Preparation Questions in English

(D) $DIBAL$-$H$ stands for diisobutylaluminium hydride,which has the formula $(i-Bu)_2AlH$.
It is a selective reducing agent that reduces esters to aldehydes at low temperatures (typically $-78 \ ^\circ C$).
Therefore,the correct reagent is $DIBAL-H$.
Hence,the correct option is $(d)$.

(D) $DIBAL-H$ (Diisobutylaluminium hydride) is a selective reducing agent.
It is specifically used for the partial reduction of esters to aldehydes at low temperatures (typically $195 \ K$ to $203 \ K$),preventing further reduction to primary alcohols.

(C) The reaction of nitriles $(R-CN)$ with stannous chloride $(SnCl_2)$ in the presence of hydrochloric acid $(HCl)$,followed by hydrolysis $(H_3O^+)$,yields an aldehyde $(R-CHO)$. This specific reduction method is known as the Stephen's reaction.
The mechanism involves the formation of an imine hydrochloride intermediate,which is subsequently hydrolyzed to the corresponding aldehyde:
$R-C \equiv N + SnCl_2 + 2HCl \rightarrow R-CH=NH \cdot HCl$
$R-CH=NH \cdot HCl + H_2O \rightarrow R-CHO + NH_4Cl$

(D) Rosenmund's reduction involves the catalytic hydrogenation of an acid chloride (acid halide) to an aldehyde using $Pd$ supported on $BaSO_4$ (Lindlar's catalyst).
The reaction is: $CH_3COCl + H_2 \xrightarrow{Pd/BaSO_4} CH_3CHO + HCl$.

(A) Reaction $(A)$ is a Friedel-Crafts acylation reaction where benzene reacts with benzoyl chloride in the presence of anhydrous $AlCl_3$ to form benzophenone $(C_6H_5COC_6H_5)$.
Reaction $(B)$ involves the reaction of benzoyl chloride with excess phenylmagnesium bromide followed by hydrolysis,which yields triphenylmethanol $( (C_6H_5)_3COH )$.
Reaction $(C)$ involves the reaction of benzoyl chloride with diphenyl cadmium $( (C_6H_5)_2Cd )$,which is a standard method to prepare ketones from acid chlorides,yielding benzophenone.
Therefore,reactions $(A)$ and $(C)$ produce benzophenone as the major product.

(A) The reaction is a Friedel-Crafts acylation,which involves the electrophilic aromatic substitution of an arene with an acylating agent (like acetic anhydride) in the presence of a Lewis acid catalyst such as anhydrous $AlCl_3$.
In isobutylbenzene,the isobutyl group $(-CH_2CH(CH_3)_2)$ is an ortho/para-directing group due to its electron-donating inductive effect.
Due to steric hindrance caused by the bulky isobutyl group at the ortho position,the electrophile (acetyl cation,$CH_3CO^+$) preferentially attacks the less hindered para position.
Therefore,the major product formed is $p-$isobutylacetophenone.

(B) Step $1$: Bromobenzene reacts with $Mg$ in dry ether to form phenylmagnesium bromide $(C_6H_5MgBr)$,which is a Grignard reagent.
Step $2$: Phenylmagnesium bromide reacts with benzonitrile $(C_6H_5CN)$ to form an intermediate imine salt.
Step $3$: Acidic hydrolysis of the intermediate imine salt yields benzophenone $(C_6H_5COC_6H_5)$ as the final product.

(B) The reaction described is the hydroformylation (or oxo process) of an alkene.
Taking the alkene as $R-CH=CH_2$,the catalyst used is $Co_2(CO)_8$.
In the first step,the alkene reacts with $CO$ and $H_2$ to form an aldehyde as a stable intermediate.
The conversion of an alkene to an aldehyde involves the addition of a formyl group $(-CHO)$,which is an oxidation process relative to the alkene.
Subsequently,the aldehyde undergoes hydrogenation to form a primary alcohol.
The reaction sequence is: $R-CH=CH_2 + CO + H_2$ $\xrightarrow{Co_2(CO)_8} R-CH_2-CH_2-CHO$ $\xrightarrow{H_2} R-CH_2-CH_2-CH_2OH$.
Thus,the stable intermediate is an aldehyde and the nature of the initial step is oxidation.

(C) Step $1$: Aniline $(C_6H_5NH_2)$ reacts with $NaNO_2 + HCl$ at $273 \ K$ to form benzene diazonium chloride $(C_6H_5N_2^+Cl^-)$.
Step $2$: Benzene diazonium chloride reacts with $H_3PO_2 + H_2O$ to form benzene $(C_6H_6)$.
Step $3$: Benzene reacts with $CO + HCl$ in the presence of anhydrous $AlCl_3/CuCl$ (Gattermann-Koch reaction) to form benzaldehyde $(C_6H_5CHO)$.
Therefore,$Z$ is $C_6H_5CHO$.

(B) The conversion of cyanobenzene (benzonitrile) to a Schiff's base involves two main steps:
$1$. Reduction of cyanobenzene to benzaldehyde using $DIBAL-H$ followed by hydrolysis $(H_2O)$.
$2$. Reaction of benzaldehyde with an amine (like aniline) to form a Schiff's base $(R-CH=N-R')$.
Given the options,the primary reagents for the reduction step are $DIBAL-H$ and $H_2O$.

(D) Step $1$: Benzyl iodide $(C_6H_5CH_2I)$ reacts with $KCN$ to form benzyl cyanide $(C_6H_5CH_2CN)$.
Step $2$: Benzyl cyanide reacts with methylmagnesium bromide $(CH_3MgBr)$ to form an imine intermediate $(C_6H_5CH_2C(CH_3)=NMgBr)$.
Step $3$: Acidic hydrolysis $(H_3O^+)$ of the imine intermediate yields phenylacetone $(C_6H_5CH_2COCH_3)$.
The molecular formula of phenylacetone is $C_{10}H_{12}O$. However,reviewing the options provided,let us re-evaluate the structure. The product is $C_6H_5-CH_2-CO-CH_3$. Counting the atoms: $6$ (ring) $+ 1$ (methylene) $+ 1$ (carbonyl) $+ 1$ (methyl) $= 9$ carbons. Wait,$C_6H_5-CH_2-CO-CH_3$ has $6+1+1+1 = 9$ carbons. Let us re-count: $C_6H_5$ $(6C, 5H)$ $+ CH_2$ $(1C, 2H)$ $+ CO$ $(1C)$ $+ CH_3$ $(1C, 3H)$. Total: $C_9H_{10}O$. Thus,the correct option is $D$.

(A) Step $1$: The reaction of benzaldehyde oxime $(C_6H_5-CH=NOH)$ with acetic anhydride $((CH_3CO)_2O)$ causes dehydration to form benzonitrile $(P = C_6H_5-CN)$.
Step $2$: The reaction of benzonitrile $(C_6H_5-CN)$ with methylmagnesium iodide $(CH_3MgI)$ followed by acid hydrolysis $(H_3O^+)$ is a standard method to prepare ketones. The Grignard reagent attacks the nitrile carbon to form an imine intermediate,which upon hydrolysis yields acetophenone $(Q = C_6H_5-CO-CH_3)$.

(C) Let us analyze each reaction:
$A$. Benzoyl chloride $(C_6H_5COCl)$ reacts with $H_2-Pd/BaSO_4$ (Rosenmund reduction) to form benzaldehyde $(C_6H_5CHO)$. This is correct.
$B$. Ethyl benzoate $(C_6H_5COOCH_2CH_3)$ reacts with $LiAlH_4$ followed by $H_2O$ to form benzyl alcohol $(C_6H_5CH_2OH)$. This is incorrect.
$C$. Toluene $(C_6H_5CH_3)$ reacts with $KMnO_4/H^+$ to form benzoic acid $(C_6H_5COOH)$. This is incorrect.
$D$. Benzonitrile $(C_6H_5CN)$ reacts with $DIBAL-H$ (or $AlH(i-Bu)_2$) followed by $H_2O$ to form benzaldehyde $(C_6H_5CHO)$. This is correct.
Therefore,the pairs $A$ and $D$ correctly form benzaldehyde.

(D) Pyridinium chlorochromate $(PCC)$ is formed by the reaction of pyridine with chlorochromic acid,which is obtained by dissolving chromium trioxide $(CrO_3)$ in aqueous $HCl$.
The chemical formula of pyridine is $C_5H_5N$.
When pyridine is protonated,it forms the pyridinium cation,$C_5H_5NH^{\oplus}$.
The chlorochromate anion is $CrO_3Cl^{\ominus}$.
Therefore,the structure of $PCC$ is $C_5H_5NH^{\oplus} CrO_3Cl^{\ominus}$.

(B) The conversion of acetylene $(HC \equiv CH)$ to but$-2-$enal $(CH_3-CH=CH-CHO)$ involves two main steps:
$1$. Hydration of acetylene using $H_2O / Hg^{2+} / H^{+}$ gives acetaldehyde $(CH_3CHO)$.
$2$. Aldol condensation of two molecules of acetaldehyde in the presence of dilute $NaOH$ yields but$-2-$enal.
Reaction: $2CH_3CHO \xrightarrow{dil. NaOH} CH_3-CH=CH-CHO + H_2O$.
Thus,the correct set of reagents is $H_2O / Hg^{2+} / H^{+}$ followed by dil. $NaOH$.
Hence,$(B)$ is the correct answer.

(D) The reaction sequence is as follows:
$1$. Allylic bromination of cyclohexene with $Br_2$ in the presence of $h\nu$ gives $3$-bromocyclohexene.
$2$. Formation of Grignard reagent: $3$-bromocyclohexene reacts with $Mg$ in dry ether to form $3$-cyclohexenylmagnesium bromide.
$3$. Reaction with $CdCl_2$: The Grignard reagent reacts with $CdCl_2$ to form an organocadmium compound,$(C_6H_9)_2Cd$.
$4$. Reaction with $CH_3COCl$: The organocadmium compound reacts with acetyl chloride $(CH_3COCl)$ to form a ketone,$1$-(cyclohex$-2-$en$-1-$yl)ethanone.
Thus,the final product is $1$-(cyclohex$-2-$en$-1-$yl)ethanone.

(D) The organic compound $X$ gives a red precipitate with Fehling's solution,which indicates that $X$ is an aliphatic aldehyde (specifically,acetaldehyde,$CH_3CHO$).
Reaction $D$ is the hydration of acetylene $(C_2H_2)$ in the presence of $40 \% H_2SO_4$ and $1 \% HgSO_4$ at $60^{\circ}C$,which yields acetaldehyde $(CH_3CHO)$ as the major product.
$C_2H_2 + H_2O \xrightarrow[1 \% HgSO_4, 60^{\circ}C]{40 \% H_2SO_4} CH_3CHO$ (Acetaldehyde)
Acetaldehyde reacts with Fehling's solution upon heating to form a red precipitate of cuprous oxide $(Cu_2O)$.

(C) The reaction of benzene with $CO$ and $HCl$ in the presence of anhydrous $AlCl_3$ is the Gattermann-Koch reaction,which produces benzaldehyde $(C_6H_5CHO)$. Thus,$X$ is benzaldehyde.
Let us analyze the given options:
$A$: Benzonitrile $(C_6H_5CN)$ on reduction with $LiAlH_4$ gives benzylamine $(C_6H_5CH_2NH_2)$.
$B$: Toluene $(C_6H_5CH_3)$ on oxidation with $KMnO_4 | OH^-$ gives benzoic acid $(C_6H_5COOH)$.
$C$: Benzoyl chloride $(C_6H_5COCl)$ on reduction with $H_2-Pd$ and $BaSO_4$ (Rosenmund reduction) gives benzaldehyde $(C_6H_5CHO)$.
$D$: Benzyl alcohol $(C_6H_5CH_2OH)$ on oxidation with $CrO_3-H_2SO_4$ (Jones reagent) gives benzoic acid $(C_6H_5COOH)$.
Therefore,benzaldehyde is obtained in reaction $C$.

(D) The reaction sequence is as follows:
$1$. The first step is the Etard reaction,where toluene is oxidized to benzaldehyde using chromyl chloride $(CrO_2Cl_2)$ in the presence of carbon disulfide $(CS_2)$,followed by hydrolysis with $H_3O^+$.
$2$. The second step is the nitration of benzaldehyde using a nitrating mixture $(HNO_3/H_2SO_4)$.
$3$. The $-CHO$ group is a meta-directing group,so the nitro group $(-NO_2)$ will be introduced at the meta-position relative to the aldehyde group.
$4$. Thus,the final product $X$ is $3$-nitrobenzaldehyde.

(A) In the Etard reaction,chromyl chloride $(CrO_2Cl_2)$ oxidises the methyl group of toluene to a chromium complex in the presence of a solvent like $CS_2$ or $CCl_4$.
This chromium complex,upon subsequent hydrolysis with water $(H_3O^+)$,yields benzaldehyde as the final product.

(A) The reaction of a Grignard reagent $(RMgBr)$ with triethyl orthoformate $(HC(OEt)_3)$ proceeds via a nucleophilic substitution mechanism.
$1$. The nucleophilic $R^-$ group from the Grignard reagent attacks the electrophilic carbon of the triethyl orthoformate, displacing one ethoxide group $(-OEt)$ to form an acetal intermediate, $RCH(OEt)_2$.
$2$. Upon acidic hydrolysis $(H_3O^{+})$, the acetal intermediate undergoes hydrolysis to yield an aldehyde $(RCHO)$ as the final product $P$.

(A) The final product is $Ph_2C=CH_2$ ($1$, $1$-diphenylethene).
The last step is a Wittig reaction: $N + Ph_3P=CH_2 \rightarrow Ph_2C=CH_2$. This implies $N$ must be benzophenone $(Ph_2C=O)$.
The step $M \xrightarrow{CrO_3 / H^{\oplus}} N$ is an oxidation of a secondary alcohol to a ketone. Thus, $M$ is diphenylmethanol $(Ph_2CHOH)$.
The first step is the Grignard reaction: $L \xrightarrow[(ii) H_3O^{\oplus}]{(i) PhMgBr} Ph_2CHOH$.
Benzaldehyde $(PhCHO)$ reacts with $PhMgBr$ followed by hydrolysis to give diphenylmethanol $(Ph_2CHOH)$.
Therefore, $L$ is benzaldehyde.

(B) The reaction of an alkyl nitrile with a Grignard reagent $(PhMgBr)$ followed by acidic hydrolysis $(H_3O^+)$ proceeds as follows:
$1$. The nucleophilic phenyl group $(Ph^-)$ from $PhMgBr$ attacks the electrophilic carbon of the nitrile group $(-CN)$ to form an imine salt intermediate.
$2$. Upon acidic hydrolysis $(H_3O^+)$, the imine salt is converted into an imine, which is unstable and further hydrolyzes to form a ketone.
$3$. The reaction sequence is: $R-CN + PhMgBr$ $\rightarrow R-C(Ph)=N-MgBr$ $\xrightarrow{H_3O^+} R-C(Ph)=NH$ $\xrightarrow{H_3O^+} R-C(=O)Ph$.
$4$. In this case, $R$ is an isopropyl group, so the final product is $3$-methyl-$1$-phenylbutan-$1$-one.

(D) The correct matches are as follows:
$A$. $H_2, Pd-BaSO_4$ is used in the Rosenmund reduction to convert acid chlorides to aldehydes $(A-II)$.
$B$. $SnCl_2, HCl$ is used in the Stephen reaction to reduce nitriles to imines, which are then hydrolyzed to aldehydes $(B-IV)$.
$C$. $CrO_2Cl_2, CS_2$ is used in the Etard reaction to oxidize toluene to benzaldehyde $(C-I)$.
$D$. $CO, HCl, \text{Anhyd. } AlCl_3$ is used in the Gatterman-Koch reaction to form benzaldehyde from benzene $(D-III)$.
Therefore, the correct sequence is $A-II, B-IV, C-I, D-III$.

(D) The given reaction is the Etard reaction, where toluene is oxidized to benzaldehyde $(P)$ using chromyl chloride $(CrO_2Cl_2)$ in $CS_2$ followed by hydrolysis.
Thus, compound $P$ is benzaldehyde $(C_6H_5CHO)$.
Let's evaluate the options:
$(A)$ Benzaldehyde does not react with $NaHCO_3$ to release $CO_2$ gas (brisk effervescence), as it is not a carboxylic acid.
$(B)$ Treating benzene with $CH_3COCl$ and anhydrous $AlCl_3$ (Friedel-Crafts acylation) yields acetophenone, not benzaldehyde.
$(C)$ Benzaldehyde does not give a white precipitate with bromine water; this is a characteristic test for phenols or anilines.
$(D)$ The Rosenmund reduction of benzoyl chloride $(C_6H_5COCl)$ using $H_2$ in the presence of $Pd$ supported on $BaSO_4$ (poisoned with sulfur or quinoline) yields benzaldehyde $(P)$.
Therefore, the correct option is $(D)$.

(A) 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.

(B) Step $1$: The Grignard reagent $(R'MgX)$ acts as a nucleophile and attacks the electrophilic carbon atom of the alkyl cyanide $(R-C \equiv N)$.
Step $2$: This forms an intermediate imine salt $(R-C(R')=NMgX)$.
Step $3$: Acid hydrolysis of the imine salt yields an imine $(R-C(R')=NH)$, which is unstable and undergoes further hydrolysis to form a ketone $(R-C(=O)R')$.
Step $4$: Therefore, the final product is a ketone.

(A) Step $1$: Reaction of $\text{Benzonitrile}$ $(C_6H_5CN)$ with $\text{Phenylmagnesium bromide}$ $(C_6H_5MgBr)$ leads to the formation of an imine intermediate $(X)$, which is $\text{N-magnesiobenzophenone imine}$ $(C_6H_5-C(=NMgBr)-C_6H_5)$.
Step $2$: Acidic hydrolysis $(H_3O^+)$ of the intermediate $X$ converts the imine group into a carbonyl group, yielding $\text{Benzophenone}$ $(C_6H_5-CO-C_6H_5)$ as the final product '$Y$', along with $MgBr(OH)$ and $NH_3$.

(D) $1$. The reaction of benzonitrile $(C_6H_5CN)$ with a Grignard reagent $(RMgX)$ followed by acid hydrolysis yields a ketone.
$2$. The reaction proceeds as follows: $C_6H_5CN + C_6H_5MgBr \rightarrow C_6H_5C(NMgBr)C_6H_5$.
$3$. Upon hydrolysis: $C_6H_5C(NMgBr)C_6H_5 + 2H_2O \rightarrow C_6H_5COC_6H_5 + NH_3 + Mg(OH)Br$.
$4$. Since the product is benzophenone $(C_6H_5COC_6H_5)$, the Grignard reagent must be phenylmagnesium bromide $(C_6H_5MgBr)$.

(C) $1$. The reaction is the Stephen reduction of a nitrile. Isopropyl cyanide is $(CH_3)_2CH-CN$.
$2$. In the first step, the nitrile is reduced by $SnCl_2/HCl$ to form an imine intermediate $(A)$, which is $(CH_3)_2CH-CH=NH$.
$3$. In the second step, the acid hydrolysis of the imine $(A)$ yields the corresponding aldehyde $(B)$.
$4$. The reaction is: $(CH_3)_2CH-CN \xrightarrow{SnCl_2, HCl} (CH_3)_2CH-CH=NH \xrightarrow{H_3O^+} (CH_3)_2CH-CHO + NH_4Cl$.
$5$. The product '$B$' is $(CH_3)_2CH-CHO$, which is $2-\text{Methylpropanal}$.

(B) The given reaction is the Rosenmund reduction, where an acid chloride is reduced to an aldehyde using hydrogen gas in the presence of a poisoned palladium catalyst.
Step $1$: The reagent used is $H_2$ gas in the presence of $Pd$ supported on $BaSO_4$.
Step $2$: $BaSO_4$ acts as a poison to prevent the further reduction of the aldehyde to an alcohol.
Step $3$: The reaction is $C_6H_5COCl + H_2 \xrightarrow{Pd/BaSO_4} C_6H_5CHO + HCl$.

(C) Step $1$: $\text{DIBAL-H}$ (Diisobutylaluminium hydride) is a selective reducing agent.
Step $2$: It reduces nitriles $(-CN)$ to imines, which upon acidic hydrolysis $(H_3O^+)$ yield aldehydes.
Step $3$: The reaction is: $CH_3-CH=CH-CH_2-CN \xrightarrow[2. H_3O^+]{1. DIBAL-H} CH_3-CH=CH-CH_2-CHO$.
Step $4$: The product formed is $\text{Pent-3-enal}$.

(C) Rosenmund reduction involves the catalytic hydrogenation of acid chlorides to aldehydes using $Pd$ supported on $BaSO_4$ (poisoned with sulfur or quinoline).
Reaction: $C_6H_5COCl + H_2 \xrightarrow{Pd/BaSO_4} C_6H_5CHO + HCl$.
Benzoyl chloride $(C_6H_5COCl)$ is reduced to Benzaldehyde $(C_6H_5CHO)$.

(A) Step $1$: Rosenmund reduction is the catalytic hydrogenation of acid chlorides to aldehydes using $H_2$ gas in the presence of $Pd$ catalyst poisoned with $BaSO_4$ (Lindlar's catalyst).
Step $2$: The reaction is $R - CO - Cl + H_2 \xrightarrow{Pd-BaSO_4} R - CHO + HCl$.
Step $3$: Option $(A)$ represents this reaction. Option $(B)$ is Stephen reduction, $(C)$ is Clemmensen reduction, and $(D)$ is Wolff-Kishner reduction.