Preparation Questions in English

(B) When calcium acetate is heated (dry distillation),it undergoes thermal decomposition to form acetone and calcium carbonate.
$(CH_3COO)_2Ca \xrightarrow{\Delta} CH_3COCH_3 + CaCO_3$

(D) The reaction of a nitrile $(-CN)$ with a Grignard reagent $(PhMgBr)$ followed by acid hydrolysis $(H_3O^+)$ is a standard method for the preparation of ketones.
$1$. The nucleophilic phenyl group $(Ph^-)$ from $PhMgBr$ attacks the electrophilic carbon of the nitrile group $(-CN)$ to form an imine magnesium salt intermediate.
$2$. Subsequent acid hydrolysis $(H_3O^+)$ converts the imine intermediate into a ketone $(C=O)$.
$3$. The other substituent on the benzene ring,$-\text{CH}(\text{CH}_3)\text{OCH}_3$,remains unaffected by these reagents.
$4$. Therefore,the final product $X$ is a ketone where the $-CN$ group is replaced by a $-COC_6H_5$ group,while the side chain $-\text{CH}(\text{CH}_3)\text{OCH}_3$ remains unchanged.

(D) $CH_3-CH_2-OH$ on oxidation with $CrO_3 - H_2SO_4$ (Jones reagent) gives acetic acid $(CH_3-COOH)$ because it is a strong oxidizing agent.
The other reactions produce acetaldehyde $(CH_3-CHO)$:
$(A)$ Dehydrogenation of ethanol: $CH_3-CH_2-OH \xrightarrow[573 \ K]{Cu} CH_3-CHO + H_2$
$(B)$ Reduction of nitrile: $CH_3-CN \xrightarrow[(ii) H_2O]{(i) DIBAL-H} CH_3-CHO$
$(C)$ Wacker process: $CH_2=CH_2 + O_2 \xrightarrow[H_2O]{Pd(II)/Cu(II)} CH_3-CHO$

(D) The reaction proceeds in three steps:
$1$. Acidic hydrolysis of propanenitrile $(CH_{3}CH_{2}CN)$ yields propanoic acid $(CH_{3}CH_{2}COOH)$.
$2$. Treatment of propanoic acid with $SOCl_{2}$ yields propanoyl chloride $(CH_{3}CH_{2}COCl)$.
$3$. Rosenmund reduction of propanoyl chloride using $Pd/BaSO_{4}$ and $H_{2}$ yields propanaldehyde $(CH_{3}CH_{2}CHO)$.

(A) The reaction is the $OXO$ process or hydroformylation of alkenes.
In this reaction,an alkene reacts with $CO$ and $H_2$ in the presence of a rhodium $(Rh)$ or cobalt catalyst to form an aldehyde.
For the substrate $CH_3CH_2CH=CH_2$ (but$-1-$ene),the hydroformylation leads to two isomeric aldehydes:
$1$. $CH_3CH_2CH_2CH_2CHO$ (pentanal,linear product)
$2$. $CH_3CH_2CH(CH_3)CHO$ ($2$-methylbutanal,branched product)
Using a rhodium catalyst,the linear aldehyde is typically the major product due to less steric hindrance during the formation of the transition state.
Therefore,the major product is $CH_3CH_2CH_2CH_2CHO$.

(A) The given reaction is the $Etard$ reaction,which is used for the oxidation of toluene to benzaldehyde using chromyl chloride $(CrO_2Cl_2)$ in the presence of a non-polar solvent like carbon disulfide $(CS_2)$.
In this reaction,the methyl group of toluene is oxidized to a chromium complex intermediate,which is $C_6H_5CH(OCrOHCl_2)_2$.
This intermediate is then hydrolyzed with water $(H_3O^+)$ to yield benzaldehyde.

(C) The reaction of ethyl ethanoate $(CH_{3}COOCH_{2}CH_{3})$ with Grignard reagent $(CH_{3}MgBr)$ proceeds in two steps:
$1$. First,$1$ equivalent of $CH_{3}MgBr$ attacks the carbonyl carbon of the ester,leading to the elimination of the ethoxide group and the formation of acetone $(CH_{3}COCH_{3})$.
$2$. Second,another equivalent of $CH_{3}MgBr$ attacks the carbonyl carbon of the formed acetone to produce a tertiary alkoxide intermediate $(CH_{3}C(OMgBr)(CH_{3})_{2})$.
$3$. Upon acidic workup (hydrolysis),this intermediate yields $2-$methylpropan$-2-$ol.
Thus,a total of $2$ equivalents of $CH_{3}MgBr$ are required for the complete conversion of $1$ mole of ethyl ethanoate to $1$ mole of $2-$methylpropan$-2-$ol.

(A) The reaction of nitriles $(R-CN)$ with $DIBAL-H$ (Diisobutylaluminium hydride) followed by hydrolysis is a standard method for the preparation of aldehydes.
The reaction proceeds as follows:
$R-C \equiv N \xrightarrow[(ii) H_2O]{(i) DIBAL-H} R-CHO$
Here,the nitrile group is reduced to an aldehyde group.
Therefore,$Y$ is $-CHO$ (Aldehyde).

(C) The conversion of $A$ ($4$-methylmethylenecyclohexane) to $B$ ($4$-methylcyclohexanecarbaldehyde) involves the transformation of an exocyclic double bond into an aldehyde group.
$1$. First,hydroboration-oxidation using $BH_3, H_2O_2 / OH^-$ is performed. This reaction follows anti-Markovnikov addition of water across the double bond to form a primary alcohol,$4$-methylcyclohexylmethanol.
$2$. Second,the primary alcohol is oxidized to an aldehyde using $PCC$ (Pyridinium chlorochromate). $PCC$ is a mild oxidizing agent that stops the oxidation at the aldehyde stage,preventing further oxidation to a carboxylic acid.

(C) Acetophenone is $C_6H_5COCH_3$.
$(A)$ $C_6H_5CHO + CH_3MgBr \rightarrow C_6H_5CH(OH)CH_3$. Oxidation with $Na_2Cr_2O_7/H^+$ gives $C_6H_5COCH_3$ (Acetophenone).
$(B)$ $CH_3CHO + C_6H_5MgBr \rightarrow CH_3CH(OH)C_6H_5$. Oxidation with $PCC$ gives $CH_3COC_6H_5$ (Acetophenone).
$(C)$ Esters react with $2$ equivalents of Grignard reagent to form tertiary alcohols. $C_6H_5COOC_2H_5 + 2CH_3MgBr \rightarrow C_6H_5C(OH)(CH_3)_2$. This does not give acetophenone.
$(D)$ Acid chlorides react with organocadmium reagents (formed in situ from $2RMgX + CdCl_2$) to give ketones. $2C_6H_5COCl + (CH_3)_2Cd \rightarrow 2C_6H_5COCH_3$. This gives acetophenone.

(B) The oxidation of toluene to benzaldehyde is known as the Etard reaction.
In this reaction,toluene is treated with chromyl chloride $(CrO_2Cl_2)$ or,as shown in the options,with $CrO_3$ in the presence of acetic anhydride.
The $CrO_3$ with acetic anhydride forms a gem-diacetate intermediate,which upon subsequent hydrolysis with $H_3O^+$ yields benzaldehyde.
The reaction sequence is:
$C_6H_5CH_3$ $\xrightarrow{CrO_3 / \text{acetic anhydride}} C_6H_5CH(OCOCH_3)_2$ $\xrightarrow{H_3O^+} C_6H_5CHO + 2CH_3COOH$
Therefore,the correct reagent is $CrO_3$ / acetic anhydride followed by $H_3O^+$.

(C) Let us analyze each reaction:
$(A)$ Benzoic acid reacts with $SOCl_2$ to form benzoyl chloride,which on Rosenmund reduction $(H_2/Pd/BaSO_4)$ yields benzaldehyde.
$(B)$ Benzyl alcohol on oxidation with $CrO_3/H_2SO_4$ (Jones reagent) yields benzoic acid,not benzaldehyde.
$(C)$ Methyl benzoate does not react with $NaBH_4$ to form benzaldehyde.
$(D)$ Toluene reacts with $CrO_3$ in the presence of acetic anhydride to form a gem-diacetate intermediate,which on hydrolysis yields benzaldehyde (Etard reaction).
Therefore,reactions $(A)$ and $(D)$ yield benzaldehyde.

(A) The correct matches are as follows:
$A$. $C_6H_5COCl + H_2 \xrightarrow{Pd-BaSO_4} C_6H_5CHO$ is the Rosenmund reaction $(IV)$.
$B$. $CH_3CN + SnCl_2 + HCl \rightarrow CH_3CHO$ is the Stephen reaction $(III)$.
$C$. $C_6H_5CH_3 + CrO_2Cl_2 \rightarrow C_6H_5CHO$ is the Etard reaction $(II)$.
$D$. $C_6H_6 + CO + HCl \xrightarrow{AlCl_3} C_6H_5CHO$ is the Gattermann-Koch reaction $(I)$.
Therefore,the correct sequence is $A-IV, B-III, C-II, D-I$.

(C) The reaction of a nitrile $(CH_3CH_2CN)$ with a Grignard reagent $(CH_3MgBr)$ followed by acidic hydrolysis $(H_3O^+)$ is a standard method for the preparation of ketones.
Step $1$: The nucleophilic methyl group from $CH_3MgBr$ attacks the electrophilic carbon of the nitrile group to form an imine salt intermediate $(CH_3CH_2C(NMgBr)CH_3)$.
Step $2$: Acidic hydrolysis of the imine salt converts it into a ketone $(CH_3CH_2COCH_3)$.
Therefore,the major final product is butan-$2$-one $(CH_3CH_2COCH_3)$.

(D) Let us analyze each reaction pathway for the synthesis of benzaldehyde:
$A$. Etard reaction: Toluene is oxidized by $CrO_2Cl_2$ in $CS_2$ followed by hydrolysis to give benzaldehyde. This is a correct method.
$B$. Rosenmund reduction: Benzoyl chloride is reduced by $H_2$ in the presence of $Pd-BaSO_4$ (Lindlar catalyst) to give benzaldehyde. This is a correct method.
$C$. Toluene is oxidized by $CrO_3$ in the presence of acetic anhydride to form a gem-diacetate intermediate,which upon hydrolysis yields benzaldehyde. This is a correct method.
$D$. Reaction of benzonitrile $(C_6H_5CN)$ with $CH_3MgBr$ (a Grignard reagent) followed by hydrolysis yields acetophenone $(C_6H_5COCH_3)$,not benzaldehyde. Therefore,this is an incorrect method for the synthesis of benzaldehyde.

(A) The reaction shown is the partial reduction of an acid chloride (benzoyl chloride) to an aldehyde (benzaldehyde) using $H_2$ in the presence of a poisoned catalyst,$Pd-BaSO_4$. This specific reaction is known as the Rosenmund reduction.

(B) The reaction involving $CrO_3$ and $(CH_3CO)_2O$ at $273-283 \ K$ is the oxidation of toluene to benzaldehyde via a gem-diacetate intermediate.
The reaction involving $CrO_2Cl_2$ (chromyl chloride) in $CS_2$ is the Etard reaction,which also oxidizes toluene to benzaldehyde via a chromium complex intermediate.
Therefore,reagent $A$ is $CrO_3$ and reagent $B$ is $CrO_2Cl_2$.

(C) The given reaction involves the treatment of benzene with carbon monoxide $(CO)$ and hydrogen chloride $(HCl)$ in the presence of anhydrous aluminum chloride $(AlCl_3)$ and cuprous chloride $(CuCl)$ to form benzaldehyde. This specific reaction is known as the Gatterman-Koch reaction.

(C) The given reaction is the Gattermann-Koch reaction. In this reaction,benzene reacts with carbon monoxide $(CO)$ and hydrogen chloride $(HCl)$ in the presence of anhydrous aluminum chloride $(AlCl_3)$ and cuprous chloride $(CuCl)$ to form benzaldehyde as the major product. The reaction is: $C_6H_6 + CO + HCl \xrightarrow{Anhydrous AlCl_3/CuCl} C_6H_5CHO$.

(A) The transformation involves the anti-Markovnikov hydration of an alkene followed by the oxidation of the resulting primary alcohol to an aldehyde.
$1$. Hydroboration-oxidation: The reaction of $Cyclohexyl-CH_2-CH=CH_2$ with $(i) BH_3$ followed by $(ii) H_2O_2 / \stackrel{\ominus}{O}H$ yields the primary alcohol,$Cyclohexyl-CH_2-CH_2-CH_2OH$.
$2$. Oxidation: The primary alcohol is then oxidized to an aldehyde using a mild oxidizing agent like $PCC$ (Pyridinium chlorochromate),which stops the oxidation at the aldehyde stage.
Therefore,the correct sequence of reagents is $(i) BH_3, (ii) H_2O_2 / \stackrel{\ominus}{O}H, (iii) PCC$.

(D) Let us analyze each reaction:
$I$. Gattermann-Koch reaction: Benzene reacts with $CO$ and $HCl$ in the presence of anhydrous $AlCl_3/CuCl$ to form benzaldehyde. This is a standard method for the preparation of benzaldehyde.
$II$. Hydrolysis of benzal chloride: Benzal chloride $(C_6H_5CHCl_2)$ on hydrolysis with water at $100 \ ^\circ C$ yields benzaldehyde $(C_6H_5CHO)$.
$III$. Rosenmund reduction: Benzoyl chloride $(C_6H_5COCl)$ is reduced to benzaldehyde using $H_2$ in the presence of $Pd-BaSO_4$ (Lindlar's catalyst).
$IV$. Reduction of ester: Methyl benzoate $(C_6H_5CO_2Me)$ is reduced to benzaldehyde using $DIBAL-H$ at $-78 \ ^\circ C$ followed by hydrolysis.
All four reactions produce benzaldehyde. Therefore,the total number of reactions is $4$.

(A) The target molecule "$x$" is $1$-acetylcyclopent$-1-$ene.
To obtain this via an intramolecular aldol condensation,the precursor must be a dicarbonyl compound,specifically $6$-oxoheptanal.
Ozonolysis of $1$-methylcyclohex$-1-$ene ($1$-methylcyclohexene) breaks the double bond to form $6$-oxoheptanal $(CH_3COCH_2CH_2CH_2CH_2CHO)$.
This dicarbonyl compound undergoes intramolecular aldol condensation in the presence of $OH^{\ominus}/\Delta$ to form the five-membered ring product,$1$-acetylcyclopent$-1-$ene.
Thus,the correct starting alkene is $1$-methylcyclohex$-1-$ene.

(A) The given reactions are matched as follows:
$A$. $RCN \xrightarrow[(ii) H_3O^+]{(i) SnCl_2, HCl} RCHO$ is the Stephen reaction $(IV)$.
$B$. $C_6H_5COCl \xrightarrow{H_2, Pd-BaSO_4} C_6H_5CHO$ is the Rosenmund reduction $(III)$.
$C$. $C_6H_5CH_3 \xrightarrow[(ii) H_3O^+]{(i) CrO_2Cl_2, CS_2} C_6H_5CHO$ is the Etard reaction $(I)$.
$D$. $C_6H_6 \xrightarrow[(ii) \text{anhydrous } AlCl_3/CuCl]{(i) CO, HCl} C_6H_5CHO$ is the Gatterman-Koch reaction $(II)$.
Therefore,the correct matching is $A-IV, B-III, C-I, D-II$.

(A) The given reaction involves the formylation of benzene using carbon monoxide $(CO)$ and hydrogen chloride $(HCl)$ in the presence of anhydrous aluminum chloride $(AlCl_3)$ to produce benzaldehyde.
This specific reaction is known as the Gatterman-Koch reaction.

(B) The reaction of $CH_3CN$ with $SnCl_2 + HCl$ (Stephen reduction) followed by hydrolysis yields an aldehyde $(CH_3CHO)$.
The reaction of $CH_3CN$ with a Grignard reagent $(CH_3MgX)$ followed by hydrolysis yields a ketone $(CH_3COCH_3)$.
Therefore,$X$ is $SnCl_2 + HCl$ and $Y$ is $CH_3MgX$.

(A) The reaction shown is the Etard reaction,which is used for the oxidation of toluene to benzaldehyde.
In this reaction,toluene reacts with chromyl chloride $(CrO_2Cl_2)$ in the presence of carbon disulfide $(CS_2)$ to form a brown chromium complex intermediate.
The structure of this chromium complex '$X$' is $C_6H_5CH(OCrOHCl_2)_2$.
Upon hydrolysis $(H_3O^+)$,this complex yields benzaldehyde.

(A) The hydrolysis of $2,2-$dichloro propane $(CH_3-CCl_2-CH_3)$ involves the replacement of two chlorine atoms with two hydroxyl groups to form $2,2-$propane diol $(CH_3-C(OH)_2-CH_3)$.
Since two hydroxyl groups are attached to the same carbon atom,the resulting gem-diol is unstable.
It readily loses a water molecule $(H_2O)$ to form a stable carbonyl compound,which is acetone $(CH_3-CO-CH_3)$.

(B) The given reaction is the $Etard$ reaction.
In this reaction,$Toluene$ is treated with $Chromyl$ $chloride$ $(CrO_2Cl_2)$ in the presence of $CS_2$ as a solvent to form a brown chromium complex.
This complex on hydrolysis with $H_3O^+$ yields $Benzaldehyde$ as the final product '$B$'.
The overall reaction is: $C_6H_5CH_3 + 2CrO_2Cl_2$ $\xrightarrow{CS_2} C_6H_5CH(OCrOHCl_2)_2$ $\xrightarrow{H_3O^+} C_6H_5CHO$.

(A) The reaction is an example of the Etard reaction modification using acetic anhydride and chromic oxide.
Step $1$: $4-$Nitrotoluene reacts with acetic anhydride and $CrO_3$ at $273-278 \ K$ to form a gem-diacetate derivative $(X)$.
Step $2$: The gem-diacetate derivative $(X)$ undergoes acid-catalyzed hydrolysis $(H_3O^{+} / \Delta)$ to yield $4-$nitrobenzaldehyde $(Y)$.

(A) The Rosenmund reduction is a hydrogenation process where an acyl chloride $(R-COCl)$ is reduced to an aldehyde $(R-CHO)$ using hydrogen gas in the presence of a poisoned palladium catalyst,specifically palladium on barium sulfate $(Pd-BaSO_4)$.
The reaction is represented as:
$R-COCl + H_2 \xrightarrow{Pd-BaSO_4} R-CHO + HCl$
Therefore,option $A$ represents the Rosenmund reduction.

(C) Rosenmund reduction involves the hydrogenation of acid chlorides to aldehydes using $H_2$ gas in the presence of palladium supported on barium sulfate $(Pd-BaSO_4)$.
In this reaction,benzoyl chloride $(C_6H_5COCl)$ is reduced to benzaldehyde $(C_6H_5CHO)$.
The chemical equation is:
$C_6H_5COCl + H_2 \xrightarrow{Pd-BaSO_4} C_6H_5CHO + HCl$
Therefore,the correct product is benzaldehyde.

(D) Let us analyze the given reactions:
$A$) Rosenmund reduction uses $H_2/Pd-BaSO_4$ to reduce acid chlorides to aldehydes. This is correct.
$B$) Stephen reaction uses $SnCl_2/HCl$ to reduce nitriles to imines,which are then hydrolyzed to aldehydes. This is correct.
$C$) Etard reaction uses $CrO_2Cl_2$ to oxidize toluene to benzaldehyde. This is correct.
$D$) Gatterman-Koch reaction uses $CO + HCl$ in the presence of anhydrous $AlCl_3$ to form benzaldehyde from benzene. The reagent given in the option $(CrO_3/(CH_3CO)_2O)$ is for the Etard-like oxidation of toluene to benzaldehyde diacetate. Thus,option $D$ is incorrectly matched.

(A) The side chain chlorination of toluene $(C_6H_5CH_3)$ with $Cl_2$ in the presence of light $(hv)$ gives benzal chloride $(C_6H_5CHCl_2)$.
Subsequent acid hydrolysis of benzal chloride at $373 \ K$ leads to the formation of benzaldehyde $(C_6H_5CHO)$ as the final product.
The reaction sequence is:
$C_6H_5CH_3$ $\xrightarrow{Cl_2/hv} C_6H_5CHCl_2$ $\xrightarrow{H_2O/H^+, 373 \ K} C_6H_5CHO$.

(D) The reaction of benzonitrile $(C_6H_5CN)$ with phenyl magnesium bromide $(C_6H_5MgBr)$ in dry ether proceeds via a nucleophilic addition to the nitrile group.
$1$. The phenyl group from the Grignard reagent attacks the electrophilic carbon of the nitrile group to form an imine magnesium bromide complex $(C_6H_5-C(C_6H_5)=NMgBr)$.
$2$. Upon subsequent acid hydrolysis $(H_3O^+)$,the imine complex is converted into a ketone.
$3$. The final product formed is benzophenone $(C_6H_5-CO-C_6H_5)$.

(B) $Ethyl \ methyl \ ketone$ is an unsymmetric ketone,so two different acids (in the form of calcium salts) are used.
The structure of $ethyl \ methyl \ ketone$ is $CH_3COCH_2CH_3$.
The reaction involves the dry distillation of a mixture of calcium acetate and calcium propionate:
$(CH_3COO)_2Ca + (C_2H_5COO)_2Ca \xrightarrow{\Delta} 2CH_3COCH_2CH_3 + 2CaCO_3$.

(A) $2, 2-$dichloropentane on hydrolysis gives $pentan-2-one$ because the geminal dichloride is at the $C-2$ position.
$3, 3-$dichloropentane on hydrolysis gives $pentan-3-one$.
$Pentan-3-ol$ on oxidation gives $pentan-3-one$.
$Pent-2-yne$ on hydration in the presence of $Hg^{2+}/H^+$ gives $pentan-2-one$ and $pentan-3-one$ as a mixture.
Therefore,$2, 2-$dichloropentane is the correct answer as it exclusively yields $pentan-2-one$.

(D) The reaction described is the $Stephen$ reduction.
$1$. Benzonitrile $(C_6H_5CN)$ reacts with stannous chloride $(SnCl_2)$ and hydrochloric acid $(HCl)$ to form an imine hydrochloride intermediate $(C_6H_5CH=NH \cdot HCl)$.
$2$. Subsequent acid hydrolysis of this intermediate yields benzaldehyde $(C_6H_5CHO)$.
$3$. The overall reaction is: $C_6H_5CN + 2[H]$ $\xrightarrow{SnCl_2/HCl} C_6H_5CH=NH$ $\xrightarrow{H_3O^+} C_6H_5CHO$.

(B) The Rosenmund reduction is a hydrogenation process used to convert acid chlorides into aldehydes.
The reagent used is $H_2$ gas in the presence of palladium $(Pd)$ catalyst supported on barium sulfate $(BaSO_4)$.
$BaSO_4$ acts as a poison to prevent the further reduction of the aldehyde to a primary alcohol.
Therefore,the correct reagent is $H_2 / Pd, BaSO_4$.

(B) The reaction of benzoyl chloride $(C_6H_5COCl)$ to benzaldehyde $(C_6H_5CHO)$ is a standard reduction reaction.
Acyl chlorides are reduced to their corresponding aldehydes by hydrogen gas in the presence of a palladium catalyst poisoned with barium sulfate $(Pd-BaSO_4)$.
This specific reaction is known as the Rosenmund reduction.
Therefore,the reagent '$R$' is $H_2, Pd-BaSO_4$.

(B) Nitriles are reduced to imine hydrochloride by stannous chloride in the presence of hydrochloric acid,which on acid hydrolysis gives corresponding aldehydes. This reaction is called the Stephen reaction.
$R-CN \xrightarrow[ii) H_3O^+]{i) SnCl_2, HCl} R-CHO + NH_4Cl$

(B) The given reaction is the Gattermann-Koch reaction.
In this reaction,benzene $(C_6H_6)$ reacts with carbon monoxide $(CO)$ and hydrogen chloride $(HCl)$ in the presence of anhydrous aluminum chloride $(AlCl_3)$ under high pressure to form benzaldehyde $(C_6H_5CHO)$.
The reaction is: $C_6H_6 + CO + HCl \xrightarrow[\text{Anhydrous } AlCl_3]{\text{high pressure}} C_6H_5CHO$.
Therefore,the correct product is benzaldehyde $(C_6H_5CHO)$.

(C) The given reaction is the Stephen reduction reaction.
In the first step,the nitrile $(R-C \equiv N)$ is reduced by $SnCl_2/HCl$ to form an imine hydrochloride intermediate $(A)$,which is $R-CH=NH \cdot HCl$.
In the second step,the hydrolysis of the imine hydrochloride $(A)$ with $H_3O^{+}$ yields an aldehyde $(B)$ as the final product,which is $R-CHO$.

(C) The reduction of nitriles to aldehydes is known as the $Stephen$ reduction.
In this reaction,nitriles $(R-C \equiv N)$ are reduced to imines using stannous chloride $(SnCl_2)$ in the presence of hydrochloric acid $(HCl)$,which are then hydrolyzed to form aldehydes $(RCHO)$.
The reaction is:
$R-C \equiv N \xrightarrow[(ii) H_3O^+]{(i) SnCl_2 / dil. HCl} RCHO + NH_4Cl$

(A) The reaction of benzonitrile $(C_6H_5CN)$ with phenylmagnesium bromide $(C_6H_5MgBr)$ in the presence of dry ether followed by acid hydrolysis $(H_3O^+)$ yields benzophenone $(C_6H_5COC_6H_5)$.
The reaction steps are:
$1$. $C_6H_5-C \equiv N + C_6H_5-MgBr \xrightarrow{\text{dry ether}} C_6H_5-C(C_6H_5)=NMgBr$
$2$. $C_6H_5-C(C_6H_5)=NMgBr \xrightarrow{H_3O^+} C_6H_5-CO-C_6H_5 + NH_3 + Mg(Br)OH$
Thus,the required reagent is $C_6H_5MgBr$.

(B) In the Etard reaction,the methyl group of toluene (methylbenzene) is oxidized using $CrO_2Cl_2$ (chromyl chloride) in the presence of $CS_2$ (carbon disulfide) as a solvent.
This process forms a chromium complex intermediate,which upon acid hydrolysis yields benzaldehyde.
Therefore,the reagent used is chromyl chloride.

(C) The reaction of toluene with chromyl chloride $(CrO_2Cl_2)$ in the presence of carbon disulfide $(CS_2)$ followed by acid hydrolysis is known as the $Etard$ reaction.
In this reaction,the methyl group of toluene is oxidized to an aldehyde group,resulting in the formation of benzaldehyde $(C_6H_5CHO)$.

(B) The given reaction is the Etard reaction.
In this reaction,Toluene is oxidized by chromyl chloride $(CrO_2Cl_2)$ in the presence of carbon disulfide $(CS_2)$ to form a brown chromium complex $(A)$.
This complex $(A)$ on hydrolysis with water $(H_3O^+)$ gives Benzaldehyde $(B)$.
The reaction is:
$C_6H_5CH_3 + 2CrO_2Cl_2$ $\xrightarrow{CS_2} C_6H_5CH(OCrOHCl_2)_2$ $\xrightarrow{H_3O^+} C_6H_5CHO$
Thus,the product '$B$' is Benzaldehyde.

(C) The given reaction is the hydrogenation of an acid chloride $(R-CO-Cl)$ to an aldehyde $(R-CHO)$ using hydrogen gas $(H_2)$ in the presence of a palladium catalyst supported on barium sulfate $(Pd-BaSO_4)$.
This specific reaction is known as the Rosenmund reduction.

(D) The Rosenmund reduction is a hydrogenation process used to convert acid chlorides into aldehydes.
It employs $H_2$ gas in the presence of a poisoned catalyst,specifically $Pd$ supported on $BaSO_4$ (palladium on barium sulfate).
The $BaSO_4$ acts as a poison to prevent further reduction of the aldehyde to an alcohol.

(C) In the Gatterman-Koch reaction,benzene or its derivatives are treated with carbon monoxide $(CO)$ and hydrogen chloride $(HCl)$ in the presence of anhydrous aluminum chloride $(AlCl_3)$ or cuprous chloride $(CuCl)$ under high pressure to form benzaldehyde or substituted benzaldehydes.
Thus,the reagents used are $CO$ and $HCl$.