Properties Questions in English

(A) Tollen's reagent is an oxidizing agent that oxidizes aldehydes to carboxylate ions.
$(A)$ is acrolein (an $\alpha, \beta$-unsaturated aldehyde),which gives a positive Tollen's test.
$(B)$ is benzaldehyde (an aromatic aldehyde),which gives a positive Tollen's test.
$(C)$ is benzoin (an $\alpha$-hydroxyketone),which is oxidized by Tollen's reagent to benzil,thus giving a positive Tollen's test.
$(D)$ is chalcone (an $\alpha, \beta$-unsaturated ketone),which does not give a positive Tollen's test.
Therefore,compounds $(A)$,$(B)$,and $(C)$ give a positive Tollen's test.

(A) The reaction sequence involves two steps:
$1$. The starting material is isobutyrophenone $(Ph-CO-CH(CH_3)_2)$. Treatment with excess $HCHO$ and $NaOH$ (base-catalyzed aldol condensation) leads to the hydroxymethylation of the $\alpha$-carbon. Since there is only one $\alpha$-hydrogen,it reacts with one equivalent of $HCHO$ to form $Ph-CO-C(CH_3)_2-CH_2OH$.
$2$. In the second step,treatment with $HCHO$ and a catalytic amount of $H^+$ leads to the formation of a cyclic acetal ($1$,$3$-dioxane derivative) by the reaction of the carbonyl group and the hydroxyl group with formaldehyde. The carbonyl oxygen and the hydroxyl oxygen react with $HCHO$ to form a six-membered ring structure.
Thus,the final product is the cyclic acetal shown in option $A$.

(D) The transformation involves the selective reduction of an aldehyde group to a primary alcohol in the presence of a carboxylic acid,an ester,and an epoxide.
$1$. $LiAlH_4$ is a strong reducing agent that reduces aldehydes,carboxylic acids,esters,and epoxides.
$2$. $BH_3$ in $THF$ is a selective reducing agent that reduces carboxylic acids and aldehydes but can also react with epoxides.
$3$. $NaBH_4$ is a mild reducing agent that selectively reduces aldehydes and ketones to alcohols without affecting carboxylic acids,esters,or epoxides.
$4$. Raney $Ni / H_2$ is a catalytic hydrogenation reagent that reduces aldehydes and ketones but generally does not reduce carboxylic acids,esters,or epoxides under mild conditions.
Therefore,both $NaBH_4$ and Raney $Ni / H_2$ can be used to achieve this selective transformation.

(C) Formation of $P$: $CH_3CH_2CH(CH_3)CH_2CN$ reacts with $PhMgBr$ to form a ketone,which further reacts with $PhMgBr$ to form $Ph_2C(OH)CH_2CH(CH_3)CH_2CH_3$. This molecule has an asymmetric carbon at the $CH(CH_3)$ position.
Formation of $Q$: $PhH + CH_3COCl$ (Friedel-Crafts acylation) gives $PhCOCH_3$. Reaction with $PhMgBr$ gives $Ph_2C(OH)CH_3$. This molecule has no asymmetric carbon because two phenyl groups are attached to the central carbon.
Formation of $R$: $CH_3CH_2COCl + (PhCH_2)_2Cd$ gives $CH_3CH_2COCH_2Ph$. Reaction with $PhMgBr$ gives $CH_3CH_2C(OH)(Ph)CH_2Ph$. The central carbon is attached to $CH_3CH_2-$,$Ph-$,$PhCH_2-$,and $-OH$. Since all four groups are different,it is an asymmetric carbon.
Formation of $S$: $PhCH_2CHO$ reacts to form a product $S$ which is $Ph-CH=C(Ph)COOH$. This molecule has no asymmetric carbon.
Conclusion: $P$ and $R$ have asymmetric carbons. $Q$ and $S$ do not. Therefore,statements $(C)$ and $(D)$ are correct.

(B) The reaction proceeds in two steps:
$1$. Reduction of the tetranitrile compound with excess $LiAlH_4$ followed by $H_2O$ yields a tetraamine: $(H_2N-CH_2)_2CH-CH(CH_2-NH_2)_2$.
$2$. The tetraamine reacts with excess acetophenone $(CH_3-CO-Ph)$ to form an imine (Schiff base) product $P$.
In the product $P$,each of the four acetophenone-derived units contains one $C=N$ carbon and six aromatic ring carbons,totaling $7$ $sp^2$ carbons per unit.
Since there are $4$ such units,the total number of $sp^2$ hybridised carbon atoms is $4 \times 7 = 28$.

(D) Analysis of the reactions:
$(I)$ Toluene + $Br_2 / h\nu$ $\rightarrow$ Benzyl bromide (Free radical substitution,$R$)
$(II)$ Acetophenone + $NaOH / Br_2$ $\rightarrow$ Sodium benzoate + $CHBr_3$ (Haloform reaction,$S$)
$(III)$ Benzaldehyde + $(CH_3CO)_2O / CH_3COOK$ $\rightarrow$ Cinnamic acid + Acetic acid (Perkin condensation,$P$)
$(IV)$ Phenol + $NaOH / CO_2$ $\rightarrow$ Salicylic acid (Kolbe's reaction,Carboxylation,$Q$)
$(1)$ Synthesis of benzoic acid: The haloform reaction of acetophenone $(II)$ with $NaOH / Br_2$ $(i)$ gives sodium benzoate,which on acidification yields benzoic acid. Thus,$(II)(i)(S)$ is correct. Option $D$ matches.
$(2)$ Radical mechanism: Toluene $(I)$ reacts with $Br_2 / h\nu$ $(ii)$ via a free radical substitution mechanism $(R)$. Thus,$(I)(ii)(R)$ is correct. Option $A$ matches.
$(3)$ Two different carboxylic acids: Perkin condensation of benzaldehyde $(III)$ with $(CH_3CO)_2O / CH_3COOK$ $(iii)$ produces cinnamic acid and acetic acid. Thus,$(III)(iii)(P)$ is correct. Option $B$ matches.
Therefore,the correct sequence is $(1)-D, (2)-A, (3)-B$.

(C) $1$. $Q$ has the molecular formula $C_8H_8O$. It undergoes the Cannizzaro reaction but not the haloform reaction. This implies $Q$ is an aldehyde without an $\alpha$-hydrogen atom (e.g.,$p$-methylbenzaldehyde).
$2$. $S$ has the molecular formula $C_8H_8O$. It undergoes the haloform reaction but not the Cannizzaro reaction. This implies $S$ is a methyl ketone (e.g.,acetophenone).
$3$. Analyzing the options:
- In option $A$,$P$ is $1$-methyl$-4-$vinylbenzene,which on ozonolysis gives $p$-methylbenzaldehyde $(Q)$. $R$ is $2$-phenylpropene,which on ozonolysis gives acetophenone $(S)$. Both match the criteria.
- In option $B$,the structures do not yield $C_8H_8O$ products as required.
- In option $C$ and $D$,the structures do not match the required molecular formula or the specific reaction criteria.

(A) The starting material is $1,2-dimethylenecyclohexane$ derivative,which upon ozonolysis ($O_3$ followed by $Zn, H_2O$) yields $Y$,which is $cyclodecane-1,6-dione$.
$Y$ $(cyclodecane-1,6-dione)$ has two equivalent carbonyl groups and equivalent $\alpha$-hydrogens.
Intramolecular aldol condensation of $cyclodecane-1,6-dione$ involves the formation of an enolate at one of the $\alpha$-positions,followed by nucleophilic attack on the other carbonyl group.
Due to the symmetry of the molecule,the attack can occur to form a $5$-membered ring (via a $5$-exo-trig cyclization) or a $7$-membered ring (via a $7$-endo-trig cyclization).
According to Baldwin's rules and thermodynamic stability,the formation of the $5$-membered ring is highly favored over the $7$-membered ring.
Thus,only $1$ major intramolecular aldol condensation product is formed.

(A) The reaction sequence is an aldol condensation followed by cyanohydrin formation and subsequent hydrolysis/cyclization.
$1.$ The reaction between isobutyraldehyde $(P)$ and formaldehyde $(Q)$ in the presence of aqueous $K_2CO_3$ is a cross-aldol condensation. $P$ is $(CH_3)_2CHCHO$ and $Q$ is $HCHO$. Thus,$P$ and $Q$ are $(B)$ and $(A)$ respectively.
$2.$ The aldol product $R$ formed is $3-hydroxy-2,2-dimethylpropanal$,which corresponds to structure $(B)$.
$3.$ Treatment of $R$ with $HCN$ yields the cyanohydrin $S$,which is $2,2-dimethyl-3-hydroxy-4-hydroxybutanenitrile$ derivative,corresponding to structure $(D)$.
Therefore,the correct sequence is $(B, B, D)$ for $(P, Q)$,$R$,and $S$ respectively. However,based on the provided options,the closest match is $(B, A, D)$.

(C) . This is an electrophilic aromatic substitution reaction,specifically a diazo coupling reaction.
$B$. This is the Pinacol-Pinacolone rearrangement,which proceeds via a carbocation intermediate.
$C$. This is a nucleophilic addition reaction of a hydride to a ketone,resulting in a chiral alcohol. Since the carbonyl carbon is planar,the hydride can attack from either side,producing a racemic mixture.
$D$. This is an intramolecular nucleophilic substitution reaction where the thiolate ion displaces the chloride ion.

(C) The reaction of $CH_3CHO$ with $4HCHO$ in the presence of $conc. aq. NaOH$ proceeds through a series of steps.
$1$. The first step is an aldol condensation between $CH_3CHO$ and $HCHO$ to form $HOCH_2-CH_2-CHO$.
$2$. The second step is another aldol condensation with $HCHO$ to form $(HOCH_2)_2CH-CHO$.
$3$. The third step is a final aldol condensation with $HCHO$ to form $(HOCH_2)_3C-CHO$.
$4$. The final step is a Cannizzaro reaction where the aldehyde group is reduced to an alcohol group,resulting in pentaerythritol $(HOCH_2)_4C$.
Thus,there are $3$ aldol reactions involved in the transformation.

(A) Step $1$: Acetophenone $(10 \ mol)$ $\xrightarrow{NaOBr, H_3O^+}$ Benzoic acid $(A)$ with $60\%$ yield. Moles of $A = 10 \times 0.6 = 6 \ mol$.
Step $2$: Benzoic acid $(A)$ $\xrightarrow{NH_3, \Delta}$ Benzamide $(B)$ with $50\%$ yield. Moles of $B = 6 \times 0.5 = 3 \ mol$.
Step $3$: Benzamide $(B)$ $\xrightarrow{Br_2/KOH}$ Aniline $(C)$ with $50\%$ yield. Moles of $C = 3 \times 0.5 = 1.5 \ mol$.
Step $4$: Aniline $(C)$ $\xrightarrow{Br_2 (3 \ equiv), AcOH}$ $2,4,6$-Tribromoaniline $(D)$ with $100\%$ yield. Moles of $D = 1.5 \times 1.0 = 1.5 \ mol$.
Molar mass of $D$ $(C_6H_4NBr_3) = (6 \times 12) + (4 \times 1) + 14 + (3 \times 80) = 72 + 4 + 14 + 240 = 330 \ g \ mol^{-1}$.
Amount of $D = 1.5 \ mol \times 330 \ g \ mol^{-1} = 495 \ g$.

(C) $1$. Friedel-Crafts acylation of $P$ with succinic anhydride gives $Q$,which is a keto-acid: $Ar-CO-CH_2-CH_2-COOH$.
$2$. Clemmensen reduction $(Zn-Hg/HCl)$ reduces the ketone to a methylene group,followed by intramolecular cyclization using $H_3PO_4$ to form the cyclic ketone $R$.
$3$. Reaction of $R$ with $CH_3MgBr$ followed by acid workup gives a tertiary alcohol.
$4$. Dehydration of the alcohol with $H_2SO_4/\Delta$ yields the alkene $S$.
$5$. Comparing the structures,$A$ represents the correct structure for $S$ and $C$ represents the correct structure for $R$. Thus,the correct combination is $A.C$.

(A) In reaction $I$ (basic medium),the haloform reaction occurs. Since the rate of reaction increases with each $\alpha$-halogenation,$1.0 \ mol$ of acetone reacts with $1.0 \ mol$ of $Br_2$ to produce $CH_3COONa$ $(T)$,$CHBr_3$ $(U)$,and some unreacted acetone.
In reaction $II$ (acidic medium),monohalogenation takes place. $1.0 \ mol$ of acetone reacts with $1.0 \ mol$ of $Br_2$ to form $P$ $(CH_3COCH_2Br)$.

(D) The molecular formula for a ketone is $C_nH_{2n}O$. Given $MW = 100$,we have $12n + 2n + 16 = 100$,which gives $14n = 84$,so $n = 6$. The formula is $C_6H_{12}O$.
The possible isomeric ketones are:
$1$. $Hexan-2-one$: Reduction gives a chiral alcohol,forming a racemic mixture.
$2$. $Hexan-3-one$: Reduction gives a chiral alcohol,forming a racemic mixture.
$3$. $4-Methylpentan-2-one$: Reduction gives a chiral alcohol,forming a racemic mixture.
$4$. $3-Methylpentan-2-one$: This has a chiral center. Reduction creates a new chiral center,resulting in diastereomers,not a racemic mixture.
$5$. $3,3-Dimethylbutan-2-one$: Reduction gives a chiral alcohol,forming a racemic mixture.
$6$. $3-Methylpentan-3-one$: Reduction gives a chiral alcohol,forming a racemic mixture.
Out of these,$5$ ketones produce a racemic mixture upon reduction with $NaBH_4$.

(C) $1$. The reaction of benzene with succinic anhydride in the presence of $AlCl_3$ (Friedel-Crafts acylation) gives $P$ ($4$-oxo$-4-$phenylbutanoic acid),which is a carboxylic acid.
$2$. Clemmensen reduction $(Zn/Hg, HCl)$ of $P$ gives $Q$ ($4$-phenylbutanoic acid),which is also a carboxylic acid. Thus,statement $A$ is correct.
$3$. $Q$ reacts with $SOCl_2$ to form $R$ ($4$-phenylbutanoyl chloride). Intramolecular Friedel-Crafts acylation of $R$ gives $S$ ($1$-tetralone).
$4$. $S$ ($1$-tetralone) is a ketone and does not contain a carbon-carbon double bond,so it does not decolorize bromine water. Statement $B$ is incorrect.
$5$. Both $P$ (a ketone) and $S$ (a ketone) react with hydroxylamine $(NH_2OH)$ to form oximes. Statement $C$ is correct.
$6$. Compound $R$ (an acid chloride) reacts with dialkylcadmium $(R'_2Cd)$ to give a ketone,not a tertiary alcohol. Statement $D$ is incorrect.
$7$. Therefore,the correct statements are $A$ and $C$.

(A) $1$. The reaction of iso-propylbenzene (cumene) with $O_2$ followed by $H_3O^{+}$ is the cumene process, which yields phenol and acetone $(P = CH_3COCH_3)$.
$2$. Acetone $(P)$ reacts with $3$ equivalents of $Cl_2$ to form trichloroacetone $(Q = Cl_3CCOCH_3)$.
$3$. Trichloroacetone $(Q)$ reacts with $Ca(OH)_2$ to produce chloroform $(R = CHCl_3)$ and calcium acetate $(S = (CH_3COO)_2Ca)$.
$4$. Statement $(A)$: $CHCl_3$ reacts with acetone in the presence of $KOH$ to form chloritone $(Cl_3CC(OH)(CH_3)_2)$. This is correct.
$5$. Statement $(B)$: $CHCl_3$ $(R)$ reacts with $O_2$ in the presence of light to form phosgene $(COCl_2)$. This is correct.
$6$. Statement $(C)$: $Q$ $(Cl_3CCOCH_3)$ reacts with $NaOH$ via the haloform reaction, not as described. This is incorrect.
$7$. Statement $(D)$: Calcium acetate $(S)$ on dry distillation gives acetone $(P)$ and $CaCO_3$. This is correct.
Therefore, the correct statements are $A, B$, and $D$.

(C) The reaction of acetaldehyde $(CH_3CHO)$ with excess formaldehyde $(HCHO)$ in the presence of conc. $NaOH$ is an aldol condensation followed by Cannizzaro reaction,yielding pentaerythritol ($P$,$C(CH_2OH)_4$) and sodium formate ($Q$,$HCOONa$).
$P$ is $C(CH_2OH)_4$,which has no aldehyde group and does not give Tollens' test.
$Q$ is $HCOONa$,which upon acidification gives formic acid $(HCOOH)$,which gives a positive Tollens' test.
Reaction of $P$ with excess cyclohexanone in the presence of $PTSA$ forms an acetal $R$ (a spiro-compound).
The structure of $R$ is $C(CH_2O)_2(C_6H_{10})_2$.
In $R$,there are $12$ methylene groups from two cyclohexanone rings and $2$ methylene groups from the central carbon's substituents,totaling $14$ methylene groups $(-CH_2-)$.
There are $4$ oxygen atoms in the two acetal linkages.
Sum $= 14 + 4 = 18$.

(A) For $(P)$: Cyclohexene undergoes ozonolysis to give hexanedial,followed by intramolecular aldol condensation to form cyclohex$-1-$enecarbaldehyde. Protection of the aldehyde with ethylene glycol,hydroboration-oxidation,deprotection,and reduction with $NaBH_4$ yields $(3)$.
For $(Q)$: $1-$Methylcyclohexene undergoes ozonolysis,aldol condensation,protection,hydroboration-oxidation,deprotection,and reduction to yield $(5)$.
For $(R)$: $3-$Methylcyclopent$-2-$enone is protected,subjected to oxymercuration-demercuration,deprotected,and reduced to yield $(4)$.
For $(S)$: $3-$Methylcyclopent$-2-$enone is protected,subjected to hydroboration-oxidation,deprotected,and reduced to yield $(1)$.
Thus,the correct matching is $P-3, Q-5, R-4, S-1$.

(C) To convert $P$ to $Q$,we need to perform the following transformations:
$1$. Reduction of the ketone and the ester/nitrile groups.
$2$. Selective reduction of the two alkyne groups to cis-alkenes using Lindlar's catalyst and $H_2$.
$3$. Reduction of the nitrile group $(-CN)$ to an aldehyde $(-CHO)$ using $SnCl_2 / HCl$ (Stephen reduction).
$4$. Reduction of the ketone group to an alcohol using $NaBH_4$.
$5$. Hydrolysis of the ester and imine intermediate using $H_3O^{+}$.
Analyzing the sequence in option $(C)$:
$i$) $NaBH_4$ reduces the ketone to an alcohol.
$ii$) $SnCl_2 / HCl$ reduces the nitrile to an imine.
$iii$) $H_3O^{+}$ hydrolyzes the imine to an aldehyde and the ester to a carboxylic acid.
$iv$) Lindlar's catalyst with $H_2$ reduces the two internal alkynes to cis-alkenes.
This sequence correctly transforms $P$ to $Q$.

(A) The compound $P$ is cyclohexane-$1,3,5$-trione,which exists in tautomeric equilibrium with phloroglucinol. It gives a positive $FeCl_3$ test due to the enol form and lacks intramolecular hydrogen bonding in the tri-keto form.
$P$ reacts with $3$ equivalents of $NH_2OH$ to form the tri-oxime $Q$.
Treatment of $P$ with excess $CH_3I$ and $KOH$ results in the exhaustive methylation of the $\alpha$-carbons,yielding $2,2,4,4,6,6$-hexamethylcyclohexane-$1,3,5$-trione $(R)$.
Reaction of $R$ with excess iso-butylmagnesium bromide $(iso-BuMgBr)$ acts as a base due to the steric hindrance of the six methyl groups on the $\alpha$-carbons,leading to the reduction of the carbonyl groups to hydroxyl groups (enolization/reduction) rather than nucleophilic addition. The final product $S$ is $2,2,4,4,6,6$-hexamethylcyclohexane-$1,3,5$-triol.
In $2,2,4,4,6,6$-hexamethylcyclohexane-$1,3,5$-triol,there are $6$ methyl groups attached to the ring carbons. Each methyl group is a $-CH_3$ unit. Thus,the total number of methyl groups is $6$.

(C) Fehling's test is given by aliphatic aldehydes. Aromatic aldehydes do not give this test.
$(A)$ $C_6H_5CHO$ is an aromatic aldehyde,so it gives a negative test.
$(B)$ $C_6H_5COCH_3$ is a ketone,so it gives a negative test.
$(C)$ $HOCH_2-CO-(CHOH)_3-CH_2OH$ is a ketose sugar (fructose). In alkaline medium,it isomerizes to aldose,thus giving a positive Fehling's test.
$(D)$ $CH_3CHO$ is an aliphatic aldehyde,so it gives a positive test.
$(E)$ $C_6H_5CH_2CHO$ is an aliphatic aldehyde (the aldehyde group is not directly attached to the benzene ring),so it gives a positive test.
Therefore,compounds $(C)$,$(D)$,and $(E)$ give a positive Fehling's test.

(D) $1$. Toluene reacts with $CrO_2Cl_2$ in $CS_2$ followed by hydrolysis $(H_3O^+)$ to form benzaldehyde $(C_6H_5CHO)$ via the Etard reaction.
$2$. Benzaldehyde then reacts with sodium bisulphite $(NaHSO_3)$ to form a crystalline addition product,which is the residue $(A)$. The structure of $(A)$ is $C_6H_5CH(OH)SO_3Na$.
$3$. When this residue $(A)$ is treated with dilute $HCl$,it undergoes hydrolysis to regenerate benzaldehyde,which is compound $(B)$.
$4$. Therefore,residue $(A)$ is $C_6H_5CH(OH)SO_3Na$ and compound $(B)$ is $C_6H_5CHO$.

(C) The reaction of $CH_3-CH_2-CHO$ with excess $HCHO$ in the presence of an alkali involves two steps:
$1$. Multiple cross-aldol condensation: The $\alpha$-hydrogens of propanal are replaced by hydroxymethyl groups $(CH_2OH)$ through reaction with $HCHO$. Since propanal has two $\alpha$-hydrogens,both are replaced to form $CH_3-C(CH_2OH)_2-CHO$.
$2$. Cross-Cannizzaro reaction: The remaining aldehyde group $(-CHO)$ is then reduced to a primary alcohol $(-CH_2OH)$ by the excess $HCHO$,which itself gets oxidized to formate.
The final product is $CH_3-C(CH_2OH)_3$.

(B) The hydration of carbonyl compounds involves the nucleophilic addition of water to the carbonyl carbon.
For formaldehyde $(HCHO)$,the $K_{eq}$ for hydration is approximately $2280$. This high value is due to the lack of steric hindrance (small substituents) and the high electrophilicity of the carbonyl carbon.
For trichloroacetaldehyde ($CCl_3CHO$,also known as chloral),the $K_{eq}$ for hydration is approximately $2000$. This high value is primarily due to the strong electron-withdrawing $-I$ effect of the three chlorine atoms,which makes the carbonyl carbon highly electrophilic and susceptible to nucleophilic attack by water.
Both statements are scientifically accurate based on standard chemical literature.
Therefore,both Statement $I$ and Statement $II$ are true.

(C) The reactivity of carbonyl compounds towards nucleophilic addition reactions depends on two factors: steric hindrance and electronic effects.
$1$. Steric hindrance: Ketones are less reactive than aldehydes due to the presence of two bulky alkyl/aryl groups. Thus,acetophenone is the least reactive among the given compounds.
$2$. Electronic effects: Electron-withdrawing groups (like $-NO_2$) increase the electrophilicity of the carbonyl carbon,thereby increasing reactivity. Electron-donating groups (like $-CH_3$) decrease the electrophilicity,thereby decreasing reactivity.
Comparing the compounds:
- Acetophenone: Ketone,least reactive.
- $p$-Tolualdehyde: Contains an electron-donating $-CH_3$ group,making it less reactive than benzaldehyde.
- Benzaldehyde: Standard reactivity.
- $p$-Nitrobenzaldehyde: Contains an electron-withdrawing $-NO_2$ group,making it the most reactive.
Therefore,the correct order of reactivity is: $acetophenone < p-tolualdehyde < benzaldehyde < p-nitrobenzaldehyde$.

(B) To determine which reactions both acetaldehyde $(CH_3CHO)$ and acetone $(CH_3COCH_3)$ undergo,we analyze each reaction:

$S.No$. Name of Reaction	Acetaldehyde $(CH_3CHO)$ | Acetone $(CH_3COCH_3)$
$1$. Iodoform reaction	Positive | Positive
$2$. Cannizzaro reaction	Negative | Negative
$3$. Aldol condensation	Positive | Positive
$4$. Tollen's test	Positive | Negative
$5$. Clemmensen reduction	Positive | Positive

$1$. Iodoform reaction $(A)$: Both contain the $CH_3CO-$ group,so both give a positive test.
$2$. Cannizzaro reaction $(B)$: Neither has an $\alpha$-hydrogen,so neither undergoes this reaction.
$3$. Aldol condensation $(C)$: Both have $\alpha$-hydrogens,so both undergo this reaction.
$4$. Tollen's test $(D)$: Only aldehydes give a positive test; acetone (a ketone) does not.
$5$. Clemmensen reduction $(E)$: Both aldehydes and ketones undergo this reduction.
Therefore,both compounds undergo reactions $A, C$,and $E$.

(A) nucleophile is a species that donates an electron pair to form a chemical bond. They are either negatively charged or neutral species with at least one lone pair of electrons or a $\pi$-bond.
$1$. $NH_3$: Has a lone pair on $N$.
$2$. $PhSH$: Has lone pairs on $S$.
$3$. $(H_3C)_2S$: Has lone pairs on $S$.
$4$. $H_2C=CH_2$: Contains a $\pi$-bond.
$5$. $\stackrel{\ominus}{O}H$: Negatively charged with lone pairs.
$H_3O^{\oplus}$ is an electrophile (or acid),$(CH_3)_2CO$ is an electrophile (carbonyl carbon),and $CH_3CH=NCH_3$ has a lone pair on $N$ but is generally considered an electrophile at the $C=N$ bond,though it can act as a weak nucleophile. However,in standard classification,the first $5$ are clearly identified as nucleophiles.
Thus,the total number of nucleophiles is $5$.

(D) The given reaction uses $Zn-Hg/HCl$,which is the reagent for Clemmensen reduction.
Clemmensen reduction specifically reduces carbonyl groups (aldehydes and ketones) to methylene groups $(-CH_2-)$.
It does not affect ester groups $(-COOC_2H_5)$.
In the given reactant,there is an acetyl group $(-COCH_3)$ and an aldehyde group $(-CHO)$ attached to the benzene ring.
Both the acetyl group and the aldehyde group will be reduced to ethyl $(-CH_2CH_3)$ and methyl $(-CH_3)$ groups respectively.
The ester group remains unchanged.
Therefore,the product $(P)$ is ethyl $3-$ethyl$-5-$methylbenzoate.

(C) The chemical reaction is: $2 \ C_6H_5CHO + CH_3COCH_3 \rightarrow C_6H_5CH=CHCOCH=CHC_6H_5 + 2H_2O$.
Moles of benzaldehyde used $= \frac{5.3 \ g}{106 \ g/mol} = 0.05 \ mol$.
According to the stoichiometry,$2 \ mol$ of benzaldehyde produces $1 \ mol$ of dibenzalacetone.
Therefore,theoretical moles of dibenzalacetone $= \frac{0.05}{2} = 0.025 \ mol$.
Molar mass of dibenzalacetone $(C_{17}H_{14}O) = (17 \times 12) + (14 \times 1) + 16 = 234 \ g/mol$.
Theoretical yield in grams $= 0.025 \ mol \times 234 \ g/mol = 5.85 \ g$.
Actual yield $= 3.51 \ g$.
Percentage yield $= \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 = \frac{3.51}{5.85} \times 100 = 60 \%$.

(C) $1$. The optically active alkyl halide $[A]$ is $sec$-butyl bromide $(CH_3CH_2CH(Br)CH_3)$.
$2$. Reaction with hot alcoholic $KOH$ (dehydrohalogenation) gives but$-2-$ene as the major product $[B]$ $(CH_3CH=CHCH_3)$.
$3$. Reaction of $[B]$ with $Br_2$ gives $2,3-$dibromobutane $[C]$ $(CH_3CH(Br)CH(Br)CH_3)$.
$4$. Reaction of $[C]$ with alcoholic $NaNH_2$ (dehydrohalogenation) gives but$-1-$yne $[D]$ $(CH_3CH_2C \equiv CH)$.
$5$. Hydration of but$-1-$yne $[D]$ with $H_2O$ in the presence of $HgSO_4$ and $H^+$ (Kucherov reaction) follows Markovnikov's rule to form an enol intermediate,which tautomerizes to form butan$-2-$one $[E]$ $(CH_3CH_2COCH_3)$.
$6$. The $IUPAC$ name of $[E]$ is butan$-2-$one.

(B) Vanillin contains a phenolic $-OH$ group,which is acidic and reacts with $NaOH$ to form a salt. It also contains an aldehyde group $(-CHO)$,which reacts with Tollen's reagent to give a silver mirror test.
Therefore,Statement $(I)$ is correct.
Vanillin does not have any $\alpha$-hydrogen atoms relative to the carbonyl group. Aldol condensation requires the presence of at least one $\alpha$-hydrogen atom. Thus,it cannot undergo self-aldol condensation.
Therefore,Statement $(II)$ is incorrect.

(B) The iodoform reaction is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$1$. Ethanol $(CH_3CH_2OH)$: Gives iodoform test.
$2$. Isopropyl alcohol $(CH_3CH(OH)CH_3)$: Gives iodoform test.
$3$. Bromoacetone $(CH_3COCH_2Br)$: Gives iodoform test.
$4$. $2-$Butanol $(CH_3CH(OH)CH_2CH_3)$: Gives iodoform test.
$5$. $2-$Butanone $(CH_3COCH_2CH_3)$: Gives iodoform test.
$6$. Butanal $(CH_3CH_2CH_2CHO)$: Does not give iodoform test.
$7$. $2-$Pentanone $(CH_3COCH_2CH_2CH_3)$: Gives iodoform test.
$8$. $3-$Pentanone $(CH_3CH_2COCH_2CH_3)$: Does not give iodoform test.
$9$. Pentanal $(CH_3CH_2CH_2CH_2CHO)$: Does not give iodoform test.
$10$. $3-$Pentanol $(CH_3CH_2CH(OH)CH_2CH_3)$: Does not give iodoform test.
The molecules that do not give the iodoform reaction are: Butanal,$3-$Pentanone,Pentanal,and $3-$Pentanol.
Total count = $4$.

(B) The given reaction is an intramolecular Cannizzaro reaction of phenylglyoxal $(Ph-CO-CHO)$.
In the presence of a base like $KOH$,the hydroxide ion attacks the more electrophilic carbonyl carbon (the aldehyde group).
This leads to a hydride shift from the aldehyde carbon to the ketone carbonyl carbon.
The aldehyde group is oxidized to a carboxylate group $(COO^-)$,and the ketone group is reduced to a secondary alcohol group $(CH(OH))$.
Thus,the product formed is $Ph-CH(OH)-COO^-K^+$.

(D) Intramolecular aldol condensation involves the formation of a ring within a single molecule containing two carbonyl groups.
$(1)$ Cyclodecane$-1,6-$dione undergoes intramolecular aldol condensation to form a bicyclic $\alpha, \beta-$unsaturated ketone.
$(2)$ $6-$Oxoheptanal undergoes intramolecular aldol condensation to form cyclohex$-2-$en$-1-$one.
$(3)$ $2-$Acetylbenzaldehyde undergoes intramolecular aldol condensation to form inden$-1-$one.
$(4)$ The reaction between $1-$tetralone and formaldehyde is an intermolecular aldol condensation (specifically a Claisen-Schmidt condensation),not an intramolecular one.
Therefore,the product shown in option $D$ is not formed via intramolecular aldol condensation.

(B) $1$. Compound $(X)$ is $C_3H_6O$. Since it is not readily oxidised,it is a ketone,$CH_3COCH_3$ (acetone).
$2$. Reduction of $(X)$ gives $(Y)$,which is $CH_3CH(OH)CH_3$ (propan$-2-$ol).
$3$. Reaction of $(Y)$ with $HBr$ gives $(Z)$,which is $CH_3CH(Br)CH_3$ ($2$-bromopropane).
$4$. $(Z)$ reacts with $Mg$ in ether to form the Grignard reagent $CH_3CH(MgBr)CH_3$.
$5$. The Grignard reagent reacts with $(X)$ $(CH_3COCH_3)$ followed by hydrolysis to yield $2,3-$dimethylbutan$-2-$ol.
$6$. Thus,$(X) = CH_3COCH_3$,$(Y) = CH_3CH(OH)CH_3$,and $(Z) = CH_3CH(Br)CH_3$.

(B) The reaction with $NaOI/NaOH$ is the iodoform test. Compounds that contain a $CH_3CO-$ group or a $CH_3CH(OH)-$ group give a positive iodoform test,resulting in a yellow solid precipitate of iodoform $(CHI_3)$.
$(A)$ $CH_3-CH(OH)-C_2H_5$ is a secondary alcohol with a $CH_3CH(OH)-$ group,so it gives a positive test.
$(B)$ $CH_3-CH_2-CH_2-OH$ is a primary alcohol that does not contain the required group,so it gives a negative test.
$(C)$ $CH_3-CO-C_2H_5$ is a ketone with a $CH_3CO-$ group,so it gives a positive test.
$(D)$ $CH_3-COOH$ is an acetic acid,which does not give the iodoform test.
$(E)$ $CH_3-CH_2-CHO$ is an aldehyde (propanal) that does not contain the $CH_3CO-$ group,so it gives a negative test.
Therefore,compounds $(A)$ and $(C)$ give a yellow solid.

(B) $1$. The molecular formula $C_6H_{12}O$ corresponds to the general formula $C_nH_{2n}O$,which indicates that the compound $P$ is either an aldehyde or a ketone.
$2$. $P$ gives a positive test with $2,4-$dinitrophenylhydrazine $(2,4-DNP)$,which confirms the presence of a carbonyl group $(C=O)$.
$3$. $P$ gives a negative test with Tollens reagent,which indicates that $P$ is a ketone,not an aldehyde.
$4$. $P$ is optically active,meaning it must contain at least one chiral carbon atom.
$5$. Let's examine the options:
- Option $A$: $CH_3-CO-CH_2-CH_2-CH_2-CH_3$ (Hexan-$2$-one) is not optically active.
- Option $B$: $CH_3-CO-CH(CH_3)-CH_2-CH_3$ ($3-$methylpentan-$2$-one) has a chiral carbon at the $C-3$ position. It is a ketone and gives a positive $2,4-DNP$ test and a negative Tollens test.
- Option $C$: This is an aldehyde,which would give a positive Tollens test.
- Option $D$: $CH_3-CO-CH_2-CH(CH_3)_2$ ($4-$methylpentan-$2$-one) is not optically active.
$6$. Therefore,the correct structure is $3-$methylpentan-$2$-one.

(B) The given compound is $CH_3COCH_2CH_2CH_2CH_2CHO$ (heptane$-2,6-$dione derivative or similar keto-aldehyde).
Intramolecular aldol condensation occurs between the $\alpha$-hydrogen of the ketone and the carbonyl carbon of the aldehyde.
The enolate formed at the $\alpha$-position of the ketone attacks the aldehyde carbonyl carbon to form a five-membered ring.
Dehydration of the resulting $\beta$-hydroxy ketone under heating $( \Delta )$ yields the conjugated enone.
The major product is $1-\text{acetylcyclopentene}$.

(B) The starting material is $3$-benzoylpropanenitrile. It contains both a ketone and a nitrile functional group.
When treated with excess $CH_3MgBr$,the Grignard reagent attacks both the carbonyl carbon of the ketone and the carbon of the nitrile group.
$1$. The ketone group reacts with one equivalent of $CH_3MgBr$ to form a tertiary alcohol after workup.
$2$. The nitrile group reacts with another equivalent of $CH_3MgBr$ to form an imine intermediate,which upon hydrolysis $(H_3O^+)$ yields a ketone.
Therefore,the final product is $5$-hydroxy-$5$-phenylhexan-$2$-one.

(B) The conversion of an ester to an aldehyde requires a selective reducing agent that stops at the aldehyde stage without further reduction to a primary alcohol.
$AlH(iBu)_2$ (also known as $DIBAL-H$) is a selective reducing agent that reduces esters to aldehydes.
$LiAlH_4$ reduces esters directly to primary alcohols.
$NaBH_4$ is generally not reactive enough to reduce esters.
$H_2 / Pd-BaSO_4$ is used for the Rosenmund reduction of acid chlorides to aldehydes.
Therefore,the correct reagent is $i. \ AlH(iBu)_2$,$ii. \ H_2 O$.

(C) Step $1$: Ozonolysis of $2,3-\text{dimethylbut-2-ene}$ $((CH_3)_2C=C(CH_3)_2)$ with $O_3$ followed by $Zn/H_2O$ yields acetone $(CH_3COCH_3)$ as product $A$.
Step $2$: Reaction of acetone $(CH_3COCH_3)$ with $CH_3MgBr$ (Grignard reagent) followed by acidic hydrolysis $(H_2O/H^+)$ results in the formation of $2-\text{methylpropan-2-ol}$ $((CH_3)_3COH)$,which is product $B$.

(B) Benzaldehyde $(C_6H_5CHO)$ does not contain an $\alpha$-hydrogen atom. Therefore,it undergoes the Cannizzaro reaction when treated with a concentrated base like $KOH$.
In the Cannizzaro reaction,one molecule of the aldehyde is oxidized to a carboxylic acid salt (potassium benzoate) and another molecule is reduced to an alcohol (benzyl alcohol).
The reaction is:
$2C_6H_5CHO + KOH_{(conc.)} \rightarrow C_6H_5COOK + C_6H_5CH_2OH$

(A) The iodoform test is given by compounds containing the $CH_3CO-$ group (methyl ketones) or the $CH_3CH(OH)-$ group (secondary alcohols with a methyl group).
$Propan-2-ol$ $(CH_3CH(OH)CH_3)$ contains the $CH_3CH(OH)-$ group,which is oxidized to acetone $(CH_3COCH_3)$ by iodine and alkali,subsequently giving a yellow precipitate of iodoform $(CHI_3)$.
Benzophenone $(C_6H_5COC_6H_5)$ lacks the methyl group attached to the carbonyl carbon.
Methyl acetate and acetamide do not give the iodoform test under standard conditions.

(B) Statement-$I$ is correct: Tollen's reagent $([Ag(NH_3)_2]^+)$ is used to distinguish between aldehydes and ketones. Aldehydes (like acetaldehyde) reduce Tollen's reagent to form a silver mirror,whereas ketones (like acetone) do not react with it.
Statement-$II$ is correct: Fehling's test is specific to aliphatic aldehydes. Aromatic aldehydes,such as benzaldehyde,do not give a positive Fehling's test because the aldehyde group is directly attached to the benzene ring,which is not easily oxidized by the mild Fehling's reagent.

(D) The reaction is a mixed (cross) Aldol condensation between acetaldehyde $(CH_3CHO)$ and propanal $(CH_3CH_2CHO)$.
Both aldehydes have $\alpha$-hydrogens,so four possible aldol products can be formed:
$1$. Self-aldol of $CH_3CHO$: $CH_3-CH=CH-CHO$ (But$-2-$enal)
$2$. Self-aldol of $CH_3CH_2CHO$: $CH_3-CH_2-CH=C(CH_3)-CHO$ ($2$-methylpent$-2-$enal)
$3$. Cross-aldol ($CH_3CHO$ as nucleophile): $CH_3-CH_2-CH=CH-CHO$ (Pent$-2-$enal)
$4$. Cross-aldol ($CH_3CH_2CHO$ as nucleophile): $CH_3-CH=C(CH_3)-CHO$ ($2$-methylbut$-2-$enal)
Option $D$ in the original image represents $CH_3-CH=CH-CH_2-CHO$,which is not an $\alpha,\beta$-unsaturated aldehyde and is not a product of this condensation reaction.

(B) The correct matches are as follows:
$(i)$. Benzophenone $\rightarrow$ Diphenylmethane: This is a Clemmensen reduction,which uses $Zn(Hg) / \text{Conc. } HCl$ to reduce the carbonyl group to a methylene group. Matches with $(c)$.
$(ii)$. Benzaldehyde $\rightarrow$ $1-$Phenylethanol: This is a Grignard reaction where $CH_3MgBr$ attacks the aldehyde,followed by hydrolysis to form a secondary alcohol. Matches with $(d)$.
$(iii)$. Cyclohexanone $\rightarrow$ Cyclohexanol: This is a reduction of a ketone to a secondary alcohol using $LiAlH_4$. Matches with $(a)$.
$(iv)$. Phenyl benzoate $\rightarrow$ Benzaldehyde: This is a partial reduction of an ester to an aldehyde using $DIBAL-H$. Matches with $(b)$.
Therefore,the correct sequence is $i-c, ii-d, iii-a, iv-b$.

(B) The given reaction involves the selective reduction of a ketone group to a secondary alcohol while keeping the ester group $(-COOCH_3)$ intact.
$LiAlH_4$ is a strong reducing agent that reduces both ketones and esters to alcohols.
$NaBH_4$ is a selective reducing agent that reduces aldehydes and ketones to alcohols but does not reduce esters under normal conditions.
$Ni, H_2$ (catalytic hydrogenation) would reduce both the ketone and the ester group.
$Zn-Hg / HCl$ (Clemmensen reduction) reduces the ketone group to a methylene $(-CH_2-)$ group,not an alcohol.
Therefore,$NaBH_4$ is the correct reagent for this selective reduction.

(D) Fehling's solution is a mild oxidizing agent that is reduced by aliphatic aldehydes and formic acid $(HCOOH)$.
Formic acid contains an aldehydic group $(-CHO)$ in its structure,which allows it to act as a reducing agent.
Ketones,such as acetone $(CH_3COCH_3)$,do not contain an aldehydic group and therefore do not reduce Fehling's solution.