Properties of Phenols Questions in English

(A) The acidity of $C_6H_5OH$ (phenol) is greater than that of $C_2H_5OH$ (ethanol) because the phenoxide ion formed after the loss of $H^+$ is stabilized by resonance with the benzene ring,whereas the ethoxide ion $(C_2H_5O^-)$ is destabilized by the electron-donating inductive effect $(+I)$ of the ethyl group.
Phenol does not react with $NaHCO_3$ because it is a weaker acid than carbonic acid $(H_2CO_3)$.
Phenol does not react with $NH_2OH$ to form an oxime; this reaction is characteristic of carbonyl compounds (aldehydes and ketones).

(B) The given reaction is known as the Gattermann reaction.
In this reaction,an aromatic compound such as phenol is treated with a mixture of $HCN$ and dry $HCl$ gas in the presence of anhydrous $ZnCl_2$ to introduce a formyl group $(-CHO)$ into the aromatic nucleus.

(A) When salicylic acid $(C_6H_4(OH)COOH)$ is heated with phenol $(C_6H_5OH)$ in the presence of phosphorus oxychloride $(POCl_3)$,an esterification reaction occurs.
This reaction results in the formation of phenyl salicylate,which is commonly known as Salol.
The chemical equation is:
$C_6H_4(OH)COOH + C_6H_5OH \xrightarrow{POCl_3} C_6H_4(OH)COOC_6H_5 + H_2O$
Therefore,the correct option is $A$.

(C) In aqueous solution,phenol ionizes to form the phenoxide ion,which strongly activates the benzene ring towards electrophilic substitution,leading to the formation of $2, 4, 6-$ tribromophenol.
In the presence of a non-polar solvent like $CS_2$,phenol does not ionize significantly. The benzene ring is only weakly activated by the $-OH$ group,resulting in the formation of monosubstituted products,specifically $o-$ bromophenol and $p-$ bromophenol.

(C) The chemical equation for the reaction is:
$C_6H_5OH + 3Br_2 \rightarrow C_6H_2Br_3OH + 3HBr$
From the stoichiometry of the reaction,$1 \ mole$ of phenol $(94 \ g/mol)$ reacts with $3 \ moles$ of $Br_2$ $(3 \times 160 = 480 \ g/mol)$.
Therefore,$94 \ g$ of phenol requires $480 \ g$ of $Br_2$.
For $2 \ g$ of phenol,the amount of $Br_2$ required is:
$\text{Amount of } Br_2 = \frac{480 \ g}{94 \ g} \times 2 \ g \approx 10.2127 \ g \approx 10.22 \ g$.

(B) The presence of an electron-donating group $(-OH)$ in Phenol significantly increases the electron density of the benzene ring through resonance,making it highly reactive toward electrophilic aromatic substitution like nitration compared to Toluene,Benzoic acid,or Nitrobenzene.

(D) $1$. Phenol reacts with $Zn$ dust upon distillation to form Benzene $(A = C_6H_6)$.
$2$. Benzene undergoes nitration with a mixture of $Conc. H_2SO_4$ and $Conc. HNO_3$ to form Nitrobenzene $(B = C_6H_5NO_2)$.
$3$. Nitrobenzene undergoes reduction in the presence of $Zn$ and $NaOH$ (alkaline medium) to form Hydrazobenzene $(C = C_6H_5-NH-NH-C_6H_5)$.
Thus,the correct sequence is $C_6H_6, C_6H_5NO_2$ and hydrazobenzene.

(B) The reaction of phenol with $CCl_4$ in the presence of an aqueous alkali (like $KOH$) is known as the Reimer-Tiemann reaction (specifically the variant using $CCl_4$).
This reaction proceeds as follows:
$C_6H_5OH + CCl_4 + 4KOH \rightarrow C_6H_4(OH)(COOK) + 4KCl + 2H_2O$
The intermediate product formed is the potassium salt of salicylic acid.
Upon subsequent acidic hydrolysis,this salt is converted into salicylic acid $(2-hydroxybenzoic \ acid)$.

(B) For $o$-nitrophenol,hydrogen bonding is intramolecular,i.e.,confined to the same molecule.
For $m$-nitrophenol and $p$-nitrophenol,hydrogen bonding is intermolecular,occurring between different molecules.
Therefore,$p$-nitrophenol has the highest melting point due to strong intermolecular hydrogen bonding and symmetrical packing,followed by $m$-nitrophenol,while $o$-nitrophenol has the lowest melting point.
The correct order of melting point is $p > m > o$.

(A) The Reimer-Tiemann reaction is a chemical reaction used for the ortho-formylation of phenols.
In this reaction,phenol is treated with chloroform $(CHCl_3)$ in the presence of an aqueous base like sodium hydroxide $(NaOH)$,followed by acidic workup $(H^+)$ to yield salicylaldehyde as the major product.

(B) The rate of electrophilic substitution reaction in phenol is faster than in benzene.
This is because the $-OH$ group is an electron-donating group due to its $+M$ (mesomeric) effect.
The lone pair of electrons on the oxygen atom is delocalized into the benzene ring,which increases the electron density at the $o-$ (ortho) and $p-$ (para) positions.
This makes the ring more susceptible to attack by an electrophile.

(A) When phenol reacts with dilute $HNO_3$ at a low temperature (around $298 \ K$),it undergoes electrophilic aromatic substitution to form a mixture of $o$-nitrophenol and $p$-nitrophenol.
The reaction is as follows:
$C_6H_5OH + HNO_3 \text{ (dilute)} \xrightarrow{298 \ K} o\text{-Nitrophenol} + p\text{-Nitrophenol}$
Therefore,the correct option is $A$.

(B) The reaction of phenol with bromine water leads to the electrophilic substitution of bromine atoms at the $2, 4,$ and $6$ positions of the phenol ring.
The balanced chemical equation for this reaction is:
$C_6H_5OH + 3Br_2 \rightarrow C_6H_2Br_3OH + 3HBr$
From the stoichiometry of the reaction,$1 \ mol$ of phenol reacts with $3 \ mol$ of bromine $(Br_2)$ to produce $1 \ mol$ of $2,4,6-$tribromophenol.
Therefore,the value of $X$ is $3$.

(A) When phenol is treated with chloroform $(CHCl_3)$ in the presence of aqueous sodium hydroxide $(NaOH)$ at $340 \ K$,it forms $2$-hydroxybenzaldehyde,also known as salicylaldehyde.
This specific chemical transformation is known as the Reimer-Tiemann reaction.

(D) Picric acid,chemically known as $2,4,6$-trinitrophenol,is a derivative of phenol.
At room temperature $(25\,^{\circ}C)$,it exists as a bright yellow crystalline solid.
The intense yellow colour is due to the presence of three electron-withdrawing nitro groups $(-NO_2)$ attached to the benzene ring,which affect the electronic transitions of the molecule.

(A) When phenol $(C_6H_5OH)$ is distilled with zinc dust,it undergoes reduction to form benzene $(C_6H_6)$ and zinc oxide $(ZnO)$.
The chemical reaction is as follows:
$C_6H_5OH + Zn \rightarrow C_6H_6 + ZnO$
Thus,the correct option is $A$.

(C) The reaction of phenol $(C_6H_5OH)$ with diazomethane $(CH_2N_2)$ in the presence of a catalyst like $HBF_4$ results in the methylation of the phenolic hydroxyl group to form anisole $(C_6H_5OCH_3)$.
During this reaction,the nitrogen atom from the diazomethane is released as nitrogen gas $(N_2)$.
The chemical equation is: $C_6H_5OH + CH_2N_2 \xrightarrow{HBF_4} C_6H_5OCH_3 + N_2 \uparrow$.

(C) The acidity of phenol is primarily determined by the stability of the phenoxide ion formed after the loss of a proton from the $-OH$ group. Ring deuteration involves the replacement of hydrogen atoms on the aromatic ring with deuterium atoms. Since deuterium is an isotope of hydrogen,it has similar electronic properties but a slightly stronger $C-D$ bond compared to $C-H$. This substitution on the aromatic ring does not significantly alter the electronic environment of the $-OH$ group or the stability of the phenoxide ion. Therefore,ring deuteration has no significant effect on the acidity of phenol.

(C) The acidity of phenols is determined by the stability of the phenoxide ion formed after the loss of a proton.
Electron-withdrawing groups $(EWG)$ like $-Cl$ increase the acidity of phenols by stabilizing the phenoxide ion through the inductive effect.
Benzyl alcohol and cyclohexanol are not acidic in nature as they do not form stable phenoxide-like ions.
Between phenol and $m$-chlorophenol,the presence of the $-Cl$ group in $m$-chlorophenol exerts an electron-withdrawing inductive effect ($-I$ effect),which stabilizes the conjugate base more than in unsubstituted phenol.
Therefore,$m$-chlorophenol is the most acidic among the given options.

(B) In Friedel-Crafts acylation,an aromatic compound (such as phenol) reacts with an acyl halide (such as $CH_3COCl$) in the presence of a Lewis acid catalyst like anhydrous $AlCl_3$ to form an acyl derivative.
The reaction is:
$\text{Phenol} + CH_3COCl \xrightarrow{AlCl_3} o-\text{acetylphenol} + p-\text{acetylphenol}$
Therefore,the correct reactants are phenol and acetyl chloride $(CH_3COCl)$.

(A) The reaction of phenol with $CHCl_3$ in the presence of aqueous $NaOH$ is known as the Reimer-Tiemann reaction.
This reaction introduces a formyl group $(-CHO)$ at the ortho position of the phenol ring to produce salicylaldehyde.
The overall reaction is:
Phenol + $CHCl_3 + aq. NaOH \rightarrow$ Intermediate $\rightarrow$ Salicylaldehyde (after acidification with $H^+$).

(A) The given reaction is an electrophilic aromatic substitution reaction known as a coupling reaction.
In this reaction,benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$ reacts with phenol $(C_6H_5OH)$ in the presence of a base.
The diazonium cation acts as an electrophile and attacks the para-position of the phenol ring to form $p$-hydroxyazobenzene,which is an orange dye.
The reaction is: $C_6H_5N_2^+Cl^- + H-C_6H_4-OH \xrightarrow{\text{base}} C_6H_5-N=N-C_6H_4-OH + HCl$.

(B) The Reimer-Tiemann reaction involves the introduction of a formyl group $(-CHO)$ into the benzene ring of phenol using chloroform $(CHCl_3)$ and aqueous $NaOH$.
This process involves the formation of a new carbon$-$carbon bond between the aromatic ring and the formyl carbon.
The reaction is represented as:
Phenol + $CHCl_3$ $\xrightarrow{NaOH, 340 K}$ Salicylaldehyde (via intermediate) $\xrightarrow{dil. HCl, -NaCl}$ Salicylaldehyde.
Therefore,the correct option is $(B)$.

(A) The reaction described is the $Reimer-Tiemann$ reaction.
In this reaction,phenol reacts with chloroform $(CHCl_3)$ in the presence of an aqueous base $(NaOH)$ to form $o$-hydroxybenzaldehyde (salicylaldehyde).
The electrophile involved in this reaction is dichlorocarbene $(:CCl_2)$,which is generated from chloroform by the action of the base.
The mechanism is as follows:
$OH^{-} + CHCl_3 \rightleftharpoons H_2O + :CCl_3^{-} \to Cl^{-} + :CCl_2$ (dichlorocarbene).

(A) The reaction sequence given is the characteristic test for phenols known as $Liebermann's \ nitroso \ reaction$.
In this reaction,phenol reacts with $NaNO_2$ and $H_2SO_4$ to form $p-nitrosophenol$ $(B)$,which further reacts with another molecule of phenol in the presence of $H_2SO_4$ to form an indophenol dye $(C)$.
Upon adding $NaOH$,the solution turns deep blue or green due to the formation of the sodium salt of the indophenol $(D)$.

(A) $o-$ and $p-$ nitrophenols are separated by steam distillation.
$o-$ nitrophenol exhibits intramolecular hydrogen bonding,making it steam volatile.
$p-$ nitrophenol exhibits intermolecular hydrogen bonding,leading to higher molecular association and lower volatility,thus it is not steam volatile.

(B) The Reimer-Tiemann reaction involves the reaction of phenol with chloroform $(CHCl_3)$ in the presence of an aqueous base ($NaOH$ or $KOH$).
In the first step,the base abstracts a proton from chloroform to form a trichloromethyl carbanion $(:CCl_3^-)$,which subsequently loses a chloride ion to generate dichlorocarbene $(:CCl_2)$ as an electrophilic intermediate.
This dichlorocarbene then attacks the phenoxide ion to form the product.
Therefore,the correct answer is $(B)$.

(D) The reaction sequence is known as Kolbe's reaction.
$1$. Phenol $(C_6H_5OH)$ reacts with $NaOH$ to form sodium phenoxide $(A)$: $C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O$.
$2$. Sodium phenoxide reacts with $CO_2$ at $140 \ ^oC$ and $4-7 \ atm$ pressure to form sodium salicylate $(B)$: $C_6H_5ONa + CO_2 \rightarrow C_6H_4(OH)COONa$.
$3$. Acidification of sodium salicylate with $HCl$ yields salicylic acid $(C)$: $C_6H_4(OH)COONa + HCl \rightarrow C_6H_4(OH)COOH + NaCl$.
Thus,the final product $C$ is salicylic acid.

(B) The preparation of salicylic acid from phenol using $CO_2$ and $NaOH$ is known as the Kolbe-Schmidt reaction.
In this reaction,phenol is first converted to sodium phenoxide,which then undergoes electrophilic substitution with $CO_2$ to form sodium salicylate,followed by acidification to yield salicylic acid.
The reaction shown in the image uses $CCl_4$ and $NaOH$,which is the mechanism for the Reimer-Tiemann reaction (specifically the form that produces salicylic acid).
Therefore,the correct answer is $(B)$.

(D) The Reimer-Tiemann reaction of phenol with $CHCl_3$ and $NaOH$ yields salicylaldehyde. However,the question refers to salicylic acid (which is obtained by the Kolbe-Schmitt reaction). When salicylic acid ($o$-hydroxybenzoic acid) is heated with $Zn$ dust,the phenolic $-OH$ group is removed,resulting in the reduction of the compound to benzoic acid.

(D) Benzoic acid $(C_6H_5COOH)$ undergoes decarboxylation when heated with sodalime $(NaOH + CaO)$ to form benzene $(C_6H_6)$. Thus,$X$ is sodalime.
Phenol $(C_6H_5OH)$ undergoes reduction when heated with zinc dust $(Zn)$ to form 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 $(D)$.

(A) The acidity of phenols increases with the presence of electron-withdrawing groups $(-NO_2)$ on the benzene ring due to the $-I$ and $-M$ effects.
Picric acid ($2,4,6-$trinitrophenol) contains three $-NO_2$ groups at the $ortho$ and $para$ positions,which exert a strong electron-withdrawing effect,significantly stabilizing the phenoxide ion.
Comparing the options:
$1$. Picric acid ($3$ $-NO_2$ groups)
$2$. $o,p-$dinitrophenol ($2$ $-NO_2$ groups)
$3$. $p-$nitrophenol ($1$ $-NO_2$ group)
$4$. $m-$nitrophenol ($1$ $-NO_2$ group)
Therefore,Picric acid is the most acidic.

(A) The coupling reaction occurs between a diazonium salt and an activated aromatic ring such as phenol or aniline.
In this process,benzene diazonium chloride reacts with phenol in a weakly alkaline medium to form an azo dye.
Therefore,the correct option is $(A)$.

(C) Aspirin is chemically known as $2$-acetoxybenzoic acid.
It is prepared by the acetylation of salicylic acid ($2$-hydroxybenzoic acid) using acetyl chloride $(CH_3COCl)$ or acetic anhydride.
The reaction is:
$C_6H_4(OH)COOH + CH_3COCl \rightarrow C_6H_4(OCOCH_3)COOH + HCl$
Thus,salicylic acid reacts with $CH_3COCl$ to form aspirin.

(B) The given reaction sequence is:
$C_6H_5 - NH_2 \xleftarrow[HNO_2]{Cu} C_6H_5N_2Cl \xrightarrow[\Delta]{Water} C_6H_5OH$.
Here,$X$ is the $-NH_2$ group (aniline),which is an $o-, p-$ directing group.
$Y$ is the $-OH$ group (phenol),which is also an $o-, p-$ directing group.
Therefore,both $X$ and $Y$ are $o-, p-$ directing groups.

(B) The reaction of chlorobenzene with aqueous $NaOH$ at high temperature $(633\,K)$ and high pressure $(300\,bar)$ is known as the Dow process.
In this process,chlorobenzene is converted into sodium phenoxide,which upon acidification with an acid catalyst (like $HCl$) yields phenol.
The chemical equation is: $C_6H_5Cl + 2NaOH \xrightarrow{633\,K, 300\,bar} C_6H_5ONa + NaCl + H_2O$,followed by $C_6H_5ONa + H^+ \rightarrow C_6H_5OH + Na^+$.

(B) The reaction of phenol with $CHCl_3$ (chloroform) and aqueous $NaOH$ at $340 \ K$ is known as the Reimer-Tiemann reaction.
In this reaction,an aldehyde 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.

(D) The given reaction involves the treatment of phenol with chloroform $(CHCl_3)$ in the presence of aqueous sodium hydroxide $(aq. NaOH)$ at $70^{\circ}C$,which results in the formation of salicylaldehyde (o-hydroxybenzaldehyde). This specific chemical transformation is known as the Reimer-Tiemann reaction.

(D) The reaction of phenol with $CHCl_3$ and $KOH$ is the Reimer-Tiemann reaction.
In this reaction,phenol is converted into $o$-hydroxybenzaldehyde (salicylaldehyde).
The reaction mechanism involves the formation of a dichlorocarbene $(:CCl_2)$ intermediate.
Therefore,statements $A$,$B$,and $C$ are correct.
Statement $D$ is incorrect because benzyl chloride is not formed in this reaction.

(A) $1$. Decarboxylation of benzoic acid $(C_6H_5COOH)$ with sodalime $(NaOH + CaO)$ yields benzene $(C_6H_6)$. Thus,$X$ is sodalime.
$2$. Reduction of phenol $(C_6H_5OH)$ with zinc dust $(Zn)$ yields benzene $(C_6H_6)$. Thus,$Y$ is zinc dust.
$3$. Therefore,$X$ and $Y$ are sodalime and zinc dust respectively.

(B) The compounds are: $(I)$ Catechol (ortho-dihydroxybenzene),$(II)$ Resorcinol (meta-dihydroxybenzene),$(III)$ Hydroquinone (para-dihydroxybenzene),and $(IV)$ Phenol.
Boiling point depends on intermolecular hydrogen bonding.
$(IV)$ Phenol has only one $-OH$ group,so it has the lowest boiling point.
Among the dihydroxybenzenes,$(I)$ Catechol exhibits intramolecular hydrogen bonding,which reduces the extent of intermolecular hydrogen bonding,leading to a lower boiling point compared to $(II)$ and $(III)$.
$(III)$ Hydroquinone has a symmetrical structure and strong intermolecular hydrogen bonding,resulting in the highest boiling point.
$(II)$ Resorcinol has an intermediate boiling point.
Thus,the order is: $(IV) < (I) < (II) < (III)$.

(A) Reaction $(I)$: Benzenesulfonic acid $(C_6H_5SO_3H)$ is a strong acid. When it reacts with sodium bicarbonate $(NaHCO_3)$,it undergoes an acid-base reaction to form sodium benzenesulfonate,water,and carbon dioxide $(CO_2)$ gas.
Reaction $(II)$: $p$-Nitrophenol is a weaker acid than carbonic acid $(H_2CO_3)$. Therefore,it does not react with sodium bicarbonate $(NaHCO_3)$ to evolve $CO_2$ gas.
However,looking at the provided options,the question implies a reaction occurs in both. If we consider the acidity,benzenesulfonic acid is stronger than $H_2CO_3$,so it releases $CO_2$. $p$-Nitrophenol is also acidic enough to react with $NaHCO_3$ to release $CO_2$ gas because its $pK_a$ is approximately $7.15$,which is comparable to or stronger than carbonic acid ($pK_a \approx 6.35$ for the first dissociation). Thus,both reactions evolve $CO_2$ gas.

(C) The acidity of phenols depends on the stability of the phenoxide ion formed after the loss of a proton. Electron-withdrawing groups $(EWG)$ increase acidity by stabilizing the phenoxide ion,while electron-donating groups $(EDG)$ decrease acidity by destabilizing it.
Let the compounds be:
$I$: $p$-chlorophenol ($-Cl$ has $-I$ effect,which increases acidity)
$II$: $p$-cresol ($-CH_3$ has $+I$ and $+H$ effects,which decrease acidity)
$III$: $p$-nitrophenol ($-NO_2$ has strong $-I$ and $-M$ effects,which significantly increase acidity)
$IV$: $p$-methoxyphenol ($-OCH_3$ has $-I$ effect but strong $+M$ effect,which decreases acidity more than $-CH_3$)
Comparing the effects:
$-NO_2$ (strong $EWG$) > $-Cl$ (weak $EWG$) > $-CH_3$ (weak $EDG$) > $-OCH_3$ (strong $EDG$).
Therefore,the order of acidity is $III > I > II > IV$.

(D) The acidity of phenols is determined by 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 $-I$ and $-M$ effects,thereby increasing acidity.
Electron-donating groups $(EDG)$ like $-CH_3$ (present in $o$-cresol) destabilize the phenoxide ion through $+I$ and hyperconjugation effects,thereby decreasing acidity.
Comparing the options:
$1$. $p$-Nitrophenol: Strongest acid due to the powerful $-M$ effect of $-NO_2$.
$2$. $p$-Chlorophenol: Acidic due to the $-I$ effect of $-Cl$.
$3$. Phenol: Baseline acidity.
$4$. $o$-Cresol: Least acidic due to the electron-donating $+I$ effect of the methyl group.
Therefore,$o$-cresol is the least acidic.

(C) The acidity of phenols increases with the presence of electron-withdrawing groups $(EWG)$ like $-NO_2$ due to their $-I$ and $-M$ effects,which stabilize the phenoxide ion. Conversely,electron-donating groups $(EDG)$ like $-OH$ decrease acidity.
$I$: $o$-Nitrophenol ($-NO_2$ at ortho position,$-I$ and $-M$ effect).
$II$: $p$-Nitrophenol ($-NO_2$ at para position,$-I$ and $-M$ effect).
$III$: $2,4,6$-Trinitrophenol (Picric acid,three $-NO_2$ groups,strongest acid).
$IV$: Resorcinol ($m$-hydroxy phenol,$-OH$ group at meta position acts as an $EDG$ via $+M$ effect,making it the least acidic).
Comparing $I$ and $II$: $p$-Nitrophenol $(II)$ is generally more acidic than $o$-Nitrophenol $(I)$ because $o$-Nitrophenol forms intramolecular hydrogen bonding,which stabilizes the molecule and makes the proton less available compared to $p$-Nitrophenol.
Thus,the order of acidity (and $K_a$) is: $IV < I < II < III$.

(A) The acidic strength of phenols depends on the stability of the phenoxide ion formed after the loss of a proton. Electron-withdrawing groups $(EWG)$ increase acidity by stabilizing the negative charge,while electron-donating groups $(EDG)$ decrease acidity by destabilizing it.
$I$: $p$-cresol ($-CH_3$ is an $EDG$,which decreases acidity).
$II$: Phenol (reference compound).
$III$: $p$-nitrophenol ($-NO_2$ is a strong $EWG$ at the para position,providing both $-I$ and $-M$ effects,making it the most acidic).
$IV$: $m$-nitrophenol ($-NO_2$ is an $EWG$ at the meta position,providing only $-I$ effect,making it less acidic than $p$-nitrophenol but more acidic than phenol).
Comparing the effects:
$p$-cresol $(I)$ < Phenol $(II)$ < $m$-nitrophenol $(IV)$ < $p$-nitrophenol $(III)$.
Thus,the correct order is $I < II < IV < III$.

(B) The acidity of a compound depends on the stability of its conjugate base.
$1$. $Cl-CH_2-CH_2-OH$ is an aliphatic alcohol,which is much less acidic than phenols.
$2$. Among the phenols,the acidity is determined by the electron-withdrawing or electron-donating nature of the substituents.
$3$. The $-NO_2$ group is a strong electron-withdrawing group ($-I$ and $-M$ effect),which stabilizes the phenoxide ion significantly.
$4$. The $-CH_3$ group is an electron-donating group ($+I$ and hyperconjugation),which destabilizes the phenoxide ion,making it less acidic than phenol.
$5$. Therefore,$o$-Nitrophenol is the most acidic among the given options.

(D) The acidity of phenols is determined by the stability of the phenoxide ion formed after the loss of a proton. Electron-withdrawing groups $(EWG)$ increase acidity by stabilizing the negative charge,while electron-donating groups $(EDG)$ decrease acidity by destabilizing it.
$(I)$ Phenol: Reference compound.
$(II)$ $p$-Methylphenol: The $-CH_3$ group is an $EDG$ (via $+I$ and hyperconjugation),which destabilizes the phenoxide ion,making it the least acidic.
$(III)$ $m$-Nitrophenol: The $-NO_2$ group is a strong $EWG$ (via $-I$ effect),which stabilizes the phenoxide ion.
$(IV)$ $p$-Nitrophenol: The $-NO_2$ group is a strong $EWG$ (via both $-I$ and $-M$ effects),which provides maximum stabilization to the phenoxide ion.
Comparing the effects: The $-M$ effect of $-NO_2$ at the para position is more effective than the $-I$ effect at the meta position. Thus,the order of acidity is $IV > III > I > II$.

(D) The acidity of phenol is directly proportional to the electron-withdrawing effect ($-I$ and $-M$ effect) of the substituent attached to the benzene ring.
All the given groups,$-NO_2$,$-CN$,and $-CHO$,are electron-withdrawing groups $(EWG)$ that stabilize the phenoxide ion by dispersing the negative charge through $-I$ and $-M$ effects.
Therefore,all of these groups increase the acidity of phenol.