Preparation Questions in English

(C) The reaction shown is the Rosenmund reduction,which is used to convert acid chlorides to aldehydes.
In this reaction,benzoyl chloride is hydrogenated using $H_2$ gas in the presence of a palladium catalyst supported on barium sulfate $(Pd-BaSO_4)$.
The $BaSO_4$ acts as a poison to prevent the further reduction of the aldehyde to an alcohol.
Therefore,the correct reagent is $H_2 / Pd-BaSO_4$.

(D) The reaction of benzoyl chloride with hydrogen in the presence of palladium $(Pd)$ catalyst supported on barium sulphate $(BaSO_4)$ is known as the Rosenmund reduction.
In this reaction,the acid chloride is selectively reduced to an aldehyde.
The chemical equation is:
$C_6H_5COCl + H_2 \xrightarrow{Pd/BaSO_4} C_6H_5CHO + HCl$
Thus,the product formed is benzaldehyde.

(C) The Rosenmund reduction is a hydrogenation process where an acid chloride $(R-CO-Cl)$ is reduced to an aldehyde $(R-CHO)$ using hydrogen gas in the presence of a poisoned palladium catalyst,typically $Pd$ supported on $BaSO_4$. The reaction is: $R-CO-Cl + \text{H}_2 \xrightarrow{\text{Pd/BaSO}_4} R-CHO + \text{HCl}$. Therefore,option $C$ represents the Rosenmund reduction.

(A) The conversion of a nitrile $(-CN)$ to an aldehyde $(-CHO)$ while preserving other functional groups like a double bond is achieved using Diisobutylaluminium hydride ($DIBAL-H$ or $AlH(i-Bu)_2$),followed by acid hydrolysis $(H_3O^{+})$.
$DIBAL-H$ is a selective reducing agent that reduces nitriles to imines,which upon hydrolysis yield aldehydes.
Therefore,the correct reagent is $AlH(i-Bu)_2 / H_3O^{+}$.

(A) Rosenmund reduction is the hydrogenation of an acid chloride $(R-COCl)$ to an aldehyde $(R-CHO)$ using a palladium catalyst poisoned with barium sulfate $(Pd/BaSO_4)$.
The reaction is:
$R-COCl + H_2 \xrightarrow{Pd/BaSO_4} R-CHO + HCl$
Therefore,option $A$ represents the Rosenmund reduction.

(D) The reaction of benzoyl chloride with $H_{2}$ in the presence of $Pd$ supported on $BaSO_{4}$ is known as the Rosenmund reduction.
In this reaction,the acid chloride is reduced to the corresponding aldehyde.
$C_{6}H_{5}COCl + H_{2} \xrightarrow{Pd/BaSO_{4}} C_{6}H_{5}CHO + HCl$
Therefore,benzaldehyde is obtained.

(C) The reaction of a nitrile $(R-C \equiv N)$ with $DIBAl-H$ (Diisobutylaluminium hydride) followed by acidic hydrolysis $(H_{3}O^{+})$ is a standard method for the preparation of aldehydes.
$DIBAl-H$ acts as a selective reducing agent that reduces the nitrile group to an imine intermediate,which upon hydrolysis yields the corresponding aldehyde $(R-CHO)$.

(B) When calcium formate $(HCOO)_2Ca$ is subjected to dry distillation,it undergoes thermal decomposition to produce formaldehyde (methanal) and calcium carbonate $(CaCO_3)$.
The chemical reaction is as follows:
$(HCOO)_2Ca \xrightarrow{\Delta} HCHO + CaCO_3$
Thus,the correct product is methanal.

(C) When toluene is treated with a solution of chromyl chloride $(CrO_2Cl_2)$ in $CS_2$,a brown chromium complex is obtained,which on acid hydrolysis gives benzaldehyde. This reaction is known as the Etard reaction. The reaction is represented as follows:
$C_6H_5CH_3 + 2CrO_2Cl_2$ $\xrightarrow{CS_2} C_6H_5CH(OCrCl_2OH)_2$ $\xrightarrow{H_3O^+} C_6H_5CHO$

(C) The given reaction is the Rosenmund reduction,where an acid chloride is hydrogenated to an aldehyde using a poisoned catalyst,$Pd$ supported on $BaSO_{4}$.
In this reaction,benzoyl chloride $(C_{6}H_{5}COCl)$ reacts with $H_{2}$ in the presence of $Pd-BaSO_{4}$ to form benzaldehyde $(C_{6}H_{5}CHO)$ and $HCl$.
Therefore,'$A$' is benzoyl chloride.

(B) The hydrogenation of an acyl chloride $(RCOCl)$ to an aldehyde $(RCHO)$ using palladium $(Pd)$ supported on barium sulphate $(BaSO_4)$ is known as the $Rosenmund \ reduction$.
The $BaSO_4$ acts as a poison to prevent the further reduction of the aldehyde to a primary alcohol.

(D) The Stephen reaction involves the reduction of nitriles $(R-C \equiv N)$ to imines using stannous chloride $(SnCl_2)$ in the presence of hydrochloric acid $(HCl)$,followed by hydrolysis to yield aldehydes $(R-CHO)$.
Option $(D)$ represents the correct chemical equation for the Stephen reaction:
$R-C \equiv N + 2[H]$ $\xrightarrow{SnCl_2/\text{dil. } HCl} R-CH=NH \cdot HCl$ $\xrightarrow{H_3O^+} R-CHO + NH_4Cl$

(A) The conversion of olefins (alkenes) into aldehydes is known as the hydroformylation or oxo process.
In this process,an alkene reacts with a mixture of $CO$ and $H_{2}$ in the presence of a metal catalyst,such as cobalt carbonyl $(Co_{2}(CO)_{8})$,to produce an aldehyde.
The reaction is represented as:
$CH_{3}-CH=CH_{2} \xrightarrow{CO/H_{2}/Co_{2}(CO)_{8}, \Delta} CH_{3}-CH_{2}-CH_{2}CHO$

(C) The Rosenmund reaction involves the hydrogenation of acid chlorides to aldehydes using a palladium catalyst supported on barium sulphate $(Pd/BaSO_4)$.
$BaSO_4$ acts as a poison to prevent the further reduction of the aldehyde to an alcohol.
The reaction is represented as:
$RCOCl + H_2 \xrightarrow{Pd/BaSO_4} RCHO + HCl$

(A) In the $Etard$ reaction,toluene is treated with chromyl chloride $(CrO_2Cl_2)$ in the presence of carbon disulfide $(CS_2)$ to form a brown chromium complex,which upon hydrolysis gives benzaldehyde. The reaction is:
$C_6H_5CH_3 + 2CrO_2Cl_2$ $\xrightarrow{CS_2} C_6H_5CH(OCrCl_2OH)_2$ $\xrightarrow{H_3O^+} C_6H_5CHO$.
Therefore,the correct option is $A$.

(B) Step $1$: Reaction of methyl magnesium bromide with cadmium chloride gives dimethyl cadmium $(A)$.
$2CH_3MgBr + CdCl_2 \rightarrow (CH_3)_2Cd + 2Mg(Br)Cl$
Step $2$: Reaction of dimethyl cadmium with acetyl chloride $(CH_3COCl)$ yields propanone $(B)$.
$(CH_3)_2Cd + 2CH_3COCl \rightarrow 2CH_3COCH_3 + CdCl_2$
Thus,the product '$B$' is propanone.

(C) The reaction of acyl chlorides with dialkyl cadmium reagents is a standard method for the preparation of ketones.
Given the reaction: $2CH_3COCl + (CH_3)_2Cd \rightarrow 2CH_3COCH_3 + CdCl_2$.
Here,$CH_3COCl$ is $Ethanoyl$ chloride,which reacts with dimethyl cadmium to form $Propanone$ $(CH_3COCH_3)$.

(C) Step $1$: Reaction of methyl magnesium bromide with cadmium chloride $(CdCl_2)$ produces dimethyl cadmium $(A)$.
$2CH_3MgBr + CdCl_2 \rightarrow (CH_3)_2Cd + 2Mg(Br)Cl$
Step $2$: Dimethyl cadmium reacts with acetyl chloride $(CH_3COCl)$ to form propanone $(B)$.
$2CH_3COCl + (CH_3)_2Cd \rightarrow 2CH_3COCH_3 + CdCl_2$
Thus,the product $B$ is propanone.

(B) The reaction of benzonitrile $(C_6H_5CN)$ with phenylmagnesium bromide $(C_6H_5MgBr)$ in the presence of dry ether,followed by acid hydrolysis,yields benzophenone $(C_6H_5COC_6H_5)$.
The chemical reaction is as follows:
$C_6H_5CN + C_6H_5MgBr$ $\xrightarrow{\text{dry ether}} C_6H_5C(NMgBr)C_6H_5$ $\xrightarrow{H_3O^+} C_6H_5COC_6H_5 + Mg(OH)Br + NH_3$

(A) The reaction of an acid chloride with a dialkylcadmium reagent is a standard method for the preparation of ketones. The general reaction is: $2RCOCl + R'_{2}Cd \rightarrow 2RCOR' + CdCl_{2}$. In the given reaction,the product is $CH_{3}-CO-CH_{2}-C_{6}H_{5}$ (benzyl methyl ketone). Comparing this with the general reaction,$R$ must be $CH_{3}$ and $R'$ must be $C_{6}H_{5}CH_{2}$. Thus,$A$ is $CH_{3}COCl$ (acetyl chloride).

(D) The reaction of propanenitrile $(CH_3CH_2CN)$ with ethylmagnesium iodide $(C_2H_5MgI)$ in the presence of dry ether proceeds as follows:
$1$. Nucleophilic attack of the ethyl group from the Grignard reagent on the carbon atom of the nitrile group leads to the formation of an imine magnesium complex: $CH_3CH_2CN + C_2H_5MgI \rightarrow CH_3CH_2C(C_2H_5)=NMgI$.
$2$. Acid hydrolysis $(H_3O^+)$ of this imine complex results in the formation of a ketone and ammonia: $CH_3CH_2C(C_2H_5)=NMgI + H_2O/H^+ \rightarrow CH_3CH_2COCH_2CH_3 + NH_3 + Mg(OH)I$.
$3$. The product formed is $CH_3CH_2COCH_2CH_3$,which is $3-$pentanone (pentan-$3-$one).
Therefore,the correct option is $D$.

(D) Organometallic compounds like dialkyl cadmium $(R_2Cd)$ react with acid chlorides $(R'COCl)$ to form ketones.
The reaction is: $2CH_3COCl + (C_6H_5CH_2)_2Cd \rightarrow 2CH_3COCH_2C_6H_5 + CdCl_2$.
Here,$Acetyl \ chloride$ $(CH_3COCl)$ reacts with $Dibenzyl \ cadmium$ $((C_6H_5CH_2)_2Cd)$ to produce $Benzyl \ methyl \ ketone$ $(CH_3COCH_2C_6H_5)$.

(B) The reagent $AlH(i-Bu)_2$ is Diisobutylaluminium hydride ($DIBAL$-$H$).
It is a selective reducing agent used to reduce nitriles $(-CN)$ to aldehydes $(-CHO)$ after acidic hydrolysis $(H_3O^+)$.
To obtain $\text{Pent-}3\text{-enal}$ $(CH_3-CH=CH-CH_2-CHO)$,the starting material must be a nitrile with the same carbon skeleton,which is $\text{Pent-}3\text{-enenitrile}$ $(CH_3-CH=CH-CH_2-CN)$.
Therefore,the substrate '$A$' is $\text{Pent-}3\text{-enenitrile}$.

(B) The Etard reaction is a chemical reaction in which an aromatic or heterocyclic bound methyl group is directly oxidized to an aldehyde using chromyl chloride $(CrO_2Cl_2)$.
Therefore,the correct reagent used in the Etard reaction is chromyl chloride.

(A) The Etard reaction is used for the oxidation of toluene to benzaldehyde.
Chromyl chloride $(CrO_2Cl_2)$ is the specific reagent used in this reaction.

(B) The Gatterman-Koch reaction is a specific formylation reaction used to introduce a formyl group $(-CHO)$ onto an aromatic ring.
In this reaction,the arene is treated with carbon monoxide $(CO)$ and hydrogen chloride $(HCl)$ in the presence of anhydrous aluminum chloride $(AlCl_3)$ and cuprous chloride $(CuCl)$ as a catalyst.
This combination generates the reactive electrophile $HCO^+$,which performs electrophilic aromatic substitution on the arene to yield an aromatic aldehyde.
Therefore,the correct reagent system is $CO, HCl$ (anhyd. $AlCl_3$).

(D) The given reaction sequence is the Stephen reduction followed by hydrolysis.
$1$. Ethanenitrile $(CH_3CN)$ reacts with $SnCl_2$ and $HCl$ to form an imine intermediate $(CH_3CH=NH)$,which is compound '$A$'.
$2$. The imine intermediate '$A$' upon acid hydrolysis $(H_3O^{+})$ yields Ethanal $(CH_3CHO)$ as the final product '$B$' along with ammonium chloride $(NH_4Cl)$.
$3$. The reaction is: $CH_3CN$ $\xrightarrow{SnCl_2, HCl} CH_3CH=NH$ $\xrightarrow{H_3O^{+}} CH_3CHO + NH_4Cl$.
Therefore,the product '$B$' is Ethanal.

(C) The reaction sequence is a Stephen reduction followed by hydrolysis.
$1$. Isopropyl cyanide $((CH_3)_2CH-CN)$ reacts with $SnCl_2$ and $HCl$ to form an imine hydrochloride intermediate $(A)$,which is $(CH_3)_2CH-CH=NH \cdot HCl$.
$2$. Subsequent acid hydrolysis $(H_3O^{+})$ of the imine hydrochloride converts the imine group into an aldehyde group,yielding $2-$Methylpropanal as product $B$ along with $NH_4Cl$.
$3$. The reaction is: $(CH_3)_2CH-CN$ $\xrightarrow[HCl]{SnCl_2} (CH_3)_2CH-CH=NH \cdot HCl$ $\xrightarrow{H_3O^{+}} (CH_3)_2CH-CHO + NH_4Cl$.

(A) The reaction of benzonitrile $(C_6H_5CN)$ with phenylmagnesium bromide $(C_6H_5MgBr)$ in the presence of dry ether forms an intermediate imine salt $(A)$: $C_6H_5-C(C_6H_5)=NMgBr$.
Upon acid hydrolysis $(H_3O^{+})$,this intermediate is converted into a ketone,specifically benzophenone $(C_6H_5-CO-C_6H_5)$,along with ammonia $(NH_3)$ and magnesium hydroxybromide $(Mg(Br)OH)$.
Therefore,product $B$ is benzophenone.

(C) The reagent $AlH(i-Bu)_2$ is Diisobutylaluminium hydride ($DIBAL$-$H$).
It is a selective reducing agent that reduces nitriles $(-CN)$ to imines,which upon acidic hydrolysis $(H_3O^+)$ yield aldehydes.
The reaction is: $CH_3-CH=CH-CH_2-CN \xrightarrow[2. H_3O^+]{1. AlH(i-Bu)_2} CH_3-CH=CH-CH_2-CHO$.
Thus,the product formed is $Pent-3-enal$.

(B) The Gatterman-Koch formylation of an arene involves the treatment of benzene or a substituted benzene with carbon monoxide $(CO)$ and hydrogen chloride $(HCl)$ under high pressure.
This reaction yields benzaldehyde or a substituted benzaldehyde.
The reaction is catalyzed by anhydrous aluminium chloride $(AlCl_3)$ or cuprous chloride $(CuCl)$.
Therefore,the correct reagent is $CO, HCl / AlCl_3$ (anhydrous).

(B) The reaction of benzonitrile $(C_6H_5CN)$ with phenylmagnesium bromide $(C_6H_5MgBr)$ in dry ether proceeds via a nucleophilic addition to the nitrile group.
First,the phenyl group from the Grignard reagent attacks the electrophilic carbon of the nitrile,forming an imine complex intermediate: $C_6H_5CN + C_6H_5MgBr \rightarrow C_6H_5-C(C_6H_5)=N-MgBr$.
Subsequently,acid-catalyzed hydrolysis $(H_3O^+)$ of the imine complex converts the $C=N-MgBr$ group into a carbonyl group $(C=O)$,yielding benzophenone $(C_6H_5-CO-C_6H_5)$ as the final product along with ammonia and magnesium salts.

(A) The reaction between an acid chloride $(RCOCl)$ and a dialkyl cadmium reagent $(R'_2Cd)$ is a standard method for the preparation of ketones.
In this reaction,$2$ moles of benzoyl chloride $(C_6H_5COCl)$ react with $1$ mole of dimethyl cadmium $((CH_3)_2Cd)$ to produce acetophenone $(C_6H_5COCH_3)$ and cadmium chloride $(CdCl_2)$.
The chemical equation is:
$2C_6H_5COCl + (CH_3)_2Cd \rightarrow 2C_6H_5COCH_3 + CdCl_2$
Thus,the product obtained is acetophenone.

(A) The dry distillation of calcium salts of carboxylic acids is a standard method for the preparation of ketones.
For calcium propionate,$(CH_3CH_2COO)_2Ca$,the reaction proceeds as follows:
$(CH_3CH_2COO)_2Ca \xrightarrow{\text{Dry Distillation}} CaCO_3 + CH_3CH_2COCH_2CH_3$
The product formed is $CH_3CH_2COCH_2CH_3$,which is diethyl ketone.
Diethyl ketone is also known as pentan-$3$-one.

(D) The reaction of a geminal dihalide with aqueous $KOH$ followed by heating leads to the hydrolysis of both halogen atoms.
$CH_3-CCl_2-CH_2-CH_3 + 2KOH_{(aq)} \rightarrow CH_3-C(OH)_2-CH_2-CH_3 + 2KCl$.
Geminal diols are unstable and readily lose a water molecule to form a carbonyl compound.
$CH_3-C(OH)_2-CH_2-CH_3 \rightarrow CH_3-CO-CH_2-CH_3 + H_2O$.
Thus,the product $A$ is butan$-2-$one $(CH_3-CO-CH_2-CH_3)$.

(C) The reaction described is the $Etard$ reaction.
In this reaction,toluene reacts with chromyl chloride $(CrO_2Cl_2)$ in the presence of a non-polar solvent like $CS_2$ or $CCl_4$ to form a brown chromium complex.
This complex,upon subsequent hydrolysis,yields benzaldehyde $(C_6H_5CHO)$.

(B) Step $1$: Reaction of Grignard reagent with cadmium chloride:
$2CH_3MgBr + CdCl_2 \rightarrow (CH_3)_2Cd + 2Mg(Cl)Br$
Here,product '$A$' is dimethyl cadmium,$(CH_3)_2Cd$.
Step $2$: Reaction of dimethyl cadmium with acetyl chloride:
$(CH_3)_2Cd + 2CH_3COCl \rightarrow 2CH_3COCH_3 + CdCl_2$
Here,product '$B$' is propanone $(CH_3COCH_3)$.

(D) This reaction is known as the Stephen reduction. The reaction proceeds as follows:
$1$. Benzonitrile $(C_6H_5-C \equiv N)$ is reduced by $SnCl_2$ and $HCl$ to form an imine hydrochloride intermediate,benzanimine hydrochloride $(C_6H_5-CH=NH \cdot HCl)$.
$2$. Subsequent acid hydrolysis $(H_3O^+)$ of the imine hydrochloride yields benzaldehyde $(C_6H_5-CHO)$ and ammonium chloride $(NH_4Cl)$.
The overall reaction is: $C_6H_5-C \equiv N + 2[H]$ $\xrightarrow{SnCl_2, HCl} C_6H_5-CH=NH \cdot HCl$ $\xrightarrow{H_3O^+} C_6H_5-CHO + NH_4Cl$.

(D) The reaction described is the $Stephen$ reduction.
Benzonitrile $(C_6H_5CN)$ reacts with $SnCl_2$ and $HCl$ to form an imine hydrochloride intermediate $(C_6H_5CH=NH \cdot HCl)$.
This intermediate,upon subsequent acid hydrolysis $(H_3O^+)$,yields benzaldehyde $(C_6H_5CHO)$ and ammonium chloride $(NH_4Cl)$.
The overall reaction is:
$C_6H_5CN + 2[H] \xrightarrow{SnCl_2, HCl} C_6H_5CH=NH \cdot HCl$
$C_6H_5CH=NH \cdot HCl + H_2O \xrightarrow{H_3O^+} C_6H_5CHO + NH_4Cl$

(B) The given reaction is the reduction of an acyl chloride $(C_6H_5COCl)$ to an aldehyde $(C_6H_5CHO)$.
This specific reaction is known as the Rosenmund reduction.
In this process,the acyl chloride is hydrogenated using hydrogen gas $(H_2)$ in the presence of a palladium catalyst that is poisoned with barium sulfate $(Pd-BaSO_4)$.
The role of $BaSO_4$ is to prevent the further reduction of the aldehyde to an alcohol.

(C) The reaction of $Toluene$ with $Chromyl chloride$ $(CrO_2Cl_2)$ in the presence of $Carbon disulfide$ $(CS_2)$ followed by acid hydrolysis $(H_3O^+)$ to form $Benzaldehyde$ is known as the $Etard reaction$.

(D) The given reaction is the hydrogenation of an acid chloride $(RCOCl)$ to an aldehyde $(RCHO)$ using a palladium catalyst poisoned with barium sulfate $(Pd-BaSO_4)$.
This specific reaction is known as the $Rosenmund$ reduction.
Therefore,the correct option is $D$.

(C) . Gattermann-Koch reaction: $C_6H_6 + CO + HCl \xrightarrow{\text{Anhy. } AlCl_3} C_6H_5CHO$ (Produces benzaldehyde).
$B$. Etard reaction: $C_6H_5CH_3 + CrO_2Cl_2 \xrightarrow{CS_2} C_6H_5CHO$ (Produces benzaldehyde).
$C$. Oxidation of toluene with $KMnO_4 + KOH$ yields potassium benzoate $(C_6H_5COOK)$,which upon acidification gives benzoic acid $(C_6H_5COOH)$,not benzaldehyde.
$D$. Rosenmund reduction: $C_6H_5COCl + H_2 \xrightarrow{Pd-BaSO_4} C_6H_5CHO$ (Produces benzaldehyde).
Therefore,the reaction that does not produce benzaldehyde is $C$.

(C) The Gatterman-Koch reaction is a specific form of Friedel-Crafts acylation used to introduce a formyl group $(-CHO)$ into an aromatic ring.
In this reaction,benzene is treated with carbon monoxide $(CO)$ and hydrogen chloride $(HCl)$ in the presence of anhydrous aluminum chloride $(AlCl_3)$ or cuprous chloride $(CuCl)$ to produce benzaldehyde.
The reaction is represented as: $C_6H_6 + CO + HCl \xrightarrow{anhydrous \ AlCl_3 / CuCl} C_6H_5CHO$.

(C) The conversion of nitriles $(R-CN)$ to aldehydes $(R-CHO)$ is typically achieved using $DIBAL-H$ (Diisobutylaluminium hydride) followed by hydrolysis.
$CH_3CN + DIBAL-H \xrightarrow{H_2O} CH_3CHO$.
$LiAlH_4$ would reduce the nitrile all the way to a primary amine $(CH_3CH_2NH_2)$.

(B) Dry distillation of a mixture of calcium salts of carboxylic acids produces carbonyl compounds.
$1. (HCOO)_{2}Ca \xrightarrow{\Delta} HCHO + CaCO_{3}$ (Methanal)
$2. (CH_{3}COO)_{2}Ca \xrightarrow{\Delta} CH_{3}COCH_{3} + CaCO_{3}$ (Propanone)
$3. (HCOO)_{2}Ca + (CH_{3}COO)_{2}Ca \xrightarrow{\Delta} 2CH_{3}CHO + 2CaCO_{3}$ (Ethanal)
Propanal $(CH_{3}CH_{2}CHO)$ is not formed in this reaction.

(D) The given reaction is: $C_6H_5CN$ $\xrightarrow[(ii) H_3O^+]{(i) SnCl_2 + HCl} C_6H_5CHO (X)$ $\xrightarrow{conc. KOH} C_6H_5COOK (Y) + C_6H_5CH_2OH (Z)$.
$1$. The conversion of nitrile $(C_6H_5CN)$ to aldehyde $(C_6H_5CHO)$ using $SnCl_2 + HCl$ followed by hydrolysis is known as the Stephen reaction.
$2$. Benzaldehyde $(C_6H_5CHO)$ lacks $\alpha$-hydrogen atoms,so it undergoes the Cannizzaro reaction in the presence of concentrated alkali $(KOH)$ to form a mixture of alcohol $(C_6H_5CH_2OH)$ and salt of carboxylic acid $(C_6H_5COOK)$.

(A) The reaction in option $(A)$ involves Clemmensen reduction conditions $(Zn-Hg/HCl)$,which are used to reduce carbonyl groups (aldehydes or ketones) to methylene groups $(-CH_2-)$. Carboxylic acids are generally inert to these conditions,so benzaldehyde cannot be prepared from benzoic acid using this method.
Option $(B)$ is the Etard reaction,which produces benzaldehyde from toluene.
Option $(C)$ is the Rosenmund reduction,which produces benzaldehyde from benzoyl chloride.
Option $(D)$ is the Gattermann-Koch reaction,which produces benzaldehyde from benzene.