Preparation Questions in English

(C) $I.$ Oxidation of $1$-phenylethanol $(C_6H_5CH(OH)CH_3)$ yields acetophenone $(C_6H_5COCH_3)$. This is correct.
$II.$ Reaction of benzaldehyde with $CH_3MgBr$ followed by oxidation yields $1$-phenylethanol,not acetophenone. This is incorrect.
$III.$ Friedel-Crafts acylation of benzene with acetyl chloride $(CH_3COCl)$ in the presence of anhydrous $AlCl_3$ yields acetophenone. This is correct.
$IV.$ Distillation of calcium benzoate yields benzophenone,not acetophenone. This is incorrect.
Therefore,statements $I$ and $III$ are correct.

(C) The Wacker process involves the oxidation of alkenes to carbonyl compounds using a palladium catalyst.
For ethene $(CH_2=CH_2)$,the reaction with $O_2$ in the presence of $PdCl_2$ and $CuCl_2$ yields acetaldehyde $(CH_3-CHO)$.
Reaction: $CH_2=CH_2 + \frac{1}{2}O_2 \xrightarrow{PdCl_2, CuCl_2} CH_3-CHO$.

(B) The reaction of a Grignard reagent $(RMgX)$ with a nitrile $(R'CN)$ followed by hydrolysis yields a ketone.
Specifically,for ethanenitrile $(CH_3CN)$:
$CH_3-C \equiv N + RMgX \to CH_3-C(R)=N-MgX$
Upon hydrolysis $(H_2O)$:
$CH_3-C(R)=N-MgX + H_2O \to CH_3-C(=O)-R + NH_3 + Mg(OH)X$
Thus,ethanenitrile reacts with a Grignard reagent to form a ketone.

(B) The reaction of benzoyl chloride with $H_2$ in the presence of $Pd$ supported on $BaSO_4$ is known as the Rosenmund reduction.
This reaction specifically reduces acid chlorides to aldehydes.
Therefore,benzoyl chloride $(C_6H_5COCl)$ is reduced to benzaldehyde $(C_6H_5CHO)$.

(A) The $SnCl_2/HCl$ reagent is specifically used in the $Stephen$ reduction to convert nitriles $(R-CN)$ into imines,which are then hydrolyzed to aldehydes $(R-CHO)$.

(C) Secondary alcohols undergo oxidation in the presence of oxidizing agents like $K_2Cr_2O_7 / H_2SO_4$ to form ketones in a single step.
The reaction is:
$R_1-CH(OH)-R_2 + [O] \xrightarrow{K_2Cr_2O_7 / H_2SO_4} R_1-CO-R_2 + H_2O$
When $R_1 = R_2$,a symmetrical ketone is obtained.

(B) The oxidation of toluene to benzaldehyde using chromyl chloride $(CrO_2Cl_2)$ is known as the Etard reaction.
In this reaction,the methyl group of toluene is oxidized to a formyl group.

(A) The reduction of acid chlorides with $H_2$ in the presence of $Pd$ supported on $BaSO_4$ is known as the Rosenmund reduction.
$CH_3COCl + H_2 \xrightarrow{Pd/BaSO_4} CH_3CHO + HCl$
Acetyl chloride is reduced to acetaldehyde.

(D) The reaction $RCOCl + H-Ar \xrightarrow{AlCl_3} R-CO-Ar + HCl$ is known as the Friedel-Crafts acylation reaction.
In this reaction,an acyl chloride $(RCOCl)$ reacts with an aromatic compound $(Ar-H)$ in the presence of a Lewis acid catalyst like $AlCl_3$ to form an aromatic ketone $(R-CO-Ar)$.

(B) $1$. Gatterman-Koch reaction: Benzene reacts with $CO$ and $HCl$ in the presence of anhydrous $AlCl_3$ to form benzaldehyde.
$2$. Etard reaction: Toluene reacts with $CrO_2Cl_2$ in $CS_2$ followed by hydrolysis to form benzaldehyde.
$3$. Rosenmund reduction: Benzoyl chloride is reduced by $H_2$ in the presence of $Pd-BaSO_4$ to form benzaldehyde.
$4$. The reaction of benzoic acid with $Zn/Hg$ and conc. $HCl$ (Clemmensen reduction) is used to reduce aldehydes and ketones to alkanes,not to prepare aldehydes from carboxylic acids. Therefore,this reaction cannot be used to prepare benzaldehyde.

(A) The given reaction is the Rosenmund reduction.
In this reaction,an acid chloride $(RCOCl)$ is hydrogenated to an aldehyde $(RCHO)$ using hydrogen 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 reaction of benzoyl chloride $(C_6H_5COCl)$ with $H_2$ in the presence of $Pd/BaSO_4$ yields benzaldehyde $(C_6H_5CHO)$.

(D) $R-CH_2-OH \xrightarrow{PCC} R-CHO$
$PCC$ (Pyridinium chlorochromate) is a mild oxidizing agent that selectively oxidizes primary alcohols to aldehydes without further oxidation to carboxylic acids.
$KMnO_4$,$K_2Cr_2O_7$,and $CrO_3$ are strong oxidizing agents that typically oxidize primary alcohols directly to carboxylic acids.

(B) Step $1$: Toluene $(C_6H_5CH_3)$ undergoes oxidation with $KMnO_4$ to form benzoic acid $(C_6H_5COOH)$,which is product $A$.
Step $2$: Benzoic acid reacts with thionyl chloride $(SOCl_2)$ to form benzoyl chloride $(C_6H_5COCl)$,which is product $B$.
Step $3$: Benzoyl chloride undergoes Rosenmund reduction with $H_2/Pd$ and $BaSO_4$ to form benzaldehyde $(C_6H_5CHO)$,which is product $C$.

(C) The reaction of nitriles $(R-CN)$ with stannous chloride $(SnCl_2)$ in the presence of hydrochloric acid $(HCl)$ reduces the nitrile to an imine $(R-CH=NH \cdot HCl)$.
Upon subsequent hydrolysis,the imine is converted into the corresponding aldehyde $(R-CHO)$.
This specific chemical transformation is known as the $Stephen$ reaction.

(B) The Rosenmund reduction involves the partial reduction of an acid chloride to an aldehyde using hydrogen gas over palladium poisoned with barium sulfate $(Pd/BaSO_4)$.
Aldehydes are carbonyl compounds.
In option $(a)$,$H_2/Pd$ reduces the acid chloride further to a primary alcohol.
In option $(c)$,$LiAlH_4$ is a strong reducing agent that reduces acid chlorides to primary alcohols.
Therefore,only the Rosenmund reduction produces an aldehyde (a carbonyl compound) as the major product.

(D) The given reactions are standard methods for the preparation of benzaldehyde from toluene:
$1$. Etard reaction: Toluene reacts with chromyl chloride $(CrO_2Cl_2)$ to form a chromium complex,which on hydrolysis gives benzaldehyde.
$2$. Oxidation with chromic oxide $(CrO_3)$ in the presence of acetic anhydride: Toluene is converted to benzylidene diacetate,which on hydrolysis yields benzaldehyde.
$3$. Side chain chlorination followed by hydrolysis: Toluene reacts with $Cl_2$ in the presence of light to form benzal chloride $(C_6H_5CHCl_2)$,which on hydrolysis with aqueous alkali gives benzaldehyde.
Since all three methods produce benzaldehyde,the correct option is $D$.

(D) Ethyl phenyl ketone $(C_6H_5-CO-CH_2-CH_3)$ is formed in all the given reactions.
$(a)$ Nitriles react with Grignard reagents followed by hydrolysis to give ketones: $C_6H_5-CN + CH_3-CH_2-MgBr$ $\rightarrow C_6H_5-C(CH_2-CH_3)=NMgBr$ $\xrightarrow{H_3O^{+}} C_6H_5-CO-CH_2-CH_3$.
$(b)$ Acid chlorides react with dialkyl/diaryl cadmium to give ketones: $2CH_3-CH_2-COCl + (C_6H_5)_2Cd \rightarrow 2C_6H_5-CO-CH_2-CH_3 + CdCl_2$.
$(c)$ Benzene undergoes Friedel-Crafts acylation with propanoyl chloride to give ethyl phenyl ketone: $C_6H_6 + CH_3-CH_2-COCl \xrightarrow{AlCl_3} C_6H_5-CO-CH_2-CH_3 + HCl$.

(C) In reaction $(c)$,a Grignard reagent reacts with a nitrile to form an imine salt,which upon hydrolysis yields a ketone:
$CH_3MgBr + CH_3-CN$ $\rightarrow CH_3-C(CH_3)=NMgBr$ $\xrightarrow{H_3O^{+}} CH_3-CO-CH_3$ (Acetone).
The other reactions are:
$(a)$ Catalytic reduction of nitrile to imine followed by hydrolysis gives $CH_3-CHO$.
$(b)$ Rosenmund reduction of acid chloride gives $CH_3-CHO$.
$(d)$ $DIBAL-H$ reduction of acid chloride gives $CH_3-CHO$.

(C) Dry distillation of a mixture of calcium benzoate and calcium acetate yields acetophenone.
The reaction is as follows:
$(C_6H_5COO)_2Ca + (CH_3COO)_2Ca \xrightarrow{\text{dry distillation}} 2C_6H_5COCH_3 + 2CaCO_3$

(C) Dry distillation of a mixture of calcium propionate and calcium formate yields propanaldehyde.
$(CH_3-CH_2-COO)_2Ca + (HCOO)_2Ca \xrightarrow{\Delta} 2 CH_3CH_2CHO + 2CaCO_3$
Calcium propionate reacts with calcium formate to produce propanaldehyde as the main product.

(C) The given reaction is a Friedel-Crafts acylation reaction.
In Friedel-Crafts acylation,benzene reacts with an acylating agent (such as an acid chloride or an acid anhydride) in the presence of a Lewis acid catalyst like $AlCl_3$ to form an aromatic ketone.
For the product $C_6H_5-C(=O)-CH_2-C_6H_5$ (phenyl benzyl ketone),the acylating group is $C_6H_5-CH_2-CO-$.
Option $(A)$ is $C_6H_5CH_2COCl$ (phenylacetyl chloride),which is an acid chloride.
Option $(B)$ is $(C_6H_5-CH_2CO)_2O$ (phenylacetic anhydride),which is an acid anhydride.
Both acid chlorides and acid anhydrides can act as acylating agents in the presence of $AlCl_3$ to produce the same ketone.
Therefore,both $(A)$ and $(B)$ are correct.

(C) The reduction of benzoyl chloride with $Pd$ and $BaSO_4/CaCO_3$ produces benzaldehyde.
$BaSO_4$ acts as a catalyst poison to prevent the further reduction of the aldehyde to an alcohol. This reaction is known as the $Rosenmund$ reduction.
$C_6H_5COCl + H_2 \xrightarrow{Pd/BaSO_4} C_6H_5CHO + HCl$

(B) The given reaction is the oxidation of toluene $(C_6H_5CH_3)$ using chromyl chloride $(CrO_2Cl_2)$ in the presence of $CS_2$ or $CCl_4$ as a solvent,followed by hydrolysis.
This reaction is known as the $Etard$ reaction.
In this process,the methyl group of toluene is oxidized to a benzaldehyde group $(-CHO)$.
Thus,$A$ is a chromium complex intermediate,and $B$ is benzaldehyde $(C_6H_5CHO)$.

(A) 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 $(CrO_2Cl_2)$.
In this reaction,toluene is treated with chromyl chloride in a suitable solvent like $CS_2$ or $CCl_4$ to form a chromium complex,which upon hydrolysis yields benzaldehyde.
The reaction is represented as: $C_6H_5CH_3 + 2CrO_2Cl_2$ $\rightarrow C_6H_5CH_3 \cdot 2CrO_2Cl_2$ $\xrightarrow{H_3O^+} C_6H_5CHO$.

(C) The reaction of a nitrile $(-CN)$ with $DIBAL-H$ (Diisobutylaluminium hydride) at $-78^{\circ}C$ followed by acid hydrolysis $(H_3O^+)$ is a selective reduction method to produce an aldehyde.
$DIBAL-H$ reduces the nitrile group to an imine intermediate,which upon hydrolysis yields the corresponding aldehyde.
Crucially,$DIBAL-H$ does not reduce the carbon-carbon double bond $(C=C)$ present in the molecule.
Therefore,the reaction is:
$CH_3-CH=CH-CH_2-CN \xrightarrow[(2) H_3O^+]{(1) DIBAL-H, -78^{\circ}C} CH_3-CH=CH-CH_2-CHO$
The correct product $A$ is $CH_3-CH=CH-CH_2-CHO$.

(D) Acetophenone $(C_6H_5COCH_3)$ is a ketone.
$1$. Oxidation of $1-$phenylethanol $(C_6H_5CH(OH)CH_3)$ using oxidizing agents like $KMnO_4$ or $CrO_3$ yields acetophenone.
$2$. Friedel-Crafts acylation of benzene with ethanoyl chloride $(CH_3COCl)$ in the presence of anhydrous $AlCl_3$ is a standard industrial method to prepare acetophenone.
Since both $A$ and $D$ are valid methods,and $D$ is the most common textbook method for aromatic ketones,$D$ is the primary answer.

(D) . $CH_3-CN$ reacts with $AlH(i-Bu)_2$ ($DIBAL$-$H$) followed by hydrolysis to give $CH_3CHO$ (acetaldehyde).
$B$. $CH_3COCl$ undergoes Rosenmund reduction with $H_2-Pd/BaSO_4$ to produce $CH_3CHO$.
$C$. $HC \equiv CH$ undergoes hydration in the presence of $HgSO_4/H_2SO_4$ to form vinyl alcohol,which tautomerizes to $CH_3CHO$.
$D$. $CH_3-CN$ is reduced by $LiAlH_4$ to form ethylamine $(CH_3-CH_2-NH_2)$,not acetaldehyde. Therefore,option $D$ is the correct answer.

(B) The given reaction involves the catalytic hydrogenation of an acid chloride $(RCOCl)$ to an aldehyde $(RCHO)$ using $H_2$ gas in the presence of palladium supported on barium sulfate $(Pd-BaSO_4)$.
This specific reaction is known as the Rosenmund reduction.
In this process,$BaSO_4$ acts as a poison to the palladium catalyst,preventing the further reduction of the aldehyde to a primary alcohol.

(A) The Gattermann-Koch reaction is a chemical reaction in which benzene is converted to benzaldehyde by treatment with carbon monoxide $(CO)$ and hydrogen chloride $(HCl)$ in the presence of a Lewis acid catalyst,typically anhydrous aluminum chloride $(AlCl_3)$ and cuprous chloride $(CuCl)$.
The reaction is represented as:
$C_6H_6 + CO + HCl \xrightarrow{Anhyd. AlCl_3/CuCl} C_6H_5CHO$
Option $A$ correctly represents this reaction.

(A) The given reaction is the Etard reaction.
In this reaction,toluene is treated with chromyl chloride $(CrO_2Cl_2)$ in the presence of carbon disulfide $(CS_2)$ as a solvent,followed by hydrolysis with $H_3O^+$.
This process selectively oxidizes the methyl group of toluene to a formyl group,resulting in the formation of benzaldehyde as the product $A$.

(D) Rosenmund reduction involves the catalytic hydrogenation of an acid chloride to an aldehyde using $H_2$ gas in the presence of $Pd/BaSO_4$.
To prepare formaldehyde $(HCHO)$,the required starting material would be formyl chloride $(HCOCl)$.
However,formyl chloride is highly unstable at room temperature and spontaneously decomposes into carbon monoxide $(CO)$ and hydrogen chloride $(HCl)$.
Therefore,$HCHO$ cannot be prepared by the Rosenmund reduction method.

(B) The reaction of a nitrile $(R-CN)$ with a Grignard reagent $(R'MgX)$ followed by acid hydrolysis $(H_3O^+)$ is a standard method to synthesize ketones.
$1$. The Grignard reagent $(CH_3MgBr)$ acts as a nucleophile and attacks the electrophilic carbon of the nitrile group $(-CN)$.
$2$. This forms an imine intermediate: $m\text{-CH}_3C_6H_4-C(CH_3)=NMgBr$.
$3$. Subsequent acid hydrolysis $(H_3O^+)$ converts the imine intermediate into a ketone.
$4$. The final product is $m\text{-methylacetophenone}$ (also known as $3\text{-methylacetophenone}$).
Therefore,the correct option is $B$.

(D) The given reaction is the Rosenmund reduction,which is used to convert acid chlorides $(RCOCl)$ into aldehydes $(RCHO)$.
In this reaction,benzoyl chloride $(C_6H_5COCl)$ is reduced to benzaldehyde $(C_6H_5CHO)$ using hydrogen gas in the presence of palladium catalyst supported on barium sulfate $(Pd/BaSO_4)$.
$BaSO_4$ acts as a poison to prevent the further reduction of the aldehyde to an alcohol.
Therefore,the reagent '$A$' is $Pd/BaSO_4 + H_2$.

(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 $(AlCl_3)$ or cuprous chloride $(CuCl)$ to form benzaldehyde.
The reaction is as follows:
$C_6H_6 + CO + HCl \xrightarrow{AlCl_3} C_6H_5CHO + HCl$
Therefore,the product is benzaldehyde.

(A) The reaction of Benzene with $Zn/Hg$ and conc. $HCl$ is the Clemmensen reduction,which is used to reduce carbonyl groups (aldehydes or ketones) to methylene $(-CH_2-)$ groups. It does not prepare Benzaldehyde from Benzene.
Option $A$ is the correct answer because this reaction is not a method for the preparation of Benzaldehyde.
Option $B$ is the Etard reaction,which converts Toluene to Benzaldehyde.
Option $C$ is the Rosenmund reduction,which converts Benzoyl chloride to Benzaldehyde.
Option $D$ is the Gattermann-Koch reaction,which converts Benzene to Benzaldehyde.

(D) Let us analyze each reaction:
$(A)$ Etard reaction: Toluene reacts with chromyl chloride $(CrO_2Cl_2)$ followed by hydrolysis to give benzaldehyde. This produces an aldehyde.
$(B)$ Dehydrogenation of primary alcohol: Propan$-1-$ol reacts with $Cu$ at $300 \ ^\circ C$ to give propanal (an aldehyde).
$(C)$ Oxidation of primary alcohol: Propan$-1-$ol reacts with Pyridinium chlorochromate $(PCC)$ to give propanal (an aldehyde).
$(D)$ Reduction of amide: Propanamide reacts with lithium aluminium hydride $(LiAlH_4)$ to give propan$-1-$amine (a primary amine),not an aldehyde.
Therefore,the reaction that does not give an aldehyde is $(D)$.

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

(C) Let us analyze each reaction:
$(A)$ $NaBH_4$ is a selective reducing agent that reduces aldehydes and ketones to alcohols but does not reduce esters. The reaction shown reduces the ester group to an alcohol while keeping the ketone intact,which is incorrect.
$(B)$ $H_2/Pd-C$ is a strong reducing agent that reduces both the $C=C$ double bond and the $C=O$ double bond. It would not stop at the alcohol stage with one equivalent; it would reduce the alkene as well.
$(C)$ The reduction of nitriles $(R-CN)$ with $DIBAL-H$ followed by acid hydrolysis $(H_3O^+)$ is a standard method to produce aldehydes $(R-CHO)$. Thus,$C_6H_5-CH_2-CN \xrightarrow{(1) DIBAL-H, (2) H_3O^+} C_6H_5-CH_2-CHO$ is correct.
$(D)$ The reduction of an internal alkyne with $Na/liq. NH_3$ (Birch reduction conditions) produces a trans-alkene,not a cis-alkene. Therefore,this is incorrect.

(C) The starting material is acetoacetic acid $(CH_3-CO-CH_2-COOH)$.
Upon heating $(\Delta)$,$\beta$-keto acids undergo decarboxylation to form a ketone.
$CH_3-CO-CH_2-COOH \xrightarrow{\Delta} CH_3-CO-CH_3 CO_2$.
Thus,$X$ is acetone $(CH_3-CO-CH_3)$.
Now,let us evaluate the options:
$(A)$ $CH_3-CH_2-CH_2-Cl \xrightarrow{LiAlH_4} CH_3-CH_2-CH_3$ (Propane,not acetone).
$(B)$ $CH_3-CH_2-COOH \xrightarrow{P HI} CH_3-CH_2-CH_3$ (Propane,not acetone).
$(C)$ $CH_3-C \equiv CH \xrightarrow{Hg^{2 }, H_2O} CH_3-CO-CH_3$ (Acetone is formed via hydration of propyne).
Therefore,the correct option is $C$.

(C) The reaction of a nitrile with a Grignard reagent followed by acid hydrolysis is a standard method for the synthesis of ketones.
$CH_3-CH_2-CH_2-CN + CH_3MgI$ $\xrightarrow{Et_2O} CH_3-CH_2-CH_2-C(CH_3)=NMgI$ $\xrightarrow{H_3O^{+}} CH_3-CH_2-CH_2-CO-CH_3 + NH_3 + Mg(OH)I$.
The product formed is $2$-pentanone. Options $(a)$,$(b)$,and $(d)$ primarily yield alcohols.

(B) The starting material is $N$-methoxy-$N$-methylbenzamide (Weinreb amide).
Weinreb amides react with Grignard reagents $(RMgX)$ to form ketones after hydrolysis.
In this reaction,$N$-methoxy-$N$-methylbenzamide reacts with $2$ equivalents of $CH_3MgBr$.
The first equivalent of $CH_3MgBr$ attacks the carbonyl carbon to form a stable tetrahedral intermediate,which prevents further reaction with the Grignard reagent.
Upon acid hydrolysis $(H_3O^+)$,this intermediate decomposes to form acetophenone $(Ph-CO-CH_3)$.
Therefore,the product $(P)$ is acetophenone $(Ph-CO-CH_3)$.

(C) When nitriles react with Grignard reagents,they form an imine salt intermediate,which upon acid hydrolysis yields a ketone.
Reaction:
$CH_3-C \equiv N + CH_3MgI \xrightarrow{Et_2O} CH_3-C(CH_3)=N-MgI$
$CH_3-C(CH_3)=N-MgI \xrightarrow{H_3O^{+}} CH_3-C(=O)-CH_3 + NH_3 + Mg(OH)I$
The major product is acetone $(CH_3-C(=O)-CH_3)$.

(A) $2$-phenyl-$2$-butanol is a tertiary alcohol with the structure $CH_3-C(OH)(Ph)-CH_2CH_3$.
It can be prepared by the nucleophilic addition of a Grignard reagent to a ketone.
The reaction of ethylmagnesium bromide $(CH_3CH_2MgBr)$ with acetophenone $(Ph-C(=O)-CH_3)$ followed by acidic hydrolysis yields $2$-phenyl-$2$-butanol.
$CH_3CH_2MgBr + Ph-C(=O)-CH_3$ $\rightarrow Ph-C(OMgBr)(CH_3)CH_2CH_3$ $\xrightarrow{H_3O^+} Ph-C(OH)(CH_3)CH_2CH_3$.

(B) The synthesis of $4-heptanol$ requires the addition of a propyl Grignard reagent to a butyraldehyde (butanal) substrate.
$(A)$ $CH_3CH_2CH_2MgBr + H_2C=O \rightarrow CH_3CH_2CH_2CH_2OH$ $(1-butanol)$.
$(B)$ $CH_3CH_2CH_2MgBr + CH_3CH_2CH_2CHO \rightarrow (CH_3CH_2CH_2)_2CHOH$ $(4-heptanol)$. This is the correct synthesis.
$(C)$ $CH_3CH_2CH_2CH_2MgBr + (CH_3)_2C=O \rightarrow 2-methyl-2-hexanol$.
$(D)$ $(CH_3CH_2CH_2)_2CHMgBr + H_2C=O \rightarrow (CH_3CH_2CH_2)_2CHCH_2OH$ $(2-propyl-1-pentanol)$.

(A) Step $1$: $CH_2=CH-CH_2Br$ undergoes nucleophilic substitution with $NaCN$ to form $CH_2=CH-CH_2CN$ $(Y)$.
Step $2$: The reaction of nitrile $(Y)$ with Grignard reagent $(C_6H_5MgBr)$ followed by acid hydrolysis $(H_3O^{+})$ yields a ketone.
$CH_2=CH-CH_2Br + NaCN \rightarrow CH_2=CH-CH_2CN$ $(Y)$
$CH_2=CH-CH_2CN + C_6H_5MgBr$ $\xrightarrow{\text{ether}} CH_2=CH-CH_2-C(=NMgBr)-C_6H_5$ $\xrightarrow{H_3O^{+}} CH_2=CH-CH_2-C(=O)-C_6H_5$ $(Z)$

(B) $PCC$ (Pyridinium chlorochromate) is a mild oxidizing agent that selectively oxidizes primary alcohols to aldehydes and secondary alcohols to ketones without affecting other functional groups like double bonds.
In the given reaction,$4-\text{tert-butylcyclohexanol}$ is a secondary alcohol.
Therefore,$PCC$ oxidizes the secondary alcohol group $(-OH)$ to a ketone group $(=O)$,resulting in $4-\text{tert-butylcyclohexanone}$.

(C) The synthesis of $2-$methyl$-3-$pentanone can be achieved by the following steps:
$1.$ Oxidation of $1-$propanol to propanal using $PCC$ in $CH_2Cl_2$.
$2.$ Nucleophilic addition of isopropyl lithium $(CH_3)_2CHLi$ to propanal followed by acidic workup $(H_3O^{+})$ to form $2-$methyl$-3-$pentanol.
$3.$ Oxidation of the resulting secondary alcohol to the ketone using $Na_2Cr_2O_7, H_2SO_4, H_2O$ (Jones reagent).
This corresponds to the sequence in option $C$.

(A) The preparation of $1$-phenylethanol involves the nucleophilic addition of a Grignard reagent to benzaldehyde.
Benzaldehyde $(C_6H_5CHO)$ reacts with methylmagnesium iodide $(CH_3MgI)$ to form an intermediate,which upon acidic hydrolysis yields $1$-phenylethanol.
The Grignard reagent $CH_3MgI$ is prepared by the reaction between $CH_3I$ and $Mg$ in the presence of dry ether.
Thus,the correct option is $A$.

(D) The synthesis of hexanal $(C_6H_{12}O)$ can be achieved by the reaction of a Grignard reagent with ethylene oxide followed by oxidation.
Step $1$: $CH_3CH_2CH_2CH_2MgBr + C_2H_4O \rightarrow CH_3CH_2CH_2CH_2CH_2CH_2OMgBr$.
Step $2$: Acidic workup $(H_3O^{+})$ yields $1$-hexanol $(CH_3CH_2CH_2CH_2CH_2CH_2OH)$.
Step $3$: Oxidation of the primary alcohol with $PCC$ (Pyridinium chlorochromate) in $CH_2Cl_2$ gives hexanal $(CH_3CH_2CH_2CH_2CH_2CHO)$.

(B) When carboxylic acids are heated with $Ca(OH)_2$,they form calcium salts of the acid.
$2CH_3COOH + Ca(OH)_2 \rightarrow (CH_3COO)_2Ca + 2H_2O$
Upon further heating,the calcium salt undergoes dry distillation to produce a ketone.
$(CH_3COO)_2Ca \xrightarrow{\Delta} CH_3COCH_3 + CaCO_3$
The product $(A)$ is acetone $(CH_3COCH_3)$,which corresponds to the structure shown in option $B$.