Properties of Phenols Questions in English

(A) The acidic strength of phenols is influenced by the nature of substituents attached to the benzene ring. Electron-withdrawing groups $(EWG)$ increase acidity,while electron-releasing groups $(ERG)$ decrease acidity.
$I$ is phenol.
$II$ is $p$-cresol,which has a methyl group $(-CH_3)$ at the para position. The methyl group is an electron-releasing group due to the $+I$ effect and hyperconjugation,which decreases the acidity compared to phenol.
$III$ is $p$-methoxyphenol,which has a methoxy group $(-OCH_3)$ at the para position. The methoxy group is a strong electron-releasing group due to the $+M$ (mesomeric) effect,which significantly decreases the acidity compared to phenol and $p$-cresol.
Therefore,the order of acidity is $I > II > III$.

(B) The acidity of nitrophenols is influenced by the inductive and resonance effects of the $-NO_2$ group.
$o$-Nitrophenol exhibits intramolecular hydrogen bonding,which stabilizes the molecule but makes the release of the proton slightly more difficult compared to $p$-nitrophenol,yet it is more acidic than phenol.
The experimental $p K_a$ values are:
$o$-Nitrophenol: $7.23$
$m$-Nitrophenol: $8.39$
$p$-Nitrophenol: $7.15$
Therefore,the correct set is $o$-nitrophenol $7.23$.

(D) The reaction sequence is as follows:
$1$. Cumene is oxidized and then hydrolyzed to form phenol $(X)$.
$2$. Phenol reacts with $NaOH$ followed by $CO_2$ and acidification (Kolbe-Schmitt reaction) to form salicylic acid $(Y)$,which is $2$-hydroxybenzoic acid.
$3$. Salicylic acid reacts with acetic anhydride $(CH_3CO)_2O$ to form acetylsalicylic acid,commonly known as aspirin $(Z)$.

(D) When phenol reacts with bromine water $(Br_2/H_2O)$,it undergoes electrophilic substitution at all ortho and para positions due to the strong activating effect of the $-OH$ group,resulting in the formation of $2,4,6$-tribromophenol $(X)$.
When phenol reacts with concentrated nitric acid $(Conc. HNO_3)$,it undergoes nitration to form $2,4,6$-trinitrophenol,commonly known as picric acid $(Y)$.

(B) The $pK_{a}$ value is inversely proportional to the acidity of the compound. $A$ higher $pK_{a}$ value indicates a weaker acid.
Among the given options,the $-NO_2$ group is a strong electron-withdrawing group $(EWG)$ that increases acidity via $-I$ and $-M$ effects.
In $2-$nitrophenol,intramolecular hydrogen bonding stabilizes the conjugate base,increasing acidity.
In $4-$nitrophenol,the $-NO_2$ group exerts a strong $-M$ effect,significantly increasing acidity.
In $2, 4-$dinitrophenol,two $-NO_2$ groups significantly increase acidity.
In $3-$nitrophenol,the $-NO_2$ group is at the meta position,where it only exerts an $-I$ effect (no resonance effect). Therefore,it is the least acidic among the choices,resulting in the highest $pK_{a}$ value.

(B) The synthesis of Aspirin (acetylsalicylic acid) involves the acetylation of salicylic acid.
Salicylic acid $(X)$ reacts with acetic anhydride $(Y)$ in the presence of an acid catalyst $(H^+)$ to form acetylsalicylic acid (Aspirin) and acetic acid $(CH_3COOH)$.
The reaction is:
$\text{Salicylic acid} + (CH_3CO)_2O \xrightarrow{H^+} \text{Aspirin} + CH_3COOH$.
Thus,$X$ is salicylic acid and $Y$ is acetic anhydride $(CH_3CO)_2O$.

(C) The oxidation of cumene (isopropylbenzene) in air gives cumene hydroperoxide,which is compound $X$.
$C_6H_5CH(CH_3)_2 + O_2 \rightarrow C_6H_5C(CH_3)_2OOH$ (Cumene hydroperoxide,$X$)
Upon treatment with dilute acid,cumene hydroperoxide undergoes rearrangement to form phenol $(Y)$ and acetone $(Z)$.
$C_6H_5C(CH_3)_2OOH \xrightarrow{H^+} C_6H_5OH (Y) + CH_3COCH_3 (Z)$
Phenol $(Y)$ contains an acidic hydroxyl group $(-OH)$ attached to the benzene ring,which reacts with sodium metal to evolve hydrogen gas.
$2C_6H_5OH + 2Na \rightarrow 2C_6H_5ONa + H_2 \uparrow$
Acetone $(Z)$ is a ketone and does not contain an acidic hydrogen atom that can react with sodium metal.
Therefore,$Z$ is acetone $(CH_3COCH_3)$.

(D) The reaction sequence is as follows:
$1$. Cumene (isopropylbenzene) reacts with $O_2$ to form cumene hydroperoxide.
$2$. Acid-catalyzed hydrolysis of cumene hydroperoxide yields phenol and acetone $(CH_3COCH_3)$.
$3$. The reaction of phenol with $Br_2$ in a solvent of low polarity like $CHCl_3$ at low temperature $(273 \ K)$ leads to monobromination.
Due to the strong activating effect of the $-OH$ group,the electrophilic substitution occurs primarily at the ortho and para positions. The para-isomer is the major product due to less steric hindrance compared to the ortho-isomer. Thus,the major product is $p$-bromophenol.

(A) The Reimer-Tiemann reaction involves the treatment of phenol with chloroform $(CHCl_3)$ in the presence of an aqueous base like sodium hydroxide $(NaOH)$.
This reaction results in the introduction of a formyl group $(-CHO)$ at the ortho position of the phenol ring,yielding salicylaldehyde (o-hydroxybenzaldehyde) as the major product.
Therefore,$X$ is salicylaldehyde,$Y$ is $CHCl_3$,and $Z$ is $Aq. NaOH$.

(A) $1$. The reaction of benzenesulfonic acid with $NaOH$ followed by acidification gives phenol $(P = C_6H_5OH)$.
$2$. The oxidation of phenol with $Na_2Cr_2O_7$ and $H_2SO_4$ (Jones reagent or similar strong oxidizing conditions) yields $p$-benzoquinone as the major product $(Q = C_6H_4O_2)$.
$3$. In $p$-benzoquinone $(C_6H_4O_2)$,the structure consists of a six-membered ring with two carbonyl groups at para positions and two double bonds in the ring.
$4$. Counting the bonds:
- $\sigma$ bonds: $6$ ($C$-$C$ in ring) + $4$ ($C$-$H$) + $2$ ($C$=$O$) + $2$ ($C$-$O$) = $14$ $\sigma$ bonds.
- $\pi$ bonds: $2$ ($C$=$C$) + $2$ ($C$=$O$) = $4$ $\pi$ bonds.
- Wait,let's re-count carefully: The structure of $p$-benzoquinone has $6$ $C$-$C$ $\sigma$ bonds,$4$ $C$-$H$ $\sigma$ bonds,$2$ $C$=$O$ $\sigma$ bonds,and $2$ $C$-$O$ $\sigma$ bonds (total $14$). The $\pi$ bonds are $2$ from $C$=$C$ and $2$ from $C$=$O$ (total $4$).
- The ratio is $14:4 = 7:2$. Let's re-examine the structure. $p$-Benzoquinone has $12$ $\sigma$ bonds and $4$ $\pi$ bonds. Ratio $12:4 = 3:1$.

(A) The reactions are as follows:
$A$. Phenol reacts with $Zn$ dust to form benzene: $C_6H_5OH + Zn \rightarrow C_6H_6 + ZnO$ $(II)$.
$B$. Phenol is oxidized by $Na_2Cr_2O_7/H_2SO_4$ to form benzoquinone $(I)$.
$C$. Reimer-Tiemann reaction of phenol with $CHCl_3/NaOH$ followed by $H^+$ gives salicylaldehyde $(IV)$.
$D$. Kolbe's reaction of phenol with $NaOH$ followed by $CO_2/H^+$ gives salicylic acid $(III)$.
Therefore,the correct matching is $A-II, B-I, C-IV, D-III$.

(B) In the Reimer-Tiemann reaction,phenol reacts with $CHCl_3$ in the presence of aqueous $NaOH$ to form salicylaldehyde as the major product.
In the Kolbe reaction,phenol reacts with $NaOH$ to form sodium phenoxide,which then reacts with $CO_2$ followed by acidification to form salicylic acid as the major product.

(B) The reaction sequence is as follows:
$1$. Phenol reacts with $NaOH$ to form sodium phenoxide.
$2$. Sodium phenoxide undergoes Kolbe's reaction with $CO_2$ followed by acidification $(H_3O^+)$ to produce salicylic acid ($2$-hydroxybenzoic acid).
$3$. Salicylic acid then reacts with acetic anhydride $((CH_3CO)_2O)$ in the presence of an acid catalyst $(H^+)$ to undergo acetylation of the phenolic $-OH$ group.
$4$. The final product is $2-$acetoxybenzoic acid,commonly known as aspirin.

(C) When phenols are heated with zinc dust,they undergo reduction (deoxygenation) to form the corresponding aromatic hydrocarbon.
Compound $A$ is benzyl alcohol,which does not undergo deoxygenation with $Zn$ dust.
Compound $B$ is $o$-cresol ($2$-methylphenol),which contains a phenolic $-OH$ group and is reduced to toluene.
Compound $C$ is $2$-methylbenzyl alcohol,which does not contain a phenolic $-OH$ group and thus does not undergo deoxygenation with $Zn$ dust.
Compound $D$ is salicylic acid ($2$-hydroxybenzoic acid),which contains a phenolic $-OH$ group and is reduced to benzoic acid.
Therefore,compounds $B$ and $D$ undergo deoxygenation when heated with $Zn$ dust.

(C) The reaction sequence is as follows:
$1$. Electrophilic aromatic substitution of $p$-cresol with $Br_2$ gives $2$-bromo-$4$-methylphenol.
$2$. Treatment with $NaOH$ forms the sodium phenoxide salt.
$3$. Williamson's ether synthesis with $CH_3I$ yields $2$-bromo-$1$-methoxy-$4$-methylbenzene.
$4$. Free radical bromination of the methyl group with $Br_2/h\nu$ gives $2$-bromo-$1$-methoxy-$4$-(bromomethyl)benzene.
$5$. Nucleophilic substitution with $KCN$ replaces the benzylic bromine with a cyano group to form $2$-bromo-$1$-methoxy-$4$-(cyanomethyl)benzene.
$6$. Acidic hydrolysis of the cyanide group $(H_3O^+, \Delta)$ yields the final product,$2$-bromo-$4$-methoxyphenylacetic acid.

(A) The reaction sequence is as follows:
$1$. The reaction of phenol with $NaOH$ followed by $CO_2$ and then $H^+/H_2O$ is the Kolbe-Schmitt reaction,which yields salicylic acid ($2$-hydroxybenzoic acid).
$2$. The subsequent reaction with $CH_3COCl$ in the presence of pyridine is an acetylation reaction of the phenolic $-OH$ group.
$3$. This converts the $-OH$ group of salicylic acid into an acetoxy group $(-OCOCH_3)$,resulting in the formation of aspirin ($2$-acetoxybenzoic acid).

(A) The reaction of resorcinol ($1,3$-dihydroxybenzene) with $3Br_2$ (bromine water) is an electrophilic aromatic substitution reaction. The $-OH$ groups are strongly activating and ortho/para directing. In resorcinol,the positions $2, 4,$ and $6$ are activated by the two $-OH$ groups. Therefore,bromination occurs at all three positions to form $2,4,6$-tribromoresorcinol.

(D) In the Reimer-Tiemann reaction,phenol is treated with chloroform $(CHCl_3)$ in the presence of aqueous sodium hydroxide $(NaOH)$.
This reaction proceeds via the formation of a dichlorocarbene $(:CCl_2)$ intermediate,which attacks the phenoxide ion to form a substituted benzal chloride intermediate.
This intermediate is subsequently hydrolyzed to form salicylaldehyde.
Therefore,the intermediate formed is a substituted benzal chloride.

(B) The oxidation of phenol with chromic acid $(Na_2Cr_2O_7 / H_2SO_4)$ yields $p$-benzoquinone as the product.
Therefore,$X$ is phenol and $Y$ is $Na_2Cr_2O_7 / H_2SO_4$.

(B) When phenol is treated with $Br_2$ in the presence of a non-polar solvent like $CS_2$ at low temperature $(0^{\circ} C)$,the reaction is controlled to yield mono-substituted products.
In a non-polar solvent,the ionization of phenol to the highly reactive phenoxide ion is suppressed.
Consequently,the electrophilic substitution occurs only once,leading to the formation of a mixture of $o$-bromophenol and $p$-bromophenol.

(B) The given reaction 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 ortho-substituted product,which is salicylaldehyde ($o$-hydroxybenzaldehyde). The mechanism involves the formation of a dichlorocarbene $(:CCl_2)$ intermediate,which acts as an electrophile and attacks the phenol ring.

(C) The conversion of $O$-acylated phenol to $C$-acylated phenol in the presence of a Lewis acid like $AlCl_3$ is known as the Fries rearrangement.
In this reaction,the acyl group $(RCO-)$ migrates from the oxygen atom of the phenolic ester to the ortho or para position of the benzene ring.
Since the atoms or groups within the same molecule rearrange to form a new isomer,this process is classified as a molecular rearrangement reaction.

(A) The acidity of phenols is determined by the stability of the phenoxide ion formed after the loss of a proton. Electron-withdrawing groups (like $-NO_2$) increase acidity (lower $pK_{a}$),while electron-donating groups (like $-CH_3$) decrease acidity (higher $pK_{a}$).
$1$. $p$-Nitrophenol has a $-NO_2$ group at the para position,which exerts a strong $-R$ (resonance) and $-I$ (inductive) effect,making it the most acidic $(pK_{a} = 7.1)$.
$2$. $m$-Nitrophenol has a $-NO_2$ group at the meta position,which exerts only a $-I$ effect,making it less acidic than $p$-nitrophenol $(pK_{a} = 8.3)$.
$3$. $p$-Cresol has a $-CH_3$ group at the para position,which is an electron-donating group ($+I$ and hyperconjugation),making it the least acidic $(pK_{a} = 10.2)$.
Thus,$X = m$-nitrophenol $(8.3)$,$Y = p$-nitrophenol $(7.1)$,and $Z = p$-cresol $(10.2)$.

(A) The reaction of alcohols with $SOCl_2$ proceeds via an $S_N2$ mechanism or $S_Ni$ mechanism to replace the $-OH$ group with $-Cl$.
In phenol,the lone pair of electrons on the oxygen atom participates in resonance with the benzene ring,giving the $C-O$ bond a partial double bond character.
This makes the $C-O$ bond stronger and shorter,preventing the cleavage of the $C-O$ bond required to replace the $-OH$ group with $-Cl$.
Therefore,chlorobenzene is not formed when phenol reacts with $SOCl_2$.
Both Assertion $(A)$ and Reason $(R)$ are correct,and $(R)$ is the correct explanation of $(A)$.