Properties of Phenols Questions in English

(N/A) In phenol, the oxygen atom of the $-OH$ group is directly attached to the $sp^{2}$ hybridized carbon atom of the benzene ring.
Due to resonance, the $C-O$ bond acquires a partial double bond character.
This makes the $C-O$ bond in phenol shorter $(136 \ pm)$ and stronger compared to the $C-O$ bond in alcohols $(142 \ pm)$, where the carbon is $sp^{3}$ hybridized and the bond is a pure single bond.

In phenol,the oxygen atom of the $-OH$ group is directly attached to an $sp^{2}$ hybridized carbon atom of the benzene ring.
Due to resonance,the $C-O$ bond acquires a partial double bond character.
This makes the $C-O$ bond shorter and stronger compared to the $C-O$ bond in alcohols,making it difficult to break.
Furthermore,the benzene ring is electron-rich,which causes repulsion between the incoming nucleophile and the electron-rich ring,thus hindering nucleophilic substitution reactions.

(N/A) The conversion of phenol to aspirin involves two main stages:
$1$. Conversion of phenol to salicylic acid (Kolbe's reaction):
Phenol is treated with $NaOH$ to form sodium phenoxide. This is then reacted with $CO_2$ at $400 \ K$ and $4-7 \ atm$ pressure,followed by acidification with $H^+$ to yield salicylic acid.
$2$. Acetylation of salicylic acid:
Salicylic acid is reacted with acetic anhydride $(CH_3CO)_2O$ in the presence of an acid catalyst $(H^+)$. This process,known as acetylation,replaces the phenolic hydrogen with an acetyl group to form aspirin ($2$-acetoxybenzoic acid) and acetic acid $(CH_3COOH)$.

(N/A) The phenoxide ion is more reactive than phenol towards electrophilic aromatic substitution. This is because the $-O^-$ group is a stronger electron-donating group than the $-OH$ group due to the presence of a full negative charge,which increases the electron density on the benzene ring. Consequently,the ring becomes more nucleophilic,facilitating the attack of $CO_2$,which is a weak electrophile.

(N/A) In phenol, the $C-O$ bond is less polar due to the electron-withdrawing $-I$ effect of the $sp^2$ hybridized carbon of the benzene ring, which pulls the electron density away from the oxygen atom.
In methanol, the $C-O$ bond is more polar due to the electron-releasing $+I$ effect of the methyl group, which pushes electron density towards the oxygen atom.
Consequently, the net dipole moment of phenol $(\mu = 1.54 \ D)$ is smaller than that of methanol $(\mu = 1.71 \ D)$.

(N/A) The acidic strength of phenols is determined by the stability of the phenoxide ion formed after the loss of a proton $(H^+)$.
In $o-$nitrophenol,an intramolecular hydrogen bond is formed between the $-OH$ group and the $-NO_2$ group,which makes the release of the proton $(H^+)$ more difficult.
In $p-$nitrophenol,the $-NO_2$ group is far from the $-OH$ group,so no intramolecular hydrogen bonding occurs.
Furthermore,the $-NO_2$ group exerts a strong electron-withdrawing $-I$ and $-M$ effect,which stabilizes the phenoxide ion by dispersing the negative charge.
Due to the absence of intramolecular hydrogen bonding and the effective stabilization of the phenoxide ion,$p-$nitrophenol is more acidic than $o-$nitrophenol.

(N/A) In phenol,the lone pair of electrons on the oxygen atom is involved in resonance with the benzene ring.
This resonance gives the $C-O$ bond a partial double bond character.
In contrast,there is no such resonance in methanol,where the $C-O$ bond is a pure single bond.
Additionally,the carbon atom attached to oxygen in phenol is $sp^2$ hybridized,while in methanol,it is $sp^3$ hybridized.
The $sp^2$ hybridized carbon is more electronegative and holds the electron pair more tightly,contributing to a shorter and stronger bond compared to the $sp^3$ hybridized carbon in methanol.

(N/A) The acidic strength of a substance depends on the stability of the conjugate base. The more stable the conjugate base,the stronger the acid.
$1$. Phenol forms a phenoxide ion,which is stabilized by resonance.
$2$. Water forms a hydroxide ion,which is less stable than the phenoxide ion but more stable than the ethoxide ion.
$3$. Ethanol forms an ethoxide ion,where the $+I$ effect of the ethyl group increases the electron density on the oxygen atom,destabilizing the conjugate base.
Therefore,the order of acidic strength is: $\text{ethanol} < \text{water} < \text{phenol}$.

(N/A) The starting material used in the industrial preparation of phenol is cumene (isopropylbenzene).
$(B)$ Bromination of phenol:
$1$. In aqueous medium (using $Br_2$ water): Phenol reacts with bromine water to give a white precipitate of $2,4,6$-tribromophenol.
$C_6H_5OH + 3Br_2(aq) \rightarrow C_6H_2(OH)Br_3 + 3HBr$
$2$. In non-aqueous medium (using $CS_2$ or $CHCl_3$ at low temperature): Phenol reacts with $Br_2$ to give a mixture of $o$-bromophenol and $p$-bromophenol.
$C_6H_5OH + Br_2 \xrightarrow{CS_2, 298 K} C_6H_4(OH)Br (ortho) + C_6H_4(OH)Br (para)$
$(C)$ $A$ Lewis acid is not required because the $-OH$ group in phenol is a strongly activating group. It increases the electron density in the benzene ring to such an extent that the electrophilic substitution reaction occurs readily even without the presence of a Lewis acid catalyst.

(N/A) Phenol is converted to aspirin through the following steps:
$1$. Kolbe reaction: Phenol is treated with $NaOH$ followed by $CO_2$ at $4-7 \ atm$ pressure and $393 \ K$ to form salicylic acid ($2$-hydroxybenzoic acid).
$2$. Acetylation: Salicylic acid is treated with acetic anhydride in the presence of an acid catalyst to form aspirin (acetylsalicylic acid).

(A) The reaction between phenol $(C_6H_5OH)$ and an arenediazonium salt $(ArN_2^+Cl^-)$ in the presence of a base $(OH^-)$ is an electrophilic aromatic substitution reaction known as coupling reaction.
In this reaction,the diazonium cation acts as an electrophile and attacks the electron-rich phenol ring.
Since the $-OH$ group is ortho/para-directing,the coupling primarily occurs at the para-position to form $p-hydroxyazobenzene$ derivatives $(Ar-N=N-C_6H_4-OH)$.

(N/A) The $-OH$ group is an electron-donating group that activates the benzene ring towards electrophilic substitution reactions due to the following reasons:
$(a)$ Resonance Effect: The lone pair of electrons on the oxygen atom of the $-OH$ group is delocalized into the benzene ring through resonance. This increases the electron density at the ortho and para positions of the ring,making them more susceptible to attack by electrophiles.
$(b)$ Inductive Effect: The oxygen atom is more electronegative than carbon,which exerts a $-I$ effect,tending to withdraw electrons. However,the $+R$ (resonance) effect of the $-OH$ group is much stronger than its $-I$ effect.
Conclusion: Since the resonance effect dominates,the overall electron density in the benzene ring of phenol increases,especially at the ortho and para positions. Consequently,phenol is more reactive than benzene towards electrophilic aromatic substitution,and the $-OH$ group acts as an activating group.

(A) The reaction sequence is as follows:
$1$. When $o$-hydroxybenzoic acid (salicylic acid) is heated with $Zn$ dust,the $-OH$ group is reduced to $H$,yielding benzoic acid.
$2$. When $o$-hydroxybenzoic acid is heated with soda lime $(NaOH + CaO)$,decarboxylation occurs,removing the $-COOH$ group and yielding phenol.
Therefore,the compound $X$ is $o$-hydroxybenzoic acid.

(A) When phenol is heated with zinc dust,it undergoes reduction to form benzene and zinc oxide.
$C_6H_5OH + Zn \rightarrow C_6H_6 + ZnO$

(A-IV, B-III, C-II, D-I) Dinitration of $4$-methylphenol ($p$-cresol) leads to the formation of $2,6$-dinitro-$4$-methylphenol,which corresponds to $(iv)$.
$(B)$ Mononitration of phenyl ethanoate gives $4$-nitrophenyl ethanoate as the major product,which corresponds to $(iii)$.
$(C)$ Dinitration of $3$-methylphenol ($m$-cresol) leads to the formation of $2,4$-dinitro-$5$-methylphenol,which corresponds to $(ii)$.
$(D)$ Mononitration of $3$-methylphenol leads to a mixture of $2$-nitro-$4$-methylphenol and $2$-nitro-$5$-methylphenol,which corresponds to $(i)$.
Therefore,the correct matching is: $(A-iv, B-iii, C-ii, D-i)$.

(A) The reaction involves the nitration of $2$-methyl$-4-$nitrophenol using concentrated $HNO_3$ and concentrated $H_2SO_4$.
The $-OH$ group is a strong activating group and is ortho/para directing.
The $-CH_3$ group is also ortho/para directing.
The $-NO_2$ group is meta directing.
In $2$-methyl$-4-$nitrophenol,the positions ortho to the $-OH$ group are position $6$ (which is vacant) and position $3$ (which is also vacant).
However,position $6$ is less sterically hindered compared to position $3$,which is adjacent to the $-CH_3$ group.
Therefore,the incoming $-NO_2^+$ electrophile attacks the $6$-position to form $2$-methyl$-4,6-$dinitrophenol as the major product.

(D) Ceric ammonium nitrate test is used to detect the presence of alcoholic $-OH$ groups. It gives a red/yellow color with alcohols.
In the given reaction,the starting material is treated with chromic anhydride (a strong oxidizing agent). The primary alcohol group at position '$a$' is oxidized to an aldehyde,and the secondary alcohol group at position '$c$' is oxidized to a ketone.
The resulting product '$P$' contains the tertiary alcohol group at position '$b$' and the phenolic $-OH$ group at position '$d$'.
However,the ceric ammonium nitrate test is specifically positive for aliphatic alcohols (primary,secondary,and tertiary). Phenols also give a positive test (often red/brown color).
Looking at the structure of '$P$',the tertiary alcohol at '$b$' and the phenol at '$d$' remain. Both are capable of giving a positive ceric ammonium nitrate test. Therefore,the presence of both '$b$' and '$d$' groups contributes to the positive test.

(A) The boiling point of compounds depends on intermolecular forces,primarily hydrogen bonding and dipole-dipole interactions. The compounds are:
$I$: $p$-cresol ($4$-methylphenol)
$II$: $p$-nitrophenol
$III$: $p$-aminophenol
$IV$: $p$-methoxyphenol
Comparing their boiling points:
$p$-cresol $(I)$ has a boiling point of $\approx 202^{\circ}C$.
$p$-methoxyphenol $(IV)$ has a boiling point of $\approx 243^{\circ}C$.
$p$-nitrophenol $(II)$ has a boiling point of $\approx 279^{\circ}C$.
$p$-aminophenol $(III)$ has a boiling point of $\approx 284^{\circ}C$.
Thus,the increasing order of boiling points is $I < IV < II < III$.

(B) The reaction of phenol with chloroform $(CHCl_3)$ in the presence of aqueous $NaOH$ is the Reimer-Tiemann reaction,which yields salicylaldehyde ($o$-hydroxybenzaldehyde) as the major product $P$.
The molecular formula of salicylaldehyde $(P)$ is $C_7H_6O_2$.
The molar mass of $C_7H_6O_2 = (7 \times 12) + (6 \times 1) + (2 \times 16) = 84 + 6 + 32 = 122 \ g/mol$.
The mass percentage of carbon in $P$ is given by:
$\text{Mass } \% C = \frac{\text{Total mass of Carbon}}{\text{Molar mass of } P} \times 100$
$\text{Mass } \% C = \frac{7 \times 12}{122} \times 100 = \frac{84}{122} \times 100 \approx 68.85\%$.
Rounding to the nearest integer,we get $69\%$.

(B) The acidity of substituted phenols depends on the electronic effects of the substituent groups attached to the benzene ring.
Electron-withdrawing groups $(EWG)$ increase acidity by stabilizing the phenoxide ion through $-I$ and $-R$ effects.
Electron-donating groups $(EDG)$ decrease acidity by destabilizing the phenoxide ion through $+I$ and $+R$ effects.
In the given options:
$(A)$ $-CH_3$ is an $EDG$ ($+I$ and hyperconjugation).
$(B)$ $-NO_2$ is a strong $EWG$ ($-I$ and $-R$ effects).
$(C)$ $-OCH_3$ is an $EDG$ ($+R$ effect dominates over $-I$).
$(D)$ $-C_2H_5$ is an $EDG$ ($+I$ effect).
Since $-NO_2$ is the only strong electron-withdrawing group among the choices,$p$-nitrophenol is the strongest acid.

(A) The reactivity of aromatic compounds towards electrophilic aromatic substitution depends on the electron density of the benzene ring.
Groups that donate electrons to the ring (via resonance or induction) increase electron density and thus increase reactivity.
Groups that withdraw electrons from the ring decrease electron density and thus decrease reactivity.
$1$. $-OH$ group: Shows a strong $+R$ (resonance) effect,which significantly increases the electron density of the benzene ring,making it highly reactive.
$2$. Benzene: Has no substituent,serving as the reference.
$3$. $-Cl$ group: Shows a $-I$ (inductive) effect,which withdraws electron density,making it less reactive than benzene.
$4$. $-NO_2$ group: Shows a strong $-R$ (resonance) effect,which strongly withdraws electron density,making it the least reactive.
Therefore,the order of reactivity is: $\text{Phenol} > \text{Benzene} > \text{Chlorobenzene} > \text{Nitrobenzene}$.
Thus,phenol is the most reactive.

(C) 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,while electron-donating groups $(-EDG)$ decrease it.
$(a)$ $p$-Nitrophenol: The $-NO_2$ group at the para position exerts a strong $-M$ (mesomeric) and $-I$ (inductive) effect,significantly stabilizing the phenoxide ion.
$(b)$ $o$-Nitrophenol: The $-NO_2$ group at the ortho position exerts $-M$ and $-I$ effects,but also forms an intramolecular hydrogen bond with the $-OH$ group,which stabilizes the neutral molecule and makes the release of $H^+$ slightly less favorable compared to $p$-nitrophenol.
$(d)$ Phenol: Has no substituent.
$(c)$ $p$-Cresol: The $-CH_3$ group at the para position exerts a $+I$ and $+H$ (hyperconjugation) effect,which destabilizes the phenoxide ion,making it the least acidic.
Thus,the correct order of acidic strength is $p$-nitrophenol $(a) > o$-nitrophenol $(b) >$ phenol $(d) > p$-cresol $(c)$.
Therefore,the correct option is $C$.

(B) The reaction of phenol with benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$ in the presence of a base $(KOH)$ is an electrophilic aromatic substitution reaction known as a coupling reaction.
In this reaction,the diazonium cation acts as an electrophile and attacks the electron-rich ring of phenol at the para-position to form $p$-hydroxyazobenzene.
Since the question mentions both phenol and aniline,but the diazonium salt is derived from aniline $(C_6H_5N_2^+Cl^-)$,the coupling occurs between the diazonium salt and the phenol to yield $p$-hydroxyazobenzene.

(A) The reaction of phenols with phthalic anhydride in the presence of concentrated $H_2SO_4$ is a condensation reaction that forms phthalein dyes.
For a phenol to form a phthalein dye (which shows a pink colour in basic medium),it must have a free para-position relative to the hydroxyl $(-OH)$ group to allow for the electrophilic aromatic substitution $(EAS)$ by the phthalic anhydride intermediate.
In $2$-propylphenol,the para-position is free,allowing the reaction to proceed to form the corresponding phthalein derivative,which turns pink upon treatment with $NaOH$.

(A) $1$. When phenol reacts with bromine water ($Br_2$ in $H_2O$),the highly activating $-OH$ group causes rapid electrophilic substitution at all ortho and para positions,resulting in the formation of $2,4,6$-tribromophenol as product $A$.
$2$. When phenol reacts with bromine in a non-polar solvent like carbon disulfide $(CS_2)$ at low temperature $(< 5^{\circ}C)$,the reaction is less vigorous,leading to monobromination. The major product $B$ is $p$-bromophenol due to steric hindrance at the ortho position.

(B) The condensation of phenols with phthalic anhydride in the presence of conc. $H_2SO_4$ is a test for the presence of a free $ortho$ or $para$ position relative to the $-OH$ group.
$p$-Cresol has the $para$ position blocked by a $-CH_3$ group and the $ortho$ positions are also hindered or do not react to form the characteristic phthalein dye color under these specific conditions.
Therefore,$p$-cresol does not give a coloured product.

(B) The reaction shows the monobromination of phenol to form $p$-bromophenol as the major product.
$(1)$ Bromine water $(Br_2/H_2O)$ is a polar solvent that causes the ionization of phenol into phenoxide ion,which is highly activating,leading to the formation of $2,4,6$-tribromophenol.
$(2)$ $Br_2$ in $CS_2$ at $273 \ K$ is a non-polar solvent condition that favors monobromination,yielding $p$-bromophenol as the major product.
$(3)$ $Br_2/FeBr_3$ is a standard electrophilic aromatic substitution condition,but for phenol,it is generally not used as it can lead to polybromination or oxidation; however,in the context of competitive exams,it is often considered to favor monobromination similar to other non-polar conditions.
$(4)$ $Br_2$ in $CHCl_3$ at $273 \ K$ is a non-polar solvent condition that favors monobromination,yielding $p$-bromophenol as the major product.
Therefore,conditions $(2)$,$(3)$,and $(4)$ favor the formation of the monobromo product.

(B) When phenol is oxidized with chromic acid $(H_2CrO_4)$,it undergoes oxidation to form $p$-benzoquinone.
$p$-Benzoquinone is a cyclic conjugated diketone.
Therefore,the correct description of the product is a conjugated diketone.

(B) Statement $I$ is correct: The $-NO_2$ group is a strong electron-withdrawing group $(EWG)$ due to both $-I$ and $-M$ effects. It stabilizes the phenoxide ion,thereby increasing the acidic strength of nitrophenols compared to phenol.
Statement $II$ is incorrect: The acidic strength of nitrophenols depends on the position of the $-NO_2$ group. The $-M$ effect is operative at ortho and para positions,while only the $-I$ effect is operative at the meta position. Therefore,their acidic strengths are different. The order of acidic strength is $p$-nitrophenol > $o$-nitrophenol > $m$-nitrophenol > phenol.

(C) The reaction of phenol with dilute $HNO_3$ produces a mixture of $o$-nitrophenol and $p$-nitrophenol.
$o$-Nitrophenol exhibits intramolecular $H$-bonding,which reduces its intermolecular attraction,making it steam volatile.
$p$-Nitrophenol exhibits intermolecular $H$-bonding,which leads to higher boiling points and lower volatility.
Therefore,steam distillation is the most effective method for separating these two isomers on a large scale.

(C) The reaction shown is the Reimer-Tiemann reaction,which involves the treatment of phenol with chloroform $(CHCl_3)$ in the presence of an aqueous base $(NaOH)$.
$1$. The base deprotonates the phenol to form a phenoxide ion.
$2$. The chloroform reacts with the base to generate the electrophile,dichlorocarbene $(:CCl_2)$.
$3$. The phenoxide ion attacks the dichlorocarbene to form an intermediate,which is sodium $2$-dichloromethylphenoxide.
$4$. This intermediate is then hydrolyzed to form salicylaldehyde.
Therefore,the intermediate $X$ is sodium $2$-dichloromethylphenoxide.

(C) The given reaction is the industrial preparation of phenol from cumene (isopropylbenzene).
$1$. Cumene is oxidized to cumene hydroperoxide $(C_6H_5C(CH_3)_2OOH)$ using $O_2$.
$2$. Upon treatment with dilute acid $(H^+/H_2O)$,the cumene hydroperoxide undergoes rearrangement to form phenol $(C_6H_5OH)$ and acetone $(CH_3COCH_3)$.
Therefore,the products $A$ and $B$ are phenol and acetone respectively.

(C) In electrophilic aromatic substitution,the incoming electrophile attacks the position that is ortho or para to the most strongly activating group (the group with the strongest electron-releasing effect).
For $2$-methylphenol ($o$-cresol),the $-OH$ group is a much stronger activating group than the $-CH_3$ group.
The $-OH$ group directs the electrophile to its ortho and para positions.
The para position with respect to the $-OH$ group is the meta position with respect to the $-CH_3$ group.
Therefore,halogenation of $2$-methylphenol yields the $m$-substituted product with respect to the methyl group as the major product.

(A) $1$. Compound $'P'$ is phenol $(C_6H_5OH)$.
$2$. Nitration of phenol with dil. $HNO_3$ gives a mixture of $o$-nitrophenol and $p$-nitrophenol.
$3$. $o$-Nitrophenol shows intramolecular $H$-bonding and is steam volatile,while $p$-nitrophenol shows intermolecular $H$-bonding and is not steam volatile.
$4$. Reaction of phenol with conc. $HNO_3$ yields $2,4,6$-trinitrophenol,commonly known as picric acid.
$5$. The chemical formula of picric acid is $C_6H_3N_3O_7$.
$6$. The number of oxygen atoms in picric acid $('C')$ is $7$.

(B) The reaction of $2$-bromophenol with $2$ equivalents of an organolithium reagent (like $n-BuLi$ or $t-BuLi$) involves two steps:
$1$. Acid-base reaction: The phenolic $-OH$ proton is acidic and is deprotonated by the first equivalent of the organolithium reagent to form a lithium phenoxide salt.
$2$. Halogen-metal exchange: The second equivalent of the organolithium reagent reacts with the aryl bromide to form an aryllithium species.
Thus,the intermediate $A$ is the dilithium salt of $2$-bromophenol,which has the structure of a benzene ring with an $-OLi$ group at the ortho position relative to a $-Li$ group. This species then reacts with $CO_2$ followed by acidic workup $(H_3O^+)$ to yield salicylic acid.

(C) Phenols are weakly acidic,so Statement $I$ is correct.
Phenols are more acidic than alcohols and water because the phenoxide ion is resonance-stabilized.
While phenols are soluble in $NaOH$ due to the formation of sodium phenoxide,the claim that they are weaker acids than alcohols and water is false.
Therefore,Statement $II$ is incorrect.

(A) The acidic strength of phenols depends on the nature of the substituents attached to the benzene ring. Electron-withdrawing groups $(EWG)$ increase acidity,while electron-donating groups $(EDG)$ decrease it.
$A$: $p$-Nitrophenol ($-NO_2$ is a strong $EWG$ at para position,strong $-I$ and $-M$ effect).
$B$: $m$-Nitrophenol ($-NO_2$ is an $EWG$ at meta position,only $-I$ effect).
$C$: $m$-Methoxyphenol ($-OCH_3$ is an $EDG$ by $+M$ effect but acts as an $EWG$ by $-I$ effect at meta position).
$D$: $p$-Methoxyphenol ($-OCH_3$ is a strong $EDG$ by $+M$ effect at para position).
Comparing the effects:
$A$ is the most acidic due to strong $-M$ effect of $-NO_2$ at para position.
$B$ is next,as $-NO_2$ at meta position only shows $-I$ effect.
$C$ is less acidic than $B$ because $-OCH_3$ at meta position acts as an $EWG$ ($-I$ effect) but is less effective than $-NO_2$.
$D$ is the least acidic due to the strong $+M$ electron-donating effect of $-OCH_3$ at para position.
Therefore,the decreasing order of acidic strength is $A > B > C > D$.

(B) The difference in the reaction products is observed due to the polarity of the solvent:
$(i)$ In a polar solvent like water,phenol ionizes to form the phenoxide ion $(C_6H_5O^-)$,which is much more reactive towards electrophilic aromatic substitution than phenol itself,leading to polybromination ($2$,$4$,$6$-tribromophenol).
$(ii)$ In a non-polar solvent like chloroform $(CHCl_3)$,phenol does not ionize significantly,and the reaction proceeds to form monobrominated products (o-bromophenol and p-bromophenol).

(B) $1$. The reaction of benzene with isopropyl chloride in the presence of $AlCl_3$ (Friedel-Crafts alkylation) gives cumene (isopropylbenzene).
$2$. Oxidation of cumene with $O_2$ followed by treatment with $H^+/H_2O$ yields phenol $(P)$ and acetone.
$3$. Phenol $(P)$ reacts with $Na_2Cr_2O_7/H_2SO_4$ (strong oxidizing agent) to form $p$-benzoquinone $(A)$.
$4$. Phenol $(P)$ reacts with $Br_2$ in $CS_2$ (a non-polar solvent) to give $p$-bromophenol $(B)$ as the major product due to the low polarity of the medium,which prevents further bromination.

(D) The reaction of an alkyl aryl ether with concentrated $HI$ involves the cleavage of the $C-O$ bond between the alkyl group and the oxygen atom. This is because the $C(aryl)-O$ bond has partial double bond character due to resonance and is stronger.
Thus,the reaction of phenyl isopropyl ether with $HI$ yields phenol $(A)$ and isopropyl iodide.
However,in the context of the given options,the reaction proceeds to form phenol $(A)$.
When phenol $(A)$ is heated with $Zn$ dust,it undergoes reduction to form benzene $(B)$.

(B) $1$. The compound $'X'$ is acidic and soluble in $NaOH$ but insoluble in $NaHCO_3$. This indicates that the compound is more acidic than water but less acidic than carbonic acid $(H_2CO_3)$. Phenols typically satisfy this condition.
$2$. The compound gives a violet colour with neutral $FeCl_3$ solution. This is a characteristic test for the presence of a phenolic group ($-OH$ group attached directly to a benzene ring).
$3$. Among the given options,Phenol $(C_6H_5OH)$ is the only compound that contains a phenolic group and satisfies all the given chemical properties.
$4$. The reaction with $FeCl_3$ is: $6C_6H_5OH + FeCl_3 \rightarrow [Fe(C_6H_5O)_6]^{3-} + 6H^+ + 3HCl$ (forming a violet complex).

(D) The reaction sequence is as follows:
$1$. Benzene reacts with oleum $(H_2SO_4 + SO_3)$ to form benzenesulfonic acid.
$2$. Benzenesulfonic acid reacts with molten $NaOH$ followed by $H_3O^+$ to form phenol $(X)$.
$3$. Phenol $(X)$ reacts with $NaOH$ to form sodium phenoxide.
$4$. Sodium phenoxide reacts with $CO_2$ under pressure followed by $H_3O^+$ (Kolbe-Schmitt reaction) to form salicylic acid $(Y)$,which is $2$-hydroxybenzoic acid.
Therefore,$X = \text{phenol}$ and $Y = \text{2-hydroxybenzoic acid}$.

(A) The reaction of phenol with $NaOH$ followed by heating with $CO_{2}$ under high pressure and subsequent acidification is the Kolbe-Schmitt reaction,which yields salicylic acid $(X)$ as the major product.
Salicylic acid $(X)$ can be purified by steam distillation due to intramolecular hydrogen bonding.
When salicylic acid $(X)$ is reacted with acetic anhydride in the presence of a trace amount of conc. $H_{2}SO_{4}$,the phenolic $-OH$ group undergoes acetylation to produce acetylsalicylic acid (aspirin) $(Y)$ as the major product.
The structure of acetylsalicylic acid $(Y)$ corresponds to option $A$.

(B) The reactant in the given reaction is salicylic acid.
When salicylic acid reacts with acetic anhydride $(CH_3CO)_2O$ in the presence of acetic acid $(CH_3COOH)$,the phenolic $-OH$ group undergoes acetylation.
This reaction produces acetylsalicylic acid,which is commonly known as aspirin.
The structure of aspirin corresponds to option $B$.

(C) The correct answer is $(C)$.
Salicylic acid $(C_6H_4(OH)COOH)$ reacts with excess bromine water to produce $2,4,6$-tribromophenol.
In this reaction,the $-COOH$ group is replaced by a bromine atom due to the high reactivity of the phenol ring,followed by electrophilic substitution at the remaining ortho and para positions.
The reaction is as follows:
$C_6H_4(OH)COOH + 2Br_2 \rightarrow C_6H_2(OH)Br_3 + CO_2 + HBr$.

(A) When phenol is treated with dilute $HNO_3$ at low temperature,it undergoes electrophilic aromatic substitution to form a mixture of ortho-nitrophenol and para-nitrophenol.
$o$-Nitrophenol $(P)$ exhibits intramolecular hydrogen bonding,which makes it steam volatile.
$p$-Nitrophenol $(Q)$ exhibits intermolecular hydrogen bonding,leading to higher boiling point and it is not steam volatile.
Thus,$P$ is $o$-nitrophenol and $Q$ is $p$-nitrophenol.

(C)
This reaction is known as the Reimer-Tiemann reaction.
In this reaction,phenol reacts with chloroform $(CHCl_3)$ in the presence of aqueous sodium hydroxide $(NaOH)$ to form salicylaldehyde (o-hydroxybenzaldehyde).
The mechanism involves the formation of dichlorocarbene $(:CCl_2)$ as an electrophilic intermediate,which attacks the phenoxide ion.

(C) The boiling point of a compound depends upon the extent of intermolecular forces,specifically hydrogen bonding.
$(I)$ Phenol has intermolecular hydrogen bonding but lacks the strong electron-withdrawing group present in the others.
$(II)$ $o$-Nitrophenol exhibits intramolecular hydrogen bonding,which reduces the extent of intermolecular association,leading to a lower boiling point compared to the para isomer.
$(III)$ $p$-Nitrophenol exhibits strong intermolecular hydrogen bonding,which leads to molecular association and results in the highest boiling point.
Therefore,the correct order of boiling points is $I < II < III$.

(B) The reaction of phenol with $HNO_2$ (nitrous acid) leads to the formation of $p$-nitrosophenol,which exists in tautomeric equilibrium with $p$-benzoquinone monoxime $(X)$.
In the presence of $H_2SO_4$,the oxime group acts as an electrophile. The phenol molecule attacks the electrophilic carbon of the oxime group at the para-position to form the final product $Y$,which is an indophenol derivative. Looking at the structures provided in the options,option $II$ correctly represents the structure of $p$-benzoquinone monoxime $(X)$ and the corresponding indophenol product $(Y)$.

(C) The reaction sequence is as follows:
$1$. Phenol reacts with $Br_2$ in $CS_2$ (a non-polar solvent) at low temperature. This electrophilic aromatic substitution reaction favors the formation of the para-isomer due to steric hindrance at the ortho-position. Thus,$X$ is $4$-bromophenol.
$2$. In the second step,$4$-bromophenol reacts with $NaOH$ to form sodium $4$-bromophenoxide,which then undergoes a nucleophilic substitution reaction $(S_N2)$ with methyl iodide $(MeI)$ to form $1$-bromo-$4$-methoxybenzene $(Y)$. This is an example of Williamson's ether synthesis.
Therefore,the correct option is $C$.