Properties of Phenols Questions in English

(C) Phenol is more acidic than ethanol because the phenoxide ion formed after the loss of a proton is resonance-stabilized by the benzene ring.
In contrast,the ethoxide ion $(CH_3CH_2O^-)$ formed from ethanol is destabilized by the electron-donating inductive effect ($+I$ effect) of the ethyl group.
Therefore,the phenoxide ion is more stable than the ethoxide ion,making phenol a stronger acid with a higher ionisation constant.

(A) Picric acid is a derivative of phenol where three nitro groups $(-NO_2)$ are attached at the $2, 4,$ and $6$ positions of the benzene ring.
Therefore,the $IUPAC$ name is $2, 4, 6-$trinitrophenol.

(B) Due to the strong $+M$ (mesomeric) effect of the $-OH$ group,the electron density on the benzene ring increases significantly,especially at the ortho and para positions. This makes the ring highly activated towards electrophilic aromatic substitution. In contrast,$-Cl$ has a $-I$ effect that deactivates the ring,and $-CH_3$ has a weaker $+I$ and hyperconjugation effect compared to the $+M$ effect of $-OH$.

(B) The acidity of a compound is determined by the stability of its conjugate base.
Electron-withdrawing groups $(EWG)$ stabilize the phenoxide ion by dispersing the negative charge through $-I$ (inductive) and $-R$ (resonance) effects,thereby increasing acidity.
$NO_2$ is a strong electron-withdrawing group.
In $o$-nitrophenol,the $-NO_2$ group is present at the ortho position,which exerts a strong $-I$ and $-R$ effect,making it the most acidic among the given options.
Therefore,$o$-nitrophenol is the most acidic compound.

(B) The rate of electrophilic aromatic substitution,such as nitration,depends on the electron density of the benzene ring.
Groups that donate electrons by resonance (like $-OH$ in phenol) strongly activate the ring towards electrophilic attack.
Although $-NH_2$ in aniline is also a strong activator,aniline forms a salt $(C_6H_5NH_3^+)$ in the acidic medium used for nitration,which is deactivating.
Therefore,$Phenol$ is the most readily nitrated among the given options.

(A) The strength of an $o, p-$directing group is determined by its ability to donate electron density to the benzene ring via the resonance effect ($+R$ or $+M$ effect).
Among the given options:
$1$. $-OH$ has a strong $+R$ effect due to the lone pair on oxygen,making it a powerful activating group.
$2$. $-Cl$ and $-Br$ are deactivating groups due to their strong $-I$ effect,although they are $o, p-$directing due to the $+R$ effect.
$3$. $-C_6H_5$ (phenyl group) is an activating group but is weaker than $-OH$ in terms of electron donation to the ring.
Therefore,$-OH$ is the strongest $o, p-$directing group among the choices provided.

(C) The reaction of phenol with $CCl_4$ in the presence of concentrated $NaOH$ is a variation of the Reimer-Tiemann reaction that yields salicylic acid.
The chemical equation is:
$C_6H_5OH + CCl_4 + 4NaOH \rightarrow C_6H_4(OH)(COOH) + 4NaCl + 2H_2O$
Thus,the correct option is $C$.

(A) The reaction of phenol with $CCl_4$ and $KOH$ followed by acidification yields salicylic acid.
This reaction is a variation of the Reimer-Tiemann reaction.
Therefore,the reactant '$A$' is phenol.

(C) The reaction of phenol with $CHCl_3$ and $KOH$ is known as the Reimer-Tiemann reaction.
In this reaction,an aldehyde group $(-CHO)$ is introduced at the ortho and para positions of the phenol ring.
The major product is $o-$hydroxybenzaldehyde (Salicylaldehyde) and the minor product is $p-$hydroxybenzaldehyde.
Therefore,both $(a)$ and $(b)$ are obtained as products.

(B) The reaction is known as the Reimer-Tiemann reaction.
In this reaction,phenol reacts with trichloromethane $(CHCl_3)$ in the presence of an aqueous base (like $NaOH$ or $KOH$) followed by acidic hydrolysis to yield salicylaldehyde as the major product.
The chemical equation is:
$C_6H_5OH + CHCl_3 + 3KOH \rightarrow C_6H_4(OH)(CHO) + 3KCl + 2H_2O$

(B) The reaction of phenol with carbon tetrachloride $(CCl_4)$ in the presence of aqueous sodium hydroxide $(NaOH)$ is known as the Reimer-Tiemann reaction (using $CCl_4$ variant).
This reaction introduces a carboxylic acid group $( -COOH )$ at the ortho position of the phenol ring.
Initially,the phenoxide ion reacts with the dichlorocarbene intermediate (generated from $CCl_4$ and $NaOH$) to form an ortho-substituted intermediate,which upon hydrolysis and acidification $(H^+)$ yields $o-$hydroxybenzoic acid (also known as salicylic acid).
Therefore,the correct product is $o-$hydroxybenzoic acid.

(B) The reaction of phenol with chloroform $(CHCl_3)$ in the presence of aqueous sodium hydroxide $(NaOH)$ at $340 \ K$ is known as the Reimer-Tiemann reaction.
In this reaction,a formyl group $(-CHO)$ is introduced at the ortho position of the benzene ring of phenol.
The product formed is $2$-hydroxybenzaldehyde,which is commonly known as Salicylaldehyde.
The chemical equation is:
$C_6H_5OH + CHCl_3 + 3NaOH \rightarrow C_6H_4(OH)(CHO) + 3NaCl + 2H_2O$
Thus,the correct option is $(B)$.

(D) $Picric$ $acid$ is the common name for the chemical compound $2, 4, 6$-trinitrophenol.
It is a yellow,crystalline solid with the chemical formula $C_6H_3N_3O_7$.

(C) Picric acid $(2,4,6-\text{trinitrophenol})$ contains a hydroxyl group directly attached to a benzene ring,making it a phenolic compound.
Phthalic acid is a dicarboxylic acid.
Phosphoric acid is an inorganic acid $(H_3PO_4)$.
Phenylacetic acid is a carboxylic acid $(C_6H_5CH_2COOH)$.
Therefore,the correct option is $(C)$.

(A) $5 \%$ aqueous solution of phenol at room temperature is known as carbolic acid.

(C) When an aqueous solution of benzene diazonium chloride is heated,it undergoes hydrolysis to form phenol,nitrogen gas,and hydrochloric acid.
The chemical reaction is:
$C_6H_5N_2Cl + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2 + HCl$
Thus,the correct option is $C$.

(B) Salicylaldehyde is prepared from phenol by 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 salicylaldehyde.
The chemical equation is:
$C_6H_5OH + CHCl_3 + 3NaOH \rightarrow C_6H_4(OH)(CHO) + 3NaCl + 2H_2O$
Therefore,the correct option is $(b)$.

(C) Phenolphthalein is prepared by the condensation reaction of phthalic anhydride with two equivalents of phenol in the presence of a dehydrating agent like concentrated $H_2SO_4$.
The reaction is as follows:
Phthalic anhydride + $2C_6H_5OH$ $\xrightarrow{Conc. H_2SO_4}$ Phenolphthalein + $H_2O$.

(C) The given reaction involves the acylation of phenol with an acid chloride $(ClCOCH_3)$ in the presence of an aqueous base $(NaOH)$ to form an ester $(C_6H_5OCOCH_3)$.
This specific type of acylation reaction,where an alcohol or phenol reacts with an acid chloride in the presence of a base,is known as the Schotten-Baumann reaction.
Dow's reaction is used to prepare phenol from chlorobenzene.
Reimer-Tiemann reaction involves the formylation of phenol using chloroform and base.
Kolbe's reaction involves the carboxylation of sodium phenoxide using $CO_2$.

(B) When phenol reacts with dilute $HNO_3$ at low temperature,it undergoes electrophilic aromatic substitution to produce a mixture of $o-$nitrophenol and $p-$nitrophenol.
$o-$nitrophenol is formed due to intramolecular hydrogen bonding,while $p-$nitrophenol is formed due to intermolecular hydrogen bonding.
Therefore,the correct option is $(B)$.

(D) The acidity of a compound depends on the stability of its conjugate base.
$1$. Acetic acid $(CH_3COOH)$ is a carboxylic acid,and its conjugate base (acetate ion) is stabilized by resonance between two equivalent oxygen atoms,making it more acidic than phenol ($pK_a \approx 4.75$ vs $pK_a \approx 10$).
$2$. $p-$nitrophenol is more acidic than phenol because the nitro group $(-NO_2)$ exerts a strong electron-withdrawing effect ($-I$ and $-M$ effects),which stabilizes the phenoxide ion.
Therefore,phenol is less acidic than both acetic acid and $p-$nitrophenol.

(C) The acidity of substituted phenols is determined by the electron-withdrawing or electron-donating nature of the substituent.
$NO_2$ is a strong electron-withdrawing group $(EWG)$ due to both $-I$ and $-M$ effects.
In $para$-nitrophenol,the $-NO_2$ group is at the $para$ position,allowing for strong $-M$ (mesomeric) effect stabilization of the phenoxide ion.
In $ortho$-nitrophenol,intramolecular hydrogen bonding stabilizes the phenol molecule itself,which slightly reduces its acidity compared to $para$-nitrophenol.
$Meta$-nitrophenol only exerts the $-I$ effect,as the $-M$ effect does not operate at the $meta$ position.
Therefore,the order of acidity is: $para$-nitrophenol > $ortho$-nitrophenol > $meta$-nitrophenol > $para$-chlorophenol.
Thus,$para$-nitrophenol is the strongest acid among the given options.

(A) The Kolbe-Schmidt reaction is a base-promoted carboxylation of phenols.
In this reaction,sodium phenoxide is treated with carbon dioxide $(CO_2)$ under pressure,followed by acidification to yield salicylic acid as the final product.

(B) The acidity of a compound depends on the stability of its conjugate base.
In phenol,the phenoxide ion is stabilized by resonance,which makes it more acidic than ethyl alcohol $(C_2H_5OH)$.
Ethyl alcohol is a very weak acid because the ethoxide ion is destabilized by the electron-donating inductive effect of the ethyl group.
Therefore,phenol is more acidic than ethyl alcohol.

(D) The reaction of $phenol$ with $phthalic \ anhydride$ in the presence of concentrated $H_2SO_4$ (acting as a dehydrating agent) is a condensation reaction.
Two molecules of $phenol$ react with one molecule of $phthalic \ anhydride$ to form $phenolphthalein$.
This is a classic synthesis of the acid-base indicator $phenolphthalein$.
Therefore,the correct option is $(D)$.

(C) The acidity of phenols is increased by the presence of electron-withdrawing groups $(EWG)$ such as the nitro group $(-NO_2)$ attached to the benzene ring,as they stabilize the phenoxide ion through inductive and resonance effects.
Conversely,electron-donating groups $(EDG)$ like the methyl group $(-CH_3)$ decrease the acidity of phenol.
Aliphatic alcohols like ethanol and methanol are significantly less acidic than phenol.
Therefore,$o$-nitrophenol is more acidic than phenol.

(B) Cresol is a methylphenol,which consists of a methyl group attached to a benzene ring along with a hydroxyl group $(-OH)$ directly attached to the benzene ring.
Since the $-OH$ group is directly bonded to the aromatic ring,it is classified as a phenolic $-OH$ group.

(A) The reaction described is the $Kolbe-Schmitt$ reaction.
Sodium phenoxide reacts with $CO_2$ at $400\,K$ and $4-7\,atm$ pressure to form sodium phenyl carbonate,which undergoes rearrangement to give sodium salicylate.
This is a standard method for the preparation of salicylic acid derivatives.

(A) The Liebermann's nitroso reaction is a test for phenols.
When phenol is treated with $NaNO_2$ and concentrated $H_2SO_4$,it forms a $p$-nitrosophenol.
This $p$-nitrosophenol reacts with excess phenol to form an indophenol dye.
The colour changes observed during the reaction are:
$1$. Initially,the solution turns red or brown.
$2$. On heating or dilution,it turns green.
$3$. Upon adding water or alkali,it turns deep blue.
Therefore,the sequence is $Red/Brown \to Green \to Deep \ Blue$.

(C) The acidity of phenols depends on the stability of the phenoxide ion formed after the loss of a proton. Electron-withdrawing groups $(EWG)$ like $-NO_2$ increase acidity by stabilizing the phenoxide ion through $-I$ and $-M$ effects,while electron-donating groups $(EDG)$ like $-CH_3$ decrease acidity through $+I$ and hyperconjugation effects.
$1$. $p-$methylphenol: The $-CH_3$ group is an $EDG$,which destabilizes the phenoxide ion,making it the least acidic.
$2$. Phenol: Acts as the reference compound.
$3$. $m-$nitrophenol: The $-NO_2$ group exerts only a $-I$ effect from the meta position,increasing acidity more than phenol.
$4$. $p-$nitrophenol: The $-NO_2$ group exerts both $-I$ and $-M$ effects from the para position,providing the strongest stabilization to the phenoxide ion,making it the most acidic.
Thus,the increasing order of acidity is: $p-$methylphenol < phenol < $m-$nitrophenol < $p-$nitrophenol.

(C) When phenol is treated with bromine water,the $-OH$ group strongly activates the benzene ring towards electrophilic substitution.
Due to this high activation,bromine atoms substitute at all available ortho and para positions,resulting in the formation of $2,4,6-$tribromophenol.
This product is obtained as a white precipitate.
The reaction is: $C_6H_5OH + 3Br_2 (aq) \rightarrow C_6H_2Br_3OH + 3HBr$.

(B) The correct answer is $(B)$.
Phenol $(C_6H_5OH)$ is acidic because the phenoxide ion formed after the loss of a proton is resonance-stabilized by the benzene ring.
Aliphatic alcohols like methanol,ethanol,and isopropyl alcohol are much less acidic than phenol because their corresponding alkoxide ions are not resonance-stabilized.

(C) When phenol is treated with excess bromine water,the highly activating $-OH$ group directs the bromine atoms to all available ortho and para positions,resulting in the formation of $2,4,6$-tribromophenol as a white precipitate.

(C) When phenol is treated with concentrated $HNO_3$,it undergoes electrophilic aromatic substitution (nitration) to form $2,4,6$-trinitrophenol,commonly known as picric acid. The hydroxyl group $(-OH)$ is a strongly activating group,which directs the incoming nitro groups to the ortho and para positions.

(A) The reaction of phenol with ammonia in the presence of anhydrous $ZnCl_2$ at $300 \ ^oC$ is known as the Bucherer reaction or a variation of the ammonolysis of phenols.
In this reaction,the hydroxyl group $(-OH)$ of phenol is replaced by an amino group $(-NH_2)$ to form aniline.
$C_6H_5OH + NH_3 \xrightarrow{ZnCl_2, 300 \ ^oC} C_6H_5NH_2 + H_2O$
Aniline $(C_6H_5NH_2)$ is a primary amine because the nitrogen atom is attached to only one carbon atom of the benzene ring.
Therefore,the correct option is $A$.

(A) The correct answer is $A$.
Phenol undergoes bromination very easily compared to the other compounds listed.
The $-OH$ group is a strongly activating group that increases the electron density of the benzene ring,particularly at the ortho and para positions,through resonance.
This makes the ring highly susceptible to electrophilic aromatic substitution,such as bromination.

(C) $o-$Nitrophenol exhibits intramolecular hydrogen bonding,which restricts the association between its molecules.
In contrast,$p-$nitrophenol and $m-$nitrophenol exhibit intermolecular hydrogen bonding,leading to molecular association and higher boiling points.
Therefore,$o-$nitrophenol has the lowest boiling point among the given options.

(C) Phenol $(C_6H_5OH)$ exhibits intermolecular hydrogen bonding due to the presence of the $-OH$ group.
Because of this hydrogen bonding,phenol has a significantly higher boiling point compared to toluene $(C_6H_5CH_3)$,which only exhibits weak van der Waals forces.
Therefore,statement $(C)$ is correct.

(B) Phenol and benzoic acid can be distinguished by aqueous $NaHCO_3$.
Phenols are very weak acids. They do not react with weak bases such as sodium hydrogen carbonate.
Benzoic acid is a stronger acid than phenol. It reacts with sodium hydrogen carbonate to release effervescence of carbon dioxide gas.
$C_6H_5OH + NaHCO_3 \rightarrow \text{No reaction}$
$C_6H_5COOH + NaHCO_3 \rightarrow C_6H_5COONa + H_2O + CO_2 \uparrow$

(D) Electrophilic substitution reactions in phenol take place at the ortho and para positions.
The $-OH$ group is an ortho-para directing group due to the resonance effect.
This increases the electron density at the ortho and para positions compared to the meta position,making them more susceptible to electrophilic attack.

(D) When phenol is reacted with $NaNO_2$ and concentrated $H_2SO_4$,it provides a deep green or blue color which changes to red on dilution with water.
The generated substance in the presence of $NaOH$ or $KOH$ restores the original green or blue color.
This reaction is termed as Liebermann's nitroso reaction.
Therefore,Liebermann's test is answered by phenol.

(A) Phenols and enols contain a hydroxyl group bonded to an $sp^2$ hybridized carbon atom. These compounds react with neutral $FeCl_3$ solution to form a violet-colored complex. Therefore,enols produce a violet colour with $FeCl_3$ solution.

(B) When phenol is heated with ammonia $(NH_3)$ in the presence of anhydrous zinc chloride $(ZnCl_2)$ at high temperature and pressure,the hydroxyl group $(-OH)$ of the phenol is replaced by an amino group $(-NH_2)$ to form aniline $(C_6H_5NH_2)$.
The reaction is as follows:
$C_6H_5OH + NH_3 \xrightarrow{ZnCl_2} C_6H_5NH_2 + H_2O$
Therefore,the correct option is $(B)$.

(A) In phenol,the lone pair of electrons on the oxygen atom of the $-OH$ group participates in resonance with the benzene ring.
As the oxygen atom donates its lone pair to form a double bond with the carbon atom of the ring,it loses electron density and acquires a positive charge in the contributing resonance structures.

(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 three available ortho and para positions.
This results in the formation of $2,4,6$-trinitrophenol,which is commonly known as Picric acid.

(C) The acidity of a compound is determined by its dissociation constant $(K_a)$.
Phenol $(C_6H_5OH)$ has a $K_a$ value in the range of $10^{-8}$ to $10^{-10}$.
Carbonic acid $(H_2CO_3)$ has a $K_a$ value of approximately $10^{-7}$.
Since the $K_a$ of phenol is lower than that of carbonic acid,phenol is a weaker acid than carbonic acid.

(A) Phenol is an aromatic organic compound which exists as a white crystalline solid at $25 \, ^\circ C$.
It is slightly volatile and often turns pink or red upon exposure to air due to oxidation.

(B) When phenol reacts with $Br_2$ in a non-polar solvent like $CS_2$ at low temperature,the ionization of phenol is suppressed. The benzene ring is only slightly activated,leading to mono-substitution,which yields a mixture of $o$-bromophenol and $p$-bromophenol.
In contrast,when phenol reacts with $Br_2$ water,it ionizes to form the phenoxide ion. The negative charge on the oxygen atom strongly activates the benzene ring,resulting in the formation of $2, 4, 6$-tribromophenol.

(A) Sodium bicarbonate $(NaHCO_3)$ reacts with acids that are stronger than carbonic acid $(H_2CO_3)$ to release $CO_2$ gas,which causes effervescence.
Phenol has a $pK_a$ value of approximately $10$,which is much higher than that of carbonic acid $(pK_a \approx 6.35)$,meaning phenol is a weaker acid and does not react with $NaHCO_3$.
Benzoic acid,$2, 4-$dinitrophenol,and $2, 4, 6-$trinitrophenol (picric acid) are stronger acids than carbonic acid due to the electron-withdrawing effects of the substituents,and thus they all give effervescence with $NaHCO_3$.

(B) Hydrogen bonding occurs when a hydrogen atom is covalently bonded to a highly electronegative atom like oxygen,nitrogen,or fluorine.
In $Phenol$ $(C_6H_5OH)$,the hydrogen atom is directly attached to an oxygen atom,which is highly electronegative. This creates a dipole,allowing the hydrogen atom of one molecule to form a hydrogen bond with the oxygen atom of another molecule.
Toluene,Chlorobenzene,and Nitrobenzene do not have a hydrogen atom bonded to an electronegative atom like $O$,$N$,or $F$ capable of forming intermolecular hydrogen bonds.
Therefore,$Phenol$ exhibits hydrogen bonding.