Preparation Questions in English

(A) Step $1$: Benzoic acid $(Ph-COOH)$ reacts with $SOCl_2$ to form benzoyl chloride $(Ph-COCl)$ as product $(A)$.
$Ph-COOH + SOCl_2 \rightarrow Ph-COCl + SO_2 + HCl$
Step $2$: Benzoyl chloride undergoes Rosenmund reduction in the presence of $H_2$ and $Pd-BaSO_4$ to form benzaldehyde $(Ph-CHO)$ as product $(B)$.
$Ph-COCl + H_2 \xrightarrow{Pd-BaSO_4} Ph-CHO + HCl$

(A) The reaction of a nitrile $(PhCN)$ with a Grignard reagent $(CH_3MgBr)$ followed by acid hydrolysis is a standard method for the synthesis of ketones.
Step $1$: The nucleophilic $CH_3^-$ group from $CH_3MgBr$ attacks the electrophilic carbon of the nitrile group $(Ph-C \equiv N)$ to form an imine magnesium salt intermediate,$Ph-C(CH_3)=NMgBr$.
Step $2$: Subsequent acid hydrolysis $(H_3O^+)$ of the imine intermediate yields the ketone,$PhCOCH_3$.

(C) Step $1$: Acidic hydrolysis of nitrile $(CH_3)_2CHCH_2C \equiv N$ yields carboxylic acid $A$,which is $(CH_3)_2CHCH_2COOH$.
Step $2$: Reduction of carboxylic acid $A$ with $LiAlH_4$ followed by water yields primary alcohol $B$,which is $(CH_3)_2CHCH_2CH_2OH$.
Step $3$: Oxidation of primary alcohol $B$ with $PCC$ (Pyridinium chlorochromate) in $CH_2Cl_2$ yields aldehyde $C$,which is $(CH_3)_2CHCH_2CHO$.

(D) . Friedel-Crafts acylation of benzene with acetyl chloride in the presence of anhydrous $AlCl_3$ yields acetophenone.
$B$. Dry distillation of a mixture of calcium benzoate and calcium acetate yields acetophenone.
$C$. Reaction of phenylmagnesium iodide (or methylmagnesium iodide with benzonitrile) followed by hydrolysis yields acetophenone.
Since all three methods are valid,the correct option is $(d)$.

(B) The given reaction is a variation of the Gattermann-Koch reaction or a form of Friedel-Crafts formylation using formyl chloride equivalent $(HN=CH-Cl)$.
$1$. The reagent $HN=CH-Cl$ in the presence of $AlCl_3$ generates an electrophilic species $HN=CH^+$.
$2$. This electrophile attacks the toluene ring. Since the methyl group $(-CH_3)$ is an ortho/para-directing group and para-position is sterically less hindered,the electrophile attacks the para-position to form an imine intermediate $(A)$.
$3$. Subsequent hydrolysis of the imine intermediate $(A)$ with $H_3O^+$ yields the corresponding aldehyde,which is $4$-methylbenzaldehyde (p-tolualdehyde).

(B) The reaction sequence proceeds as follows:
$1$. Nitration of benzene with $HNO_3/H_2SO_4$ gives nitrobenzene $(C_6H_5NO_2)$.
$2$. Reduction of nitrobenzene with $H_2/Pd/C$ gives aniline $(C_6H_5NH_2)$.
$3$. Diazotization of aniline with $NaNO_2/HCl$ at $0-5^{\circ}C$ gives benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$.
$4$. Reaction with $CuCN$ (Sandmeyer reaction) replaces the diazonium group with a cyano group,yielding benzonitrile $(C_6H_5CN)$.
$5$. Grignard reaction with $MeMgBr$ followed by acidic hydrolysis $(H_3O^+)$ converts the nitrile group into a ketone group,yielding acetophenone $(C_6H_5COCH_3)$.

(C) The reagent $AlH(i-Bu)_2$ is known as $DIBAL-H$ (Diisobutylaluminium hydride).
It is a selective reducing agent that reduces nitriles $(R-C \equiv N)$ to imines,which upon subsequent hydrolysis with $H_2O$ yield aldehydes $(RCHO)$.
The reaction mechanism involves the nucleophilic attack of the hydride from $DIBAL-H$ onto the carbon of the nitrile group,followed by hydrolysis of the resulting imine intermediate to form the aldehyde.

(A) The given reaction is the $Stephen$ reaction.
In this reaction,nitriles $(R-CN)$ are reduced to imines using stannous chloride $(SnCl_2)$ in the presence of hydrochloric acid $(HCl)$.
The intermediate imine ($A$,which is $R-CH=NH \cdot HCl$) is then hydrolyzed with water $(H_3O^+)$ to yield the corresponding aldehyde $(RCHO)$.

(C) The reaction shown is the decarboxylation of acetoacetic acid $(CH_3COCH_2COOH \xrightarrow{\Delta} CH_3COCH_3 + CO_2)$.
Thus,$X$ is acetone $(CH_3COCH_3)$.
Now,let us evaluate the options:
Option $A$: $CH_3-CH_2-CH_2-Cl \xrightarrow{LiAlH_4} CH_3-CH_2-CH_3$ (Propane).
Option $B$: $CH_3-CH_2-COOH \xrightarrow{P + HI} CH_3-CH_2-CH_3$ (Propane).
Option $C$: $CH_3-C \equiv CH \xrightarrow[H_2O]{Hg^{+2}} CH_3-CO-CH_3$ (Acetone).
Since $X$ is acetone,option $C$ is the correct method to form $X$.

(A) The given reaction involves the reduction of a nitrile $(R-CN)$ to an imine using stannous chloride $(SnCl_2)$ and hydrochloric acid $(HCl)$,followed by hydrolysis to yield an aldehyde $(R-CHO)$.
This specific chemical transformation is known as the Stephen's reaction.
Therefore,the correct option is $A$.

(D) Let us analyze each reaction:
$A$. Gattermann-Koch reaction: Benzene reacts with $CO + HCl$ in the presence of anhydrous $AlCl_3$ or $CuCl$ to give benzaldehyde. The reaction shown with $HCN + HCl$ (Gattermann reaction) also yields benzaldehyde.
$B$. Etard reaction: Toluene reacts with chromyl chloride $(CrO_2Cl_2)$ followed by hydrolysis to give benzaldehyde.
$C$. Toluene reacts with chromic oxide $(CrO_3)$ in acetic anhydride followed by hydrolysis to give benzaldehyde (this is a variation of the Etard reaction).
$D$. Reimer-Tiemann reaction: Phenol reacts with chloroform $(CHCl_3)$ and aqueous $KOH$ to give salicylaldehyde (o-hydroxybenzaldehyde) as the major product,not benzaldehyde.
Therefore,option $D$ does not yield benzaldehyde.

(A) The reaction of an alkyne with $Hg^{2+}/H^+$ (Kuccherov reaction) involves the hydration of the triple bond to form an enol intermediate,which then tautomerizes to a more stable ketone.
For the unsymmetrical alkyne $Ph-C \equiv C-CH_3$,the hydration follows Markovnikov's rule.
The $OH$ group attaches to the carbon atom that can better stabilize the positive charge in the transition state (the carbon attached to the phenyl group).
This forms the enol $Ph-C(OH)=CH-CH_3$,which tautomerizes to the ketone $Ph-CO-CH_2-CH_3$ ($1$-phenylpropan$-1-$one).

(B) The reaction is an example of the Pinacol-Pinacolone rearrangement.
In the presence of an acid $(H^{+})$,$2,3$-dimethylbutane-$2,3$-diol (Pinacol) undergoes dehydration and rearrangement to form $3,3$-dimethylbutan-$2$-one (Pinacolone).
The mechanism involves the protonation of one of the hydroxyl groups,followed by the loss of a water molecule to form a carbocation.
Then,a methyl group shifts to the adjacent carbon atom,followed by the loss of a proton to form the final ketone product,$CH_3-CO-C(CH_3)_3$.

(C) The given reaction involves the hydrogenation of an acid chloride $(C_6H_5COCl)$ to an aldehyde $(C_6H_5CHO)$ using hydrogen gas in the presence of palladium $(Pd)$ supported on barium sulfate $(BaSO_4)$.
This specific reaction is known as the Rosenmund reduction.
The $BaSO_4$ acts as a poison to prevent the further reduction of the aldehyde to an alcohol.

(C) In the Etard reaction,toluene is oxidized to benzaldehyde using chromyl chloride $(CrO_2Cl_2)$ in $CS_2$. The reaction given in option $(c)$ uses $CrO_3$ in acetic anhydride,which is a different synthetic route to benzaldehyde via a gem-diacetate intermediate.

(B) The reaction of toluene with chromyl chloride $(CrO_2Cl_2)$ followed by hydrolysis is the Etard reaction,which produces benzaldehyde $(P = C_6H_5CHO)$.
To convert benzaldehyde $(C_6H_5CHO)$ to benzoic acid $(C_6H_5COOH)$,an oxidizing agent is required.
Among the given options,the reaction sequence $P$ (benzaldehyde) $\xrightarrow{R}$ benzoic acid requires an oxidation step.
However,looking at the options provided,there is a mismatch in the standard reagents. The oxidation of benzaldehyde to benzoic acid is typically done using $KMnO_4$ or $K_2Cr_2O_7$.
Given the options,if we assume the question asks for the product $P$ and a reagent $R$ that could potentially oxidize it,none of the reagents listed ($HI$,$H_2+Pd$,$Zn-Hg/HCl$) are standard oxidants.
Re-evaluating the provided options: Option $B$ identifies $P$ correctly as benzaldehyde. If the question implies a typo in the reagent $R$ or a specific context,$P$ is definitely benzaldehyde.

(C) The oxidation of primary alcohols to aldehydes is best achieved using mild oxidizing agents that do not further oxidize the aldehyde to a carboxylic acid.
$P.C.C.$ (Pyridinium chlorochromate) is a selective oxidizing agent that stops the oxidation at the aldehyde stage.
$CrO_3$ in acidic medium or $KMnO_4$ are strong oxidizing agents that typically oxidize primary alcohols directly to carboxylic acids.
$[Ag(NH_3)_2]^+$ (Tollens' reagent) is used to oxidize aldehydes to carboxylic acids,not alcohols to aldehydes.

(A) The reaction of $4$-chlorobenzoyl chloride with benzene in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts acylation reaction.
In this reaction,the $Cl$ atom of the acid chloride group is replaced by the phenyl group from benzene,resulting in the formation of $4$-chlorobenzophenone $(Cl-C_6H_4-CO-C_6H_5)$.

(C) In the first step,chloroform $(CHCl_3)$ is oxidized by air in the presence of light to form phosgene $(COCl_2)$:
$2CHCl_3 + O_2 \xrightarrow{hv} 2COCl_2 + 2HCl$
In the second step,phosgene $(x)$ reacts with excess benzene in the presence of anhydrous $AlCl_3$ (Friedel-Crafts acylation) to form benzophenone $(y)$:
$COCl_2 + 2C_6H_6 \xrightarrow{Anhy. AlCl_3} C_6H_5-CO-C_6H_5 + 2HCl$

(A) The reaction of benzene with carbon monoxide $(CO)$ and hydrogen chloride $(HCl)$ in the presence of anhydrous aluminum chloride $(AlCl_3)$ or cuprous chloride $(CuCl)$ to form benzaldehyde $(C_6H_5CHO)$ is known as the Gatterman-Koch reaction.

(C) The catalytic oxidation of toluene with air in the presence of $V_2O_5$ (Vanadium pentoxide) is a standard industrial process to produce benzaldehyde.
The reaction is: $C_6H_5CH_3 + [O] \xrightarrow{V_2O_5} C_6H_5CHO + H_2O$.

(A) The given reaction involves the reduction of a nitrile $(R-CN)$ to an imine using stannous chloride $(SnCl_2)$ in the presence of hydrochloric acid $(HCl)$,followed by acid hydrolysis to yield an aldehyde $(R-CHO)$.
This specific chemical transformation is known as the Stephen's reaction (or Stephen reduction).
Therefore,the correct option is $A$.

(D) Let us analyze each reaction:
$A$. Gattermann-Koch reaction: Benzene reacts with $CO + HCl$ in the presence of $Anhydrous \ AlCl_3 / CuCl$ to give benzaldehyde. The reaction with $HCN + HCl$ (Gattermann reaction) also yields benzaldehyde.
$B$. Etard reaction: Toluene reacts with chromyl chloride $(CrO_2Cl_2)$ followed by hydrolysis to give benzaldehyde.
$C$. Toluene reacts with chromic oxide $(CrO_3)$ in acetic anhydride to form benzylidene diacetate,which on hydrolysis gives benzaldehyde.
$D$. Reimer-Tiemann reaction: Phenol reacts with $CHCl_3$ and $KOH$ to give salicylaldehyde (o-hydroxybenzaldehyde),not benzaldehyde.

(D) The conversion of primary alcohols $(R-CH_2OH)$ to aldehydes $(RCHO)$ requires a mild oxidizing agent that stops the oxidation at the aldehyde stage without further oxidizing it to a carboxylic acid.
$PCC$ (Pyridinium chlorochromate) is a selective oxidizing agent that performs this specific transformation efficiently.
Strong oxidizing agents like $KMnO_4$,$K_2Cr_2O_7$,and $CrO_3$ (in acidic medium) typically oxidize primary alcohols directly to carboxylic acids.

(A) The reaction is a standard method for the reduction of nitriles to aldehydes known as Stephen's reduction.
Step $1$: The nitrile $(CH_3CH_2C \equiv N)$ is reduced by $SnCl_2 / HCl$ to form an imine intermediate $(CH_3CH_2CH = NH)$.
Step $2$: The imine intermediate is then hydrolyzed by boiling with water $(H_2O)$ to yield the corresponding aldehyde $(CH_3CH_2CHO)$ and ammonium chloride $(NH_4Cl)$.

(N/A) $(i)$ $\text{Pyridinium chlorochromate } (PCC)$
$(ii)$ $\text{Jones reagent } (CrO_{3} / H_{2}SO_{4}) \text{ or } PCC$
$(iii)$ $\text{Chromyl chloride } (CrO_{2}Cl_{2}) \text{ followed by hydrolysis } (Etard \text{ reaction})$
$(iv)$ $\text{Diisobutylaluminium hydride } (DIBAL-H) \text{ followed by } H_{3}O^{+}$
$(v)$ $\text{Pyridinium chlorochromate } (PCC)$
$(vi)$ $\text{Ozonolysis } (1. O_{3}, 2. Zn / H_{2}O)$

(N/A) $(i)$ Benzene + $C_{2}H_{5}COCl$ $\xrightarrow{\text{Anhyd. } AlCl_{3}, CS_{2}}$ Propiophenone $(C_{6}H_{5}COC_{2}H_{5})$ + $HCl$
$(ii)$ $(C_{6}H_{5}CH_{2})_{2}Cd + 2 CH_{3}COCl \rightarrow 2 C_{6}H_{5}CH_{2}COCH_{3}$ ($1$-Phenylpropan$-2-$one) + $CdCl_{2}$
$(iii)$ $H_{3}C - C \equiv C - H + H_{2}O$ $\xrightarrow{Hg^{2+}, H_{2}SO_{4}}$ $H_{3}C - CO - CH_{3}$ (Propanone)
$(iv)$ $p$-Nitrotoluene $\xrightarrow{1. CrO_{2}Cl_{2}, 2. H_{3}O^{+}}$ $p$-Nitrobenzaldehyde $(O_{2}NC_{6}H_{4}CHO)$

(N/A) Preparation of aldehydes and ketones by oxidation of alcohols:
Aldehydes and ketones are generally prepared by the oxidation of primary and secondary alcohols,respectively.
These reactions involve the cleavage and formation of bonds. They are also known as dehydrogenation reactions because they involve the loss of dihydrogen from an alcohol molecule.
Depending on the oxidising agent used,a primary alcohol is oxidised to an aldehyde,which can be further oxidised to a carboxylic acid.
$RCH_2OH \xrightarrow{(K_2Cr_2O_7 + H_2SO_4), CrO_3, (O), -H_2O} RCHO$
$1^\circ$-Alcohol $\rightarrow$ Aldehyde
$CH_3CH_2OH \xrightarrow{(K_2Cr_2O_7 + H_2SO_4), CrO_3, (O), -H_2O} CH_3CHO$
$(b)$ $A$ better reagent for the oxidation of primary alcohols to aldehydes in good yield is pyridinium chlorochromate $(PCC)$,a complex of chromium trioxide with pyridine and $HCl$.
$(c)$ Secondary alcohols are oxidised to ketones by chromic anhydride $(CrO_3)$ in anhydrous medium.
$R-CH(OH)-R' \xrightarrow{CrO_3, (O), -H_2O} R-CO-R'$
Examples:
$(i) CH_3CH(OH)CH_3 \xrightarrow{CrO_3, (O), -H_2O} CH_3COCH_3$
$(ii) CH_3CH(OH)CH_2CH_3 \xrightarrow{CrO_3, (O), -H_2O} CH_3COCH_2CH_3$
$(iii) C_6H_5CH(OH)CH_3 \xrightarrow{CrO_3, (O), -H_2O} C_6H_5COCH_3$
Under strong reaction conditions such as strong oxidising agents $(KMnO_4)$ and elevated temperatures,cleavage of various $C-C$ bonds takes place,and a mixture of carboxylic acids containing a lesser number of carbon atoms is formed.

(N/A) This method is suitable for volatile alcohols and is of industrial application. In this method,alcohol vapors are passed over heavy metal catalysts ($Ag$ or $Cu$) at $573 \ K$.
$1^{\circ}$ Alcohols give aldehydes:
$RCH_2OH \xrightarrow{Cu, 573 \ K} RCHO + H_2$
Example: $CH_3CH_2OH \xrightarrow{Cu, 573 \ K} CH_3CHO + H_2$
$2^{\circ}$ Alcohols give ketones:
$R-CH(OH)-R' \xrightarrow{Cu, 573 \ K} R-CO-R' + H_2$
Example: $CH_3-CH(OH)-CH_3 \xrightarrow{Cu, 573 \ K} CH_3-CO-CH_3 + H_2$

(N/A) $i$. By ozonolysis of alkenes: Ozonolysis of alkenes followed by reaction with zinc dust and water gives aldehydes,ketones,or a mixture of both,depending on the substitution pattern of the alkene.
$1$. Preparation of methanal from ethene:
$CH_2=CH_2 + O_3$ $\rightarrow \text{Ethylene ozonide}$ $\xrightarrow{Zn, H_2O} 2HCHO + ZnO$
$2$. Preparation of ethanal and methanal from propene:
$CH_3-CH=CH_2 + O_3$ $\rightarrow \text{Propene ozonide}$ $\xrightarrow{Zn, H_2O} CH_3CHO + HCHO + ZnO$
$3$. Preparation of propanone and methanal from $2$-methylpropene:
$(CH_3)_2C=CH_2 + O_3$ $\rightarrow \text{Ozonide}$ $\xrightarrow{Zn, H_2O} CH_3COCH_3 + HCHO + ZnO$
$ii$. By hydration of alkynes: Alkynes react with water in the presence of $H_2SO_4$ and $HgSO_4$ to form carbonyl compounds.
$e.g.$,Ethyne on hydration gives ethanal:
$HC \equiv CH + H_2O$ $\xrightarrow{H_2SO_4, HgSO_4} [CH_2=CHOH]$ $\rightarrow CH_3CHO$

(N/A) Aldehyde from acyl chloride: Acyl chloride is hydrogenated over a catalyst,palladium on barium sulphate $(Pd/BaSO_4)$. This reaction is known as the Rosenmund reduction.
$(i)$ Preparation of ethanal (acetaldehyde) from ethanoyl chloride (acetyl chloride):
$CH_3COCl + H_2 \xrightarrow{Pd/BaSO_4} CH_3CHO + HCl$
$(ii)$ Preparation of benzaldehyde from benzoyl chloride:
$C_6H_5COCl + H_2 \xrightarrow{Pd/BaSO_4} C_6H_5CHO + HCl$
$(b)$ Ketone from acyl chloride: Ketones can be prepared by the reaction of acyl chlorides with dialkylcadmium $(R_2Cd)$,which is prepared by the reaction of Grignard reagent with cadmium chloride $(CdCl_2)$.
$2RMgX + CdCl_2 \rightarrow R_2Cd + 2Mg(X)Cl$
$2R'COCl + R_2Cd \rightarrow 2R'COR + CdCl_2$

(N/A) Aldehydes and ketones can be prepared from Grignard reagents $(RMgX)$ by reaction with nitriles $(R'CN)$ followed by acid hydrolysis.
$1.$ Preparation of Aldehydes:
Reaction of Grignard reagent with hydrogen cyanide $(HCN)$ followed by hydrolysis yields an aldehyde.
$H-C \equiv N + RMgX$ $\rightarrow [R-CH=NMgX]$ $\xrightarrow{H_2O, H^+} R-CHO + NH_3 + Mg(OH)X$
Example: $H-C \equiv N + CH_3MgI$ $\rightarrow CH_3-CH=NMgI$ $\xrightarrow{H_2O, H^+} CH_3CHO + NH_3 + Mg(OH)I$
$2.$ Preparation of Ketones:
Reaction of Grignard reagent with alkyl nitriles $(R'CN)$ followed by hydrolysis yields a ketone.
$R'C \equiv N + RMgX$ $\rightarrow [R'C(R)=NMgX]$ $\xrightarrow{H_2O, H^+} R'COR + NH_3 + Mg(OH)X$
Example: $CH_3C \equiv N + CH_3MgI$ $\rightarrow CH_3C(CH_3)=NMgI$ $\xrightarrow{H_2O, H^+} CH_3COCH_3 + NH_3 + Mg(OH)I$

(N/A) Preparation of aldehyde from nitrile compounds by Stephen reduction reaction:
Nitriles are reduced to the corresponding imine with stannous chloride $(SnCl_2)$ in the presence of hydrochloric acid $(HCl)$,which on hydrolysis gives the corresponding aldehyde. This reaction is called the Stephen reaction.
$RC \equiv N + SnCl_2 + HCl$ $\rightarrow RCH = NH$ $\xrightarrow{H_3O^+} RCHO$
$(b)$ Formation of aldehyde by reduction reaction of nitrile compounds with $DIBAL-H$ followed by hydrolysis:
Nitriles are selectively reduced by diisobutylaluminium hydride $(DIBAL-H)$ to imines,followed by hydrolysis to aldehydes:
$RC \equiv N \xrightarrow[(ii) H_2O]{(i) AlH(i-Bu)_2} RCHO$
$(c)$ Preparation of ketones from nitrile compounds:
Ketones are prepared by the reaction of nitriles with Grignard reagents $(R'MgX)$ followed by hydrolysis.
$RCN + R'MgX$ $\rightarrow R-C(R')=NMgX$ $\xrightarrow{H_3O^+} R-C(=O)R' + NH_3 + Mg(OH)X$

(N/A) $DIBAL-H$ stands for diisobutylaluminium hydride,$AlH(i-Bu)_2$.
$(a)$ Preparation of aldehyde from nitriles:
Nitriles are selectively reduced by $DIBAL-H$ to imines,which upon hydrolysis yield the corresponding aldehydes.
$R-CN \xrightarrow[(ii) H_2O]{(i) DIBAL-H} R-CHO$
$(b)$ Preparation of aldehyde from esters:
Esters are selectively reduced by $DIBAL-H$ to aldehydes.
$R-COOR' \xrightarrow[(ii) H_2O]{(i) DIBAL-H} R-CHO + R'OH$
Example reaction for ester:
$CH_3(CH_2)_9-CO-OC_2H_5 \xrightarrow[(ii) H_2O]{(i) DIBAL-H} CH_3(CH_2)_9-CHO$

(N/A) $1.$ $2$-Chloropropane to Propanone:
$CH_3-CHCl-CH_3$ $\xrightarrow{NaOH(aq)} CH_3-CH(OH)-CH_3$ $\xrightarrow{CrO_3} CH_3-CO-CH_3$
$2.$ Propene to Propanone:
$CH_3-CH=CH_2$ $\xrightarrow{H_2O/H^+} CH_3-CH(OH)-CH_3$ $\xrightarrow{CrO_3 \text{ or } Cu, 573K} CH_3-CO-CH_3$

(A) $PCC$ (Pyridinium chlorochromate) is a mild oxidizing agent that selectively oxidizes primary alcohols to aldehydes without further oxidation to carboxylic acids.
The reaction is:
$CH_3-CH=CH-CH_2OH \xrightarrow{PCC} CH_3-CH=CH-CHO$
The product formed is $But-2-en-1-al$.

(N/A) Benzaldehyde can be prepared by the following methods:
$1$. From acyl chloride (Rosenmund reduction):
Benzoyl chloride is hydrogenated over catalyst palladium on barium sulphate $(Pd/BaSO_4)$ to form benzaldehyde. The reaction is poisoned by sulphur or quinoline to prevent further reduction to alcohol.
$C_6H_5COCl + H_2 \xrightarrow{Pd/BaSO_4} C_6H_5CHO + HCl$
$2$. From aromatic hydrocarbons (Toluene):
$a$. Oxidation by chromyl chloride (Etard reaction): Toluene is oxidised by chromyl chloride $(CrO_2Cl_2)$ to a chromium complex,which on hydrolysis gives benzaldehyde.
$C_6H_5CH_3 + 2CrO_2Cl_2$ $\rightarrow C_6H_5CH(OCrOHCl_2)_2$ $\xrightarrow{H_3O^+} C_6H_5CHO$
$b$. Side chain chlorination followed by hydrolysis: Side chain chlorination of toluene gives benzal chloride,which on hydrolysis gives benzaldehyde.
$C_6H_5CH_3 + Cl_2$ $\xrightarrow{hv} C_6H_5CHCl_2$ $\xrightarrow{H_2O, 373K} C_6H_5CHO$
$c$. Gatterman-Koch reaction: Benzene is treated with carbon monoxide and hydrogen chloride in the presence of anhydrous aluminium chloride $(AlCl_3)$ or cuprous chloride $(CuCl)$ to give benzaldehyde.
$C_6H_6 + CO + HCl \xrightarrow{anhydrous AlCl_3/CuCl} C_6H_5CHO$

(N/A) From Acyl Chlorides: Reaction of $RCOCl$ with dialkylcadmium $(R'_2Cd)$,prepared from Grignard reagent $(R'MgX)$ and $CdCl_2$,gives ketones: $2RCOCl + R'_2Cd \rightarrow 2RCOR' + CdCl_2$.
$(b)$ From Nitriles: Treatment of nitriles with Grignard reagent followed by hydrolysis yields ketones: $RCN + R'MgX$ $\rightarrow RR'C=NMgX$ $\xrightarrow{H_3O^+} RCOR' + NH_3 + Mg(OH)X$.
$(c)$ Friedel-Crafts Acylation: Benzene or substituted benzenes react with acid chlorides $(RCOCl)$ or acid anhydrides $(RCO)_2O$ in the presence of anhydrous $AlCl_3$ to form aromatic ketones: $C_6H_6 + RCOCl \xrightarrow{Anhydrous \ AlCl_3} C_6H_5COR + HCl$.

(N/A) Aldehyde compounds can be prepared by the following methods:
$i$. Oxidation of primary alcohols using $PCC$ or $CrO_3$.
$ii$. Dehydrogenation of primary alcohols over $Cu$ catalyst at $573 \ K$.
$iii$. Ozonolysis of alkenes followed by reduction with $Zn/H_2O$.
$iv$. Rosenmund reduction: Hydrogenation of acyl chlorides using $Pd/BaSO_4$.
$v$. Stephen reaction: Reduction of nitriles with $SnCl_2/HCl$ followed by hydrolysis.
$vi$. Reduction of nitriles and esters using $DIBAL-H$ followed by hydrolysis.
$vii$. Etard reaction: Oxidation of toluene using $CrO_2Cl_2$ followed by hydrolysis.
$viii$. Oxidation of toluene with $CrO_3$ in acetic anhydride.
$ix$. Side chain chlorination of toluene followed by hydrolysis.
$x$. Gatterman-Koch reaction: Formylation of benzene using $CO$ and $HCl$ in the presence of anhydrous $AlCl_3$ or $CuCl$.

(N/A) Aldehydes are reduced to primary alcohols by using reducing agents like sodium borohydride $(NaBH_4)$ or lithium aluminium hydride $(LiAlH_4)$,as well as by catalytic hydrogenation.
General reaction:
$RCHO + 2[H] \xrightarrow{NaBH_4 / LiAlH_4 / (Pd/H_2)} RCH_2OH$
Example:
$(i) CH_3CHO + 2[H] \xrightarrow{NaBH_4 / LiAlH_4 / (Pd/H_2)} CH_3CH_2OH$
$(\text{Ethanal}) \qquad (\text{Ethanol})$

(N/A) $(i)$ Conversion of ethanal to lactic acid:
$CH_3CHO + HCN$ $\rightarrow CH_3CH(OH)CN$ $\xrightarrow{H_3O^+} CH_3CH(OH)COOH$ (Lactic acid)
$(ii)$ Conversion of $p-$fluorotoluene to $p-$fluorobenzaldehyde:
$p-F-C_6H_4-CH_3$ $\xrightarrow[273-278 \ K]{CrO_3, (CH_3CO)_2O} p-F-C_6H_4-CH(OCOCH_3)_2$ $\xrightarrow{H_3O^+} p-F-C_6H_4-CHO$ ($p-$fluorobenzaldehyde)

(N/A) Step $1$: Preparation of $benzal$ chloride from $toluene$:
$Toluene$ reacts with chlorine $(Cl_2)$ in the presence of light $(h\nu)$ at $383 \ K$ to form $benzal$ chloride (also known as $benzylidene$ chloride).
$C_6H_5CH_3 + 2Cl_2 \xrightarrow{h\nu, 383 \ K} C_6H_5CHCl_2 + 2HCl$
Step $2$: Preparation of $benzaldehyde$ from $benzal$ chloride:
$Benzal$ chloride undergoes hydrolysis with water at $373 \ K$ to yield $benzaldehyde$.
$C_6H_5CHCl_2 + H_2O \xrightarrow{373 \ K} C_6H_5CHO + 2HCl$

(N/A) Direct alkylation of benzene (Friedel-Crafts alkylation) with ethyl chloride $(CH_3CH_2Cl)$ in the presence of anhydrous $AlCl_3$ leads to polyalkylation because the alkyl group is electron-donating,which activates the benzene ring towards further electrophilic substitution.
Additionally,primary alkyl halides can undergo rearrangement.
To avoid these issues,ethylbenzene is prepared by Friedel-Crafts acylation of benzene with acetyl chloride $(CH_3COCl)$ to form acetophenone,followed by Clemmensen reduction $(Zn-Hg/HCl)$ to yield ethylbenzene.
The reaction sequence is:
$C_6H_6 + CH_3COCl$ $\xrightarrow{Anhydrous \ AlCl_3} C_6H_5COCH_3$ $\xrightarrow{Zn-Hg/HCl} C_6H_5CH_2CH_3$.

(N/A) Yes,the Gatterman-Koch reaction is considered a special case of Friedel-Crafts acylation.
In standard Friedel-Crafts acylation,benzene (or an arene) reacts with an acid chloride (e.g.,$CH_3COCl$) in the presence of anhydrous $AlCl_3$ to form an aryl ketone.
In the Gatterman-Koch reaction,benzene is treated with $CO$ and $HCl$ in the presence of anhydrous $AlCl_3$. Since formyl chloride $(HCOCl)$ is unstable,it is generated in situ from $CO$ and $HCl$.
This in situ generated $HCOCl$ then acts as the acylating agent,similar to the acid chloride in Friedel-Crafts acylation,to produce benzaldehyde. Thus,the mechanism is analogous.

(N/A) $(1)$ Acetone from propene:
$CH_3CH=CH_2$ $\xrightarrow{Br_2, CCl_4} CH_3-CHBr-CH_2Br$ $\xrightarrow{alc. KOH, \Delta} CH_3C \equiv CH$ $\xrightarrow{HgSO_4, H_2SO_4} CH_3COCH_3$
$(2)$ Ethanal from $2-$bromobutane:
$CH_3CH(Br)CH_2CH_3$ $\xrightarrow{alc. KOH, \Delta} CH_3CH=CHCH_3$ $\xrightarrow{(i) O_3, CH_2Cl_2, (ii) Zn, H_2O} 2CH_3CHO$

(A) The correct matching is as follows:
$(A).$ Rosenmund reaction uses $H_2, Pd, BaSO_4$ to convert Benzoyl chloride to benzaldehyde. So,$(A \to i \to q)$.
$(B).$ Stephen reaction uses $SnCl_2, HCl$ to convert $RCN$ to $RCHO$. So,$(B \to iii \to s)$.
$(C).$ Etard reaction uses $CrO_2Cl_2, CS_2$ to convert Toluene to a chromium complex. So,$(C \to iv \to r)$.
$(D).$ Gatterman-Koch reaction uses $CO, HCl$ with anhydrous $AlCl_3/CuCl$ to convert Benzene to benzaldehyde. So,$(D \to ii \to p)$.

(A-I-Q, B-III-Q, C-IV-P, D-II-Q) The correct matches are:
$A$ $\rightarrow i$ $\rightarrow q$: $CH_3CN + C_6H_5MgBr$ $\xrightarrow{Ether} CH_3C(NMgBr)C_6H_5$ $\xrightarrow{H_3O^+} CH_3COC_6H_5$ (Acetophenone).
$B$ $\rightarrow iii$ $\rightarrow q$: Friedel-Crafts acylation of benzene with $CH_3COCl$ in the presence of anhydrous $AlCl_3$ gives Acetophenone.
$C$ $\rightarrow iv$ $\rightarrow p$: Side-chain chlorination of toluene followed by hydrolysis gives Benzaldehyde.
$D$ $\rightarrow ii$ $\rightarrow q$: Hydration of phenyl ethyne in the presence of $Hg^{2+}$ and $H_2SO_4$ gives Acetophenone.

(C) The correct matches are as follows:
$1$. Benzene $(I)$ reacts with $CO, HCl$ and $AlCl_3$ (Gattermann-Koch reaction) to form benzaldehyde. Thus,$I-R$.
$2$. Benzonitrile $(II)$ reacts with $SnCl_2, HCl$ followed by $H_3O^{+}$ (Stephen reduction) to form benzaldehyde. Thus,$II-P$.
$3$. Benzoyl chloride $(III)$ reacts with $H_2, Pd-BaSO_4, S$ and quinoline (Rosenmund reduction) to form benzaldehyde. Thus,$III-Q$.
Therefore,the correct match is $I-R, II-P$ and $III-Q$.

(D) The reaction sequence is the industrial preparation of benzaldehyde from toluene.
$1$. Toluene reacts with chlorine in the presence of light $(Cl_2/h\nu)$ to undergo side-chain chlorination.
$2$. The reaction proceeds through the formation of benzyl chloride,then benzal chloride $(C_6H_5CHCl_2)$,and finally benzotrichloride.
$3$. To obtain benzaldehyde,the reaction is controlled to produce benzal chloride $(C_6H_5CHCl_2)$ as the intermediate $X$.
$4$. Benzal chloride $(C_6H_5CHCl_2)$ on hydrolysis with water at $373 \ K$ yields benzaldehyde $(C_6H_5CHO)$.
Therefore,compound $X$ is benzal chloride $(C_6H_5CHCl_2)$.

(B) The given reaction is the Rosenmund reduction.
In this reaction,an acid chloride $(RCOCl)$ is hydrogenated to an aldehyde $(RCHO)$ using $H_2$ gas in the presence of a palladium catalyst supported on barium sulfate $(Pd/BaSO_4)$.
The reaction is:
$C_6H_5COCl + H_2 \xrightarrow{Pd/BaSO_4} C_6H_5CHO + HCl$
Here,compound $(A)$ is benzoyl chloride $(C_6H_5COCl)$.