Properties Questions in English

(C) In the Beckmann rearrangement,the group anti to the $-OH$ group migrates to the nitrogen atom.
For the first reactant,the $Ph$ group is anti to the $-OH$ group,so it migrates to form $Ph-C(=O)-NH-CH_3$ as product $(A)$.
For the second reactant,the $CH_3$ group is anti to the $-OH$ group,so it migrates to form $CH_3-C(=O)-NH-Ph$ as product $(B)$.
Thus,the correct option is $(C)$.

(A) $1$. Phthalic anhydride reacts with $PCl_5$ to form phthaloyl chloride $(A)$.
$2$. Phthaloyl chloride is reduced by $LiAlH_4$ to form benzene$-1,2-$dimethanol $(B)$.
$3$. Benzene$-1,2-$dimethanol is oxidized by $PCC$ to form phthalaldehyde $(C)$.
$4$. Phthalaldehyde undergoes an intramolecular Cannizzaro reaction in the presence of $OH^-$ and $\Delta$ to form the phthalide anion,which is the salt of $2-$(hydroxymethyl)benzoic acid $(D)$.

(B) Acetophenone is a ketone $(C_6H_5COCH_3)$.
$1$. Ketones do not react with Tollen's reagent,so they do not form a silver mirror.
$2$. Acetophenone contains a methyl ketone group $(-COCH_3)$,so it gives a positive iodoform test with $I_2/NaOH$.
$3$. It reacts with $2,4$-dinitrophenylhydrazine to form a hydrazone derivative.
$4$. Oxidation of acetophenone with alkaline $KMnO_4$ followed by hydrolysis yields benzoic acid.
Therefore,the statement that it reacts with Tollen's reagent to form a silver mirror is false.

(A) The reaction is a mixed aldol condensation between propanal $(CH_3CH_2CHO)$ and benzaldehyde $(C_6H_5CHO)$.
Propanal has $\alpha$-hydrogens,while benzaldehyde does not.
Possible products are:
$1$. Self-aldol of propanal: $CH_3CH_2CH(OH)CH(CH_3)CHO$.
$2$. Cross-aldol (propanal as nucleophile,benzaldehyde as electrophile): $C_6H_5CH(OH)CH(CH_3)CHO$.
Since benzaldehyde cannot form an enolate,it cannot act as a nucleophile,so no other cross-aldol or self-aldol products are formed.
Thus,there are $2$ aldol products obtained.

(A) $1$. The reaction of phenylglyoxal $(C_6H_5-CO-CHO)$ with concentrated $OH^-$ undergoes an intramolecular Cannizzaro reaction to form the mandelate ion $(A)$,which is $C_6H_5-CH(OH)-COO^-$.
$2$. Acidification with $H_3O^+$ yields mandelic acid $(B)$,which is $C_6H_5-CH(OH)-COOH$.
$3$. Oxidation with $KMnO_4$ cleaves the side chain to form benzoic acid $(C_6H_5-COOH)$.
$4$. Decarboxylation of benzoic acid with soda lime $(NaOH/CaO/\Delta)$ yields benzene $(C_6H_6)$.

(D) The reaction of acetaldehyde $(CH_3CHO)$ with excess formaldehyde $(HCHO)$ in the presence of concentrated $NaOH$ is a classic example of a sequence involving aldol condensation followed by a crossed Cannizzaro reaction.
$1$. Acetaldehyde has three $\alpha$-hydrogens. It undergoes three successive aldol additions with three molecules of formaldehyde to form $CH_3-C(CH_2OH)_3-CHO$.
$2$. This intermediate product,$CH_3-C(CH_2OH)_3-CHO$,has no $\alpha$-hydrogens remaining.
$3$. In the presence of excess formaldehyde and concentrated $NaOH$,this intermediate undergoes a crossed Cannizzaro reaction where the aldehyde group $(-CHO)$ is reduced to a primary alcohol $(-CH_2OH)$ and the formaldehyde is oxidized to sodium formate.
$4$. The final product is pentaerythritol,$C(CH_2OH)_4$.

(B) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$HCHO$ (Formaldehyde) does not contain either of these groups.
Therefore,$HCHO$ does not give a yellow precipitate of iodoform $(CHI_3)$ with $I_2/NaOH$.

(C) Aldol condensation occurs in aldehydes or ketones that possess at least one $\alpha$-hydrogen atom.
$CH_3CHO$ has $3 \ \alpha$-hydrogens.
$CH_3CH_2CHO$ has $2 \ \alpha$-hydrogens.
$C_6H_5CH_2CHO$ has $2 \ \alpha$-hydrogens.
$(CH_3)_3CCHO$ (pivalaldehyde) has no $\alpha$-hydrogen atom because the $\alpha$-carbon is bonded to three methyl groups and the aldehyde group,leaving no valency for hydrogen.
Therefore,$(CH_3)_3CCHO$ will not undergo aldol condensation.

(D) The starting material is $2$-chlorocyclohexanone. When it reacts with the Grignard reagent $CH_3MgBr$,the nucleophilic $CH_3^-$ group attacks the carbonyl carbon to form an intermediate alkoxide. Upon hydrolysis with $H_2O$,this forms $2$-chloro-$1$-methylcyclohexanol. Under the reaction conditions,an intramolecular nucleophilic substitution occurs where the hydroxyl oxygen attacks the carbon bearing the chlorine atom,leading to the elimination of $HCl$ and the formation of a cyclic ether,$2,2$-dimethyltetrahydrofuran (or a related bicyclic structure depending on the specific ring strain). Given the options,the final product is $2,2$-dimethyltetrahydrofuran.

(A) $1$. The molecular formula $C_8H_8O$ corresponds to several aromatic isomers.
$2$. The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group. Among the isomers of $C_8H_8O$,$Acetophenone$ $(C_6H_5COCH_3)$ contains the $CH_3CO-$ group and thus gives a positive iodoform test. Therefore,$X = \text{Acetophenone}$.
$3$. The Benedict's test is given by aldehydes. Among the isomers of $C_8H_8O$,$Phenylacetaldehyde$ $(C_6H_5CH_2CHO)$ is an aldehyde and gives a positive Benedict's test. Therefore,$Y = \text{Phenylacetaldehyde}$.
$4$. Thus,$X = \text{Acetophenone}$ and $Y = \text{Phenylacetaldehyde}$.

(A) Pentan$-2-$one is a methyl ketone $(CH_3-CO-CH_2-CH_2-CH_3)$,which gives a positive iodoform test with $I_2$ and $NaOH$ to form a yellow precipitate of iodoform $(CHI_3)$.
Pentan$-3-$one $(CH_3-CH_2-CO-CH_2-CH_3)$ is not a methyl ketone and does not give the iodoform test.
Both ketones react with $NaHSO_3$ to form bisulfite addition products,so it cannot differentiate them.
Tollen's reagent is used to distinguish aldehydes from ketones,not between two ketones.
Therefore,the iodoform test is the correct method to differentiate these two compounds.

(B) When two different aldehydes,both containing $\alpha$-hydrogen atoms,undergo aldol condensation,a mixture of $4$ products is formed. These include two self-aldol products and two cross-aldol products.
$1.$ Self-aldol of $CH_3CHO$: $CH_3-CH(OH)-CH_2-CHO$
$2.$ Self-aldol of $CH_3CH_2CHO$: $CH_3-CH_2-CH(OH)-CH(CH_3)-CHO$
$3.$ Cross-aldol (Ethanal as acceptor + Propanal as enolate): $CH_3-CH(OH)-CH(CH_3)-CHO$
$4.$ Cross-aldol (Propanal as acceptor + Ethanal as enolate): $CH_3-CH_2-CH(OH)-CH_2-CHO$.

(A) The reaction sequence involves the conversion of an ester $(CH_3COOCH_3)$ to an aldehyde $(CH_3CHO)$ and vice versa.
$1$. For the conversion of ester to aldehyde $(P)$,$DIBAL-H$ (Diisobutylaluminium hydride) is the standard reagent used at low temperature $(-78^{\circ}C)$ followed by hydrolysis $(H_2O)$.
$2$. For the conversion of aldehyde to ester $(Q)$,the Tishchenko reaction is used. The Tishchenko reaction involves the disproportionation of an aldehyde to an ester in the presence of an alkoxide catalyst,such as aluminium isopropoxide,$Al(OCH(CH_3)_2)_3$.

(B, D) $X$ is Benzaldehyde $(Ph-CHO)$ because it gives a positive Tollen's test but a negative Fehling test.
The reaction of Benzaldehyde with acetic anhydride in the presence of sodium acetate is the Perkin reaction,which produces Cinnamic acid $(Y)$: $Ph-CHO + (CH_3CO)_2O \xrightarrow{CH_3COONa} Ph-CH=CH-COOH$ $(Y)$.
Then,$Ph-CH=CH-COOH$ $\xrightarrow{SOCl_2} Ph-CH=CH-COCl$ $\xrightarrow{Me_2NH} Ph-CH=CH-CON(CH_3)_2$ $(Z)$.
Thus,options $B$ and $D$ are correct.

(A) $1$. $n$-hexane undergoes aromatization in the presence of $Al_2O_3/\Delta$ to form benzene $(A)$.
$2$. Benzene $(A)$ reacts with $CO + HCl$ in the presence of $AlCl_3$ (Gattermann-Koch reaction) to form benzaldehyde $(B)$.
$3$. Statement $A$: Benzaldehyde $(B)$ does not form a silver mirror with Tollen's reagent $([Ag(NH_3)_2]OH)$ because it is an aromatic aldehyde and does not reduce it under standard conditions.
$4$. Statement $B$: The Gattermann-Koch reaction involves $CO$ and $HCl$ forming $HCOCl$ in situ,which reacts with $AlCl_3$ to form the formyl cation $(HCO^+)$. The reaction mechanism is bimolecular $(A + E^+)$,not trimolecular.
$5$. Statement $C$: Benzene diazonium chloride reacts with $H_3PO_2$ and $H_2O$ to form benzene $(A)$. This is correct.
$6$. Statement $D$: Benzaldehyde does not give a grey color with $HgCl_2$ (which is a test for formic acid or specific reducing agents).
$7$. The question asks for the incorrect statement. Both $A$ and $B$ are technically incorrect,but in the context of standard chemistry problems,$A$ is the most commonly cited incorrect property.

(D) $1$. The reaction of cyclopentanone with hydroxylamine $(H_2NOH)$ at $pH = 4.5$ is a nucleophilic addition-elimination reaction that forms an oxime,which is compound $(A)$.
$2$. Compound $(A)$ is cyclopentanone oxime.
$3$. When cyclopentanone oxime $(A)$ is treated with an acid catalyst $(H^+ / \Delta)$,it undergoes the Beckmann rearrangement.
$4$. The Beckmann rearrangement of a cyclic oxime results in the expansion of the ring by inserting a nitrogen atom,forming a cyclic amide (lactam).
$5$. For cyclopentanone oxime,the ring expansion leads to the formation of a six-membered lactam,which is piperidin$-2-$one (also known as $\delta$-valerolactam),which is compound $(B)$.

(B) In reaction $(A)$,$2$-methylcyclohexanone reacts with $Cl_2$ in the presence of a base $(NaOH)$. Base-catalyzed halogenation of ketones proceeds via the formation of an enolate ion. The enolate forms at the less substituted $\alpha$-carbon (the one with more hydrogen atoms) to minimize steric hindrance. Thus,chlorination occurs at the $C-6$ position,yielding $6$-chloro-$2$-methylcyclohexanone.
In reaction $(B)$,$2$-methylcyclohexanone reacts with $Cl_2$ in the presence of an acid $(CH_3COOH)$. Acid-catalyzed halogenation proceeds via the formation of an enol. The more substituted enol is more stable (Zaitsev's rule),which forms at the more substituted $\alpha$-carbon (the $C-2$ position). However,since the $C-2$ position is already substituted with a methyl group,the enolization occurs at the $C-6$ position. Wait,let's re-evaluate: for $2$-methylcyclohexanone,the $C-2$ position has one methyl group and one hydrogen. The $C-6$ position has two hydrogens. Acid-catalyzed halogenation typically occurs at the more substituted $\alpha$-carbon. Since $C-2$ has one hydrogen and $C-6$ has two,the enol forms at $C-2$. But $C-2$ is already substituted. Actually,for $2$-methylcyclohexanone,acid-catalyzed chlorination occurs at the $C-6$ position because the $C-2$ position is sterically hindered. Therefore,both reactions yield $6$-chloro-$2$-methylcyclohexanone as the major product. Looking at the options,option $B$ shows $6$-chloro-$2$-methylcyclohexanone for both,which is the correct chemical outcome.

(D) The reaction of a ketone with a peroxyacid $(MCPBA)$ is known as the Baeyer-Villiger oxidation.
In this reaction,an oxygen atom is inserted next to the carbonyl group to form an ester.
The migratory aptitude of groups in the Baeyer-Villiger oxidation follows the order: $3^o \text{ alkyl} > \text{cyclohexyl} > \text{phenyl} > 2^o \text{ alkyl} > 1^o \text{ alkyl} > \text{methyl}$.
In the given reactant,$Ph-C(=O)-Cy$ (where $Cy$ is a cyclohexyl group),the cyclohexyl group has a higher migratory aptitude than the phenyl group.
Therefore,the oxygen atom inserts between the carbonyl carbon and the cyclohexyl group,resulting in the formation of phenyl cyclohexanecarboxylate $(Ph-C(=O)-O-Cy)$.

(D) $1$. The reaction of $3\text{-methylbut-1-ene}$ with $Hg(OCOCH_3)_2/H_2O$ followed by $NaBH_4$ is oxymercuration-demercuration,which follows Markovnikov's rule to give $A = 3\text{-methylbutan-2-ol}$.
$2$. Dehydration of $3\text{-methylbutan-2-ol}$ with $H_2SO_4/\Delta$ gives the major alkene $B = 2\text{-methylbut-2-ene}$ via carbocation rearrangement.
$3$. Ozonolysis of $2\text{-methylbut-2-ene}$ $(CH_3-C(CH_3)=CH-CH_3)$ gives $C = \text{acetone} (CH_3COCH_3)$ and $D = \text{acetaldehyde} (CH_3CHO)$.
$4$. Both acetone and acetaldehyde are carbonyl compounds (ketone and aldehyde) and thus give a positive $NaHSO_3$ test (forming crystalline bisulfite addition products).
$5$. Acetone gives a positive Iodoform test,while acetaldehyde also gives a positive Iodoform test. Thus,option $D$ is the correct statement.

(D) The starting material is $3$-(hydroxymethyl)cyclopentan-$1$-one.
$1$. The reaction with ethylene glycol in the presence of $H^+$ forms a cyclic acetal (protecting group) at the ketone position,yielding $3$-(hydroxymethyl)cyclopentan-$1$-one ethylene acetal.
$2$. Treatment with $PCC$ (Pyridinium chlorochromate) oxidizes the primary alcohol group $(-CH_2OH)$ to an aldehyde group $(-CHO)$,resulting in $3$-oxocyclopentane-$1$-carbaldehyde ethylene acetal.
$3$. Addition of $EtMgBr$ (Grignard reagent) followed by $H_3O^+$ workup attacks the aldehyde carbonyl,converting it into a secondary alcohol,and simultaneously deprotects the cyclic acetal back to the ketone.
$4$. The final product is $3$-($1$-hydroxypropyl)cyclopentan-$1$-one.

(B) The reaction involves the nucleophilic attack of the Grignard reagent $ClMg(CH_2)_4MgCl$ on the carbonyl group of the cyclic ester (lactone),$\gamma$-butyrolactone.
$1$. The Grignard reagent acts as a nucleophile,attacking the carbonyl carbon of the lactone.
$2$. The ring opens,and after acidic workup $(H_3O^+)$,the product formed is a diol.
$3$. Specifically,the reaction of $\gamma$-butyrolactone with a bifunctional Grignard reagent like $ClMg(CH_2)_4MgCl$ leads to the formation of a diol where the chain is extended.
$4$. The structure corresponding to the product is a cyclopentane ring substituted with a hydroxy group and a hydroxypropyl chain,which matches the structure shown in option $B$.

(C) The rate of hydrate formation depends on the electrophilicity of the carbonyl carbon.
Greater electrophilicity leads to a faster rate of nucleophilic addition.
Electrophilicity is decreased by electron-donating groups (like $-CH_3$) and increased by steric hindrance reduction.
Compound $II$ is formaldehyde $(HCHO)$,which has no electron-donating groups and minimal steric hindrance.
Compound $I$ is acetaldehyde $(CH_3CHO)$,which has one electron-donating methyl group.
Compound $III$ is acetone $((CH_3)_2CO)$,which has two electron-donating methyl groups and greater steric hindrance.
Therefore,the order of reactivity is $II > I > III$.

(A) The reactivity of carbonyl compounds towards nucleophilic addition depends on the electrophilicity of the carbonyl carbon.
$1.$ Acid chloride $(RCOCl)$: The highly electronegative chlorine atom exerts a strong $-I$ effect,significantly increasing the electrophilicity of the carbonyl carbon.
$2.$ Aldehyde $(RCHO)$: Has one alkyl group which is electron-donating ($+I$ effect),making it less reactive than acid chloride.
$3.$ Ketone $(RCOR)$: Has two alkyl groups,which increase electron density on the carbonyl carbon more than in an aldehyde,reducing its electrophilicity.
$4.$ Ester $(RCOOR')$: The alkoxy group $(-OR')$ shows a $+M$ effect (resonance),which donates electron density to the carbonyl carbon,making it the least reactive towards nucleophilic addition.
Thus,the decreasing order of reactivity is $I > II > III > IV$.
The correct option is $A$.

(D) An acetal or ketal is a gem-dialkoxy compound because two alkoxy groups are present on the same carbon atom.
For example,in an acetal,the structure is $R-CH(OR')_2$,where two alkoxy groups $(-OR')$ are attached to the same carbon atom.

(B) The starting material contains both a ketone and an ester group.
$LiAlH_4$ is a strong reducing agent that reduces both groups.
To selectively reduce the ester group to an alcohol while keeping the ketone group intact,the ketone must first be protected as a cyclic ketal using ethylene glycol $(HO-CH_2-CH_2-OH)$ in the presence of an acid catalyst $(H^{+})$.
The subsequent acidic workup $((ii) H_2O/H^{+})$ after reduction removes the protecting group and restores the ketone functionality.

(A) Carbonyl compounds react with primary amines to form substituted imines,which are known as Schiff's bases.
The general reaction is:
$R_2C=O + R'-NH_2 \xrightarrow{H^+} R_2C=N-R' + H_2O$

(B) Schiff's reagent is a diagnostic test used to distinguish between aldehydes and ketones.
Aldehydes react with Schiff's reagent to restore its pink/magenta color.
Ketones generally do not react with Schiff's reagent.
In option $(b)$,$CH_3CHO$ (acetaldehyde) is an aldehyde,while $CH_3COCH_3$ (acetone) is a ketone.
Therefore,Schiff's reagent can be used to differentiate between them.

(C) $SeO_2$ (selenium dioxide) is an oxidizing agent that specifically oxidizes the $CH_2$ group adjacent to a carbonyl group to a $CO$ group,forming a $1,2-$dicarbonyl compound.
Ethyl methyl ketone $(CH_3-CO-CH_2-CH_3)$ reacts with $SeO_2$ to form dimethyl glyoxal $(CH_3-CO-CO-CH_3)$.

(A) The reaction of a ketone with a peroxyacid (such as $C_6H_5COOOH$) to form an ester is known as the Baeyer-Villiger oxidation.
In this reaction,an oxygen atom is inserted between the carbonyl carbon and an adjacent carbon atom.

(A) The reduction of acetone to pinacol is a bimolecular reduction reaction known as the pinacol coupling reaction.
This reaction is typically carried out using a magnesium-mercury amalgam $(Mg/Hg)$ in the presence of water.
The reaction involves the coupling of two acetone molecules to form $2,3-dimethylbutane-2,3-diol$,commonly known as pinacol.
Therefore,the correct reagent is $Mg/Hg/H_2O$.

(B) The product $CH_3-CH(OH)-CH(CH_3)-CHO$ is $3$-hydroxy-$2$-methylbutanal.
This is a cross-aldol condensation product.
In this reaction,the enolate ion formed from propanal $(CH_3-CH_2-CHO)$ acts as the nucleophile and attacks the carbonyl carbon of acetaldehyde $(CH_3-CHO)$.
Thus,$(X)$ is $CH_3-CHO$ and $(Y)$ is $CH_3-CH_2-CHO$.

(D) The given reaction is a cross-aldol condensation between benzaldehyde $(C_6H_5CHO)$ and acetaldehyde $(CH_3CHO)$.
Since both carbonyl compounds have $\alpha$-hydrogens (only acetaldehyde has $\alpha$-hydrogens,benzaldehyde does not),the reaction can proceed in the following ways:
$1$. Self-aldol condensation of acetaldehyde ($2$ molecules of $CH_3CHO$ react).
$2$. Cross-aldol condensation ($1$ molecule of $C_6H_5CHO$ and $1$ molecule of $CH_3CHO$ react).
Since benzaldehyde cannot undergo self-aldol condensation due to the absence of $\alpha$-hydrogens,there are $2$ possible aldol products formed (one from self-condensation of acetaldehyde and one from cross-condensation). However,if we consider the dehydration products (aldol condensation products),there are $4$ potential products: $2$ $\beta$-hydroxy aldehydes and $2$ $\alpha,\beta$-unsaturated aldehydes. In standard chemistry problems of this type,the total number of products including both aldol and condensation products is considered to be $4$.

(D) The given reaction is an aldol condensation of cyclohexanone in the presence of a base $(\text{OH}^-)$ and heat $(\Delta)$.
$1$. Two molecules of cyclohexanone undergo self-aldol condensation to form $\beta$-hydroxy ketone (aldol product).
$2$. Under heating $(\Delta)$,the aldol product undergoes dehydration to form an $\alpha,\beta$-unsaturated ketone.
$3$. The initial aldol product is $2-(1-\text{hydroxycyclohexyl})\text{cyclohexanone}$.
$4$. Upon heating,it loses a water molecule to form $2-\text{cyclohexylidenecyclohexanone}$.
$5$. Since the reaction includes heating $(\Delta)$,the final product $[X]$ is the dehydrated product,$2-\text{cyclohexylidenecyclohexanone}$.

(C) The given reaction is the Reformatsky reaction.
Benzaldehyde reacts with an $\alpha$-halo ester (ethyl bromoacetate) in the presence of $Zn$ to form a $\beta$-hydroxy ester.
$C_6H_5CHO + Br-CH_2-COOC_2H_5 \xrightarrow{(i) Zn, (ii) H_3O^{+}} C_6H_5-CH(OH)-CH_2-COOC_2H_5$
Thus,$[X]$ is $Br-CH_2-COOC_2H_5$.

(C) The reaction of an aldehyde or ketone with an $\alpha$-bromo ester in the presence of metallic $Zn$ followed by hydrolysis and dehydration is known as the Reformatsky reaction.
In this reaction,an organozinc intermediate is formed,which then reacts with the carbonyl group to yield a $\beta$-hydroxy ester,which upon heating $(\Delta)$ undergoes dehydration to form an $\alpha,\beta$-unsaturated ester.
Therefore,the given reaction is the Reformatsky reaction.

(C) The Cannizzaro reaction is a specific type of disproportionation reaction where an aldehyde lacking an $\alpha$-hydrogen atom undergoes self-oxidation and reduction in the presence of a concentrated base.
In this process,one molecule of the aldehyde is oxidized to a carboxylic acid salt,while another molecule is reduced to an alcohol.
Since both oxidation and reduction occur simultaneously in the same reaction,it is classified as a redox reaction.
Therefore,it is both a redox reaction and a disproportionation reaction.

(C) The reaction of acetaldehyde $(CH_3CHO)$ with excess formaldehyde $(HCHO)$ in the presence of a base like $KOH$ proceeds via a series of aldol additions followed by a crossed Cannizzaro reaction.
$1$. $CH_3CHO + 3HCHO \xrightarrow{OH^-} (HOCH_2)_3C-CHO$
$2$. $(HOCH_2)_3C-CHO + HCHO + OH^- \rightarrow (HOCH_2)_3C-CH_2OH + HCOO^-$
The final product is pentaerythritol,$C(CH_2OH)_4$.

(B) Metaformaldehyde is the trimer of formaldehyde $(HCHO)$.
It is also known as $1,3,5-$trioxane,trioxane,or trioxin.

(D) The Tollens reagent test is used to detect the presence of an aldehyde group or a reducing agent.
$1$. $C_6H_5CHO$ (Benzaldehyde) is an aromatic aldehyde and gives a positive silver mirror test.
$2$. $CH_3-CHO$ (Acetaldehyde) is an aliphatic aldehyde and gives a positive silver mirror test.
$3$. $HCOOH$ (Formic acid) contains an aldehyde group $(-CHO)$ attached to a hydrogen atom,which acts as a reducing agent and gives a positive silver mirror test.
Since all the given compounds react with Tollens reagent to form a silver mirror,the correct option is $D$.

(A) The starting material is $4$-methylenecyclohexan-$1$-one.
$1$. Reaction with $NaBH_4$:
$NaBH_4$ is a selective reducing agent that reduces aldehydes and ketones to alcohols but does not reduce isolated carbon-carbon double bonds. Therefore,the ketone group is reduced to a secondary alcohol,while the exocyclic double bond remains intact. Thus,$(X)$ is $4$-methylenecyclohexan-$1$-ol.
$2$. Reaction with $(i) BH_3, (ii) H_2O_2/OH^-$:
This is a hydroboration-oxidation reaction. Hydroboration-oxidation is a regioselective and stereospecific reaction that adds $H$ and $OH$ across a carbon-carbon double bond in an anti-Markovnikov fashion. The $BH_3$ reagent will selectively react with the exocyclic double bond,while the ketone group remains unaffected. Thus,$(Y)$ is $4$-(hydroxymethyl)cyclohexan-$1$-one.
Comparing these results with the given options,the correct products are $4$-methylenecyclohexan-$1$-ol and $4$-(hydroxymethyl)cyclohexan-$1$-one,which corresponds to option $A$.

(C) The reaction proceeds as follows:
$1$. Acetaldehyde $(CH_3CHO)$ reacts with $HCN$ to form acetaldehyde cyanohydrin $(CH_3CH(OH)CN)$.
$2$. Subsequent acid-catalyzed hydrolysis of the cyanohydrin group $(-CN)$ converts it into a carboxylic acid group $(-COOH)$,yielding lactic acid $(CH_3CH(OH)COOH)$.

(D) Step $1$: Glycerol $(CH_2OH-CHOH-CH_2OH)$ on heating with $KHSO_4$ undergoes dehydration to form acrolein $(X)$.
$CH_2OH-CHOH-CH_2OH \xrightarrow{KHSO_4, \Delta} CH_2=CH-CHO + 2H_2O$
Step $2$: Acrolein $(X)$ reacts with aluminum ethoxide $(C_2H_5O)_3Al$ via the Tishchenko reaction to form allyl acrylate $(Y)$.
$2CH_2=CH-CHO \xrightarrow{(C_2H_5O)_3Al, \Delta} CH_2=CH-C(=O)-O-CH_2-CH=CH_2$

(B) The reaction involving $NaCN$ in the presence of ethanol $(C_2H_5OH)$ and water $(HOH)$ is the Benzoin condensation.
In this reaction,two molecules of benzaldehyde undergo self-condensation in the presence of cyanide ions to form benzoin.
The reaction is represented as:
$2 C_6H_5CHO \xrightarrow{NaCN, C_2H_5OH, H_2O} C_6H_5CH(OH)COC_6H_5$

(D) The reaction of an aldehyde (furfural) with an alcohol (ethanol) initially leads to the formation of a hemiacetal.
In $Figure$ $1$,the structure shows a carbon atom bonded to a hydroxyl group $(-OH)$ and an alkoxy group $(-OCH_2CH_3)$.
This functional group arrangement $(R-CH(OH)(OR'))$ is characteristic of a hemiacetal.
Therefore,the compound shown in $Figure$ $1$ is a hemiacetal.

(C) $1$. Ozonolysis of $(A)$ $C_{16}H_{16}$ gives only one product $(B)$ $C_8H_8O$. This implies $(A)$ is $C_6H_5-CH=CH-C_6H_5$ (stilbene derivative) or similar. Given $(B)$ is obtained from cyanobenzene $(C_6H_5CN)$ and $CH_3MgBr$,$(B)$ is acetophenone $(C_6H_5COCH_3)$.
$2$. Acetophenone $(C_6H_5COCH_3)$ contains a methyl ketone group,so it gives the iodoform test (Option $A$ is correct).
$3$. Reduction of acetophenone with $LiAlH_4$ gives $1$-phenylethanol $(C_6H_5CH(OH)CH_3)$,which is a secondary alcohol with a $CH_3CH(OH)-$ group,thus it gives the iodoform test (Option $B$ is correct).
$4$. Acetophenone is a ketone,not an aldehyde,so it does not give Tollen's test (Option $C$ is incorrect).
$5$. Wolff-Kishner reduction of acetophenone using $NH_2NH_2, EtO^-$ gives ethylbenzene (Option $D$ is correct).
$6$. Therefore,the incorrect statement is $C$.

(B) $LiAlH_4$ (Lithium aluminium hydride) is a strong reducing agent that reduces aldehydes,ketones,carboxylic acids,and esters to their corresponding alcohols.
However,it does not reduce isolated carbon-carbon double bonds $(C=C)$.
In the given reaction,the reactant is cinnamaldehyde $(C_6H_5-CH=CH-CHO)$.
$LiAlH_4$ selectively reduces the aldehyde group $(-CHO)$ to a primary alcohol group $(-CH_2OH)$ while leaving the conjugated double bond $(C=C)$ intact.
Therefore,the product formed is cinnamyl alcohol,which is $C_6H_5-CH=CH-CH_2OH$.