,which is sodium salicylate.

(D) The phenolic group can be identified by several characteristic tests:
$1$. Liebermann's nitroso reaction: Phenols react with $NaNO_2$ and conc. $H_2SO_4$ to give a deep blue or green color.
$2$. Coupling reaction: Phenols react with diazonium salts in alkaline medium to form azo dyes.
$3$. Aqueous $FeCl_3$ test: Phenols react with neutral $FeCl_3$ to form a violet-colored complex.
Therefore,all the given options are characteristic tests for the phenolic group.

(A) Phenol dissociates to form a phenoxide ion and a proton $(H^+)$.
The phenoxide ion is stabilized by resonance,where the negative charge on the oxygen atom is delocalized into the benzene ring.
Since the phenoxide ion is more stable than phenol,the equilibrium shifts to the right,making phenol acidic.

(B) Phenolphthalein is prepared by the condensation reaction of phthalic anhydride with two equivalents of phenol in the presence of concentrated $H_2SO_4$ (acting as a dehydrating agent). The reaction is as follows: $\text{Phthalic anhydride} + 2 \text{Phenol} \xrightarrow{\text{conc. } H_2SO_4} \text{Phenolphthalein} + H_2O$.

(D) The sulfonation of phenol with concentrated $H_2SO_4$ is a temperature-dependent reaction. At low temperatures (around $293 \ K$),the reaction yields $o-$ phenol sulfonic acid as the major product. At higher temperatures (around $373 \ K$),the reaction yields $p-$ phenol sulfonic acid as the major product. Therefore,the overall reaction produces a mixture of both $o-$ and $p-$ isomers.

(B) Phenol has the highest boiling point due to the presence of intermolecular hydrogen bonding.

(D) When salicylic acid $(C_6H_4(OH)COOH)$ is heated with zinc dust,the phenolic $-OH$ group is reduced to a hydrogen atom.
This reaction involves the removal of oxygen from the phenolic group,resulting in the formation of benzoic acid $(C_6H_5COOH)$ and zinc oxide $(ZnO)$.
The reaction is: $C_6H_4(OH)COOH + Zn \rightarrow C_6H_5COOH + ZnO$.

(A) Benzoic acid is a stronger acid than phenol and reacts with aqueous $NaHCO_3$ to evolve $CO_2$ gas,forming sodium benzoate.
Phenol is a weaker acid and does not react with aqueous $NaHCO_3$.
Therefore,aqueous $NaHCO_3$ can be used to separate them.

(A) When salicylic acid $(C_6H_4(OH)COOH)$ is heated with sodalime $(NaOH + CaO)$,it undergoes decarboxylation. The $-COOH$ group is removed as $CO_2$ (which reacts with $NaOH$ to form $Na_2CO_3$),resulting in the formation of phenol $(C_6H_5OH)$.

(C) Picric acid is a phenol derivative ($2,4,6$-trinitrophenol) and contains a phenolic $-OH$ group.
Phenols react with neutral aqueous $FeCl_3$ to form a characteristic colored complex (often violet or green),whereas benzoic acid does not give this test.
Therefore,aqueous $FeCl_3$ can be used to distinguish between them.

(A) When phenol $(C_6H_5OH)$ is heated with ammonia $(NH_3)$ in the presence of anhydrous zinc chloride $(ZnCl_2)$ at high temperature and pressure,the hydroxyl group $(-OH)$ is replaced by an amino group $(-NH_2)$ to form aniline $(C_6H_5NH_2)$.
This reaction is known as the Bucherer reaction or ammonolysis of phenol.
The chemical equation is: $C_6H_5OH + NH_3 \xrightarrow{ZnCl_2} C_6H_5NH_2 + H_2O$.

(B) $o-$Nitrophenol exhibits intramolecular hydrogen bonding,which reduces the intermolecular forces of attraction between its molecules.
In contrast,phenol,$m-$nitrophenol,and $p-$nitrophenol exhibit intermolecular hydrogen bonding,leading to higher boiling points due to molecular association.
Therefore,$o-$nitrophenol has the lowest boiling point among the given compounds.

(C) Phenol is slightly soluble in water due to the presence of a large non-polar hydrocarbon part $(-C_6H_5)$.

(B) The phthalein test is a characteristic test for $Phenols$. In this test,a phenol reacts with phthalic anhydride in the presence of concentrated $H_2SO_4$ to form a phthalein dye,which shows a characteristic color change in alkaline medium.

(C) The Reimer-Tiemann reaction involves the treatment of phenol with $CHCl_3$ and aqueous $KOH$ to form salicylaldehyde. Option $C$ is incorrect because the reaction with $CCl_4$ and $KOH$ is a variation that produces salicylic acid,not the standard Reimer-Tiemann reaction. Option $D$ is also incorrect in the context of the Reimer-Tiemann reaction,as the reaction of phenol with formaldehyde is a polymerization reaction used to produce Bakelite,which is unrelated to the Reimer-Tiemann mechanism.

(C) The acidity of phenols is determined by the stability of the phenoxide ion formed after the loss of a proton. \\ Electron-withdrawing groups $(EWG)$ stabilize the phenoxide ion through $-I$ and $-M$ effects,thereby increasing acidity. \\ The $-NO_2$ group in $p-$nitrophenol is a strong electron-withdrawing group (both $-I$ and $-M$ effects),making it the most acidic among the given options. \\ The $-CH_3$ group in $p-$cresol is electron-donating ($+I$ and hyperconjugation),$-NH_2$ in $p-$aminophenol is strongly electron-donating $(+M)$,and $-Cl$ in $p-$chlorophenol is weakly electron-withdrawing ($-I$ effect but $+M$ effect). \\ Thus,the order of acidity is $p-$nitrophenol > $p-$chlorophenol > $p-$cresol > $p-$aminophenol.

(A) Phenols react with neutral $FeCl_3$ solution to form a violet or purple colored complex,which is a characteristic test for the presence of a phenolic group.
Cyclohexanol is an aliphatic alcohol and does not give this test.
Therefore,$FeCl_3$ is used to distinguish between phenol and cyclohexanol.

(A) $o-$Nitrophenol exhibits intramolecular hydrogen bonding,which reduces its intermolecular attraction.
This makes it more volatile compared to $p-$nitrophenol and $m-$nitrophenol,which exhibit intermolecular hydrogen bonding leading to higher boiling points.

(C) The hydrogenation of phenol $(C_6H_5OH)$ in the presence of a nickel $(Ni)$ catalyst at $160\,^oC$ involves the addition of $3$ moles of hydrogen $(H_2)$ to the aromatic ring.
This process saturates the benzene ring,converting phenol into cyclohexanol $(C_6H_{11}OH)$.
The reaction is: $C_6H_5OH + 3H_2 \xrightarrow{Ni, 160\,^oC} C_6H_{11}OH$.

(D) Thymol is a white crystalline phenol derivative found in many essential oils. It is used as a perfume and as an antiseptic.

(B) The reaction of phenol with methyl bromide $(CH_3Br)$ in the presence of a base like $NaOH$ leads to the formation of anisole $(C_6H_5OCH_3)$.
This process involves the replacement of the hydrogen atom of the hydroxyl group in phenol with a methyl group,which is a type of etherification reaction.
The chemical equation is: $C_6H_5OH + CH_3Br \xrightarrow{NaOH} C_6H_5OCH_3 + HBr$.

(B) The reaction of phenol with $CHCl_3$ (chloroform) and aqueous $NaOH$ (sodium hydroxide) is known as the Reimer-Tiemann reaction.
In this reaction,a formyl group $(-CHO)$ is introduced at the ortho position of the phenol ring.
The product formed is $2$-hydroxybenzaldehyde,commonly known as Salicylaldehyde.
The reaction is represented as:
$C_6H_5OH + CHCl_3 + 3NaOH \rightarrow C_6H_4(OH)(CHO) + 3NaCl + 2H_2O$

(B) In $Kolbe's$ reaction,phenol is treated with sodium hydroxide to form sodium phenoxide,which then reacts with carbon dioxide $(CO_2)$ under pressure followed by acidification to yield salicylic acid ($2-hydroxybenzoic$ acid). The reaction is as follows: $Phenol + CO_2 \xrightarrow{1. NaOH, 2. H^+} Salicylic \ acid$.

(A) Salol (phenyl salicylate) is prepared by the esterification reaction between salicylic acid $(C_6H_4(OH)COOH)$ and phenol $(C_6H_5OH)$ in the presence of a dehydrating agent like $POCl_3$ or $H_2SO_4$.
The reaction is: $C_6H_4(OH)COOH + C_6H_5OH \rightarrow C_6H_4(OH)COOC_6H_5 + H_2O$.
Thus,it is obtained from salicylic acid and phenol.

(A) The $-OH$ group is an ortho-para directing group due to the resonance effect,which increases electron density at the ortho and para positions. Therefore,electrophilic substitution reactions in phenol primarily occur at the $o-$ and $p-$ positions.

(C) $2, 4, 6-$ Trinitrophenol is known as picric acid.

(A) The reaction of phenol with chloroform $(CHCl_3)$ in the presence of aqueous alkali $(NaOH)$ is known as the Reimer-Tiemann reaction.
In this reaction,a formyl group $(-CHO)$ is introduced at the ortho position of the phenol ring.
The major product formed is $o$-hydroxybenzaldehyde,which is commonly known as Salicylaldehyde.

(C) The acidity of a compound depends on the stability of its conjugate base.
Electron-withdrawing groups $(EWG)$ like $-NO_2$ stabilize the phenoxide ion through the inductive effect and resonance,thereby increasing the acidity of the phenol.
Electron-donating groups like $-CH_3$ destabilize the phenoxide ion,decreasing the acidity.
Among the given options,$o$-Nitrophenol contains a strong electron-withdrawing $-NO_2$ group,making it the most acidic compound.

(B) The reaction of phenol with an acid chloride (like $CH_3COCl$) in the presence of an aqueous base $(NaOH)$ is known as the Schotten-Baumann reaction.
This process involves the introduction of an acyl group into the phenol,which is a specific type of acylation known as acetylation.
Since the reaction specifically uses an aqueous base to facilitate the acylation of a phenol,it is classified as a Schotten-Baumann reaction.

(A) The Liebermann nitroso reaction involves the reaction of phenol with sodium nitrite $(NaNO_2)$ in the presence of concentrated sulfuric acid $(H_2SO_4)$.
The sequence of color changes observed is: Red or brown $\rightarrow$ Green $\rightarrow$ Deep blue.

(A) Phenol reacts with sodium metal to form sodium phenoxide and releases hydrogen gas. The chemical reaction is as follows:
$C_6H_5OH + Na \rightarrow C_6H_5ONa + \frac{1}{2} H_2(g)$

(B) The given reactant contains both a phenolic $-OH$ group and a benzylic $-OH$ group.
$K_2CO_3$ is a mild base that selectively deprotonates the more acidic phenolic $-OH$ group to form a phenoxide ion,while the benzylic $-OH$ group remains unaffected.
Subsequent reaction with $CH_3I$ (a methylating agent) leads to the methylation of the phenoxide ion via an $S_N2$ mechanism,resulting in the formation of an ether linkage at the phenolic position.
Thus,the product $X$ is $3$-hydroxymethylanisole.

(C) Phenols react with neutral $FeCl_3$ solution to form a complex,which gives a characteristic violet or dark color.
The reaction is:
$3C_6H_5OH + FeCl_3 \rightarrow [Fe(OC_6H_5)_6]^{3-} + 3HCl$ (or similar complex formation).
Alcohols do not give this test.

(B) When phenol $(C_6H_5OH)$ is heated with zinc dust,the phenol is reduced to benzene $(C_6H_6)$.
The chemical reaction is as follows:
$C_6H_5OH + Zn \xrightarrow{\Delta} C_6H_6 + ZnO$
In this reaction,zinc acts as a reducing agent,removing the oxygen atom from the hydroxyl group of phenol to form zinc oxide $(ZnO)$.

(A) When phenol is treated with concentrated $HNO_3$ in the presence of concentrated $H_2SO_4$,it undergoes nitration to form $2,4,6$-trinitrophenol,which is commonly known as picric acid.

(B) The reaction of phenol with carbon tetrachloride $(CCl_4)$ in the presence of an aqueous base like potassium hydroxide $(KOH)$ followed by acidification leads to the formation of salicylic acid ($2$-hydroxybenzoic acid). This is a variation of the Reimer-Tiemann reaction,which typically uses chloroform $(CHCl_3)$ to produce salicylaldehyde. Both are commonly referred to under the Reimer-Tiemann reaction category.

(A) The reaction described is the Kolbe-Schmitt reaction.
When sodium phenoxide is treated with carbon dioxide $(CO_2)$ at a temperature of $400 \ K$ and a pressure of $4.7 \ atm$,it undergoes carboxylation to form sodium salicylate as the primary product.
This is a classic electrophilic aromatic substitution reaction where the phenoxide ion acts as a nucleophile attacking the $CO_2$ molecule.

(C) The acidity of substituted phenols depends on the electron-withdrawing or electron-donating nature of the substituent.
$-NO_2$ is a strong electron-withdrawing group $(EWG)$ due to both $-I$ and $-M$ effects.
$p$-Nitrophenol is more acidic than $o$-nitrophenol because $o$-nitrophenol exhibits intramolecular hydrogen bonding,which stabilizes the molecule and makes the release of the $H^+$ ion more difficult.
$p$-Chlorophenol is less acidic than $p$-nitrophenol because $-Cl$ is a weaker electron-withdrawing group than $-NO_2$.
Therefore,$p$-nitrophenol is the strongest acid among the given options.