Preparation of Carboxylic Acids and Their Derivatives Questions in English

(D) Let us analyze each reaction:
$(A)$ Nitriles on partial hydrolysis under mild conditions yield amides $(R-CONH_2)$,not carboxylic acids.
$(B)$ Grignard reagents react with $CO_2$ followed by acidic hydrolysis to form carboxylic acids $(R-COOH)$. This is a standard method for preparing carboxylic acids.
$(C)$ This is the Stephen reduction,which converts nitriles to aldehydes $(R-CHO)$.
$(D)$ $PCC$ (Pyridinium chlorochromate) is a mild oxidizing agent that oxidizes primary alcohols to aldehydes $(R-CHO)$.
$(E)$ Benzoyl chloride undergoes Rosenmund reduction with $H_2/Pd-BaSO_4$ to form benzaldehyde,which is then oxidized to benzoic acid by $Br_2/H_2O$ (acting as a mild oxidizing agent in this context). Thus,it yields a carboxylic acid.
Therefore,reactions $(B)$ and $(E)$ produce carboxylic acids as major products.

(C) Acetic anhydride is prepared by the dehydration of acetic acid using a dehydrating agent like phosphorus pentoxide $(P_2O_5)$ upon heating.
The chemical reaction is as follows:
$2CH_3COOH \xrightarrow{P_2O_5, \Delta} (CH_3CO)_2O + H_2O$
Thus,acetic anhydride is obtained by the reaction of $P_2O_5$ and acetic acid.

(B) The reaction of a Grignard reagent $(RMgX)$ with dry ice $(CO_2)$ followed by acid hydrolysis yields a carboxylic acid with one carbon atom more than the alkyl group of the Grignard reagent.
In this reaction,$CH_3CH_2MgBr$ (Ethyl magnesium bromide) reacts with $CO_2$ to form an intermediate complex,$CH_3CH_2COOMgBr$.
Upon subsequent hydrolysis with dilute $HCl$,the complex is converted into $CH_3CH_2COOH$,which is Propanoic acid.
The overall reaction is: $CH_3CH_2MgBr + CO_2$ $\rightarrow CH_3CH_2COOMgBr$ $\xrightarrow{H_3O^+} CH_3CH_2COOH + Mg(OH)Br$.

(C) The reaction of acid chlorides with dialkyl cadmium is a standard method for the preparation of ketones.
The general reaction is: $2RCOCl + R'_2Cd \rightarrow 2RCOR' + CdCl_2$.
Here,the product is propanone $(CH_3COCH_3)$ and the reagent is dimethyl cadmium $((CH_3)_2Cd)$.
Comparing the general reaction,$R'$ is a methyl group $(CH_3)$.
Thus,$2RCOCl + (CH_3)_2Cd \rightarrow 2CH_3COCH_3 + CdCl_2$.
For the product to be propanone,$R$ must be a methyl group $(CH_3)$.
Therefore,the substrate $S$ is ethanoyl chloride $(CH_3COCl)$.

(B) Aldehydes $(R-CHO)$ undergo oxidation when treated with oxidizing agents like dilute nitric acid $(HNO_3)$.
This reaction converts the aldehyde group into a carboxylic acid group.
The chemical equation is: $R-CHO \xrightarrow{\text{Dil. } HNO_3} R-COOH$ (Carboxylic acid).

(D) Acid hydrolysis of ethanenitrile $(CH_3CN)$ involves the reaction with water in the presence of an acid catalyst $(H^+)$.
Initially,it forms acetamide,which further hydrolyzes to form acetic acid $(CH_3COOH)$ and ammonium salt $(NH_4^+)$.
The overall reaction is:
$CH_3CN + 2H_2O \xrightarrow{H^+} CH_3COOH + NH_4^+$

(B) The reaction of Grignard reagent $(CH_3MgBr)$ with dry ice $(CO_2)$ in the presence of dry ether forms an intermediate magnesium salt,$CH_3COOMgBr$ $(A)$.
The reaction is: $CH_3MgBr + CO_2 \xrightarrow{\text{dry ether}} CH_3COOMgBr$.
Upon acidic hydrolysis $(H_3O^{+})$,the intermediate $A$ yields ethanoic acid $(B)$.
The reaction is: $CH_3COOMgBr + H_3O^{+} \rightarrow CH_3COOH + Mg(OH)Br$.

(B) The reaction of a Grignard reagent with carbon dioxide is a standard method for the preparation of carboxylic acids.
Step $1$: $(CH_3)_2CHMgBr + CO_2 \rightarrow (CH_3)_2CH-COO-MgBr$ (Intermediate $A$).
Step $2$: $(CH_3)_2CH-COO-MgBr + H_2O / H^+ \rightarrow (CH_3)_2CH-COOH + Mg(OH)Br$.
The product $B$ is $(CH_3)_2CH-COOH$,which is named $2-$methylpropanoic acid (also known as isobutyric acid).

(B) The reaction of a Grignard reagent $(RMgX)$ with dry ice $(CO_2)$ followed by acidic hydrolysis is a standard method for the preparation of carboxylic acids.
In this reaction,the nucleophilic alkyl group $(CH_3CH_2^-)$ of the Grignard reagent attacks the electrophilic carbon atom of $CO_2$ to form a magnesium carboxylate salt $(CH_3CH_2COO^-Mg^+Br^-)$.
Subsequent hydrolysis with dilute $HCl$ converts the salt into the corresponding carboxylic acid.
$CH_3CH_2MgBr + CO_2 \rightarrow CH_3CH_2COOMgBr$
$CH_3CH_2COOMgBr + H_2O/H^+ \rightarrow CH_3CH_2COOH + Mg(OH)Br$
The product formed is $CH_3CH_2COOH$,which is Propanoic acid.

(D) The reaction of cumene (isopropylbenzene) with alkaline $KMnO_4$ followed by heating results in the oxidation of the alkyl side chain to a carboxylate group.
In the first step,cumene is oxidized to potassium benzoate $(A)$.
In the second step,acidification of potassium benzoate with $H_3O^+$ yields benzoic acid $(B)$.

(D) The reaction of an alkylbenzene with an alkyl group having at least one benzylic hydrogen atom with alkaline $KMnO_4$ followed by acidic hydrolysis $(H_3O^+)$ results in the oxidation of the alkyl side chain to a carboxylic acid group.
In the given reaction,ethylbenzene $(C_6H_5-CH_2-CH_3)$ is oxidized to benzoic acid $(C_6H_5-COOH)$.
The reaction is: $C_6H_5-CH_2-CH_3 \xrightarrow[\text{(ii) } H_3O^{+}]{\text{(i) alk. } KMnO_4} C_6H_5-COOH$.

(A) Aromatic carboxylic acids can be prepared from alkyl benzenes by vigorous oxidation of the alkyl side chain using strong oxidizing agents.
Alkaline potassium permanganate $(KMnO_4/OH^-)$ and chromic acid $(H_2CrO_4)$ are well-known strong oxidizing agents used for this purpose.
Dilute nitric acid $(HNO_3)$ is also capable of oxidizing the alkyl side chain of alkyl benzenes to a carboxylic acid group.
Diborane $(B_2H_6)$ is a reducing agent,typically used for hydroboration-oxidation reactions to convert alkenes into alcohols,and it is not used for the oxidation of alkyl benzenes to carboxylic acids.
Therefore,the correct answer is $A$.

(C) The reaction of a Grignard reagent $(R-Mg-X)$ with solid carbon dioxide ($CO_2$,dry ice) followed by acidic hydrolysis is a standard method for the preparation of carboxylic acids.
Step $1$: $R-Mg-X + CO_2 \xrightarrow{\text{dry ether}} R-COOMgX$ (Carboxylato magnesium halide).
Step $2$: $R-COOMgX + H_2O \xrightarrow{\text{dil } HCl} R-COOH + Mg(X)OH$ (Carboxylic acid).
Therefore,compound $A$ is $CO_2$ (Solid).

(A) The reaction of $Cumene$ (isopropylbenzene) with alkaline $KMnO_4$ followed by acidification is a standard method for the oxidation of alkylbenzenes.
$1$. In the first step,$Cumene$ is oxidized to potassium benzoate $(A)$ by alkaline $KMnO_4$ under heating.
$2$. In the second step,acidification with $H_3O^{+}$ converts potassium benzoate into $Benzoic \ acid$ $(B)$.
Therefore,the final product $B$ is $Benzoic \ acid$.

(D) Formic acid $(HCOOH)$ cannot be prepared from methyl magnesium bromide $(CH_{3}MgBr)$ because Grignard reagents react with $CO_{2}$ to form carboxylic acids with at least two carbon atoms (e.g.,acetic acid).
Formic acid can be prepared from the other reagents as follows:
$(a)$ $CH_{3}OH$ $\xrightarrow{[O]} HCHO$ $\xrightarrow{[O]} HCOOH$
$(b)$ $CO + NaOH$ $\longrightarrow HCOONa$ $\xrightarrow{H_{2}SO_{4}} HCOOH + NaHSO_{4}$
$(c)$ Glycerol + Oxalic acid $\xrightarrow{383 \ K} HCOOH + \text{glycerol}$

(C) The reaction of ethanoyl chloride $(CH_3COCl)$ with water $(H_2O)$ is a hydrolysis reaction.
In this reaction,the chlorine atom is replaced by a hydroxyl group $(-OH)$ from water.
The chemical equation is: $CH_3COCl + H_2O \rightarrow CH_3COOH + HCl$.
The product $X$ formed is ethanoic acid $(CH_3COOH)$.

(C) When two molecules of carboxylic acid $(R-COOH)$ are heated in the presence of a dehydrating agent like phosphorus pentoxide $(P_2O_5)$,they undergo a condensation reaction to form an acid anhydride by the elimination of a water molecule $(H_2O)$.
The reaction is represented as:
$2R-COOH \xrightarrow{P_2O_5, \Delta} (RCO)_2O + H_2O$
Thus,option $C$ is the correct reaction for the formation of acid anhydride.

(B) Step $1$: Protection of the $-OH$ group of $m$-hydroxybenzaldehyde with benzyl chloride $(C_6H_5CH_2Cl)$ gives $m$-benzyloxybenzaldehyde $(A)$.
Step $2$: Oxidation of the aldehyde group in $A$ using an oxidizing agent $[O]$ yields $m$-benzyloxybenzoic acid $(B)$.
Step $3$: Deprotection of the benzyl group (usually via catalytic hydrogenation or acid-catalyzed cleavage) removes the protecting group to yield $m$-hydroxybenzoic acid $(C)$.

(C) The reaction is as follows:
$CH_3Cl + KCN_{(alc)} \xrightarrow{\Delta} CH_3CN (A) + KCl$
$CH_3CN + 2H_2O \xrightarrow{HCl} CH_3COOH (B) + NH_4Cl$
In the first step,chloromethane reacts with alcoholic $KCN$ to form ethanenitrile $(CH_3CN)$,which is compound $A$.
In the second step,the acid hydrolysis of ethanenitrile yields ethanoic acid $(CH_3COOH)$,which is compound $B$.
Therefore,$A =$ Ethanenitrile and $B =$ Ethanoic acid.

(C) The reaction of $Cumene$ (isopropylbenzene) with alkaline $KMnO_4$ followed by acidic hydrolysis is a standard method for the oxidation of alkylbenzenes.
Step $1$: Oxidation of $Cumene$ with $KMnO_4$ and $KOH$ yields $Potassium \ benzoate$ $(A)$.
Step $2$: Acidification of $Potassium \ benzoate$ with $H_3O^+$ yields $Benzoic \ acid$ $(B)$.
Therefore,the final product $B$ is $Benzoic \ acid$.

(D) The reaction of an alkylbenzene with an alkyl group having at least one benzylic hydrogen atom with alkaline $KMnO_4$ followed by acidic hydrolysis $(H_3O^+)$ results in the oxidation of the alkyl side chain to a carboxylic acid group.
In the given reactant,$C_6H_5-CH_2-CH_3$ (ethylbenzene),the benzylic carbon is the $CH_2$ group.
Upon oxidation,the entire alkyl side chain is converted into a carboxyl group $(-COOH)$ attached to the benzene ring.
Therefore,the product formed is benzoic acid,$C_6H_5-COOH$.

(D) When ethyl benzene is treated with strong oxidizing agents like $HNO_3$ or $KMnO_4$,the alkyl side chain attached to the benzene ring is oxidized to a carboxylic acid group $(-COOH)$.
Thus,ethyl benzene undergoes strong oxidation to form benzoic acid.

(B) The reaction of dry ice $(CO_2)$ with a Grignard reagent $(CH_3MgBr)$ in the presence of dry ether forms an intermediate addition product $(A)$,which is $CH_3COOMgBr$.
$CO_2 + CH_3MgBr \xrightarrow{Dry \ ether} CH_3COOMgBr (A)$
This intermediate $(A)$ upon hydrolysis with dilute $HCl$ yields a carboxylic acid $(B)$,which is ethanoic acid $(CH_3COOH)$.
$CH_3COOMgBr + H_2O \xrightarrow{dil. \ HCl} CH_3COOH (B) + Mg(OH)Br$
Therefore,the product $B$ is ethanoic acid.

(B) The reaction $R-CHO \rightarrow R-CH_2OH$ is the reduction of an aldehyde to a primary alcohol. Both $NaBH_4$ and $H_2 | Pd$ can perform this reduction.
The reaction $R-COOH \rightarrow R-CH_2OH$ is the reduction of a carboxylic acid to a primary alcohol. Carboxylic acids are resistant to reduction by $NaBH_4$ and $H_2 | Pd$. $A$ strong reducing agent like $LiAlH_4$ followed by acidic workup $(H_3O^+)$ is required for this transformation.
Comparing the options,option $B$ provides a valid set of reagents for both steps: $H_2 | Pd$ can reduce the aldehyde,and $LiAlH_4$ is the standard reagent for reducing carboxylic acids to primary alcohols.

(A) Cyanides $(-CN)$ undergo hydrolysis in the presence of an alkali (like $NaOH$) or an acid to produce a carboxylic acid group $(-COOH)$.
The reaction is as follows:
$R-CN + 2H_2O \xrightarrow{NaOH} R-COOH + NH_3$

(C) Step $1$: Reaction of chloromethane with potassium cyanide $(KCN)$ is a nucleophilic substitution reaction where the cyanide ion $(CN^-)$ replaces the chloride ion $(Cl^-)$ to form methyl cyanide $(CH_3CN)$,which is compound $(A)$.
$CH_3Cl + KCN \rightarrow CH_3CN + KCl$
Step $2$: Acidic hydrolysis of methyl cyanide $(CH_3CN)$ leads to the formation of an amide intermediate,which further hydrolyzes to form acetic acid $(CH_3COOH)$ as the final product $(B)$.
$CH_3CN + 2H_2O \xrightarrow{H^+} CH_3COOH + NH_3$

(A) The reaction sequence is as follows:
$1.$ Addition of $HBr$ to ethene $(CH_2=CH_2)$ follows Markovnikov's rule to give ethyl bromide $(A)$:
$CH_2=CH_2 + HBr \rightarrow CH_3-CH_2-Br$
$2.$ Nucleophilic substitution of ethyl bromide with $KCN$ gives ethyl cyanide $(B)$:
$CH_3-CH_2-Br + KCN \rightarrow CH_3-CH_2-CN + KBr$
$3.$ Acidic hydrolysis of ethyl cyanide gives propanoic acid $(C)$:
$CH_3-CH_2-CN \xrightarrow{H^{+}/H_2O} CH_3-CH_2-COOH$
Therefore,the correct option is $A$.

(C) The reaction sequence is as follows:
$1$. $CH_3OH + PCl_3 \rightarrow CH_3Cl (A) + H_3PO_3$
$2$. $CH_3Cl + KCN \rightarrow CH_3CN (B) + KCl$
$3$. $CH_3CN + 2H_2O \xrightarrow{H_3O^+} CH_3COOH (C) + NH_3$
Therefore,$C$ is $CH_3COOH$.

(C) The alkyl bromide $X$ $(C_5H_{11}Br)$ reacts with $Mg$ in dry ether to form a Grignard reagent $(R-MgBr)$.
Reaction with $CO_2$ followed by acidification yields a carboxylic acid with one additional carbon atom $(R-COOH)$.
Since the final product $Y$ is a carboxylic acid derived from a $C_5$ alkyl bromide,it must contain $5+1=6$ carbon atoms.
Among the options,$3$-methylbutanoic acid $(C_5H_{10}O_2)$ has $5$ carbons,hexanoic acid $(C_6H_{12}O_2)$ has $6$ carbons,$3$-methylpentanoic acid $(C_6H_{12}O_2)$ has $6$ carbons,and $2,2$-dimethylbutanoic acid $(C_6H_{12}O_2)$ has $6$ carbons.
Given the structure of the options provided,the reaction of $1$-bromo$-3-$methylbutane with $Mg$ followed by $CO_2$ and $H_3O^+$ gives $4$-methylpentanoic acid. However,looking at the provided options,option $A$ represents $3$-methylbutanoic acid,$B$ represents hexanoic acid,$C$ represents $3$-methylpentanoic acid,and $D$ represents $2,2$-dimethylbutanoic acid.
Based on the standard synthesis of carboxylic acids from alkyl halides via Grignard reagents,the chain length increases by one. The correct structure for $Y$ corresponding to the Grignard reagent derived from $1$-bromo$-3-$methylbutane is $4$-methylpentanoic acid. Since this is not explicitly listed,but $3$-methylpentanoic acid is a common isomer,we identify the correct structure based on the provided image $C$.

(D) $1$. The ozonolysis of $C_4H_8$ gives acetone $(CH_3COCH_3)$ and formaldehyde $(HCHO)$. This confirms that $C_4H_8$ is isobutylene,$(CH_3)_2C=CH_2$.
$2$. Reaction of isobutylene with $HBr$ in the presence of peroxide follows anti-Markovnikov addition: $(CH_3)_2C=CH_2 + HBr \xrightarrow{\text{peroxide}} (CH_3)_2CH-CH_2Br$ ($X$ is $1-$bromo$-2-$methylpropane).
$3$. $X$ reacts with $Mg$ in dry ether to form the Grignard reagent: $(CH_3)_2CH-CH_2MgBr$.
$4$. The Grignard reagent reacts with $CO_2$ followed by acid hydrolysis to form a carboxylic acid: $(CH_3)_2CH-CH_2MgBr + CO_2$ $\rightarrow (CH_3)_2CH-CH_2COOMgBr$ $\xrightarrow{H_3O^+} (CH_3)_2CH-CH_2COOH$.
$5$. The final product $Y$ is $3-$methylbutanoic acid.

(A) The oxidation of acetaldehyde $(CH_3CHO)$ to acetic acid $(CH_3COOH)$ is an industrial process.
This reaction is typically catalyzed by manganese acetate $(Mn(CH_3COO)_2)$ in the presence of air or oxygen.
The reaction is: $2CH_3CHO + O_2 \xrightarrow{Mn(CH_3COO)_2} 2CH_3COOH$.

(B) The reaction of Grignard reagent $(CH_3MgBr)$ with carbon dioxide $(CO_2)$ in the presence of dry ether forms an intermediate magnesium carboxylate complex $(Y = CH_3COOMgBr)$.
Upon subsequent acid hydrolysis $(H_3O^{\oplus})$,this complex yields a carboxylic acid.
Since the Grignard reagent used is methylmagnesium bromide $(CH_3MgBr)$,the resulting carboxylic acid is acetic acid $(CH_3COOH)$.

(A) Cyanides $(-CN)$ undergo hydrolysis in the presence of an alkali (like $NaOH$) or an acid to yield a carboxylic acid group $(-COOH)$.
The reaction is as follows:
$R-CN + 2H_2O \xrightarrow{NaOH} R-COOH + NH_3$

(C) The reaction proceeds as follows:
$1$. The first step is the anti-Markovnikov addition of $HBr$ to methylenecyclohexane in the presence of a peroxide,$(C_6H_5CO)_2O_2$. This yields (cyclohexylmethyl) bromide,$C_6H_{11}CH_2Br$.
$2$. The second step is a nucleophilic substitution reaction with $KCN$,where the bromide ion is replaced by a cyanide group,yielding cyclohexylacetonitrile,$C_6H_{11}CH_2CN$.
$3$. The third step is the acid-catalyzed hydrolysis of the nitrile group followed by heating,which converts the $-CN$ group into a carboxylic acid group,$-COOH$. The final product is cyclohexylacetic acid,$C_6H_{11}CH_2COOH$.

(A) The reaction sequence is as follows:
$1$. $C_2H_5Cl + KCN \xrightarrow{C_2H_5OH} C_2H_5CN (X) + KCl$
This is a nucleophilic substitution reaction where $CN^{-}$ replaces $Cl^{-}$.
$2$. $C_2H_5CN + 2H_2O \xrightarrow{H_3O^{\oplus}, \Delta} C_2H_5COOH (Y) + NH_3$
Acidic hydrolysis of the nitrile $(X)$ yields the corresponding carboxylic acid $(Y)$,which is propanoic acid $(C_2H_5COOH)$.
The molecular formula of propanoic acid $(C_2H_5COOH)$ is $C_3H_6O_2$.

(A) $1$. Toluene reacts with $CrO_2Cl_2$ in $CS_2$ followed by hydrolysis ($Etard$ reaction) to form benzaldehyde $(X)$.
$2$. Benzaldehyde $(X)$ contains a carbonyl group,so it forms $2,4-$dinitrophenylhydrazone and acts as a reducing agent towards Tollen's reagent (ammonical silver nitrate).
$3$. Toluene reacts with $KMnO_4 | OH^-, \Delta$ followed by acidic workup to form benzoic acid $(Y)$.
$4$. Benzoic acid $(Y)$ is acidic enough to react with $NaHCO_3$ to evolve $CO_2$ gas.
$5$. Therefore,$A$ is $CrO_2Cl_2 | CS_2, H_3O^+$ and $B$ is $KMnO_4 | OH^-, \Delta, H_3O^+$.

(A, C, D) To synthesize benzoic acid from benzene:
Option $(A)$: Benzene reacts with $Br_2/Fe$ to form bromobenzene,which forms a Grignard reagent $(PhMgBr)$ with $Mg/\text{dry ether}$. This reacts with $CO_2$ followed by acid hydrolysis $(H_3O^{\oplus})$ to yield benzoic acid.
Option $(C)$: Friedel-Crafts alkylation of benzene with $CH_3Cl/AlCl_3$ gives toluene. Oxidation of toluene with alkaline $KMnO_4$ followed by acidification gives benzoic acid.
Option $(D)$: Friedel-Crafts acylation of benzene with $CH_3COCl/AlCl_3$ gives acetophenone. The haloform reaction with $Br_2/NaOH$ followed by acidification gives benzoic acid.
Option $(B)$ is incorrect as the reaction sequence does not lead to benzoic acid.
Therefore,the correct sets are $(A), (C),$ and $(D)$.

(D) The preparation of $Me_{3}CCOOH$ (pivalic acid) from a Grignard reagent involves the reaction of a Grignard reagent with dry ice $(CO_{2})$.
The reaction is as follows:
$O=C=O + (CH_{3})_{3}CMgI \rightarrow (CH_{3})_{3}C-COOMgI$
Followed by hydrolysis:
$(CH_{3})_{3}C-COOMgI + H_{2}O \rightarrow (CH_{3})_{3}C-COOH + Mg(OH)I$
Thus,treating $1 \ mol$ of dry ice with $1 \ mol$ of $Me_{3}CMgI$ yields $Me_{3}CCOOH$.

(A) At $STP$,$22.4 \ dm^3$ of a gas corresponds to $1 \ mole$.
Given that $1.4 \ dm^3$ of gas $(Q)$ weighs $1 \ g$.
Therefore,the molar mass of $(Q)$ is $\frac{22.4 \ dm^3/mol \times 1 \ g}{1.4 \ dm^3} = 16 \ g/mol$.
Since the gas $(Q)$ is formed from $RMgBr$ and water,it is an alkane. The alkane with molar mass $16 \ g/mol$ is methane $(CH_4)$.
Thus,the alkyl group $R$ in $RMgBr$ is a methyl group $(CH_3)$,and $(P)$ is $CH_3MgBr$.
Reaction of $CH_3MgBr$ with dry ice $(CO_2)$ followed by acidic hydrolysis $(H_3O^{+})$ yields acetic acid $(CH_3COOH)$ as compound $(Z)$.
The molar mass of $CH_3COOH$ is $12 + 3(1) + 12 + 2(16) + 1 = 60 \ g/mol$.
Weight of $0.1 \ mole$ of $(Z)$ = $0.1 \ mol \times 60 \ g/mol = 6 \ g$.