Properties of Phenols Questions in English

(D) The reaction of phenol with an aromatic acyl chloride (like benzoyl chloride) in the presence of an aqueous base such as $NaOH$ or pyridine is known as the Schotten-Baumann reaction. This reaction leads to the formation of an ester (phenyl benzoate).

(D) When phenol is treated with excess bromine water,it undergoes electrophilic substitution at all available ortho and para positions.
This results in the formation of a white precipitate of $2,4,6-$tribromophenol.

(B) The reaction proceeds via the formation of a carbocation intermediate. When $H^+$ adds to the double bond,it forms a carbocation at the benzylic position,which is stabilized by resonance due to the electron-donating $-OH$ group at the para position. The carbocation $I$ $(CH_3-CH_2-CH^+-C_6H_4-OH)$ is more stable than other possible carbocations. Subsequently,the $Br^-$ ion attacks this stable carbocation to form the final product: $CH_3-CH_2-CH(Br)-C_6H_4-OH$.

(A) $1$. Picric acid is $2,4,6-\text{trinitrophenol}$. Its structure contains a phenolic $-OH$ group and three $-NO_2$ groups,but no $-COOH$ group.
$2$. Benzoic acid is $C_6H_5COOH$,which contains a $-COOH$ group.
$3$. Aspirin is acetylsalicylic acid,which contains a $-COOH$ group.
$4$. Therefore,picric acid is the substance that does not contain a $-COOH$ group.

(A) Phenols react with acid chlorides $(R-COCl)$ in the presence of a base (like pyridine) to form esters. The reaction is known as esterification or acylation of phenols.
The reaction is as follows:
$C_6H_5OH + R-COCl \xrightarrow{\text{base}} C_6H_5O-CO-R + HCl$
Thus,phenol is the compound that reacts with acid chloride to form an ester.

(C) The reaction of phenol with potassium persulphate $(K_2S_2O_8)$ in an alkaline medium is known as the Elbs persulphate oxidation.
This reaction results in the hydroxylation of phenol to form a mixture of ortho- and para-dihydroxybenzenes (catechol and quinol).
Therefore,the reagent $X$ is $K_2S_2O_8$ in an alkaline solution.

(C) The boiling point of a compound depends on the extent of intermolecular hydrogen bonding.
$1.$ Hydroxybenzene $(IV)$ has only one $-OH$ group,so it has the lowest boiling point.
$2.$ Among the dihydroxybenzenes,$1, 2-$dihydroxybenzene $(I)$ exhibits intramolecular hydrogen bonding,which reduces the extent of intermolecular hydrogen bonding,leading to a lower boiling point compared to the $1, 3-$ and $1, 4-$ isomers.
$3.$ $1, 3-$dihydroxybenzene $(II)$ and $1, 4-$dihydroxybenzene $(III)$ exhibit strong intermolecular hydrogen bonding. $1, 4-$dihydroxybenzene has a more symmetrical structure and stronger intermolecular forces,resulting in the highest boiling point.
Thus,the increasing order of boiling points is: $IV < I < II < III$.

(B) The reaction shown is the Reimer-Tiemann reaction.
In this reaction,the electrophile involved is dichlorocarbene $(:CCl_2)$.
It is generated from chloroform $(CHCl_3)$ by the action of a base $(OH^-)$:
$OH^- + CHCl_3 \rightleftharpoons H_2O + :CCl_3^-$
$:CCl_3^- \to Cl^- + :CCl_2$

(C) The reaction of phenol with $NaOH$ followed by $CO_2$ is known as the Kolbe-Schmitt reaction.
In this reaction,phenol is first converted to sodium phenoxide,which then undergoes electrophilic substitution with $CO_2$ at high pressure and temperature to form sodium salicylate.
Upon acidification,this yields salicylic acid as the major product.

(C) $1$. Phenol reacts with $Zn$ dust to form Benzene $(X)$.
$2$. Benzene undergoes Friedel-Crafts alkylation with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ to form Toluene $(Y)$.
$3$. Toluene is oxidized by alkaline $KMnO_4$ to form Benzoic acid $(Z)$.

(D) Benzoic acid $(C_6H_5COOH)$ reacts with sodalime $(NaOH + CaO)$ to undergo decarboxylation,yielding benzene $(C_6H_6)$ and sodium carbonate $(Na_2CO_3)$. Thus,$X$ is sodalime.
Phenol $(C_6H_5OH)$ reacts with zinc dust $(Zn)$ to undergo reduction,yielding benzene $(C_6H_6)$ and zinc oxide $(ZnO)$. Thus,$Y$ is zinc dust.
Therefore,$X$ is sodalime and $Y$ is zinc dust. The correct option is $D$.

(B) Salicylic acid contains a phenolic $-OH$ group attached to the benzene ring.
Compounds containing a phenolic group react with neutral $FeCl_3$ to form a violet-colored complex.
This is a characteristic test for phenols.

(A) The acidity of the given compounds follows the order: $HCOOH > C_6H_5COOH > CH_3COOH > C_6H_5OH$.
Carboxylic acids are generally stronger acids than phenols because the carboxylate ion is more effectively stabilized by resonance than the phenoxide ion.
Among the given options,phenol $(C_6H_5OH)$ is the weakest acid.

(C) Ethanoic acid $(CH_3COOH)$ is a stronger acid than phenol $(C_6H_5OH)$.
Ethanoic acid reacts with sodium carbonate $(Na_2CO_3)$ to evolve $CO_2$ gas,which is observed as effervescence.
Phenol is a weaker acid and does not react with sodium carbonate to evolve $CO_2$ gas.
Therefore,sodium carbonate solution is used to distinguish between them.

(D) Salicylic acid $(C_7H_6O_3)$ contains one carboxylic group $(-COOH)$ and one phenolic hydroxyl group $(-OH)$. It acts as a monobasic acid due to the $-COOH$ group. It reacts with neutral $FeCl_3$ to give a violet color (due to the phenolic group) and with $NaHCO_3$ to release $CO_2$ gas (effervescence). Methyl salicylate is an ester known as 'oil of wintergreen' and is found in several natural oils. Therefore,the statement that it is not found in natural oils is incorrect.

(D) The methoxy group $(-OCH_3)$ exerts a $+I$ effect and $+M$ effect. The $+I$ effect increases the electron density on the oxygen atom of the $-OH$ group,which destabilizes the phenoxide ion and decreases the acidity of phenol.
$o-$Methoxy phenol is less acidic than phenol due to the electron-donating nature of the methoxy group.
Acetylene $(C_2H_2)$ is a terminal alkyne with a $pK_a$ of approximately $25$,whereas phenol has a $pK_a$ of approximately $10$. Thus,phenol is significantly more acidic than acetylene.
$p-$Nitrophenol is more acidic than phenol due to the strong $-I$ and $-M$ effects of the nitro group,which stabilize the phenoxide ion.

(B) The given reaction is the Kolbe-Schmitt reaction.
In this reaction,sodium phenoxide reacts with carbon dioxide $(CO_2)$ at a temperature of $390 \ K$ and high pressure to form sodium phenyl carbonate as an intermediate,which then undergoes rearrangement to form sodium salicylate $(X)$.
Finally,acidification with $HCl$ yields salicylic acid.
Thus,the compound $X$ is sodium salicylate.

(A) The reaction of phenol with formaldehyde $(CH_2O)$ in the presence of an acid or base catalyst leads to the formation of ortho- and para-hydroxybenzyl alcohols.
This specific electrophilic aromatic substitution reaction is known as the Lederer-Manasse reaction.
Therefore,the correct option is $(A)$.

(D) The reaction described is the Reimer-Tiemann reaction.
$1$. $NaOH$ reacts with $CHCl_3$ to generate the electrophilic intermediate dichlorocarbene,$:CCl_2$.
$2$. Phenol reacts with $NaOH$ to form the phenoxide ion,which is more reactive towards electrophilic substitution.
$3$. The phenoxide ion undergoes electrophilic attack by $:CCl_2$ to form a dichloromethyl substituted intermediate.
$4$. Since $:CCl_2$ and the phenoxide ion are both involved as intermediates in the reaction mechanism,and the dichloromethyl substituted species is also an intermediate,all the listed species are involved.
Therefore,the correct option is $D$.

(D) The acidity of phenols depends on the nature of the substituents attached to the benzene ring.
$1$. Electron-withdrawing groups $(EWG)$ like $-NO_2$ increase acidity by stabilizing the phenoxide ion through inductive and resonance effects.
$2$. Electron-releasing groups $(ERG)$ like $-CH_3$ decrease acidity by destabilizing the phenoxide ion.
$3$. In $p$-nitrophenol $(IV)$,the $-NO_2$ group at the para position exerts both $-I$ and $-M$ effects,making it the most acidic.
$4$. In $m$-nitrophenol $(III)$,the $-NO_2$ group exerts only the $-I$ effect,making it less acidic than $p$-nitrophenol but more acidic than phenol.
$5$. Phenol $(I)$ is the reference compound.
$6$. In $p$-cresol $(II)$,the $-CH_3$ group exerts $+I$ and hyperconjugation effects,which decrease the acidity,making it the least acidic.
Therefore,the correct order of acidity is $IV > III > I > II$.

(A) The acidity of these compounds depends on the stability of their conjugate bases.
$1$. $ClCH_2COOH$ is the strongest acid due to the strong electron-withdrawing effect of the chlorine atom,which stabilizes the carboxylate anion.
$2$. $HCOOH$ is stronger than $CH_3COOH$ because the methyl group in $CH_3COOH$ is electron-donating,which destabilizes the carboxylate anion.
$3$. Phenol $(C_6H_5OH)$ is a much weaker acid than carboxylic acids because the negative charge on the phenoxide ion is less effectively delocalized compared to the resonance-stabilized carboxylate ion.
Therefore,phenol is the weakest acid among the given options.

(A) The reactivity of an aromatic ring towards electrophilic substitution depends on the nature of the substituent attached to it.
Groups that donate electrons to the ring (activating groups) increase the electron density on the ring,making it more reactive towards electrophiles. Examples include $-OH$,$-NH_2$,etc.
Groups that withdraw electrons from the ring (deactivating groups) decrease the electron density on the ring,making it less reactive. Examples include $-NO_2$,$-Cl$ (which is deactivating due to its strong $-I$ effect despite its $+M$ effect).
In the given options:
$(A)$ Phenol ($-OH$ group) is strongly activating due to the $+M$ effect.
$(B)$ Chlorobenzene ($-Cl$ group) is deactivating due to the strong $-I$ effect.
$(C)$ Nitrobenzene ($-NO_2$ group) is strongly deactivating due to both $-I$ and $-M$ effects.
$(D)$ Benzyl alcohol ($-CH_2OH$ group) has a weak activating effect due to hyperconjugation,but it is less activating than the $-OH$ group directly attached to the ring.
Therefore,phenol is the most reactive towards electrophilic attack.

(C) The acidity of phenols depends on the stability of the phenoxide ion formed after the loss of a proton. Electron-withdrawing groups ($-I$ and $-M$ effects) increase acidity by stabilizing the phenoxide ion,while electron-donating groups (like $-CH_3$) decrease acidity.
$1$. $2,4,6$-Trinitrophenol has three $-NO_2$ groups,which exert strong $-I$ and $-M$ effects,making it the most acidic.
$2$. $p$-Nitrophenol has one $-NO_2$ group,which is acidic but less than picric acid.
$3$. Phenol has no substituent.
$4$. $p$-Cresol has a $-CH_3$ group,which is electron-donating ($+I$ and hyperconjugation),making it the least acidic.
Therefore,the order of acidity is: $2,4,6$-Trinitrophenol > $p$-Nitrophenol > Phenol > $p$-Cresol.

(C) The reaction of phenol with chloroform $(CHCl_3)$ in the presence of aqueous sodium hydroxide $(NaOH)$ is known as the Reimer-Tiemann reaction.
This reaction introduces an aldehyde group $(-CHO)$ at the $ortho$ position of the phenol ring.
The overall transformation is: $C_6H_5OH + CHCl_3 + 3NaOH \rightarrow C_6H_4(OH)(CHO) + 3NaCl + 2H_2O$.

(C) The solubility of an organic acid in $NaHCO_3$ depends on its acidity. An acid will react with $NaHCO_3$ to form a salt and evolve $CO_2$ gas if it is stronger than carbonic acid $(H_2CO_3)$.
The general reaction is: $\text{Acid} + NaHCO_3 \longrightarrow \text{Salt} + H_2O + CO_2$.
$1$. $2,4,6-$Trinitrophenol (picric acid),benzoic acid,and benzenesulphonic acid are stronger acids than $H_2CO_3$ and thus dissolve in $NaHCO_3$ solution.
$2$. $o-$Nitrophenol is a weaker acid than $H_2CO_3$ due to intramolecular hydrogen bonding,which stabilizes the molecule and hinders the release of the proton. Therefore,it does not react with $NaHCO_3$ and is not soluble in it.

(A) The acidic strength depends on the stability of the conjugate base formed after the loss of a proton $(H^+)$.
$2,4,6-$trinitrophenol $(III)$ is the most acidic because the phenoxide ion is highly stabilized by the strong $-I$ and $-M$ effects of three $-NO_2$ groups.
Acetic acid $(II)$ is more acidic than phenol $(IV)$ because the carboxylate ion is stabilized by resonance between two electronegative oxygen atoms,whereas the phenoxide ion is stabilized by resonance within the aromatic ring.
Phenol $(IV)$ is more acidic than cyclohexanol $(I)$ because the phenoxide ion is resonance-stabilized,while the cyclohexoxide ion has no such stabilization.
Cyclohexanol $(I)$ is the least acidic as the alkyl group is electron-donating ($+I$ effect),which destabilizes the alkoxide ion.
Therefore,the order of decreasing acidic strength is $III > II > IV > I$.

(B) The acidity of a compound depends on the stability of its conjugate base.
$1$. Phenol $(C_6H_5OH)$ forms a phenoxide ion $(C_6H_5O^-)$ upon losing a proton. This phenoxide ion is highly resonance-stabilized by the benzene ring,making phenol the most acidic among the given options.
$2$. Benzyl alcohol $(C_6H_5CH_2OH)$ forms a benzyloxide ion $(C_6H_5CH_2O^-)$,where the negative charge is localized on the oxygen atom and is not resonance-stabilized by the benzene ring.
$3$. Cyclohexanol and dicyclohexylmethanol are aliphatic alcohols. The alkyl groups attached to the carbon bearing the $-OH$ group are electron-releasing ($+I$ effect),which destabilizes the alkoxide ion and decreases acidity.
Therefore,phenol is the most acidic compound.

(A) The acidity of phenols is determined by the stability of the phenoxide ion formed after the loss of a proton.
Electron-withdrawing groups $(EWG)$,such as $-NO_2$,stabilize the phenoxide ion through $-I$ and $-M$ effects,thereby increasing the acidity.
Electron-releasing groups $(ERG)$,such as $-CH_3$,destabilize the phenoxide ion through $+I$ and $+H$ effects,thereby decreasing the acidity.
Comparing the compounds:
$1$. $(iv)$ Para-nitrophenol: $-NO_2$ at para position exerts both $-I$ and $-M$ effects,providing maximum stabilization.
$2$. $(iii)$ Meta-nitrophenol: $-NO_2$ at meta position exerts only $-I$ effect,providing less stabilization than para.
$3$. $(i)$ Phenol: Reference compound.
$4$. $(ii)$ Methyl phenol: $-CH_3$ group exerts $+I$ and $+H$ effects,destabilizing the phenoxide ion.
Thus,the correct order of acidity is $(iv) > (iii) > (i) > (ii)$.

(B) Step $1$: Phenol reacts with $Zn$ dust to form benzene $(X)$.
$C_6H_5OH + Zn \rightarrow C_6H_6 + ZnO$
Step $2$: Benzene undergoes Friedel-Crafts alkylation with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ to form toluene $(Y)$.
$C_6H_6 + CH_3Cl \xrightarrow{Anhyd.\,AlCl_3} C_6H_5CH_3 + HCl$
Step $3$: Toluene $(Y)$ is oxidized by alkaline $KMnO_4$ to form benzoic acid $(Z)$.
$C_6H_5CH_3 \xrightarrow{Alkaline\,KMnO_4} C_6H_5COOH$
Therefore,the product $Z$ is benzoic acid.

(B) Due to intramolecular $H-$bonding,the $-OH$ group is not available to form a hydrogen bond with water molecules.
Hence,$o-$nitrophenol is sparingly soluble in water,while $m-$ and $p-$nitrophenols are more soluble due to their ability to form intermolecular $H-$bonding with water.

(C) When phenols are treated with bromine water,the $-OH$ group strongly activates the benzene ring,leading to electrophilic substitution at all available ortho and para positions.
For $m$-cresol ($3$-methylphenol),the positions ortho and para to the $-OH$ group are available for substitution.
Specifically,the positions ortho to the $-OH$ group (at $C2$ and $C6$) and the position para to the $-OH$ group (at $C4$) are all activated,leading to the formation of $2,4,6-$tribromo$-3-$methylphenol.
Therefore,$m$-cresol gives a tribromo derivative.

(D) The given reaction is the $Reimer-Tiemann$ reaction.
In this reaction,phenol reacts with chloroform $(CHCl_3)$ in the presence of aqueous sodium hydroxide $(NaOH)$ to form salicylaldehyde.
The mechanism involves the formation of dichlorocarbene $(:\,CCl_2)$ as the reactive intermediate,which acts as an electrophile.
Therefore,the correct option is $D$.

(B) When phenol reacts with concentrated $H_2SO_4$,it undergoes sulphonation to form phenol-$2, 4-$disulphonic acid.
Subsequently,when this product reacts with concentrated $HNO_3$,the sulphonic acid groups are replaced by nitro groups to yield $2, 4, 6-$trinitrophenol,which is commonly known as picric acid.
Therefore,option $B$ is correct.

(B) The reaction of phenol with $NaOH$ followed by $CO_2$ at $6 \ atm$ and $140 \ ^\circ C$,and subsequent acidification,is known as the Kolbe-Schmitt reaction.
In this reaction,phenol is first converted to sodium phenoxide.
Sodium phenoxide then reacts with $CO_2$ to form sodium salicylate,which upon acidification yields salicylic acid as the major product.

(D) When phenol is treated with a mixture of $KBr$ and $KBrO_3$ in an acidic medium,$Br_2$ is generated in situ.
$5KBr + KBrO_3 + 6H^+ \rightarrow 3Br_2 + 6K^+ + 3H_2O$
Due to the highly activating nature of the $-OH$ group,phenol undergoes electrophilic aromatic substitution with bromine water to form a white precipitate of $2, 4, 6-$ tribromophenol.

(C) The acidity of substituted phenols depends on the nature of the substituent group attached to the benzene ring.
Electron-withdrawing groups $(EWG)$ stabilize the phenoxide ion through $-I$ and $-M$ effects,thereby increasing acidity.
Electron-releasing groups $(ERG)$ destabilize the phenoxide ion through $+I$ and $+M$ effects,thereby decreasing acidity.
Comparing the substituents:
$III$ $(NO_2)$: Strong $-M$ and $-I$ effect (Strongest acid).
$I$ $(Cl)$: $-I$ effect (Weakly acidic).
$II$ $(CH_3)$: $+I$ effect and hyperconjugation (Electron releasing).
$IV$ $(OCH_3)$: Strong $+M$ effect (Strongest electron releasing group,weakest acid).
Thus,the order of decreasing acidity is $III > I > II > IV$.

(A) The reaction of sodium phenoxide with $CO_2$ at $125\,^{\circ}C$ and $5 \ atm$ pressure is the Kolbe-Schmidt reaction,which yields salicylic acid ($2$-hydroxybenzoic acid) as the intermediate $B$.
In the second step,salicylic acid undergoes acetylation with acetic anhydride $(Ac_2O)$ to form acetylsalicylic acid,commonly known as Aspirin,which is the major product $C$.
The structure of $C$ is $2$-acetoxybenzoic acid.

(D) Phenol reacts with neutral $FeCl_3$ to form a violet-colored complex.
Benzoic acid reacts with neutral $FeCl_3$ to form a buff-colored (pale dull yellow) precipitate of ferric benzoate.
Therefore,neutral $FeCl_3$ can be used to distinguish between them.

(C) When phenol is distilled with $Zn$ dust,it undergoes reduction to form benzene as the major product.
The chemical reaction is as follows:
$C_6H_5OH + Zn \rightarrow C_6H_6 + ZnO$

(A) The $pK_a$ value is inversely proportional to the acidity of the compound. $A$ stronger acid has a lower $pK_a$ value.
Among the given compounds,the nitro group $(-NO_2)$ is a strong electron-withdrawing group ($-I$ and $-M$ effects),which increases the acidity of phenol by stabilizing the phenoxide ion.
$p$-nitrophenol is more acidic than $o$-nitrophenol because $o$-nitrophenol exhibits intramolecular hydrogen bonding,which stabilizes the undissociated molecule and reduces its acidity compared to $p$-nitrophenol.
Therefore,$p$-nitrophenol is the most acidic and has the least $pK_a$ value.

(D) Acids that are stronger than carbonic acid $(H_2CO_3)$ react with sodium bicarbonate $(NaHCO_3)$ to evolve carbon dioxide $(CO_2)$ gas.
Benzene sulfonic acid $(C_6H_5SO_3H)$ is a strong acid $(pK_a \approx -2.8)$,which is much stronger than $H_2CO_3$ $(pK_a \approx 6.35)$,so it releases $CO_2$.
$p-$Nitrophenol $(O_2N-C_6H_4-OH)$ has a $pK_a \approx 7.15$,which is slightly weaker than $H_2CO_3$. However,in many standard chemistry contexts,it is considered acidic enough to react with $NaHCO_3$ to release $CO_2$ gas due to the electron-withdrawing effect of the nitro group.
Therefore,both compounds release $CO_2$ gas.

(D) The reaction of $p$-cresol ($4$-methylphenol) with $Br_2$ in the presence of $FeCl_3$ (a Lewis acid catalyst) is an electrophilic aromatic substitution reaction.
Since the $-OH$ group is a strong activating group and ortho/para directing,and the para position is already occupied by a $-CH_3$ group,the electrophile $(Br^+)$ will attack the ortho position relative to the $-OH$ group.
Therefore,the major product is $2$-bromo-$4$-methylphenol.

(D) The reaction sequence is as follows:
$1$. Phenol reacts with $CO_2$ in the presence of $NaOH$ (Kolbe-Schmitt reaction) to form sodium salicylate,which upon acidification gives salicylic acid $(A)$,which is $2$-hydroxybenzoic acid.
$2$. Salicylic acid $(A)$ reacts with acetic anhydride $(Ac_2O)$ in the presence of an acid catalyst to undergo acetylation of the phenolic $-OH$ group,forming $2$-acetoxybenzoic acid $(B)$,commonly known as aspirin.
$3$. The structure of $(B)$ is a benzene ring with a $-COOH$ group at position $1$ and a $-OCOCH_3$ group at position $2$.

(A) In the Reimer-Tiemann reaction,phenol $(C_6H_5OH)$ reacts with chloroform $(CHCl_3)$ in the presence of aqueous $NaOH$ to form salicylaldehyde $(C_6H_4(OH)CHO)$.
The chemical equation is: $C_6H_5OH + CHCl_3 + 3NaOH \rightarrow C_6H_4(OH)CHO + 3NaCl + 2H_2O$.
The molecular weight of phenol $(C_6H_5OH)$ is $6 \times 12 + 6 \times 1 + 16 = 94 \ g/mol$.
The molecular weight of salicylaldehyde $(C_6H_4(OH)CHO)$ is $7 \times 12 + 6 \times 1 + 2 \times 16 = 84 + 6 + 32 = 122 \ g/mol$.
The increase in molecular weight is $122 - 94 = 28$.
This corresponds to the replacement of one $H$ atom (mass $1$) with one $CHO$ group (mass $29$),resulting in a net increase of $29 - 1 = 28$.

(D) Phenol $(C_6H_5OH)$ contains an electron-rich aromatic ring which undergoes electrophilic substitution with bromine water $(Br_2 / H_2O)$ to form $2,4,6$-tribromophenol,causing the reddish-brown colour of bromine to disappear.
Additionally,phenol reacts with neutral $FeCl_3$ to form a violet-coloured iron-phenoxide complex,which is a characteristic test for phenolic groups.

(B) The reaction of $4$-hydroxybenzenesulphonic acid with bromine water involves electrophilic aromatic substitution. The $-OH$ group is a strong activating group and is ortho/para directing. The $-SO_3H$ group is at the para position. Bromination occurs at the ortho positions relative to the $-OH$ group (positions $3$ and $5$). Since the $-SO_3H$ group is a good leaving group in the presence of water,it can be displaced by bromine,but under standard conditions with bromine water,the primary product is $3, 5$-dibromo-$4$-hydroxybenzenesulphonic acid. If excess bromine is used,the $-SO_3H$ group is replaced to form $2, 4, 6$-tribromophenol. Given the options,$3, 5$-dibromo-$4$-hydroxybenzenesulphonic acid is the specific product of the substitution.

(D) The process of benzoylation of compounds containing active hydrogen such as phenol,aniline,or alcohol with benzoyl chloride in the presence of aqueous $NaOH$ is known as the Schotten-$Baumann$ reaction.

(C) The reaction of phenol with $CO_2$ in the presence of $NaOH$ followed by acidification is the Kolbe-Schmidt reaction,which produces salicylic acid as product $(A)$.
Salicylic acid then undergoes acetylation with acetic anhydride to form aspirin (acetylsalicylic acid) as product $(B)$.
In this reaction,the phenolic $-OH$ group is acetylated to form an ester linkage,resulting in $2$-acetoxybenzoic acid.

(A) The reaction sequence is as follows:
$1.$ $p\text{-toluidine}$ $(4\text{-methylaniline})$ reacts with $NaNO_2/HCl$ at $0\,^{\circ}C$ to form the diazonium salt $(A)$,which is $p\text{-toluenediazonium chloride}$.
$2.$ Hydrolysis of $(A)$ with $H_2O$ at high temperature yields $p\text{-cresol}$ $(4\text{-methylphenol})$ as $(B)$.
$3.$ The Reimer-Tiemann reaction of $p\text{-cresol}$ with $CHCl_3$ and $KOH$ followed by acidification introduces a formyl group $(-CHO)$ at the ortho position relative to the $-OH$ group,resulting in $2\text{-hydroxy-5-methylbenzaldehyde}$ as $(C)$.

(D) The reaction of phenol with $Br_2$ in the presence of $OH^{\ominus}$ is an electrophilic aromatic substitution reaction.
First,phenol reacts with $OH^{\ominus}$ to form the phenoxide ion,which is highly activated towards electrophilic substitution.
The phenoxide ion then reacts with $Br_2$ to form ortho-bromophenoxide ($2$-bromophenoxide) and para-bromophenoxide ($4$-bromophenoxide) intermediates.
Further reaction with $Br_2$ leads to the formation of $2,4,6-$tribromophenol.
Since all the given species ($2$-bromophenoxide,$4$-bromophenoxide,and $2,4-$dibromophenoxide) are formed as intermediates during the stepwise bromination process,the correct answer is 'All of these'.