Properties of Phenols Questions in English

(C) The correct acidity order is $III > IV > I > II$.
In substituted phenols,electron-withdrawing groups $(EWG)$ like the nitro $(-NO_2)$ group increase acidity by stabilizing the phenoxide ion through delocalization of the negative charge.
$p$-Nitrophenol $(III)$ is more acidic than $m$-nitrophenol $(IV)$ because the $-NO_2$ group at the para position exerts both $-I$ and $-R$ effects,whereas at the meta position,it only exerts the $-I$ effect.
Phenol $(I)$ is less acidic than nitrophenols but more acidic than $p$-methoxyphenol $(II)$.
$p$-Methoxyphenol $(II)$ contains a methoxy $(-OCH_3)$ group,which is an electron-donating group $(EDG)$ due to the $+R$ effect,destabilizing the phenoxide ion and decreasing the acidity.

(D) The reaction of phenol with $CO_2$ in the presence of $NaOH$ followed by acidification is the Kolbe-Schmitt reaction,which yields salicylic acid as compound $X$.
Treatment of salicylic acid $(X)$ with acetic anhydride in the presence of $H_2SO_4$ (acetylation) yields acetylsalicylic acid,commonly known as Aspirin,as compound $Y$.
Aspirin is a non-narcotic analgesic. It exhibits the following properties:
$(i)$ Antipyretic (reduces fever)
$(ii)$ Anti-inflammatory (reduces inflammation)
$(iv)$ Antiplatelet (prevents blood clotting)
It is not a narcotic analgesic.
Therefore,the correct properties are $(i), (ii)$ and $(iv)$.

(A) The acidity of phenols is determined by the stability of the phenoxide ion formed after the loss of a proton.
Electron-withdrawing groups ($-I$ and $-M$ effects) stabilize the phenoxide ion and increase acidity,thereby decreasing $pK_{a}$.
Electron-donating groups ($+I$ and $+M$ effects) destabilize the phenoxide ion and decrease acidity,thereby increasing $pK_{a}$.
Comparing the substituents:
$1$. $2, 4-$Dinitrophenol: Two strong $-M$ and $-I$ groups (most acidic).
$2$. $4-$Nitrophenol: One strong $-M$ and $-I$ group.
$5$. $3-$Chlorophenol: One $-I$ group (weakly acidic).
$4$. Phenol: No substituent.
$3$. $2, 4, 5-$Trimethylphenol: Three $+I$ groups (least acidic).
Decreasing order of acidity: $(1) > (2) > (5) > (4) > (3)$.
Since $pK_{a} = -\log(K_{a})$,the increasing order of $pK_{a}$ is the reverse of the acidity order: $(3) < (4) < (5) < (2) < (1)$.

(B) $1$. Compound '$A$' is trisubstituted and gives a positive neutral $FeCl_3$ test,indicating it is a phenol derivative.
$2$. The formula $C_{10}H_{12}O_2$ with a benzene ring ($6$ carbons) leaves $4$ carbons for substituents. The degree of unsaturation is $10 - 12/2 + 1 = 5$. Since the benzene ring accounts for $4$ degrees of unsaturation ($3$ $\pi$ bonds + $1$ ring),there is $1$ additional $\pi$ bond or ring present in the side chains.
$3$. The compound decolorises alkaline $KMnO_4$,confirming the presence of an alkene or an oxidizable side chain.
$4$. The reaction with $HI$ gives methyl iodide,indicating the presence of a methoxy group $(-OCH_3)$ or a similar ether linkage. However,the initial test with $NaOH$ and $CH_3Br$ suggests the presence of a phenolic $-OH$ group which gets methylated.
$5$. Given the structure is trisubstituted on a benzene ring,the total $\pi$ bonds are $3$ (from the benzene ring) + $1$ (from the alkene side chain) = $4$ $\pi$ bonds.

(B) 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,while electron-donating groups $(-EDG)$ decrease acidity.
$1$. Compound $(c)$ has a $-NO_{2}$ group,which exerts a strong $-M$ and $-I$ effect,making it the most acidic.
$2$. Compound $(a)$ is phenol,which has no substituent.
$3$. Compound $(d)$ has an isopropyl group,which exerts a $+I$ effect and hyperconjugation,making it less acidic than phenol.
$4$. Compound $(b)$ has a $-N(CH_{3})_{2}$ group,which exerts a strong $+M$ effect,making it the least acidic.
Therefore,the order of acidic strength is $(c) > (a) > (d) > (b)$.
Since $pK_{a} \propto \frac{1}{\text{acidic strength}}$,the order of $pK_{a}$ values is $(b) > (d) > (a) > (c)$.

(A) The empirical formula $C_6H_6O$ corresponds to phenol $(C_6H_5OH)$.
Phenol is an aromatic compound,which burns with a sooty flame.
When phenol reacts with bromine $(Br_2)$ in a low polarity solvent like $CS_2$ or $CHCl_3$ at low temperature,the electrophilic substitution occurs primarily at the para-position due to steric hindrance at the ortho-position.
Thus,the major product $B$ is $4$-bromophenol.

(C) The reactions of phenol are as follows:
$a$. Phenol reacts with $(i)$ $NaOH$,$(ii)$ $CO_2$,$(iii)$ $H^{+}$ to form Salicylic acid (Kolbe's reaction). Thus,$a-iv$.
$b$. Phenol reacts with $(i)$ Aqueous $NaOH + CHCl_3$,$(ii)$ $H^{+}$ to form Salicylaldehyde (Reimer-Tiemann reaction). Thus,$b-iii$.
$c$. Phenol reacts with $Zn$ dust upon heating to form Benzene. Thus,$c-ii$.
$d$. Phenol reacts with $Na_2Cr_2O_7$ and $H_2SO_4$ (chromic acid) to form Benzoquinone. Thus,$d-i$.
Therefore,the correct matching is $a-iv, b-iii, c-ii, d-i$.

(A) The acidity of substituted phenols depends on the electronic effects of the substituents. Electron-withdrawing groups $(EWG)$ increase acidity by stabilizing the phenoxide ion,while electron-donating groups $(EDG)$ decrease acidity.
$1$. $-NO_2$ is a strong electron-withdrawing group ($-I$ and $-M$ effect).
$2$. $-Cl$ is an electron-withdrawing group ($-I$ effect).
$3$. $-CH_3$ is an electron-donating group ($+I$ and hyperconjugation).
Comparing the substituents,the $-NO_2$ group has the strongest electron-withdrawing effect,making $3-$nitrophenol the most acidic among the given options.

(A) The reaction of phenol (also known as carbolic acid) with phthalic anhydride in the presence of concentrated $H_2SO_4$ is a condensation reaction.
Two molecules of phenol react with one molecule of phthalic anhydride to form phenolphthalein,which is a well-known acid-base indicator.
Therefore,'$X$' is phenol (carbolic acid) and '$Y$' is phenolphthalein.

(A) The acidity of hydroxyl compounds depends on the stability of the conjugate base formed after the loss of a proton $(H^+)$.
$1$. $E$ ($p$-nitrophenol) is the most acidic because the $-NO_2$ group is a strong electron-withdrawing group ($-I$ and $-M$ effect),which stabilizes the phenoxide ion significantly.
$2$. $C$ (phenol) is more acidic than aliphatic alcohols due to resonance stabilization of the phenoxide ion.
$3$. $D$ ($p$-methoxyphenol) is less acidic than phenol because the $-OCH_3$ group exerts a $+M$ effect,which destabilizes the phenoxide ion.
$4$. Between $A$ $(CH_3OH)$ and $B$ $((CH_3)_3COH)$,$A$ is more acidic because the three methyl groups in $B$ exert a strong $+I$ effect,which destabilizes the alkoxide ion.
Thus,the correct order of acidity is $E > C > D > A > B$.

(B) The reaction sequence is as follows:
$1$. The starting material is $2$-isopropyl-$5$-methylphenol (thymol).
$2$. Treatment with $NaNO_2/HCl$ performs electrophilic aromatic substitution (nitrosation) at the para-position relative to the $-OH$ group,forming $A$ (a nitroso compound).
$3$. Treatment of the nitroso compound with $NH_4SH$ (ammonium hydrosulfide) reduces the $-NO$ group to an $-NH_2$ group,yielding $B$ as $4$-amino-$2$-isopropyl-$5$-methylphenol.

(A) 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 negative charge,while electron-donating groups $(-EDG)$ decrease acidity by destabilizing it.
$1$. $E$ ($2,4$-dinitrophenol): Has two $-NO_2$ groups (strong $-M$ and $-I$ effect),making it the most acidic.
$2$. $B$ ($p$-nitrophenol): Has one $-NO_2$ group ($-M$ and $-I$ effect),making it more acidic than phenol.
$3$. $D$ (Phenol): The reference compound.
$4$. $C$ ($p$-methoxyphenol): The $-OCH_3$ group has a $+M$ effect,which destabilizes the phenoxide ion,making it less acidic than phenol.
$5$. $A$ $(Bu-OH)$: Aliphatic alcohols are significantly less acidic than phenols because the alkoxide ion is not resonance-stabilized.
Thus,the order of acidity is: $A < C < D < B < E$.

(A) The Phthalein dye test is used for the identification of phenols. When a phenol is heated with phthalic anhydride in the presence of concentrated $H_2SO_4$,a phthalein dye is formed.
$1$. Carbylamine test: Used for the identification of primary amines.
$2$. Lucas test: Used for the differentiation between $1^{\circ}$,$2^{\circ}$,and $3^{\circ}$ alcohols.
$3$. Tollen's test: Used for the identification of aldehydes.

(C) Assertion $(A)$ is true. The $C-O$ bond in phenol has partial double bond character due to resonance,making it very strong and difficult to break by nucleophilic substitution with halogen acids.
Reason $(R)$ is false. Phenols do not react with halogen acids under normal conditions because the $C-O$ bond is strong and the formation of a phenyl carbocation is highly unfavorable.
Therefore,$A$ is true but $R$ is false.

(A) The $pK_a$ values for the given compounds are as follows:
$A$. Ethanol: $15.9$
$B$. Phenol: $10.0$
$C$. $m$-Nitrophenol: $8.3$
$D$. $p$-Nitrophenol: $7.1$
Therefore,the correct matching is $A-II, B-I, C-IV, D-III$.

(D) The reaction described is the Reimer-Tiemann reaction.
In this reaction,phenol reacts with chloroform $(CHCl_3)$ in the presence of an aqueous base like sodium hydroxide $(NaOH)$ to form an intermediate,which upon acidic hydrolysis yields $2$-hydroxybenzaldehyde (also known as salicylaldehyde) as the major product.
The chemical equation is:
$C_6H_5OH + CHCl_3 + 3NaOH \rightarrow C_6H_4(OH)(CHO) + 3NaCl + 2H_2O$

(A) The reaction of phenol with $CHCl_3$ (chloroform) 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,resulting in the formation of salicylaldehyde ($2$-hydroxybenzaldehyde).

(A) The $pK_a$ value is inversely proportional to the acid strength. $A$ lower $pK_a$ value indicates a stronger acid.
Phenol has a $pK_a$ of $10.0$,whereas ethanol has a $pK_a$ of $15.9$. Thus,phenol is a stronger acid than ethanol.
This is because the phenoxide ion (conjugate base of phenol) is stabilized by resonance,whereas the ethoxide ion (conjugate base of ethanol) is destabilized by the $+I$ effect of the ethyl group.
Therefore,$Assertion$ $A$ is true,but $Reason$ $R$ is false.

(A) The reactant is $4-hydroxybenzaldehyde$,which contains both an aldehyde group $(-CHO)$ and a phenolic hydroxyl group $(-OH)$.
When $PhMgBr$ (a Grignard reagent) is added,it acts as a strong base and reacts with the acidic phenolic $-OH$ group first.
$4-hydroxybenzaldehyde + PhMgBr \rightarrow 4-formylphenoxide magnesium bromide + Benzene$
The $Ph^-$ from $PhMgBr$ abstracts the acidic proton from the phenolic $-OH$ group to form benzene $(C_6H_6)$ and the magnesium salt of the phenol.
Upon workup with $aq. NH_4Cl$,the magnesium salt is protonated back to the original $4-hydroxybenzaldehyde$.
Therefore,the final product $P$ is $4-hydroxybenzaldehyde$,which contains $1$ hydroxyl group.

(D) The reactivity of a benzene ring towards an electrophilic substitution reaction depends on the electron density of the ring.
Groups that donate electrons to the ring (via resonance or inductive effect) increase the electron density,making it more susceptible to electrophilic attack.
- In $C_6H_6$ (Benzene),there is no substituent.
- In $C_6H_5CH_3$ (Toluene),the $-CH_3$ group is electron-donating via the inductive effect and hyperconjugation.
- In $C_6H_5Cl$ (Chlorobenzene),the $-Cl$ group is electron-withdrawing via the inductive effect,although it donates electrons via resonance.
- In $C_6H_5OH$ (Phenol),the $-OH$ group is strongly electron-donating via resonance (+$M$ effect),which significantly increases the electron density of the benzene ring.
Therefore,phenol is the most reactive towards electrophilic attack.

(C) The reactions are as follows:
$A$. Phenol reacts with $Zn$ dust upon heating to form Benzene $(III)$.
$B$. Phenol reacts with $CHCl_3$ and $NaOH$ followed by acidification $(HCl)$ to form Salicylaldehyde $(I)$ (Reimer-Tiemann reaction).
$C$. Phenol reacts with $CO_2$ and $NaOH$ followed by acidification $(HCl)$ to form Salicylic acid $(II)$ (Kolbe-Schmidt reaction).
$D$. Phenol reacts with concentrated $HNO_3$ to form $2,4,6$-trinitrophenol,commonly known as Picric acid $(IV)$.
Therefore,the correct matching is $A-III, B-I, C-II, D-IV$.

(C) The reaction sequence is as follows:
$1$. Chlorobenzene reacts with $NaOH$ at $623 \ K$ and $300 \ atm$ pressure to form sodium phenoxide $(X)$.
$2$. Sodium phenoxide $(X)$ on acidification with $HCl$ gives phenol $(Y)$.
$3$. Phenol $(Y)$ on treatment with concentrated $HNO_3$ undergoes electrophilic aromatic substitution (nitration) to form $2,4,6$-trinitrophenol,commonly known as picric acid $(Z)$.

(D) Phenol is a highly activated aromatic ring due to the strong electron-donating effect of the $-OH$ group.
Because of this high activation,phenol undergoes electrophilic substitution (bromination) even in non-polar solvents like $CHCl_3$ or $CS_2$ without the need for a Lewis acid catalyst.
Therefore,Statement $I$ is false.
Statement $II$ correctly describes the general role of a Lewis acid in halogenation reactions (polarizing $Br_2$ to generate $Br^+$),but since Statement $I$ is false,the correct choice is that Statement $I$ is false but Statement $II$ is true.

(D) The reaction of phenol with concentrated $H_2SO_4$ at $373 \ K$ yields phenol-$2,4$-disulphonic acid (Product $A$).
Chemical formula of Product $A$ $(C_6H_6O_7S_2)$: It contains $7$ oxygen atoms.
Reaction of phenol-$2,4$-disulphonic acid with concentrated $HNO_3$ yields $2,4,6$-trinitrophenol (Picric acid) (Product $B$).
Chemical formula of Product $B$ $(C_6H_3N_3O_7)$: It contains $7$ oxygen atoms.
Total sum of oxygen atoms = $7$ (in $A$) + $7$ (in $B$) = $14$.

(A) Statement $I$ is incorrect because picric acid is $2,4,6$-trinitrophenol,not $2,4,6$-trinitrotoluene.
Statement $II$ is correct. Phenol is first treated with conc. $H_2SO_4$ to form phenol-$2,4$-disulphonic acid,which is then treated with conc. $HNO_3$ to yield picric acid ($2,4,6$-trinitrophenol) via electrophilic substitution.

(D) The Reimer-Tiemann reaction involves the treatment of phenol with chloroform $(CHCl_3)$ in the presence of an aqueous base (like $NaOH$).
$1$. The base deprotonates the phenol to form a phenoxide ion.
$2$. The chloroform reacts with the base to generate a dichlorocarbene $(:CCl_2)$ intermediate,which acts as an electrophile.
$3$. The phenoxide ion attacks the dichlorocarbene to form an ortho-dichloromethyl phenoxide intermediate,which is represented as a benzene ring with an $-O^-Na^+$ group and a $-CHCl_2$ group at the ortho position.
$4$. This intermediate then undergoes hydrolysis with $NaOH$ followed by acidification to yield salicylaldehyde.

(D) $1$. Phenetole $(C_6H_5OC_2H_5)$ undergoes nitration with $HNO_3$ and $H_2SO_4$ to form $p$-nitrophenetole as the major product $(P)$.
$2$. $p$-Nitrophenetole then undergoes bromination with $Br_2$ in the presence of $Fe$ to form $2,6$-dibromo-$4$-nitrophenetole as the major product $(Q)$.
$3$. The molecular formula of $Q$ is $C_8H_7Br_2NO_3$.
$4$. In $Q$,the number of oxygen atoms is $3$ and the number of bromine atoms is $2$.
$5$. The ratio of oxygen atoms to bromine atoms is $\frac{3}{2} = 1.5$.
$6$. Expressing $1.5$ as $.... \times 10^{-1}$,we get $15 \times 10^{-1}$.
$7$. Therefore,the correct option is $D$.

(D) Phenol $(C_6H_5OH)$ is more acidic than water and aliphatic alcohols. Therefore,it reacts with a strong base like $NaOH$ to form sodium phenoxide and water.
$C_6H_5OH + NaOH \rightarrow C_6H_5O^-Na^+ + H_2O$
Aliphatic alcohols ($C_6H_5CH_2OH$,$C_2H_5OH$,$(CH_3)_3COH$) are less acidic than water and do not react with dilute $NaOH$.

(A) The acidic strength of phenols is increased by electron-withdrawing groups $(-NO_2)$ and decreased by electron-donating groups $(-OCH_3)$.
$pK_{a}$ is inversely proportional to acidic strength.
$(A)$ Phenol
$(B)$ $p$-Nitrophenol (Strongest acid due to $-M$ and $-I$ effect of $-NO_2$)
$(C)$ $m$-Methoxyphenol ($-I$ effect of $-OCH_3$ increases acidity compared to phenol)
$(D)$ $o$-Nitrophenol (Strong acid due to $-I$ and $-M$ effect,but slightly weaker than $p$-nitrophenol due to intramolecular $H$-bonding)
$(E)$ $p$-Methoxyphenol (Strongest electron-donating effect $+M$ of $-OCH_3$ decreases acidity the most)
Acidic strength order: $(B) > (D) > (A) > (C) > (E)$
Therefore,the increasing order of $pK_{a}$ is: $(E) < (C) < (A) < (D) < (B)$.

(D) Benzene sulfonic acid $(C_6H_5SO_3H)$ is a strong acid,stronger than carbonic acid $(H_2CO_3)$. Therefore,it reacts with $NaHCO_3$ to release $CO_2$ gas.
$p$-Nitrophenol is more acidic than phenol due to the electron-withdrawing effect of the $-NO_2$ group. Its $pK_a$ value is approximately $7.15$,which is lower than that of carbonic acid ($pK_a \approx 6.35$ for $H_2CO_3$ in water). However,in many contexts,$p$-nitrophenol is considered acidic enough to react with $NaHCO_3$ to evolve $CO_2$ gas,as shown in the provided reaction scheme.
Both compounds react with $NaHCO_3$ to release $CO_2$ gas.

(B) The boiling point of phenolic compounds depends on intermolecular hydrogen bonding.
$IV$ (Hydroxybenzene or phenol) has only one $-OH$ group,so it has the lowest boiling point.
Among the dihydroxybenzenes $(I, II, III)$,the boiling point increases with the extent of intermolecular hydrogen bonding.
$I$ ($1,2$-dihydroxybenzene or catechol) exhibits strong intramolecular hydrogen bonding,which reduces its ability to form intermolecular hydrogen bonds,leading to a lower boiling point compared to $II$ and $III$.
$II$ ($1,3$-dihydroxybenzene or resorcinol) has a higher boiling point than $I$ due to more effective intermolecular hydrogen bonding.
$III$ ($1,4$-dihydroxybenzene or hydroquinone) has the highest boiling point because its symmetrical structure allows for the most efficient intermolecular hydrogen bonding in the solid state.
Therefore,the increasing order is $IV < I < II < III$.

(C) $1.$ The Reimer-Tiemann reaction uses chloroform $(CHCl_3)$ and an aqueous base $(NaOH)$ to generate the electrophile. Thus,the correct reagent is $aq. NaOH + CHCl_3$ (Option $C$).
$2.$ The active electrophile generated in the Reimer-Tiemann reaction is dichlorocarbene $(:CCl_2)$,which is formed by the dehydrohalogenation of chloroform (Option $C$).
$3.$ The intermediate $I$ formed by the attack of the dichlorocarbene electrophile on the phenoxide ion is the dichloromethyl-substituted phenoxide,which corresponds to structure $B$ in the provided image.

(B) $1$. Cumene is oxidized to phenol $(P)$ via cumene hydroperoxide.
$2$. Reimer-Tiemann reaction of phenol $(P)$ with $CHCl_3/NaOH$ gives salicylaldehyde ($Q$,major) and $p$-hydroxybenzaldehyde ($R$,minor).
$3$. $Q$ (salicylaldehyde) has intramolecular hydrogen bonding,making it steam volatile.
$4$. $Q$ contains a phenolic $-OH$ group,so it gives a violet coloration with $FeCl_3$.
$5$. $S$ is formed by the reaction of $Q$ with $PhCH_2Br$ in the presence of $NaOH$ (Williamson ether synthesis),resulting in $2-(benzyloxy)benzaldehyde$.
$6$. $S$ contains an aldehydic group,so it gives a yellow precipitate with $2, 4-DNP$.
$7$. $S$ does not have a free phenolic $-OH$ group,so it does not give a violet coloration with $FeCl_3$.
Therefore,statements $(B)$ and $(C)$ are correct.

(A) The reaction of phenol with $NaOH(aq)$ forms the phenoxide ion,which is highly activated towards electrophilic aromatic substitution.
$Br_2$ acts as an electrophile.
The phenoxide ion undergoes electrophilic attack by $Br_2$ at the ortho and para positions.
Intermediate $C$ represents the $p$-bromophenoxide ion.
Intermediate $A$ represents the $o$-bromophenoxide ion.
Intermediate $B$ represents a dienone intermediate formed during the bromination process.
Thus,the intermediates involved in the reaction pathway are $A$,$B$,and $C$.

(B) Compounds that are acidic enough to react with aqueous $NaOH$ (a base) to form water-soluble salts are soluble in aqueous $NaOH$.
$1$. Cyclohexanecarboxylic acid (contains $-COOH$ group,acidic).
$2$. Phenol (contains phenolic $-OH$ group,acidic).
$3$. $4$-(Dimethylamino)phenol (contains phenolic $-OH$ group,acidic).
$4$. $1$-Naphthoic acid (contains $-COOH$ group,acidic).
Other compounds like $N,N$-dimethylaniline,$2$-ethoxybenzyl alcohol,nitrobenzene,and $1,2$-diethylbenzene are not sufficiently acidic to react with $NaOH$.
Therefore,the total number of compounds soluble in aqueous $NaOH$ is $4$.

(A) The molecular formula $C_8H_{10}O_2$ and the positive $FeCl_3$ test indicate that the compound is a phenol derivative.
Since the compound rotates plane-polarized light,it must be optically active,meaning it contains a chiral center.
The structure consists of a benzene ring with an $-OH$ group and a side chain of $-CH(OH)CH_3$.
There are three possible positions for the side chain relative to the $-OH$ group on the benzene ring: ortho,meta,and para.
For each of these three positions,the side chain $-CH(OH)CH_3$ contains a chiral carbon atom,which results in two enantiomers ($R$ and $S$ configurations).
Therefore,the total number of optically active isomers is $3 \times 2 = 6$.

(D) Sodium bicarbonate $(NaHCO_3)$ reacts with acids that are stronger than carbonic acid $(H_2CO_3)$ to liberate $CO_2$ gas.
Benzoic acid,benzenesulphonic acid,and salicylic acid are stronger acids than $H_2CO_3$ and thus liberate $CO_2$.
Phenol (carbolic acid) is a weaker acid than $H_2CO_3$ and therefore does not react with $NaHCO_3$ to liberate $CO_2$.

(A) The reaction is the Reimer-Tiemann reaction. $p$-cresol reacts with $CHCl_3$ and $OH^-$ to form an intermediate dichloromethyl cyclohexadienone derivative.
$1$. The phenoxide ion attacks the dichlorocarbene $(:CCl_2)$ generated from $CHCl_3$.
$2$. This leads to the formation of a cyclohexadienone intermediate ($Q$ and $R$ are related to these intermediates).
$3$. Hydrolysis and subsequent rearrangement lead to the formation of the major product $S$ ($2$-hydroxy$-5-$methylbenzaldehyde).
$4$. Some side products like $Q$ ($4$-methyl$-4-$dichloromethyl-cyclohexa$-2,5-$dienone) are also formed as minor products during the reaction process.
Thus,the major product is $S$ and the minor product is $Q$.

(A) The $-OH$ group is strongly activating and $o, p$-directing due to its powerful $+M$ effect.
$1$. The tert-butyl group $(-C(CH_3)_3)$ is a bulky group,which creates significant steric hindrance at the ortho position $(B)$ relative to it.
$2$. The halogen electrophile $(X^{\oplus})$ size increases in the order $Cl^{\oplus} < Br^{\oplus} < I^{\oplus}$.
$3$. With $I_2$,the electrophile is very large,so it only substitutes at the least hindered position $(A)$.
$4$. With $Br_2$,the electrophile is smaller,allowing substitution at $A$ and $C$.
$5$. With $Cl_2$,the electrophile is the smallest,allowing substitution at all three positions $(A, B, C)$.
Thus,the pattern is explained by the steric effect of the halogen $(A)$,the steric effect of the tert-butyl group $(B)$,and the electronic effect of the phenolic group $(C)$ which activates the ring. The correct combination is $A, B, C$.

(C) Step $1$: Cyclotrimerization of $15$ moles of acetylene $(HC \equiv CH)$ gives benzene $(A)$. Since $3$ moles of acetylene form $1$ mole of benzene,$15$ moles of acetylene would theoretically form $5$ moles of benzene. Given an $80\%$ yield,the actual amount of benzene $(A)$ formed is $5 \times 0.8 = 4$ moles.
Step $2$: Friedel-Crafts alkylation of $4$ moles of benzene with isopropyl chloride gives cumene $(B)$. Given a $50\%$ yield,the amount of cumene $(B)$ formed is $4 \times 0.5 = 2$ moles.
Step $3$: Cumene hydroperoxide rearrangement of $2$ moles of cumene gives phenol $(C)$. Given a $50\%$ yield,the amount of phenol $(C)$ formed is $2 \times 0.5 = 1$ mole.
Step $4$: Acetylation of $1$ mole of phenol with acetyl chloride $(CH_3COCl)$ in the presence of pyridine gives phenyl acetate $(D)$. Given a $100\%$ yield,the amount of phenyl acetate $(D)$ formed is $1$ mole.
The molar mass of phenyl acetate $(C_8H_8O_2)$ is $(8 \times 12) + (8 \times 1) + (2 \times 16) = 96 + 8 + 32 = 136 \ g/mol$.
Therefore,the mass of $1$ mole of $D$ is $136 \ g$.

(A) The reaction of phenol with bromine water is: $C_6H_5OH + 3Br_2 \rightarrow C_6H_2Br_3OH + 3HBr$
Molar mass of phenol $(C_6H_5OH) = (6 \times 12) + (6 \times 1) + 16 = 72 + 6 + 16 = 94 \ g \ mol^{-1}$.
Moles of phenol = $\frac{2 \ g}{94 \ g \ mol^{-1}} = 0.02128 \ mol$.
According to the stoichiometry,$1 \ mol$ of phenol reacts with $3 \ mol$ of $Br_2$.
Moles of $Br_2$ required = $3 \times 0.02128 = 0.06384 \ mol$.
Molar mass of $Br_2 = 2 \times 80 = 160 \ g \ mol^{-1}$.
Mass of $Br_2$ required = $0.06384 \ mol \times 160 \ g \ mol^{-1} = 10.2144 \ g \approx 10.22 \ g$.

(C) The reaction of chlorobenzene with $NaOH$ at $623 \ K$ and $300 \ atm$ followed by acidification with $H^+$ gives phenol as product $[A]$.
Phenol on oxidation with chromic acid $(Na_2Cr_2O_7 / H_2SO_4)$ yields $p$-benzoquinone as product $[B]$.

(D) The reaction with aqueous $NaHCO_3$ to release $CO_2$ gas occurs only if the compound is more acidic than carbonic acid ($H_2CO_3$,$pK_a \approx 6.35$).
$1$. $2,4,6$-Trinitrophenol (Picric acid) is a very strong acid $(pK_a \approx 0.38)$ and readily reacts with $NaHCO_3$.
$2$. $4$-Nitrobenzoic acid is a strong organic acid $(pK_a \approx 3.4)$ and reacts with $NaHCO_3$.
$3$. $PhNH_3^+ Cl^-$ (Anilinium chloride) is the salt of a weak base and a strong acid. In aqueous solution,it exists as $PhNH_3^+$ and $Cl^-$. $PhNH_3^+$ is a weak acid $(pK_a \approx 4.6)$,which is stronger than $H_2CO_3$,so it can react with $NaHCO_3$ to release $CO_2$.
$4$. $3$-Nitrophenol is a weak acid $(pK_a \approx 8.39)$,which is less acidic than $H_2CO_3$ $(pK_a \approx 6.35)$. Therefore,it does not react with $NaHCO_3$ to give effervescence of $CO_2$.

(A) The reaction of phenol with $CO_2$ in the presence of $NaOH$ followed by acidification is the Kolbe-Schmitt reaction,which yields salicylic acid ($2$-hydroxybenzoic acid) as product $(A)$.
Salicylic acid then reacts with acetic anhydride $(CH_3CO)_2O$ in the presence of an acid catalyst to undergo acetylation of the phenolic $-OH$ group,forming acetylsalicylic acid (aspirin) as product $(B)$.

(B) $1$. Phenol $(C_6H_5OH)$ reacts with bromine water $(Br_2/H_2O)$ to form $2,4,6$-tribromophenol,which is a white precipitate.
$2$. Phenol reacts with zinc dust $(Zn)$ upon heating to undergo reduction,forming benzene $(C_6H_6)$ and zinc oxide $(ZnO)$.
$3$. The reaction is: $C_6H_5OH + Zn \rightarrow C_6H_6 + ZnO$.
$4$. Therefore,the compound $x$ is phenol.

(A) The acidic strength of phenols depends on the nature of substituents attached to the benzene ring. Electron-withdrawing groups $(EWG)$ increase acidity by stabilizing the phenoxide ion,while electron-donating groups $(EDG)$ decrease acidity by destabilizing it.
$(1)$ $2,4,6$-trichlorophenol: Contains three $-Cl$ groups,which are electron-withdrawing by the inductive effect $(-I)$.
$(2)$ $2,4,6$-trimethylphenol: Contains three $-CH_3$ groups,which are electron-donating by the inductive effect $(+I)$ and hyperconjugation.
$(3)$ $2,4,6$-trinitrophenol (picric acid): Contains three $-NO_2$ groups,which are strong electron-withdrawing groups by both $-I$ and $-M$ (mesomeric) effects.
$(4)$ Phenol: Has no substituent.
Comparing the effects:
- $-CH_3$ $(EDG)$ makes the compound less acidic than phenol.
- $-Cl$ $(EWG)$ makes the compound more acidic than phenol.
- $-NO_2$ (strong $EWG$) makes the compound significantly more acidic than phenol.
Order of increasing acidity: $(2) < (4) < (1) < (3)$.
Thus,the correct option is $A$.

(A) $o$-Nitrophenol $(B)$ exhibits intramolecular hydrogen bonding,which restricts intermolecular hydrogen bonding between its molecules. This makes it more volatile.
$p$-Nitrophenol $(A)$ exhibits intermolecular hydrogen bonding,which leads to the association of molecules,resulting in a higher boiling point and lower volatility.
Since vapour pressure is inversely related to boiling point,$o$-nitrophenol $(B)$ will have a higher vapour pressure than $p$-nitrophenol $(A)$.

(D) The solubility of organic compounds in water depends on their ability to form hydrogen bonds with water molecules.
$p$-Nitrophenol has a $-NO_2$ group which is a strong electron-withdrawing group.
This group increases the acidity of the phenolic $-OH$ group,making it more capable of forming strong intermolecular hydrogen bonds with water.
Additionally,$o$-Nitrophenol exhibits intramolecular hydrogen bonding,which reduces its ability to form hydrogen bonds with water,thereby decreasing its solubility.
$p$-Cresol contains a hydrophobic methyl group,which decreases its solubility compared to phenol.
Therefore,$p$-Nitrophenol is the most soluble among the given options.

(D) Alcohols are generally neutral or very weakly acidic,whereas phenols are more acidic than alcohols.
Electron-withdrawing groups,such as the nitro group $(-NO_2)$,stabilize the phenoxide ion through $-I$ and $-M$ effects,thereby increasing the acidity of the phenol.
Since $p-$nitrophenol contains a strong electron-withdrawing $-NO_2$ group at the para position,it exhibits the highest acidic strength among the given compounds.