Preparation of Carboxylic Acids and Their Derivatives Questions in English

(B) The conversion of methanol $(CH_3OH)$ to acetic acid $(CH_3COOH)$ is an industrial process known as the Monsanto process or carbonylation of methanol.
In this reaction,methanol reacts with carbon monoxide $(CO)$ in the presence of a rhodium $(Rh)$ catalyst to produce acetic acid.
The chemical equation is: $CH_3OH + CO \xrightarrow{Rh} CH_3COOH$.

(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 $(C_6H_5COOH)$.

(C) Acetic acid is prepared by the oxidation of acetaldehyde using an oxidizing agent like potassium dichromate $(K_2Cr_2O_7)$ in the presence of sulfuric acid $(H_2SO_4)$.
The chemical reaction is:
$CH_3CHO + [O] \xrightarrow{K_2Cr_2O_7 + H_2SO_4} CH_3COOH$
Acetaldehyde $\rightarrow$ Acetic acid

(A) The $Hell-Volhard-Zelinsky$ $(HVZ)$ reaction is the most suitable method for the preparation of $\alpha$-halocarboxylic acids,such as $\alpha$-chloroacetic acid,by reacting a carboxylic acid containing an $\alpha$-hydrogen with $Cl_2$ or $Br_2$ in the presence of a small amount of red phosphorus.

(C) Toluene can be oxidized to benzoic acid using strong oxidizing agents like alkaline $KMnO_4$ (followed by acidification) or acidic $KMnO_4$ or acidic $K_2Cr_2O_7$.
Both alkaline $KMnO_4$ and acidic $KMnO_4$ are effective reagents for this transformation.
Therefore,the correct answer is both $A$ and $B$.

(C) During the oxidation of alkylbenzenes with alkaline $KMnO_4$,the entire side chain is oxidized to a carboxyl group $(-COOH)$,regardless of the length of the side chain,provided that the benzylic carbon has at least one hydrogen atom. In this reaction,ethylbenzene is oxidized to benzoic acid.

(D) $1$. Chlorination of toluene in the presence of light $(h\nu)$ and heat $(\Delta)$ leads to the formation of benzotrichloride $(C_6H_5CCl_3)$.
$2$. When benzotrichloride is treated with aqueous $NaOH$,the three chlorine atoms are replaced by hydroxyl groups to form an unstable intermediate,phenylmethanetriol $(C_6H_5C(OH)_3)$.
$3$. This intermediate loses a water molecule to form benzoic acid $(C_6H_5COOH)$,which exists as sodium benzoate $(C_6H_5COONa)$ in the presence of excess $NaOH$. Upon acidification,it yields benzoic acid.

(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: $R-CH_2-CH_2OH \xrightarrow{PBr_3} R-CH_2-CH_2Br$
$2$. Nucleophilic substitution with cyanide: $R-CH_2-CH_2Br \xrightarrow{KCN} R-CH_2-CH_2CN$
$3$. Acidic hydrolysis of nitrile to carboxylic acid: $R-CH_2-CH_2CN \xrightarrow{H_3O^+} R-CH_2-CH_2COOH$
Thus,the correct sequence of reagents is $PBr_3, KCN, H_3O^+$.

(B) The freezing point of pure acetic acid is $16.6 \ ^{\circ}C$.
It is obtained by cooling acetic acid until it crystallizes,separating the crystals,and then melting them to obtain pure glacial acetic acid.

(A) Step $1$: Reduction of acetonitrile $(CH_3CN)$ with $Na/C_2H_5OH$ (Mendius reduction) gives ethylamine $(A)$:
$CH_3CN + 4[H] \xrightarrow{Na/C_2H_5OH} CH_3CH_2NH_2$ $(A)$.
Step $2$: Reaction of ethylamine $(A)$ with nitrous acid $(HNO_2)$ gives ethanol $(B)$:
$CH_3CH_2NH_2 + HNO_2 \rightarrow CH_3CH_2OH + N_2 + H_2O$ $(B)$.
Step $3$: Oxidation of ethanol $(B)$ with potassium dichromate $(Cr_2O_7^{2-}/H^+)$ gives acetic acid $(C)$:
$CH_3CH_2OH \xrightarrow{[O]} CH_3COOH$ $(C)$.

(A) The synthesis of pyruvic acid involves the reaction of acetaldehyde with hydrogen cyanide to form acetaldehyde cyanohydrin.
$CH_3CHO + HCN \to CH_3CH(OH)CN$
Hydrolysis of the cyanohydrin yields lactic acid.
$CH_3CH(OH)CN + 2H_2O \to CH_3CH(OH)COOH + NH_3$
Finally,the oxidation of the secondary alcohol group in lactic acid gives pyruvic acid.
$CH_3CH(OH)COOH + [O] \to CH_3COCOOH + H_2O$

(D) When alkyl benzenes (like ethyl benzene) are treated with strong oxidizing agents such as alkaline $KMnO_4$,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,provided that the benzylic carbon has at least one hydrogen atom.
In the case of ethyl benzene $(C_6H_5-CH_2-CH_3)$,the benzylic carbon has two hydrogen atoms,so it undergoes oxidation to form benzoic acid $(C_6H_5-COOH)$.

(A) The conversion of an alcohol to a carboxylic acid with one additional carbon atom involves the following steps:
$1$. Conversion of alcohol to alkyl bromide: $RCH_2CH_2OH \xrightarrow{PBr_3} RCH_2CH_2Br$
$2$. Nucleophilic substitution with cyanide: $RCH_2CH_2Br \xrightarrow{KCN} RCH_2CH_2CN$
$3$. Acidic hydrolysis of nitrile to carboxylic acid: $RCH_2CH_2CN \xrightarrow{H_3O^{+}} RCH_2CH_2COOH$
Thus,the correct sequence is $X$ $\xrightarrow{PBr_3}$ $\xrightarrow{KCN}$ $\xrightarrow{H_3O^{+}} Y$.

(C) The reaction sequence is as follows:
$1$. $X$ is $Toluene$ $(C_6H_5CH_3)$.
$2$. $Toluene$ reacts with $Cl_2$ in the presence of light (free radical chlorination) to form $Benzotrichloride$ $(C_6H_5CCl_3)$.
$3$. $Benzotrichloride$ undergoes hydrolysis to form $Benzoic \ acid$ $(C_6H_5COOH)$.
Therefore,$X$ is $Toluene$ and $Y$ is $Benzoic \ acid$.

(B) $1.$ $\alpha$-Bromination of acetic acid using $Br_2$ and $\text{Red } P$ ($HVZ$ reaction): $CH_3COOH \xrightarrow{Br_2/\text{Red } P} BrCH_2COOH$.
$2.$ Substitution of the bromine atom with a cyano group using $KCN$: $BrCH_2COOH \xrightarrow{KCN} NCCH_2COOH$.
$3.$ Acidic hydrolysis of the cyano group to a carboxyl group: $NCCH_2COOH \xrightarrow{H_3O^{+}} HOOCCH_2COOH$ (Malonic acid).

(B) $2,2-$dimethylpropanoic acid is $(CH_3)_3CCOOH$.
To synthesize this,we need a tertiary alkyl halide precursor,specifically $2-$bromo$-2-$methylpropane $(CH_3)_3CBr$.
Starting with acetone $(CH_3)_2C=O$:
$1.$ Reaction with $CH_3MgBr$ followed by $H^+$ gives $2-$methylpropan$-2-$ol $(CH_3)_3COH$.
$2.$ Reaction with $SOBr_2$ converts the alcohol to $2-$bromo$-2-$methylpropane $(CH_3)_3CBr$.
$3.$ Reaction with $Mg$ in $Et_2O$ forms the Grignard reagent $(CH_3)_3CMgBr$.
$4.$ Reaction with $CO_2$ followed by $H^+$ yields $2,2-$dimethylpropanoic acid $(CH_3)_3CCOOH$.
Thus,option $B$ is the correct synthesis.

(D) The reaction sequence is as follows:
$1.$ Reduction: $CH_3-CH(CH_3)-COOH \xrightarrow{LiAlH_4/H_3O^{+}} CH_3-CH(CH_3)-CH_2OH$ $(I)$
$2.$ Bromination: $CH_3-CH(CH_3)-CH_2OH \xrightarrow{PBr_3} CH_3-CH(CH_3)-CH_2Br$ $(II)$
$3.$ Grignard formation: $CH_3-CH(CH_3)-CH_2Br \xrightarrow{Mg/ether} CH_3-CH(CH_3)-CH_2MgBr$ $(III)$
$4.$ Carboxylation: $CH_3-CH(CH_3)-CH_2MgBr \xrightarrow{CO_2} CH_3-CH(CH_3)-CH_2COOMgBr$ $(IV)$
$5.$ Hydrolysis: $CH_3-CH(CH_3)-CH_2COOMgBr \xrightarrow{H_3O^{+}} CH_3-CH(CH_3)-CH_2COOH$ $(V)$
The final product $V$ is $3$-methylbutanoic acid.

(D) The conversion of a primary alcohol $(R-CH_2OH)$ to a carboxylic acid $(R-COOH)$ requires a strong oxidizing agent.
$PCC$ (Pyridinium chlorochromate) is a mild oxidizing agent that stops the oxidation at the aldehyde stage.
Anhydrous $CrO_3$ in $CH_2Cl_2$ also typically oxidizes primary alcohols to aldehydes.
$Cu/573 \ K$ is used for the dehydrogenation of primary alcohols to aldehydes.
$KMnO_4/H^{\oplus}$ is a strong oxidizing agent that effectively converts primary alcohols directly into carboxylic acids.

(C) The reaction of propyne $(CH_3-C \equiv CH)$ with Grignard reagent $(CH_3MgBr)$ is an acid-base reaction.
Terminal alkynes have acidic hydrogen,which reacts with the Grignard reagent to form an alkynyl magnesium bromide and methane $(CH_4)$.
$CH_3-C \equiv CH + CH_3MgBr \rightarrow CH_3-C \equiv C-MgBr + CH_4$
Here,$(A)$ is $CH_3-C \equiv C-MgBr$.
Next,the reaction of $(A)$ with $CO_2$ followed by acidic hydrolysis $(H_2O/H^+)$ is a standard method to prepare carboxylic acids from Grignard reagents.
$CH_3-C \equiv C-MgBr + CO_2 \rightarrow CH_3-C \equiv C-COOMgBr$
$CH_3-C \equiv C-COOMgBr + H_2O/H^+ \rightarrow CH_3-C \equiv C-COOH + Mg(OH)Br$
Thus,$(B)$ is but$-2-$ynoic acid $(CH_3-C \equiv C-COOH)$.

(C) The substrate is $1$-bromoadamantane,a tertiary bridgehead halide.
Method $2$ involves an $S_N2$ reaction with $NaCN$. $S_N2$ reactions are extremely difficult at bridgehead carbons because the nucleophile cannot approach from the backside,and the transition state would require an $sp^2$-hybridized bridgehead carbon,which is forbidden by Bredt's rule.
Method $1$ involves the formation of a Grignard reagent $(RMgBr)$. While formation of Grignard reagents at bridgehead carbons is slow,it is generally possible. The subsequent reaction of the Grignard reagent with $CO_2$ is a standard method for preparing carboxylic acids.
Therefore,Method $1$ is appropriate,while Method $2$ is not.

(B) The reaction of benzotrichloride $(Ph-CCl_3)$ with milk of lime $(Ca(OH)_2)$ involves the hydrolysis of the trichloromethyl group.
First,the three chlorine atoms are replaced by hydroxyl groups to form an unstable intermediate,phenylmethanetriol $(Ph-C(OH)_3)$.
This intermediate undergoes dehydration by losing a water molecule $(-H_2O)$ to form benzoic acid $(Ph-COOH)$.

(B) Step $1$: The alcohol $CH_3-(CH_2)_3-OH$ reacts with methanesulfonyl chloride $(CH_3SO_2Cl)$ in the presence of triethylamine $(Et_3N)$ to form a good leaving group,the mesylate $(A)$,which is $CH_3-(CH_2)_3-OSO_2CH_3$.
Step $2$: The mesylate $(A)$ undergoes an $S_N2$ reaction with $K^{14}CN$. The $^{14}CN^-$ nucleophile attacks the primary carbon,displacing the mesylate group to form the nitrile $(B)$,which is $CH_3-(CH_2)_3-^{14}CN$.
Step $3$: The nitrile $(B)$ undergoes acid-catalyzed hydrolysis $(H_3O^{\oplus})$ to form the corresponding carboxylic acid $(C)$,which is $CH_3-(CH_2)_3-^{14}COOH$.

(B) The reaction of acetophenone $(PhCOCH_3)$ with $NaNO_2/HCl$ leads to the formation of an isonitroso compound (oxime),$PhCOCH=NOH$ (Product $A$).
Treatment of the oxime with acetic anhydride $(Ac_2O)$ followed by heating causes dehydration to form the nitrile,$PhCOCN$ (Product $B$).
Finally,the acid-catalyzed hydrolysis of the nitrile group $(CN)$ yields the carboxylic acid group $(COOH)$,resulting in the final product $PhCOCO_2H$ (Product $C$).

(D) When $1, 1, 1-$trichloropropane $(CH_3-CH_2-CCl_3)$ is treated with aqueous $KOH$,a nucleophilic substitution reaction occurs. The three chlorine atoms are replaced by three hydroxyl $(-OH)$ groups to form an unstable intermediate gem-triol. This intermediate immediately loses a water molecule to form propionic acid (propanoic acid).
Reaction:
$CH_3-CH_2-CCl_3 + 3KOH_{(aq)}$ $\rightarrow [CH_3-CH_2-C(OH)_3] + 3KCl$ $\rightarrow CH_3-CH_2-COOH + H_2O$

(D) The reaction of nitriles $(CH_3-CN)$ with dilute acid $(H_2O/H^{+})$ and heat $(\Delta)$ results in complete hydrolysis to form carboxylic acids $(CH_3-COOH)$ and ammonium ions $(NH_4^+)$.
The formation of an amide $(CH_3-CONH_2)$ is an intermediate step and is not the final product under these conditions.
Therefore,the reaction is incorrectly written.

(A) The starting material is $1-$bromo-$3-$nitrobenzene.
Step $1$: Reaction with $Mg$ in dry ether forms the Grignard reagent,$3-$nitrophenylmagnesium bromide $(A)$.
Step $2$: The Grignard reagent reacts with $CO_2$ followed by acid hydrolysis $(H_3O^+)$ to form a carboxylic acid group at the position where $MgBr$ was attached.
Thus,the product $B$ is $3-$nitrobenzoic acid.

(B) The reaction of cane sugar $(C_{12}H_{22}O_{11})$ with concentrated nitric acid $(HNO_3)$ acts as an oxidizing agent to produce oxalic acid.
The balanced chemical equation is:
$C_{12}H_{22}O_{11} + 18[O] \xrightarrow{conc. HNO_3} 6(COOH)_2 + 5H_2O$
Thus,the main product formed is oxalic acid.

(D) The reaction proceeds as follows:
$1$. Cyclopropyl bromide reacts with $Mg$ in the presence of $Et_2O$ to form the Grignard reagent,cyclopropylmagnesium bromide $(C_3H_5MgBr)$.
$2$. The Grignard reagent then undergoes a nucleophilic addition reaction with $CO_2$ to form the magnesium salt of the carboxylic acid,which is cyclopropylmagnesium carboxylate.
$3$. Finally,acid hydrolysis with $H_3O^+$ converts the magnesium salt into the final product,cyclopropanecarboxylic acid $(C_3H_5COOH)$.

(B) The hydrolysis of acetonitrile $(CH_3CN)$ in an acidic medium proceeds in two steps:
$1$. Partial hydrolysis to form acetamide $(CH_3CONH_2)$.
$2$. Complete hydrolysis to form acetic acid $(CH_3COOH)$.
In the presence of an acidic medium,the reaction is:
$CH_3-CN + 2H_2O \xrightarrow{H^+} CH_3-COOH + NH_4^+$

(C) The reaction proceeds as follows:
$1$. $R - X + KCN \xrightarrow{\text{alcoholic}} R - CN + KX$ (Nucleophilic substitution to form an alkyl cyanide $A$).
$2$. $R - CN + 2H_2O \xrightarrow{H^\oplus/\Delta} R - COOH + NH_3$ (Acidic hydrolysis of cyanide to form a carboxylic acid $B$).
Therefore,the final product $(B)$ is a carboxylic acid.

(B) $1$. The reaction of $1$-bromo-$4$-chlorobenzene with $1 \text{ mole}$ of $Mg$ in $THF$ selectively forms the Grignard reagent at the $Br$ position because $C-Br$ bond is weaker than $C-Cl$ bond. Thus,$A$ is $4$-chlorophenylmagnesium bromide.
$2$. The Grignard reagent $(A)$ reacts with $CO_2$ to form a magnesium carboxylate intermediate $(B)$,which is $4$-chlorophenylmagnesium carboxylate.
$3$. Acidic hydrolysis $(H^+)$ of the intermediate $B$ yields $4$-chlorobenzoic acid as the final product $C$.

(A) The oxidation of toluene with acidic $KMnO_4$ is a standard reaction used to convert the alkyl side chain of an aromatic ring into a carboxylic acid group.
The methyl group $(-CH_3)$ attached to the benzene ring is oxidized to a carboxyl group $(-COOH)$.
The reaction is: $C_6H_5CH_3 + [O] \xrightarrow{KMnO_4/H^+} C_6H_5COOH$ (Benzoic acid).

(C) The reaction proceeds as follows:
$1$. $R-X + KCN (alc.) \rightarrow R-CN + KX$. Here,$A$ is an alkyl cyanide (nitrile).
$2$. $R-CN + 2H_2O \xrightarrow{dil. HCl} R-COOH + NH_4Cl$. The partial or complete hydrolysis of nitriles in the presence of dilute acid yields carboxylic acids.
Therefore,$B$ is a carboxylic acid.

(C) $1$. The reaction of ethyl bromide $(C_2H_5Br)$ with $KCN$ is a nucleophilic substitution reaction.
$2$. $C_2H_5Br + KCN \rightarrow C_2H_5CN + KBr$. Here,$A$ is ethyl cyanide $(C_2H_5CN)$.
$3$. The hydrolysis of alkyl cyanides $(R-CN)$ yields carboxylic acids $(R-COOH)$.
$4$. $C_2H_5CN + 2H_2O \xrightarrow{H^+} C_2H_5COOH + NH_3$.
$5$. The product $B$ is propionic acid $(C_2H_5COOH)$.

(A) The conversion of an alcohol to a carboxylic acid with one additional carbon atom involves the following steps:
$1$. Conversion of alcohol to alkyl bromide using $PBr_3$: $R-CH_2CH_2OH + PBr_3 \rightarrow R-CH_2CH_2Br + H_3PO_3$.
$2$. Nucleophilic substitution with $KCN$ to form a nitrile: $R-CH_2CH_2Br + KCN \rightarrow R-CH_2CH_2CN + KBr$.
$3$. Acidic hydrolysis of the nitrile to form a carboxylic acid: $R-CH_2CH_2CN + 2H_2O + H^+ \rightarrow R-CH_2CH_2COOH + NH_4^+$.
Thus,the correct sequence is $PBr_3, KCN, H_3O^+$.

(C) The reaction proceeds as follows:
$1$. $NaBH_4$ reduces the aldehyde group $(-CHO)$ to a primary alcohol $(-CH_2OH)$ without affecting the triple bond. The product is $hex-3-yn-1-ol$.
$2$. $PBr_3$ converts the primary alcohol into an alkyl bromide: $CH_3-CH_2-C \equiv C-CH_2-CH_2-Br$.
$3$. $Mg / ether$ forms the Grignard reagent: $CH_3-CH_2-C \equiv C-CH_2-CH_2-MgBr$.
$4$. The Grignard reagent reacts with $CO_2$ followed by acidic workup $(H_3O^+)$ to form a carboxylic acid with one additional carbon atom.
$5$. The final product is $hept-4-ynoic$ acid,which corresponds to the structure $CH_3-CH_2-C \equiv C-CH_2-CH_2-COOH$.

(N/A) Oxidation of primary alcohol compounds: Primary alcohols can be oxidized to carboxylic acids using strong oxidizing agents like potassium permanganate $(KMnO_4)$ in acidic,basic,or neutral media,or potassium dichromate $(K_2Cr_2O_7)$ in acidic medium,or Jones reagent $(CrO_3 + H_2SO_4)$.
General reaction:
$RCH_2OH \xrightarrow{[O]} RCOOH$
Examples:
$1$. $CH_3(CH_2)_8CH_2OH \xrightarrow{CrO_3-H_2SO_4} CH_3(CH_2)_8COOH$
$2$. $CH_3CH_2OH \xrightarrow{K_2Cr_2O_7-H_2SO_4} CH_3COOH$
$(b)$ Oxidation of aldehyde compounds: Aldehydes are easily oxidized to carboxylic acids using mild oxidizing agents like Tollen's reagent or stronger agents like $KMnO_4$ or $HNO_3$.
General reaction:
$R-CHO \xrightarrow{[O]} RCOOH$
Examples:
$1$. $CH_3CHO \xrightarrow{[O]} CH_3COOH$
$2$. $C_6H_5CHO \xrightarrow{[O]} C_6H_5COOH$

(N/A) Aromatic carboxylic acids can be prepared by the vigorous oxidation of alkyl benzenes with chromic acid or acidic/alkaline potassium permanganate $(KMnO_4)$.
$(a)$ The entire side chain is oxidized to the carboxyl group $(-COOH)$ irrespective of the length of the side chain,provided that the benzylic carbon atom has at least one hydrogen atom.
$(b)$ Primary and secondary alkyl groups are oxidized to the carboxyl group in this manner.
$(c)$ If the benzylic carbon atom does not have any hydrogen atom (i.e.,tertiary alkyl group),then oxidation is not possible.
$(d)$ The reaction proceeds via the formation of a potassium salt (e.g.,potassium benzoate),which upon acidic hydrolysis $(H_3O^+)$ yields the corresponding aromatic carboxylic acid.
General reaction:
$C_6H_5-CH_2-R$ $\xrightarrow[\Delta]{KMnO_4, KOH} C_6H_5-COOK$ $\xrightarrow{H_3O^+} C_6H_5-COOH$

(N/A) Carboxylic acids can be prepared from nitriles $(R-CN)$ through hydrolysis.
$1.$ Hydrolysis of Nitriles:
Nitriles are first hydrolyzed to amides in the presence of $H^{+}$ or $OH^{-}$ catalysts,and then further hydrolyzed to carboxylic acids.
General Reaction:
$R-C \equiv N + H_2O$ $\xrightarrow{H^{+} \text{ or } OH^{-}} R-CONH_2$ $\xrightarrow{H^{+} \text{ or } OH^{-}, H_2O} R-COOH + NH_3$
$2.$ Preparation from Alcohols (via Nitriles):
Alcohols can be converted to alkyl halides,then to nitriles,and finally to carboxylic acids with one additional carbon atom.
Example:
$CH_3OH$ $\xrightarrow{PCl_3} CH_3Cl$ $\xrightarrow{KCN} CH_3CN$ $\xrightarrow{H^{+}, \Delta, H_2O} CH_3COOH$ (Ethanoic acid)
Note: The reaction can be stopped at the amide stage by controlling the reaction conditions.

(N/A) Amides can be converted into carboxylic acids through hydrolysis in the presence of an acid or base catalyst followed by heating.
General reaction:
$RCONH_2 \xrightarrow{H_3O^{+}, \Delta} RCOOH + NH_3$
Example:
$CH_3CONH_2 \xrightarrow{H_3O^{+}, \Delta} CH_3COOH + NH_3$
Another example:
$C_6H_5CONH_2 \xrightarrow{H_3O^{+}, \Delta} C_6H_5COOH + NH_3$

(N/A) Grignard reagents react with carbon dioxide (dry ice) to form salts of carboxylic acids,which upon acidification with mineral acid yield the corresponding carboxylic acids.
General reaction:
$R-Mg-X + O=C=O \xrightarrow{\text{dry ether}} R-COO^-Mg^+X$
$R-COO^-Mg^+X \xrightarrow{H_3O^+} RCOOH + Mg(OH)X$
Example $(i)$: Preparation of ethanoic acid from methyl bromide:
$CH_3Br \xrightarrow{Mg, \text{dry ether}} CH_3MgBr$
$CH_3MgBr + CO_2 \xrightarrow{\text{dry ether}} CH_3COOMgBr$
$CH_3COOMgBr \xrightarrow{H_3O^+} CH_3COOH + Mg(OH)Br$
Example $(ii)$: Preparation of benzoic acid from bromobenzene:
$C_6H_5Br \xrightarrow{Mg, \text{dry ether}} C_6H_5MgBr$
$C_6H_5MgBr + CO_2 \xrightarrow{\text{dry ether}} C_6H_5COOMgBr$
$C_6H_5COOMgBr \xrightarrow{H_3O^+} C_6H_5COOH + Mg(OH)Br$

(N/A) Carboxylic acid from acid chloride compounds:
$(i)$ Acid chlorides when hydrolysed with water give carboxylic acids.
$RCOCl + H_{2}O \longrightarrow RCOOH + HCl$
$(ii)$ Acyl chlorides are more readily hydrolysed with aqueous base to give carboxylate ions,which on acidification provide corresponding carboxylic acids.
$RCOCl$ $\xrightarrow{OH^{-}, H_{2}O} RCOO^{-}$ $\xrightarrow{H_{3}O^{+}} RCOOH$
$(b)$ Carboxylic acid from acid anhydrides:
Anhydrides are hydrolysed to corresponding acid$(s)$ with water.
$(i)$ $(CH_{3}CO)_{2}O + H_{2}O \longrightarrow 2CH_{3}COOH$ (Acetic anhydride $\longrightarrow$ Ethanoic acid)
$(ii)$ $(C_{6}H_{5}CO)_{2}O + H_{2}O \longrightarrow 2C_{6}H_{5}COOH$ (Benzoic anhydride $\longrightarrow$ Benzoic acid)
$(iii)$ $C_{6}H_{5}CO-O-COCH_{3} + H_{2}O \longrightarrow C_{6}H_{5}COOH + CH_{3}COOH$ (Benzoic ethanoic anhydride $\longrightarrow$ Benzoic acid + Acetic acid)

(N/A) $(i)$ Acidic hydrolysis of esters gives directly carboxylic acids while basic hydrolysis gives carboxylates,which on acidification give corresponding carboxylic acids.
$(ii)$ General reaction:
$R-CO-OR' + H_2O \xrightarrow{H^+} RCOOH + R'OH$
$(iii)$ Examples:
Example-$(i)$:
$C_6H_5COOC_2H_5 + H_2O \xrightarrow{H_3O^+} C_6H_5COOH + C_2H_5OH$
(Ethyl benzoate $\rightarrow$ Benzoic acid + Ethanol)
Example-$(ii)$:
$CH_3COOC_2H_5 + H_2O \xrightarrow{H_3O^+} CH_3COOH + C_2H_5OH$
(Ethyl acetate $\rightarrow$ Ethanoic acid + Ethanol)

(C) The reaction involves the oxidation of an alkyl group attached to a benzene ring using alkaline $KMnO_4$ followed by acidification $(H^+)$.
Alkaline $KMnO_4$ is a strong oxidizing agent that oxidizes the alkyl side chain (specifically,a methyl group attached to the benzene ring) to a carboxylic acid group $(-COOH)$.
In the given reactant,$1$-methoxy-$4$-methylbenzene,the methyl group $(-CH_3)$ at the para position is oxidized to a carboxylic acid group $(-COOH)$,while the methoxy group $(-OCH_3)$ remains unaffected.
Therefore,the product $X$ is $4$-methoxybenzoic acid.

(C) The reaction shows the oxidation of an alkyl group (specifically a methyl group attached to a benzene ring) to a carboxylic acid group $(-COOH)$.
Alkaline $KMnO_4$ followed by acidic workup $(H^+)$ is a strong oxidizing agent that converts alkyl side chains on aromatic rings into carboxylic acid groups,regardless of the length of the alkyl chain,provided there is at least one benzylic hydrogen atom.
Therefore,the reagent $A$ is alkaline $KMnO_4$ followed by $H^+$.

(D) Let us analyze each reaction:
$(A)$ $CH_{3}CH_{2}COCH_{3}$ undergoes the haloform reaction with $OI^{-}$ to yield propanoic acid $(CH_{3}CH_{2}COOH)$.
$(B)$ Oxidation of $CH_{3}CH_{2}CH_{3}$ with $KMnO_{4}$ can yield propanoic acid.
$(C)$ Hydrolysis of $CH_{3}CH_{2}CCl_{3}$ with $OH^{-}$ followed by acidification yields propanoic acid $(CH_{3}CH_{2}COOH)$.
$(D)$ $CH_{3}CH_{2}CH_{2}Br$ reacts with $Mg$ to form a Grignard reagent $(CH_{3}CH_{2}CH_{2}MgBr)$,which on reaction with $CO_{2}$ followed by hydrolysis yields butanoic acid $(CH_{3}CH_{2}CH_{2}COOH)$,not propanoic acid.
Therefore,the correct option is $D$.

(D) The reaction of a Grignard reagent $(RMgX)$ with carbon dioxide $(CO_2)$ is a nucleophilic addition reaction.
The nucleophilic alkyl group $(R^-)$ of the Grignard reagent attacks the electrophilic carbon atom of $CO_2$.
This results in the formation of an intermediate carboxylate salt,which is $RCOO^{-}Mg^{+}X$ (often written as $RCOOMgX$).
Subsequent acid hydrolysis $(H_3O^{+})$ of this intermediate yields the corresponding carboxylic acid $(RCOOH)$.

(C) The reaction sequence is as follows:
$1$. $4-$Bromotoluene reacts with $Cl_{2}$ in the presence of heat (free radical substitution) to form $4-$bromobenzyl chloride $(Br-C_{6}H_{4}-CH_{2}Cl)$.
$2$. This product then reacts with $CN^{-}$ (nucleophilic substitution) to form $4-$bromophenylacetonitrile $(Br-C_{6}H_{4}-CH_{2}CN)$.
$3$. Finally,acid-catalyzed hydrolysis of the nitrile group yields $4-$bromophenylacetic acid $(Br-C_{6}H_{4}-CH_{2}COOH)$.
Therefore,the starting material '$A$' is $4-$bromotoluene.

(C) The correct option is $C$.
Bromobenzene reacts with $Mg$ in the presence of dry ether to form a Grignard reagent (phenylmagnesium bromide).
This Grignard reagent then reacts with $CO_{2}$ to form an intermediate,which upon hydrolysis with $\text{dil. } HCl$ gives benzoic acid.

(B) The reaction sequence is as follows:
$1$. $C_2H_5Br$ reacts with alcoholic $KOH$ (dehydrohalogenation) to form ethene $(CH_2=CH_2)$ as product $A$.
$2$. Ethene $(A)$ reacts with $Br_2$ in $CCl_4$ (electrophilic addition) to form $1,2-$dibromoethane $(BrCH_2-CH_2Br)$ as product $B$.
$3$. $1,2-$dibromoethane $(B)$ reacts with excess $KCN$ (nucleophilic substitution) to form succinonitrile $(NC-CH_2-CH_2-CN)$ as product $C$.
$4$. Succinonitrile $(C)$ undergoes acid hydrolysis $(H_3O^+)$ to form succinic acid $(HOOC-CH_2-CH_2-COOH)$ as product $D$.