Properties Questions in English

(A) The ease of deprotonation depends on the stability of the resulting carbanion. The carbanion is stabilized by the electron-withdrawing effect of the adjacent carbonyl groups.
In compound $b$ (dimethyl malonate),the central carbon is flanked by two ester groups $(-COOCH_3)$. The ester group is less electron-withdrawing than a ketone group due to the $+M$ effect of the oxygen atom.
In compound $a$ (acetylacetone),the central carbon is flanked by two ketone groups $(-COCH_3)$. Ketones are more electron-withdrawing than esters.
In compound $c$ (methyl acetoacetate),the central carbon is flanked by one ketone and one ester group.
The acidity of the $\alpha$-hydrogens follows the order: $a > c > b$.
Therefore,compound $a$ will undergo deprotonation most readily.

(A) Assertion $A$ is true: Acetals and ketals are essentially cyclic or acyclic ethers. Ethers are stable in basic medium because the nucleophilic attack of $OH^-$ on the carbon atom is not favored due to the lack of a good leaving group and the electron-rich nature of the oxygen atom.
Reason $R$ is false: Alkoxide ions $(RO^-)$ are strong bases and are considered poor leaving groups. The stability of acetals/ketals in basic medium is due to the fact that they do not undergo hydrolysis in the presence of bases,not because of the high leaving tendency of the alkoxide ion.

(A) The reaction sequence is as follows:
$1$. Dehydrogenation of heptane $(P)$ using $Cr_2O_3$ at $770 \ K$ and $20 \ atm$ yields toluene.
$2$. Etard reaction of toluene with $CrO_2Cl_2$ followed by hydrolysis gives benzaldehyde.
$3$. Benzaldehyde undergoes Cannizzaro reaction with $NaOH$ to form sodium benzoate and benzyl alcohol.
$4$. Acidification with $H_3O^{+}$ converts sodium benzoate to benzoic acid $(Q)$ and leaves benzyl alcohol $(R)$.

(B) The reaction involves the acid-catalyzed condensation of an aldehyde (pivalaldehyde) with an $\alpha$-hydroxy acid ($2$-hydroxyisobutyric acid).
$1$. The carbonyl oxygen of the aldehyde gets protonated by $H^+$.
$2$. The hydroxyl group of the $\alpha$-hydroxy acid acts as a nucleophile and attacks the electrophilic carbonyl carbon of the protonated aldehyde.
$3$. Subsequently,the carboxylic acid group of the $\alpha$-hydroxy acid reacts with the newly formed hemiacetal hydroxyl group to form a cyclic acetal structure.
$4$. The final product is a cyclic acetal containing a carbonyl group,which corresponds to the structure shown in option $B$.

(C) . Hoffmann Bromamide Degradation uses $Br_2$ and $NaOH$ to convert amides to amines $(A-III)$.
$B$. Clemmensen reduction uses $Zn-Hg / HCl$ to reduce carbonyl groups to methylene groups $(B-IV)$.
$C$. Cannizzaro reaction uses concentrated $KOH$ (or $NaOH$) and heat for aldehydes lacking $\alpha$-hydrogen $(C-I)$.
$D$. Reimer-Tiemann reaction uses $CHCl_3$ and $NaOH$ followed by acid workup to introduce a formyl group into phenol $(D-II)$.
Therefore,the correct match is $A-III, B-IV, C-I, D-II$.

(A) Assertion $A$ is false because the molecule contains an alcohol group which is acid-sensitive. In the presence of $HCl$ (a strong acid),the $-OH$ group will undergo dehydration or substitution,making Clemmensen reduction unsuitable for this specific conversion.
Reason $R$ is true because $Zn-Hg / HCl$ (Clemmensen reduction) is indeed a standard reagent used to reduce carbonyl groups to methylene $(-CH_2-)$ groups in acid-stable compounds.

(B) $1$. Cyclohexylamine $(C_6H_{11}NH_2)$ reacts with nitrous acid $(HNO_2)$ to form cyclohexanol $(P)$ $(C_6H_{11}OH)$.
$2$. Cyclohexanol $(P)$ on oxidation with $PCC$ (Pyridinium chlorochromate) gives cyclohexanone $(Q)$ $(C_6H_{10}O)$.
$3$. Cyclohexanone $(Q)$ undergoes an aldol condensation reaction when heated with dilute $NaOH$ to form the final product $(R)$. The reaction involves the formation of a $\beta$-hydroxy ketone which then undergoes dehydration to form an $\alpha,\beta$-unsaturated ketone. The product $(R)$ is $2$-cyclohexylidenecyclohexanone.

(D) The haloform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
Let us analyze the given molecules:
$1$. $2,4$-dimethylacetophenone: Contains $CH_3CO-$ group attached to an aromatic ring. It gives a positive haloform test.
$2$. Ethyl acetoacetate: Contains $CH_3CO-$ group. It gives a positive haloform test.
$3$. Acetophenone: Contains $CH_3CO-$ group attached to a phenyl ring. It gives a positive haloform test.
$4$. Pentan-$3$-ol: Does not contain $CH_3CH(OH)-$ group. It does not give the haloform test.
$5$. Butan-$2$-ol: Contains $CH_3CH(OH)-$ group. It gives a positive haloform test.
$6$. $3,3,5,5$-tetramethylcyclohexanone: Does not contain the required group. It does not give the haloform test.
Thus,there are $4$ molecules that give a positive haloform test.

(B) The reaction involves the nucleophilic attack of the $-NH_2$ group on the electrophilic carbonyl carbon of diethyl carbonate,followed by the loss of an ethanol molecule to form a carbamate intermediate.
Subsequently,the lone pair of the oxygen atom from the $-CH_2OH$ group attacks the carbonyl carbon of the carbamate,leading to the elimination of another molecule of ethanol and the formation of a cyclic five-membered oxazolidin$-2-$one derivative.

(D) $1$. The reaction of the aldehyde with $NaCN$ and $H_2O$ (cyanohydrin formation) yields the cyanohydrin product $[A]$. The structure of $[A]$ is a $2$-hydroxy-$3$-methylhexanenitrile derivative.
$2$. The reduction of the nitrile group $(-CN)$ in $[A]$ using $LiAlH_4$ produces a primary amine $(-CH_2NH_2)$. Thus, $[B]$ is the corresponding amino-alcohol.
$3$. The acid-catalyzed hydrolysis of the nitrile group $(-CN)$ in $[A]$ using $HCl/H_2O$ and heat $(\Delta)$ converts it into a carboxylic acid group $(-COOH)$. Thus, $[C]$ is the corresponding hydroxy-acid.
$4$. Comparing these structures with the given options, option $D$ correctly represents the structures of $[A]$, $[B]$, and $[C]$.

(B) The reaction sequence is as follows:
$1$. Benzaldehyde reacts with $HCN$ to form benzaldehyde cyanohydrin: $C_6H_5CHO + HCN \rightarrow C_6H_5CH(OH)CN$.
$2$. Acidic hydrolysis of the cyanohydrin yields mandelic acid: $C_6H_5CH(OH)CN + H_3O^+ \rightarrow C_6H_5CH(OH)COOH$.
$3$. Heating mandelic acid with $NaOH$ and $CaO$ (sodalime) leads to decarboxylation. However,the provided reaction sequence and options suggest a reduction pathway. Given the options,the final product is $C_6H_5CH_2OH$ (Benzyl alcohol).

(A) Nucleophilic addition-elimination reactions involve the addition of a nucleophile to the carbonyl carbon followed by the elimination of a water molecule.
In the reaction of $CH_3CHO$ with $NaHSO_3$,the bisulfite ion acts as a nucleophile and adds to the carbonyl carbon to form a bisulfite addition product.
This is a pure nucleophilic addition reaction,not an addition-elimination reaction,as no water molecule is eliminated.
Options $B$,$C$,and $D$ represent the formation of oximes,hydrazones,and imines respectively,which are classic examples of nucleophilic addition-elimination reactions.

(A) The reaction involves a conjugate addition ($1,4$-addition) of an organocopper reagent (Gilman reagent,formed in situ from $MeMgBr$ and $CuI$) to the $\alpha,\beta$-unsaturated ketone.
$1$. The $Me^-$ group attacks the $\beta$-carbon of the $3$-methylcyclohex$-2-$en$-1-$one,resulting in the formation of an enolate intermediate.
$2$. This enolate is then alkylated by $n$-propyl iodide $(nPrI)$ at the $\alpha$-position.
$3$. The final product is $2$-propyl-$3,3$-dimethylcyclohexanone.

(C) Compounds that give a positive iodoform test must contain either a methyl ketone group $(CH_3CO-)$ or a methyl carbinol group $(CH_3CH(OH)-)$.
$(a)$ $1-$Phenylbutan$-2-$one: Structure is $Ph-CH_2-CO-CH_2-CH_3$. It does not contain a $CH_3CO-$ group. (Negative)
$(b)$ $2-$Methylbutan$-2-$ol: Structure is $CH_3-C(OH)(CH_3)-CH_2-CH_3$. It does not contain a $CH_3CH(OH)-$ group. (Negative)
$(c)$ $3-$Methylbutan$-2-$ol: Structure is $CH_3-CH(OH)-CH(CH_3)_2$. It contains a $CH_3CH(OH)-$ group. (Positive)
$(d)$ $1-$Phenylethanol: Structure is $Ph-CH(OH)-CH_3$. It contains a $CH_3CH(OH)-$ group. (Positive)
$(e)$ $3,3-$dimethylbutan$-2-$one: Structure is $CH_3-CO-C(CH_3)_3$. It contains a $CH_3CO-$ group. (Positive)
$(f)$ $1-$Phenylpropan$-2-$ol: Structure is $Ph-CH_2-CH(OH)-CH_3$. It contains a $CH_3CH(OH)-$ group. (Positive)
Thus,the compounds $(c)$,$(d)$,$(e)$,and $(f)$ give a positive iodoform reaction. The total number of such compounds is $4$.

(A) The reaction of $2$-substituted cyclohexanone with $KMnO_4$ involves oxidative cleavage of the ring.
$KMnO_4$ acts as a strong oxidizing agent that cleaves the $C-C$ bond adjacent to the carbonyl group,resulting in the formation of a keto-acid,$R-CO-(CH_2)_4-COOH$ (which is $A$).
Subsequently,the Wolff-Kishner reduction,using $NH_2NH_2$ and $KOH$ followed by $H_3O^+$,reduces the carbonyl group $(C=O)$ to a methylene group $(-CH_2-)$.
Therefore,the final product $B$ is $R-CH_2-(CH_2)_4-COOH$,which is $R-(CH_2)_5-COOH$.

(D) The reactant is $4$-hydroxybutanal,which contains both an aldehyde group $(-CHO)$ and a hydroxyl group $(-OH)$.
$1$ mole of $MeMgBr$ reacts with the acidic hydroxyl group $(-OH)$ to form an alkoxide,releasing methane gas.
Another $1$ mole of $MeMgBr$ performs a nucleophilic attack on the carbonyl carbon of the aldehyde group.
Thus,for $y = 1$ mole of the reactant,$x = 2$ moles of $MeMgBr$ are required.
The ratio $x / y = 2 / 1 = 2$.

(C) Assertion $A$ is false. Wolff-Kishner reduction involves the use of a strong base (like $KOH$ or $EtO^-K^+$) at high temperatures. The substrate $3$-chloro-$2$-methylpentan-$3$-one contains a chlorine atom at the $\alpha$-position relative to the carbonyl group. In the presence of a strong base,this compound undergoes dehydrohalogenation (elimination of $HCl$) to form an $\alpha,\beta$-unsaturated ketone rather than the desired reduction product.
Reason $R$ is true. Wolff-Kishner reduction is indeed a standard method used to reduce carbonyl groups $(C=O)$ to methylene groups $(CH_2)$.

(B) The given reaction involves the treatment of a $\beta$-hydroxy aldehyde with Clemmensen reduction conditions $(Zn(Hg)/HCl, \Delta)$.
$1$. Under these acidic conditions,the aldehyde group $(-CHO)$ is reduced to a methyl group $(-CH_3)$.
$2$. Simultaneously,the $\beta$-hydroxy group $(-OH)$ undergoes dehydration due to the acidic medium and heat,leading to the formation of a stable conjugated alkene.
$3$. The reaction proceeds through the reduction of the aldehyde to a hydrocarbon chain,followed by the elimination of water to form the most stable alkene,which is $C_6H_5-CH=C(CH_3)-C_2H_5$.

(B) The reaction involves the formation of a Grignard reagent from the alkyl bromide part of the molecule: $CH_3COCH_2CH(CH_3)CH_2Br + Mg \rightarrow CH_3COCH_2CH(CH_3)CH_2MgBr$.
This Grignard reagent acts as a nucleophile and attacks the carbonyl group of another molecule of the same reactant (intermolecular reaction).
The nucleophilic carbon of the Grignard reagent attacks the electrophilic carbonyl carbon of another molecule.
After hydrolysis with $H_2O$,the final product '$A$' is formed,which is a dimer containing both a ketone and an alcohol group.

(C) The reaction is an intramolecular aldol condensation of heptane$-2,6-$dione.
$1$. The base $OEt^-$ abstracts an $\alpha$-hydrogen from one of the terminal methyl groups to form an enolate.
$2$. The enolate carbon attacks the other carbonyl carbon,forming a five-membered ring (intramolecular aldol addition).
$3$. This results in a $\beta$-hydroxy ketone intermediate.
$4$. Upon heating $(\Delta)$,the intermediate undergoes dehydration (elimination of water) to form the stable $\alpha,\beta$-unsaturated ketone,which is $3$-methylcyclopent-$2$-en-$1$-one derivative.

(B) The chemical reaction for the oxidation of cyclohexanecarbaldehyde with Tollen's reagent is:
$C_6H_{11}CHO + 2[Ag(NH_3)_2]OH \rightarrow C_6H_{11}COONH_4 + 2Ag + 3NH_3 + H_2O$
Molar mass of cyclohexanecarbaldehyde $(C_7H_{12}O)$ = $(7 \times 12) + (12 \times 1) + 16 = 84 + 12 + 16 = 112 \ g/mol$.
From the stoichiometry,$1 \ mol$ of cyclohexanecarbaldehyde produces $3 \ mol$ of $NH_3$.
Moles of cyclohexanecarbaldehyde = $\frac{131.8 \times 10^3 \ g}{112 \ g/mol} \approx 1176.78 \ mol$.
Moles of $NH_3$ produced = $3 \times 1176.78 = 3530.34 \ mol$.
Mass of $NH_3$ = $3530.34 \ mol \times 17 \ g/mol = 60015.78 \ g \approx 60 \ kg$.

(D) The starting material is ninhydrin ($1,2,3$-indantrione). When it reacts with water $(H_2O)$,the central carbonyl group,which is highly electrophilic due to the electron-withdrawing effect of the two adjacent carbonyl groups,undergoes nucleophilic addition by water to form a gem-diol. This gem-diol is stabilized by intramolecular hydrogen bonding between the hydroxyl groups and the adjacent carbonyl oxygen atoms. The product is $2,2$-dihydroxy-$1,3$-indanedione.

(B) The given reaction uses $Zn-Hg$ and $conc. HCl$,which are the reagents for the Clemmensen reduction.
Clemmensen reduction converts carbonyl groups $(>C=O)$ into methylene groups $(-CH_2-)$.
In the given reactant,there are two ketone groups: one attached to the cyclohexane ring and one attached to the benzene ring.
Both ketone groups are reduced to methylene groups.
Specifically,the acetyl group $(-COCH_3)$ is reduced to an ethyl group $(-CH_2CH_3)$.
Therefore,the product $(A)$ will have two ethyl groups attached to the respective rings.

(A) The given reaction involves the oxidation of an aldehyde group using Tollen's reagent $([Ag(NH_3)_2]^+)$ in a basic medium $(OH^-)$.
Aldehydes are oxidized to carboxylate ions $(COO^-)$ by Tollen's reagent,while ketones remain unaffected under these conditions.
In the given reactant,$2$-acetylbenzaldehyde,the aldehyde group $(-CHO)$ is oxidized to a carboxylate group $(-COO^-)$,while the acetyl group $(-COCH_3)$ remains unchanged.
Thus,the major product is the $2-$acetylbenzoate ion.

(A) Step $1$: Hydration of styrene $(C_6H_5CH=CH_2)$ with $H_2O, H^+$ follows Markovnikov's rule to yield $1$-phenylethanol $(C_6H_5CH(OH)CH_3)$.
Step $2$: Oxidation of $1$-phenylethanol with $CrO_3$ yields acetophenone $(C_6H_5COCH_3)$.
Step $3$: Wolff-Kishner reduction of acetophenone using $H_2N-NH_2, KOH$ and heating reduces the carbonyl group to a methylene group,yielding ethylbenzene $(C_6H_5CH_2CH_3)$ as the final product $A$.

(A) Fehling's test is given by aliphatic aldehydes only.
$1$. Benzaldehyde: Aromatic aldehyde (Negative)
$2$. Acetaldehyde $(CH_3CHO)$: Aliphatic aldehyde (Positive)
$3$. Acetone: Ketone (Negative)
$4$. Acetophenone: Aromatic ketone (Negative)
$5$. Methanal $(HCHO)$: Aliphatic aldehyde (Positive)
$6$. $4$-nitrobenzaldehyde: Aromatic aldehyde (Negative)
$7$. Cyclohexanecarbaldehyde: Aliphatic aldehyde (Positive)
Thus,the compounds that give a positive Fehling's test are Acetaldehyde,Methanal,and Cyclohexanecarbaldehyde.
The total number of such compounds is $3$.

(D) Step $A$: $DIBAL-H$ is used for the selective reduction of the ester group to an aldehyde group without affecting the existing aldehyde or hydroxyl groups.
Step $B$: $NaOH_{(alc)}$ is used for the intramolecular aldol condensation of the dialdehyde to form the cyclic $\alpha,\beta$-unsaturated aldehyde.
Step $C$: $Zn/HCl$ (Clemmensen reduction) is used to reduce the aldehyde group to a methyl group.
Therefore,the correct reagents are $A=DIBAL-H, B=NaOH_{(alc)}, C=Zn/HCl$.

(C) The reaction between ethanal $(CH_3CHO)$ and semicarbazide $(NH_2NHCONH_2)$ is a nucleophilic addition-elimination reaction.
The reaction is as follows:
$CH_3CHO + NH_2NHCONH_2 \rightarrow CH_3CH=NNHCONH_2 + H_2O$
In the product,ethanal semicarbazone $(CH_3CH=NNHCONH_2)$,we can count the nitrogen atoms:
There is one $N$ atom in the $C=N$ group,one $N$ atom in the $NH$ group,and one $N$ atom in the $NH_2$ group.
Total number of nitrogen atoms = $1 + 1 + 1 = 3$.

(B) $m$-chlorobenzaldehyde does not have any $\alpha$-hydrogen atom. Therefore,it undergoes the Cannizzaro reaction in the presence of a concentrated base like $50 \% KOH$. In this reaction,one molecule of the aldehyde is oxidized to the corresponding carboxylate ion ($m$-chlorobenzoate) and another molecule is reduced to the corresponding alcohol ($m$-chlorobenzyl alcohol). The reaction is as follows:
$2 \text{ } m\text{-Cl-C}_6\text{H}_4\text{CHO} + KOH$ $\rightarrow m\text{-Cl-C}_6\text{H}_4\text{COO}^- \text{K}^+ + m\text{-Cl-C}_6\text{H}_4\text{CH}_2\text{OH}$

(B) The correct matches are as follows:
$(A)$ $CH_3(CH_2)_5COOC_2H_5 \xrightarrow{DIBAL-H, H_2O} CH_3(CH_2)_5CHO$. This is a partial reduction of an ester to an aldehyde using $DIBAL-H$. Thus,$A-(IV)$.
$(B)$ $C_6H_5COC_6H_5 \xrightarrow{Zn(Hg) \& conc. HCl} C_6H_5CH_2C_6H_5$. This is a Clemmensen reduction of a ketone to an alkane. Thus,$B-(II)$.
$(C)$ $C_6H_5CHO \xrightarrow[H_3O^+]{CH_3MgBr} C_6H_5CH(OH)CH_3$. This is a Grignard reaction where an aldehyde is converted to a secondary alcohol. Thus,$C-(I)$.
$(D)$ $CH_3COCH_2COOC_2H_5 \xrightarrow{NaBH_4, H^+} CH_3CH(OH)CH_2COOC_2H_5$. $NaBH_4$ selectively reduces the ketone group to an alcohol in the presence of an ester. Thus,$D-(III)$.
Therefore,the correct sequence is $A-(IV), B-(II), C-(I), D-(III)$.

(D) Statement $I$ is correct: The presence of acidic $\alpha$-hydrogens in aldehydes and ketones allows them to form enolate ions in the presence of a base,which is the key step in the Aldol reaction.
Statement $II$ is incorrect: Benzaldehyde (which lacks $\alpha$-hydrogens) and ethanal (which has $\alpha$-hydrogens) undergo a Claisen-Schmidt condensation,which is a type of Cross-Aldol reaction,to form cinnamaldehyde $(C_6H_5CH=CHCHO)$.
Therefore,Statement $I$ is correct but Statement $II$ is incorrect.

(D) The given reaction is a Clemmensen reduction,which uses $Zn-Hg$ and $HCl$ to reduce carbonyl groups (aldehydes and ketones) to methylene $(-CH_2-)$ groups.
The reactant is $3$-acetylbenzaldehyde,which contains both an aldehyde group $(-CHO)$ and a ketone group $(-COCH_3)$ attached to a benzene ring.
Clemmensen reduction reduces both the aldehyde and the ketone groups to their corresponding alkyl groups.
Therefore,the aldehyde group $(-CHO)$ is reduced to a methyl group $(-CH_3)$ and the ketone group $(-COCH_3)$ is reduced to an ethyl group $(-CH_2CH_3)$.
The final product is $1$-ethyl-$3$-methylbenzene.

(B) Let us analyze each reaction:
$(A)$ Friedel-Crafts acylation of benzene with benzoyl chloride in the presence of anhydrous $AlCl_3$ yields benzophenone $(C_6H_5COC_6H_5)$,not diphenylmethane. Thus,this reaction is incorrect.
$(B)$ Rosenmund reduction of benzoyl chloride with $H_2$ in the presence of $Pd-BaSO_4$ yields benzaldehyde $(C_6H_5CHO)$,not benzoic acid. Thus,this reaction is incorrect.
$(C)$ Gattermann-Koch reaction of benzene with $CO$ and $HCl$ in the presence of anhydrous $AlCl_3/CuCl$ yields benzaldehyde $(C_6H_5CHO)$. This reaction is correct.
$(D)$ Acidic hydrolysis of benzamide $(C_6H_5CONH_2)$ yields benzoic acid $(C_6H_5COOH)$ and ammonium ions,not aniline. Thus,this reaction is incorrect.
Therefore,only one reaction $(C)$ is correct. The number of correct reactions is $1$.

(A) $1$. The first step involves the oxidative cleavage of $1$-methylcyclohexene using ozone $(O_3)$ followed by reductive workup with $Zn / H_2O$. This reaction is known as ozonolysis,which breaks the double bond to form a dicarbonyl compound ($6$-oxoheptanal). Thus,"$A$" is $O_3, Zn / H_2O$.
$2$. The second step involves the reaction of the resulting ketone group with $NaOH$ and $I_2$. This is the iodoform test,which specifically reacts with methyl ketones to produce a carboxylate salt and iodoform $(CHI_3)$. Therefore,"$B$" represents the reagents for the iodoform test,which are $NaOH_{(aq)} / I_2$ (often written as $NaOH / I_2$).
$3$. Comparing these with the given options,option $A$ is the correct choice.

(B) The given reaction is the Wolff-Kishner reduction.
In this reaction,a carbonyl group $(C=O)$ of an aldehyde or ketone is reduced to a methylene group $(CH_2)$ using hydrazine $(N_2H_4)$ followed by heating with a strong base like $KOH$ in a high-boiling solvent such as ethylene glycol.
The starting material is butan$-2-$one $(CH_3-CO-CH_2-CH_3)$.
Upon treatment with $N_2H_4$ and ethylene glycol/$KOH$,the ketone group is reduced to a methylene group,yielding butane $(CH_3-CH_2-CH_2-CH_3)$.

(A) The reaction sequence is as follows:
$1$. Benzene reacts with succinic anhydride in the presence of anhydrous $AlCl_3$ (Friedel-Crafts acylation) to form $A$,which is $4\text{-oxo-}4\text{-phenylbutanoic acid}$ (also known as $\beta\text{-benzoylpropionic acid}$).
$2$. $A$ undergoes Clemmensen reduction $(Zn-Hg/HCl)$ to reduce the ketone group to a methylene group,forming $B$,which is $4\text{-phenylbutanoic acid}$.
$3$. $4\text{-phenylbutanoic acid}$ then undergoes intramolecular Friedel-Crafts acylation in the presence of $H^+$ to form $\alpha\text{-tetralone}$.

(B) The chemical equation for the Claisen-Schmidt reaction is:
$2C_6H_5CHO + CH_3COCH_3 \xrightarrow{NaOH} C_6H_5CH=CHCOCH=CHC_6H_5 + 2H_2O$
From the stoichiometry,$1 \ mole$ of acetone reacts with $2 \ moles$ of benzaldehyde to produce $1 \ mole$ of dibenzalacetone.
Molar mass of dibenzalacetone $(C_{17}H_{14}O) = (17 \times 12) + (14 \times 1) + 16 = 204 + 14 + 16 = 234 \ g/mol$.
Number of moles of dibenzalacetone produced $= \frac{351 \ g}{234 \ g/mol} = 1.5 \ moles$.
According to the stoichiometry,$1 \ mole$ of dibenzalacetone requires $2 \ moles$ of benzaldehyde.
Therefore,$1.5 \ moles$ of dibenzalacetone requires $1.5 \times 2 = 3 \ moles$ of benzaldehyde.
Molar mass of benzaldehyde $(C_6H_5CHO) = (7 \times 12) + (6 \times 1) + 16 = 84 + 6 + 16 = 106 \ g/mol$.
Mass of benzaldehyde required $= 3 \ moles \times 106 \ g/mol = 318 \ g$.

(A) The reactivity of carbonyl compounds towards nucleophilic addition reactions depends on two factors:
$1$. Steric hindrance: Smaller groups around the carbonyl carbon facilitate the attack of the nucleophile.
$2$. Electronic effect: Electron-donating groups (like alkyl groups) decrease the electrophilicity of the carbonyl carbon by donating electron density,thereby reducing reactivity.
$HCHO$ (formaldehyde) has the smallest hydrogen atoms attached to the carbonyl carbon,resulting in the least steric hindrance and the highest electrophilicity at the carbonyl carbon.
Therefore,$HCHO$ is the most reactive towards nucleophilic addition reactions.

(A) The reaction between two moles of benzaldehyde $(C_6H_5CHO)$ and one mole of acetone $(CH_3COCH_3)$ in the presence of aqueous $NaOH$ and heat is a Claisen-Schmidt condensation reaction.
The product $x$ formed is dibenzylideneacetone,which has the structure: $C_6H_5-CH=CH-CO-CH=CH-C_6H_5$.
To calculate the number of $\pi$ bonds in dibenzylideneacetone:
$1$. Each phenyl ring $(C_6H_5)$ contains $3$ $\pi$ bonds.
$2$. There are two phenyl rings,so $2 \times 3 = 6$ $\pi$ bonds.
$3$. There are two $C=C$ double bonds,contributing $2$ $\pi$ bonds.
$4$. There is one $C=O$ double bond,contributing $1$ $\pi$ bond.
Total $\pi$ bonds = $6 + 2 + 1 = 9$.

(D) The reaction sequence is as follows:
$1$. Friedel-Crafts acylation of benzene with succinic anhydride in the presence of $AlCl_3$ gives $4$-oxo-$4$-phenylbutanoic acid $(A)$.
$2$. Clemmensen reduction of $A$ using $Zn-Hg/HCl$ reduces the keto group to a methylene group,yielding $4$-phenylbutanoic acid $(B)$.
$3$. Intramolecular Friedel-Crafts acylation of $B$ in the presence of $conc. H_2SO_4$ leads to the formation of $\alpha$-tetralone $(C)$,which is $1,2,3,4$-tetrahydronaphthalen-$1$-one.

(A) The silver mirror test (Tollens' test) is given by aldehydes and $\alpha$-hydroxy ketones.
Among the given compounds:
$1$. Formic acid $(HCOOH)$ contains an aldehydic hydrogen atom attached to the carbonyl group,which allows it to be oxidized to $CO_2$ and $H_2O$ by Tollens' reagent,thus giving a positive silver mirror test.
$2$. Formaldehyde $(HCHO)$ is an aldehyde and gives a positive silver mirror test.
$3$. Benzaldehyde $(C_6H_5CHO)$ is an aromatic aldehyde and gives a positive silver mirror test.
$4$. Acetone $(CH_3COCH_3)$ is a ketone and does not give a silver mirror test.
Therefore,compounds $A$,$B$,and $C$ will give a silver mirror test.

(D) Fehling's solution '$A$' consists of aqueous copper sulphate $(CuSO_4 \cdot 5H_2O)$.
Fehling's solution '$B$' consists of an alkaline solution of sodium potassium tartrate,also known as Rochelle's salt.

(B) The correct matches are as follows:
$A$. The reaction is the reductive ozonolysis of an alkene,which breaks the double bond to form two carbonyl compounds. Thus,$A-IV$.
$B$. The reaction of benzene with benzoyl chloride in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts acylation reaction. Thus,$B-I$.
$C$. The oxidation of secondary alcohols to ketones is typically performed using $CrO_3$. Thus,$C-II$.
$D$. The oxidation of alkylbenzenes with $KMnO_4 / KOH, \Delta$ results in the formation of potassium benzoate. Thus,$D-III$.
Therefore,the correct matching is $A-IV, B-I, C-II, D-III$.

(B) The reaction of $2$-formylbenzoic acid (phthalaldehyde acid) with concentrated $NaOH$ is a classic example of the Cannizzaro reaction.
Since the molecule contains both an aldehyde group $(-CHO)$ and a carboxylic acid group $(-COOH)$ in an ortho-position,the aldehyde group undergoes an intramolecular Cannizzaro reaction.
The aldehyde group is oxidized to a carboxylate ion $(-COO^-)$ and reduced to a primary alcohol group $(-CH_2OH)$.
This results in the formation of a hydroxy-acid intermediate,$2$-hydroxymethylbenzoic acid.
Upon acidification,the carboxylic acid group and the alcohol group undergo intramolecular esterification (cyclization) to form the stable five-membered lactone,phthalide.

(B) The smallest ketone is acetone $(CH_3COCH_3)$. Its reaction with $NH_2OH$ gives acetone oxime,which is $(CH_3)_2C=NOH$. This molecule has identical groups on the carbon atom,so it does not show geometrical isomerism.
The next homologue is butan$-2-$one $(CH_3COCH_2CH_3)$. Its reaction with $NH_2OH$ gives butan$-2-$one oxime,which is $CH_3(C_2H_5)C=NOH$. This molecule has different groups on the carbon atom,so it exhibits geometrical isomerism,resulting in two geometrical isomers ($syn$ and $anti$ forms).
Thus,in total,there are $1 + 2 = 3$ different oximes formed.

(A) $1$. Ozonolysis of cyclohexene followed by reductive workup $(Zn/H_2O)$ yields hexane$-1,6-$dial (compound $E$).
$2$. Hexane$-1,6-$dial contains two aldehyde groups and $\alpha$-hydrogens,which undergo intramolecular aldol condensation in the presence of aqueous $KOH$.
$3$. The reaction involves the formation of a five-membered ring with an $\alpha,\beta$-unsaturated aldehyde group,resulting in cyclopent$-1-$ene$-1-$carbaldehyde (compound $F$).
$4$. Thus,the correct option is $(A)$.

(D) . $C_6H_5CHO$ (benzaldehyde) is an aldehyde,so it reacts with $2, 4-DNP$ to give a precipitate $(p)$ and acts as an electrophile (not a nucleophile). It does not form cyanohydrin directly without $CN^-$,but $CN^-$ is the nucleophile involved in that reaction. Wait,let us re-evaluate:
$A$. $C_6H_5CHO$: Reacts with $2, 4-DNP$ $(p)$.
$B$. $CH_3C \equiv CH$: Terminal alkyne,reacts with ammoniacal $AgNO_3$ to give white precipitate $(q)$.
$C$. $CN^-$: Is a nucleophile $(r)$ and is involved in cyanohydrin formation $(s)$. It also reacts with $AgNO_3$ to give $AgCN$ precipitate $(q)$.
$D$. $I^-$: Is a nucleophile $(r)$ and reacts with $AgNO_3$ to give $AgI$ precipitate $(q)$.
Correct matches:
$A: p$
$B: q$
$C: q, r, s$
$D: q, r$
Thus,option $D$ is the correct match.

(A) The starting material is $3-oxo-3-phenylpropanoic$ acid with a $^{13}C$ label on the methylene carbon.
Upon heating,it undergoes decarboxylation to form acetophenone $(E)$ where the methyl group is $^{13}C$ labelled: $Ph-CO-CH_2^*COOH \xrightarrow{\Delta} Ph-CO-CH_3^* + CO_2$.
Acetophenone $(E)$ then reacts with $I_2$ in the presence of $NaOH$ (haloform reaction) to produce sodium benzoate $(F)$ and iodoform $(G)$: $Ph-CO-CH_3^* + 3I_2 + 4NaOH \rightarrow Ph-COONa + CHI_3^* + 3NaI + 3H_2O$.
Thus,the $^{13}C$ label is present in the methyl group of $E$ and in the iodoform $(G)$.

(C) Step $1$: The reaction of $PhCOCH_3$ (acetophenone) with $PhCH_2MgBr$ (benzylmagnesium bromide) yields the tertiary alcohol $H$ $(Ph-C(OH)(CH_3)(CH_2Ph))$.
Step $2$: Acid-catalysed dehydration of $H$ gives the alkene $I$ $(Ph-C(CH_3)=CH-Ph)$.
Step $3$: Ozonolysis of $I$ gives $J$ ($PhCHO$,benzaldehyde) and $K$ ($PhCOCH_3$,acetophenone).
Step $4$: $J$ $(PhCHO)$ undergoes Cannizzaro reaction with $KOH$ to give benzyl alcohol $(PhCH_2OH)$ and $L$ $(PhCOO^{-}K^{+})$.
Step $5$: $K$ $(PhCOCH_3)$ undergoes aldol condensation with $KOH$ to give $M$.
Thus,$H$ is formed from $PhCOCH_3 + PhCH_2MgBr$ (Option $B$),$I$ is the structure shown in option $C$,and $J, K, L$ are $PhCHO, PhCOCH_3, PhCOO^{-}K^{+}$ (Option $D$). The correct sequence is $(B, C, D)$. However,based on the provided options,the closest match is $(B, C, B)$ if $L$ is interpreted differently or $(B, A, D)$ if $I$ is $A$. Re-evaluating: $H$ is $PhCOCH_3 + PhCH_2MgBr$ $(B)$,$I$ is $A$,$J, K, L$ are $PhCHO, PhCOCH_3, PhCOO^-K^+$ $(D)$. Thus,$(B, A, D)$ is the correct sequence.

(B) $1$. Compound $P$ must be $4-$phenylbut$-3-$en$-2-$one (structure $B$ in the first set) because it contains a methyl ketone group (gives positive iodoform test) and has the correct carbon skeleton to form the subsequent products.
$2$. Reaction of $P$ with $MeMgBr$ followed by dehydration gives $Q$ ($1$,$1$-dimethyl-1H-indene,structure $A$ in the second set). Ozonolysis of $Q$ gives $R$ ($2$-($2$-acetylphenyl)$-2-$methylpropanal,structure $B$ in the second set).
$3$. Intramolecular aldol condensation of $R$ gives $S$ ($3$,$3$-dimethyl$-3,4-$dihydronaphthalen-$1$(2H)-one,structure $B$ in the third set).
Thus,the correct sequence is $P=B, Q=A, R=B, S=B$. The answer is $(B, A, B)$.