Properties Questions in English

(D) symmetrical ketone is one in which the two alkyl or aryl groups attached to the carbonyl carbon $(C=O)$ are identical.
$A$. Acetophenone: $C_6H_5-CO-CH_3$ (Unsymmetrical)
$B$. Ethyl methyl ketone: $CH_3CH_2-CO-CH_3$ (Unsymmetrical)
$C$. Ethylphenyl ketone: $C_6H_5-CO-CH_2CH_3$ (Unsymmetrical)
$D$. Benzophenone: $C_6H_5-CO-C_6H_5$ (Symmetrical,as both groups are phenyl groups)
Therefore,Benzophenone is a symmetrical ketone.

(D) Chain isomerism in ketones requires a minimum of $5$ carbon atoms in the main chain to allow for branching.
For example,pentan-$2$-one $(CH_3COCH_2CH_2CH_3)$ can form a chain isomer,$3$-methylbutan-$2$-one $(CH_3COCH(CH_3)_2)$,by branching the carbon chain.
Thus,the minimum number of carbon atoms required is $5$.

(D) The Wolff-Kishner reduction is a reaction used to convert carbonyl groups (aldehydes or ketones) into alkanes.
It involves the formation of a hydrazone intermediate by reacting the carbonyl compound with hydrazine $(H_2N-NH_2)$,followed by base-catalyzed decomposition (using $KOH$ in ethylene glycol) at high temperatures.
Option $D$ correctly represents this process: $RCOR' \xrightarrow[\text{ii) } KOH, HO(CH_2)_2OH]{\text{i) } H_2N-NH_2} RCH_2R'$.
Option $A$ is Rosenmund reduction,option $B$ is Stephen reduction,and option $C$ is Clemmensen reduction.

(C) The Clemmensen reduction is a chemical reaction that reduces aldehydes or ketones to their corresponding alkanes using zinc amalgam $(Zn-Hg)$ and concentrated hydrochloric acid $(HCl)$.
The general reaction is:
$R_1-CO-R_2 \xrightarrow{Zn(Hg), HCl} R_1-CH_2-R_2$
Option $C$ represents the reduction of an aldehyde to an alkane using $Zn-Hg$ and concentrated $HCl$,which is the characteristic condition for Clemmensen reduction.

(C) In the Wolff-Kishner reduction,the first step involves the reaction of a carbonyl compound (aldehyde or ketone) with hydrazine $(H_2N-NH_2)$.
This reaction results in the formation of a hydrazone intermediate with the loss of a water molecule $(H_2O)$.
The general reaction is: $R_2C=O + H_2N-NH_2 \rightarrow R_2C=N-NH_2 + H_2O$.
Therefore,the compound obtained in the first step is a hydrazone.

(B) The reaction of an aldehyde or ketone with $Zn-Hg$ amalgam and concentrated $HCl$ is known as the Clemmensen reduction.
The general reaction is:
$R-C(=O)-R' \xrightarrow[\text{conc. } HCl]{Zn/Hg} R-CH_2-R' + H_2O$
This process reduces the carbonyl group $(C=O)$ to a methylene group $(CH_2)$.

(D) The Wolff-Kishner reduction is a chemical reaction used to convert carbonyl groups (aldehydes or ketones) into methylene groups $(-CH_2-)$ using hydrazine $(H_2N-NH_2)$ and a strong base (like $KOH$) in a high-boiling solvent such as ethylene glycol $(HO(CH_2)_2OH)$.
Option $D$ represents this reaction: $R-CO-R \xrightarrow[(ii) KOH, HO(CH_2)_2OH]{(i) H_2N-NH_2, \Delta} R-CH_2-R + N_2$.
Option $A$ is the Rosenmund reduction.
Option $B$ is the Stephen reduction.
Option $C$ is the Clemmensen reduction.

(A) The reaction described is the Wolff-Kishner reduction.
In this reaction,aldehydes and ketones are treated with hydrazine $(NH_2NH_2)$ to form a hydrazone intermediate.
This hydrazone is then heated with a strong base like potassium hydroxide $(KOH)$ in a high-boiling solvent such as ethylene glycol $(HO-CH_2-CH_2-OH)$.
This process results in the reduction of the carbonyl group $(>C=O)$ to a methylene group $(-CH_2-)$ with the evolution of nitrogen gas $(N_2)$.

(C) The Clemmensen reduction is a chemical reaction that reduces aldehydes or ketones to their corresponding alkanes using zinc amalgam $(Zn-Hg)$ and concentrated hydrochloric acid $(HCl)$.
Option $C$ represents this reaction: $RCHO \xrightarrow{Zn-Hg, \text{Conc. } HCl, \Delta} RCH_3 + H_2O$.

(A) The given reaction involves the reduction of an aldehyde $(CH_3CHO)$ to an alkane $(CH_3CH_3)$ using zinc amalgam $(Zn-Hg)$ and concentrated hydrochloric acid $(HCl)$.
This specific reduction of carbonyl groups (aldehydes and ketones) to methylene groups $(-CH_2-)$ using $Zn-Hg$ and $conc. HCl$ is known as the Clemmensen reduction.

(A) The reaction involves the catalytic hydrogenation of an unsaturated aldehyde $(CH_3-CH=CH-CH_2-CHO)$ using $Ni$ as a catalyst.
$Ni$ is a strong reducing agent that reduces both the carbon-carbon double bond $(C=C)$ and the carbonyl group $(C=O)$ in the presence of excess hydrogen.
$1$. The $C=C$ double bond is reduced to a single bond: $CH_3-CH=CH-CH_2-CHO \rightarrow CH_3-CH_2-CH_2-CH_2-CHO$.
$2$. The aldehyde group $(-CHO)$ is reduced to a primary alcohol $(-CH_2OH)$: $CH_3-CH_2-CH_2-CH_2-CHO \rightarrow CH_3-CH_2-CH_2-CH_2-CH_2-OH$.
Therefore,the final product is pentan$-1-$ol.

(C) The reaction that converts the carbonyl group $(>C=O)$ of aldehydes and ketones into a methylene group $(>CH_2)$ using zinc amalgam $(Zn-Hg)$ and concentrated hydrochloric acid $(HCl)$ is known as the Clemmensen reduction.
The general reaction is:
$R-CO-R' \xrightarrow{Zn-Hg/HCl} R-CH_2-R'$
Therefore,the correct option is $C$.

(A) The Wolff-Kishner reduction is a chemical reaction used to convert carbonyl groups (aldehydes or ketones) into methylene groups $(-CH_2-)$.
The reagents used for this reduction are hydrazine $(NH_2NH_2)$ and a strong base like potassium hydroxide $(KOH)$ in a high-boiling solvent such as ethylene glycol.

(C) Step $1$: Reaction of $2,2-$dichloropropane with aqueous $KOH$ gives a gem-diol intermediate,which loses a water molecule to form acetone $(CH_3COCH_3)$ as product $A$.
Step $2$: Clemmensen reduction of acetone $(CH_3COCH_3)$ using $Zn(Hg)/HCl$ reduces the carbonyl group to a methylene group $(-CH_2-)$,yielding propane $(CH_3CH_2CH_3)$ as product $B$.
Therefore,the correct option is $C$.

(A) Clemmensen's reduction involves the reduction of carbonyl compounds (aldehydes and ketones) to alkanes.
The reagent used is zinc amalgam $(Zn-Hg)$ in the presence of concentrated hydrochloric acid $(HCl)$.
This reaction converts the $C=O$ group into a $CH_2$ group.

(A) The reaction between formaldehyde $(HCHO)$ and ethyl magnesium bromide $(CH_3CH_2MgBr)$ is a Grignard reaction.
Step $1$: Nucleophilic attack of the ethyl group on the carbonyl carbon of formaldehyde forms an intermediate alkoxide: $HCHO + CH_3CH_2MgBr \rightarrow CH_3CH_2CH_2OMgBr$.
Step $2$: Acidic hydrolysis of the intermediate yields the final alcohol: $CH_3CH_2CH_2OMgBr + H_2O \rightarrow CH_3CH_2CH_2OH + Mg(OH)Br$.
The product formed is $CH_3CH_2CH_2OH$,which is Propan-$1$-ol.
Therefore,the correct option is $A$.

(D) Reduction of $Propanoic \ acid$ $(CH_3CH_2COOH)$ gives $Propan-1-ol$ (primary alcohol).
Reduction of $Propanal$ $(CH_3CH_2CHO)$ gives $Propan-1-ol$ (primary alcohol).
Reduction of $Methyl \ propanoate$ $(CH_3CH_2COOCH_3)$ gives $Propan-1-ol$ and $Methanol$ (both primary alcohols).
Reduction of $Propan-2-one$ $(CH_3COCH_3)$ gives $Propan-2-ol$,which is a secondary alcohol.
Therefore,$Propan-2-one$ does not give a primary alcohol on reduction.

(B) The reduction of $Butanone$ $(CH_3COCH_2CH_3)$ using a reducing agent like $NaBH_4$ or $LiAlH_4$ yields $Butan-2-ol$ $(CH_3CH(OH)CH_2CH_3)$.
In $Butan-2-ol$,the carbon atom attached to the $-OH$ group is bonded to four different groups: $-H$,$-OH$,$-CH_3$,and $-CH_2CH_3$.
Since this carbon is a chiral center,$Butan-2-ol$ exists as a pair of enantiomers and is optically active.
The other options yield primary alcohols $(Butan-1-ol)$ which are achiral.

(D) The Cannizzaro reaction is given by aldehydes that do not possess an $\alpha$-hydrogen atom.
$HCHO$ (formaldehyde) has no $\alpha$-hydrogen.
Benzaldehyde $(C_6H_5CHO)$ has no $\alpha$-hydrogen.
$1$-methylcyclohexanecarbaldehyde has no $\alpha$-hydrogen at the carbonyl carbon position.
$CH_3CHO$ (acetaldehyde) possesses three $\alpha$-hydrogen atoms,so it undergoes aldol condensation instead of the Cannizzaro reaction.
Therefore,the correct option is $D$.

(B) The given reaction is a $Cross-Aldol$ condensation reaction between $benzaldehyde$ $(C_6H_5CHO)$ and $acetaldehyde$ $(CH_3CHO)$ in the presence of a dilute base $(OH^-)$ and heat $(\Delta)$.
$1.$ $Acetaldehyde$ contains $\alpha$-hydrogens,so it forms an enolate ion in the presence of a base.
$2.$ This enolate ion attacks the carbonyl carbon of $benzaldehyde$ (which has no $\alpha$-hydrogens) to form a $\beta$-hydroxy aldehyde intermediate $(C_6H_5CH(OH)CH_2CHO)$.
$3.$ Upon heating $(\Delta)$,this intermediate undergoes dehydration (loss of $H_2O$) to form the stable $\alpha,\beta$-unsaturated aldehyde,which is $cinnamaldehyde$ $(C_6H_5CH=CHCHO)$.
Therefore,the main product is $C_6H_5CH=CHCHO$.

(D) The "Iodoform" test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$A$. $CH_3CHO$ (Acetaldehyde) contains the $CH_3CO-$ group,so it gives a positive "Iodoform" test.
$B$. $CH_3CH(OH)CH_2CH_3$ (Butan-$2$-ol) contains the $CH_3CH(OH)-$ group,which is oxidized to a $CH_3CO-$ group by "Sodium hypoiodite",hence it gives a positive "Iodoform" test.
$C$. $CH_3CH_2COCH_3$ (Butan-$2$-one) contains the $CH_3CO-$ group,so it gives a positive "Iodoform" test.
$D$. $CH_3CH_2COCH_2CH_3$ (Pentan-$3$-one) does not contain the $CH_3CO-$ group or the $CH_3CH(OH)-$ group. Therefore,it will not give the "Iodoform" test.
Thus,the correct option is $D$.

(D) Aldol condensation occurs in aldehydes or ketones that possess at least one $\alpha$-hydrogen atom.
$A$. Trimethyl acetaldehyde $(CH_3)_3CCHO$ has no $\alpha$-hydrogen.
$B$. Formaldehyde $(HCHO)$ has no $\alpha$-hydrogen.
$C$. Trichloro acetaldehyde $(CCl_3CHO)$ has no $\alpha$-hydrogen.
$D$. Acetaldehyde $(CH_3CHO)$ has three $\alpha$-hydrogen atoms attached to the $\alpha$-carbon.
Therefore,acetaldehyde undergoes aldol condensation.

(D) The reactivity of carbonyl compounds towards nucleophilic addition depends on steric hindrance and electronic effects.
$1$. Steric hindrance: As the number and size of alkyl or aryl groups attached to the carbonyl carbon increase,the attack of the nucleophile becomes more difficult.
$2$. Electronic effects: Electron-donating groups decrease the electrophilicity of the carbonyl carbon.
Formaldehyde $(HCHO)$ is the most reactive due to the absence of bulky groups.
Benzaldehyde $(C_6H_5CHO)$ is more reactive than ketones.
Among ketones,Acetophenone $(C_6H_5COCH_3)$ has one phenyl and one methyl group,while Benzophenone $(C_6H_5COC_6H_5)$ has two bulky phenyl groups.
Due to maximum steric hindrance and the resonance stabilization of the carbonyl group by two phenyl rings,Benzophenone is the least reactive.

(A) The cross aldol condensation of ethanal $(CH_3CHO)$ and propanal $(CH_3CH_2CHO)$ involves four possible products:
$1$. Self-aldol of ethanal: $CH_3CH=CHCHO$ (But$-2-$enal)
$2$. Self-aldol of propanal: $CH_3CH_2CH=C(CH_3)CHO$ ($2-$Methylpent$-2-$enal)
$3$. Cross-aldol (ethanal as nucleophile): $CH_3CH_2CH=CHCHO$ (Pent$-2-$enal)
$4$. Cross-aldol (propanal as nucleophile): $CH_3CH=C(CH_3)CHO$ ($2-$Methylbut$-2-$enal)
Comparing these with the options,$3-$Methylbut$-2-$enal is not formed.

(D) The cross-aldol condensation between benzene carbaldehyde (benzaldehyde,$C_6H_5CHO$) and $1$-phenyl ethane$-1$-one (acetophenone,$C_6H_5COCH_3$) in the presence of a base leads to the formation of $1,3$-diphenylprop$-2$-en$-1$-one,commonly known as chalcone.
The reaction is as follows:
$C_6H_5CHO + CH_3COC_6H_5 \xrightarrow{OH^-} C_6H_5CH=CHCOC_6H_5 + H_2O$
Here,the enolate ion formed from acetophenone attacks the carbonyl carbon of benzaldehyde,followed by dehydration to form the $\alpha,\beta$-unsaturated ketone,which is represented by option $D$.

(C) The reaction described is the $Cannizzaro$ reaction.
Aldehydes that do not contain an $\alpha$-hydrogen atom undergo self-oxidation and reduction (disproportionation) when treated with concentrated alkali.
$Benzaldehyde$ $(C_6H_5CHO)$ has no $\alpha$-hydrogen.
$Trimethylacetaldehyde$ $((CH_3)_3CCHO)$ has no $\alpha$-hydrogen.
$Formaldehyde$ $(HCHO)$ has no $\alpha$-hydrogen.
$Dimethyl$ $acetaldehyde$ (isobutyraldehyde,$(CH_3)_2CHCHO$) contains one $\alpha$-hydrogen atom attached to the $\alpha$-carbon.
Therefore,$Dimethyl$ $acetaldehyde$ does not undergo the $Cannizzaro$ reaction.
The correct option is $C$.

(C) Fehling's solution is prepared by mixing two solutions,Fehling $A$ and Fehling $B$,in equal amounts just before use.
Fehling $A$ is an aqueous copper$(II)$ sulfate solution.
Fehling $B$ is an alkaline solution of sodium potassium tartrate,also known as Rochelle salt.
Therefore,the correct option is $C$.

(A) The $Wolff-Kishner$ reduction is a chemical reaction used to convert carbonyl groups (aldehydes or ketones) into methylene groups $(-CH_2-)$ using hydrazine $(NH_2NH_2)$ and a strong base like $KOH$ in a high-boiling solvent.
$Acetone$ $(CH_3COCH_3)$ has $3$ carbon atoms. Upon $Wolff-Kishner$ reduction,it is converted into $Propane$ $(CH_3CH_2CH_3)$,which also contains $3$ carbon atoms.
Therefore,the correct option is $A$.

(B) The reaction of acetaldehyde with sodium bisulphite $(NaHSO_3)$ is a classic example of a nucleophilic addition reaction.
In this reaction,the bisulphite ion $(HSO_3^-)$ acts as a nucleophile and attacks the electrophilic carbonyl carbon of the acetaldehyde to form a crystalline bisulphite addition compound.
This reaction is reversible and is commonly used for the purification and separation of aldehydes and ketones.

(B) The Cannizzaro reaction is given by aldehydes that do not contain an $\alpha$-hydrogen atom.
In $CH_3CHO$,there are three $\alpha$-hydrogens.
In $CH_2ClCHO$,there are two $\alpha$-hydrogens.
In $CHCl_2CHO$,there is one $\alpha$-hydrogen.
In $CCl_3CHO$ (trichloroacetaldehyde),there are no $\alpha$-hydrogen atoms attached to the carbonyl carbon.
Therefore,$CCl_3CHO$ undergoes the Cannizzaro reaction.

(B) The reaction between benzaldehyde $(C_6H_5CHO)$ and acetophenone $(C_6H_5COCH_3)$ in the presence of a dilute base $(OH^-)$ is a Claisen-Schmidt condensation reaction.
In this reaction,the enolate ion formed from acetophenone attacks the carbonyl carbon of benzaldehyde.
The initial product is a $\beta$-hydroxy ketone,which subsequently undergoes dehydration to form an $\alpha,\beta$-unsaturated ketone.
The reaction is:
$C_6H_5CHO + CH_3COC_6H_5$ $\xrightarrow{OH^-} C_6H_5CH(OH)CH_2COC_6H_5$ $\xrightarrow{-H_2O} C_6H_5CH=CHCOC_6H_5$
The final product is benzalacetophenone (also known as chalcone),which corresponds to option $B$.

(B) Fehling reagent is a mild oxidizing agent that is used to distinguish between aliphatic aldehydes and ketones.
Aliphatic aldehydes,such as $CH_3CHO$ (Acetaldehyde),are easily oxidized by Fehling reagent to their corresponding carboxylate ions,resulting in a reddish-brown precipitate of $Cu_2O$.
Ketones (like Acetophenone and Acetone) and aromatic aldehydes (like Benzaldehyde) do not reduce Fehling reagent.
Therefore,the correct answer is $B$ (Acetaldehyde).

(B) The Iodoform reaction is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$1$. Propanal $(CH_3CH_2CHO)$ does not contain the $CH_3CO-$ group.
$2$. Acetone $(CH_3COCH_3)$ contains the $CH_3CO-$ group and thus gives a positive Iodoform test.
$3$. Pent-$3$-one $(CH_3CH_2COCH_2CH_3)$ does not contain a methyl group attached to the carbonyl carbon.
$4$. Benzophenone $(C_6H_5COC_6H_5)$ does not contain a methyl group attached to the carbonyl carbon.
Therefore,the correct answer is $B$.

(A) The reaction of a ketone with hydroxylamine $(NH_2OH)$ in an acidic medium is a nucleophilic addition-elimination reaction (condensation reaction).
Cyclopentanone reacts with hydroxylamine to form cyclopentanone oxime and water.
The reaction is represented as:
$C_5H_8O + NH_2OH \xrightarrow{H^+} C_5H_8NOH + H_2O$.
Therefore,the correct product is cyclopentanone oxime.

(D) The Cannizzaro reaction is a disproportionation reaction given by aldehydes that do not possess an $\alpha$-hydrogen atom.
$HCHO$ (formaldehyde) has no $\alpha$-hydrogen.
$C_6H_5CHO$ (benzaldehyde) has no $\alpha$-hydrogen.
$1\text{-methylcyclohexanecarbaldehyde}$ has no $\alpha$-hydrogen at the carbonyl carbon position.
$CH_3CHO$ (acetaldehyde) possesses three $\alpha$-hydrogen atoms,so it undergoes aldol condensation instead of the Cannizzaro reaction.

(C) The reaction of formaldehyde $(HCHO)$ with a Grignard reagent $(RMgX)$ followed by acid hydrolysis yields a primary alcohol.
In this case,the product is ethanol $(C_{2}H_{5}OH)$,which contains two carbon atoms.
Since formaldehyde provides one carbon atom,the Grignard reagent must provide the other methyl group $(CH_{3})$.
Therefore,the reagent $X$ is methylmagnesium iodide $(CH_{3}MgI)$.
The reaction is: $HCHO + CH_{3}MgI$ $\rightarrow CH_{3}CH_{2}OMgI$ $\xrightarrow{H_{3}O^{+}} CH_{3}CH_{2}OH + Mg(OH)I$.

(B) In the reactant $CH_3-CH=CH-CH_2OH$,the carbon atom at position $C-1$ is bonded to two hydrogen atoms,one oxygen atom,and one carbon atom. It is $sp^3$ hybridised.
In the product $CH_3-CH=CH-CHO$,the carbon atom at position $C-1$ is part of the aldehyde group $(-CHO)$,where it forms a double bond with oxygen. It is $sp^2$ hybridised.
Therefore,the hybridisation of $C-1$ changes from $sp^3$ to $sp^2$.

(A) The reactivity of carbonyl compounds towards nucleophilic addition reaction is governed by two factors: electronic effects (positive charge on the carbonyl carbon) and steric hindrance.
Aldehydes are generally more reactive than ketones because they have less steric hindrance and a higher partial positive charge on the carbonyl carbon.
Among the given compounds:
$1$. $CH_3CHO$ (Acetaldehyde): One alkyl group attached to the carbonyl carbon.
$2$. $CH_3COCH_3$ (Acetone): Two methyl groups attached.
$3$. $CH_3COC_2H_5$ (Butanone): One methyl and one ethyl group attached.
As the size of the alkyl groups increases,steric hindrance increases,decreasing reactivity. Thus,the order is $CH_3CHO > CH_3COCH_3 > CH_3COC_2H_5$.

(C) The given reaction involves the oxidation of a methyl ketone group $(-COCH_3)$ to a carboxylic acid group $(-COOH)$ while keeping the carbon-carbon double bond intact.
This transformation is characteristic of the haloform reaction.
When a methyl ketone is treated with $I_2$ and $NaOH$,it undergoes the iodoform reaction to form a carboxylate salt,which upon subsequent acidification yields the corresponding carboxylic acid.
The reaction is:
$CH_3-CH=CH-CH_2-COCH_3 \xrightarrow[(ii) \text{ Acidification}]{(i) I_2/NaOH} CH_3-CH=CH-CH_2-COOH + CHI_3$

(A) The given reaction proceeds as follows:
$1$. The cumene hydroperoxide process yields phenol $(A)$ and acetone $(B)$.
$2$. Phenol $(A)$ gives a white precipitate with bromine water due to the formation of $2,4,6-$tribromophenol.
$3$. Acetone $(B)$ undergoes aldol condensation in the presence of barium hydroxide $(Ba(OH)_2)$ to form $4-$hydroxy$-4-$methylpentan$-2-$one $(C)$.
$4$. Upon strong heating,$C$ undergoes dehydration to form $4-$methylpent$-3-$en$-2-$one $(D)$,which is mesityl oxide.
Therefore,the product $D$ is $4-$methylpent$-3-$en$-2-$one.

(C) The reaction between benzaldehyde $(C_6H_5CHO)$ and acetaldehyde $(CH_3CHO)$ in the presence of dilute $NaOH$ followed by heating $(\Delta)$ is a cross-aldol condensation reaction.
$1$. The enolate ion formed from acetaldehyde attacks the carbonyl carbon of benzaldehyde to form a $\beta$-hydroxy aldehyde.
$2$. Upon heating,this $\beta$-hydroxy aldehyde undergoes dehydration to form an $\alpha,\beta$-unsaturated aldehyde,which is cinnamaldehyde $(C_6H_5CH=CHCHO)$.

(C) Any carbonyl compound having $\alpha$-hydrogen will undergo aldol condensation.
In $CCl_3CHO$ (trichloro acetaldehyde),the $\alpha$-carbon is bonded to three chlorine atoms and has no $\alpha$-hydrogen atom attached to it.
Therefore,it does not show the aldol condensation reaction.

(A) The starting material is $HOCH_{2}-CH=CH-CHO$.
Step $A$: The product is $HOCH_{2}-CH=CH-CH_{2}OH$. This indicates the reduction of the aldehyde group to a primary alcohol while leaving the $C=C$ double bond intact. $NaBH_{4}$ is a selective reducing agent that reduces aldehydes/ketones but not $C=C$ double bonds.
Step $B$: The product is $HOCH_{2}-CH=CH-CHO$. This indicates that the aldehyde group remains unchanged. $PCC$ (Pyridinium chlorochromate) is an oxidizing agent used to oxidize alcohols to aldehydes,but it does not affect the aldehyde group itself.
Step $C$: The product is $HOCH_{2}-CH_{2}-CH_{2}-CH_{2}OH$. This indicates the reduction of both the aldehyde group and the $C=C$ double bond. $H_{2} / Pd$ is a catalytic hydrogenation reagent that reduces both aldehydes and $C=C$ double bonds.
Therefore,the reagents are $A = NaBH_{4}$,$B = PCC$,$C = H_{2} / Pd$.

(B) The transformation involves the reduction of a ketone group $(-C=O)$ to a methylene group $(-CH_2-)$ in the presence of an alcohol group $(-OH)$.
$Zn-Hg / HCl$ (Clemmensen reduction) is acidic and would cause the dehydration of the alcohol group.
$H_2N-NH_2, KOH / \text{ethylene glycol}$ (Wolff-Kishner reduction) is a basic condition,which is compatible with the presence of an alcohol group.
Therefore,the appropriate reagent is $H_2N-NH_2, KOH / \text{ethylene glycol}$.

(B) This reaction is an example of the Cannizzaro reaction,which occurs in aldehydes lacking $\alpha$-hydrogen atoms.
$2 C_6H_5CHO + NaOH (conc.) \rightarrow C_6H_5COONa (A) + C_6H_5CH_2OH (B)$
Product $A$ is Sodium benzoate,which is commonly used as a food preservative.
Product $B$ is Phenyl methanol (also known as Benzyl alcohol),which is an aromatic hydroxy compound where the $OH$ group is attached to an $sp^3$ hybridized carbon atom $(-CH_2-)$ adjacent to the benzene ring.
Therefore,the products $A$ and $B$ are Sodium benzoate and Phenyl methanol.

(B) The formation of cyanohydrin from a ketone by reaction with $HCN$ is an example of a nucleophilic addition reaction.
In this reaction,the nucleophile $CN^-$ attacks the electrophilic carbonyl carbon of the ketone,leading to the formation of an intermediate alkoxide ion,which is then protonated to form the cyanohydrin.

(A) The Cannizzaro reaction mechanism involves the following steps:
Step $I$: Nucleophilic attack of $OH^{-}$ on the carbonyl carbon of the first aldehyde molecule to form a dihydroxyalkoxide intermediate.
Step $II$: Transfer of a hydride ion $(H^{-})$ from the intermediate to the carbonyl carbon of a second aldehyde molecule,resulting in the formation of a carboxylate ion and an alkoxide ion.
Step $III$: Rapid proton transfer $(H^{+})$ from the carboxylic acid (or solvent) to the alkoxide ion to form the alcohol.

(B) $Step \ 1$: Reduction of acetic acid $(CH_3COOH)$ with $LiAlH_4$ gives ethanol $(CH_3CH_2OH)$ as $X$.
$Step \ 2$: Dehydrogenation of ethanol $(X)$ over $Cu$ at $300^{\circ}C$ gives acetaldehyde $(CH_3CHO)$ as $Y$.
$Step \ 3$: Acetaldehyde $(Y)$ undergoes aldol condensation in the presence of dilute $NaOH$ to form $3-hydroxybutanal$ $(CH_3CH(OH)CH_2CHO)$,which is commonly known as aldol,as $Z$.
Therefore,the correct option is $B$.

(A) Aldehydes react with alcohols in the presence of dry $HCl$ gas to form hemiacetals,which further react with another molecule of alcohol to form acetals.
The reaction is as follows:
$R-CHO + R'-OH \rightleftharpoons R-CH(OH)(OR')$ (Hemiacetal)
$R-CH(OH)(OR') + R'-OH \rightleftharpoons R-CH(OR')_2 + H_2O$ (Acetal)

(C) $\text{Pentan-2-one}$ contains a $CH_3CO-$ group,which gives a positive iodoform test when reacted with $I_2$ and $NaOH$,resulting in yellow precipitates of iodoform $(CHI_3)$.
$\text{Pentan-3-one}$ does not contain a $CH_3CO-$ group and therefore does not give the iodoform test.
The reaction for $\text{pentan-2-one}$ is:
$CH_3CH_2CH_2COCH_3 + 3NaOI \rightarrow CH_3CH_2CH_2COONa + CHI_3 \downarrow + 2NaOH$