Properties of Phenols Questions in English

(B) The melting point of a compound is significantly influenced by the nature of hydrogen bonding.
$o-$Nitrophenol exhibits intramolecular hydrogen bonding,which restricts intermolecular association,leading to a lower melting point.
In $p-$Cresol,the presence of the non-polar $-CH_3$ group does not significantly enhance intermolecular forces compared to the nitro group.
$p-$Nitrophenol exhibits strong intermolecular hydrogen bonding,which leads to the association of molecules in the solid state,resulting in a much higher melting point $(114 \ ^\circ C)$ compared to the others.

(B) $p-$Nitrophenol has the highest melting point among the given compounds.
This is due to the presence of strong intermolecular hydrogen bonding in $p-$nitrophenol,which leads to the association of molecules.
In contrast,$o-$nitrophenol exhibits intramolecular hydrogen bonding,which reduces the intermolecular forces of attraction,resulting in a lower melting point.

(C) $2,4,6-$trinitrophenol is commonly known as picric acid. The presence of three electron-withdrawing $-NO_2$ groups at the ortho and para positions significantly increases the acidity of the phenolic $-OH$ group by stabilizing the phenoxide ion through resonance and inductive effects. Therefore,it is strongly acidic in nature.

(C) Carbolic acid (also known as phenol) is oxidised by acidified sodium dichromate $(Na_2Cr_2O_7 / H_2SO_4)$ to give $p$-benzoquinone.
The reaction is as follows:
$C_6H_5OH + [O] \xrightarrow{Na_2Cr_2O_7 / H_2SO_4} C_6H_4O_2$ (Benzoquinone)
In the presence of air,phenols are slowly oxidised to dark-coloured mixtures containing quinones.

(D) Step $1$: Aniline reacts with $NaNO_2 + HCl$ at $273-278 \ K$ to form $A$,which is Benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$.
Step $2$: When Benzenediazonium chloride is heated with water $(\Delta, H_2O)$,it undergoes hydrolysis to form $B$,which is Phenol $(C_6H_5OH)$,along with the evolution of Nitrogen gas $(N_2 \uparrow)$ and Hydrochloric acid $(HCl)$.
Therefore,product $B$ is Phenol.

(C) The reaction of $2-\text{hydroxybenzoic acid}$ (Salicylic acid) with acetic anhydride in the presence of an acid catalyst $(H^{+})$ is an acetylation reaction.
In this reaction,the phenolic $-OH$ group of salicylic acid is acetylated to form $2-\text{acetoxybenzoic acid}$,which is commonly known as Aspirin,along with the byproduct acetic acid.
Therefore,$A$ is Salicylic acid.

(D) The reaction between a phenol $(Ar-OH)$ and an acid chloride $(R-CO-Cl)$ in the presence of a base like pyridine is known as acylation.
Pyridine acts as a base to neutralize the $HCl$ produced during the reaction,which helps in driving the reaction forward.
The reaction is:
$Ar-OH + R-CO-Cl \xrightarrow{\text{Pyridine}} R-CO-OAr + HCl$
Thus,the product formed is an ester,specifically $R-CO-OAr$.

(A) Carbolic acid is the common name for phenol,which has the chemical formula $C_6H_5OH$.
To determine the number of $\sigma$ bonds,we look at its structure:
$1$. There are $6$ carbon atoms in the ring,with $3$ $C-C$ $\sigma$ bonds and $3$ $C=C$ bonds (each containing $1$ $\sigma$ bond),totaling $6$ $\sigma$ bonds in the ring.
$2$. There are $5$ $C-H$ $\sigma$ bonds attached to the ring.
$3$. There is $1$ $C-O$ $\sigma$ bond.
$4$. There is $1$ $O-H$ $\sigma$ bond.
Total $\sigma$ bonds = $6 + 5 + 1 + 1 = 13$.

(A) The reaction between phenol $(C_{6}H_{5}OH)$ and acetic acid $(CH_{3}COOH)$ in the presence of an acid catalyst $(H^{+})$ is an esterification reaction.
In this reaction,the hydroxyl group of the phenol reacts with the carboxyl group of the acetic acid to form an ester,phenyl acetate $(C_{6}H_{5}OCOCH_{3})$,and water $(H_{2}O)$.
The reaction is: $C_{6}H_{5}OH + CH_{3}COOH \xrightarrow{H^{+}} C_{6}H_{5}OCOCH_{3} + H_{2}O$.
Therefore,the product $(X)$ is $C_{6}H_{5}OCOCH_{3}$.

(B) $2,4-D$ichlorophenoxyacetic acid (commonly known as $2,4-D$) is a synthetic auxin used as a selective herbicide.
It mimics the plant growth hormone auxin,causing uncontrolled and disorganized growth in broadleaf weeds,which leads to their death while leaving grasses and crops relatively unaffected.

(B) The structure of $Benzene-1,3-diol$ consists of a benzene ring with two hydroxyl $(-OH)$ groups attached at the $1$ and $3$ positions.
This compound is commonly known as $Resorcinol$.
$Catechol$ is $Benzene-1,2-diol$.
$Quinol$ (or $Hydroquinone$) is $Benzene-1,4-diol$.
$Pyrogallol$ is $Benzene-1,2,3-triol$.

(B) trihydric phenol contains three hydroxyl $(-OH)$ groups attached to the benzene ring.
$1$. Catechol is $1,2$-dihydroxybenzene (dihydric).
$2$. Pyrogallol is $1,2,3$-trihydroxybenzene (trihydric).
$3$. Resorcinol is $1,3$-dihydroxybenzene (dihydric).
$4$. Quinol (Hydroquinone) is $1,4$-dihydroxybenzene (dihydric).
Therefore,Pyrogallol is the correct answer.

(D) Dihydric phenols are compounds containing two hydroxyl $(-OH)$ groups attached to a benzene ring.
$1$. Catechol has two $-OH$ groups at the $1,2$-positions.
$2$. Resorcinol has two $-OH$ groups at the $1,3$-positions.
$3$. Quinol (or hydroquinone) has two $-OH$ groups at the $1,4$-positions.
$4$. Phloroglucinol has three $-OH$ groups at the $1,3,5$-positions (trihydric phenol).
$5$. Pyrogallol has three $-OH$ groups at the $1,2,3$-positions (trihydric phenol).
Therefore,the pair of dihydric phenols is Catechol and Quinol.

(C) Hydroquinone is a dihydroxy benzene derivative where the two hydroxyl $(-OH)$ groups are attached at the $1$ and $4$ positions of the benzene ring.
According to $IUPAC$ nomenclature,this structure is named as Benzene-$1, 4$-diol.

(C) The $IUPAC$ names for the given compounds are as follows:
$1$. Catechol is $Benzene-1,2-diol$.
$2$. Resorcinol is $Benzene-1,3-diol$.
$3$. $o-Cresol$ is $2-Methylphenol$.
$4$. Quinol is $Benzene-1,4-diol$.
Comparing these with the options,$o-Cresol$ is incorrectly matched with $Benzene-1,2,3-triol$. Therefore,option $C$ is the correct answer.

(B) The common name for $2, 4, 6-$trinitrophenol is picric acid.
It consists of a phenol ring where three nitro $(-NO_2)$ groups are attached at the $2, 4,$ and $6$ positions of the benzene ring.
Therefore,the correct $IUPAC$ name is $2, 4, 6-$trinitrophenol.

(A) Hydroquinone is a phenolic compound with two hydroxyl $(-OH)$ groups attached to the benzene ring at the para positions ($1$ and $4$ positions).
Therefore,the $IUPAC$ name of hydroquinone is Benzene-$1,4$-diol.

(C) dihydric alcohol (or diol) contains two hydroxyl $(-OH)$ groups attached to the carbon chain.
$1$. Catechol is $1,2$-dihydroxybenzene (dihydric).
$2$. Hydroquinone is $1,4$-dihydroxybenzene (dihydric).
$3$. Resorcinol is $1,3$-dihydroxybenzene (dihydric).
$4$. Phloroglucinol is $1,3,5$-trihydroxybenzene,which is a trihydric phenol,not a dihydric alcohol.
Therefore,the correct answer is $C$.

(A) Phenol is more acidic than alcohol because the phenoxide ion formed after the loss of a proton $(H^+)$ is stabilised by resonance. The negative charge on the oxygen atom is delocalised over the benzene ring,which increases the stability of the phenoxide ion compared to the alkoxide ion formed from alcohols.

(C) Phenols react with neutral $FeCl_{3}$ solution to form a complex,which results in a characteristic blue,violet,or green colouration. This is a standard test for the presence of the phenolic group.

(A) The acidic strength of phenols depends on the nature of the substituent attached to the benzene ring.
Electron-withdrawing groups $(EWG)$ like $-NO_2$ increase the acidic strength by stabilizing the phenoxide ion through $-I$ and $-M$ effects.
Electron-donating groups $(EDG)$ like $-CH_3$ decrease the acidic strength by destabilizing the phenoxide ion through $+I$ and hyperconjugation effects.
Compound $(ii)$ has a $-NO_2$ group (strong $EWG$),which makes it the most acidic.
Compound $(i)$ is phenol (no substituent).
Compound $(iii)$ has a $-CH_3$ group $(EDG)$,which makes it the least acidic.
Therefore,the decreasing order of acidic strength is $(ii) > (i) > (iii)$.

(B) The conversion of phenol to salicylic acid is achieved using the $Kolbe$ reaction.
In this reaction,phenol is treated with sodium hydroxide $(NaOH)$ to form sodium phenoxide,which then reacts with carbon dioxide $(CO_2)$ under pressure followed by acidification to yield $2$-hydroxybenzoic acid,commonly known as salicylic acid.

(D) . $C_6H_5COONa + NaOH/CaO \xrightarrow{\Delta} C_6H_6 + Na_2CO_3$ (Benzene is formed).
$B$. $C_6H_5N_2^{+}Cl^{-} + H_3PO_2 + H_2O \longrightarrow C_6H_6 + H_3PO_3 + HCl + N_2$ (Benzene is formed).
$C$. $C_6H_5OH + Zn \xrightarrow{\Delta} C_6H_6 + ZnO$ (Benzene is formed).
$D$. $C_6H_5OH + H_2CrO_4 \xrightarrow{[O]} \text{Benzoquinone}$ (Benzene is $NOT$ formed).
Therefore,the correct option is $D$.

(D) The $pK_a$ value is inversely proportional to the acidity of the compound. Higher acidity corresponds to a lower $pK_a$ value.
Nitro groups $(-NO_2)$ are strong electron-withdrawing groups $(EWG)$ that increase the acidity of phenol by stabilizing the phenoxide ion through $-I$ and $-M$ effects.
In $p-O_2N-C_6H_4-OH$,$m-O_2N-C_6H_4-OH$,and $o-O_2N-C_6H_4-OH$,the $-NO_2$ group increases acidity compared to phenol $(C_6H_5OH)$.
Since $C_6H_5OH$ lacks any electron-withdrawing substituent,it is the least acidic among the given compounds.
Therefore,$C_6H_5OH$ has the maximum $pK_a$ value.

(A) The compound must satisfy two conditions:
$1$. It must react with ethylmagnesium bromide $(C_2H_5MgBr)$,which is a Grignard reagent. This requires the presence of an acidic hydrogen (like in $-OH$,$-COOH$,etc.) or a carbonyl/cyano group.
$2$. It must decolourize bromine water,which indicates the presence of an unsaturated bond (alkene or alkyne).
Option $A$ is $2$-vinylphenol. It contains a phenolic $-OH$ group (acidic hydrogen) which reacts with $C_2H_5MgBr$ to form ethane gas. It also contains a vinyl group $(-CH=CH_2)$ which reacts with bromine water,causing decolourisation.
Therefore,the correct answer is $A$.

(D) The oxidation of phenol with chromic acid $(Na_2Cr_2O_7 / H_2SO_4)$ results in the formation of a conjugated diketone known as benzoquinone.
The chemical structure of benzoquinone is $Cyclohexa-2,5-diene-1,4-dione$.
Therefore,the correct $IUPAC$ name is $Cyclohexa-2,5-diene-1,4-dione$.

(C) The Reimer-Tiemann reaction of phenol with $CHCl_3$ and $NaOH$ yields salicylaldehyde ($2$-hydroxybenzaldehyde) as the final product.
The chemical formula of salicylaldehyde is $C_7H_6O_2$.
Structure: $A$ benzene ring with an $-OH$ group at position $1$ and a $-CHO$ group at position $2$.
Counting $\sigma$ bonds:
- $6$ $C-C$ bonds in the ring.
- $4$ $C-H$ bonds on the ring.
- $1$ $C-O$ bond (phenolic).
- $1$ $O-H$ bond.
- $1$ $C-C$ bond (between ring and $-CHO$).
- $1$ $C-H$ bond (in $-CHO$).
- $1$ $C-O$ bond (in $-CHO$).
Total $\sigma$ bonds = $6 + 4 + 1 + 1 + 1 + 1 + 1 = 15$.
Counting $\pi$ bonds:
- $3$ $\pi$ bonds in the benzene ring.
- $1$ $\pi$ bond in the $C=O$ group.
Total $\pi$ bonds = $3 + 1 = 4$.
Thus,the number of $\sigma$ and $\pi$ bonds are $15$ and $4$ respectively.

(B) Phenol is a very weak acid $(K_a \approx 10^{-10})$ and it does not react with sodium carbonate $(Na_2CO_3)$ to form salt and evolve $CO_2$ gas.
Therefore,the statement that phenol is neutralised by sodium carbonate is incorrect.
$o$-Nitrophenol exhibits intramolecular hydrogen bonding,whereas $p$-nitrophenol exhibits intermolecular hydrogen bonding,making the boiling point of $o$-nitrophenol lower than that of $p$-nitrophenol.
Phenol is used in the synthesis of drugs like aspirin (an analgesic).
Phenol is more soluble in water than chlorobenzene due to its ability to form hydrogen bonds with water molecules.

(B) The chemical formula of $Picric \ Acid$ is $2,4,6-trinitrophenol$.
It contains a phenolic $-OH$ group and three nitro $(-NO_2)$ groups attached to the benzene ring,but it does not contain a carboxylic acid $(-COOH)$ group.
Ethanoic acid $(CH_3COOH)$,Benzoic acid $(C_6H_5COOH)$,and Salicylic acid $(C_6H_4(OH)COOH)$ all contain the $-COOH$ group.
Therefore,the correct option is $B$.

(C) The conversion of chlorobenzene to phenol is known as the $Dow$ process,where chlorobenzene reacts with $NaOH$ at high temperature and pressure.
The conversion of phenol to salicylaldehyde is known as the $Reimer-Tiemann$ reaction,which involves treatment with $CHCl_3$ and $NaOH$ followed by hydrolysis.
Therefore,$X$ is the $Dow$ process and $Y$ is the $Reimer-Tiemann$ reaction.

(D) In phenol $(C_6H_5OH)$,the carbon atom attached to the $-OH$ group is part of the benzene ring and is bonded to two other carbons and one oxygen atom via double and single bonds,giving it $sp^2$ hybridisation.
The oxygen atom in the $-OH$ group is bonded to one carbon atom and one hydrogen atom,and it possesses two lone pairs of electrons. Therefore,the steric number of oxygen is $4$ ($2$ bond pairs + $2$ lone pairs),which corresponds to $sp^3$ hybridisation.
Thus,the hybridisation of carbon and oxygen is $sp^2$ and $sp^3$ respectively.

(C) The reaction of phenol with $H_2CrO_4$ (chromic acid) is an oxidation reaction,not an electrophilic substitution reaction.
In this reaction,phenol is oxidized to $p$-benzoquinone.
Other reactions listed:
$(A)$ Reimer-Tiemann reaction (electrophilic substitution),
$(B)$ Nitration of phenol (electrophilic substitution),
$(D)$ Bromination of phenol (electrophilic substitution).

(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 leads to the introduction of a formyl group $(-CHO)$ at the ortho position of the phenol ring.
The final product formed is $2$-hydroxybenzaldehyde,commonly known as Salicylaldehyde.
Therefore,the correct option is $A$.

(A) The $pK_{a}$ value is inversely proportional to the acidity of the compound. Higher $pK_{a}$ means lower acidity.
$1$. $p-$cresol contains an electron-donating methyl group $(-CH_{3})$,which decreases the acidity of phenol.
$2$. Phenol has no substituent.
$3$. $o-$nitrophenol and $m-$nitrophenol contain electron-withdrawing nitro groups $(-NO_{2})$,which significantly increase the acidity of phenol.
Comparing these,$p-$cresol is the least acidic,therefore it has the highest $pK_{a}$ value.

(B) When salicylaldehyde $(2-hydroxybenzaldehyde)$ is heated with zinc dust,the phenolic $-OH$ group is reduced to a hydrogen atom,and the aldehyde group is also reduced to a methyl group,resulting in the formation of toluene $(C_6H_5CH_3)$. However,in the context of standard textbook reactions where zinc dust is used to reduce phenolic groups,the product is often identified as benzaldehyde if the reaction conditions specifically target the $-OH$ group. Given the options provided,the most chemically accurate transformation for salicylaldehyde with zinc dust is the reduction of the phenolic $-OH$ to $H$,yielding benzaldehyde $(C_6H_5CHO)$.

(D) The reaction of phenyl methyl ether (anisole) with $HI$ involves the cleavage of the $C-O$ bond,yielding phenol $(A)$ and methyl iodide $(CH_3I)$.
Phenol $(A)$ reacts with concentrated $HNO_3$ (nitration) to form $2,4,6-$trinitrophenol,commonly known as picric acid $(B)$.

(C) The acidity of a compound depends on the stability of its conjugate base.
$o$-Nitrophenol is the most acidic among the given options because the $-NO_2$ group is a strong electron-withdrawing group ($-I$ and $-M$ effect).
This group stabilizes the phenoxide ion formed after the loss of a proton $(H^+)$ by dispersing the negative charge through resonance.
In contrast,$-CH_3$ (in $o$-cresol) is an electron-donating group ($+I$ and hyperconjugation),which destabilizes the phenoxide ion,making it less acidic than phenol.
$Cl-CH_2-CH_2-OH$ is an aliphatic alcohol,which is significantly less acidic than phenolic compounds.

(C) The acidity of phenols is due to the greater resonance stabilization of the phenoxide ion relative to phenol. Electron withdrawing groups $(EWG)$ like $-NO_{2}$ stabilize the phenoxide ion by dispersing the negative charge,which increases the acidity of phenols.
Conversely,electron donating groups $(EDG)$ like alkyl groups destabilize the phenoxide ion by intensifying the negative charge,which decreases the acidic strength of phenols.
Since the methyl group has a $+I$ effect,which is stronger at the $o$-position than at the $p$-position (as the $+I$ effect decreases with distance),$o$-cresol is a weaker acid than $p$-cresol.
Therefore,the order of acidic strength is: $p$-nitrophenol $>$ phenol $>$ $p$-cresol $>$ $o$-cresol.

(A) Phenol reacts with bromine water to form a white precipitate of $2,4,6$-tribromophenol due to the high reactivity of the benzene ring towards electrophilic substitution.
Propanol does not undergo this reaction and does not form a white precipitate with bromine water.
Therefore,bromine water is an effective reagent to distinguish between phenol and propanol.

(B) The acidic strength order is: $\text{Ethyl alcohol} < \text{Phenol} < p-\text{Nitrophenol} < \text{Picric acid}$.
Since $pK_a = -\log(K_a)$,the $pK_a$ values follow the inverse order of acidic strength:
$(i)$ Phenol: $10$
$(ii)$ $p-$Nitrophenol: $7.1$
$(iii)$ Ethyl alcohol: $16$
$(iv)$ Picric acid: $0.78$
Therefore,the correct matching is: $(i)-a, (ii)-b, (iii)-c, (iv)-d$.

(D) In Kolbe's reaction,phenol is first converted to sodium phenolate by reaction with $NaOH$.
Then,sodium phenolate reacts with $CO_2$ under pressure,followed by acidification to form salicylic acid.
The reaction is as follows:
$C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O$
$C_6H_5ONa + CO_2 \xrightarrow{H^+} \text{Salicylic acid}$
Thus,the reacting substances for the carboxylation step are sodium phenolate and $CO_2$.

(A) The Reimer-Tiemann reaction involves the treatment of phenol with $CHCl_3$ in the presence of aqueous $NaOH$.
In this reaction,$CHCl_3$ reacts with $OH^-$ to generate the electrophile,dichlorocarbene $(:CCl_2)$.
The dichlorocarbene then attacks the phenoxide ion to form an intermediate,which upon hydrolysis yields salicylaldehyde.

(A) The acidic strength 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.
$1$. $m-$cresol $(ii)$: The $-CH_3$ group is an $EDG$ due to $+I$ effect and hyperconjugation,making it less acidic than phenol.
$2$. Phenol $(iii)$: The reference compound.
$3$. $m-$chlorophenol $(iv)$: The $-Cl$ group is an $EWG$ due to its strong $-I$ effect,which outweighs its $+R$ effect at the meta position,making it more acidic than phenol.
$4$. $m-$nitrophenol $(i)$: The $-NO_2$ group is a strong $EWG$ due to both $-I$ and $-R$ effects,making it the most acidic among the given compounds.
Thus,the increasing order of acidic strength is: $m-$cresol $(ii)$ < phenol $(iii)$ < $m-$chlorophenol $(iv)$ < $m-$nitrophenol $(i)$.
The correct option is $A$.

(B) $1$. The reaction of phenol with $NaOH$ gives sodium phenoxide $(A)$.
$2$. The reaction of sodium phenoxide with $CO_2$ followed by acidification gives salicylic acid $(B)$ (Kolbe's reaction).
$3$. The acetylation of salicylic acid $(B)$ with acetic anhydride $(CH_3CO)_2O$ in the presence of an acid catalyst gives acetylsalicylic acid,which is commonly known as aspirin $(Y)$.

(B) $1$. The compound $A$ $(C_{7}H_{8}O)$ is insoluble in $NaHCO_{3}$ (meaning it is not a carboxylic acid) but soluble in $NaOH$ and gives a characteristic color with neutral $FeCl_{3}$ solution,which indicates the presence of a phenolic $-OH$ group.
$2$. The reaction with bromine water to form a tribromo derivative $(C_{7}H_{5}OBr_{3})$ is a characteristic reaction of phenols where the ortho and para positions are substituted by bromine atoms.
$3$. Among the given options,$m$-cresol ($3$-methylphenol) is a phenol that reacts with bromine water to form $2,4,6$-tribromo-$3$-methylphenol $(C_{7}H_{5}OBr_{3})$.
$4$. Therefore,compound $A$ is $m$-cresol.

(A) Phenol reacts with bromine water to give a white precipitate of $2,4,6$-tribromophenol.
Ethanol does not react with bromine water.
The reaction is:
$C_6H_5OH + 3Br_2(aq) \rightarrow C_6H_2Br_3OH(s) + 3HBr(aq)$
($2,4,6$-Tribromophenol,White precipitate)

(C) The acidic strength of phenols depends on the nature of the substituent attached to the benzene ring.
Electron-withdrawing groups $(EWG)$ like $-NO_2$ stabilize the phenoxide ion through resonance and inductive effects,thereby increasing acidic strength.
Electron-donating groups $(EDG)$ like $-OCH_3$ and $-CH_3$ destabilize the phenoxide ion by increasing electron density.
The $-OCH_3$ group acts as an $EDG$ via the $+M$ effect,which is stronger than the $+I$ effect of the $-CH_3$ group,making $p-$methoxy phenol less acidic than $p-$methyl phenol.
Thus,the order of acidic strength is: $p-$methoxy phenol $(i)$ < $p-$methyl phenol $(ii)$ < $p-$nitro phenol $(iii)$.
The correct increasing order is $i < ii < iii$.