Preparation of Phenols Questions in English

(A) The given reaction is the industrial method for the preparation of phenol from chlorobenzene,which is known as the $Dow's$ process.
In this process,chlorobenzene is fused with $NaOH$ at a temperature of $623 \ K$ and a pressure of $300 \ atm$ to form sodium phenoxide $(C_6H_5ONa)$,which upon acidification gives phenol.

(D) When benzene diazonium chloride is boiled with dilute sulphuric acid $(H_2SO_4)$,it undergoes hydrolysis to form phenol.
The reaction is as follows:
$C_6H_5N_2Cl + H_2O \xrightarrow{dil. H_2SO_4, \Delta} C_6H_5OH + N_2 + HCl$
Therefore,the correct option is $(D)$.

(A) The reaction of sodium benzene sulphonate with molten $NaOH$ followed by acidification is a standard industrial method for the preparation of phenol.
Step $1$: Sodium benzene sulphonate reacts with $NaOH$ to form sodium phenoxide $(C_6H_5ONa)$ and sodium sulphite $(Na_2SO_3)$.
Step $2$: The resulting sodium phenoxide is then subjected to acidic hydrolysis $(H^+ / H_2O)$ to yield phenol $(C_6H_5OH)$.
Therefore,the correct option is $A$.

(B) Phenol is prepared by boiling an aqueous solution of benzene diazonium chloride with water. The reaction is as follows:
$C_6H_5N_2Cl + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2 + HCl$
Thus,the correct option is $(b)$.

(B) The catalytic oxidation of benzene $(C_6H_6)$ by air in the presence of vanadium pentoxide $(V_2O_5)$ at high temperature is a method for the preparation of phenol $(C_6H_5OH)$.
The reaction is as follows:
$2C_6H_6 + O_2 \xrightarrow{V_2O_5} 2C_6H_5OH$
Therefore,the correct option is $(B)$.

(C) In the Dow process,chlorobenzene is fused with $NaOH$ at a high temperature of $623 \ K$ $(350^{\circ}C)$ and a high pressure of $320 \ atm$ to form sodium phenoxide.
This sodium phenoxide is then acidified with $HCl$ to yield phenol $(C_6H_5OH)$.
The reaction is:
$C_6H_5Cl + 2NaOH \xrightarrow{623 \ K, 320 \ atm} C_6H_5ONa + NaCl + H_2O$
$C_6H_5ONa + HCl \rightarrow C_6H_5OH + NaCl$

(A) The cumene process involves the oxidation of cumene to cumene hydroperoxide,followed by acid-catalyzed cleavage to produce phenol and acetone. Cumene is isopropylbenzene,which is systematically named as $1$-methylethylbenzene.

(C) The given reaction is the Dow process,which is used for the industrial preparation of phenol from chlorobenzene.
In this process,chlorobenzene is fused with $NaOH$ at a high temperature of $623 \ K$ and a high pressure of $300 \ atm$.
This reaction produces sodium phenoxide as an intermediate,which upon acidification with $H^+$ yields phenol $(C_6H_5OH)$ as the final product $A$.

(D) The reaction sequence describes the industrial preparation of phenol from benzene and propene via the cumene process.
$1$. Benzene reacts with propene in the presence of $H_3PO_4$ to form cumene (isopropylbenzene),which is product $A$.
$2$. Cumene is oxidized by $O_2$ to form cumene hydroperoxide.
$3$. Upon treatment with $H_3O^{+}$ and heat,cumene hydroperoxide undergoes rearrangement to yield phenol $(B)$ and acetone $(C)$.

(A) The industrial process for the production of phenol involves the oxidation of $Cumene$ (isopropylbenzene) with $O_2$ in the presence of an acid catalyst.
This reaction forms $Cumene$ hydroperoxide as an intermediate.
$Cumene$ hydroperoxide then undergoes acid-catalyzed rearrangement to yield phenol and acetone as the final products.

(B) $1$. Nitrobenzene reacts with $Sn+HCl$ (a reducing agent) to form aniline $(A)$:
$C_6H_5NO_2 + 6[H] \xrightarrow{Sn+HCl} C_6H_5NH_2 + 2H_2O$
$2$. Aniline $(A)$ reacts with nitrous acid $(HNO_2)$ at $0^{\circ}C-5^{\circ}C$ to form benzene diazonium chloride $(B)$:
$C_6H_5NH_2 + HNO_2 + HCl \xrightarrow{0^{\circ}C-5^{\circ}C} C_6H_5N_2^+Cl^- + 2H_2O$
$3$. Benzene diazonium chloride $(B)$ on heating with water undergoes hydrolysis to form phenol $(C)$:
$C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2 + HCl$
Therefore,the product $C$ is phenol.

(A) The given reaction is the industrial preparation of phenol from cumene (isopropylbenzene).
$1$. In the first step,cumene reacts with oxygen $(O_2)$ to form cumene hydroperoxide $(A)$.
$2$. In the second step,cumene hydroperoxide is treated with dilute acid $(H_2O/H^+)$ to yield phenol $(C_6H_5OH)$ and acetone $(CH_3COCH_3)$.
Therefore,$A$ is cumene hydroperoxide,which has the structure $C_6H_5-C(CH_3)_2-OOH$.

(A) The reaction of chlorobenzene with aqueous $NaOH$ at room temperature does not occur because the $C-Cl$ bond in chlorobenzene has partial double bond character due to resonance,making it resistant to nucleophilic substitution under mild conditions. High temperature and pressure are required for this reaction (Dow process). Therefore,option $A$ will not yield phenol.

(D) The air oxidation of isopropylbenzene (cumene) in the presence of air followed by treatment with dilute acid is the industrial process for the production of phenol and acetone.
$1$. Isopropylbenzene reacts with $O_2$ to form cumene hydroperoxide.
$2$. Cumene hydroperoxide,upon treatment with dilute acid $(H^+)$,undergoes rearrangement to yield phenol $(C_6H_5OH)$ and acetone $(CH_3COCH_3)$.
The reaction is:
$C_6H_5CH(CH_3)_2 + O_2 \xrightarrow{H^+} C_6H_5OH + CH_3COCH_3$
Therefore,the correct option is $D$.

(N/A) The preparation of phenol from cumene involves two steps:
$1$. Oxidation of cumene to cumene hydroperoxide:
$C_6H_5-CH(CH_3)_2 + O_2 \xrightarrow{368-408 \ K} C_6H_5-C(CH_3)_2-OOH$
$2$. Acid-catalyzed cleavage of cumene hydroperoxide to phenol and acetone:
$C_6H_5-C(CH_3)_2-OOH \xrightarrow{H^+, 323-363 \ K} C_6H_5OH + CH_3COCH_3$

(N/A) Chlorobenzene is fused with $NaOH$ (at $623 \, K$ and $320 \, atm$ pressure) to produce sodium phenoxide,which gives phenol on acidification.
The reaction is as follows:
$C_6H_5Cl + 2NaOH \xrightarrow{623 \, K, 320 \, atm} C_6H_5ONa + NaCl + H_2O$
$C_6H_5ONa + HCl \rightarrow C_6H_5OH + NaCl$

(N/A) $1.$ Sulfonation of benzene: $C_6H_6 + conc. H_2SO_4 \xrightarrow{\Delta} C_6H_5SO_3H + H_2O$
$2.$ Fusion with sodium hydroxide: $C_6H_5SO_3H + 2NaOH \xrightarrow{\Delta} C_6H_5ONa + Na_2SO_3 + H_2O$
$3.$ Acidification: $C_6H_5ONa + H^+ \rightarrow C_6H_5OH + Na^+$
(Note: $H_2SO_4$ can also be used for the final acidification step instead of $HCl$ or $H^+$).

(A) Phenol was first isolated from coal tar in the early nineteenth century. Currently,phenol is produced commercially through synthetic methods.
In the laboratory,phenol is prepared from benzene derivatives using the following methods:
$(i)$ From haloarenes: Chlorobenzene is fused with $NaOH$ at $623 \ K$ and $300 \ atm$ pressure to form sodium phenoxide,which upon acidification gives phenol.
$(ii)$ From benzenesulphonic acid: Benzene is sulphonated with oleum $(H_2S_2O_7)$ to form benzenesulphonic acid,which is then converted to sodium phenoxide by fusion with molten $NaOH$,followed by acidification to yield phenol.
$(iii)$ From diazonium salts: $A$ diazonium salt is prepared by treating an aromatic primary amine with nitrous acid $(HNO_2)$ at $273-278 \ K$. The diazonium salt is then hydrolysed to phenol by warming with water or by treating with dilute acids.
$(iv)$ From cumene: Cumene (isopropylbenzene) is oxidised in the presence of air to cumene hydroperoxide,which is then converted to phenol and acetone by treatment with dilute acid.

(N/A) When benzene diazonium chloride $(C_6H_5N_2^+Cl^-)$ is heated with water,it undergoes hydrolysis to form phenol $(C_6H_5OH)$,nitrogen gas $(N_2)$,and hydrochloric acid $(HCl)$.
The chemical reaction is as follows:
$C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2 + HCl$

(B) Carbolic acid is the common name for phenol $(C_6H_5OH)$.
When benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$ is treated with water (hydrolysis),it yields phenol,nitrogen gas $(N_2)$,and hydrochloric acid $(HCl)$.
The reaction is as follows:
$C_6H_5N_2^+Cl^- + H_2O \rightarrow C_6H_5OH + N_2 + HCl$

(D) The reaction shown is the industrial preparation of phenol from cumene-like hydroperoxidation. The starting material $X$ is cyclohexylbenzene. Upon oxidation with $O_2$ in the presence of a catalyst and heat,it forms a hydroperoxide intermediate. Subsequent treatment with aqueous acid leads to the cleavage of the hydroperoxide into phenol and cyclohexanone $(Y)$.
Thus,$X$ is cyclohexylbenzene and $Y$ is cyclohexanone.

(D) The reaction sequence represents the industrial preparation of phenol from cumene (isopropylbenzene).
$1$. Cumene is oxidized in the presence of air $(O_2)$ to form cumene hydroperoxide $(X)$.
$2$. Cumene hydroperoxide is then treated with dilute acid $(H^+, H_2O)$ to undergo rearrangement,yielding phenol $(Y)$ and acetone $(Z)$.

(A) In the industrial preparation of phenol,cumene $(isopropylbenzene)$ is oxidized by air to form cumene hydroperoxide as an intermediate.
This intermediate is then hydrolyzed in the presence of an acid to form phenol and acetone as the final products.

(C) The preparation of phenol from chlorobenzene requires harsh conditions (high temperature and high pressure,e.g.,$623 \ K$ and $300 \ atm$) because the $C-Cl$ bond in chlorobenzene has partial double bond character due to resonance.
Under $STP$ conditions,chlorobenzene does not react with $NaOH$ to form phenol.
Therefore,the reaction sequence involving chlorobenzene and $NaOH$ under $STP$ conditions is incorrect for the preparation of phenol.

(B) The reaction sequence is as follows:
$1$. Aniline reacts with $NaNO_2 + HCl$ at $0-5 \ ^\circ C$ to form benzenediazonium chloride.
$2$. Benzenediazonium chloride reacts with $Cu_2Cl_2$ (Sandmeyer reaction) to form chlorobenzene.
$3$. Chlorobenzene reacts with $NaOH$ at $623 \ K$ and $300 \ atm$ (Dow's process),followed by acidification $(H^+)$,to form phenol.
Therefore,the final product $(A)$ is phenol.

(D) The given reaction is the industrial preparation of phenol from cumene (isopropyl benzene).
$1.$ Cumene (isopropyl benzene) reacts with $O_2$ (air) in the presence of cobalt naphthenate at $423 \ K$ to form cumene hydroperoxide.
$2.$ The cumene hydroperoxide is then treated with dilute $HCl$ and heated to yield phenol and acetone as by-products.
Therefore,the substrate $X$ is isopropyl benzene.

(B) The preparation of phenol from aniline involves two main steps:
$1$. Aniline is treated with $NaNO_2$ and $HCl$ at $0-5 \ ^\circ C$ to form benzene diazonium chloride.
$2$. Benzene diazonium chloride is then hydrolyzed by boiling with water to produce phenol.
Therefore,the intermediate product is benzene diazonium chloride.

(D) The preparation of phenol involves specific industrial and laboratory methods.
$A$. Aniline can be converted to phenol via benzene diazonium chloride (diazotization followed by hydrolysis). This is a standard method.
$B$. Cumene oxidation is the most common industrial method for producing phenol.
$C$. The Dow process involves the reaction of chlorobenzene with $NaOH$ at high temperature and pressure to form sodium phenoxide,which upon acidification yields phenol. This is a standard method.
$D$. Benzene reacts with concentrated $H_2SO_4$ to form benzene sulfonic acid,not phenol directly. Phenol is obtained from benzene sulfonic acid by fusion with $NaOH$ followed by acidification. Therefore,the reaction as written is incorrect.

(A) The air oxidation of cumene (isopropylbenzene) in the presence of cobalt naphthenate as a catalyst leads to the formation of cumene hydroperoxide.
This is the first step in the industrial preparation of phenol and acetone from cumene.
The reaction is:
$C_6H_5-CH(CH_3)_2 + O_2 \xrightarrow{\text{Co-naphthenate}} C_6H_5-C(CH_3)_2OOH$
Thus,the product formed is cumene hydroperoxide.

(A) In the Raschig process,chlorobenzene is hydrolyzed to phenol by reacting it with steam at a high temperature of $698 \ K$ in the presence of a catalyst.
The catalyst used in this reaction is calcium phosphate,represented by the chemical formula $Ca_3(PO_4)_2$.
The reaction is as follows:
$C_6H_5Cl + H_2O \xrightarrow{698 \ K, \ Ca_3(PO_4)_2} C_6H_5OH + HCl$
Therefore,the correct option is $A$.

(D) In the commercial preparation of phenol from cumene (isopropylbenzene),the process involves oxidation to cumene hydroperoxide followed by acid-catalyzed cleavage. However,in the specific industrial oxidation of toluene to benzoic acid and subsequently to phenol,or in the oxidation of benzene,$Cobalt \ naphthenate$ is commonly used as a catalyst.

(B) In Dow's process,chlorobenzene is treated with $NaOH$ at $573-623 \ K$ and $200 \ atm$ pressure to form sodium phenoxide,which upon subsequent acid hydrolysis with $HCl$ yields phenol as the final product.
The reaction sequence is:
$C_6H_5Cl + 2NaOH \xrightarrow{573-623 \ K, 200 \ atm} C_6H_5ONa + NaCl + H_2O$
$C_6H_5ONa + HCl \rightarrow C_6H_5OH + NaCl$
Thus,the starting raw material is chlorobenzene.

(D) The reaction sequence is the industrial preparation of phenol from cumene.
Step $1$: Benzene reacts with propene in the presence of $AlCl_3/\text{ether}$ and heat to form isopropylbenzene,which is also known as cumene $(X)$.
Step $2$: Cumene $(X)$ is then oxidized by $O_2$ at $130^{\circ}C$ to form cumene hydroperoxide,which upon further treatment with acid yields phenol and acetone $(Z)$.
Therefore,$X$ is isopropylbenzene (cumene) and $Z$ represents the final products,phenol and acetone.

(D) Cumene (isopropylbenzene) is oxidised in the presence of air to form cumene hydroperoxide.
On treating with dilute acid,it undergoes rearrangement to form phenol and acetone (propan$-2-$one) as a by-product.

(C) The process described is the industrial preparation of phenol from cumene (isopropyl benzene).
$1$. Isopropyl benzene (cumene) undergoes aerial oxidation to form cumene hydroperoxide.
$2$. Cumene hydroperoxide,upon treatment with dilute acid (acid hydrolysis),decomposes to give phenol $(C_6H_5OH)$ and acetone $((CH_3)_2CO)$.

(A) The given reaction sequence is the industrial method for the preparation of phenol,known as the cumene process.
Step $(i)$: Benzene reacts with isopropyl chloride in the presence of anhydrous $AlCl_3$ (Friedel-Crafts alkylation) to form cumene (isopropylbenzene).
Step (ii): Cumene is oxidized by $O_2$ to form cumene hydroperoxide.
Step (iii): Acid-catalyzed hydrolysis of cumene hydroperoxide yields phenol and acetone $(CH_3COCH_3)$ as products.
Therefore,the major organic product formed is phenol.

(D) The industrial preparation of phenol is known as the cumene process.
In this process,cumene (isopropylbenzene) is oxidized in the presence of air to form cumene hydroperoxide.
This hydroperoxide is then treated with dilute acid to yield phenol and acetone as a byproduct.
The reaction is:
$C_6H_5CH(CH_3)_2$ $\xrightarrow{O_2} C_6H_5C(CH_3)_2OOH$ $\xrightarrow{H^+, H_2O} C_6H_5OH + CH_3COCH_3$
Thus,the compound $X$ is cumene.

(A) The reaction sequence represents the industrial preparation of phenol from cumene (isopropylbenzene).
$1$. Cumene reacts with $O_2$ (air) to form cumene hydroperoxide $(A)$.
$2$. Cumene hydroperoxide,upon acidic hydrolysis $(H^+ / H_2O)$,undergoes rearrangement to yield phenol $(B)$ and acetone $(C)$.
Reaction:
$C_6H_5-CH(CH_3)_2 + O_2 \rightarrow C_6H_5-C(CH_3)_2-OOH (A)$
$C_6H_5-C(CH_3)_2-OOH \xrightarrow{H^+ / H_2O} C_6H_5OH (B) + CH_3COCH_3 (C)$

(B) Step $1$: Aniline reacts with $NaNO_2$ and $HCl$ at $273-278 \ K$ to form benzene diazonium chloride.
Step $2$: Benzene diazonium chloride,when treated with warm water,undergoes hydrolysis to form phenol,$N_2$ gas,and $HCl$.

(C) The reaction of benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$ with boiling water is a standard method for the preparation of phenol.
The reaction proceeds as follows:
$C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2 \uparrow + HCl$
Here,the diazonium group is replaced by a hydroxyl group $(-OH)$,resulting in the formation of phenol.