Properties of Phenols Questions in English

(D) Methoxybenzene (Structure $I$) is an ether and cannot form intermolecular hydrogen bonds,so it has the lowest boiling point.
Both $m$-cresol (Structure $II$) and $o$-cresol (Structure $III$) are phenols and can form intermolecular hydrogen bonds,leading to higher boiling points than $I$.
Between $m$-cresol and $o$-cresol,$o$-cresol exhibits intramolecular hydrogen bonding,which reduces the extent of intermolecular hydrogen bonding compared to $m$-cresol.
Therefore,$m$-cresol has a higher boiling point than $o$-cresol.
The correct order of boiling points is $II > III > I$.

(C) The reaction sequence proceeds as follows:
$1$. Aniline reacts with $NaNO_2/HCl$ at $273-278 \ K$ to form benzene diazonium chloride.
$2$. Hydrolysis with warm $H_2O$ converts the diazonium salt into phenol.
$3$. Phenol reacts with excess $Br_2$ (bromine water) to undergo electrophilic aromatic substitution,yielding $2,4,6$-tribromophenol.
$4$. Treatment with $NaOH$ converts the phenol into sodium phenoxide ($2,4,6$-tribromophenoxide).
$5$. Finally,reaction with $CH_3I$ (Williamson ether synthesis) yields $2,4,6$-tribromoanisole as the final product.

(C) The reaction of phenol with $CHCl_3$ in the presence of aqueous $NaOH$ is known as the $Reimer-Tiemann$ reaction.
In this reaction,$CHCl_3$ reacts with $NaOH$ to generate a dichlorocarbene intermediate $(:CCl_2)$.
This electrophilic carbene attacks the phenoxide ion at the ortho position.
The intermediate formed undergoes hydrolysis to yield salicylaldehyde ($2$-hydroxybenzaldehyde) in its salt form,which is sodium $2-$formylphenolate.

(B) The reaction shown is a Fries rearrangement.
When phenyl benzoate is treated with $AlCl_3$ (a Lewis acid),it undergoes rearrangement to form a mixture of ortho-hydroxybenzophenone and para-hydroxybenzophenone.
The product '$X$' is ortho-hydroxybenzophenone,which contains a phenolic hydroxyl group $(-OH)$ and a ketone carbonyl group $(-C=O)$.
Therefore,the functional groups present are $-OH$ and $-C=O$ (ketone).

(B) $2-$Hydroxybenzoic acid (salicylic acid) reacts with acetic anhydride in the presence of conc. $H_2SO_4$ to form $2-$acetoxybenzoic acid,which is commonly known as aspirin. The reaction is an acetylation reaction of the phenolic $-OH$ group.

(A) The acidity of phenols is inversely proportional to their $pK_{a}$ values $(Acidity \propto \frac{1}{pK_{a}})$.
Electron-withdrawing groups (like $-NO_{2}$) increase the acidity of phenol by stabilizing the phenoxide ion through $-I$ and $-M$ effects.
$1$. Phenol $(A)$ has no electron-withdrawing group,so it is the least acidic (highest $pK_{a}$).
$2$. meta-Nitrophenol $(C)$ exhibits only the $-I$ effect,making it more acidic than phenol but less acidic than ortho and para isomers.
$3$. ortho-Nitrophenol $(B)$ and para-Nitrophenol $(D)$ exhibit both $-I$ and $-M$ effects. para-Nitrophenol $(D)$ is the most acidic due to the strong $-M$ effect of the $-NO_{2}$ group at the para position.
ortho-Nitrophenol $(B)$ is less acidic than para-Nitrophenol $(D)$ due to intramolecular hydrogen bonding,which stabilizes the undissociated molecule.
Thus,the order of acidity is: $D > B > C > A$.
Since $pK_{a}$ is inversely proportional to acidity,the decreasing order of $pK_{a}$ is: $A > C > B > D$.

(D) The acidic strength of compounds is determined by the stability of their conjugate bases. More stable conjugate bases correspond to stronger acids.
$A$: $H_2O$ (Conjugate base: $OH^-$)
$B$: Phenol $(C_6H_5OH)$ (Conjugate base: Phenoxide ion,$C_6H_5O^-$,which is resonance stabilized)
$C$: Isopropyl alcohol $((CH_3)_2CHOH)$ (Conjugate base: Isopropoxide ion,$(CH_3)_2CHO^-$,destabilized by two electron-donating methyl groups)
$D$: Ethanol $(CH_3CH_2OH)$ (Conjugate base: Ethoxide ion,$CH_3CH_2O^-$,destabilized by one electron-donating ethyl group)
Stability order of conjugate bases: $C_6H_5O^- > OH^- > CH_3CH_2O^- > (CH_3)_2CHO^-$.
Therefore,the order of acidic strength is: $B > A > D > C$.
Thus,option $D$ is correct.

(C) Nitration is an electrophilic aromatic substitution reaction. The rate of this reaction depends on the electron density of the aromatic ring.
Groups that increase electron density (electron-donating groups) activate the ring,while groups that decrease electron density (electron-withdrawing groups) deactivate the ring.
In phenol $(I)$,the $-OH$ group exerts a strong $+R$ (resonance) effect,which significantly increases the electron density of the ring,making it highly reactive.
In benzene $(III)$,there is no substituent to alter the electron density.
In nitrobenzene $(II)$,the $-NO_2$ group exerts a strong $-R$ effect,which significantly decreases the electron density of the ring,making it the least reactive.
Therefore,the order of reactivity is $(I) > (III) > (II)$.

(D) $pK_a = -\log K_a$. As $K_a$ increases,$pK_a$ decreases.
Benzoic acid $(pK_a \approx 4.2)$ is a stronger acid than phenol $(pK_a \approx 10)$,so the $pK_a$ of benzoic acid is smaller than that of phenol.
Therefore,Assertion $(A)$ is false.
The phenoxide ion is stabilized by non-equivalent resonance structures,while the benzoate ion is stabilized by two equivalent resonance structures,which makes the benzoate ion more stable and benzoic acid a stronger acid.
Thus,Reason $(R)$ is true.

(B) The $-OH$ group is an electron-donating group by resonance effect,which makes the benzene ring highly activated towards electrophilic substitution.
It is ortho/para directing in nature.
Therefore,ortho/para substitution in phenol by an electrophile is very facile compared to the other options,which contain electron-withdrawing groups.

(A) The acidity of phenols depends on the stability of the phenoxide ion formed after the loss of a proton. Electron-withdrawing groups $(EWG)$ stabilize the phenoxide ion,increasing acidity,while electron-donating groups $(EDG)$ destabilize it,decreasing acidity.
$1$. $A$ (Cyclohexanol): The conjugate base is an alkoxide ion,which is less stable than the phenoxide ion. Thus,$A$ is the least acidic.
$2$. $B$ (Phenol): The reference compound.
$3$. $C$ (p-Methoxyphenol): $-OCH_3$ is an $EDG$ by resonance (+$M$ effect),which destabilizes the phenoxide ion. Thus,$C < B$.
$4$. $D$ (p-Hydroxyacetophenone): $-COCH_3$ is an $EWG$ (-$M$ effect),which stabilizes the phenoxide ion. Thus,$D > B$.
$5$. $E$ (m-Nitrophenol): $-NO_2$ is a strong $EWG$ (-$I$ and -$M$ effects). At the meta position,only the -$I$ effect operates. Thus,$E > B$.
$6$. $F$ (p-Nitrophenol): $-NO_2$ at the para position exerts both -$I$ and -$M$ effects,providing maximum stabilization to the phenoxide ion. Thus,$F > E$.
Comparing the $EWG$ strengths: $p-NO_2$ $(F)$ > $m-NO_2$ $(E)$ > $p-COCH_3$ $(D)$.
Comparing the $EDG$ strengths: $p-OCH_3$ $(C)$ destabilizes more than $B$. $A$ is the least acidic.
The correct order is $F > E > D > B > C > A$.

(D) Step $1$: Phenol reacts with zinc dust $(Zn)$ under heating $(\Delta)$ to undergo reduction,resulting in the formation of benzene.
$C_6H_5OH Zn \xrightarrow{\Delta} C_6H_6 ZnO$
Step $2$: Benzene then undergoes electrophilic aromatic substitution (chlorination) in the presence of anhydrous ferric chloride $(FeCl_3)$ and chlorine $(Cl_2)$ to form chlorobenzene.
$C_6H_6 Cl_2 \xrightarrow{\text{anhyd. } FeCl_3} C_6H_5Cl HCl$
Therefore,the major product is chlorobenzene.

(B) Phenol is acidic enough to react with strong bases like $NaOH$ to form sodium phenoxide $(C_{6}H_{5}ONa)$,which is water-soluble.
Phenol is a weaker acid than carbonic acid $(H_{2}CO_{3})$,therefore it cannot displace $CO_{2}$ from $NaHCO_{3}$.

(C) $1$. The reaction of benzene with $CH_3CH_2CH_2Cl$ in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation. The primary carbocation formed $(CH_3CH_2CH_2^+)$ undergoes a $1,2-H^-$ shift to form a more stable secondary carbocation $(CH_3CH^+CH_3)$.
$2$. This secondary carbocation attacks the benzene ring to form isopropylbenzene (cumene) as the major product '$P$'.
$3$. Cumene undergoes oxidation with $O_2/OH^-, \Delta$ to form cumene hydroperoxide,which upon treatment with dilute $H_2SO_4$ yields phenol and acetone $(CH_3COCH_3)$ as the major product '$Q$'.
$4$. Thus,'$P$' is isopropylbenzene and '$Q$' is acetone.

(A) Compounds that are stronger acids than carbonic acid $(H_2CO_3)$ react with sodium bicarbonate $(NaHCO_3)$ to release carbon dioxide $(CO_2)$ gas,which causes effervescence.
$2,4,6-$Trinitrophenol (also known as picric acid) is a strong acid due to the electron-withdrawing effect of three nitro $(-NO_2)$ groups,which stabilize the phenoxide ion.
Since the $pK_a$ of picric acid is lower than the $pK_a$ of $H_2CO_3$,it reacts with $NaHCO_3$ to produce effervescence.
Other options like acetone,ethanol,and diethyl ether are not acidic enough to react with $NaHCO_3$.

(C) $1$. $\text{Phenol}$ reacts with dilute $HNO_3$ to form a mixture of $o$-nitrophenol and $p$-nitrophenol. The major product is $p$-nitrophenol $(X)$.
$2$. $p$-nitrophenol $(X)$ is reduced using $Zn/HCl$ to form $p$-aminophenol.
$3$. $p$-aminophenol reacts with $1 \ equiv$ of acetic anhydride $(CH_3CO)_2O$. The amino group $(-NH_2)$ is more nucleophilic than the hydroxyl group $(-OH)$, so acetylation occurs at the nitrogen atom to form $p$-acetamidophenol $(Y)$.

(D) The reaction is a Kolbe-Schmitt reaction.
$1$. $4$-Methoxyphenol reacts with $NaOH$ to form sodium $4$-methoxyphenoxide.
$2$. The phenoxide ion is highly reactive towards electrophilic aromatic substitution.
$3$. The $-O^-$ group is a strong ortho/para directing group. Since the para position is already occupied by the $-OMe$ group,the electrophile $(CO_2)$ attacks the ortho position relative to the $-O^-$ group.
$4$. Subsequent acidification with $H_3O^+$ yields the final product,$2$-hydroxy-$5$-methoxybenzoic acid.

(C) Phenol is a weaker acid than carbonic acid $(H_2CO_3)$.
Therefore,it cannot displace carbonic acid from its salts like $Na_2CO_3$.
Consequently,phenol does not liberate $CO_2$ gas from $Na_2CO_3$ solution.
Thus,the statement in option $C$ is incorrect.

(B) Electron-withdrawing groups $(EWG)$ like $-CHO$ increase the acidity of phenols by stabilizing the phenoxide ion.
In $4-$hydroxybenzaldehyde $(II)$,the $-CHO$ group is at the para position and exerts both $-M$ (mesomeric) and $-I$ (inductive) effects.
In $3-$hydroxybenzaldehyde $(III)$,the $-CHO$ group is at the meta position and exerts only the $-I$ effect (the mesomeric effect does not operate at the meta position).
Phenol $(I)$ has no such electron-withdrawing group.
Therefore,the acidity order is $I < III < II$.

(C) Volatility is inversely proportional to the strength of intermolecular forces.
$o$-Hydroxybenzoic acid (Salicylic acid) exhibits intramolecular hydrogen bonding,which reduces the extent of intermolecular hydrogen bonding compared to its isomers ($p$- and $m$-hydroxybenzoic acid).
This results in weaker intermolecular forces,making $o$-hydroxybenzoic acid more volatile than the others.

(C) The acidity of phenols depends on the stability of the phenoxide ion formed after the loss of a proton. Electron-withdrawing groups $(EWG)$ stabilize the phenoxide ion and increase acidity,while electron-donating groups $(EDG)$ destabilize it and decrease acidity.
$1$. Compound $(IV)$ has a $-NO_2$ group,which is a strong electron-withdrawing group ($-R$ and $-I$ effect),making it the most acidic.
$2$. Compound $(III)$ has a $-CH=CH_2$ group,which is an electron-withdrawing group ($-R$ effect),making it more acidic than $(I)$ and $(II)$.
$3$. Compound $(I)$ has a $-CH_3$ group,which is an electron-donating group ($+I$ and hyperconjugation),making it less acidic than $(III)$ but more acidic than $(II)$.
$4$. Compound $(II)$ has a $-NMe_2$ group,which is a very strong electron-donating group ($+R$ effect),making it the least acidic.
Therefore,the correct order of acidity is $(IV) > (III) > (I) > (II)$.

(A) The acidity of nitrophenols is determined by the electron-withdrawing effect ($-I$ and $-R$ effects) of the $-NO_2$ group.
$p$-Nitrophenol is the most acidic because the $-NO_2$ group at the para-position exerts both strong $-I$ and $-R$ effects,stabilizing the phenoxide ion significantly.
In $o$-nitrophenol,the $-NO_2$ group is at the ortho-position,which also exerts strong $-I$ and $-R$ effects,but the formation of intramolecular $H$-bonding between the $-OH$ group and the $-NO_2$ group makes the release of the proton slightly more difficult compared to $p$-nitrophenol.
In $m$-nitrophenol,the $-NO_2$ group only exerts a $-I$ effect (no resonance effect),making it the least acidic among the three.
Thus,the correct order of decreasing acidity is: $p$-Nitrophenol $ > $ $o$-Nitrophenol $ > $ $m$-Nitrophenol.

(C) The acidic strength of substituted phenols depends on the electron-withdrawing effect ($-I$ and $-M$ effects) of the substituent group attached to the benzene ring.
Stronger electron-withdrawing groups stabilize the phenoxide ion more effectively,thereby increasing the acidic strength.
The electron-withdrawing tendency of the given substituents follows the order: $-F < -Cl < -NO_2$.
Therefore,the correct order of acidic strength is $p$-fluorophenol $ < $ $p$-chlorophenol $ < $ $p$-nitrophenol.

(A) In an acidic medium,phenol exists in equilibrium with its protonated form,and the $-OH$ group activates the benzene ring towards electrophilic substitution.
The $-OH$ group is an electron-donating group by resonance,which increases the electron density at the $ortho$ and $para$ positions.
When treated with $D_{2}SO_{4} / D_{2}O$,the $D^+$ acts as an electrophile.
The electrophile $(D^+)$ attacks the electron-rich $ortho$ and $para$ positions of the benzene ring.
Additionally,the phenolic hydrogen is acidic and readily exchanges with $D$ in $D_{2}O$ to form $-OD$.
Therefore,the hydrogen atoms at the $ortho$ and $para$ positions are replaced by deuterium atoms,and the $-OH$ group becomes $-OD$.
The final product is $2,4,6-trideuterophenol-d_1$ (where the phenolic hydrogen is also replaced).

(A) The $FeCl_3$ test is used to detect the presence of a phenolic group ($-OH$ group attached directly to the benzene ring).
Compounds containing a phenolic group react with neutral $FeCl_3$ to form a violet-colored complex.
Among the given options,$o-$cresol ($2-$methylphenol) is a phenol and gives a violet color with $FeCl_3$.
Benzyl alcohol $(C_6H_5CH_2OH)$ does not contain a phenolic group (the $-OH$ is on the side chain),so it does not give a violet color with $FeCl_3$.
Therefore,$o-$cresol and benzyl alcohol can be distinguished using this test.

(C) The reaction is the Kolbe-Schmitt reaction. The molecular weight of compound $(y)$ is $2 \times \text{vapour density} = 2 \times 47 = 94$.
Given the percentages: $C = 76.6\%$,$H = 6.38\%$,$O = 100 - (76.6 + 6.38) = 17.02\%$.
Moles of $C = 76.6/12 = 6.38$,$H = 6.38/1 = 6.38$,$O = 17.02/16 = 1.06$.
Ratio $C:H:O = 6:6:1$. The empirical formula is $C_6H_6O$,which corresponds to phenol $(C_6H_5OH)$.
Thus,compound $(x)$ is phenol and compound $(y)$ is salicylic acid ($2$-hydroxybenzoic acid).
Phenol $(x)$ is acidic and dissolves in $NaOH$. Salicylic acid $(y)$ is more acidic than phenol due to the electron-withdrawing $-COOH$ group and intramolecular hydrogen bonding,so it also dissolves in $NaOH$.
Salicylic acid $(y)$ reacts with $NaHCO_3$ to evolve $CO_2$ gas.
Both are aromatic compounds and burn with a sooty flame.
Statement $(c)$ is incorrect because salicylic acid $(y)$ is more acidic than phenol $(x)$.

(D) The reaction of phenol with $CHCl_3$ and $aq. KOH$ is the Reimer-Tiemann reaction.
In this reaction,$o$-hydroxybenzaldehyde (salicylaldehyde) is the major product due to intramolecular hydrogen bonding,while $p$-hydroxybenzaldehyde is the minor product.
Therefore,Statement $I$ is false.
$o$-hydroxybenzaldehyde exhibits intramolecular hydrogen bonding,making it volatile with steam,whereas $p$-hydroxybenzaldehyde exhibits intermolecular hydrogen bonding,making it less volatile.
Thus,they can be separated by steam distillation.
Therefore,Statement $II$ is true.

(C) The reaction sequence is the industrial cumene process for the preparation of phenol.
Step $1$: Benzene $(C_6H_6)$ reacts with isopropyl chloride $(CH_3CH(Cl)CH_3)$ in the presence of anhydrous $AlCl_3$ (Friedel-Crafts alkylation) to form isopropylbenzene (cumene).
Step $2$: Cumene is oxidized by $O_2$ to form cumene hydroperoxide,which upon acid hydrolysis $(H^+)$ yields phenol $(C_6H_5OH)$ as $Y$ and acetone $(CH_3COCH_3)$ as $Z$.
Therefore,$X = CH_3CH(Cl)CH_3$,$Y = C_6H_5OH$,and $Z = CH_3COCH_3$.

(A) The coupling reaction involves a diazonium salt attacking a phenol.
Phenol is a weak acid,and it forms a phenoxide ion in an alkaline medium.
The phenoxide ion is much more strongly activated towards electrophilic aromatic substitution than neutral phenol,making the reaction with the diazonium electrophile faster and more efficient.

(A) The Reimer-Tiemann reaction is specifically used to introduce a formyl group $(-\text{CHO})$ into the ortho position of a phenol by using chloroform $(CHCl_3)$ and aqueous sodium hydroxide $(NaOH)$.
This reaction results in the formation of $2$-hydroxybenzaldehyde,commonly known as salicylaldehyde.

(D) Phenol $(C_6H_5OH)$ reacts with dilute $HNO_3$ to form a mixture of $o$-nitrophenol and $p$-nitrophenol.
$o$-Nitrophenol is steam volatile due to intramolecular hydrogen bonding,while $p$-nitrophenol is not due to intermolecular hydrogen bonding.
Thus,the steam volatile compound $(X)$ is $o$-nitrophenol $(C_6H_5NO_3)$.
Molar mass of phenol $(C_6H_6O)$ = $(6 \times 12) + (6 \times 1) + 16 = 94 \ g \ mol^{-1}$.
Percentage of oxygen in phenol = $(16 / 94) \times 100 \approx 17.021 \ \%$.
Molar mass of $o$-nitrophenol $(C_6H_5NO_3)$ = $(6 \times 12) + (5 \times 1) + 14 + (3 \times 16) = 72 + 5 + 14 + 48 = 139 \ g \ mol^{-1}$.
Percentage of oxygen in $o$-nitrophenol = $(48 / 139) \times 100 \approx 34.532 \ \%$.
Increase in percentage of oxygen = $34.532 - 17.021 = 17.511 \ \%$.
Expressing as $175.11 \times 10^{-1} \% \approx 175 \times 10^{-1} \%$. Given the options provided,the closest integer value is $175$ (Note: The provided options in the prompt seem to be placeholders; based on standard calculation,the result is $175$).

(C) The chemical reaction for the reduction of phenol with zinc dust is: $C_6H_5OH + Zn \rightarrow C_6H_6 + ZnO$.
First,calculate the molar mass of phenol $(C_6H_5OH)$: $(6 \times 12) + (6 \times 1) + 16 = 72 + 6 + 16 = 94 \text{ g mol}^{-1}$.
Calculate the initial moles of phenol: $\text{Moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{4.7 \text{ g}}{94 \text{ g mol}^{-1}} = 0.05 \text{ moles}$.
Since the reaction goes to $60\%$ completion,the moles of product $X$ (benzene) formed is: $0.05 \times 0.60 = 0.03 \text{ moles}$.
Expressing this in the required format: $0.03 = 3 \times 10^{-2}$.
Therefore,the value is $3$.