Preparation of Carboxylic Acids and Their Derivatives Questions in English

(A) The oxidation of alkylbenzenes (like ethylbenzene) with alkaline $KMnO_4$ followed by acidification results in the oxidation of the alkyl side chain to a carboxylic acid group,regardless of the length of the alkyl chain,provided that at least one benzylic hydrogen is present.
In the case of ethylbenzene $(C_6H_5CH_2CH_3)$,the side chain is oxidized to a carboxyl group $(-COOH)$,yielding benzoic acid $(C_6H_5COOH)$.

(C) The oxidation of alkylbenzenes (like toluene) to benzoic acid is a standard reaction in organic chemistry.
When toluene $(C_6H_5CH_3)$ is treated with an alkaline or acidic solution of potassium permanganate $(KMnO_4)$,the methyl group is oxidized to a carboxyl group $(-COOH)$,resulting in the formation of benzoic acid $(C_6H_5COOH)$.
Therefore,the correct reagent is $KMnO_4$.

(D) The oxidation of toluene with alkaline $KMnO_4$ followed by acidification results in the oxidation of the methyl group $(-CH_3)$ attached to the benzene ring into a carboxylic acid group $(-COOH)$.
Thus,toluene is converted into benzoic acid $(C_6H_5COOH)$.

(D) When toluene reacts with $Cl_2$ in the presence of light and heat (free radical substitution),it undergoes side-chain chlorination to form benzotrichloride $(C_6H_5CCl_3)$.
When benzotrichloride is treated with aqueous $NaOH$,the $CCl_3$ group is hydrolyzed to a carboxylic acid group,resulting in the formation of benzoic acid $(C_6H_5COOH)$.

(B) The $Koch$ reaction is the industrial process for the production of carboxylic acids from alkenes using carbon monoxide $(CO)$ and water $(H_2O)$ in the presence of a strong acid catalyst (like $H_2SO_4$ or $H_3PO_4$).

(D) Toluene $(C_6H_5CH_3)$ can be oxidized to benzoic acid $(C_6H_5COOH)$ using strong oxidizing agents such as alkaline potassium permanganate $(KMnO_4)$ or acidic potassium dichromate $(K_2Cr_2O_7)$.
Both reagents are capable of oxidizing the alkyl side chain attached to the benzene ring to a carboxylic acid group.
Therefore,the correct option is $(d)$.

(B) The reaction of ethylene $(C_2H_4)$ with carbon monoxide $(CO)$ and water $(H_2O)$ at high temperature and pressure is known as the hydrocarboxylation reaction.
The chemical equation is: $C_2H_4 + CO + H_2O \xrightarrow{\text{High temp.}} C_2H_5COOH$.
The product formed is propionic acid $(C_2H_5COOH)$.

(C) When phenol reacts with $CCl_4$ in the presence of aqueous $NaOH$,the reaction is a variation of the Riemer-Tiemann reaction.
Instead of forming an aldehyde group (as with $CHCl_3$),the use of $CCl_4$ results in the introduction of a carboxylic acid group $(-COOH)$ at the ortho position of the phenol.
The reaction is:
$C_6H_5OH + CCl_4 + 4NaOH \rightarrow C_6H_4(OH)(COOH) + 4NaCl + 2H_2O$.
The product formed is Salicylic acid.

(B) The reaction proceeds as follows:
$CH_3CH_2Br + KCN \rightarrow CH_3CH_2CN + KBr$
The intermediate $CH_3CH_2CN$ (propanenitrile) undergoes acid hydrolysis $(HOH/H^+)$ to form a carboxylic acid.
$CH_3CH_2CN + 2H_2O \xrightarrow{H^+} CH_3CH_2COOH + NH_3$
The product $X$ is $CH_3CH_2COOH$,which is propionic acid.

(A) When glycerol is heated with oxalic acid at $110^\circ C$,it forms glycerol monooxalate.
This intermediate undergoes decarboxylation to produce glycerol monoformate.
Finally,the hydrolysis of glycerol monoformate yields formic acid $(HCOOH)$ and regenerates glycerol.

(D) The oxidation of ethyl alcohol $(C_2H_5OH)$ in the presence of air and $Acetobacter$ $aceti$ bacteria leads to the formation of acetic acid $(CH_3COOH)$.
$C_2H_5OH + O_2 \xrightarrow{\text{Acetobacter aceti}} CH_3COOH + H_2O$
Therefore,the correct option is $D$.

(B) Aldehydes $(R-CHO)$ are easily oxidized to carboxylic acids $(R-COOH)$ containing the same number of carbon atoms. This is because the hydrogen atom attached to the carbonyl carbon in an aldehyde can be easily oxidized to a hydroxyl group.

(D) The hydrolysis of nitroalkanes with concentrated mineral acids like $85\%$ $H_2SO_4$ is a standard method for the preparation of carboxylic acids.
The reaction proceeds as follows:
$CH_3CH_2NO_2 + H_2O \xrightarrow{H_2SO_4} CH_3COOH + NH_2OH$
Thus,the product obtained is acetic acid $(CH_3COOH)$.

(A) In the laboratory,the reaction of $CH_3COOH$ with $SOCl_2$ (thionyl chloride) is preferred for the preparation of acetyl chloride $(CH_3COCl)$.
This is because the by-products,$SO_2$ and $HCl$,are gases that escape easily,leaving behind pure $CH_3COCl$.
The reaction is: $CH_3COOH + SOCl_2 \to CH_3COCl + SO_2 \uparrow + HCl \uparrow$.

(C) The oxidation of acetaldehyde $(CH_3CHO)$ using potassium dichromate $(K_2Cr_2O_7)$ in the presence of sulphuric acid $(H_2SO_4)$ yields acetic acid $(CH_3COOH)$.
Reaction: $CH_3CHO \xrightarrow{K_2Cr_2O_7 / H_2SO_4} CH_3COOH$

(A) The industrial production of acetic acid via fermentation involves the oxidation of ethanol using bacteria such as $Acetobacter \text{ } aceti$.
$C_2H_5OH + O_2 \xrightarrow{\text{Acetobacter}} CH_3COOH + H_2O$
Therefore,the correct option is $A$.

(A) The hydrolysis of nitriles $(R-CN)$ in the presence of an acid or base catalyst yields carboxylic acids.
$CH_3CN + 2H_2O \xrightarrow{H^+} CH_3COOH + NH_3$
Therefore,$CH_3CN$ (acetonitrile) on hydrolysis forms acetic acid $(CH_3COOH)$.

(D) Benzyl chloride $(C_6H_5CH_2Cl)$ on oxidation with strong oxidizing agents like acidic $K_2Cr_2O_7$ undergoes oxidation to form benzoic acid $(C_6H_5COOH)$.
The reaction is as follows:
$C_6H_5CH_2Cl + [O] \xrightarrow{K_2Cr_2O_7 / H_2SO_4} C_6H_5COOH + HCl$
Therefore,the correct option is $(D)$.

(C) The correct option is $(C)$. When glycerol is heated with oxalic acid at $110^\circ C$,formic acid is produced.
The reaction proceeds via the formation of glycerol monooxalate and glycerol monoformate intermediates.
The overall reaction is:
$C_3H_5(OH)_3 + (COOH)_2 \xrightarrow{110^\circ C} HCOOH + CO_2 + C_3H_5(OH)_3$

(B) The reaction of an alkene with $CO$ and $H_2O$ in the presence of an acid catalyst (like $H_3PO_4$) at high temperature and pressure to form a carboxylic acid is known as the Koch reaction.

(B) The reaction of a Grignard reagent $(C_6H_5MgBr)$ with carbon dioxide $(CO_2)$ followed by acid hydrolysis $(H_3O^+)$ is a standard method for the preparation of carboxylic acids.
The reaction proceeds as follows:
$C_6H_5MgBr + CO_2 \rightarrow C_6H_5COOMgBr$
$C_6H_5COOMgBr + H_3O^+ \rightarrow C_6H_5COOH + Mg(OH)Br$
Thus,the product $P$ is benzoic acid $(C_6H_5COOH)$.

(B) Acetic acid freezes at $16.6 \ ^oC$,while water freezes at $0 \ ^oC$.
Thus,glacial acetic acid is obtained by cooling acetic acid until it crystallizes,separating the solid crystals from the liquid,and then melting the pure crystals.

(A) Propyl magnesium bromide $(C_3H_7MgBr)$ reacts with carbon dioxide $(CO_2)$ to form an intermediate adduct,which upon acidic hydrolysis yields butanoic acid $(C_3H_7COOH)$.
$C_3H_7MgBr + CO_2 \rightarrow C_3H_7COOMgBr$
$C_3H_7COOMgBr + H_3O^{+} \rightarrow C_3H_7COOH + Mg(OH)Br$

(B) The reaction of a Grignard reagent $(R-MgX)$ with carbon dioxide $(CO_2)$ followed by acid hydrolysis yields a carboxylic acid with one additional carbon atom.
Ethyl magnesium bromide is $CH_3CH_2MgBr$.
The reaction is:
$CH_3CH_2MgBr + CO_2 \rightarrow CH_3CH_2COOMgBr$
$CH_3CH_2COOMgBr + H_2O/H^+ \rightarrow CH_3CH_2COOH + Mg(OH)Br$
The product formed is propanoic acid $(CH_3CH_2COOH)$.

(B) $CO_2$ reacts with Grignard reagent to yield carboxylic acids.
$CO_2 + CH_3MgI \rightarrow CH_3COOMgI$
$CH_3COOMgI + H_2O \xrightarrow{H^+} CH_3COOH + Mg(OH)I$
Therefore,the compound $X$ is $CO_2$.

(A) The reaction of an amide with nitrous acid $(HNO_2)$,generated in situ by $NaNO_2/HCl$,results in the formation of a carboxylic acid,nitrogen gas,and water.
The chemical equation is: $CH_3CONH_2 + HNO_2 \rightarrow CH_3COOH + N_2 + H_2O$.
Therefore,the product $X$ is $CH_3COOH$.

(A) The oxidation of primary alcohols with strong oxidizing agents like $KMnO_4$ or $K_2Cr_2O_7$ yields carboxylic acids with the same number of carbon atoms.
Ethanol $(C_2H_5OH)$ contains two carbon atoms.
Upon oxidation,it forms acetic acid $(CH_3COOH)$,which also contains two carbon atoms.
The reaction is: $C_2H_5OH + 2[O] \xrightarrow{Oxidation} CH_3COOH + H_2O$.

(C) The reaction sequence is as follows:
$1$. $R - X + KCN \rightarrow R - CN + KX$ (Nucleophilic substitution to form an alkyl cyanide $A$).
$2$. $R - CN + NaOH + H_2O \rightarrow R - COONa + NH_3$ (Alkaline hydrolysis of cyanide to form the sodium salt of carboxylic acid $B$).

(C) The reaction between acetic acid $(CH_3COOH)$ and phosphorus trichloride $(PCl_3)$ is a standard method for the preparation of acid chlorides.
The balanced chemical equation is:
$3CH_3COOH + PCl_3 \to 3CH_3COCl + H_3PO_3$
Thus,the product obtained is acetyl chloride $(CH_3COCl)$.

(A) The conversion of an alcohol to a carboxylic acid with an additional carbon atom involves the following steps:
$1$. Conversion of alcohol to alkyl bromide using $PBr_3$: $R-CH_2-CH_2OH \xrightarrow{PBr_3} R-CH_2-CH_2-Br$
$2$. Nucleophilic substitution with $KCN$ to form a nitrile: $R-CH_2-CH_2-Br \xrightarrow{KCN} R-CH_2-CH_2-CN$
$3$. Acidic hydrolysis of the nitrile to form a carboxylic acid: $R-CH_2-CH_2-CN \xrightarrow{H_3O^{+}} R-CH_2-CH_2-COOH + NH_3$
Therefore,the correct sequence is $PBr_3, KCN, H_3O^{+}$.

(C) Benzoic acid reacts with thionyl chloride $(SOCl_2)$ to form benzoyl chloride. The reaction is as follows:
$C_6H_5COOH + SOCl_2 \rightarrow C_6H_5COCl + SO_2 + HCl$
This method is preferred because the by-products ($SO_2$ and $HCl$) are gases,which escape,leaving the pure product behind.

(C) The reaction of a Grignard reagent $(RMgX)$ with carbon dioxide $(CO_2)$ followed by acid hydrolysis yields a carboxylic acid with one more carbon atom than the original alkyl group $(R)$:
$RMgX + CO_2$ $\rightarrow RCOOMgX$ $\xrightarrow{H_3O^+} RCOOH + Mg(OH)X$
To prepare formic acid $(HCOOH)$,the $R$ group would need to be a hydrogen atom $(H)$. However,a Grignard reagent cannot be formed with a hydrogen atom ($HMgX$ does not exist). Therefore,formic acid cannot be prepared using this method. Acetic acid $(CH_3COOH)$ can be prepared using methylmagnesium bromide $(CH_3MgBr)$,and succinic acid can be prepared using appropriate di-Grignard reagents.

(C) The hydrolysis of alkyl cyanides $(R-CN)$ in the presence of an acid or base catalyst proceeds as follows:
$R-CN + 2H_2O \rightarrow R-COOH + NH_3$
During this reaction,the nitrogen atom of the cyanide group is released as ammonia $(NH_3)$ gas.

(D) The hydrolysis of $methyl \ cyanide$ $(CH_3CN)$ in the presence of an alkali (like $NaOH$) proceeds through the formation of $acetamide$ $(CH_3CONH_2)$ as an intermediate,which further undergoes hydrolysis to form $acetate$ ions. Upon acidification,this yields $acetic \ acid$ $(CH_3COOH)$.
The reaction is:
$CH_3-C \equiv N + 2H_2O \xrightarrow{OH^-} CH_3COOH + NH_3$

(C) $CH_3CN \xrightarrow{Na + C_2H_5OH} CH_3CH_2NH_2$ ($X$ is ethylamine).
$CH_3CH_2NH_2 \xrightarrow{HNO_2} CH_3CH_2OH$ ($Y$ is ethanol).
$CH_3CH_2OH \xrightarrow[H_2SO_4]{K_2Cr_2O_7} CH_3COOH$ ($Z$ is acetic acid).

(A) The acid-catalyzed hydrolysis of methyl cyanide $(CH_3CN)$ proceeds as follows:
$CH_3CN + 2H_2O \xrightarrow{H^+} CH_3COOH + NH_3$
Thus,the product formed is acetic acid.

(B) The hydrolysis of acetonitrile $(CH_3CN)$ in an acidic medium proceeds as follows:
$CH_3CN + 2H_2O \xrightarrow{H^+} CH_3COOH + NH_4^+$
Thus,the product formed is acetic acid $(CH_3COOH)$.

(B) $C_6H_5NH_2 \xrightarrow{NaNO_2/HCl, 0-5^{\circ}C} \underset{(X)}{C_6H_5N_2^{+}Cl^{-}}$ (Benzenediazonium chloride)
$\underset{(X)}{C_6H_5N_2^{+}Cl^{-}} \xrightarrow{Cu_2(CN)_2} \underset{(Y)}{C_6H_5CN}$ (Benzonitrile)
$\underset{(Y)}{C_6H_5CN} \xrightarrow{H_2O/H^{+}} \underset{(Z)}{C_6H_5COOH}$ (Benzoic acid)
Thus,the final product $Z$ is $C_6H_5 - COOH$.

(D) The partial hydrolysis of nitriles $(R-CN)$ with cold concentrated $HCl$ or $H_2SO_4$ yields amides $(R-CONH_2)$.
For acetonitrile $(CH_3CN)$: $CH_3CN + H_2O \xrightarrow{conc. HCl, \text{cold}} CH_3CONH_2$ (Acetamide).
Complete hydrolysis at higher temperatures would yield acetic acid $(CH_3COOH)$.

(A) The hydrolysis of methyl cyanide $(CH_3CN)$ in the presence of an acid proceeds as follows:
$CH_3CN + 2H_2O \xrightarrow{H^+} CH_3COOH + NH_3$
Thus,the product obtained is acetic acid $(CH_3COOH)$.

(A) The hydrolysis of cyanide $(R-CN)$ proceeds in two steps.
In the first step,partial hydrolysis occurs to form an amide $(R-CONH_2)$.
In the second step,further hydrolysis leads to the formation of a carboxylic acid $(R-COOH)$ and ammonia $(NH_3)$.
The reaction is: $R-CN$ $\xrightarrow{H_2O, H^+} R-CONH_2$ $\xrightarrow{H_2O, H^+} R-COOH + NH_3$.
Thus,the initial product is a primary amide.

(B) The given reaction is known as the Koch reaction (also known as the Koch carbonylation of alkenes). In this process,an alkene reacts with carbon monoxide and water in the presence of an acid catalyst (like $H_3PO_4$ or $H_2SO_4$) to form a carboxylic acid. The reaction is: $CH_3CH=CH_2 + CO + H_2O \xrightarrow{H^+} CH_3-CH(COOH)-CH_3$.

(C) When alkylbenzenes (like ethylbenzene) are treated with strong oxidizing agents such as alkaline $KMnO_4$ followed by acidic hydrolysis,the alkyl side chain is oxidized to a carboxylic acid group $(-COOH)$ attached to the benzene ring,regardless of the length of the alkyl chain.
Therefore,ethylbenzene $(C_6H_5CH_2CH_3)$ is oxidized to benzoic acid $(C_6H_5COOH)$.

(B) The reaction of an alkene with carbon monoxide $(CO)$ and water $(H_2O)$ in the presence of an acid catalyst $(H^+)$ to form a carboxylic acid is known as the Koch reaction (or hydrocarboxylation).
In this specific case,propene reacts to form isobutyric acid.

(B) The reaction of a Grignard reagent $(C_6H_5MgBr)$ with carbon dioxide $(CO_2)$ followed by acid hydrolysis $(H_3O^+)$ is a standard method for the preparation of carboxylic acids.
Step $1$: The nucleophilic phenyl group $(C_6H_5^-)$ from the Grignard reagent attacks the electrophilic carbon atom of $CO_2$ to form a magnesium carboxylate intermediate $(C_6H_5COOMgBr)$.
Step $2$: Subsequent acid hydrolysis $(H_3O^+)$ converts the magnesium carboxylate into benzoic acid $(C_6H_5COOH)$.

(C) In the reactant $CH_3CN$ (acetonitrile),the functional carbon atom is the nitrile carbon $(C \equiv N)$. The hybridization of this carbon is $sp$ because it forms two sigma bonds and two pi bonds.
In the product $CH_3COOH$ (acetic acid),the functional carbon atom is the carboxyl carbon $(C=O)$. The hybridization of this carbon is $sp^2$ because it forms three sigma bonds and one pi bond.
Therefore,the hybridization changes from $sp$ to $sp^2$.

(A) The conversion of an alcohol to a carboxylic acid with an additional carbon atom involves the following steps:
$1$. Conversion of alcohol to alkyl bromide using $PBr_3$: $R-CH_2-CH_2-OH \xrightarrow{PBr_3} R-CH_2-CH_2-Br$
$2$. Nucleophilic substitution with $KCN$ to form a nitrile: $R-CH_2-CH_2-Br \xrightarrow{KCN} R-CH_2-CH_2-CN$
$3$. Acidic hydrolysis of the nitrile to a carboxylic acid: $R-CH_2-CH_2-CN \xrightarrow{H_3O^+} R-CH_2-CH_2-COOH$
Thus,the correct sequence is $PBr_3, KCN, H_3O^+$.

(B) When glycerol is heated with oxalic acid at $110\,^{\circ}C$,it forms glycerol mono-oxalate,which upon further heating loses $CO_2$ to form glycerol mono-formate. Hydrolysis of glycerol mono-formate yields glycerol and formic acid ($HCOOH$,also known as methanoic acid).
Reaction sequence:
$Glycerol + Oxalic \, acid \xrightarrow{110\,^{\circ}C, -H_2O} Glycerol \, mono-oxalate$
$Glycerol \, mono-oxalate \xrightarrow{\Delta, -CO_2} Glycerol \, mono-formate$
$Glycerol \, mono-formate \xrightarrow{H_2O} Glycerol + HCOOH$ (Methanoic acid)

(B) $1$. $CH \equiv CH + CH_3MgBr \rightarrow CH \equiv C-MgBr + CH_4$
$2$. $CH \equiv C-MgBr + CO_2 \xrightarrow{H_3O^+} CH \equiv C-COOH$ (Propiolic acid)
$3$. $CH \equiv C-COOH + H_2O \xrightarrow{HgSO_4/H_2SO_4} [CH_2=C(OH)-COOH] \rightleftharpoons OHC-CH_2-COOH$ (Formylacetic acid)
$4$. $OHC-CH_2-COOH \xrightarrow{Ag_2O, \Delta} HOOC-CH_2-COOH$ (Malonic acid)
Thus,the final product is $CH_2(COOH)_2$.

(C) Formic acid is prepared by heating glycerol with oxalic acid at $110\,^{\circ}C$.
$C_3H_5(OH)_3 + (COOH)_2 \xrightarrow{110\,^{\circ}C} HCOOH + CO_2 + C_3H_5(OH)_3$
In this reaction,glycerol acts as a catalyst and is regenerated at the end of the process.