Preparation Questions in English

(A) The given reactions are standard methods for the preparation of acetaldehyde $(CH_3CHO)$:
$1$. $A$ is ethyne $(HC \equiv CH)$. Hydration of ethyne in the presence of $HgSO_4 / H_2SO_4$ gives vinyl alcohol $(CH_2=CH-OH)$,which tautomerises to form acetaldehyde $(CH_3CHO)$.
$2$. $B$ is ethanol $(CH_3CH_2OH)$. $PCC$ (Pyridinium chlorochromate) is a mild oxidising agent that oxidises primary alcohols to aldehydes. Thus,$CH_3CH_2OH \xrightarrow{PCC} CH_3CHO$.
$3$. $C$ is ethane nitrile $(CH_3CN)$. The Stephen reduction of nitriles using $SnCl_2 / HCl$ followed by acid hydrolysis $(H_3O^+)$ yields acetaldehyde $(CH_3CHO)$.
Therefore,$A$ is ethyne,$B$ is ethanol,and $C$ is ethane nitrile.

(B) Statement-$I$ is true: Esters are reduced to aldehydes using $DIBAL-H$ at low temperature followed by hydrolysis.
Statement-$II$ is false: Oxidation of benzyl alcohol with aqueous $KMnO_4$ (a strong oxidizing agent) leads to the formation of benzoic acid,not benzaldehyde.

(A) Acetophenone $(C_{6}H_{5}COCH_{3})$ is a ketone.
$1$. $C_{6}H_{6}$ (Benzene) can undergo Friedel-Crafts acylation with acetyl chloride to form acetophenone.
$2$. $C_{6}H_{5}CH(OH)CH_{3}$ ($1$-phenylethanol) can be oxidized to form acetophenone.
$3$. $C_{6}H_{5}C \equiv CH$ (Phenylacetylene) can undergo hydration to form acetophenone.
$4$. $C_{6}H_{5}CH_{3}$ (Toluene) cannot be directly converted to acetophenone easily because oxidation of the methyl group typically yields benzoic acid,and Friedel-Crafts acylation of toluene would result in a mixture of ortho and para isomers,not the desired acetophenone.

(A) Dry distillation of a mixture of calcium formate and another calcium salt of a carboxylic acid yields an aldehyde.
Specifically,the reaction between calcium formate $(HCOO)_2Ca$ and calcium acetate $(CH_3COO)_2Ca$ produces acetaldehyde $(CH_3CHO)$ along with calcium carbonate $(CaCO_3)$.
The reaction is: $(HCOO)_2Ca + (CH_3COO)_2Ca \xrightarrow{400^{\circ}C} 2CH_3CHO + 2CaCO_3$.

(C) When acetic acid reacts with slaked lime,calcium acetate is formed,which on dry distillation produces a ketone.
$2 CH_3COOH + Ca(OH)_2 \rightarrow (CH_3COO)_2Ca + 2 H_2O$
$(CH_3COO)_2Ca \xrightarrow{\Delta} CH_3COCH_3 + CaCO_3$
The final product obtained is propanone $(CH_3COCH_3)$.

(B) The dry distillation of a mixture of calcium acetate $(CH_3COO)_2Ca$ and calcium formate $(HCOO)_2Ca$ results in the formation of acetaldehyde $(CH_3CHO)$ and calcium carbonate $(CaCO_3)$.
The chemical reaction is as follows:
$(CH_3COO)_2Ca + (HCOO)_2Ca \xrightarrow{\text{dry distillation}} 2CH_3CHO + 2CaCO_3$

(A) The oxidation of $Toluene$ to $Benzaldehyde$ using chromyl chloride $(CrO_2Cl_2)$ in a suitable solvent like $CS_2$ or $CCl_4$,followed by hydrolysis,is a well-known named reaction in organic chemistry.
This specific transformation is called the $Etard$ reaction.
The reaction proceeds through the formation of a brown chromium complex intermediate,which upon hydrolysis yields $Benzaldehyde$.

(B) The reaction of toluene $(C_6H_5CH_3)$ with chromyl chloride $(CrO_2Cl_2)$ in the presence of carbon disulfide $(CS_2)$ followed by hydrolysis is known as the Etard reaction.
This reaction specifically oxidizes the methyl group of toluene to a formyl group,resulting in the formation of benzaldehyde $(C_6H_5CHO)$.
Therefore,the compound $Z$ is benzaldehyde.

(D) The conversion of benzene to acetophenone is achieved by reacting benzene with acetyl chloride $(CH_3COCl)$ in the presence of an anhydrous Lewis acid catalyst like $AlCl_3$.
This reaction is a specific type of electrophilic aromatic substitution known as Friedel-Crafts acylation,where an acyl group $(-COCH_3)$ is introduced into the benzene ring.
The reaction is represented as:
$C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_6H_5COCH_3 + HCl$

(D) The final product is $4\text{-methylpent-3-en-2-one}$,which is an $\alpha,\beta\text{-unsaturated ketone}$ formed by the aldol condensation of $2$ molecules of acetone $(CH_3COCH_3)$.
Thus,$R$ must be acetone $(CH_3COCH_3)$.
The reaction of $P$ with $CH_3MgBr$ followed by hydrolysis yields acetone $(R)$.
$CH_3CN + CH_3MgBr$ $\rightarrow CH_3C(CH_3)=NMgBr$ $\xrightarrow{H_3O^+} CH_3COCH_3 + NH_3 + Mg(OH)Br$.
Therefore,$P$ is $CH_3CN$ (Ethanenitrile).

(D) The reaction sequence is:
$HC \equiv CH + H_2O \xrightarrow{Hg^{2+}/H^+, 333 \ K} [CH_2=CH-OH] \rightleftharpoons CH_3CHO$
Here,$Y$ is $CH_3CHO$ (Ethanal).
$A$: Rosenmund reduction of acetyl chloride gives ethanal $(CH_3COCl + H_2 \xrightarrow{Pd/BaSO_4} CH_3CHO + HCl)$.
$B$: Dehydrogenation of ethanol gives ethanal $(CH_3CH_2OH \xrightarrow{Cu, 573 \ K} CH_3CHO + H_2)$.
$C$: Stephen reduction of acetonitrile gives ethanal $(CH_3CN + SnCl_2 + HCl$ $\rightarrow CH_3CH=NH$ $\xrightarrow{H_3O^+} CH_3CHO)$.
$D$: Reduction of acetic acid with $LiAlH_4$ gives ethanol $(CH_3CH_2OH)$,not ethanal $(CH_3CHO)$.
Therefore,$Y$ cannot be obtained from reaction $D$.

(B) The reaction of toluene with $Cl_2$ in the presence of $hv$ (photochemical chlorination) leads to the formation of benzal chloride $(C_6H_5CHCl_2)$ as the intermediate $X$,which upon hydrolysis at $373 \ K$ gives benzaldehyde.
The reaction of toluene with chromyl chloride $(CrO_2Cl_2)$ in $CS_2$ is the Etard reaction. This reaction proceeds through the formation of a brown chromium complex intermediate $Y$,which is $C_6H_5CH[OCr(OH)Cl_2]_2$. This complex upon acidic hydrolysis $(H_3O^+)$ yields benzaldehyde.
Therefore,$X = C_6H_5CHCl_2$ and $Y = C_6H_5CH[OCr(OH)Cl_2]_2$.

(C) The conversion of toluene to benzaldehyde can be achieved by the following methods:
$1$. Side chain chlorination followed by hydrolysis $(A)$: Toluene reacts with $Cl_2$ in the presence of light $(h\nu)$ to form benzal chloride $(C_6H_5CHCl_2)$,which upon hydrolysis with $H_2O$ at high temperature $(\Delta)$ yields benzaldehyde.
$2$. Etard reaction $(D)$: Toluene reacts with chromyl chloride $(CrO_2Cl_2)$ in $CS_2$ solvent followed by acidic hydrolysis $(H_3O^{+})$ to give benzaldehyde.
$3$. Reagent $C$ $(KMnO_4 / OH^{-} ; H^{+})$ is an oxidizing agent that converts toluene directly to benzoic acid,not benzaldehyde.
Therefore,the correct sets are $A$ and $D$.

(A) The conversion of ethyl bromide $(CH_3CH_2Br)$ to propanal $(CH_3CH_2CHO)$ involves the following steps:
$1$. Reaction of ethyl bromide with silver acetate $(CH_3COOAg)$ to form ethyl ethanoate $(CH_3COOCH_2CH_3)$:
$CH_3COOAg + CH_3CH_2Br \rightarrow CH_3COOCH_2CH_3 + AgBr \downarrow$
$2$. Reduction of the ester (ethyl ethanoate) to an aldehyde using $DIBAL-H$ (Diisobutylaluminium hydride) followed by hydrolysis $(H_2O)$:
$CH_3COOCH_2CH_3 \xrightarrow{DIBAL-H, H_2O} CH_3CH_2CHO$
Thus,the correct sequence of reagents is $CH_3COOAg, DIBAL-H, H_2O$.

(C) The Etard reaction $(I)$ involves the oxidation of toluene to benzaldehyde using chromyl chloride $(CrO_2Cl_2)$ in the presence of $CCl_4$ as a solvent. The reaction proceeds through the formation of a brown chromium complex,which is subsequently hydrolyzed to give benzaldehyde.
The Stephen reaction $(II)$ involves the reduction of nitriles $(R-CN)$ to imines using stannous chloride $(SnCl_2)$ in the presence of hydrochloric acid $(HCl)$,followed by hydrolysis to yield the corresponding aldehyde $(R-CHO)$.

(D) The reaction of benzoyl chloride with hydrogen in the presence of $Pd$ supported on $BaSO_4$ is known as the Rosenmund reduction reaction.
In this reaction,acid chlorides are selectively reduced to their corresponding aldehydes.
The chemical equation is:
$C_6H_5COCl + H_2 \xrightarrow{Pd/BaSO_4} C_6H_5CHO + HCl$

(D) The given reaction involves the oxidation of toluene $(C_6H_5CH_3)$ to benzaldehyde $(C_6H_5CHO)$ using chromyl chloride $(CrO_2Cl_2)$ in the presence of carbon disulfide $(CS_2)$ as a solvent,followed by hydrolysis.
This specific chemical transformation is known as the Etard reaction.

(C) The final product is $4\text{-methylpent-3-en-2-one}$,which is an $\alpha,\beta\text{-unsaturated ketone}$ formed by the aldol condensation of $2$ moles of acetone $(propanone)$.
Thus,$(R)$ must be $propanone$ $(CH_3COCH_3)$.
The reaction of $(P)$ with $CH_3MgBr$ followed by $H_3O^{+}$ yields $propanone$. This is a characteristic reaction of nitriles.
$CH_3CN + CH_3MgBr$ $\rightarrow CH_3C(CH_3)=NMgBr$ $\xrightarrow{H_3O^{+}} CH_3COCH_3 + NH_3 + Mg(OH)Br$.
Therefore,$(P)$ is $CH_3CN$ $(Ethane\ nitrile)$.

(C) The reaction of a Grignard reagent $(RMgX)$ with cadmium chloride $(CdCl_2)$ produces an organocadmium compound $(R_2Cd)$,which is represented as $X$.
This organocadmium compound $(R_2Cd)$ reacts with an acid chloride $(R^{\prime}COCl)$,represented as $Y$,to form a ketone $(R-CO-R^{\prime})$.
The reaction is:
$2RMgX + CdCl_2 \rightarrow R_2Cd + 2MgXCl$
$R_2Cd + 2R^{\prime}COCl \rightarrow 2R-CO-R^{\prime} + CdCl_2$

(C) The reagent $AlH(i-Bu)_2$ is Diisobutylaluminium hydride,commonly known as $DIBAL-H$.
$DIBAL-H$ is a selective reducing agent that reduces nitriles $(-CN)$ to aldehydes $(-CHO)$ upon hydrolysis.
In the given reaction,the nitrile group $(-CN)$ is converted to an aldehyde group $(-CHO)$ while the double bond $(C=C)$ remains unaffected.
Therefore,the reaction is: $CH_3-CH=CH-CH_2-CH_2-CN \xrightarrow[\text{2) } H_2O]{\text{1) } DIBAL-H} CH_3-CH=CH-CH_2-CH_2-CHO$.
Thus,the correct option is $C$.

(D) Step $1$: The conversion of benzene diazonium chloride $(C_6H_5N_2^+Cl^-)$ to benzonitrile $(C_6H_5CN)$ is a Sandmeyer reaction,which uses $CuCN \mid KCN$ as the reagent $X$.
Step $2$: The reaction of benzonitrile with a Grignard reagent $(CH_3MgBr)$ followed by acid hydrolysis $(H_2O)$ is a standard method to prepare ketones.
Step $3$: The mechanism involves the nucleophilic attack of $CH_3^-$ on the nitrile carbon to form an imine intermediate,which upon hydrolysis yields acetophenone $(C_6H_5COCH_3)$.
Therefore,$X = CuCN \mid KCN$ and $Y = C_6H_5COCH_3$.

(C) The reaction sequence is as follows:
$1$. $C_2H_5Br + KCN \rightarrow C_2H_5CN + KBr$ ($A$ is Propanenitrile).
$2$. $C_2H_5CN + 2H_2O + H_3O^{+} \rightarrow C_2H_5COOH + NH_4^{+}$ ($B$ is Propanoic acid).
$3$. $C_2H_5COOH \xrightarrow{LiAlH_4} C_2H_5CH_2OH$ ($C$ is Propan$-1-$ol).
$4$. $C_2H_5CH_2OH \xrightarrow[573 \ K]{Cu} C_2H_5CHO + H_2$ ($D$ is Propionaldehyde or Propanal).

(C) The reaction involves the nucleophilic addition of a Grignard reagent $(PhMgBr)$ to a nitrile $(PhCN)$.
$1$. The phenyl group $(Ph^-)$ from $PhMgBr$ attacks the electrophilic carbon of the nitrile group $(-CN)$,forming an intermediate imine salt: $Ph-C(Ph)=NMgBr$.
$2$. Upon hydrolysis with $H_3O^+$,the imine salt is converted into an imine $(Ph_2C=NH)$,which is further hydrolyzed to the final carbonyl compound,benzophenone $(Ph_2C=O)$.

(C) The reaction of a nitrile with a Grignard reagent $(RMgX)$ followed by acid hydrolysis yields a ketone.
For the formation of propiophenone $(CH_3CH_2-CO-C_6H_5)$,the starting material is propionitrile $(CH_3CH_2-CN)$.
The Grignard reagent $(A)$ must be phenylmagnesium bromide $(C_6H_5MgBr)$.
The reaction proceeds as follows:
$CH_3CH_2-C \equiv N + C_6H_5MgBr \longrightarrow CH_3CH_2-C(C_6H_5)=NMgBr$ $(B)$
$CH_3CH_2-C(C_6H_5)=NMgBr + 2H_2O \xrightarrow{H_3O^{+}} CH_3CH_2-CO-C_6H_5 + NH_3 + Mg(OH)Br$
Thus,$A$ is $C_6H_5MgBr$ and $B$ is $CH_3CH_2C(NMgBr)C_6H_5$.

(A) $C_2H_4Br_2$ undergoes dehydrohalogenation with alc. $KOH$ followed by $NaNH_2$ to form ethyne $(X)$: $C_2H_4Br_2 \xrightarrow{\text{alc. } KOH, NaNH_2} HC \equiv CH (X)$.
Ethyne undergoes cyclic polymerization in the presence of red hot $Fe$ tube at $873 \ K$ to form benzene $(Y)$: $3C_2H_2 \xrightarrow{\text{Red hot Fe tube}} C_6H_6 (Y)$.
Benzene reacts with acetic anhydride in the presence of anhydrous $AlCl_3$ (Friedel-Crafts acylation) to form acetophenone $(Z)$: $C_6H_6 + (CH_3CO)_2O \xrightarrow{AlCl_3} C_6H_5COCH_3 (Z)$.

(A) The correct reagent for the Etard reaction is $CrO_2Cl_2$ followed by hydrolysis $(H_3O^+)$.
In the Etard reaction,toluene is oxidized to benzaldehyde using chromyl chloride $(CrO_2Cl_2)$.
The reaction proceeds through the formation of a chromium complex intermediate which upon hydrolysis yields the aldehyde.
The reaction is represented as:
$C_6H_5CH_3 \xrightarrow{CrO_2Cl_2, CCl_4 / H_3O^+} C_6H_5CHO$.

(D) Fehling's solution is a test for aliphatic aldehydes. The compound $X$ that gives a red precipitate with Fehling's solution is acetaldehyde $(CH_3CHO)$.
The reaction of acetylene $(C_2H_2)$ with water in the presence of $40\% H_2SO_4$ and $1\% HgSO_4$ (Kucherov reaction) yields acetaldehyde $(CH_3CHO)$ as the major product.
$C_2H_2 + H_2O \xrightarrow[1\% HgSO_4]{40\% H_2SO_4} CH_3CHO$
Acetaldehyde then reacts with Fehling's solution to give a red precipitate of cuprous oxide $(Cu_2O)$.

(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 $(CrO_2Cl_2)$. The reaction involves the formation of a brown chromium complex intermediate,which upon hydrolysis yields the corresponding aldehyde. The correct representation is: $C_6H_5CH_3 + 2CrO_2Cl_2$ $\rightarrow C_6H_5CH(OCrOHCl_2)_2$ $\xrightarrow{H_3O^{+}} C_6H_5CHO$.

(A) The reaction described is the Etard reaction or a variation of the oxidation of toluene to benzaldehyde using chromyl chloride or chromic oxide in acetic anhydride.
Specifically,the reaction of toluene with $CrO_3$ and acetic anhydride $(CH_3CO)_2O$ at $273-283 \ K$ forms a gem-diacetate intermediate,$C_6H_5CH(OCOCH_3)_2$,which is $C$.
Upon subsequent hydrolysis with $H_3O^{\oplus}$,this intermediate yields benzaldehyde.
Therefore,$A = CrO_3$,$B = (CH_3CO)_2O$,and $C = C_6H_5CH(OCOCH_3)_2$.

(C) Acetophenone is prepared by the Friedel-Crafts acylation of benzene.
In this reaction,benzene $(C_6H_6)$ reacts with acetyl chloride $(CH_3COCl)$ in the presence of an anhydrous Lewis acid catalyst,anhydrous $AlCl_3$.
The reaction is:
$C_6H_6 + CH_3COCl \xrightarrow{\text{anhydrous } AlCl_3} C_6H_5COCH_3 + HCl$
Thus,the chemicals required are $C_6H_6$ $(A)$,$CH_3COCl$ $(C)$,and anhydrous $AlCl_3$ $(D)$.

(A) The reaction sequence is as follows:
$1$. Dehydrogenation of $C_5H_{12}O$ (a primary or secondary alcohol) with $Cu$ at $573 \ K$ gives an alkene $(C_5H_{10})$.
$2$. Ozonolysis of the alkene $(C_5H_{10})$ followed by reductive workup $(Zn/H_2O)$ yields carbonyl compounds $X$ and $Y$.
$3$. Considering the options,if the alkene is $2$-methylbut-$2$-ene $(CH_3-C(CH_3)=CH-CH_3)$,its ozonolysis gives acetone $(CH_3COCH_3)$ and acetaldehyde $(CH_3CHO)$.
$4$. Thus,$X$ and $Y$ are acetone and acetaldehyde.

(C) Primary alcohols,when passed over heated copper at $300^{\circ}C$,undergo dehydrogenation to form aldehydes.
The reaction is: $C_2H_5OH \xrightarrow[300^{\circ}C]{Cu} CH_3CHO + H_2$.
The product $X$ is acetaldehyde,which has the molecular formula $C_2H_4O$.

(C) The Gatterman-Koch reaction involves the formylation of benzene or its derivatives using carbon monoxide $(CO)$ and hydrogen chloride $(HCl)$ in the presence of anhydrous aluminum chloride $(AlCl_3)$ and cuprous chloride $(CuCl)$ as a catalyst to produce benzaldehyde.
Option $A$ represents the Rosenmund reduction.
Option $B$ represents the Etard reaction.
Option $C$ represents the Gatterman-Koch reaction.
Option $D$ represents the Friedel-Crafts acylation reaction.

(B) The dry distillation of a mixture of calcium acetate $(CH_3COO)_2Ca$ and calcium formate $(HCOO)_2Ca$ results in the formation of ethanal $(CH_3CHO)$ and calcium carbonate $(CaCO_3)$.
The reaction is as follows:
$(CH_3COO)_2Ca + (HCOO)_2Ca \xrightarrow{\text{dry distillation}} 2CH_3CHO + 2CaCO_3$
Thus,the correct product is ethanal.

(A) The correct matches are as follows:
$(A)$ Acid chloride to aldehyde is the Rosenmund reduction,which uses $H_2, Pd-BaSO_4$ $(IV)$.
$(B)$ Benzene to benzaldehyde is the Gatterman-Koch reaction,which uses $CO, HCl, \text{anhyd. } AlCl_3$ $(II)$.
$(C)$ Acetylene to aldehyde involves the hydration of alkyne,which uses $HgSO_4, H_2SO_4$ $(III)$.
$(D)$ Ester to aldehyde is achieved using $DIBAL-H$ $(I)$.
Therefore,the correct matching is $A-IV, B-II, C-III, D-I$.

(B) In reaction $(A)$,toluene is converted to benzaldehyde using $CrO_2Cl_2$ in $CS_2$ followed by hydrolysis. This is the Etard reaction,so $X = CrO_2Cl_2$.
In reaction $(B)$,benzyl alcohol is converted to benzaldehyde using $Cu$ at $573 \ K$. This is a catalytic dehydrogenation reaction,so $Y = Cu, 573 \ K$.
Therefore,the correct option is $B$.