Properties Questions in English

(A) The Cannizzaro reaction is a disproportionation reaction given by aldehydes that do not possess an $\alpha$-hydrogen atom.
$HCHO$,$Me_{3}CCHO$ (pivalaldehyde),and $C_{6}H_{5}CHO$ (benzaldehyde) all lack $\alpha$-hydrogens and thus undergo the Cannizzaro reaction.
$Cl_{3}CCHO$ (chloral) also lacks an $\alpha$-hydrogen,but it does not undergo the standard Cannizzaro reaction.
Instead,when treated with alkali,it undergoes a haloform-type cleavage because the $CCl_{3}^{-}$ group is an excellent leaving group due to the strong electron-withdrawing effect of the three chlorine atoms,which stabilizes the resulting carbanion.
Therefore,$Cl_{3}CCHO$ reacts with alkali to produce chloroform $(CHCl_{3})$ and formate $(HCOO^-)$ rather than undergoing the typical disproportionation of the Cannizzaro reaction.

(C) The given reaction is an example of an intramolecular Cannizzaro reaction. In this reaction,phthalaldehyde reacts with a base $(OH^-)$ to undergo disproportionation,where one aldehyde group is reduced to a primary alcohol $(-CH_2OH)$ and the other is oxidized to a carboxylate ion $(-COO^-)$. The final product is the ortho-hydroxymethylbenzoate ion.

(C) Aldehydes lacking $\alpha-H$ atoms undergo the Cannizzaro reaction when treated with concentrated alkali $(NaOH)$. In this reaction,one part of the molecule is reduced to an alcohol,and the other part is oxidized to a salt of a carboxylic acid.
Glyoxal $(CHO-CHO)$ lacks $\alpha-H$ atoms and undergoes an intramolecular Cannizzaro reaction.
Upon treatment with $NaOH$ followed by acidification with $H^+$,one aldehyde group is reduced to a $-CH_2OH$ group,and the other is oxidized to a $-CO_2H$ group,resulting in glycolic acid $(CH_2OH-CO_2H)$.

(A) The Wittig reaction involves the reaction of a phosphonium ylide with a carbonyl compound to form an alkene.
$1$. The phosphonium salt reacts with a strong base like $n-BuLi$ to form the phosphorus ylide.
$2$. The ylide then undergoes a $[2+2]$ cycloaddition with the carbonyl compound (formaldehyde,$CH_2=O$) to form a four-membered ring intermediate known as an oxaphosphetane.
$3$. This oxaphosphetane intermediate $J$ subsequently decomposes to form the alkene and triphenylphosphine oxide.
$4$. The structure of the oxaphosphetane intermediate $J$ is a four-membered ring containing carbon,oxygen,and phosphorus atoms,where the carbon is also attached to the cyclopentyl group. Option $A$ correctly represents this oxaphosphetane structure.

(A) The reaction of $CH_3CHO$ (acetaldehyde) with $Al(OEt)_3$ (aluminum ethoxide) is known as the Tishchenko reaction.
This reaction involves the disproportionation of two molecules of an aldehyde to form an ester.
Specifically,$2CH_3CHO \xrightarrow{Al(OEt)_3} CH_3COOCH_2CH_3$ (ethyl acetate).
Therefore,the product formed is ethyl acetate $(CH_3COOCH_2CH_3)$.

(C) When acetone $(CH_3COCH_3)$ is saturated with dry $HCl$ gas,it undergoes self-aldol condensation followed by dehydration to form mesityl oxide $(CH_3COCH=C(CH_3)_2)$.
The reaction is as follows:
$2CH_3COCH_3 \xrightarrow{HCl(g)} CH_3COCH=C(CH_3)_2 + H_2O$
Note: Phorone is formed as a minor product,but the major product is mesityl oxide.

(A) The reaction between formaldehyde $(HCHO)$ and ammonia $(NH_3)$ results in the formation of hexamethylene tetramine,also known as urotropine.
The balanced chemical equation is:
$6 HCHO + 4 NH_3 \rightarrow (CH_2)_6 N_4 + 6 H_2 O$

(B) The iodoform reaction of acetone involves the sequential replacement of hydrogen atoms on the alpha-carbon by iodine atoms in the presence of a base $(OI^-)$.
Step $1$: $CH_3COCH_3 + OI^- \rightarrow CH_3COCH_2I + OH^-$
Step $2$: $CH_3COCH_2I + OI^- \rightarrow CH_3COCH_2I + OH^-$
Step $3$: $CH_3COCH_2I + OI^- \rightarrow CH_3COCI_3 + OH^-$
Step $4$: $CH_3COCI_3 + OH^- \rightarrow CH_3COO^- + CHI_3$
Comparing these intermediates with the given options,$ICH_2COCH_2I$ is not formed during the reaction as the iodination occurs specifically on the methyl group that is adjacent to the carbonyl group.

(D) $2$-pentanone $(CH_3COCH_2CH_2CH_3)$ contains a $CH_3CO-$ group,which makes it capable of undergoing the iodoform test with $I_2 / NaOH$.
$3$-pentanone $(CH_3CH_2COCH_2CH_3)$ does not contain a $CH_3CO-$ group and therefore does not give the iodoform test.
Since the iodoform test is not listed among the options $(A)$,$(B)$,or $(C)$,the correct choice is $(D)$.

(C) $2,2,2-$trichloroacetophenone reacts with aqueous $KOH$ via a haloform-like cleavage reaction. The hydroxide ion attacks the carbonyl carbon,followed by the departure of the stable $CHCl_3$ (chloroform) leaving group to form the potassium salt of benzoic acid. Subsequent acidification with $H_3O^+$ yields benzoic acid as the final product $P$.
Thus,option $(c)$ is the correct answer.

(A) In the first step,acetone $(CH_3COCH_3)$ reacts with $Br_2$ in an acidic medium $(AcOH)$. Acidic halogenation of ketones typically results in monohalogenation at the $\alpha$-carbon,yielding $X = CH_3COCH_2Br$.
In the second step,the monobromoacetone $(CH_3COCH_2Br)$ reacts with $Br_2$ in a basic medium $(NaOH)$. This is a haloform reaction. The base promotes further halogenation of the $\alpha$-carbon followed by cleavage to form bromoform $(CHBr_3)$ and the corresponding carboxylate salt,which is sodium acetate $(CH_3CO_2Na)$.

(C) The reaction proceeds in two steps:
$1$. The Grignard reagent $CH_3MgBr$ acts as a nucleophile and attacks the carbonyl carbon of $5-chloropentan-2-one$. This forms an alkoxide intermediate.
$2$. Upon workup with $H_3O^+$,the alkoxide oxygen is protonated to form an alcohol. However,in the presence of the internal chloride group,the alkoxide oxygen performs an intramolecular nucleophilic substitution $(S_N2)$ reaction,displacing the chloride ion to form a five-membered cyclic ether,$2,2-dimethyltetrahydrofuran$.

(A, B, C) The given reaction is the reduction of a ketone (specifically $4$-methylcyclohexanone) to an alkane ($1$-methylcyclohexane). This involves the reduction of the carbonyl group $(-C=O)$ to a methylene group $(-CH_2-)$.
$A$. $Zn-Hg / \text{Conc. } HCl$ is the reagent for Clemmensen reduction.
$B$. $i. H_2NNH_2, ii. NaOH \text{ in ethylene glycol, } \Delta$ is the reagent for Wolff-Kishner reduction.
$C$. $i. HSCH_2CH_2SH / H^{\oplus}, ii. H_2 / Ni$ is the Mozingo reduction method.
All three methods $(A, B, C)$ are standard chemical reactions used to reduce a carbonyl group to a methylene group. Therefore,all these reagents can carry out the given conversion.

(A, B) The haloform reaction is given by compounds containing the $CH_{3}CO-$ group or compounds that can be oxidized to this group (like $CH_{3}CH(OH)-$ derivatives).
$1$. $ICH_{2}COCH_{2}Ph$: This compound contains an $\alpha$-hydrogen at the $ICH_{2}$ position. Under basic conditions with $I_{2}$,the $ICH_{2}$ group can be further iodinated to $CI_{3}CO-$,which then undergoes cleavage to form iodoform $(CHI_{3})$.
$2$. $PhCOCH(OH)CH_{3}$: This is a secondary alcohol with a $CH_{3}CH(OH)-$ group. It is oxidized by $I_{2}/KOH$ to the corresponding ketone $PhCOCOCH_{3}$,which contains the $CH_{3}CO-$ group and thus gives the haloform reaction.
Therefore,both compounds $(a)$ and $(b)$ respond to the haloform reaction.

(B) The reaction of benzene with $4$-oxopentanoyl chloride in the presence of anhydrous $AlCl_3$ (Friedel-Crafts acylation) gives compound $(P)$,which is $1$-phenylpentane-$1,4$-dione.
Intramolecular aldol condensation of $(P)$ in the presence of aqueous $NaOH$ and heat gives compound $(Q)$,which is $3$-phenylcyclopent-$2$-en-$1$-one.
The molecular formula of $(Q)$ is $C_{11}H_{10}O$.
The molecular mass of $(Q) = (11 \times 12) + (10 \times 1) + (1 \times 16) = 132 + 10 + 16 = 158 \ g/mol$.
The percentage of oxygen in $(Q) = \frac{\text{Mass of oxygen}}{\text{Total molecular mass}} \times 100 = \frac{16}{158} \times 100 \approx 10.12 \%$.
The nearest integer value is $10 \%$.

(C) Statement $A$ is true: $NaOCl + KI \rightarrow NaCl + KOI$.
Statement $B$ is false: $KOI$ is an oxidizing agent,not a reducing agent.
Statement $C$ is true: $\alpha, \beta$-unsaturated methylketones (like $CH_3-CH=CH-CO-CH_3$) contain the $CH_3CO-$ group and thus give the iodoform reaction.
Statement $D$ is false: Isopropyl alcohol $(CH_3CH(OH)CH_3)$ contains the $CH_3CH(OH)-$ group and gives a positive iodoform test.
Statement $E$ is false: Methanoic acid $(HCOOH)$ does not contain the $CH_3CO-$ or $CH_3CH(OH)-$ group and does not give the iodoform test.
Therefore,only statements $A$ and $C$ are true.

(A) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$A$. Methanol $(CH_3OH)$ does not give the test,while Ethanol $(CH_3CH_2OH)$ gives the test. Thus,they can be differentiated.
$B$. Both $CH_3COOH$ and $CH_3CH_2COOH$ do not contain the required group,so they cannot be differentiated.
$C$. Cyclohexene does not give the test,while cyclohexanone (if it had a methyl group at the alpha position) would,but simple cyclohexanone does not have the $CH_3CO-$ group. However,in many contexts,this is considered a differentiation between a ketone and an alkene,but strictly speaking,cyclohexanone does not give a positive iodoform test.
$D$. Diethyl ether does not give the test,while Pentan$-3-$one $(CH_3CH_2COCH_2CH_3)$ does not contain the $CH_3CO-$ group and thus does not give the test.
$E$. Anisole $(C_6H_5OCH_3)$ does not give the test,while acetone $(CH_3COCH_3)$ gives the test. Thus,they can be differentiated.
Therefore,pairs $A$ and $E$ can be differentiated by the iodoform test.

(A) Statement $I$ is false because cross aldol condensation between two different aldehydes can produce a mixture of $2$ or $4$ products depending on whether both aldehydes possess $\alpha$-hydrogens.
Statement $II$ is false because both benzaldehyde (an aldehyde) and acetophenone (a ketone) react with semicarbazide to form their respective semicarbazones under optimum $pH$ conditions. Benzaldehyde is generally more reactive towards nucleophilic addition than acetophenone due to less steric hindrance and electronic factors.

(D) The acidic hydrolysis of the given unsaturated ether yields '$P$' (propanal, $CH_3CH_2CHO$) and '$Q$' (acetone, $CH_3COCH_3$).
'$P$' is an aldehyde, while '$Q$' is a ketone.
Fehling's solution is a mild oxidizing agent that reacts with aliphatic aldehydes to form a red precipitate of $Cu_2O$, but it does not react with ketones.
Therefore, Fehling's solution can be used to distinguish between '$P$' and '$Q$'.
'$P$' gives a positive Fehling's test, while '$Q$' gives a negative Fehling's test.

(D) $1$. Propanenitrile $(CH_3CH_2CN)$ on reduction with $SnCl_2/HCl$ followed by hydrolysis gives propanal $(CH_3CH_2CHO)$,which is $x$.
$2$. But$-2-$ene $(CH_3CH=CHCH_3)$ on ozonolysis followed by hydrolysis gives two moles of ethanal $(CH_3CHO)$,which is $y$.
$3$. The reaction between propanal $(x)$ and ethanal $(y)$ in the presence of alkali $(NaOH)$ and heat is a crossed aldol condensation reaction.
$4$. The possible products are:
- Self-aldol of propanal: $CH_3CH_2CH(OH)CH(CH_3)CHO \xrightarrow{\Delta} CH_3CH_2CH=C(CH_3)CHO$ ($2-$Methylpent$-2-$enal).
- Self-aldol of ethanal: $CH_3CH(OH)CH_2CHO \xrightarrow{\Delta} CH_3CH=CHCHO$ (But$-2-$enal).
- Crossed-aldol ($CH_3CHO$ as nucleophile): $CH_3CH_2CH(OH)CH_2CHO \xrightarrow{\Delta} CH_3CH_2CH=CHCHO$ (Pent$-2-$enal).
- Crossed-aldol ($CH_3CH_2CHO$ as nucleophile): $CH_3CH(OH)CH(CH_3)CHO \xrightarrow{\Delta} CH_3CH=C(CH_3)CHO$ ($2-$Methylbut$-2-$enal).
$5$. $3-$Methylbut$-2-$enal is not formed in this reaction.

(A) The structures are: $P$ ($1$-phenylprop$-2-$en$-1-$ol),$Q$ ($3$-phenylpropanal),$R$ ($1$-phenylpropan$-2-$one),$S$ ($1$-phenylpropan$-1-$one).
$A$. $2,4-DNP$ reacts with aldehydes and ketones. $Q$ (aldehyde),$R$ (ketone),and $S$ (ketone) give the test. Statement $A$ is correct.
$B$. Bayer's test is for unsaturation. Only $P$ has a double bond. Statement $B$ is incorrect.
$C$. Aromatic compounds give a sooty flame. All four compounds contain a phenyl ring,so all give a sooty flame. Statement $C$ is technically correct,but we evaluate based on the options provided.
$D$. $I_2/NaOH$ (Iodoform test) requires a $CH_3CO-$ group or $CH_3CH(OH)-$ group. $R$ has $CH_3CO-$ and $S$ does not. Statement $D$ is incorrect.
$E$. Tollen's reagent is for aldehydes. Only $Q$ is an aldehyde. Statement $E$ is correct.
Combining the correct statements $A, C, E$,the correct option is $A$.

(C) Tollen's reagent is an oxidizing agent that reacts with aldehydes to form a silver mirror.
Compounds that exist in equilibrium with an open-chain aldehyde form will give a positive Tollen's test.
Hemiacetals are in equilibrium with their open-chain aldehyde form in the presence of a base (present in Tollen's reagent).
Compound $C$ is a cyclic hemiacetal,which can open to form an aldehyde,thus giving a positive Tollen's test.
Compounds $A$ and $B$ are acetals (ethers),which do not open to form aldehydes under these conditions.
Compound $D$ is a cyclic alcohol,which does not contain the hemiacetal functional group required to form an aldehyde.
Therefore,the correct compound is $C$.

(B) Compound $A$ $(C_{8}H_{8}O_{2})$ undergoes cross-Aldol condensation with acetophenone to form a single product. This implies $A$ is an aldehyde without $\alpha$-hydrogens,such as $4-$methoxybenzaldehyde $(p-CH_3OC_6H_4CHO)$.
When $4-$methoxybenzaldehyde reacts with conc. $NaOH$,it undergoes the Cannizzaro reaction because it lacks $\alpha$-hydrogens.
The reaction produces $4-$methoxybenzyl alcohol $(p-CH_3OC_6H_4CH_2OH)$ and $4-$methoxybenzoate ion $(p-CH_3OC_6H_4COO^-)$.
Therefore,compound $A$ is $4-$methoxybenzaldehyde.

(A) The correct matches are as follows:
$A$. $NH_2-NH_2, KOH$ is used in the Wolff-Kishner reduction $(III)$.
$B$. $[Ag(NH_3)_2]OH$ is the reagent for Tollen's test $(I)$.
$C$. Aqueous $CuSO_4$,Sodium Potassium tartarate,and $KOH$ constitute the Fehling's solution used in Fehling's test $(IV)$.
$D$. $Zn-Hg, HCl$ is used in the Clemmensen reduction $(II)$.
Thus,the correct sequence is $A-III, B-I, C-IV, D-II$.

(B) The reaction of a Grignard reagent $(R'''MgX)$ with a ketone $(R'R''C=O)$ produces a tertiary alcohol.
The structure of $2-methylbutane-2-ol$ is $CH_3-C(OH)(CH_3)-CH_2-CH_3$.
This tertiary alcohol is formed by the nucleophilic addition of a Grignard reagent to a ketone.
If we react butan$-2-$one $(CH_3-CO-C_2H_5)$ with methylmagnesium bromide $(CH_3MgBr)$,the methyl group $(CH_3)$ attacks the carbonyl carbon,resulting in $2-methylbutane-2-ol$.
Comparing this with the general reaction $R'R''C=O + R'''MgX \rightarrow R'R''C(OH)R'''$,we identify $R'=CH_3$,$R''=C_2H_5$,and $R'''=CH_3$.

(D) Acetophenone $(C_6H_5COCH_3)$ contains a methyl ketone group $(-COCH_3)$,which undergoes the iodoform reaction with $NaOI$ (sodium hypoiodite) to form a yellow precipitate of iodoform $(CHI_3)$.
Benzophenone $(C_6H_5COC_6H_5)$ lacks a methyl group attached to the carbonyl carbon; therefore,it does not give the iodoform test.
Thus,$NaOI$ is the correct reagent to distinguish between these two compounds.

(D) The Cannizzaro reaction is a characteristic reaction of aldehydes that do not possess any $\alpha$-hydrogen atoms.
$1$. $HCHO$ (Formaldehyde) has no $\alpha$-hydrogen.
$2$. $CCl_3CHO$ (Trichloroacetaldehyde) has no $\alpha$-hydrogen.
$3$. Benzaldehyde $(C_6H_5CHO)$ has no $\alpha$-hydrogen.
All of the above compounds undergo the Cannizzaro reaction.
$4$. Propanal $(CH_3CH_2CHO)$ contains two $\alpha$-hydrogen atoms attached to the $\alpha$-carbon. Therefore,it undergoes aldol condensation in the presence of a dilute base,not the Cannizzaro reaction.
Thus,the correct answer is $CH_3CH_2CHO$.

(B) Formaldehyde $(HCHO)$ is the simplest aldehyde. When it reacts with a Grignard reagent $(RMgX)$,it forms an addition product which,upon acidic hydrolysis,yields a primary alcohol $(RCH_2OH)$.
Other aldehydes $(R'CHO)$ react with Grignard reagents to form secondary alcohols $(R'CH(OH)R)$.
Ketones $(R'COR'')$ react with Grignard reagents to form tertiary alcohols $(R'R''C(OH)R)$.

(C) Let the empirical formula be $(COH)_{n}$.
The molar mass is $12n + 16n + n = 29n = 58 \Rightarrow n = 2$.
Thus,the molecular formula is $C_{2}H_{2}O_{2}$,which corresponds to glyoxal $(CHO-CHO)$.
Glyoxal contains two aldehyde groups and lacks $\alpha$-hydrogen atoms,so it undergoes the Cannizzaro reaction in the presence of $50\% \ KOH$.
In this reaction,one aldehyde group is oxidized to a carboxylate group and the other is reduced to a primary alcohol group.
$CHO-CHO + KOH \rightarrow HO-CH_{2}-COOK$.
Upon acidification,the salt converts to glycolic acid: $HO-CH_{2}-COOH$.
Therefore,the correct structure of $y$ is $HO-CH_{2}-COOH$.

(A) Statement $I$: Acetaldehyde $(CH_{3}CHO)$ reacts with semicarbazide $(H_{2}NNHCONH_{2})$ via a nucleophilic addition-elimination reaction to form a semicarbazone. The product $CH_{3} - CH = N - NH - C(=O) - NH_{2}$ is correctly represented.
Statement $II$: The molecule $Ph - CH(OH)(OCH_{3})$ is a hemiacetal. In the presence of dilute acid,hemiacetals undergo hydrolysis to regenerate the corresponding aldehyde $(Ph-CHO)$ and alcohol $(CH_{3}OH)$. Thus,both statements are correct.

(B) Step $1$: Formation of 'x'. Hydrolysis of prop$-1-$yne $(CH_{3}C \equiv CH)$ in the presence of $HgSO_{4}/H^{+}$ at $333 \ K$ follows Kucherov's reaction to yield acetone $(CH_{3}COCH_{3})$.
Step $2$: Formation of 'y'. Reaction of ethanenitrile $(CH_{3}CN)$ with methylmagnesium bromide $(CH_{3}MgBr)$ in dry ether followed by hydrolysis yields acetone $(CH_{3}COCH_{3})$.
Step $3$: Aldol condensation. Both 'x' and 'y' are acetone. The reaction of two molecules of acetone in the presence of $Ba(OH)_{2}$ undergoes self-aldol condensation to form $4-$hydroxy$-4-$methylpentan$-2-$one $(CH_{3}COCH_{2}C(OH)(CH_{3})_{2})$.
Step $4$: Dehydration. Heating the aldol product causes dehydration (loss of $H_{2}O$) to form the $\alpha,\beta$-unsaturated ketone,$4$-methylpent$-3-$en$-2-$one $(CH_{3}COCH=C(CH_{3})_{2})$.

(D) The reaction sequence is as follows:
$1$. Glucose reacts with $HI/\Delta$ to form $n$-hexane.
$2$. $n$-hexane undergoes aromatization in the presence of $V_2O_5$ at $773 \ K$ and high pressure to form benzene.
$3$. Benzene reacts with benzoyl chloride in the presence of anhydrous $AlCl_3$ (Friedel-Crafts acylation) to form benzophenone $(Ph-CO-Ph)$.
In the product,benzophenone $(C_{13}H_{10}O)$:
- Each phenyl ring has $3$ $\pi$ bonds,so $2$ rings have $6$ $\pi$ bonds.
- The $C=O$ group has $1$ $\pi$ bond.
- Total $\pi$ bonds = $6 + 1 = 7$. Each $\pi$ bond contains $2$ electrons,so total $\pi$ electrons = $7 \times 2 = 14$.
- The oxygen atom in the $C=O$ group has $2$ lone pairs of electrons,which equals $2 \times 2 = 4$ electrons.
- Total number of electrons = $14 + 4 = 18$.

(D) The molecule $(X)$ is an enol ether. Acidic hydrolysis of an enol ether involves the protonation of the double bond followed by the attack of water,leading to the cleavage of the $C-O$ bond to produce a carbonyl compound (aldehyde or ketone) and an alcohol.
In this specific case,the hydrolysis of the given enol ether yields propanal $(CH_3CH_2CHO)$ and acetone $(CH_3COCH_3)$.
$(Y)$ and $(Z)$ are propanal and acetone.
Statement $(A)$ is correct as both products have distinct molar masses,but in general,hydrolysis products can be analyzed for mass.
Statement $(D)$ is correct because both propanal (an aldehyde) and acetone (a ketone) contain a carbonyl group $(C=O)$,which reacts with $2,4$-dinitrophenylhydrazine $(2,4-DNP)$ to form a hydrazone derivative (addition-elimination reaction).

(C) The reaction of $n$-butyl alcohol $(CH_3CH_2CH_2CH_2OH)$ with $HBr$ involves the formation of a primary carbocation,which undergoes a $1,2$-hydride shift to form a more stable secondary carbocation,leading to a rearranged product ($2$-bromobutane).
In the reaction with conc. $H_2SO_4$ at $413 \ K$,the primary carbocation intermediate can undergo rearrangement before elimination to form the alkene.
In the reaction with $PCl_5$ or $SOCl_2$ (in the presence of pyridine),the reaction proceeds via an $S_N2$ mechanism. In an $S_N2$ mechanism,the nucleophile attacks the carbon atom from the backside,and the leaving group departs simultaneously. Since there is no carbocation intermediate formed in an $S_N2$ reaction,no rearrangement can occur.
Therefore,the reaction of $n$-butyl alcohol with $PCl_5$ or $SOCl_2$ (in the presence of pyridine) does not involve a rearrangement reaction.

(C) Statement $I$: Metamers are isomers that have the same molecular formula but differ in the distribution of alkyl groups attached to the same polyvalent functional group. Both $2, 6$-diethylcyclohexanone and $6$-methyl-$2$-$n$-propylcyclohexanone have the same molecular formula $(C_{10}H_{18}O)$ but different alkyl substituents attached to the carbonyl group. Thus,they are metamers. Statement $I$ is true.
Statement $II$: Keto-enol tautomerism requires the presence of at least one $\alpha$-hydrogen atom adjacent to the carbonyl group. In $2, 2, 6, 6$-tetramethylcyclohexanone,all four $\alpha$-positions are substituted by methyl groups,meaning there are no $\alpha$-hydrogens available. Therefore,it cannot exhibit keto-enol tautomerism. Statement $II$ is false.

(A) The optically active alkyl bromide $C_4H_9Br$ is $2$-bromobutane.
Reaction of $2$-bromobutane with ethanolic $KOH$ (dehydrohalogenation) gives but-$2$-ene as the major product $[A]$.
Reaction of but-$2$-ene $[A]$ with $Br_2$ yields $2,3$-dibromobutane $[B]$.
Reaction of $2,3$-dibromobutane $[B]$ with alcoholic $KOH$ followed by $NaNH_2$ (dehydrohalogenation) yields but-$2$-yne $[C]$.
Hydration of but-$2$-yne $[C]$ in the presence of $HgSO_4/H_2SO_4$ (Kucherov's reaction) yields butanone $(CH_3COCH_2CH_3)$,which is a methyl ketone.
Methyl ketones are confirmed by the Haloform test (specifically the Iodoform test).

(C) $1$. Reaction of benzene with $CO$ and $HCl$ in the presence of $CuCl$ is the Gattermann-Koch reaction,which yields Benzaldehyde $(x)$.
$2$. Reaction of benzene with ethanoyl chloride in the presence of anhydrous $AlCl_3$ is Friedel-Crafts acylation,which yields Acetophenone $(y)$.
$3$. Heating Benzaldehyde $(x)$ and Acetophenone $(y)$ in the presence of an alkali is a Claisen-Schmidt condensation reaction,which yields Benzalacetophenone (Chalcone) as product $(z)$.
$4$. The structure of Chalcone $(C_6H_5-CH=CH-CO-C_6H_5)$ contains two benzene rings,one $C=C$ double bond,and one $C=O$ double bond.
$5$. Each benzene ring contributes $6$ $\pi$ electrons $(6 \times 2 = 12)$.
$6$. The $C=C$ bond contributes $2$ $\pi$ electrons.
$7$. The $C=O$ bond contributes $2$ $\pi$ electrons.
$8$. Total $\pi$ electrons = $12 + 2 + 2 = 16$.

(B) $1$. Compound $P$ $(C_{8}H_{8}O)$ reacts with $2,4-DNP$ reagent,which confirms the presence of a carbonyl group (aldehyde or ketone).
$2$. It does not reduce Fehling's reagent,which indicates that $P$ is a ketone,not an aldehyde.
$3$. The molecular formula $C_{8}H_{8}O$ corresponds to acetophenone $(C_{6}H_{5}COCH_{3})$.
$4$. On drastic oxidation with chromic acid $(H_{2}CrO_{4})$,the alkyl group attached to the benzene ring is oxidized to a carboxylic acid group.
$5$. Acetophenone $(C_{6}H_{5}COCH_{3})$ on oxidation yields benzoic acid $(C_{6}H_{5}COOH)$ as product $Q$.
$6$. Benzoic acid $(Q)$ reacts with aqueous $NaHCO_{3}$ to produce effervescence due to the evolution of $CO_{2}$ gas.
$7$. Therefore,$P$ is acetophenone and $Q$ is benzoic acid.