Properties of Phenols Questions in English

(C) Acidity is inversely proportional to the $pKa$ value.
$1$. Picric acid ($2,4,6$-trinitrophenol) is the strongest acid among the given options due to the presence of three electron-withdrawing $-NO_2$ groups,so it has the lowest $pKa$ value $(0.78)$.
$2$. $p$-Nitrophenol is more acidic than phenol due to the $-I$ and $-M$ effect of the $-NO_2$ group $(pKa = 7.1)$.
$3$. Phenol is more acidic than ethanol because the phenoxide ion is resonance stabilized $(pKa = 10)$.
$4$. Ethanol is the least acidic $(pKa = 16)$.
Therefore,the correct match is: $a-iii, b-iv, c-i, d-ii$.

(B) The reaction sequence is as follows:
$1$. Phenol $(C_6H_5OH)$ reacts with benzoyl chloride $(C_6H_5COCl)$ in the presence of a base to form phenyl benzoate $(X)$ via the Schotten-Baumann reaction.
$2$. Phenyl benzoate $(C_6H_5COOC_6H_5)$ undergoes electrophilic aromatic substitution (nitration). The ester group $(-OCOC_6H_5)$ is ortho/para directing. Due to steric hindrance,the para-substituted product is the major product.
$3$. The nitration occurs on the ring attached to the oxygen atom (the phenoxy ring) because it is more activated than the benzoyl ring. Thus,the major product $Y$ is $4-$nitrophenyl benzoate.

(B) The acidity of phenols depends on the stability of the phenoxide ion formed after the loss of a proton.
Electron-withdrawing groups $(EWG)$ like $-NO_{2}$ and $-Cl$ stabilize the phenoxide ion through the inductive effect and resonance,thereby increasing acidity.
Electron-donating groups $(EDG)$ like $-CH_{3}$ (as in $o-$cresol) destabilize the phenoxide ion by increasing the electron density on the oxygen atom,thereby decreasing acidity.
Comparing the given compounds:
$1$. $o-$nitrophenol: Contains $-NO_{2}$ (strong $EWG$),most acidic.
$2$. $o-$chlorophenol: Contains $-Cl$ $(EWG)$,acidic.
$3$. Phenol: Standard reference.
$4$. $o-$cresol: Contains $-CH_{3}$ $(EDG)$,least acidic.
Therefore,the correct order of acidic strength is $o-$nitrophenol > $o-$chlorophenol > phenol > $o-$cresol.

(B) The reaction sequence is the industrial preparation of phenol from cumene:
$1$. Benzene reacts with isopropyl chloride in the presence of anhydrous $AlCl_3$ to form cumene $(A)$.
$2$. Cumene $(A)$ is oxidized by $O_2$ at $130^{\circ}C$ to form cumene hydroperoxide $(B)$.
$3$. Cumene hydroperoxide $(B)$ is treated with dilute $H_2SO_4$ at $100^{\circ}C$ to yield phenol and propanone $(C)$.
Therefore,$C$ is propanone.

(B) The molecular formula $C_7H_8O$ corresponds to cresols or benzyl alcohol.
Since the compound dissolves in $NaOH$,it must be acidic in nature,which indicates it is a phenol (cresol).
It gives a characteristic violet color with neutral $FeCl_3$,which is a test for phenolic groups.
Treatment with bromine water leads to a tribromo derivative,indicating the presence of three reactive positions (ortho and para) relative to the $-OH$ group.
All three isomers of cresol ($o$-,$m$-,and $p$-) satisfy these conditions. However,in many textbook contexts,$m$-cresol is often cited for such reactions.
Specifically,the reaction with $Br_2$ water on $m$-cresol yields $2,4,6$-tribromo-$m$-cresol.

(B) When phenol is treated with bromine water $(Br_2/H_2O)$,it undergoes electrophilic substitution at all available ortho and para positions due to the highly activating nature of the $-OH$ group.
This reaction results in the formation of $2,4,6$-tribromophenol as a white precipitate.
Therefore,the reagent "$X$" is bromine in water.

(A) The conversion of $p$-nitrophenol to quinol proceeds through the following steps:
$1$. Reduction: $p$-nitrophenol is reduced using $Sn/HCl$ to form $p$-aminophenol.
$2$. Diazotization: $p$-aminophenol is treated with $NaNO_2/HCl$ at $0-5 \ ^\circ C$ to form the corresponding diazonium salt.
$3$. Hydrolysis: The diazonium salt is hydrolyzed by warming with water to yield quinol ($p$-dihydroxybenzene).

(B) The acidity of substituted phenols is significantly affected by the nature of the substituents.
Electron-withdrawing groups $(EWG)$ such as $-NO_2$ increase the acidic strength by stabilizing the phenoxide ion through the $-I$ and $-M$ effects.
Conversely,electron-donating groups $(EDG)$ such as $-CH_3$ (methyl group in cresol) decrease the acidic strength due to the $+I$ effect and hyperconjugation,which destabilizes the phenoxide ion.
Therefore,the acidic strength follows the order: $o-cresol < phenol < o-nitrophenol$.

(A) In $(A)$ (para-nitrophenol),intermolecular $H$-bonding exists,which leads to association of molecules and higher boiling point.
In $(B)$ (ortho-nitrophenol),intramolecular $H$-bonding exists,which restricts association and results in a lower boiling point.
Since $(B)$ has a lower boiling point,it is more volatile than $(A)$.
Therefore,the vapour pressure of $(B)$ is higher than that of $(A)$ at a particular temperature.

(A) The acidity of compounds depends on the stability of the conjugate base formed after the loss of a proton $(H^+)$.
$1$. $p-$Nitrophenol $(A)$: The nitro group $(-NO_2)$ is a strong electron-withdrawing group ($-I$ and $-M$ effect),which stabilizes the phenoxide ion significantly,making it the most acidic. Its $pK_a$ is $7.1$ $(I)$.
$2$. Phenol $(B)$: Phenol is more acidic than ethanol but less acidic than $p-$nitrophenol. Its $pK_a$ is $10.0$ $(II)$.
$3$. Ethanol $(C)$: Aliphatic alcohols are much less acidic than phenols due to the electron-donating alkyl group ($+I$ effect) which destabilizes the alkoxide ion. Its $pK_a$ is $15.9$ $(III)$.
$4$. $p-$Cresol $(D)$: The methyl group $(-CH_3)$ is an electron-donating group ($+I$ and hyperconjugation),which destabilizes the phenoxide ion compared to phenol,making it less acidic. Its $pK_a$ is $10.2$ $(IV)$.
Therefore,the correct matching is $A-I, B-II, C-III, D-IV$.

(A) $1$. The reaction of benzenediazonium chloride $(C_6H_5N_2Cl)$ with water at $283 \ K$ yields phenol $(X)$.
$2$. Phenol $(X)$ reacts with concentrated $HNO_3$ to undergo electrophilic aromatic substitution,resulting in the formation of $2,4,6-$trinitrophenol,also known as picric acid $(Y)$.
$3$. Phenol $(X)$ reacts with bromine water $(Br_2/H_2O)$ to undergo rapid electrophilic substitution at all available ortho and para positions,yielding $2,4,6-$tribromophenol $(Z)$.

(A) $C_6H_7N$ corresponds to aniline $(A)$.
Aniline undergoes diazotisation reaction with $NaNO_2 / HCl$ at $273-278 \ K$ to form benzene diazonium chloride,which on warming with water produces phenol $(B)$.
Phenol reacts with concentrated nitric acid $(HNO_3)$ to undergo electrophilic aromatic substitution,yielding $2,4,6-$trinitrophenol $(C)$,which is commonly known as picric acid.

(C) The reaction sequence is as follows:
$1$. $p-chloronitrobenzene$ reacts with $NaOH$ at $443 \ K$ to form $p-nitrophenol$ $(X)$.
$2$. $p-nitrophenol$ is reduced using $Sn + HCl$ to form $p-aminophenol$ $(Y)$.
$3$. $p-aminophenol$ reacts with $NaNO_2 / HCl$ at $0-5^{\circ}C$ to form a diazonium salt intermediate.
$4$. Hydrolysis of this diazonium salt with $H_3O^{+}$ at $10^{\circ}C$ replaces the diazonium group with an $-OH$ group,resulting in the formation of $hydroquinone$ $(Z)$.

(C) The given reaction is the industrial preparation of phenol from cumene (isopropylbenzene).
$1$. Cumene reacts with oxygen $(O_2)$ to form cumene hydroperoxide $(A)$:
$C_6H_5-CH(CH_3)_2 + O_2 \rightarrow C_6H_5-C(CH_3)_2-O-OH$
$2$. Cumene hydroperoxide $(A)$ on treatment with dilute acid $(H_3O^+)$ undergoes rearrangement to produce phenol $(B)$ and acetone $(C)$:
$C_6H_5-C(CH_3)_2-O-OH \xrightarrow{H_3O^+} C_6H_5OH + (CH_3)_2CO$
Thus,$A$ is cumene hydroperoxide,$B$ is phenol,and $C$ is acetone. The correct option is $C$.

(B) $1$. Reaction with $Zn$ dust and heat: $p-Cresol$ $(p-methylphenol)$ reacts with $Zn$ dust to undergo reduction,where the $-OH$ group is removed and replaced by a hydrogen atom,resulting in the formation of $Toluene$ $(X)$.
$2$. Reaction with $(CH_3CO)_2O$ followed by $H^+$: $p-Cresol$ reacts with acetic anhydride $(CH_3CO)_2O$ to undergo acetylation of the phenolic $-OH$ group,forming $p-acetoxy-toluene$ $(Y)$.

(D) The reaction sequence is as follows:
$1$. Benzene reacts with $Oleum$ $(H_2S_2O_7)$ to form benzene sulphonic acid.
$2$. Benzene sulphonic acid reacts with $NaOH$ followed by $H^+$ to form phenol $(X)$.
$3$. Phenol $(X)$ reacts with concentrated $HNO_3$ (nitration) to form $2,4,6-trinitrophenol$,commonly known as $Picric \ acid$ $(Y)$.

(A) $1$. The reaction of benzene with oleum $(H_2S_2O_7)$ gives benzene sulfonic acid $(X)$.
$2$. The fusion of benzene sulfonic acid with $NaOH$ followed by acidification gives phenol $(Y)$.
$3$. The reaction of phenol with $CHCl_3$ and $NaOH$ is the Reimer-Tiemann reaction,which introduces a formyl group $(-CHO)$ at the ortho position to the $-OH$ group.
$4$. Thus,the major product '$Z$' is $o-$hydroxy benzaldehyde (salicylaldehyde).

(A) When $p$-cresol ($4$-methylphenol) is heated with zinc dust,it undergoes a reduction reaction.
In this reaction,the phenolic $-OH$ group is removed and replaced by a hydrogen atom.
This process converts the phenol derivative into the corresponding hydrocarbon.
Thus,$p$-cresol reacts with zinc dust to form toluene $(C_6H_5CH_3)$ and zinc oxide $(ZnO)$.
The reaction is: $CH_3-C_6H_4-OH + Zn \rightarrow CH_3-C_6H_5 + ZnO$.

(C) When phenol is treated with concentrated nitric acid $(Conc. HNO_3)$,it undergoes electrophilic aromatic substitution to form $2,4,6$-trinitrophenol,which is commonly known as picric acid.

(B) The reaction sequence is the $Kolbe-Schmitt$ reaction.
$1$. Phenol reacts with $NaOH$ to form sodium phenoxide.
$2$. Sodium phenoxide reacts with $CO_2$ under pressure to form an intermediate,which upon acidification $(H_3O^+)$ yields salicylic acid as the major product.
$3$. The ortho-substitution is favored due to the formation of a stable chelated intermediate involving the sodium ion and the oxygen atoms of the phenoxide and the carboxylate group.

(B) When phenol is treated with bromine water,it undergoes electrophilic aromatic substitution at all ortho and para positions due to the strong activating effect of the $-OH$ group.
This reaction results in the formation of $2, 4, 6-$tribromophenol,which appears as a white precipitate.
The chemical reaction is:
$C_6H_5OH + 3Br_2(aq) \rightarrow C_6H_2Br_3OH + 3HBr$

(A) The given reaction is the Reimer-Tiemann reaction.
Phenol reacts with chloroform $(CHCl_3)$ in the presence of aqueous sodium hydroxide $(NaOH)$ to form an intermediate,which upon hydrolysis with acid $(H^+)$ yields $2$-hydroxybenzaldehyde (salicylaldehyde) as the major product.

(A) The Reimer-Tiemann reaction is a chemical reaction used for the ortho-formylation of phenols.
In this reaction,phenol reacts with chloroform $(CHCl_3)$ in the presence of an aqueous base like $KOH$ to form salicylaldehyde $(X)$.
The reaction proceeds through the formation of a dichlorocarbene $(:CCl_2)$ intermediate,which attacks the phenoxide ion to form an ortho-dichloromethyl substituted phenoxide intermediate $(Y)$.
This intermediate $Y$ then undergoes hydrolysis to yield the final product,salicylaldehyde $(X)$.

(D) The reaction of phenol with $CHCl_3$ and aqueous $NaOH$ followed by acidification is the Reimer-Tiemann reaction,which yields $2$-hydroxybenzaldehyde (salicylaldehyde) as product $X$.
The reaction of phenol with $NaOH$ followed by $CO_2$ and then acidification is the Kolbe-Schmitt reaction,which yields $2$-hydroxybenzoic acid (salicylic acid) as product $Y$.
Therefore,$X$ is $2$-hydroxybenzaldehyde and $Y$ is $2$-hydroxybenzoic acid.

(A) When phenol is exposed to air,it undergoes slow oxidation to form $p$-benzoquinone (commonly referred to as quinone). The reaction is as follows:
$C_6H_5OH + [O] \rightarrow C_6H_4O_2$ (quinone)
This process is responsible for the pinkish color that phenol develops upon standing in air.

(B) The oxidation of phenol with chromic acid ($Na_2Cr_2O_7$ in the presence of $H_2SO_4$) yields $p$-benzoquinone.
Therefore,$X$ is phenol and $Y$ is $Na_2Cr_2O_7 / H_2SO_4$.

(B) The acidic strength of phenols is determined by the stability of the corresponding phenoxide ion. Electron-withdrawing groups $(EWG)$ increase acidity by stabilizing the phenoxide ion,while electron-donating groups $(EDG)$ decrease acidity by destabilizing it.
$1.$ $IV$ ($p$-nitrophenol): The $-NO_2$ group at the $p$-position exerts both a strong $-I$ effect and a strong $-M$ (mesomeric) effect,which significantly stabilizes the phenoxide ion.
$2.$ $III$ ($m$-nitrophenol): The $-NO_2$ group at the $m$-position exerts only a $-I$ effect (no $-M$ effect),providing less stabilization than at the $p$-position.
$3.$ $I$ (Phenol): This is the reference compound.
$4.$ $II$ ($p$-methylphenol): The $-CH_3$ group at the $p$-position exerts a $+I$ effect and hyperconjugation $(+H)$,which donates electron density and destabilizes the phenoxide ion,making it the least acidic.
Therefore,the correct order of acidic strength is $IV > III > I > II$.

(A) Acidic character $\propto$ stability of conjugate base (phenoxide ion).
The stability of the phenoxide ion is increased by electron-withdrawing groups ($-M$ and $-I$ effects) and decreased by electron-donating groups ($+M$,$+H$,and $+I$ effects).
The substituents are:
$(I)$ $p$-methylphenol: $-CH_3$ group shows $+H$ and $+I$ effects,which destabilize the phenoxide ion.
$(II)$ $m$-nitrophenol: $-NO_2$ group shows only $-I$ effect ($-M$ effect does not operate at $m$-position).
$(III)$ $p$-nitrophenol: $-NO_2$ group shows both $-M$ and $-I$ effects,providing strong stabilization.
$(IV)$ Phenol: No substituent.
Comparing these,the order of stability of the conjugate base is: $p$-methylphenoxide < phenoxide < $m$-nitrophenoxide < $p$-nitrophenoxide.
Therefore,the increasing order of acidic character is: $I < IV < II < III$.

(D) In reaction $(i)$,Aniline reacts with $NaNO_2/HCl$ at $273 \ K$ to form benzene diazonium chloride $(A = C_6H_5N_2^+Cl^-)$. This compound on hydrolysis with warm water $(B = \text{Warm } H_2O)$ yields phenol.
In reaction $(ii)$,Cumene (isopropylbenzene) undergoes oxidation with $O_2$ to form cumene hydroperoxide $(C = C_6H_5C(CH_3)_2OOH)$. This hydroperoxide on treatment with dilute acid $(D = H^+/H_2O)$ undergoes rearrangement to give phenol and acetone.
Thus,the correct set is option $D$.

(D) $1$. The reaction of benzenediazonium salt $(C_6H_5N_2^+X^-)$ with ethanol $(C_2H_5OH)$ reduces the diazonium group to benzene,but in the presence of water or ethanol,it typically forms phenol $(C_6H_5OH)$ as the major product. Given the subsequent reaction,$X$ is phenol $(C_6H_5OH)$.
$2$. The reaction of phenol $(C_6H_5OH)$ with $CO$ and $HCl$ in the presence of anhydrous $AlCl_3$ is the Gattermann-Koch formylation reaction,which introduces a formyl group $(-CHO)$ at the ortho position of the phenol ring,resulting in salicylaldehyde ($2$-hydroxybenzaldehyde).
$3$. Therefore,$X$ is phenol and $Y$ is salicylaldehyde.

(A) Key Idea: The Fries rearrangement is the rearrangement of a phenolic ester to a hydroxy aryl ketone in the presence of a Lewis acid catalyst.
In this reaction,the acyl group $(RCO-)$ migrates from the oxygen atom of the ester to the ortho or para position of the benzene ring.
Thus,it represents the conversion of an $O$-acylated phenol (phenolic ester) to a $C$-acylated phenol (hydroxy aryl ketone).
Therefore,option $(A)$ is correct.
The reaction is as follows:
$C_6H_5OCOR \xrightarrow{AlCl_3} o-HOC_6H_4COR + p-HOC_6H_4COR$

(A) The reaction sequence is as follows:
$1$. Phenol reacts with $NaOH$ to form sodium phenoxide $(X)$.
$2$. Sodium phenoxide reacts with $CO_2$ followed by acid hydrolysis to form salicylic acid $(Y)$,which is $o$-hydroxybenzoic acid.
$3$. Salicylic acid reacts with acetic anhydride $((CH_3CO)_2O)$ in the presence of acid to form acetylsalicylic acid $(Z)$,commonly known as Aspirin,and acetic acid $(CH_3COOH)$.
$4$. Aspirin $(Z)$ is a non-narcotic analgesic and an antipyretic.
Therefore,statements $II$ and $III$ are correct.