Properties of Phenols Questions in English

(B) $p$-Nitrophenol has the highest melting point due to strong intermolecular hydrogen bonding in the solid state,which leads to a more stable crystal lattice structure compared to the other isomers. $o$-Nitrophenol exhibits intramolecular hydrogen bonding,which reduces its melting point.

(A) dihydric phenol is a compound containing two hydroxyl $(-OH)$ groups attached directly to a benzene ring.
$1$. $\alpha$-naphthol is a monohydric phenol (contains one $-OH$ group).
$2$. Resorcinol ($1,3$-dihydroxybenzene),Hydroquinone ($1,4$-dihydroxybenzene),and Catechol ($1,2$-dihydroxybenzene) are all dihydric phenols.
Therefore,$\alpha$-naphthol is not a dihydric phenol.

(D) The reaction of $Phenol$ with chromic acid $(CrO_3)$ is an oxidation reaction.
$Phenol$ undergoes oxidation in the presence of $CrO_3$ to form $p-Benzoquinone$ as the major product.
This is a characteristic reaction of phenols where the aromatic ring is oxidized to a conjugated diketone.

(B) The process is known as the cumene process for the industrial production of phenol.
Step $1$: Cumene (isopropylbenzene) is oxidized by air in the presence of cobalt naphthenate to form cumene hydroperoxide.
Step $2$: Cumene hydroperoxide is then treated with dilute acid (like $dil. H_2SO_4$) to yield phenol and acetone as the final products.
The reaction is: $C_6H_5CH(CH_3)_2 + O_2$ $\xrightarrow{Co-naphthenate} C_6H_5C(CH_3)_2OOH$ $\xrightarrow{dil. H_2SO_4} C_6H_5OH + CH_3COCH_3$.

(C) The defensive froth emitted by the grasshopper $Romalea \ microptera$ contains several odorous compounds,including a chlorinated aromatic compound,$2, 5-$dichlorophenol.
This compound acts as a repellent to ants and is defensively useful to the grasshopper.
It is believed that this compound is derived from herbicides or herbicide derivatives ingested by the insect through its diet.

(A) dihydric phenol contains two hydroxyl $(-OH)$ groups attached to the benzene ring.
$1$. Resorcinol is benzene-$1,3$-diol,which is a dihydric phenol.
$2$. $m$-Cresol is a monohydric phenol ($3$-methylphenol).
$3$. Phloroglucinol is benzene-$1,3,5$-triol,which is a trihydric phenol.
$4$. Pyrogallol is benzene-$1,2,3$-triol,which is a trihydric phenol.
Therefore,the correct option is Resorcinol.

(B) The reaction described is the $Kolbe-Schmitt$ reaction.
In this reaction,$Sodium \ phenoxide$ reacts with $CO_2$ at $398 \ K$ and $6 \ atm$ pressure to form $Sodium \ salicylate$.
Upon acidification with $H_3O^{+}$,$Sodium \ salicylate$ is converted into $Salicylic \ acid$.
Therefore,the substrate '$A$' is $Sodium \ phenoxide$.

(A) The reaction between a phenol $(Ar-OH)$ and an acid chloride $(R-COCl)$ in the presence of a base like pyridine is an acylation reaction.
Pyridine acts as a base to neutralize the $HCl$ produced during the reaction,which helps in shifting the equilibrium towards the product side.
The reaction proceeds as follows:
$Ar-OH + R-COCl \xrightarrow{\text{Pyridine}} Ar-O-CO-R + HCl$
Here,the phenolic oxygen attacks the carbonyl carbon of the acid chloride,leading to the formation of an ester $(Ar-O-CO-R)$.

(D) When phenol reacts with bromine water (aqueous solution of bromine),it undergoes electrophilic substitution at all available ortho and para positions.
Due to the strong activating effect of the $-OH$ group,the reaction proceeds rapidly to form $2,4,6-$tribromophenol as a white precipitate.

(A) When phenol $(C_6H_5OH)$ is heated with hydrogen gas in the presence of a nickel catalyst at $433 \ K$,the benzene ring undergoes catalytic hydrogenation.
The reaction is: $C_6H_5OH + 3H_2 \xrightarrow{Ni, 433 \ K} C_6H_{11}OH$.
The product formed is cyclohexanol.

(C) The given reaction is the $Kolbe-Schmitt$ reaction.
In this reaction,$Sodium \ phenoxide$ reacts with $CO_2$ at $398 \ K$ and $6 \ atm$ pressure to form an intermediate $A$,which is $Sodium \ salicylate$.
Upon acidification with $H_3O^{+}$,$Sodium \ salicylate$ is converted into $Salicylic \ acid$ $(B)$.
The reaction sequence is:
$Sodium \ phenoxide + CO_2$ $\xrightarrow[6 \ atm]{398 \ K} Sodium \ salicylate (A)$ $\xrightarrow{H_3O^{+}} Salicylic \ acid (B)$.

(D) When phenol reacts with dilute nitric acid at low temperature $(298 \ K)$,it undergoes electrophilic aromatic substitution to form a mixture of $o-$nitrophenol and $p-$nitrophenol.
The ortho and para isomers can be separated by steam distillation because $o-$nitrophenol is steam volatile due to intramolecular hydrogen bonding,while $p-$nitrophenol is not steam volatile due to intermolecular hydrogen bonding.

(B) Salicylic acid is a phenolic acid that occurs naturally in the bark of the $Willow$ plant (genus $Salix$). The name 'salicylic' is derived from the Latin word '$salix$',meaning willow tree. Historically,extracts from willow bark have been used for their medicinal properties,specifically for pain relief and fever reduction.

(D) The reaction of $Chlorobenzene$ with $NaOH$ at $623 \ K$ and $150 \ atm$ followed by acidification $(H_3O^+)$ is the $Dow$ process,which yields $Phenol$ $(A)$.
When $Phenol$ reacts with $Br_2$ water,it undergoes electrophilic aromatic substitution to form $2,4,6-$tribromophenol $(B)$ as the final product due to the high reactivity of the $-OH$ group.

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

(D) When phenol is treated with chromic acid $(CrO_3)$,it undergoes oxidation to form a conjugated diketone known as $p-$benzoquinone.

(B) When phenol reacts with chloroform $(CHCl_3)$ in the presence of an aqueous base ($NaOH$ or $KOH$),an aldehyde $(-CHO)$ group is introduced into the benzene ring at the ortho position to the phenolic group,resulting in the formation of salicylaldehyde. This reaction is known as the Reimer-Tiemann reaction. The chemical equation is: $C_6H_5OH + CHCl_3 + 3KOH \rightarrow C_6H_4(OH)CHO + 3KCl + 3H_2O$.

(D) When phenol $(C_6H_5OH)$ is treated with chromic acid $(CrO_3)$,it undergoes oxidation to form a conjugated diketone known as $p$-benzoquinone.
This is a specific oxidation reaction of phenol.

(D) The conversion of phenol to picric acid $(2,4,6-\text{trinitrophenol})$ is an electrophilic aromatic substitution reaction known as nitration.
Phenol reacts with concentrated nitric acid $(HNO_3)$ in the presence of concentrated sulphuric acid $(H_2SO_4)$ to form picric acid.
The concentrated $H_2SO_4$ acts as a catalyst and helps in the generation of the nitronium ion $(NO_2^+)$,which is the electrophile.

(C) When phenol $(C_6H_5OH)$ is heated with zinc dust $(Zn)$,it undergoes a reduction reaction.
In this process,the oxygen atom of the hydroxyl group is removed by zinc to form zinc oxide $(ZnO)$,and phenol is converted into benzene $(C_6H_6)$.
The chemical equation is: $C_6H_5OH + Zn \xrightarrow{\Delta} C_6H_6 + ZnO$.

(D) The conversion of phenol to picric acid ($2,4,6$-trinitrophenol) is an electrophilic aromatic substitution reaction.
Phenol reacts with concentrated nitric acid $(HNO_3)$ in the presence of concentrated sulphuric acid $(H_2SO_4)$ to undergo nitration at the $ortho$ and $para$ positions,resulting in the formation of $2,4,6$-trinitrophenol,commonly known as picric acid.

(C) When phenol is treated with concentrated nitric acid in the presence of concentrated $H_2SO_4$,it undergoes electrophilic aromatic substitution to form $2,4,6-\text{trinitrophenol}$,which is commonly known as picric acid.
This reaction involves the nitration of the phenol ring at all available ortho and para positions due to the strong activating effect of the $-\text{OH}$ group.

(C) When phenol $(C_6H_5OH)$ is heated with zinc dust,it undergoes a reduction reaction.
Zinc dust acts as a reducing agent and removes the oxygen atom from the phenol group.
The reaction is as follows:
$C_6H_5OH + Zn \xrightarrow{\Delta} C_6H_6 + ZnO$
Thus,phenol is converted into benzene.

(B) When phenol is treated with bromine water,it undergoes electrophilic substitution to form a white precipitate of $2,4,6$-tribromophenol.
The reaction is as follows:
$C_6H_5OH + 3Br_2 \xrightarrow{H_2O} C_6H_2Br_3OH + 3HBr$
Thus,water is the solvent used.

(C) When phenol reacts with dilute nitric acid $(HNO_3)$ at low temperature $(298 \ K)$,it undergoes electrophilic aromatic substitution to form a mixture of $ortho$-nitrophenol and $para$-nitrophenol.
$C_6H_5OH + dil. HNO_3 \rightarrow o-Nitrophenol + p-Nitrophenol$

(A) The Reimer-Tiemann reaction involves the treatment of phenol with chloroform $(CHCl_3)$ in the presence of an aqueous base like $KOH$ or $NaOH$.
This reaction results in the introduction of a formyl group $(-CHO)$ at the ortho position of the phenol ring.
The major product formed is $2-$hydroxybenzaldehyde,commonly known as Salicylaldehyde,along with a small amount of the para-isomer.

(B) When phenol $(C_6H_5OH)$ vapours are passed over heated zinc dust,it undergoes reduction to form benzene $(C_6H_6)$ and zinc oxide $(ZnO)$.
The chemical reaction is as follows:
$C_6H_5OH + Zn \rightarrow C_6H_6 + ZnO$

(A) The reaction of phenol with concentrated $H_2SO_4$ at $373 \ K$ is an electrophilic aromatic substitution reaction known as sulphonation.
At higher temperatures $(373 \ K)$,the $-SO_3H$ group is predominantly directed to the para-position due to steric hindrance at the ortho-position,resulting in the formation of $p-$phenol sulphonic acid as the major product.

(D) The reaction described is the $Kolbe-Schmitt$ reaction.
$1$. Phenol reacts with $NaOH$ to form sodium phenoxide.
$2$. Sodium phenoxide reacts with $CO_2$ at high pressure and temperature $(140 \ ^\circ C)$ to form sodium salicylate.
$3$. Subsequent acid hydrolysis $(H^+)$ converts the intermediate into salicylic acid.
Thus,the final product is salicylic acid.

(A) In Kolbe's reaction,sodium phenoxide $(C_6H_5ONa)$ is treated with carbon dioxide $(CO_2)$ at a temperature of $398\ K$ and a pressure of $6\ atm$,followed by acidification to yield salicylic acid.
Therefore,the correct reactants and conditions are $C_6H_5ONa, CO_2$ and $398\ K, 6\ atm$.

(B) When phenol reacts with concentrated sulphuric acid at a low temperature of $293 \ K$,the electrophilic substitution reaction occurs primarily at the ortho position.
This results in the formation of $o-$phenol sulphonic acid as the major product.
At higher temperatures (e.g.,$373 \ K$),the para isomer becomes the major product due to thermodynamic stability.

(D) When phenol is treated with concentrated $HNO_3$ in the presence of concentrated $H_2SO_4$,it undergoes electrophilic aromatic substitution (nitration) at all available ortho and para positions.
This reaction results in the formation of $2, 4, 6-$trinitrophenol,which is commonly known as picric acid.
Therefore,the correct product $(A)$ is $2, 4, 6-$trinitrophenol.

(D) When phenol $(C_6H_5OH)$ is hydrogenated in the presence of a nickel catalyst at $433 \ K$,the aromatic ring undergoes reduction to form a saturated cyclic alcohol.
The reaction is as follows:
$C_6H_5OH + 3H_2 \xrightarrow{Ni, 433 \ K} C_6H_{11}OH$
Thus,the product obtained is cyclohexanol.

(B) Carbolic acid is another name for phenol $(C_6H_5OH)$.
When phenol is heated with a mixture of concentrated nitric acid $(HNO_3)$ and concentrated sulphuric acid $(H_2SO_4)$,it undergoes electrophilic aromatic substitution (nitration) at the $2, 4,$ and $6$ positions.
This reaction results in the formation of $2,4,6$-trinitrophenol,which is commonly known as picric acid.
The chemical reaction is as follows:
$C_6H_5OH + 3HNO_3 \xrightarrow{conc. H_2SO_4} C_6H_2(NO_2)_3OH + 3H_2O$

(A) When phenol is treated with chloroform $(CHCl_3)$ in the presence of aqueous sodium hydroxide $(NaOH)$,a formyl group $(-CHO)$ is introduced at the ortho position of the benzene ring.
This reaction is known as the Reimer-Tiemann reaction.
The reaction proceeds through the formation of a dichlorocarbene intermediate $(:CCl_2)$,which acts as an electrophile.
The final product obtained after acidification is salicylaldehyde (o-hydroxybenzaldehyde).

(D) When phenol reacts with dilute $HNO_3$ at low temperature,it undergoes electrophilic aromatic substitution to form a mixture of $o$-nitrophenol and $p$-nitrophenol.
$o$-nitrophenol is formed as the major product due to the presence of intra-molecular hydrogen bonding,which stabilizes the ortho isomer.

(A) When phenol reacts with bromine water $(Br_2/H_2O)$,it undergoes electrophilic aromatic substitution.
Due to the highly activating nature of the $-OH$ group,the reaction occurs at all available ortho and para positions simultaneously.
This results in the formation of a white precipitate of $2,4,6$-tribromophenol.

(A) When phenol is heated with zinc dust,it undergoes reduction to form benzene and zinc oxide.
$C_{6}H_{5}OH + Zn \xrightarrow{\Delta} C_{6}H_{6} + ZnO$

(B) The reaction of phenol with chloroform $(CHCl_3)$ in the presence of an aqueous base like sodium hydroxide $(NaOH)$ followed by acidification leads to the introduction of a formyl group $(-CHO)$ at the ortho position of the benzene ring,resulting in the formation of salicylaldehyde.
This specific chemical transformation is known as the Reimer-Tiemann reaction.

(D) The acidity of phenolic compounds is determined by the stability of the phenoxide ion formed after the loss of a proton. $e-$ withdrawing groups $(EWG)$ stabilize the phenoxide ion through $-I$ and $-M$ effects,thereby increasing acidity.
$p-$Nitrophenol contains a nitro group at the para position,which exerts both $-I$ and $-M$ effects,significantly stabilizing the phenoxide ion.
$m-$Nitrophenol only exerts a $-I$ effect because the $-M$ effect does not operate at the meta position.
$p-$Aminophenol contains an amino group,which is an $e-$ donating group ($+M$ effect),decreasing acidity.
Therefore,the correct order of acidity is: $p-$Nitrophenol $ > m-$Nitrophenol $ > $ Phenol $ > p-$Aminophenol.

(B) Phenol reacts with bromine water $(Br_2/H_2O)$ to undergo electrophilic aromatic substitution,yielding a white precipitate of $2,4,6-$tribromophenol.
The reaction is:
$C_6H_5OH + 3Br_2(aq) \rightarrow C_6H_2Br_3OH(s) + 3HBr(aq)$

(C) The reaction shown is the $Kolbe-Schmitt$ reaction.
In the first step,sodium phenoxide reacts with $CO_2$ at $398 \text{ K}$ and $6 \text{ atm}$ pressure to form sodium salicylate $(A)$.
In the second step,acidification with $H_3O^{+}$ converts sodium salicylate $(A)$ into salicylic acid $(B)$.
Therefore,the product $B$ is salicylic acid.

(C) dihydric compound contains two hydroxyl $(-OH)$ groups attached to the benzene ring.
$1$. Catechol ($1,2$-dihydroxybenzene) is a dihydric phenol.
$2$. Resorcinol ($1,3$-dihydroxybenzene) is a dihydric phenol.
$3$. Hydroquinone ($1,4$-dihydroxybenzene) is a dihydric phenol.
$4$. Phloroglucinol ($1,3,5$-trihydroxybenzene) contains three hydroxyl groups,so it is a trihydric phenol.
Therefore,Phloroglucinol is $NOT$ a dihydric compound.

(D) Dihydric phenols are compounds that contain two hydroxyl $(-OH)$ groups attached to a benzene ring.
Based on the provided structures:
$1$. Catechol: Contains two $-OH$ groups (dihydric).
$2$. Resorcinol: Contains two $-OH$ groups (dihydric).
$3$. Quinol: Contains two $-OH$ groups (dihydric).
$4$. Phloroglucinol: Contains three $-OH$ groups (trihydric).
$5$. Pyrogallol: Contains three $-OH$ groups (trihydric).
Therefore,the pair consisting of two dihydric phenols is Catechol and Quinol.

(C) dihydric phenol contains two hydroxyl $(-OH)$ groups attached to the benzene ring.
$1$. $p$-Cresol is a monohydric phenol with a methyl group at the para position.
$2$. Phloroglucinol ($1,3,5$-trihydroxybenzene) is a trihydric phenol.
$3$. Catechol ($1,2$-dihydroxybenzene) is a dihydric phenol.
$4$. Pyrogallol ($1,2,3$-trihydroxybenzene) is a trihydric phenol.
Therefore,Catechol is the correct answer.

(D) dihydric phenol contains two hydroxyl $(-OH)$ groups attached to the benzene ring.
$1$. Catechol ($1,2$-dihydroxybenzene) is a dihydric phenol.
$2$. Resorcinol ($1,3$-dihydroxybenzene) is a dihydric phenol.
$3$. Quinol or Hydroquinone ($1,4$-dihydroxybenzene) is a dihydric phenol.
$4$. Hydroxyquinol ($1,2,4$-trihydroxybenzene) contains three hydroxyl groups,making it a trihydric phenol.
Therefore,Hydroxyquinol is not a dihydric phenol.

(C) $Benzene-1,4-diol$ is also known as $Hydroquinone$ or $Quinol$.
$Catechol$ is $Benzene-1,2-diol$.
$Resorcinol$ is $Benzene-1,3-diol$.
$Pyrogallol$ is $Benzene-1,2,3-triol$.

(C) Solubility in water depends on the ability of a compound to form hydrogen bonds with water molecules.
$o$-Nitrophenol exhibits intramolecular hydrogen bonding,which prevents it from forming effective intermolecular hydrogen bonds with water molecules.
In contrast,$p$-Nitrophenol,$p$-Cresol,and Phenol can form intermolecular hydrogen bonds with water,making them more soluble.
Therefore,$o$-Nitrophenol has the lowest solubility in water.

(C) The conversion of phenol into picric acid ($2,4,6$-trinitrophenol) is an electrophilic aromatic substitution reaction known as nitration.
When phenol is treated with concentrated nitric acid $(HNO_3)$ in the presence of concentrated sulfuric acid $(H_2SO_4)$,it undergoes nitration at the $2, 4,$ and $6$ positions to form $2,4,6$-trinitrophenol,commonly known as picric acid.

(B) Phenol $(C_6H_5OH)$ is a white,crystalline,low-melting solid $(m.p. = 41 \ ^\circ C)$.
It has a characteristic medicinal odour and is toxic in nature.
Therefore,the statement that pure phenol is odourless,nontoxic,and a high-melting solid is incorrect.