Metamerism arises due to the different distribution of alkyl groups on either side of the same polyvalent functional group while retaining the same molecular formula.

In 2,6-diethylcyclohexanone, the carbonyl group is associated with two ethyl substituents, whereas in 6-methyl-2-(n)-propylcyclohexanone, it is associated with a methyl group and an (n)-propyl group. Both compounds have the same molecular formula but differ in the distribution of alkyl groups around the carbonyl functionality. Hence, they are metamers.

Therefore, Statement I is true.

Keto-enol tautomerism requires the presence of at least one (\alpha)-hydrogen atom adjacent to the carbonyl group.

In 2,2,6,6-tetramethylcyclohexanone, both (\alpha)-carbon atoms are fully substituted by two methyl groups each and do not possess any (\alpha)-hydrogen atoms. Consequently, the compound cannot undergo keto-enol tautomerism.

Therefore, Statement II is false.

Hence, Statement I is true, Statement II is false, and the correct answer is Option C.

Step 1:

Let the molar ratios of the elements in compound $$x$$ be equal.
So its empirical formula is $$C_nH_nO_n$$.

Molar mass calculation:
Atomic masses: C = 12, H = 1, O = 16.
Molar mass of $$C_nH_nO_n = 12n + 1n + 16n = 29n$$.

Given molar mass of $$x$$ is 58 g mol$$^{-1}$$, 

Therefore,
$$29n = 58 \;\Longrightarrow\; n = 2$$.

Hence the molecular formula of $$x$$ is $$C_2H_2O_2$$.

Step 2:

Saturation check (index of hydrogen deficiency):
For two carbons the saturated formula would be $$C_2H_{2\times2+2}=C_2H_6$$.

Degrees of unsaturation $$=\dfrac{(2C+2-H)}{2}=\dfrac{(4+2-2)}{2}=2$$.

Two degrees of unsaturation can be satisfied by two carbonyl groups, giving the structure $$OHC{-}CHO$$ (glyoxal). Glyoxal has no $$\alpha$$-hydrogen, so it undergoes Cannizzaro reaction.

Reaction in 50 % KOH, reflux:

Two molecules of glyoxal disproportionate (Cannizzaro reaction): one carbonyl group is reduced to an alcohol and the other is oxidised to a carboxylate within the same carbon skeleton, giving potassium glycolate $$HO{-}CH_2{-}COOK$$.

Subsequent acidification converts the salt to its free acid:

$$HO{-}CH_2{-}COOK + HCl \longrightarrow HO{-}CH_2{-}COOH + KCl$$.

Thus product $$y$$ is glycolic acid, $$HO{-}CH_2{-}COOH$$.

Option C $$\longrightarrow\ $$ $$O = C(OH){-}CH_2{-}OH$$ (glycolic acid)

Statement I - Self-condensation of acetaldehyde

Acetaldehyde is represented as $$CH_3CHO$$ (the abbreviated notation $$CH_3{-}CH{=}O$$). When the pH of the medium is adjusted to a mild value of about 7-8 (weakly basic), an enolate ion can be generated without causing extensive dehydration. Two molecules of the aldehyde then undergo the aldol reaction:

$$2\,CH_3CHO \;\xrightarrow[\text{pH }7\text{-}8]{OH^-}\; CH_3CH(OH)CH_2CHO$$

The product is 3-hydroxybutanal, the normal “aldol’’ obtained before any loss of water. Thus the condensation described in Statement I is correctly stated, so Statement I is true.

Statement II - Acidic hydrolysis of an acetal

Acetals (dialkoxy derivatives) revert to the parent carbonyl compound in the presence of dilute mineral acid. For benzaldehyde dimethyl acetal the reaction is

$$PhCH(OMe)_2 + H_2O \;\xrightarrow{\text{dil.\ }H^+}\; PhCHO + 2\,MeOH$$

Hence the molecule named in Statement II indeed generates $$PhCHO$$ (benzaldehyde) on treatment with dilute acid, so Statement II is also true.

Since both statements are correct, the option that matches this situation is:

Option A which is: Both statements are true

The given compound $$(X)$$ is an acetal. 

Under mild acidic conditions, acetals undergo hydrolysis to give back the two carbonyl compounds from which they were originally obtained.
On hydrolysis, $$(X)$$ breaks at the two $$C-O$$ bonds adjacent to the acetal carbon and furnishes the following products:

$$\displaystyle (X)\; \xrightarrow[\text{H}_2O]{\text{dil.\,H}^+}\; (Y)\;+\; (Z)$$

The two fragments obtained are
$$\text{propanal :}\; CH_3CH_2CHO \; (Y)$$
$$\text{acetone :}\; (CH_3)_2CO \; (Z)$$

Both have the same molecular formula $$C_3H_6O$$, so their molar mass is identical:

$$M_r(CH_3CH_2CHO)=58\;{\rm g\,mol^{-1}}$$
$$M_r((CH_3)_2CO)=58\;{\rm g\,mol^{-1}}$$

Let us now analyse each statement:

Statement A: “Both (Y) and (Z) have same molar mass.”
As shown above, this is correct.

Statement B: “(Y) and (Z) can be distinguished from each other by NaHCO$$$_3$$.”
$$NaHCO_3$$ gives effervescence only with acids that are stronger than $$H_2CO_3$$ (mainly carboxylic acids and sulfonic acids). Aldehydes and ketones do not react. Hence neither propanal nor acetone shows any reaction; this test cannot distinguish them. Statement B is false.

Statement C: “(Y) and (Z) react with HCN with same rates.”
The rate of cyanohydrin formation depends on both the electronic nature of the carbonyl carbon and steric hindrance. Aldehydes (propanal) are less hindered and more polarised than ketones (acetone); therefore propanal reacts faster with $$HCN$$ than acetone. Rates are not the same, making Statement C false.

Statement D: “(Y) and (Z) undergo addition reaction with 2,4-DNP.”
2,4-Dinitrophenylhydrazine adds to any aldehyde or ketone to form a yellow/orange hydrazone precipitate. Both propanal and acetone give a positive 2,4-DNP test. Statement D is correct.

Thus only Statements A and D are correct.

Option D $$\longrightarrow\ $$ A and D Only

We first identify the products $$x$$ and $$y$$.

Finding x: Propanenitrile ($$CH_3CH_2CN$$) is treated with stannous chloride ($$SnCl_2$$) and hydrochloric acid, followed by hydrolysis. This is the Stephen reduction, which converts a nitrile to an aldehyde. Therefore $$x = \text{propanal}\;(CH_3CH_2CHO)$$.

Finding y: But-2-ene ($$CH_3CH{=}CHCH_3$$) undergoes ozonolysis followed by hydrolysis. The double bond is cleaved to produce two molecules of ethanal (acetaldehyde, $$CH_3CHO$$). Since we take one mole of the ozonolysis product, $$y = \text{ethanal}\;(CH_3CHO)$$.

When one mole of propanal and one mole of ethanal react in the presence of alkali followed by heating, aldol condensation followed by dehydration takes place. Four $$\alpha,\beta$$-unsaturated aldehyde products are possible:

Product 1 — Self-aldol of ethanal: The enolate of one ethanal molecule attacks the carbonyl of another ethanal. After dehydration this gives $$CH_3CH{=}CHCHO$$, which is but-2-enal (crotonaldehyde).

Product 2 — Cross-aldol (ethanal enolate + propanal): The enolate of ethanal ($${}^{-}CH_2CHO$$) attacks the carbonyl carbon of propanal. The resulting aldol product dehydrates to give $$CH_3CH_2CH{=}CHCHO$$ = pent-2-enal.

Product 3 — Cross-aldol (propanal enolate + ethanal): The enolate of propanal ($$CH_3\overset{-}{C}HCHO$$) attacks the carbonyl carbon of ethanal. The aldol product dehydrates to give $$CH_3CH{=}C(CH_3)CHO$$ = 2-methylbut-2-enal.

Product 4 — Self-aldol of propanal: The enolate of one propanal attacks the carbonyl of another propanal. Dehydration gives $$CH_3CH_2CH{=}C(CH_3)CHO$$ = 2-methylpent-2-enal.

Now consider the options. 3-Methylbut-2-enal has the structure $$(CH_3)_2C{=}CHCHO$$. Forming this would require a branched reactant such as acetone or isobutyraldehyde as one of the starting materials, but neither is present in our reaction mixture. Therefore, 3-methylbut-2-enal cannot be obtained from this reaction.

Hence, the correct answer is Option 2.

Benzene reacts with carbon monoxide and hydrogen chloride in the presence of cuprous chloride to give benzaldehyde by the Gattermann-Koch reaction.

$$x=C_6H_5OH$$

Benzene reacts with ethanoyl chloride $$(\mathrm{CH_3COCl})$$ in presence of anhydrous $$\mathrm{AlCl_3}$$​ to give acetophenone by Friedel-Crafts acylation.

$$y=C_{6}H_{5}COCH_3$$

On heating benzaldehyde and acetophenone in presence of alkali, Claisen-Schmidt condensation occurs to form chalcone:

Now count total $$\pi$$-electrons in Chalcone:

• Two benzene rings: 6+6=12
• One $$C=C$$ bond: 2
• One $$C=O$$ bond: 2

Total $$\pi$$ bonds is 16.

We need to identify reactions that both acetaldehyde ($$CH_3CHO$$) and acetone ($$CH_3COCH_3$$) undergo individually.

A. Iodoform Reaction:

Both acetaldehyde and acetone have the $$CH_3CO-$$ group (methyl ketone/methyl aldehyde). Both undergo the iodoform reaction to give $$CHI_3$$. YES for both.

B. Cannizzaro Reaction:

Cannizzaro reaction requires aldehydes without alpha-hydrogen. Acetaldehyde has alpha-hydrogens ($$CH_3$$ group), so it does NOT undergo Cannizzaro reaction. Instead, it undergoes aldol condensation. Acetone also has alpha-hydrogens and does not undergo Cannizzaro reaction. NO for both.

C. Aldol Condensation:

Both acetaldehyde and acetone have alpha-hydrogens and undergo aldol condensation in the presence of dilute base. YES for both.

D. Tollen's Test:

Tollen's test (silver mirror test) is specific to aldehydes. Acetaldehyde gives a positive test, but acetone does not. NO (not both).

E. Clemmensen Reduction:

Clemmensen reduction ($$Zn-Hg/HCl$$) reduces the carbonyl group to $$CH_2$$. Both aldehydes and ketones undergo this reduction. YES for both.

The reactions common to both are A, C, and E. The answer is Option C) A, C and E Only.

$$\text{}$$

• Option A: Cleaving the double bond at the ring junction yields a single open-chain diketone molecule. Therefore, it is an intramolecular aldol product.
  
  
• Option B: Cleaving the double bond in the 5-membered ring yields a single aromatic dialdehyde/diketone precursor (2-formylbenzaldehyde derivative). Therefore, it is an intramolecular aldol product.
  
  
• Option C: Cleaving the double bond in the cyclohexenone ring yields a single open-chain dicarbonyl precursor. Therefore, it is an intramolecular aldol product.
  
  
• Option D: This compound features a terminal exocyclic double bond ($$=\text{CH}_2$$) attached to the $\alpha$-carbon. Cleaving this bond breaks the molecule into two separate components: a cyclic ketone and formaldehyde ($$\text{HCHO}$$). Because it requires two independent starting molecules, it cannot be formed via an intramolecular pathway.
  

Correct Answer

Option D (The compound with the terminal exocyclic double bond, $$=\text{CH}_2$$)

1. Retrosynthetic Analysis of Product (x):

• Retro-Aldol:
  Product (x) is 1-acetylcyclopentene, an $$\alpha,\beta$$-unsaturated ketone. Cleaving the double bond between the ring carbon and the carbonyl carbon adds two hydrogens to the ring carbon and an oxygen to the acetyl group carbon, yielding the dicarbonyl intermediate:
  $$\text{7-oxooctanal: } \text{CH}_3\text{-CO-CH}_2\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-CHO}$$
• Retro-Ozonolysis:
  Connecting the two carbonyl carbons ($$-\text{C7}=\text{C1}-$$) of 7-oxooctanal with a double bond re-establishes the starting cyclic alkene:
  $$\text{1-methylcyclohexene}$$

2. Forward Reaction Steps:

• Step 1: Ozonolysis ($$\text{O}_3, \text{Zn/H}_2\text{O}$$)
  Ozonolysis of 1-methylcyclohexene cleaves its endocyclic double bond to produce 7-oxooctanal containing a keto group ($$-\text{COCH}_3$$) and an aldehyde group ($$-\text{CHO}$$).
• Step 2: Intramolecular Aldol Condensation ($\text{OH}^-, \Delta$)
  The base abstracts an $$\alpha$$-proton from the $$\text{C5}$$ position (adjacent to the keto group) to form a stable enolate. This enolate attacks the more reactive aldehyde carbon ($$\text{C1}$$), forming a stable 5-membered ring. Dehydration ($$-\text{H}_2\text{O}$$) yields 1-acetylcyclopentene (x).

Conclusion:

The starting alkene required to form product (x) is 1-methylcyclohexene (Option C).

Answer: Option C

To determine the structure of compound P, we need to interpret each of the chemical tests given in the question.

The compound gives a positive test with 2,4-dinitrophenylhydrazine (2,4-DNP).

A positive 2,4-DNP test indicates the presence of a carbonyl group. Therefore, compound P must be either an aldehyde or a ketone.

The compound gives a negative Tollens' test.

Tollens' reagent oxidizes aldehydes but does not react with ketones. Since the test is negative, compound P must be a ketone.

This immediately eliminates Option (C), which is an aldehyde.

The compound is also optically active.

For a molecule to be optically active, it must be chiral. In this case, it should contain at least one carbon atom attached to four different groups.

Now let us examine the remaining ketone options.

Option (A): Hexan-2-one

$$CH_3-C(=O)-CH_2-CH_2-CH_2-CH_3$$

None of the carbon atoms are bonded to four different groups. Therefore, the molecule is achiral and optically inactive.

Option (D): 4-methylpentan-2-one

$$CH_3-C(=O)-CH_2-CH(CH_3)-CH_3$$

The branched carbon is attached to one hydrogen atom, one side chain, and two identical $$CH_3$$ groups. Since two substituents are identical, it is not a chiral center.

Hence, this compound is also achiral.

Option (B): 3-methylpentan-2-one

$$CH_3-C(=O)-CH(CH_3)-CH_2-CH_3$$

The carbon at position 3 is attached to:

1. A hydrogen atom.
2. A methyl group $$(-CH_3)$$.
3. An ethyl group $$(-CH_2CH_3)$$.
4. An acetyl group $$(-C(=O)CH_3)$$.

Since all four groups are different, this carbon is a chiral center.

Therefore, this compound is optically active.

Hence, Option (B) is the only compound that satisfies all the given conditions. It is a ketone, gives a positive 2,4-DNP test, gives a negative Tollens' test, and contains a chiral center.

Therefore, the correct answer is Option (B).

The reaction shown is the classical “cyanohydrin formation” in which hydrogen cyanide adds to the carbonyl group of an aldehyde or a ketone:

$$\mathrm{R{-}C(=O){-}R'} + HCN \;\longrightarrow\; \mathrm{R{-}C(OH)(CN){-}R'}$$

Here the substrate is $$CH_3CHO$$ (ethanal). Hence the product is $$CH_3CH(OH)CN$$.

Statement I: “The product obtained in the above reaction possesses a chiral centre.”
  • In $$CH_3CH(OH)CN$$ the carbon marked with an asterisk below is attached to four different groups: $$CH_3{-},\; {-}OH,\; {-}CN,$$ and $$H{-}$$.
  • Because the four substituents are different, that carbon is a stereogenic (asymmetric) centre, making the molecule chiral.
Therefore Statement I is true.

Statement II: “HCN adds to the carbonyl group by a nucleophilic-addition mechanism in which $$CN^-$$ first attacks the electrophilic carbonyl carbon and the resulting alkoxide is then protonated.”
  • Mechanism steps:
    1. Generation of the nucleophile: $$HCN \rightleftharpoons H^+ + CN^-$$ (in the presence of a base such as $$NaCN$$).
    2. Nucleophilic attack: $$CN^-$$ attacks the partially positive carbon of the $$C=O$$ bond, giving an alkoxide ion.
    3. Protonation: the alkoxide abstracts a proton (from $$HCN$$ or another source) to yield the cyanohydrin.
  • These are the standard steps of nucleophilic addition to a carbonyl, so Statement II is also true.

Since both statements correctly describe the reaction and its product, the correct choice is:

Option D which is: Both Statement I and Statement II are true

The starting material is phenylglyoxal, $$\text{Ph-C(=O)-CHO},$$ which is an $\alpha$-ketoaldehyde. It contains two carbonyl groups but does not possess any $\alpha$-hydrogen atoms.

The reaction is carried out in the presence of $$KOH$$ and heat $$\Delta.$$

When an aldehyde lacking $\alpha$-hydrogens is treated with a strong base, it undergoes the Cannizzaro reaction. Since phenylglyoxal contains both an aldehyde and a ketone group within the same molecule, it undergoes an intramolecular Cannizzaro reaction.

In a Cannizzaro reaction, one carbonyl group is oxidized while the other is reduced. Here, the aldehyde portion is oxidized to a carboxylate ion, whereas the ketone portion is reduced to an alcohol.

The mechanism proceeds as follows:

1. The hydroxide ion $$OH^-$$ attacks the more electrophilic and less sterically hindered aldehyde carbonyl carbon.
2. The resulting alkoxide intermediate undergoes a hydride shift. The negatively charged oxygen reforms the carbonyl group, causing the adjacent hydrogen atom to migrate as a hydride ion $$H^-$$ to the neighboring ketone carbonyl carbon.
3. This produces a carboxylate group and an alkoxide ion. A rapid proton transfer then converts the alkoxide into a secondary alcohol while the oxidized portion remains as the potassium carboxylate salt.

Thus,

• the aldehyde group $$-CHO$$ is converted into $$-COO^-K^+,$$
• and the ketone group $$-C(=O)-$$ is converted into $$-CH(OH)-.$$

The final product obtained is potassium mandelate, having the structure

$$\boxed{\text{Ph-CH(OH)-COO}^-,\text{K}^+.}$$

Therefore, the correct answer is

$$\boxed{\text{Option (B).}}$$

The iodoform test ($$I_2/NaOH$$) gives a yellow precipitate of $$CHI_3$$ whenever the organic compound contains either of the following structural units:

• a methyl ketone group $$\;-\!COCH_3$$, or
• an $$\;-\!CH(OH)CH_3$$ group (because it is oxidised by the reagent to the corresponding methyl ketone).

Let us check each molecule.

Case 1: Ethanol, $$CH_3CH_2OH$$.
Ethanol is oxidised by $$I_2/NaOH$$ to ethanal $$\;(CH_3CHO)$$, which has the $$CH_3CO-$$ group. Hence ethanol gives the iodoform test.

Case 2: Isopropyl alcohol (propan-2-ol), $$CH_3CH(OH)CH_3$$.
It already contains the $$-\!CH(OH)CH_3$$ arrangement, so it gives the test.

Case 3: Bromo-acetone, $$CH_3COCH_2Br$$.
The carbonyl carbon is directly attached to a $$CH_3$$ group, i.e. it has a $$-\!COCH_3$$ fragment. The presence of the bromine atom on the other side does not affect the reaction; therefore it gives the iodoform test.

Case 4: Butan-2-ol, $$CH_3CH(OH)CH_2CH_3$$.
Contains the $$-\!CH(OH)CH_3$$ unit, so it is iodoform-positive.

Case 5: Butan-2-one, $$CH_3COCH_2CH_3$$.
A clear methyl-ketone, hence positive.

Case 6: Butanal, $$CH_3CH_2CH_2CHO$$.
The carbonyl carbon is attached to hydrogen (aldehyde) and $$CH_2CH_2CH_3$$, not to a $$CH_3$$ group. Thus no $$-\!COCH_3$$ fragment is present, and it cannot be oxidised to one under the test conditions. Therefore butanal does NOT give the iodoform test.

Case 7: Pentan-2-one, $$CH_3COCH_2CH_2CH_3$$.
Has the required $$-\!COCH_3$$ group, so it is positive.

Case 8: Pentan-3-one, $$CH_3CH_2COCH_2CH_3$$.
On both sides of the carbonyl carbon there are $$CH_2$$ groups; no $$CH_3$$ group is directly bonded to the carbonyl carbon. Hence the $$-\!COCH_3$$ fragment is absent and the compound is iodoform-negative.

Case 9: Pentanal, $$CH_3CH_2CH_2CH_2CHO$$.
Like butanal, it lacks the $$-\!COCH_3$$ fragment, so it does not give the test.

Case 10: Pentan-3-ol, $$CH_3CH_2CH(OH)CH_2CH_3$$.
The alcohol carbon is attached to two $$CH_2$$ groups, not to a $$CH_3$$ group. After oxidation it would form pentan-3-one (just discussed), which is iodoform-negative. Hence pentan-3-ol is also iodoform-negative.

Collecting the results:

Positive (give iodoform): Ethanol, Isopropyl alcohol, Bromoacetone, 2-Butanol, 2-Butanone, 2-Pentanone  ⇒ 6 compounds.

Negative (do not give iodoform): Butanal, 3-Pentanone, Pentanal, 3-Pentanol  ⇒ 4 compounds.

Therefore, the number of molecules that cannot give the iodoform reaction is $$4$$, which matches Option B.

(The answer indicated as “2” in the prompt appears to be a typographical error; the correct count is 4.)

In the Claisen-Schmidt reaction between benzaldehyde and acetone, represented by $$ 2C_6H_5CHO + CH_3COCH_3 \to C_6H_5CH=CHCOCH=CHC_6H_5 + 2H_2O $$, if 5.3 g of benzaldehyde are used and 3.51 g of dibenzalacetone are obtained, the percentage yield can be calculated as follows.

The molar mass of benzaldehyde is $$6(12) + 5(1) + 12 + 16 + 1 = 106$$ g/mol, and the molar mass of dibenzalacetone is $$17(12) + 14(1) + 16 = 204 + 14 + 16 = 234$$ g/mol.

The number of moles of benzaldehyde is $$\frac{5.3}{106} = 0.05$$ mol. Since 2 mol of benzaldehyde produce 1 mol of dibenzalacetone, the theoretical yield in moles is $$0.05/2 = 0.025$$ mol.

Multiplying by the molar mass of dibenzalacetone gives the theoretical mass as $$0.025 \times 234 = 5.85$$ g.

Therefore, the percentage yield is $$\frac{3.51}{5.85} \times 100 = 60\%$$, so the yield of the reaction is 60%.

The reaction sequence finally furnishes acetophenone, $$C_6H_5COCH_3$$, as product $$D$$.

Hybridisation check:
• The six ring-carbons of the benzene nucleus are all $$sp^2$$.
• The carbonyl carbon of the $$C=O$$ group is also $$sp^2$$.
• The three hydrogens on the methyl carbon come from an $$sp^3$$ carbon, so that carbon is not counted.

Hence the total number of $$sp^2$$-hybridised carbon atoms in product $$D$$ is $$6 + 1 = 7$$.

Answer: 7

The acidity of a hydrogen atom depends on the stability of the conjugate base formed after its removal. Greater stabilization of the resulting carbanion leads to higher acidity.

In the alpha-diketone and gamma-diketone structures, the acidic hydrogens are influenced by only one carbonyl group at a time or are too far from the second carbonyl group to experience significant stabilization.

In the monoketone, the alpha hydrogen is stabilized by resonance with only a single carbonyl group. Therefore, its acidity is moderate and comparable to that of ordinary ketones.

In the beta-diketone structure, the central $$-CH_2-$$ group is located between two carbonyl groups. When a proton is removed from this carbon, the resulting negative charge is delocalized through resonance over both adjacent carbonyl groups.

This extensive resonance stabilization distributes the negative charge onto both oxygen atoms, making the conjugate base exceptionally stable. The combined electron-withdrawing effect of the two carbonyl groups greatly enhances the acidity of the central methylene hydrogens.

Therefore, the hydrogen atoms of the active methylene group present in the beta-diketone are the most acidic among the given compounds.

Hence, the correct answer is, $$\text{The }\beta\text{-diketone compound (Option D)}$$

The cumene (Hock) process converts iso-propylbenzene (cumene) to phenol and acetone by way of the cumene-hydroperoxide rearrangement.
Hence, in the first step the by-product is

$$P = (CH_3)_2C=O \;(\text{acetone}).$$

Preparation of Q
Each molecule of $$Cl_2$$ supplies two chlorine atoms. Three “equivalents” (three moles) of $$Cl_2$$ therefore furnish six chlorine atoms - exactly the number needed to replace all six α-hydrogens of acetone. Chlorination in sunlight thus converts both methyl groups into trichloromethyl groups:

$$CH_3COCH_3 \;\xrightarrow[\;h\nu\;]{3\,Cl_2}\; CCl_3COCCl_3$$

so

$$Q = CCl_3COCCl_3\;(\text{hexachloroacetone}).$$

Formation of R and S
A 1,1,1-trihalo ketone undergoes the haloform reaction even in simple alkali. Treating Q with slaked lime gives one molecule of chloroform (haloform) together with the calcium salt of the residual acid:

$$CCl_3COCCl_3 + 2\,Ca(OH)_2 \;\longrightarrow\; 2\,COCl_3^-Ca^{2+} + 2\,CHCl_3 + 2\,H_2O$$

Thus

$$R = CHCl_3\;(\text{chloroform}),\qquad S = Ca(CCl_3COO)_2\;(\text{calcium trichloroacetate}).$$

Checking the statements

Option A Acetone (P) condenses with chloroform (R) in strong base (KOH) to give chlorobutanol (also called chloretone). Acidification merely neutralises the medium: the product remains unchanged. Hence statement A is correct.

Option B For the slow photo-oxidation of chloroform moisture is essential; dry $$CHCl_3$$ exposed only to $$O_2$$ and light does not accumulate phosgene. Therefore statement B is not accepted as correct for the conditions written.

Option C Alkaline hydrolysis of Q liberates chloroform and the trichloroacetate ion; it never produces $$Cl_3CCH_2OH$$. Hence statement C is incorrect.

Option D On heating, calcium trichloroacetate (S) does not regenerate acetone. Therefore statement D is incorrect.

Only statement A is correct.

Final answer: Option A which is: Reaction of P with R in the presence of KOH followed by acidification gives chlorobutanol.

The extent of enolization depends on the stability of the enol form. Enol content increases when the enol is stabilized by conjugation, intramolecular hydrogen bonding, or aromaticity.

In 1,3-cyclohexanedione, enolization produces a conjugated system stabilized by intramolecular hydrogen bonding. Hence, it has a significant enol content.

In 1,4-cyclohexanedione, enolization does not generate an extended conjugated system and lacks substantial stabilization. Therefore, its enol content is relatively low.

In 1,3,5-cyclohexanetrione, complete enolization converts the molecule into 1,3,5-trihydroxybenzene (phloroglucinol). The resulting structure contains a fully conjugated aromatic benzene ring.

The aromatic enol form is exceptionally stable due to aromatic resonance stabilization, which is much greater than the stabilization provided by simple conjugation or intramolecular hydrogen bonding. As a result, the keto-enol equilibrium is shifted strongly toward the enol form.

Therefore, 1,3,5-cyclohexanetrione exhibits the highest enol content among the given compounds.

Hence, the correct answer is $$\text{1,3,5-cyclohexanetrione}$$

The reaction sequence is a standard industrial route to the poly-alcohol pentaerythritol and its ketal derivatives. We analyse every step.

Step 1 : Reaction of acetaldehyde with excess formaldehyde in conc. NaOH
• Formaldehyde has no $$\alpha$$-hydrogen, so in strong alkali it undergoes the Cannizzaro reaction: $$2\,HCHO + OH^- \;\longrightarrow\; HCOO^- + CH_3OH$$

• In the presence of acetaldehyde, a cross-Cannizzaro/Aldol sequence occurs.   1 mol acetaldehyde first gives an enolate which adds to two molecules of formaldehyde (cross-aldol).   The aldehydic group thus produced is then reduced by hydride transfer from a third molecule of formaldehyde (Cannizzaro step).
The overall stoichiometry is

$$CH_3CHO + 3\,HCHO + 3\,NaOH \; \xrightarrow{\;\;\text{heat}\;\;} C(CH_2OH)_4 \;+\; 3\,NaHCOO$$

• The poly-alcohol obtained is pentaerythritol, $$P = C(CH_2OH)_4$$.   It possesses no -CHO group, hence it does not respond to Tollens’ reagent.

• The salt $$Q = NaHCOO$$ on acidification furnishes formic acid, $$HCOOH$$, which does reduce Tollens’ reagent. Thus the statements regarding $$P$$ and $$Q$$ are consistent.

Step 2 : Ketal formation between $$P$$ and excess cyclohexanone in the presence of catalytic p-toluenesulfonic acid (PTSA)
Acid catalyses the reaction of a ketone with two alcohol functions to give a ketal. Each molecule of cyclohexanone requires two -OH groups:

$$\text{cyclohexanone}+2\,ROH \;\xrightarrow{\;\;H^+\;\;} \text{cyclohexanone ketal}\;(\;C=O \text{ is replaced by }C(OR)_2\;)$$

Pentaerythritol has four -OH groups, therefore it can furnish the necessary two alcohol units twice. Consequently, with excess cyclohexanone it forms a diketal:

$$R = \bigl[\;C(CH_2O{-})_4\bigr]$$ Each pair $$\;{-}OCH_2{-}$$ originates from one molecule of cyclohexanone, so $$R$$ contains two cyclohexanone rings attached through ketal linkages.

Counting -CH2- groups in $$R$$
• Central pentaerythritol carbon: four $$CH_2$$ groups → $$4$$
• Each cyclohexanone ring (after ketal formation) supplies five $$CH_2$$ groups   (the carbonyl carbon becomes quaternary, carries no hydrogens).
  Two rings ⇒ $$2 \times 5 = 10$$
Total methylene groups $$= 4 + 10 = 14$$

Counting oxygen atoms in $$R$$
Every ketal linkage contains the two original alcohol oxygens; thus the four -OH groups of pentaerythritol become four ether oxygens. (The carbonyl oxygens of the ketones are lost as water during ketalisation).
Total oxygen atoms $$= 4$$

Required sum
$$\text{(number of }{-}CH_2{-}\text{ groups)} + \text{(number of O atoms)} = 14 + 4 = 18$$

Therefore, the asked sum is 18.

{image}

We need to identify the given organic reduction reaction where benzoyl chloride ($$C_6H_5COCl$$) is converted into benzaldehyde ($$C_6H_5CHO$$).

Statement (A): Rosenmund reduction. The Rosenmund reduction is a hydrogenation process where an acyl chloride is selectively reduced to an aldehyde using hydrogen gas ($$H_2$$) over palladium on barium sulfate ($$Pd\text{-}BaSO_4$$). The barium sulfate acts as a catalyst poison to prevent further reduction of the aldehyde into a primary alcohol. This matches the given reaction perfectly. This is TRUE.

Statement (B): Wolff-Kishner reduction. The Wolff-Kishner reduction is used to convert a carbonyl group (aldehyde or ketone) into a methylene group ($$\text{-}CH_2\text{-}$$) using hydrazine ($$NH_2NH_2$$) and a strong base like $$KOH$$. This is FALSE.

Statement (C): Stephen reduction. The Stephen reduction involves the preparation of aldehydes from nitrites ($$R\text{-}CN$$) using tin(II) chloride ($$SnCl_2$$) and hydrochloric acid ($$HCl$$), followed by hydrolysis. This is FALSE.

Statement (D): Etard reduction. The Etard reduction is a reaction that oxidizes an aromatic methyl group to an aldehyde group using chromyl chloride ($$CrO_2Cl_2$$). This is FALSE.

The given reaction represents the Rosenmund reduction.

Answer: Option A — Rosenmund reduction

We need to evaluate the truth of two statements about the Aldol reaction.

Statement I: "Acidity of alpha-hydrogens of aldehydes and ketones is responsible for the Aldol reaction."

The Aldol reaction proceeds through the following mechanism:

1. A base abstracts an $$\alpha$$-hydrogen from one carbonyl compound, forming a resonance-stabilised enolate ion.

2. This enolate ion (a nucleophile) attacks the electrophilic carbonyl carbon of another molecule.

3. The resulting alkoxide is protonated to give the $$\beta$$-hydroxy aldehyde (aldol product).

The entire reaction depends on the ability to form the enolate, which in turn requires that the $$\alpha$$-hydrogen be sufficiently acidic ($$pK_a \approx 17{-}20$$ for typical aldehydes and ketones). Without acidic $$\alpha$$-hydrogens, the enolate cannot form and the Aldol reaction cannot proceed. Therefore, Statement I is correct.

Statement II: "Reaction between benzaldehyde and ethanal will NOT give a Cross-Aldol product."

Let us analyse the two reactants:

- Benzaldehyde ($$C_6H_5CHO$$): This aldehyde has no $$\alpha$$-hydrogen (the carbon adjacent to $$C=O$$ is part of the benzene ring). Therefore, benzaldehyde cannot form an enolate and cannot act as the donor component.

- Ethanal ($$CH_3CHO$$): This aldehyde has three $$\alpha$$-hydrogens on the methyl group. It can readily form an enolate ion.

In a cross-Aldol reaction between these two, ethanal provides the enolate (nucleophile), which attacks the carbonyl carbon of benzaldehyde (electrophile). This produces 3-hydroxy-3-phenylpropanal, which upon dehydration gives cinnamaldehyde ($$C_6H_5CH=CHCHO$$). This is a well-known reaction called the Claisen-Schmidt reaction.

Since the cross-Aldol product is formed, Statement II is incorrect.

The correct answer is Option 1: Statement I is correct but Statement II is incorrect.

The reaction involves benzene reacting with $$CO$$ and $$HCl$$ in the presence of anhydrous $$AlCl_3$$ and $$CuCl$$.

Under these conditions, $$CO$$ and $$HCl$$ generate a formyl electrophile in situ. This electrophile undergoes electrophilic aromatic substitution with benzene, introducing a $$-CHO$$ group onto the aromatic ring.

As a result, benzene is converted into benzaldehyde.

This method of direct formylation of an aromatic ring using $$CO$$ and $$HCl$$ in the presence of $$AlCl_3$$ and $$CuCl$$ is known as the Gatterman-Koch reaction.

Stephen reaction converts nitriles into aldehydes and does not involve direct formylation of benzene.

Etard reaction oxidizes the methyl group of toluene to form an aldehyde using chromyl chloride.

Rosenmund reduction converts acid chlorides into aldehydes using hydrogen over a poisoned palladium catalyst.

Hence, the given reaction is the $$\text{Gatterman-Koch reaction}$$

The iodoform test is highly specific for compounds containing either a methyl ketone group ($$\text{CH}_3\text{--C}(=\text{O})--$$) or a secondary methyl alcohol group ($$\text{CH}_3\text{--CH(OH)}--$$) that can easily oxidize to a methyl ketone.

When starting with a geminal dihalide complex, treatment with a strong base like aqueous $$\text{KOH}$$ causes nucleophilic substitution to replace the halogen atoms with hydroxyl ($$-\text{OH}$$) groups. If this intermediate step generates a stable or unstable gem-diol that eliminates water to form a methyl ketone structure, it will give a positive iodoform reaction upon adding $$\text{KOI}$$.

Step-by-Step Analysis of Compound A:

1. Nucleophilic Substitution with Base:

   Compound A is 2,2-dichlorobutane ($$\text{CH}_3\text{--CH}_2\text{--C(Cl)}_2\text{--CH}_3$$). When treated with aqueous $$\text{KOH}$$, the two chlorine atoms on carbon-2 are substituted by two hydroxyl groups, forming an unstable gem-diol intermediate:

   $$\text{CH}_3\text{--CH}_2\text{--C(Cl)}_2\text{--CH}_3 \xrightarrow{\text{aq. KOH}} \text{CH}_3\text{--CH}_2\text{--C(OH)}_2\text{--CH}_3$$

2. Dehydration to Form a Functional Carbonyl:

   Since two hydroxyl groups cannot comfortably sit on the same carbon atom, the intermediate immediately loses a water molecule ($$-\text{H}_2\text{O}$$) to form 2-butanone:

   $$\text{CH}_3\text{--CH}_2\text{--C(OH)}_2\text{--CH}_3 \xrightarrow{-\text{H}_2\text{O}} \text{CH}_3\text{--CH}_2\text{--C}(=\text{O})\text{--CH}_3$$

3. Iodoform Conversion:

   The resulting 2-butanone contains the required methyl ketone ($$\text{--C}(=\text{O})\text{CH}_3$$) structural component. When potassium hypoiodite ($$\text{KOI}$$) is added, it undergoes oxidative cleavage to produce a bright yellow precipitate of iodoform ($$\text{CHI}_3$$) along with a potassium carboxylate salt:

   $$\text{CH}_3\text{--CH}_2\text{--C}(=\text{O})\text{--CH}_3 \xrightarrow{\text{KOI}} \text{CH}_3\text{--CH}_2\text{--COOK} + \text{CHI}_3 \downarrow$$

Conclusion:

As clearly depicted in the reaction scheme, 2,2-dichlorobutane converts directly into a methyl ketone intermediate under basic conditions, making it perfectly suited to deliver a positive iodoform response.

Answer: Option A

The four reagents that are routinely used to identify and distinguish aldehydes and ketones are Fehling’s solution, the Iodoform reagent, 2,4-DNP (Brady’s reagent) and Schiff’s reagent. Each of them gives a very characteristic observation that the examiner places in List-II. We recall the basic chemistry of every reagent first, then match.

Case A : Fehling’s solution
Fehling’s solution contains $$Cu^{2+}$$ complexed with tartrate ion in an alkaline medium. Only aliphatic aldehydes can reduce $$Cu^{2+}$$ to $$Cu_2O$$, giving a brick-red precipitate. Neither ketones nor aromatic aldehydes respond.
⇒ Observation matched with Fehling’s solution is “formation of a red (brick-red) precipitate”. This statement is numbered IV.

Case B : Iodoform test ($$I_2/NaOH$$)
The haloform reaction is given by carbonyl compounds that contain the $$CH_3CO$$- group (i.e. methyl ketones) and by acetaldehyde (which is readily oxidised to that group). The tell-tale observation is a pale-yellow precipitate having the characteristic smell of iodoform, $$CHI_3$$.
⇒ “Formation of yellow precipitate of $$CHI_3$$ (iodoform)” is statement II.

Case C : 2,4-DNP (Brady’s reagent)
Both aldehydes and ketones add to 2,4-dinitrophenylhydrazine giving intensely coloured yellow to orange crystalline hydrazones. This test is used merely to confirm the presence of the carbonyl (>C=O) group.
⇒ “Formation of yellow/orange crystalline precipitate” is statement I.

Case D : Schiff’s reagent
Schiff’s reagent is a decolourised solution of fuchsin dye in $$SO_2$$. With aldehydes the magenta colour reappears almost instantaneously, whereas ketones do not restore the colour under ordinary conditions.
⇒ “Solution turns colourless to magenta (positive with aldehydes)” is statement III.

Collecting all the matches:
A - IV (B - II) C - I D - III

Therefore the correct choice is
Option B which is: A - IV, B - II, C - I, D - III.

An aldehyde that possesses at least one $$\alpha$$-hydrogen undergoes the base-catalysed aldol reaction. In the given problem the reactant is ethanal (acetaldehyde) and the reagent is dilute aqueous $$NaOH$$ kept below about $$40^{\circ}\text{C}$$, so dehydration does not take place. Only the aldol addition is observed.

Step 1 - Enolate formation
$$CH_3CHO + OH^- \;\longrightarrow\; CH_2CHO^- + H_2O$$

Step 2 - Nucleophilic attack on a second carbonyl molecule
The enolate ion formed above attacks the carbonyl carbon of another acetaldehyde molecule:
$$CH_2CHO^- + CH_3CHO \;\longrightarrow\; CH_3CH(OH)CH_2CHO + OH^-$$

Step 3 - Protonation
The alkoxide produced in step 2 is protonated by water, regenerating the hydroxide ion used in step 1. The overall stoichiometry is therefore
$$2CH_3CHO \xrightarrow[\;dil.\;]{NaOH,\; 295\,\text{K}} CH_3CH(OH)CH_2CHO$$

The product is $$\beta$$-hydroxyaldehyde, specifically 3-hydroxybutanal (often called “aldol”). Among the options given, this structure corresponds to Option A.

Hence the correct answer is:
Option A which is: 3-hydroxybutanal (CH3CH(OH)CH2CHO).

We need to identify which compounds give a silver mirror test (Tollens' test) with ammoniacal silver nitrate ($$[Ag(NH_3)_2]^+$$).

Tollens' reagent (ammoniacal silver nitrate) acts as a mild oxidising agent that oxidises aldehydes and certain other reducing compounds while itself being reduced. In this process silver ions ($$Ag^+$$) are reduced to metallic silver ($$Ag$$), which deposits as a mirror on the test tube wall:
$$RCHO + 2[Ag(NH_3)_2]^+ + 2OH^- \rightarrow RCOO^- + 2Ag \downarrow + 4NH_3 + H_2O$$

Formic acid has the structure $$H-\underset{\displaystyle \|}{\overset{\displaystyle O}{C}}-OH$$. Unlike other carboxylic acids, it contains an aldehyde-like $$H-C=O$$ group in addition to the $$-OH$$ group. This structural feature allows it to act as a reducing agent, so formic acid gives a positive silver mirror test.

Formaldehyde (HCHO) is an aldehyde and readily reduces Tollens' reagent. It is oxidised all the way to formate or carbonate because both hydrogen atoms on the carbon can be oxidised, and thus formaldehyde gives a positive silver mirror test.

Benzaldehyde ($$C_6H_5CHO$$) is an aromatic aldehyde containing the $$-CHO$$ group, and all aldehydes give a positive Tollens' test whether they are aliphatic or aromatic. Consequently, benzaldehyde gives a positive silver mirror test.

Acetone ($$CH_3COCH_3$$) is a ketone that lacks the $$-CHO$$ group and is not easily oxidised by the mild oxidising agent in Tollens' reagent. Therefore, acetone does not give a silver mirror test.

Consequently, formic acid, formaldehyde, and benzaldehyde give a positive silver mirror test, whereas acetone does not. The correct answer is Option 1: A, B and C only.

The Claisen-Schmidt reaction to prepare dibenzalacetone involves the condensation of acetone with benzaldehyde:

$$ CH_3COCH_3 + 2C_6H_5CHO \xrightarrow{NaOH} C_6H_5CH=CHCOCH=CHC_6H_5 + 2H_2O $$

The molar mass of acetone (CH_3COCH_3) is 12×3 + 6 + 16 = 58 g/mol, that of benzaldehyde (C_6H_5CHO) is 12×7 + 6 + 16 = 106 g/mol, and that of dibenzalacetone (C_{17}H_{14}O) is 12×17 + 14 + 16 = 234 g/mol.

To prepare 351 g of dibenzalacetone:

$$ \text{Moles of dibenzalacetone} = \frac{351}{234} = 1.5 \text{ mol} $$

The available 87 g of acetone corresponds to:

$$ \text{Moles of acetone} = \frac{87}{58} = 1.5 \text{ mol} $$

Since the reaction requires 2 mol of benzaldehyde per mole of acetone:

$$ \text{Moles of benzaldehyde} = 2 \times 1.5 = 3 \text{ mol} $$

Thus, the mass of benzaldehyde needed is:

$$ \text{Mass} = 3 \times 106 = 318 \text{ g} $$

Therefore, 318 g of benzaldehyde is required.

We need to count the number of compounds that give a positive Fehling's test from: Benzaldehyde, Acetaldehyde, Acetone, Acetophenone, Methanal, 4-nitrobenzaldehyde, Cyclohexane carbaldehyde.

Recall what Fehling's test detects.

Fehling's test uses Fehling's reagent (an alkaline solution of cupric tartrate, deep blue in colour) to detect aliphatic aldehydes. Aldehydes reduce $$Cu^{2+}$$ (blue) to $$Cu_2O$$ (red-brown precipitate). Important: aromatic aldehydes do NOT give a positive Fehling's test because the benzene ring reduces the reactivity of the aldehyde group. Ketones also do not give a positive test.

Analyse each compound.

- Benzaldehyde ($$C_6H_5CHO$$): Aromatic aldehyde. Does NOT give positive Fehling's test.

- Acetaldehyde ($$CH_3CHO$$): Aliphatic aldehyde. Positive.

- Acetone ($$CH_3COCH_3$$): Ketone. Does NOT give positive test.

- Acetophenone ($$C_6H_5COCH_3$$): Aromatic ketone. Does NOT give positive test.

- Methanal ($$HCHO$$): Aliphatic aldehyde (simplest one). Positive.

- 4-Nitrobenzaldehyde ($$O_2N-C_6H_4-CHO$$): Aromatic aldehyde. Does NOT give positive Fehling's test.

- Cyclohexane carbaldehyde ($$C_6H_{11}CHO$$): Aliphatic aldehyde (cyclohexane ring is not aromatic). Positive.

Count.

Positive Fehling's test: Acetaldehyde, Methanal, Cyclohexane carbaldehyde = 3.

The answer is 3.

Ethanal ($$CH_3CHO$$) reacts with semicarbazide ($$NH_2NHCONH_2$$) to form a semicarbazone through a condensation reaction.

The reaction is:

$$ CH_3CHO + NH_2NHCONH_2 \rightarrow CH_3CH=NNHCONH_2 + H_2O $$

The product is acetaldehyde semicarbazone: $$CH_3CH=N-NH-CO-NH_2$$.

Counting the nitrogen atoms in the product:

1. The $$=N-$$ nitrogen

2. The $$-NH-$$ nitrogen

3. The $$-NH_2$$ nitrogen

Total nitrogen atoms = $$3$$.

Therefore, the answer is $$\boxed{3}$$.

Step 1 - Index of hydrogen deficiency for $$P$$
For an open-chain compound $$C_9H_{18}O_2$$ the index is
$$\text{IHD}= \dfrac{2\times 9+2-18}{2}=1$$
So $$P$$ possesses one and only one π bond (no ring).

Step 2 - Nature of functional groups in $$P$$
$$P$$ decolorises bromine water ⇒ the single IHD corresponds to a $$C=C$$ bond.
$$P$$ also gives a positive iodoform test; without any carbonyl present (IHD already used up), this can only arise from a secondary alcohol of the type $$CH_3-C^*(OH)-$$. Hence both oxygen atoms of $$P$$ are present as non-carbonyl oxygens (alcohol/ether). In short, $$P$$ is an >C=C< containing di-oxygenated alcohol (or alcohol-ether).

Step 3 - Oxidative ozonolysis of the double bond
Ozonolysis followed by $$H_2O_2/H_2O$$ cleaves the $$C=C$$, converting each double-bond carbon into a -COOH group while leaving every pre-existing -OH intact. Thus both fragments, $$Q$$ and $$R$$, are $$\beta$$-hydroxy carboxylic acids.

Step 4 - Why $$Q$$ shows, but $$R$$ lacks, the iodoform test
The iodoform test in acids is positive only when the acid contains a $$CH_3CO-$$ or $$CH_3CH(OH)-$$ unit. Therefore the -OH in $$Q$$ is on the carbon that already bears a methyl group, giving $$CH_3-C(OH)(\ldots )-COOH$$; $$R$$ lacks this arrangement.

Step 5 - Oxidation with PCC followed by heating
(i) PCC oxidises the secondary -OH in each $$\beta$$-hydroxy acid to a carbonyl group, producing a $$\beta$$-keto acid.
(ii) A classic property of $$\beta$$-keto acids is thermal decarboxylation:
$$R-CO-CH_2-COOH \xrightarrow{\Delta} R-CO-CH_3 + CO_2$$
Thus $$Q$$ → $$S$$ and $$R$$ → $$T$$, where both $$S$$ and $$T$$ are methyl ketones.

Step 6 - Counting oxygen atoms in $$S$$ and $$T$$
A methyl ketone contains only one oxygen atom (the carbonyl oxygen).
Hence $$n_O(S)=1$$ and $$n_O(T)=1$$.

Step 7 - Required sum
$$n_O(S)+n_O(T)=1+1=2$$

Therefore, the sum of the number of oxygen atoms present in $$S$$ and $$T$$ is 2.

Assertion A: Acyl chlorides can be subjected to Wolff-Kishner reduction.

This is False. Wolff-Kishner reduction requires treatment with hydrazine (NH$$_2$$NH$$_2$$) and strong base (KOH/NaOH) at high temperature. Acyl chlorides would react with hydrazine to form acid hydrazides rather than undergo the intended carbonyl reduction.

Reason R: Wolff-Kishner reduction converts C=O to CH$$_2$$.

This is True. The Wolff-Kishner reduction specifically reduces the carbonyl group to a methylene group (-CH$$_2$$-) through formation and decomposition of a hydrazone intermediate.

A is false but R is true.

Explanation:

The reagents are:

$$Zn(Hg),\ \mathrm{conc.}\ HCl,\ \Delta$$

These are the conditions for Clemmensen reduction.

Clemmensen reduction converts an aldehyde or ketone carbonyl group into a methylene group.

The given compound is:

$$C_6H_5-CH(OH)-CH(CH_3)-CH_2-CHO$$

The aldehyde group:

$$-CHO$$

is reduced to:

$$-CH_3$$

Hence,

$$-CH_2-CHO\longrightarrow-CH_2-CH_3$$

The benzylic $$-OH$$ group is not affected under the expected transformation.

Therefore, the product formed is:

$$C_6H_5-CH(OH)-CH(CH_3)-C_2H_5$$

This corresponds to option B.

Molecule (a), acetylacetone, is the most acidic because it lacks the electron-donating alkoxy groups found in the esters. Therefore, it will undergo deprotonation most readily in a basic medium.

Statement I: True

• Reactant: A keto-amide ($$\text{H}_2\text{N-C}(=\text{O})-\text{CH}_2-\text{CH}_2-\text{C}(=\text{O})-\text{CH}_3$$).
• Reagent: Clemmensen reduction utilizes zinc amalgam in concentrated hydrochloric acid ($$\text{Zn(Hg)} / \text{HCl}$$).
• Reaction:
  1. The strongly acidic medium ($$\text{HCl}$$) hydrolyzes the primary amide group ($$\text{-CONH}_2$$) into a carboxylic acid group ($$\text{-COOH}$$).
  2. The ketone group ($$\text{-C=O-}$$) is reduced to a methylene group ($$\text{-CH}_2-$$).
• Product: Pentanoic acid ($$\text{HOOC-CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3$$), which matches the product shown in Statement I.

Statement II: False

• Reactant: A halo-ketone ($$\text{Cl-CH}_2-\text{CH}_2-\text{CO-CH}_3$$).
• Reagent: Wolff-Kishner reduction uses hydrazine and a strong base ($$\text{NH}_2\text{NH}_2 / \text{KOH}$$) at high temperatures.
• Reaction: While the carbonyl group is targeted for reduction to an alkane, the presence of a strong base ($$\text{KOH}$$) induces a highly favorable competing side reaction—dehydrohalogenation (elimination of $$\text{HCl}$$). Instead of obtaining the simple reduced haloalkane ($$\text{Cl-CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3$$), elimination reactions dominate, making the statement false.
  

• Statement I is true.
• Statement II is false.

Correct Answer: [C] Statement I is true but Statement II is false

Analysis of Assertion (A):

The reactant provided is a keto-alcohol (contains both a carbonyl group, $$\text{C=O}$$, and a hydroxyl group, $$\text{--OH}$$). When treated with Clemmensen reducing agent (amalgamated zinc and concentrated hydrochloric acid, $$\text{Zn--Hg/HCl}$$), two distinct chemical transformations happen simultaneously:

1. Reduction of the Carbonyl Group: The ketone group ($$\text{C=O}$$) is successfully reduced to a methylene group ($$\text{--CH}_2\text{--}$$).
2. Substitution of the Hydroxyl Group: Because the reaction happens under highly acidic conditions using concentrated $$\text{HCl}$$, the secondary alcohol ($$\text{--OH}$$) undergoes a nucleophilic substitution reaction where it is replaced by a chlorine atom ($$\text{--Cl}$$).

Since the final product doesn't accurately shows the carbonyl converted to a alkane linkage and the hydroxyl replaced by chlorine, Assertion A is FALSE.

Analysis of Reason (R):

Reason (R) states that $$\text{Zn--Hg/HCl}$$ is standardly employed to reduce a carbonyl group down to a active methylene ($$\text{--CH}_2\text{--}$$) group. This is the fundamental definition of a Clemmensen reduction, making Reason R perfectly TRUE as an independent chemical statement.

Assertion A: Acetal/Ketal is stable in basic medium. — TRUE. Acetals and ketals are stable under basic conditions but hydrolyze under acidic conditions.

Reason R: The high leaving tendency of alkoxide ion gives stability to acetal/ketal in basic medium. — FALSE. Alkoxide ions are actually poor leaving groups (high pKa ~16). The stability in basic medium is because the first step of hydrolysis requires protonation of the oxygen (acid catalysis), which doesn't occur in basic conditions.

Since A is true and R is false, the correct answer is Option A: $$\boxed{\text{A is true but R is false}}$$.

1. Step 1: Cyanohydrin Formation and Alcoholysis ($$\text{1) NaCN}$$, $$\text{2) EtOH, H}_2\text{O, H}^\oplus$$)

   The starting material is a ketone, 4-chloroacetophenone ($$\text{Cl--C}_6\text{H}_4\text{--COCH}_3$$).

   • First, nucleophilic addition of cyanide ($$\text{CN}^\ominus$$) to the carbonyl group yields a cyanohydrin intermediate: $$\text{Cl--C}_6\text{H}_4\text{--C(OH)(CN)(CH}_3)$$.
   • Second, treating the cyanohydrin with ethanol ($$\text{EtOH}$$) under acidic aqueous conditions hydrolyzes and esterifies the nitrile group ($$\text{--CN}$$) into an ethyl ester group ($$\text{--CO}_2\text{Et}$$).

   This gives the first intermediate shown in the scheme: ethyl 2-(4-chlorophenyl)-2-hydroxypropanoate.




2. Step 2: Grignard Reaction with Excess Methylmagnesium Bromide ($$\text{MeMgBr (excess)}$$ followed by $$\text{H}_3\text{O}^\oplus$$)

   The intermediate contains two key functional groups: a tertiary alcohol ($$\text{--OH}$$) and an ester ($$\text{--CO}_2\text{Et}$$).

   • The first equivalent of the Grignard reagent ($$\text{MeMgBr}$$) acts as a base and deprotonates the acidic hydroxyl proton ($$\text{--OH}$$), which is later regenerated during acidic workup.
   • Another equivalent of $$\text{MeMgBr}$$ undergoes nucleophilic acyl substitution at the ester carbonyl carbon, displacing the ethoxide leaving group ($$\text{--OEt}$$) to form a methyl ketone intermediate ($$\text{--COCH}_3$$).
   • Because the Grignard reagent is added in excess, a final equivalent immediately attacks this newly formed ketone group to yield a tertiary alkoxide.

   After acidic workup ($$\text{H}_3\text{O}^\oplus$$), the final product contains a highly branched tertiary alcohol unit attached to the original aromatic framework.

Structure of the Final Product:

The carbon atom alpha to the benzene ring preserves its original hydroxyl group ($$\text{--OH}$$) and methyl group ($$\text{--Me}$$). The ester group ($$\text{--CO}_2\text{Et}$$) is converted completely into a 2-hydroxypropan-2-yl group ($$\text{--C(OH)Me}_2$$).

Therefore, the structural formula of the final major product is:

$$\text{Cl--C}_6\text{H}_4\text{--C(OH)(Me)--C(OH)(Me)}_2$$

Conclusion:

The reaction sequence effectively introduces multiple methyl groups by transforming the ester functionality into a tertiary alcohol while maintaining the structural integrity of the chloro-substituted benzene ring.

Answer: The molecular structure labeled as 'C' in the reaction scheme, which corresponds to $$\text{Cl--C}_6\text{H}_4\text{--C(OH)(Me)--C(OH)(Me)}_2$$.

Step 1: Conjugate Addition (1,4-addition)

• The combination of MeMgBr and CuI forms a Gilman reagent (organocopper species), which selectively undergoes 1,4-addition (conjugate addition) rather than 1,2-addition.
• The methyl group $$(\text{Me}^{-})$$ attacks the $$\beta$$-carbon of the 3-methylcyclohex-2-en-1-one.
• This creates a quaternary center at the 3-position and generates a nucleophilic enolate intermediate.

Step 2: Alpha-Alkylation

• Instead of quenching with water, the reaction adds n-PrI (n-propyl iodide).
• The enolate oxygen-carbon double bond reforms, and the $$\alpha $$-carbon acts as a nucleophile to attack the n-propyl iodide via an $$S_{N}2$$mechanism.
• The n-propyl group is added to the 2-position (the $$\alpha $$-carbon).

The major product of this reaction sequence is 3,3-dimethyl-2-propylcyclohexanone.

Propanal undergoes crossed aldol condensation with methanal in the presence of dilute $$NaOH$$ to form a $$\beta$$-hydroxy aldehyde.

On heating, dehydration occurs to give the $$\alpha,\beta$$-unsaturated aldehyde:

$$CH_2=C(CH_3)CH_2CHO$$

Addition of $$HCN$$ generated from $$NaCN/H_3O^+$$ across the carbonyl group forms a cyanohydrin.

Acidic hydrolysis converts the $$-CN$$ group into a $$-COOH$$ group, giving an $$\alpha$$-hydroxy carboxylic acid with molecular formula:

$$C_5H_8O_3$$

The carbon bearing the $$-OH$$ group becomes a new chiral center.

Since the carbonyl group is planar, cyanide attacks from both faces with equal probability.

Hence, the product is obtained as a racemic mixture.

The presence of the $$-COOH$$ group causes the compound to react with saturated $$NaHCO_3$$, liberating carbon dioxide.

$$RCOOH+NaHCO_3\rightarrow RCOONa+CO_2+H_2O$$

Hence, the correct answer is:

$$\boxed{\text{c}}$$

Step 1: Reduction with $$\text{LiAlH}_4$$ (excess)

The starting material is $$1,1,2,2\text{-tetracyanoethane}$$:

$$\text{(NC)}_2\text{CH}-\text{CH(CN)}_2$$

Lithium aluminum hydride ($$\text{LiAlH}_4$$) is a strong reducing agent that reduces all four nitrile ($$-\text{C}\equiv\text{N}$$) groups completely into primary amine ($$-\text{CH}_2\text{NH}_2$$) groups.

The intermediate tetra-amine formed after workup with $$\text{H}_2\text{O}$$ is:

$$\text{(H}_2\text{NCH}_2)_2\text{CH}-\text{CH}(\text{CH}_2\text{NH}_2)_2$$

Step 2: Reaction with Acetophenone (excess)

Acetophenone ($$\text{Ph}-\text{CO}-\text{CH}_3$$) reacts with primary amines ($$-\text{NH}_2$$) via a nucleophilic addition-elimination reaction to form imines ($$-\text{N}=\text{C}\langle$$):

$$\text{R}-\text{NH}_2 + \text{O}=\text{C}(\text{CH}_3)\text{Ph} \longrightarrow \text{R}-\text{N}=\text{C}(\text{CH}_3)\text{Ph} + \text{H}_2\text{O}$$

Since acetophenone is in excess, all four primary amine groups will react to form four imine groups in the final product $$P$$.

Step 3: Counting $$\text{sp}^2$$ Hybridised Carbon Atoms

Let's analyze the hybridization of carbons in one attached acetophenone-derived unit ($$-\text{N}=\text{C}(\text{CH}_3)\text{Ph}$$):

1. Imine Carbon ($$-\text{N}=\underline{\text{C}}-$$): $$1\text{ sp}^2$$ carbon
2. Methyl Carbon ($$-\text{CH}_3$$): $$0\text{ sp}^2$$ carbons ($$\text{sp}^3$$)
3. Phenyl Ring ($$-\text{C}_6\text{H}_5$$): All $$6$$ carbons in the benzene ring are $$\text{sp}^2$$ hybridised.

$$\text{Number of sp}^2\text{ carbons per unit} = 1 + 6 = 7$$

$$\text{Total number of sp}^2\text{ hybridised carbons} = 4 \times 7 = \mathbf{28}$$

The starting material is a 2-alkylcyclohexanone (a cyclic ketone with an alkyl group $$R$$ at the $$\alpha$$-position).

When treated with a strong oxidizing agent like $$\text{KMnO}_4$$, cyclic ketones undergo oxidative cleavage of the $$\text{C}-\text{C}$$ bond adjacent to the carbonyl group.

Regioselectivity: The bond cleavage preferentially occurs between the carbonyl carbon ($$\text{C}_1$$) and the more substituted $$\alpha$$-carbon ($$\text{C}_2$$, which bears the alkyl group $$R$$).

The carbonyl carbon ($$\text{C}_1$$) is oxidized to a carboxylic acid ($$-\text{CO}_2\text{H}$$), and the substituted $$\alpha$$-carbon ($$\text{C}_2$$) is oxidized to a ketone ($$-\text{C}(=\text{O})R$$) group because it holds the alkyl chain.

Thus, the ring opens up to form a keto-acid.

Product 'A' is then treated with hydrazine and a strong base ($$\text{NH}_2\text{NH}_2, \text{KOH}$$) followed by acidic workup ($$\text{H}_3\text{O}^+$$). This reagent combination describes a Wolff-Kishner Reduction.

The Wolff-Kishner reduction selectively converts carbonyl groups ($$\text{C}=\text{O}$$) of aldehydes and ketones into methylene groups ($$\text{CH}_2$$).

Selectivity: It does not reduce carboxylic acid ($$-\text{CO}_2\text{H}$$) groups.

The ketone carbonyl group ($$\text{C}=\text{O}$$) attached to $$R$$ is completely reduced to a $$-\text{CH}_2-$$ group.

Correct Option B.

$$LiAlH_4$$ is a powerful reducing agent so, it will completely reduces primary amides down to primary amines by converting the carbonyl carbon into a methylene group ($$-CH2-$$). AND The ketone carbonyl group is cleanly reduced to a secondary alcohol group.

$$LiAlH_4$$ is a nucleophilic reducing agent, meaning it attacks electron-deficient carbon atoms (like carbonyls). It does not reduce ordinary isolated carbon-carbon double bonds because alkenes are electron-rich and repel nucleophilic attack. Therefore, the central double bond remains completely unaffected.

Hence Final structure would be C

Match the reactions with their reagents.

(A) Hoffmann Degradation: Conversion of amide to amine using $$Br_2/NaOH$$ — matches (III).

(B) Clemmensen Reduction: Reduction of $$C=O$$ to $$CH_2$$ using $$Zn-Hg/HCl$$ — matches (IV).

(C) Cannizzaro Reaction: Disproportionation of aldehydes (without $$\alpha$$-H) using concentrated KOH — matches (I).

(D) Reimer-Tiemann Reaction: Introduction of $$CHO$$ group to phenol using $$CHCl_3/NaOH$$ — matches (II).

Matching: A-III, B-IV, C-I, D-II

This matches Option 3.

The answer is $$\boxed{A-III, B-IV, C-I, D-II}$$ (Option 3).

Firstly CN- nucleophile gets attached to the carbonyl reducing it to alcohol. In the next step EtOH in acidic medium on reaction with a Cyanide group results in the formation of Ethyl Ester. At last the 2 CH3- from Grignard Reagent get attached to the ester, firstly removing -OEt since it is a good leaving group then reducing the carbonyl carbon to -OH.

Aldol Addition: Isobutyraldehyde reacts with Formaldehyde in the presence of KOH. Since Isobutyraldehyde has only one alpha-hydrogen, it adds one HCHO unit to form 2,2-dimethyl-3-hydroxypropanal.

(CH3)2CHCHO + HCHO  ->  HO-CH2-C(CH3)2-CHO

2. Cyanohydrin Formation: The aldehyde group reacts with $KCN$ to form a cyanohydrin.

$$HO-CH_2-C(CH_3)_2-CHO -> HO-CH2-C(CH3)2-CH(OH)CN

3. Hydrolysis & Cyclization: Acidic hydrolysis  converts the $-CN$ to $-COOH$. The gamma-hydroxy group then attacks the carboxylic acid to form the stable 5-membered Lactone (Y).

1. Formation of Product [A]

The starting material is 2-methylpentanal. It is treated with $$\text{NaHSO}_3$$, $$\text{dil. HCl}$$, $$\text{NaCN}$$, and $$\text{H}_2\text{O}$$, which are the standard reagents for cyanohydrin formation via the bisulfite addition pathway.

• Mechanism: The aldehyde group ($$-\text{CHO}$$) undergoes nucleophilic addition by the cyanide ion ($$\text{CN}^-$$), followed by protonation to yield a cyanohydrin.
• Structure of [A]:
  $$A = \text{CH}_3\text{CH}_2\text{CH}_2\text{CH(CH}_3)\text{CH(OH)CN}$$
  

2. Formation of Product [B]

Product [A] (the cyanohydrin) is treated with Lithium Aluminum Hydride ($$\text{LiAlH}_4$$), a powerful reducing agent.

• Mechanism: $$\text{LiAlH}_4$$ reduces the nitrile group ($$-\text{CN}$$) down to a primary amine group ($$-\text{CH}_2\text{NH}_2$$). The alcohol group ($$-\text{OH}$$) remains unaffected after workup.
• Structure of [B]:
  $$B = \text{CH}_3\text{CH}_2\text{CH}_2\text{CH(CH}_3)\text{CH(OH)CH}_2\text{NH}_2$$

3. Formation of Product [C]

Product [A] is treated with $$\text{HCl}/\text{H}_2\text{O}$$ and heated ($$\Delta$$).

• Mechanism: Acidic hydrolysis of a nitrile group ($$-\text{CN}$$) converts it into a carboxylic acid group ($$-\text{COOH}$$). Because it is a cyanohydrin, this yields a $$\alpha$$-hydroxy acid.
• Structure of [C]:$$C = \text{CH}_3\text{CH}_2\text{CH}_2\text{CH(CH}_3)\text{CH(OH)COOH}$$

Molecules having 

give positive iodoform test

C is the only option with a di-ketone being in conjugation with double bonds, the other options are not conjugated. Therefore, C is the correct option.

Direct oxidation of toluene usually produces benzoic acid because strong oxidizing agents over-oxidize the methyl group.

Controlled oxidation using $$\mathrm{CrO_3}$$ in acetic anhydride $$\mathrm{((CH_3CO)_2O)}$$ prevents this over-oxidation.

The methyl group is first converted into the stable intermediate benzylidene diacetate.

This intermediate resists further oxidation.

Acid hydrolysis using $$\mathrm{H_3O^+}$$ then converts the intermediate into benzaldehyde $$\mathrm{(C_6H_5CHO)}$$.

Reagents such as $$\mathrm{KMnO_4}$$ directly oxidize toluene into benzoic acid $$\mathrm{(C_6H_5COOH)}$$.

The Gattermann-Koch reaction using $$\mathrm{CO/HCl/AlCl_3}$$ introduces a formyl group onto the aromatic ring rather than oxidizing the side chain.

Hence, the correct reagent sequence for conversion of toluene to benzaldehyde is Option $$\mathrm{(B)}$$.

Thus, D is the right option.

We need to find the final product A in the reaction sequence where butanone reacts with HCN, followed by treatment with 95% $$H_2SO_4$$ and heat.

Butanone ($$CH_3CH_2COCH_3$$) undergoes nucleophilic addition with HCN to form a cyanohydrin:

$$CH_3CH_2COCH_3 + HCN \rightarrow CH_3CH_2C(OH)(CN)CH_3$$

The product is 2-hydroxy-2-methylbutanenitrile.

Concentrated $$H_2SO_4$$ (95%) acts as both a dehydrating agent and a hydrolysis catalyst:

First, the nitrile group ($$-CN$$) is hydrolyzed to a carboxylic acid group ($$-COOH$$) in the acidic medium:

$$CH_3CH_2C(OH)(CN)CH_3 \xrightarrow{H_2SO_4} CH_3CH_2C(OH)(COOH)CH_3$$

Then, dehydration occurs — the hydroxyl group and a hydrogen from the adjacent carbon are eliminated:

$$CH_3CH_2C(OH)(COOH)CH_3 \xrightarrow{-H_2O} CH_3CH=C(CH_3)COOH$$

The final product is $$CH_3-CH=C(CH_3)-COOH$$, which is 2-methylbut-2-enoic acid (also known as tiglic acid or $$\alpha,\beta$$-unsaturated acid).

Therefore, the correct answer is Option A.

The structure of compound P is pent-3-en-2-ol, $$CH_3CH=CHCH(OH)CH_3$$. It contains one carbon-carbon double bond and one hydroxyl (-OH) group.

Step 1 (Cleavage of the C=C bond)
Under complete ozonolysis followed by reductive work-up (O$$\_3$$, Zn/H$$\_2$$O), every carbon of the C=C bond is converted into a carbonyl centre. Hence

$$CH_3CH=CHCH(OH)CH_3 \xrightarrow[H_2O]{O_3/Zn} CH_3CHO + CH_3CH(OH)CHO$$

Therefore, one molecule of P gives two organic fragments:

• Acetaldehyde, $$CH_3CHO$$ (achiral, because the carbonyl carbon is attached to two identical atoms—oxygen on one side and nothing else asymmetric).
• 2-Hydroxypropanal, $$CH_3CH(OH)CHO$$.

Step 2 (Checking chirality of each product)

For acetaldehyde there is no stereogenic (asymmetric) carbon, so it is achiral.

In 2-hydroxypropanal the middle carbon (shown in bold) $$CH_3\mathbf{C}H(OH)CHO$$ is attached to four different groups: $$CH_3$$, $$H$$, $$OH$$ and $$CHO$$. Thus it is a stereogenic centre and the molecule exists as two non-superimposable mirror images (enantiomers), labelled R and S.

Step 3 (Total count of distinct chiral molecules)

Because 2-hydroxypropanal forms one pair of enantiomers, the number of different chiral molecules obtained from a single molecule of P is 2.

Hence, the total number of chiral molecules produced is 2.

To determine which reactions produce benzaldehyde, the reagents and reaction conditions must be analysed.

In reaction $$\mathrm{(A)}$$, benzoic acid first reacts with $$\mathrm{SOCl_2}$$ to form benzoyl chloride.

Subsequent treatment with $$\mathrm{H_2/Pd/BaSO_4}$$ performs the:

$$\mathrm{Rosenmund\ Reduction}$$

which selectively converts the acid chloride into benzaldehyde.

Hence, reaction $$\mathrm{(A)}$$ gives benzaldehyde.

In reaction $$\mathrm{(B)}$$, benzyl alcohol reacts with $$\mathrm{CrO_3/H_2SO_4}$$ (Jones reagent).

This strong oxidising agent converts the primary alcohol completely into benzoic acid.

Hence, benzaldehyde is not obtained.

In reaction $$\mathrm{(C)}$$, methyl benzoate is treated with $$\mathrm{NaBH_4}$$.

$$\mathrm{NaBH_4}$$ is too mild to reduce esters.

Therefore, no aldehyde is formed and PCC has no effect.

Hence, benzaldehyde is not obtained.

In reaction $$\mathrm{(D)}$$, toluene reacts with $$\mathrm{CrO_3/(CH_3CO)_2O}$$ to form a protected gem-diacetate intermediate.

Acid hydrolysis using $$\mathrm{H_3O^+,\ \Delta}$$ converts this intermediate into benzaldehyde.

Hence, reaction $$\mathrm{(D)}$$ gives benzaldehyde.

Therefore, the reactions producing benzaldehyde are: $$\mathrm{(A)\ and\ (D)}$$
Thus, the right option is C.

The given molecule contains:

1. An ester group $$\mathrm{(-COOEt)}$$

2. An aldehyde group $$\mathrm{(-CHO)}$$

3. A carbon-carbon double bond inside the cyclopentene ring

The reagent used is:

$$\mathrm{DIBAL\text{-}H \ followed\ by\ H_2O}$$

$$\mathrm{DIBAL\text{-}H}$$ is a mild and selective reducing agent.

It partially reduces esters to aldehydes.

Unlike $$\mathrm{LiAlH_4}$$, it does not reduce esters completely to primary alcohols.

General reaction:

$$\mathrm{RCOOR' \xrightarrow[\ H_2O\ ]{DIBAL\text{-}H} RCHO}$$

The reaction occurs through formation of a tetrahedral aluminium intermediate.

At low temperature, the reduction stops at the aldehyde stage.

Also:

$$\mathrm{DIBAL\text{-}H}$$ does not reduce isolated $$\mathrm{C=C}$$ bonds.

Therefore, the double bond in the cyclopentene ring remains unchanged.

Applying this to the given compound:

$$\mathrm{-COOEt \longrightarrow -CHO}$$

while the existing aldehyde group and alkene remain unaffected.

Thus, the final product contains aldehyde groups at both ends of the molecule.

Therefore, the correct answer is:

$$\boxed{\mathrm{Option\ A}}$$

The starting compound is:

$$\mathrm{p\text{-}Aminobenzonitrile}$$

It reacts with:

$$\mathrm{DIBAL\text{-}H}$$

followed by hydrolysis.

DIBAL-H selectively reduces the nitrile group:

$$\mathrm{-CN}$$

to an aldehyde.

$$\mathrm{H_2N-C_6H_4-CN \xrightarrow[DIBAL\text{-}H]{H_2O} H_2N-C_6H_4-CHO}$$

Thus, compound $$\mathrm{A}$$ is:

$$\mathrm{p\text{-}Aminobenzaldehyde}$$

Compound $$\mathrm{A}$$ then reacts with acetaldehyde:

$$\mathrm{CH_3CHO}$$

in presence of:

$$\mathrm{NaOH/\Delta}$$

This is a:

$$\mathrm{Cross\ Aldol\ Condensation}$$

Compound $$\mathrm{A}$$ has no:

$$\mathrm{\alpha\text{-}hydrogen}$$

Hence, it acts only as the electrophile.

Acetaldehyde possesses:

$$\mathrm{\alpha\text{-}hydrogens}$$

and forms an enolate ion.

The enolate attacks the carbonyl carbon of compound $$\mathrm{A}$$.

Subsequent dehydration forms an:

$$\mathrm{\alpha,\beta\text{-}unsaturated\ aldehyde}$$

The final product $$\mathrm{B}$$ is:

$$\mathrm{H_2N-C_6H_4-CH=CH-CHO}$$

Correct option:

$$\mathrm{B}$$

We need to follow the sequence of reactions starting from hex-4-en-2-ol, which has the structure: $$CH_3-CH(OH)-CH_2-CH=CH-CH_3$$.

PCC (Pyridinium Chlorochromate) is a mild oxidizing agent that converts a secondary alcohol to a ketone without further oxidation. $$CH_3-CH(OH)-CH_2-CH=CH-CH_3 \xrightarrow{PCC} CH_3-CO-CH_2-CH=CH-CH_3$$ Compound A is hex-4-en-2-one.

Sodium hypoiodite is the iodoform reagent. It reacts with methyl ketones ($$R-CO-CH_3$$) to give a carboxylate salt and iodoform ($$CHI_3$$): $$CH_3-CO-CH_2-CH=CH-CH_3 + 3NaOI \to CHI_3 + CH_3-CH=CH-CH_2-COONa$$. Compound B is sodium pent-3-enoate ($$CH_3CH=CHCH_2COONa$$).

Heating sodium pent-3-enoate with soda lime ($$NaOH + CaO$$) causes decarboxylation, removing $$CO_2$$ and replacing the $$-COONa$$ group with $$-H$$: $$CH_3-CH=CH-CH_2-COONa \xrightarrow{NaOH/CaO} CH_3-CH=CH-CH_3 + Na_2CO_3$$. Compound C is 2-butene ($$CH_3CH=CHCH_3$$).

The correct answer is Option C: 2-butene.

Reaction $$\mathrm{A}$$ shows the conversion of a nitrile to an aldehyde using:

$$\mathrm{SnCl_2/HCl}$$

followed by hydrolysis.

This reaction is known as the Stephen reaction.

$$\mathrm{R-CN \xrightarrow[\ H_2O\ ]{SnCl_2/HCl} R-CHO}$$

Therefore:

$$\mathrm{A \rightarrow III}$$

Reaction $$\mathrm{B}$$ shows oxidation of toluene using chromyl chloride:

$$\mathrm{CrO_2Cl_2}$$

to form benzaldehyde.

This reaction is known as the Etard reaction.

$$\mathrm{C_6H_5CH_3 \xrightarrow{CrO_2Cl_2} C_6H_5CHO}$$

Therefore:

$$\mathrm{B \rightarrow II}$$

Reaction $$\mathrm{C}$$ shows reduction of an acid chloride to an aldehyde using:

$$\mathrm{H_2/Pd-BaSO_4}$$

This reaction is known as the Rosenmund reduction.

$$\mathrm{R-COCl \xrightarrow{H_2/Pd-BaSO_4} R-CHO}$$

Therefore:

$$\mathrm{C \rightarrow IV}$$

Reaction $$\mathrm{D}$$ shows formylation of benzene using:

$$\mathrm{CO + HCl/AlCl_3}$$

This reaction is known as the Gatterman-Koch reaction.

$$\mathrm{C_6H_6 \xrightarrow{CO/HCl,\ AlCl_3} C_6H_5CHO}$$

Therefore:

$$\mathrm{D \rightarrow I}$$

Correct matching:

$$\boxed{\mathrm{A-III,\ B-II,\ C-IV,\ D-I}}$$

Thus, the right option is A.

Benzaldehyde $$\mathrm{(C_6H_5CHO)}$$ reacts with $$\mathrm{CH_3MgBr}$$ to form the secondary alcohol $$\mathrm{C_6H_5CH(OH)CH_3}$$, which on oxidation gives acetophenone $$\mathrm{(C_6H_5COCH_3)}$$.

Acetaldehyde $$\mathrm{(CH_3CHO)}$$ reacts with $$\mathrm{C_6H_5MgBr}$$ to form the same secondary alcohol $$\mathrm{C_6H_5CH(OH)CH_3}$$, which on oxidation again gives acetophenone $$\mathrm{(C_6H_5COCH_3)}$$.

In the reaction of ethyl benzoate $$\mathrm{(C_6H_5COOC_2H_5)}$$ with excess $$\mathrm{CH_3MgBr}$$, the first addition temporarily forms acetophenone $$\mathrm{(C_6H_5COCH_3)}$$.

However, ketones readily react with Grignard reagents. The second equivalent of $$\mathrm{CH_3MgBr}$$ attacks the ketone immediately to form a tertiary alcohol $$\mathrm{C_6H_5C(OH)(CH_3)_2}$$.

Thus, acetophenone is not obtained as the final major product.

Reaction of benzoyl chloride $$\mathrm{(C_6H_5COCl)}$$ with dialkylcadmium $$\mathrm{((CH_3)_2Cd)}$$ stops at the ketone stage because dialkylcadmium reagents are less reactive than Grignard reagents.

Hence, the sequence that does not yield acetophenone as the major product is Option $$\mathrm{(C)}$$.

Option $$\mathrm{C}$$ is incorrect.

In reaction $$\mathrm{(C)}$$, the starting compound is a primary alcohol.

Treatment with $$\mathrm{Na_2Cr_2O_7/H_2SO_4}$$ forms a strong oxidising medium.

Primary alcohols under these conditions are completely oxidised into carboxylic acids.

Hence, the actual product formed is:

$$\mathrm{p\text{-}Toluic\ Acid}$$

and not $$\mathrm{p\text{-}Tolualdehyde}$$.

Reaction $$\mathrm{(A)}$$ is the $$\mathrm{Rosenmund\ Reduction}$$, where $$\mathrm{H_2/Pd-BaSO_4}$$ selectively reduces an acid chloride into an aldehyde.

Reaction $$\mathrm{(B)}$$ uses $$\mathrm{DIBAL\text{-}H}$$, which reduces a nitrile into an aldehyde after hydrolysis.

Reaction $$\mathrm{(D)}$$ also uses $$\mathrm{DIBAL\text{-}H}$$, which selectively reduces an ester into an aldehyde.

Therefore, the incorrect reaction is:

$$\mathrm{C}$$

The reaction sequence begins with propanenitrile (CH₃-CH₂-CN) reacting with methylmagnesium bromide (CH₃MgBr), a Grignard reagent, in the presence of dry ether. The nucleophilic methyl group from the Grignard reagent attacks the electrophilic carbon of the nitrile group, forming an intermediate imine salt designated as compound A (CH₃-CH₂-C(CH₃)=NMgBr).

This intermediate is then immediately subjected to acidic hydrolysis (H₃O⁺), which cleaves the carbon-nitrogen double bond to yield a ketone. Consequently, compound B is butan-2-one (CH₃-CH₂-CO-CH₃).

In the final step, butan-2-one undergoes a Clemmensen reduction using zinc amalgam (Zn-Hg) and concentrated hydrochloric acid (HCl). This reduction explicitly targets the carbonyl group, completely deoxygenating it and converting the >C=O group into a methylene (-CH₂-) group. Therefore, the final product, the correct structure of C, is the straight-chain alkane butane (CH₃-CH₂-CH₂-CH₃).

Explanation: Enamine formation requires removal of an $$\mathrm{\alpha}$$-hydrogen from the carbon adjacent to the carbonyl group in the final dehydration step.

Therefore, a ketone must contain at least one $$\mathrm{\alpha}$$-hydrogen to form an enamine.

In di-$$\mathrm{tert}$$-butyl ketone, the $$\mathrm{\alpha}$$-carbon is bonded to three methyl groups and the carbonyl carbon, $$\mathrm{-C(CH_3)_3}$$.

Thus, the $$\mathrm{\alpha}$$-carbon contains no hydrogen atoms.

Because no $$\mathrm{\alpha}$$-hydrogen is available, formation of the required $$\mathrm{C=C}$$ bond is impossible.

Hence, di-$$\mathrm{tert}$$-butyl ketone cannot form an enamine.

Ketones such as diethyl ketone and cyclohexanone possess $$\mathrm{\alpha}$$-hydrogens and therefore readily form enamines.

Correct Option: $$\mathrm{(B)}$$

We need to find which reactant gives the target alcohol on reaction with one mole of phenyl magnesium bromide ($$PhMgBr$$) followed by acidic hydrolysis.

Grignard reagents ($$RMgBr$$) react with various carbonyl compounds as follows:

(i) With formaldehyde ($$HCHO$$): Gives a primary alcohol ($$RCH_2OH$$)

(ii) With other aldehydes ($$R'CHO$$): Gives a secondary alcohol ($$RR'CHOH$$)

(iii) With ketones ($$R'COR''$$): Gives a tertiary alcohol

(iv) With nitriles ($$R'C \equiv N$$): After hydrolysis, gives a ketone

Based on the question context, the target alcohol is a secondary alcohol with one phenyl group. When $$PhMgBr$$ reacts with acetaldehyde ($$CH_3CHO$$):

$$PhMgBr + CH_3CHO \rightarrow CH_3CH(OMgBr)Ph \xrightarrow{H_3O^+} CH_3CH(OH)Ph$$

This gives 1-phenylethanol, which is a secondary alcohol with the structure $$C_6H_5CH(OH)CH_3$$.

Option A: $$CH_3C \equiv N$$ (acetonitrile) + $$PhMgBr$$ gives an imine intermediate, which on hydrolysis gives acetophenone ($$CH_3COC_6H_5$$), a ketone — not the target alcohol.

Option B: $$PhC \equiv N$$ (benzonitrile) + $$PhMgBr$$ would give benzophenone imine, which on hydrolysis gives benzophenone — not the target alcohol.

Option C: Formaldehyde ($$HCHO$$) + $$PhMgBr$$ gives benzyl alcohol ($$C_6H_5CH_2OH$$), a primary alcohol — not a secondary alcohol.

Option D: Acetaldehyde ($$CH_3CHO$$) + $$PhMgBr$$ gives 1-phenylethanol — this matches the target alcohol.

Hence, the correct answer is Option D: Acetaldehyde.

The reaction sequence (given in the paper) finally produces the compound $$\textbf{P}$$ which contains an aldehydic (>CHO) carbonyl group.

Checking Statement A
Sodium borohydride is a mild hydride donor that readily reduces aldehydes to primary alcohols and ketones to secondary alcohols:
$$R{-}CHO \;\xrightarrow[\;]{\;NaBH_4/\,ROH\;}\;R{-}CH_2OH$$
Because $$\textbf{P}$$ is an aldehyde, treatment with $$NaBH_4$$ will indeed give a primary alcohol.
Hence, Statement A is correct.

Checking Statement B
Aldehydes react with concentrated aqueous ammonia to give imines (Schiff bases):
$$R{-}CHO + NH_3 \;\rightleftharpoons\; R{-}CH{=}NH + H_2O$$
Simple acidification of this imine merely regenerates the parent aldehyde; no new compound $$\textbf{Q}$$ is obtained. Formation of a primary amine from an aldehyde would require an additional reducing step (reductive amination), which is not mentioned here.
Therefore, Statement B is false.

Checking Statement C
Statement C presupposes that $$\textbf{Q}$$ is a primary amine (so that diazotisation with $$NaNO_2/HCl$$ would evolve $$N_2$$ gas). As Statement B itself is incorrect, such a primary amine $$\textbf{Q}$$ is never formed in the stated sequence. Consequently, diazotisation and the liberation of $$N_2$$ cannot occur.
Hence, Statement C is false.

Checking Statement D
Acid strength is compared through $$pK_a$$ values. Typical values are
$$pK_a\bigl(CH_3CH_2COOH\bigr)\approx 4.9$$,   $$pK_a\bigl(R{-}CHO\bigr)\approx 17{-}18$$.
The lower the $$pK_a$$, the stronger the acid. A carboxylic acid is therefore far more acidic than an aldehyde or ketone.
So, $$\textbf{P}$$ (an aldehyde) is \emph{less} acidic than $$CH_3CH_2COOH$$.
Thus, Statement D is false.

Only Statement A survives careful scrutiny.

Final Answer: Option A which is: P can be reduced to a primary alcohol using $$NaBH_4$$.

Toluene:

$$\mathrm{C_7H_8}$$

Molar mass of toluene:

$$\mathrm{= (7\times12) + (8\times1)}$$

$$\mathrm{= 84 + 8 = 92\ g\ mol^{-1}}$$

Benzaldehyde:

$$\mathrm{C_7H_6O}$$

Molar mass of benzaldehyde:

$$\mathrm{= (7\times12) + (6\times1) + 16}$$

$$\mathrm{= 84 + 6 + 16 = 106\ g\ mol^{-1}}$$

Moles of toluene used:

$$\mathrm{= \frac{5}{92}\ mol}$$

Reaction stoichiometry:

$$\mathrm{1\ mol\ Toluene \rightarrow 1\ mol\ Benzaldehyde}$$

Therefore, theoretical moles of benzaldehyde:

$$\mathrm{= \frac{5}{92}\ mol}$$

Theoretical mass of benzaldehyde:

$$\mathrm{= \frac{5}{92}\times106\ g}$$

Given yield:

$$\mathrm{92\%}$$

Actual mass produced:

$$\mathrm{= \left(\frac{5}{92}\times106\right)\times\frac{92}{100}}$$

$$\mathrm{= \frac{5\times106}{100}}$$

$$\mathrm{= \frac{530}{100}}$$

$$\mathrm{= 5.3\ g}$$

Expressing in the form $$\mathrm{\times10^{-2}\ g}$$:

$$\mathrm{5.3\ g = 530\times10^{-2}\ g}$$

$$\boxed{\mathrm{530}}$$

Cyclohexanone contains $$\alpha$$-hydrogens. In the presence of $$OH^-$$ followed by heating, it undergoes self-aldol condensation.

In the first step, hydroxide ion removes an $$\alpha$$-hydrogen from cyclohexanone to form an enolate ion.

The enolate ion then attacks the carbonyl carbon of another cyclohexanone molecule, forming a new carbon-carbon bond and producing a $$\beta$$-hydroxy ketone (aldol).

On heating, the $$\beta$$-hydroxy ketone undergoes dehydration to form an $$\alpha,\beta$$-unsaturated ketone.

The major product $$P$$ is 2-(cyclohexylidene)cyclohexan-1-one.

The product contains:

1. One carbonyl double bond, $$C=O$$

   $$1\ \pi\text{-bond} = 2\ \pi\text{-electrons}$$

2. One carbon-carbon double bond, $$C=C$$

   $$1\ \pi\text{-bond} = 2\ \pi\text{-electrons}$$

Therefore, the total number of $$\pi$$ electrons in product $$P$$ is

$$2+2=4$$

Hence, the number of $$\pi$$ electrons present in the major product is

$$\boxed{4}$$

We have the molecular formula $$\mathrm{C_3H_6}$$. In open-chain form, the only alkene possible is propene, written structurally as $$\mathrm{CH_3-CH=CH_2}$$.

The first reagent is aqueous acid, written as $$\mathrm{H^+/H_2O}$$. In the presence of dilute acid, an alkene undergoes electrophilic addition of water. The governing principle is Markovnikov’s rule: “When HX or H-OH adds to an unsymmetrical alkene, the hydrogen attaches to the doubly-bonded carbon already holding the larger number of hydrogens, and the other part (here -OH) goes to the other carbon.” Applying that rule to propene, the -OH group attaches to the middle carbon while the extra hydrogen attaches to the terminal carbon.

Writing every step symbolically,

$$$\mathrm{CH_3-CH=CH_2 \;+\; H^+ \xrightarrow{\;\;}\; CH_3-CH^+-CH_3}$$$ $$$\mathrm{CH_3-CH^+-CH_3 \;+\; H_2O \longrightarrow CH_3-CH(OH)-CH_3 + H^+}$$$

Thus the product after the first step is $$\mathrm{CH_3-CH(OH)-CH_3}$$, i.e. propan-2-ol (commonly called isopropyl alcohol). Let us designate it as compound $$A$$.

Now $$A$$ is treated with “$$\mathrm{KIO/dil\;KOH}$$”. In the laboratory the iodoform reaction is normally carried out with iodine and alkali ($$\mathrm{I_2/KOH}$$); $$\mathrm{I_2}$$ in base actually produces the active species hypoiodite $$\mathrm{IO^-}$$, so writing the reagent as $$\mathrm{KIO}$$ in dilute $$\mathrm{KOH}$$ conveys the same chemistry. The iodoform reaction occurs whenever the substrate contains either

(i) a methyl ketone group $$\mathrm{ -CO-CH_3}$$, or
(ii) an alcohol that can be oxidised in situ to such a ketone, specifically $$\mathrm{R-CH(OH)-CH_3}$$.

Propan-2-ol fits criterion (ii) because it has the fragment $$\mathrm{ -CH(OH)-CH_3}$$. The reaction sequence is:

1. Oxidation of the secondary alcohol to the corresponding methyl ketone (propan-2-one, acetone).

$$$\mathrm{CH_3-CH(OH)-CH_3 + [O] \longrightarrow CH_3-CO-CH_3 + H_2O}$$$

2. Successive iodination of the $$\mathrm{ -CO-CH_3}$$ group and base-promoted cleavage, giving yellow iodoform and a carboxylate anion whose carbon skeleton is the original ketone minus the iodine-bearing methyl group.

For acetone the detailed stoichiometry can be summarised as

$$$\mathrm{CH_3-CO-CH_3 + 3\,I_2 + 4\,KOH \longrightarrow CHI_3 \downarrow + CH_3COOK + 3\,KI + 3\,H_2O}$$$

Therefore the two organic products obtained from the iodoform reaction of propan-2-ol are

$$$\mathrm{CHI_3} \quad\text{(iodoform, a yellow solid)}$$$

and

$$$\mathrm{CH_3COOK} \quad\text{(potassium acetate)}$$$

Denoting them in the order given in the problem statement, we can identify

$$B = \mathrm{CHI_3}, \qquad C = \mathrm{CH_3COOK}$$

Among the alternatives, only Option D lists the pair “$$\mathrm{CHI_3}$$, $$\mathrm{CH_3COOK}$$”.

Hence, the correct answer is Option D.

Mesityl oxide is a well-known organic compound formed by the aldol condensation of acetone. When two molecules of acetone undergo aldol condensation, they first form diacetone alcohol, which then undergoes dehydration to give mesityl oxide.

The product has the molecular formula $$C_6H_{10}O$$ and its structure is $$CH_3\text{-}CO\text{-}CH=C(CH_3)_2$$. To name it by IUPAC rules, we identify the longest carbon chain containing both the carbonyl group and the double bond. The longest such chain has 5 carbons (pent-), so the parent name is pentanone.

Numbering from the end nearer to the carbonyl group: carbon 1 is $$CH_3$$, carbon 2 carries the $$C=O$$ (keto group), carbon 3 is $$CH=$$, carbon 4 is $$=C(CH_3)$$, and carbon 5 is the terminal $$CH_3$$ attached to carbon 4. The double bond is between carbons 3 and 4, and there is a methyl branch at carbon 4.

Putting it together, the IUPAC name is 4-methylpent-3-en-2-one. This matches option (D).

Option A represents only the intermediate tertiary alcohol formed after Grignard addition and ignores the subsequent reaction with $$HCl$$.

Option B incorrectly assumes hydrogenation of the double bond, which cannot occur with the given reagents.

Option C represents the correct final product formed after electrophilic addition of $$HCl$$ across the double bond according to Markovnikov’s rule.

The proton adds to the terminal carbon, while $$Cl^-$$ attaches to the more substituted carbon.

Option D represents the anti-Markovnikov product, which is not formed because $$HCl$$ normally follows Markovnikov addition in the absence of peroxide conditions.

Therefore, Option C is the major product.

The 2,4-DNP (2,4-dinitrophenylhydrazine) test is a standard qualitative test used to detect the presence of the carbonyl group ($$C=O$$) in aldehydes and ketones.

When 2,4-dinitrophenylhydrazine reacts with an aldehyde or ketone, it forms a yellow, orange, or red precipitate of 2,4-dinitrophenylhydrazone. This confirms the presence of the carbonyl functional group.

Among the given options — aldehyde, amine, halogens, and ether — the 2,4-DNP test is specifically used to identify aldehydes (and ketones). Amines, halogens, and ethers do not contain a carbonyl group and will not give a positive 2,4-DNP test.

Therefore, the correct answer is aldehyde.

Product $$\mathrm{A}$$ is formed by acetalisation of the aldehyde group using ethanol and dry $$\mathrm{HCl}$$.

The primary bromide remains unaffected under these acidic conditions.

Product $$\mathrm{A}$$ is:

$$\mathrm{Br-CH_2-CH(OEt)_2}$$

Product $$\mathrm{B}$$ is formed by $$\mathrm{E2}$$ dehydrohalogenation using potassium tert-butoxide.

The strong bulky base removes $$\mathrm{HBr}$$ to form a double bond.

Product $$\mathrm{B}$$ is:

$$\mathrm{H_2C=C(OEt)_2}$$

Correct Option: $$\mathrm{A}$$

The reaction sequence involves a crossed aldol addition followed by the haloform reaction.

First, $$\mathrm{NaOH}$$ deprotonates the only available $$\mathrm{\alpha}$$-position of $$\mathrm{2,2\text{-}Dimethylcyclopentan\text{-}1\text{-}one}$$ to form an enolate ion.

This enolate attacks the carbonyl carbon of acetaldehyde, producing intermediate $$\mathrm{P}$$, a $$\mathrm{\beta}$$-hydroxy ketone containing the group $$\mathrm{(-CH(OH)-CH_3)}$$.

Intermediate $$\mathrm{P}$$ is then treated with $$\mathrm{I_2/NaOH}$$, initiating the iodoform reaction.

Sodium hypoiodite first oxidises the secondary alcohol into a methyl ketone.

The methyl group undergoes tri-iodination followed by cleavage to form yellow precipitate:

$$\mathrm{CHI_3}$$

and a sodium carboxylate salt.

Filtration removes the iodoform precipitate.

Acidification with $$\mathrm{HCl}$$ protonates the carboxylate ion to form the corresponding carboxylic acid.

The final major product $$\mathrm{X}$$ is:

$$\mathrm{2,2\text{-}Dimethyl\text{-}5\text{-}carboxycyclopentan\text{-}1\text{-}one}$$

We need an oxidising agent, thus the option is C.

Step 1 : Formation of the acyl chloride (compound A)
Acetic acid reacts with thionyl chloride as
$$CH_3COOH + SOCl_2 \rightarrow CH_3COCl + SO_2 + HCl$$
Thus $$A = CH_3COCl$$ (ethanoyl chloride or acetyl chloride).

Step 2 : Friedel-Crafts acylation on benzene (compound B)
In the presence of $$AlCl_3$$, acetyl chloride furnishes the acylium ion $$CH_3CO^+$$, which electrophilically substitutes a hydrogen atom of benzene:
$$C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_6H_5COCH_3 + HCl$$
Hence $$B = C_6H_5COCH_3$$, i.e. acetophenone (phenyl-methyl ketone).

Step 3 : Cyanohydrin formation with $$KCN/OH^-$$ (compound C)
Under basic conditions, $$CN^-$$ acts as a nucleophile. It adds to the carbonyl carbon of acetophenone and the resulting alkoxide ion is protonated to give a cyanohydrin:
$$C_6H_5COCH_3 + CN^- + H_2O \rightarrow C_6H_5C(OH)(CN)CH_3$$

Therefore compound C possesses a hydroxyl group and a cyano group on the same carbon, directly attached to the benzene-carbonyl carbon. Its condensed structure is
$$C_6H_5C(OH)(CN)CH_3$$.

Among the given drawings, this corresponds to Option C.

Final answer: Option C which is: $$C_6H_5C(OH)(CN)CH_3$$ (cyanohydrin of acetophenone).

The reaction undergoes an intramolecular aldol condensation.

The starting compound is a dialdehyde containing acidic $$\mathrm{\alpha}$$-hydrogens adjacent to both aldehyde groups.

In the presence of sodium hydroxide:

$$\mathrm{NaOH}$$

one $$\mathrm{\alpha}$$-hydrogen is removed to form a resonance-stabilised enolate ion.

$$\mathrm{R-CH_2-CHO + OH^- \longrightarrow R-CH^- -CHO + H_2O}$$

The enolate ion acts as a nucleophile.

Since another aldehyde group is present within the same molecule, intramolecular nucleophilic attack occurs on the second carbonyl carbon.

$$\mathrm{Enolate + CHO \longrightarrow Cyclic\ \beta\text{-}Hydroxy\ Aldehyde}$$

Because the two side chains are ortho-substituted on the benzene ring, they remain close together, favouring cyclisation.

This forms a new seven-membered ring fused to the benzene ring.

On heating:

$$\mathrm{\Delta}$$

the cyclic $$\mathrm{\beta}$$-hydroxy aldehyde undergoes dehydration.

$$\mathrm{\beta\text{-}Hydroxy\ Aldehyde \longrightarrow \alpha,\beta\text{-}Unsaturated\ Aldehyde + H_2O}$$

Dehydration of this product gives us the required answer.

The final product contains:

1. A fused seven-membered ring

2. An $$\mathrm{\alpha,\beta}$$-unsaturated double bond

3. Conjugation with the remaining aldehyde group

Therefore, the major product $$\mathrm{A}$$ is the conjugated cyclic $$\mathrm{\alpha,\beta}$$-unsaturated aldehyde.

To determine the major product of this reaction, we first identify the reactive sites present in the starting compound.

The given molecule is a tri-carbonyl compound containing a cyclohexanone ring. The carbon adjacent to the ketone group of the ring is a quaternary carbon attached to:

1. An aldehyde group

2. A $$\mathrm{3\text{-}oxobutyl}$$ side chain containing a methyl ketone

Step 1: Formation of the Enolate Ion

Alcoholic potassium hydroxide acts as a base and removes an acidic alpha-hydrogen to generate an enolate ion.

Among the available alpha-hydrogens, deprotonation occurs preferentially at the terminal methyl group of the side chain because it is less sterically hindered and forms a stable enolate.

$$\mathrm{CH_3COCH_2CH_2-C^*(CHO)-Cyclohexanone \xrightarrow{KOH/Alcohol} CH_2^-COCH_2CH_2-C^*(CHO)-Cyclohexanone}$$

Step 2: Intramolecular Aldol Condensation

The generated enolate undergoes intramolecular nucleophilic attack on the aldehyde carbonyl group.

Aldehydes are more reactive than ketones because they possess less steric hindrance and fewer electron-donating alkyl groups.

Thus, the enolate attacks the aldehyde carbonyl carbon selectively.

$$\mathrm{CH_2^-COCH_2CH_2-C^*(CHO)-Cyclohexanone \longrightarrow \beta\text{-}Hydroxy\ Spiro\ Intermediate}$$

This intramolecular cyclisation forms a new six-membered ring.

Since the newly formed ring and the original cyclohexanone ring share only one common carbon atom, the product obtained is a spiro compound.

Step 3: Dehydration

Under acidic conditions and heat, the beta-hydroxy ketone undergoes dehydration.

A molecule of water is eliminated, producing an alpha,beta-unsaturated ketone.

$$\mathrm{\beta\text{-}Hydroxy\ Spiro\ Intermediate \xrightarrow{H^+,\ \Delta} Spiro\ Enone + H_2O}$$

The elimination occurs between:

1. The hydroxyl group formed from the aldehyde carbonyl

2. A neighbouring hydrogen atom on the attacking carbon

This produces a conjugated double bond, increasing the stability of the final product.

Overall Transformation:

$$\mathrm{Tri\text{-}carbonyl\ Compound \xrightarrow[\ H^+,\ \Delta\ ]{KOH/Alcohol} Spiro[5.5]undec\text{-}en\text{-}one\ Derivative}$$

Final Major Product:

The final major product is a $$\mathrm{spiro[5.5]undecane}$$ derivative consisting of:

1. The original cyclohexanone ring

2. A newly formed cyclohex-2-en-1-one ring

Both rings are connected through a single spiro carbon atom.

The newly formed ring contains an $$\mathrm{\alpha,\beta}$$-unsaturated ketone system, making the final product highly stabilised through conjugation.

Clemmensen Reduction and Wolff-Kishner reduction reduce ketones to hydrocarbons. The required reagent is that of Wolff-Kishner reduction.

Upon dehydration(due to heating) gives product as per C. Thus, C is the right option

The reactant $$\mathrm{4\text{-}(Hydroxymethyl)benzaldehyde}$$ contains an aldehyde group without:

$$\mathrm{\alpha\text{-}Hydrogens}$$

Hence, in presence of $$\mathrm{NaOH}$$ and heat, it undergoes the:

$$\mathrm{Cannizzaro\ Reaction}$$

In this reaction, one molecule undergoes oxidation while another undergoes reduction.

The aldehyde group $$\mathrm{(-CHO)}$$ of one molecule is oxidised into:

$$\mathrm{(-COOH)}$$

forming:

$$\mathrm{4\text{-}(Hydroxymethyl)benzoic\ Acid}$$

The original hydroxymethyl group $$\mathrm{(-CH_2OH)}$$ remains unaffected.

The aldehyde group of another molecule is reduced into:

$$\mathrm{(-CH_2OH)}$$

forming:

$$\mathrm{1,4\text{-}Benzenedimethanol}$$

Thus, the products formed contain:

$$\mathrm{Monocarboxylic\ Acid}$$

and

$$\mathrm{Diol}$$

No product containing two carboxylic acid groups is formed.

Correct Option: $$\mathrm{D}$$

For any carbonyl compound, that is, an aldehyde $$\mathrm{R{-}CHO}$$ or a ketone $$\mathrm{R_2C{=}O}$$, the carbon of the carbonyl group carries a partial positive charge because the double-bonded oxygen is more electronegative and withdraws electron density. Hence this carbon is susceptible to attack by nucleophiles.

First we examine Statement I. In the sodium hydrogen sulphite reaction the actual nucleophile is the bisulphite ion $$\mathrm{HSO_3^-}$$. According to the general mechanism of nucleophilic addition to a carbonyl group, the very first step is:

$$$\mathrm{R_2C{=}O + HSO_3^- \;\longrightarrow\; R_2C(OH)(SO_3H)^{-}}$$$

Here the $$\mathrm{HSO_3^-}$$ ion donates its lone pair to the electrophilic carbon, the $$\pi$$ bond shifts onto oxygen and we obtain an alkoxide ion $$\mathrm{O^-}$$ attached to $$\mathrm{SO_3H}$$. Because we have generated a basic (negatively charged) oxygen and at the same time possess an acidic hydrogen on the $$\mathrm{SO_3H}$$ group, an intramolecular proton transfer immediately follows:

$$$\mathrm{R_2C(} \underset{\small{-}}{O}\mathrm{)(SO_3H) \;+\; H\,(SO_3) \;\longrightarrow\; R_2C(OH)(SO_3H)}$$$

This proton transfer neutralises the alkoxide and produces a stable bisulphite addition product, which is often isolated as a crystalline solid. Because the success of the reaction relies precisely on this proton shift that converts an unstable alkoxide into a neutral alcohol function, the description given in Statement I is accurate. Hence, Statement I is true.

Now we turn to Statement II. The classical nucleophilic addition of hydrogen cyanide to a carbonyl compound is described by the following sequence. First, in the presence of a weak base such as $$\mathrm{NaCN}$$, a small amount of $$\mathrm{HCN}$$ dissociates to give the actual nucleophile $$\mathrm{CN^-}$$:

$$\mathrm{HCN \;\rightleftharpoons\; H^+ + CN^-}$$

The cyanide ion then attacks the carbonyl carbon exactly as any nucleophile would:

$$$\mathrm{R_2C{=}O + CN^- \;\longrightarrow\; R_2C(} \underset{\small{-}}{O}\mathrm{)CN}$$$

The resulting alkoxide immediately abstracts a proton from $$\mathrm{HCN}$$ (or from any other proton donor in the medium) to give what is known as a cyanohydrin:

$$$\mathrm{R_2C(} \underset{\small{-}}{O}\mathrm{)CN + HCN \;\longrightarrow\; R_2C(OH)CN + CN^-}$$$

Thus the direct product of the reaction is $$\mathrm{R_2C(OH)CN}$$, which contains both a hydroxyl (-OH) and a nitrile (-CN) group. No amine (-NH2) function is produced in this step. An amine could appear only after a subsequent hydrogenation of the nitrile group, but that is a completely separate reaction. Therefore Statement II, which claims that the nucleophilic addition of hydrogen cyanide yields amine as the final product, is false.

Summarising, Statement I is true while Statement II is false. Among the given options, this corresponds to Option D.

Hence, the correct answer is Option D.

The reactant is a $$\mathrm{\beta\text{-}Keto\ Ester}$$, specifically $$\mathrm{Ethyl\ Acetoacetate}$$, containing both a ketone and an ester group.

When treated with ethylene glycol in presence of acid $$\mathrm{(H^+)}$$, the reaction forms a cyclic ketal.

Ketones are more reactive toward nucleophilic addition than esters because the ester carbonyl is resonance stabilised by the adjacent oxygen atom.

Hence, the reaction occurs selectively at the ketone carbonyl group.

Ethylene glycol condenses with the ketone group, eliminating water and forming a five-membered cyclic ketal $$\mathrm{(1,3\text{-}Dioxolane)}$$ ring.

The ester group remains unaffected.

Thus, product $$\mathrm{A}$$ contains a cyclic ketal at the original ketone position with the ester group unchanged.

Correct Option: $$\mathrm{B}$$

DiBAL-H selectively reduces acid anhydrides into aldehyde and alcohol.

Tollens' reagent (ammoniacal silver nitrate, $[Ag(NH_3)_2]^+$) is a mild oxidizing agent prepared in a basic solution.

• Positive Test (Forms Silver Mirror): It successfully oxidizes aldehydes into carboxylic acid salts, reducing the silver ions into solid metallic silver (the "mirror"). It also reacts with hemiacetals (because they open up to form aldehydes in solution) and $\alpha$-hydroxy ketones.
• Negative Test (No Silver Mirror): It is not strong enough to oxidize ketones, standard alcohols, or acetals/ketals.

Because Tollens' reagent is basic, any compound capable of tautomerizing (shifting between a keto and enol form) will rapidly do so in the solution. We must evaluate the most stable form of each molecule.

Analysis of the Compounds

• Compound (I): Isobutyraldehyde
  • Structure: This molecule clearly contains a free aldehyde group ($-CHO$).
  • Reaction: Since it is a standard aliphatic aldehyde, it will readily be oxidized.
  • Result: Forms a silver mirror.
• Compound (II): Prop-1-en-2-ol
  • Structure: This is an enol (an alkene and an alcohol on the same carbon).
  • Reaction: Enols are generally unstable compared to their carbonyl counterparts. In the basic medium of Tollens' reagent, this molecule will rapidly undergo tautomerization to its more stable keto form, which is acetone ($CH_3-CO-CH_3$). Acetone is a ketone. Since ketones cannot be oxidized by Tollens' reagent, no reaction occurs.
  • Result: Does NOT form a silver mirror.
• Compound (III): (Cyclohexylidene)methanol
  • Structure: This is also an enol (the double bond is right next to the $-OH$ group).
  • Reaction: Just like Compound II, this will tautomerize in the basic Tollens' solution. However, when the double bond shifts into the ring and the proton moves to the carbon, it forms cyclohexanecarbaldehyde (a cyclohexane ring attached to a $-CHO$ group). Because its keto form is an aldehyde, it will successfully react with the reagent.
  • Result: Forms a silver mirror.
• Compound (IV): 2-Hydroxytetrahydropyran
  • Structure: This is a cyclic hemiacetal (a carbon bonded to both an $-OH$ group and an $-OR$ ether linkage within a ring).
  • Reaction: In an aqueous solution, cyclic hemiacetals exist in a dynamic equilibrium with their open-chain forms. When this ring opens, it forms 5-hydroxypentanal, revealing a free aldehyde group at the end of the chain. This free aldehyde will then react with the Tollens' reagent.
  • Result: Forms a silver mirror.

Conclusion

Because it tautomerizes into a ketone rather than an aldehyde, Compound (II) is the only molecule in this lineup that will not give a positive Tollens' test.

Step 1: Identify the type of reaction.
The conversion of propane to propanal involves removal of hydrogen and introduction of one oxygen atom. This is called oxidative dehydrogenation (ODH).

Step 2: Write the balanced chemical equation for ODH of propane.
Using one‐half mole of oxygen per mole of propane yields propanal and water:
$$CH_3CH_2CH_3 + \frac{1}{2}O_2 \rightarrow CH_3CH_2CHO + H_2O$$
$$-(1)$$

Step 3: State the requirement for selective partial oxidation.
We need a catalyst that activates O-H and C-H bonds without over-oxidizing the aldehyde to an acid. Many strong oxidants or high-temperature metals give complete oxidation or cracking.

Step 4: Examine each option.
Option A: Copper at high temperature and pressure tends to crack alkanes or give CO/CO_2 rather than stop at an aldehyde.
Option B: Manganese acetate is used in radical allylic oxidations (Kharasch-Sosnovsky), not for propane.
Option D: Potassium permanganate is a very strong oxidant and would oxidize propanal further to propionic acid.

Step 5: Select the correct catalyst.
Molybdenum trioxide ($$MoO_3$$) is known to promote selective partial oxidation (ODH) of light alkanes to aldehydes with minimal over-oxidation. It activates $$O_2$$ gently and removes hydrogen without decomposing the aldehyde.

Therefore, the reagent used is Option C: Molybdenum oxide ($$MoO_3$$).

When an alkyne undergoes hydration in the presence of $$HgSO_4$$ and $$H_2SO_4$$, the addition of water follows Markovnikov's rule. The $$-OH$$ group preferentially adds to the more substituted (internal) carbon of the triple bond, and the resulting enol tautomerises to a carbonyl compound.

Consider acetylene ($$HC \equiv CH$$): since it is a symmetric alkyne, the hydration gives the enol $$CH_2 = CHOH$$, which tautomerises to acetaldehyde ($$CH_3CHO$$). So option (3), acetaldehyde, can be prepared by this method.

Consider 1-butyne ($$CH_3CH_2C \equiv CH$$): by Markovnikov addition, the $$-OH$$ adds to the internal carbon ($$C_2$$ of the triple bond), giving the enol $$CH_3CH_2C(OH) = CH_2$$. This tautomerises to methyl ethyl ketone ($$CH_3COCH_2CH_3$$). So option (2) can be prepared.

Cyclohexanone can be obtained by the acid-catalysed hydration of cyclohex-1-yne. The triple bond in this cyclic alkyne undergoes Markovnikov hydration to give the corresponding enol, which tautomerises to cyclohexanone. So option (4) can be prepared.

Now consider propanal ($$CH_3CH_2CHO$$): this is an aldehyde with the carbonyl at the terminal position ($$C_1$$). The only three-carbon alkyne is propyne ($$CH_3C \equiv CH$$). By Markovnikov hydration, the $$-OH$$ adds to the internal carbon ($$C_2$$), forming the enol $$CH_3C(OH) = CH_2$$, which tautomerises to acetone ($$CH_3COCH_3$$), a ketone, not propanal. In general, Markovnikov hydration of any terminal alkyne gives a methyl ketone, and only acetylene (being symmetric) can give an aldehyde. Since there is no alkyne whose Markovnikov hydration yields propanal, it cannot be prepared by this method.

The correct answer is Option (1): $$CH_3CH_2CHO$$ (propanal).

The $$\mathrm{2,4\text{-}DNP}$$ ($$\mathrm{2,4\text{-}dinitrophenylhydrazine}$$) test, also known as Brady’s Reagent test, is a qualitative chemical test used to detect the presence of a carbonyl group.

Carbonyl compounds such as aldehydes and ketones give a positive test.

How It Works:

1. The Reagent

$$\mathrm{2,4\text{-}DNP}$$ reacts with the carbonyl carbon $$\mathrm{(C=O)}$$ of aldehydes and ketones through a nucleophilic addition-elimination reaction.

$$\mathrm{R_2C=O + H_2N\text{-}NH\text{-}C_6H_3(NO_2)_2 \longrightarrow R_2C=NNH\text{-}C_6H_3(NO_2)_2 + H_2O}$$

2. The Result

The reaction produces a crystalline precipitate called a hydrazone derivative.

3. Visual Confirmation

A positive test is indicated by the formation of a yellow, orange, or red precipitate.

The only ketone/aldehyde is option D.

DIBAL-H (diisobutylaluminium hydride) is a selective reducing agent that converts nitriles into aldehydes under controlled reaction conditions.

In the first step, the nitrile $$\mathrm{R-CN}$$ is partially reduced by DIBAL-H to form an imine intermediate.

Upon subsequent hydrolysis with water, the imine is converted into the corresponding aldehyde.

The overall transformation is

$$\mathrm{R-CN \xrightarrow[\text{2. }H_2O]{\text{1. DIBAL-H}} R-CHO}$$

Thus, the functional group $$Y$$ is

$$\boxed{\mathrm{-CHO}}$$

Hence, the correct answer is

$$\boxed{\text{Option (C)}}$$

The Reaction

Hydroformylation, or the Oxo Process, is an industrial catalytic reaction that converts alkenes into aldehydes. Its most important feature is that it lengthens the carbon chain, always producing an aldehyde that contains exactly one more carbon atom than the starting alkene.


$$\text{R-CH=CH}_{2}+\text{CO}+\text{H}_{2}\xrightarrow{\text{Catalyst}}\text{R-CH}_{2}\text{-CH}_{2}\text{-CHO\ (Linear)}+\text{R-CH(CHO)-CH}_{3}\text{\ (Branched)}$$

The Requirements

To achieve this, the alkene is reacted with "syngas"—a mixture of carbon monoxide (CO) and hydrogen gas ($$H_2$$). This process requires high temperatures, high pressures, and a specialized transition metal catalyst, typically containing cobalt or rhodium, to force the reaction to occur.

The Products

The catalyst breaks the alkene's double bond, adding a hydrogen atom to one side and a formyl group ($$-CHO$$) to the other. Because the formyl group can attach to either of the two carbons from the double bond, the reaction produces a mixture of two structural isomers: a straight-chain linear aldehyde (which is highly desired in industry) and a branched aldehyde.

The molecule contains two functional groups:

• A nitrile group, $$-C\equiv N$$
• An ether group, $$-OCH_3$$

Step (i)

Phenylmagnesium bromide, $$C_6H_5MgBr$$, is a Grignard reagent and acts as a source of the nucleophilic phenyl group, $$C_6H_5^-$$.

Since only $$1.0$$ equivalent of the Grignard reagent is used, it reacts with the more reactive functional group.

The ether group remains unaffected because ethers are inert towards Grignard reagents.

The electrophilic carbon of the nitrile group undergoes nucleophilic attack by $$C_6H_5^-$$ to form an imine magnesium salt.

$$R-C\equiv N \xrightarrow{C_6H_5MgBr} R-C(C_6H_5)=NMgBr$$

Step (ii)

On acidic hydrolysis, the imine magnesium salt is converted into a ketone.

$$R-C(C_6H_5)=NMgBr \xrightarrow{H_3O^+} R-COC_6H_5$$

Thus, the nitrile group is converted into a phenyl ketone, while the ether group remains unchanged.

Therefore, the major product $$X$$ is the structure containing the group

$$-COC_6H_5$$

with the $$-CH(CH_3)OCH_3$$ group unchanged.

Hence, the correct answer is Option (D).

First, the two equivalents of the Grignard reagent (CH₃MgBr) react with the two separate electrophilic centers in the starting molecule.

One equivalent of CH₃MgBr nucleophilically attacks the ketone carbon on the non-aromatic ring. Following the acidic workup (H₃O⁺), this functional group is converted into a tertiary alcohol, placing a new methyl group at that position.

The second equivalent of CH₃MgBr attacks the nitrile group (-CN) located on the aromatic ring. The acidic workup hydrolyzes the resulting imine intermediate, converting the nitrile group into a methyl ketone, also known as an acetyl group (-COCH₃).

Finally, the addition of sulfuric acid (H₂SO₄) and heat induces the acid-catalyzed dehydration of the newly formed tertiary alcohol. A molecule of water is eliminated to create a carbon-carbon double bond. The elimination occurs preferentially between the alcohol carbon and the adjacent benzylic carbon (the carbon directly attached to the aromatic ring). This specific pathway is favored because it yields an alkene that is in direct conjugation with the aromatic benzene ring, providing significant thermodynamic stability through resonance.

Thus, the correct option is C.

We need to identify which reaction does not form acetaldehyde ($$CH_3CHO$$).

Option 1 uses $$CrO_3$$ in $$H_2SO_4$$, which is known as Jones reagent. This is a strong oxidising agent that oxidises primary alcohols ($$CH_3CH_2OH$$) all the way to carboxylic acids ($$CH_3COOH$$), not stopping at the aldehyde stage. Therefore, this reaction does not produce acetaldehyde.

Option 2 is the Wacker process, where ethylene is oxidised by $$O_2$$ in the presence of $$PdCl_2/CuCl_2$$ catalyst in water. This reaction is a standard industrial method for producing acetaldehyde.

Option 3 involves reduction of $$CH_3CN$$ (methyl cyanide) with DIBAL-H (diisobutylaluminium hydride) followed by hydrolysis. DIBAL-H reduces nitriles to aldehydes at low temperature, yielding $$CH_3CHO$$.

Option 4 is the catalytic dehydrogenation of ethanol using copper at 573 K, which is a standard laboratory method for preparing acetaldehyde from primary alcohols.

Therefore, the reaction that will not form acetaldehyde is Option 1.

We have benzonitrile whose molecular formula is $$\mathrm{C_6H_5CN}$$. When a nitrile reacts with one equivalent of a Grignard reagent ($$\mathrm{R-MgX}$$) in dry ether, the general rule is that the Grignard reagent adds to the carbon of the $$\mathrm{C\equiv N}$$ group and converts the nitrile into an imine magnesium halide. Later, on acidic hydrolysis, this imine is transformed into a ketone. Symbolically the rule can be stated as

$$\mathrm{R'-C\equiv N \;+\; R-MgX \;\longrightarrow\; R'-C(=N^-MgX)-R \xrightarrow[{\,\text{heat}\,}]{\text{H}_3\text{O}^+} R'-CO-R}$$

Now we apply this rule to the given reagents. In benzonitrile, $$\mathrm{R'} = \mathrm{C_6H_5}$$ (phenyl) and the Grignard reagent supplied is $$\mathrm{CH_3MgBr}$$ with $$\mathrm{R = CH_3}$$. Substituting these groups in the rule gives

$$\mathrm{C_6H_5-C\equiv N \;+\; CH_3MgBr \;\longrightarrow\; C_6H_5-C(=N^-MgBr)-CH_3 \xrightarrow[{\,\text{hydrolysis}\,}]{\text{H}_3\text{O}^+} C_6H_5-CO-CH_3}$$

Therefore the yellow liquid “P” obtained after hydrolysis is $$\mathrm{C_6H_5COCH_3}$$, whose common name is acetophenone and whose structural unit is

$$\mathrm{Ph-CO-CH_3}$$

We next decide which qualitative test is answered by this molecule. Acetophenone contains the $$\mathrm{-CO-CH_3}$$ functional group, i.e. the carbonyl carbon is directly attached to a methyl group. Organic chemistry tells us that

• The Iodoform test is positive for all compounds containing the $$\mathrm{-CO-CH_3}$$ (methyl ketone) group or $$\mathrm{CH_3-CH(OH)-}$$ group.
• Schiff’s, Ninhydrin’s and Tollen’s tests are meant for aldehydes or amino acids and are not given by simple methyl ketones.

Because acetophenone is a methyl ketone, it will liberate a yellow precipitate of $$\mathrm{CHI_3}$$ (iodoform) in the iodoform test, while it will fail the other three tests listed. Hence the liquid “P” gives a positive iodoform test.

Hence, the correct answer is Option A.

Reaction $$\mathrm{(a)}$$ converts an acid chloride $$\mathrm{(RCOCl)}$$ into an aldehyde $$\mathrm{(RCHO)}$$. This reaction is called the $$\mathrm{Rosenmund\ Reduction}$$ and uses $$\mathrm{H_2/Pd-BaSO_4}$$. Hence, $$\mathrm{(a)\rightarrow(ii)}$$.

Reaction $$\mathrm{(b)}$$ involves $$\mathrm{\alpha}$$-halogenation of a carboxylic acid $$\mathrm{(RCH_2COOH \rightarrow RCH(Cl)COOH)}$$. This is the $$\mathrm{Hell\text{-}Volhard\text{-}Zelinsky\ Reaction}$$ and uses $$\mathrm{Cl_2/Red\ P,\ H_2O}$$. Hence, $$\mathrm{(b)\rightarrow(iv)}$$.

Reaction $$\mathrm{(c)}$$ converts a primary amide $$\mathrm{(RCONH_2)}$$ into a primary amine $$\mathrm{(RNH_2)}$$ with loss of one carbon atom. This is the $$\mathrm{Hoffmann\ Bromamide\ Degradation}$$ and uses $$\mathrm{Br_2/NaOH}$$. Hence, $$\mathrm{(c)\rightarrow(i)}$$.

Reaction $$\mathrm{(d)}$$ reduces a ketone carbonyl group $$\mathrm{(RCOCH_3)}$$ into $$\mathrm{RCH_2CH_3}$$. This is the $$\mathrm{Clemmensen\ Reduction}$$ and uses $$\mathrm{Zn(Hg)/Conc.\ HCl}$$. Hence, $$\mathrm{(d)\rightarrow(iii)}$$.

Therefore, the correct matching is $$\mathrm{(a)\rightarrow(ii),\ (b)\rightarrow(iv),\ (c)\rightarrow(i),\ (d)\rightarrow(iii)}$$.

Correct Option: $$\mathrm{A}$$

We look at transformation (a). The starting compound is $$\mathrm{CH_3COOCH_2CH_3}$$, an ester known as ethyl acetate. The desired product is $$\mathrm{CH_3CH_2OH}$$, ethanol. We recall the general hydrolysis reaction of an ester in acidic medium, which is stated as

$$$\mathrm{RCOOR' \;+\; H_2O \;\xrightarrow[\;]{H^+}\; RCOOH \;+\; R'OH}$$$

Here $$\mathrm{R = CH_3}$$ and $$\mathrm{R' = CH_2CH_3}$$, so hydrolysis will furnish acetic acid and ethanol. The reagent that supplies both water and acid is the mixture $$\mathrm{H_2SO_4/H_2O}$$, listed as (ii). Hence, for part (a) we must choose (ii).

Now we examine transformation (b). The substrate is $$\mathrm{CH_3COOCH_3}$$ (methyl acetate) and the target is $$\mathrm{CH_3CHO}$$ (acetaldehyde). Converting an ester directly to an aldehyde requires partial reduction that stops at the aldehyde stage. The reagent famous for this selective one-step reduction is DIBAL-H followed by aqueous work-up. The overall reaction may be represented as

$$$\mathrm{RCOOR' \;\xrightarrow[\;]{\text{DIBAL-H,}\; -78^{\circ}\text{C}} \; RCHO \;\xrightarrow{\;H_2O\;} RCHO}$$$

DIBAL-H/H$$_2$$O is option (iii), so (b) corresponds to (iii).

Next, consider transformation (c): $$\mathrm{CH_3C\equiv N \;\rightarrow\; CH_3CHO}$$. Converting a nitrile to an aldehyde is achieved by the Stephen reduction. The textbook Stephen reduction uses stannous chloride in the presence of hydrochloric acid, followed by hydrolysis:

$$$\mathrm{RCN \;+\; 2\,SnCl_2 \;+\; 2\,HCl \;\xrightarrow[\;]{\;} RCH=NH\cdot HCl \;\xrightarrow{\;H_2O\;} RCHO}$$$

This reagent set is given as (iv) $$\mathrm{SnCl_2,\,HCl/H_2O}$$. Therefore, (c) matches (iv).

Finally, we look at transformation (d): $$\mathrm{CH_3C\equiv N \;\rightarrow\; CH_3COCH_3}$$, i.e. conversion of acetonitrile to acetone. A single equivalent of a Grignard reagent adds an alkyl group to the electrophilic carbon of the nitrile, giving an imine magnesium salt; hydrolysis of the imine then produces a ketone. The general formula is

$$$\mathrm{RCN \;+\; R''MgBr \;\xrightarrow[\;]{\;} RC(R'')=NMgBr \;\xrightarrow{\;H_3O^+\;} RC(O)R''}$$$

With $$\mathrm{R = CH_3}$$ and $$\mathrm{R'' = CH_3}$$, we indeed obtain $$\mathrm{CH_3COCH_3}$$. The required reagent is $$\mathrm{CH_3MgBr/H_3O^+}$$, listed as (i). Thus, (d) corresponds to (i).

Collecting all the matches we have:

$$$\text{(a)} \rightarrow (ii), \qquad \text{(b)} \rightarrow (iii), \qquad \text{(c)} \rightarrow (iv), \qquad \text{(d)} \rightarrow (i)$$$

This sequence appears exactly in Option A.

Hence, the correct answer is Option A.

Mesityl oxide is a well-known organic compound with the molecular formula $$\text{C}_6\text{H}_{10}\text{O}$$ and the IUPAC name 4-methylpent-3-en-2-one. Its structure is $$(\text{CH}_3)_2\text{C}=\text{CH}-\text{CO}-\text{CH}_3$$.

Let us write out the carbon skeleton and count all C-C sigma bonds. The structure has six carbon atoms arranged as: $$\text{CH}_3 - \text{C}(\text{CH}_3) = \text{CH} - \text{C}(=\text{O}) - \text{CH}_3$$.

The C-C sigma bonds are: (1) the first $$\text{CH}_3$$ to the central carbon $$\text{C}$$, (2) the branching $$\text{CH}_3$$ to the same central carbon, (3) the $$\text{C}=\text{C}$$ double bond contains one sigma and one pi bond, so there is one C-C sigma bond here, (4) between $$=\text{CH}$$ and the carbonyl carbon $$\text{C}(=\text{O})$$, and (5) between the carbonyl carbon and the terminal $$\text{CH}_3$$.

This gives a total of 5 C-C sigma bonds.

The answer is 5.

In the Tollens' test, the silver ion in the Tollens' reagent $$[\text{Ag}(\text{NH}_3)_2]^+$$ is reduced from $$\text{Ag}^+$$ to $$\text{Ag}^0$$ (metallic silver that forms the silver mirror). Each silver ion gains one electron: $$\text{Ag}^+ + e^- \to \text{Ag}$$.

The overall balanced reaction is: $$\text{RCHO} + 2[\text{Ag}(\text{NH}_3)_2]^+ + 2\text{OH}^- \to \text{RCOO}^- + 2\text{Ag} + 4\text{NH}_3 + \text{H}_2\text{O}$$. The aldehyde carbon goes from oxidation state +1 (in $$\text{RCHO}$$) to +3 (in $$\text{RCOO}^-$$), losing 2 electrons overall. These 2 electrons are transferred to two $$[\text{Ag}(\text{NH}_3)_2]^+$$ ions (one electron each), producing 2 atoms of silver.

The total number of electrons transferred to the Tollens' reagent per aldehyde group is $$2$$.

Semicarbazone is formed when a ketone (or aldehyde) reacts with semicarbazide (H₂N-NH-CO-NH₂) in a condensation reaction, releasing water.

Acetone is (CH₃)₂CO. The reaction of acetone with semicarbazide is:

$$(CH_3)_2CO + H_2N-NH-CO-NH_2 \rightarrow (CH_3)_2C=N-NH-CO-NH_2 + H_2O$$

The semicarbazone of acetone is: $$(CH_3)_2C=N-NH-CO-NH_2$$

Now count the nitrogen atoms in this molecule: the first N is in the C=N group (imine/azomethine nitrogen), the second N is in the -NH- group, and the third N is in the terminal -NH₂ group of the carbamoyl part.

Total nitrogen atoms = 3.

Therefore, the number of nitrogen atoms in the semicarbazone of acetone is $$\boxed{3}$$.

For any carbonyl compound, the acidity of an $$\alpha$$-hydrogen depends on how well the conjugate base (the enolate ion) is stabilised. Two main factors decide this stabilisation:

(i)  Inductive (-I) effect of the carbonyl group.
(ii)  Resonance delocalisation of the negative charge after deprotonation.

More electron-withdrawing carbonyl groups and better delocalisation make the enolate ion more stable, so the parent carbonyl compound becomes a stronger acid.

The four compounds given in the question are:

(A) $$CH_3CHO$$          (acetaldehyde)
(B) $$CH_3COCH_2COCH_3$$   (pentane-2,4-dione; a $$\beta$$-diketone)
(C) $$CH_3COCH_3$$        (acetone, a ketone)
(D) $$CH_3COOCH_3$$       (methyl acetate, an ester)

Case 1 - Compound B (β-diketone)
After removal of an $$\alpha$$-H, the negative charge is delocalised over two carbonyl groups by resonance: $$\ce{-CH^-}$$ ⇌ $$\ce{-COCH=CO-}$$. This extensive delocalisation gives very high stability to the conjugate base. Hence B is the most acidic of the four (pKa ≈ 9-10).

Case 2 - Compound A (aldehyde)
In an aldehyde, the carbonyl carbon is bonded to only one electron-releasing alkyl group, so its -I effect is stronger than that of a ketone. The resulting enolate $$CH_3CH^-O$$ is therefore better stabilised than the ketone enolate. Aldehydes are thus more acidic than simple ketones (pKa ≈ 17).

Case 3 - Compound C (ketone)
A ketone has two electron-releasing alkyl groups attached to the carbonyl carbon. These reduce the overall -I effect, so the enolate ion is less stabilised than that from an aldehyde. Hence ketones are weaker acids than aldehydes (pKa ≈ 19-20).

Case 4 - Compound D (ester)
In an ester, the carbonyl is attached to an -OR group. Although the carbonyl itself has a -I effect, the -OR group can donate electron density to the carbonyl carbon by +R (mesomeric effect). This electron donation notably decreases the withdrawing ability of the carbonyl, making the ester enolate the least stabilised of the four. Consequently, esters have the least acidic $$\alpha$$-hydrogen (pKa ≈ 24-25).

Putting these facts together:

Least acidic           Most acidic
$$\text{(D) } \lt \text{(C) } \lt \text{(A) } \lt \text{(B)}$$

Therefore, the increasing order of acidity of the $$\alpha$$-hydrogen is indeed:

Option A which is: $$(D) < (C) < (A) < (B)$$

• Step 1: Reaction with excess $$\text{HBr}$$ and heat ($$\Delta$$)

  • The starting cyclic ether is protonated by $$\text{HBr}$$, followed by a nucleophilic attack by $$\text{Br}^-$$ to cleave the ether ring. This forms a phenolic group ($$-\text{OH}$$) on the benzene ring and an alkyl bromide intermediate.
  • Since $$\text{HBr}$$ is in excess, it undergoes an addition reaction across the double bond of the side chain according to Markovnikov's rule.
  • This gives the intermediate product A, which is a vicinal dibromide with a phenolic group: $$\text{C}_6\text{H}_4(\text{OH})(\text{CH(Br)CH(Br)CH}_3)$$



• Step 2: Reaction with alcoholic $$\text{KOH}$$ followed by acidic workup ($$\text{H}^+$$)

  • Alcoholic $$\text{KOH}$$ is a strong dehydrohalogenating agent. It carries out a double elimination ($$-2\text{HBr}$$) on the vicinal dibromide chain to form a triple bond (alkyne).
  • The resulting intermediate B contains a terminal or internal alkyne attached to the benzene ring alongside the phenol group. Under these conditions, the conjugated structure stabilizes as an allylic/propargyl double bond system: $$\text{C}_6\text{H}_4(\text{OH})(\text{CH}=\text{CH}\text{--CH}_3)$$



• Step 3: Ozonolysis with (i) $$\text{O}_3$$ and (ii) $$\text{Zn/H}_3\text{O}^+$$

  • This is a reductive ozonolysis step that selectively cleaves the carbon-carbon double bond ($$\text{C}=\text{C}$$) in the side chain.
  • The double bond is broken and replaced with carbonyl ($$\text{C}=\text{O}$$) groups on both fragments: $$\text{C}_6\text{H}_4(\text{OH})(\text{CH}=\text{CH}\text{--CH}_3) \xrightarrow{\text{O}_3, \ \text{Zn/H}_3\text{O}^+} \text{C}_6\text{H}_4(\text{OH})(\text{CHO}) + \text{CH}_3\text{CHO}$$
  • The side chain is chopped down to an aldehyde group ($$-\text{CHO}$$) attached directly to the benzene ring at the position adjacent (ortho) to the hydroxyl group. This aromatic product is known as Salicylaldehyde (2-hydroxybenzaldehyde).

Conclusion:

The sequence converts the cyclic core through opening, double elimination, and ozonolytic cleavage to yield a substituted benzene ring featuring both an explicit phenol and a newly formed aldehyde function. This perfectly matches the molecule highlighted in the solution panel.

Answer: Option A — Salicylaldehyde (2-hydroxybenzaldehyde)

We have the sequence $$A \xrightarrow{\,(i)\;CH_3MBr\;(ii)\;H_3O^+}\;B \xrightarrow{Cu,\;573\,K}\;2\text{-methyl-2-butene}.$$

The second arrow tells us that on heating with copper at $$573\;K,$$ the compound $$B$$ loses a molecule of water to give the alkene 2-methyl-2-butene. It is well known (and you might have seen in your NCERT book) that tertiary alcohols on such heating undergo dehydration to alkenes. Therefore, $$B$$ must be a tertiary alcohol whose dehydration product is 2-methyl-2-butene.

2-methyl-2-butene has the skeleton

$$CH_3-C(CH_3)=CH-CH_3.$$ Removing the double bond and inserting an $$OH$$ group at the doubly bonded carbon gives the corresponding tertiary alcohol

$$CH_3-C(OH)(CH_3)-CH_2-CH_3,$$

which is 2-methyl-2-butan-2-ol (commonly written as 2-methyl-2-butanol). Hence

$$B = \text{2-methyl-2-butanol}.$$

Now we look at the first arrow. The reagent sequence $$(i)\;CH_3MBr\;(ii)\;H_3O^+$$ is the standard method for adding a methyl group from a Grignard reagent to a carbonyl compound and then protonating the alkoxide to an alcohol. In words:

Ketone/Aldehyde $$+$$ CH3MgBr $$\longrightarrow$$ alkoxide $$\xrightarrow{H_3O^+}$$ alcohol.

So, to obtain the tertiary alcohol $$B$$ we must start from a ketone whose carbonyl carbon already bears the two other groups seen in $$B,$$ namely $$CH_3$$ and $$CH_2CH_3.$$ Let us denote the unknown ketone by $$A.$$ If the carbonyl carbon of $$A$$ is attached to $$CH_3$$ and $$CH_2CH_3,$$ then the structure of $$A$$ is

$$CH_3-CO-CH_2-CH_3,$$

which is butan-2-one (methyl ethyl ketone). Addition of $$CH_3^-$$ from the Grignard reagent followed by acid work-up places an extra $$CH_3$$ on that carbon, giving exactly the tertiary alcohol $$B.$$ Therefore

$$A = \text{butan-2-one (CH}_3COCH_2CH_3).$$

Next we calculate the mass percentage of carbon in $$A.$$ First write its molecular formula:

Butan-2-one : $$C_4H_8O.$$

Molar mass calculation:
$$\text{Mass of }C = 4 \times 12\;\text{g mol}^{-1} = 48\;\text{g mol}^{-1},$$
$$\text{Mass of }H = 8 \times 1\;\text{g mol}^{-1} = 8\;\text{g mol}^{-1},$$
$$\text{Mass of }O = 1 \times 16\;\text{g mol}^{-1} = 16\;\text{g mol}^{-1}.$$

Total molar mass of $$A:$$ $$M = 48 + 8 + 16 = 72\;\text{g mol}^{-1}.$$

The formula for mass percentage is stated as $$\%\text{ element} = \frac{\text{mass of that element in 1 mole of compound}}{\text{molar mass of compound}}\times 100\%.$$

Applying it for carbon,

$$ \%C = \frac{48}{72}\times 100\% = 66.67\%. $$

So, the answer is $$66.67\%.$$

The phenomenon we are examining is keto-enol tautomerism, which is the equilibrium

$$\text{keto form}\; \rightleftharpoons \;\text{enol form}$$

The percentage of the enol form present at equilibrium depends mainly on two factors.

1. Acidity of the $$\alpha$$-hydrogen: The more acidic this hydrogen is, the more easily the enolate ion is formed, and the greater is the conversion to the enol after protonation.

2. Stabilization of the enol formed: If the enol can be stabilized by conjugation and / or by intramolecular hydrogen bonding (chelation), its concentration rises.

We now analyse each option one by one, keeping both points in mind.

Option D: $$\text{CH}_3\text{COCH}_3$$ (acetone)

This molecule has only one carbonyl group. The $$\alpha$$-hydrogens (those on the carbon between the carbonyl carbon and the methyl group) have a pKa of about $$19$$. Such hydrogens are not very acidic, so very little enolate—and therefore very little enol—is produced. Moreover, any enol that does form has no special stabilization beyond ordinary C=C-O conjugation. Hence the enol content is extremely small, usually < 0.1 %.

Option A: $$\text{CH}_3\text{COCH}_2\text{CONH}_2$$

This is a $$1,3$$-dicarbonyl system in which the second carbonyl is an amide. The central $$\alpha$$-hydrogen lies between a ketone carbonyl and an amide carbonyl. Because two electron-withdrawing groups flank it, its pKa drops to roughly $$13$$, so deprotonation is easier than in acetone.

However, when the enol form is drawn

$$\text{CH}_3\text{CO-CH}=\text{CONH}_2 \quad\text{with}\quad \text{OH on the left carbonyl}$$

the amide carbonyl is a poor hydrogen-bond acceptor; its oxygen is less basic because it is involved in resonance with the nitrogen (-C=O↔-C-O--N+). As a result, the expected six-membered intramolecular O-H···O hydrogen bond is not very strong. Conjugation is still present, but the reduced hydrogen-bond energy lowers stabilization. Hence the enol content is higher than acetone’s but still moderate.

Option C: $$\text{CH}_3\text{COCH}_2\text{COOC}_2\text{H}_5$$ (a $$\beta$$-keto ester)

Here the central hydrogen is between a ketone carbonyl and an ester carbonyl. Its pKa is about $$11$$, making it even more acidic than in Option A, so enolate formation is easier.

The enol obtained can be represented as

$$\text{CH}_3\text{C(OH)=CH-COOC}_2\text{H}_5$$

and this enol can indeed form a six-membered ring via intramolecular hydrogen bonding between the newly formed -OH and the ester carbonyl oxygen. Conjugation extends over the C=C-C=O system. Nevertheless, an ester carbonyl is less basic than a ketone carbonyl because the -OR group donates electron density by resonance: $$-C(=O)OR \leftrightarrow -C(-O^{-})=O-R^{+}$$. Consequently the hydrogen bond is weaker than it is with a ketone carbonyl, so the net stabilization—though appreciable—is not maximal.

Option B: $$\text{CH}_3\text{COCH}_2\text{COCH}_3$$ (acetylacetone)

This is a true $$\beta$$-diketone. Both flanking groups are ketone carbonyls, each strongly electron withdrawing. Hence the pKa of the central hydrogen is around $$9$$, the smallest among all the given molecules, so the enolate forms most readily.

The corresponding enol is

$$\text{CH}_3\text{C(OH)=CH-COCH}_3$$

In this structure we have two powerful stabilizing effects simultaneously:

(i) Conjugation: The C=C is directly conjugated with a carbonyl group, giving the usual enol resonance

$$\text{C(OH)=C-C=O} \;\;\rightleftharpoons\;\; \mathrm{C(O}^{-}\mathrm{)-C}^+\mathrm{-C=O}$$

(ii) Intramolecular hydrogen bonding (chelation): The enolic -OH can donate a hydrogen to the oxygen of the second ketone carbonyl, creating a six-membered chelate ring. A ketone carbonyl oxygen is more basic than an ester or amide oxygen, so this hydrogen bond is very strong. The extra stabilization may be depicted as

$$\text{C(OH)…O=C}$$

The combined resonance and hydrogen-bond energies make the enol of a $$\beta$$-diketone so stable that, in solvents like benzene, the equilibrium mixture is as high as $$70$$-$$80\,\%$$ enol.

We now compare the options quantitatively in the order of the two criteria discussed:

$$\text{Enol content} \propto \dfrac{\text{ease of enolate formation}}{\text{pK}_a} \times \text{stabilization of the enol}$$

Option B: lowest pKa (≈9) & strongest chelation ⇒ highest enol content.

Option C: slightly higher pKa (≈11) & somewhat weaker chelation ⇒ less than B.

Option A: pKa (≈13) & poor chelation ⇒ still lower.

Option D: pKa (≈19) & no special stabilization ⇒ the least.

Putting all observations together, the $$\beta$$-diketone $$\text{CH}_3\text{COCH}_2\text{COCH}_3$$ in Option B possesses the maximum enol content.

Hence, the correct answer is Option B.

• Ferric Chloride (FeCl3) Test: A negative result means the compound is not a phenol 
• Fehling Solution Test: A negative result means the compound is not an aliphatic aldehyde (like acetaldehyde). Since it doesn't reduce Fehling's solution, it is likely a ketone.

Reacts with Grignard reagent

• This confirms the presence of a carbonyl group . Grignard reagents perform nucleophilic attack on the carbonyl carbon of ketones to form alcohols.

Positive Iodoform test

• This is the most specific clue. For a compound to give a positive iodoform test, it must contain a methyl ketone group or a methyl carbinol group

We are dealing with the acid-catalysed conversion of a carbonyl compound to an acetal in methanol. The simplified mechanism proceeds through these key elementary steps:

$$\text{(1) Protonation of the carbonyl: }R_{2}C=O + H^+ \longrightarrow R_{2}C=OH^+$$

$$\text{(2) Nucleophilic attack by methanol: }R_{2}C=OH^+ + MeOH \longrightarrow R_{2}C(OMe)OH + H^+$$

The overall rate, to a very good approximation in the initial stages, is proportional to the concentrations of the species actually colliding in the slow step. After the fast proton-transfer of step (1), the slow or rate-determining step is generally step (2). Therefore we can write

$$\text{Rate} \propto [\text{protonated carbonyl}]\,[\text{MeOH}].$$

Now we compare the options on the basis of two independent factors that control this rate expression:

Factor 1: Nature of the carbonyl electrophile.

An aldehyde is more electrophilic than a ketone because:

(i) it has only one electron-donating alkyl group instead of two, so the carbonyl carbon bears a larger partial positive charge, and

(ii) it suffers less steric hindrance to nucleophilic attack.

Hence $$\text{Rate}_{\text{aldehyde}} > \text{Rate}_{\text{ketone}}.$$

Between the substrates listed, propanal (an aldehyde) is thus inherently faster than acetone (a ketone).

Factor 2: Concentration of the nucleophile MeOH.

The rate expression contains $$[\text{MeOH}],$$ so increasing the concentration of methanol directly raises the rate. Using methanol in excess means a higher $$[\text{MeOH}]$$ than in a stoichiometric amount.

Therefore $$\text{Rate}_{\text{excess MeOH}} > \text{Rate}_{\text{stoichiometric MeOH}}.$$

Let us rank the four options by systematically combining these two factors.

Step 1 - Substrate effect:

Propanal options (B and C) > Acetone options (A and D).

Step 2 - Methanol concentration effect (within each substrate type):

Excess MeOH (B and D) > Stoichiometric MeOH (A and C).

Combining both steps:

The single option that has the more reactive aldehyde and methanol in excess is Option B.

Mathematically, if we denote the proportionality constant as $$k$$ and use subscripts to distinguish situations, we can write

$$\text{Rate}_{\text{B}} = k[\text{propanal}\!-\!\text{H}^+][\text{MeOH}]_{\text{excess}},$$

$$\text{Rate}_{\text{C}} = k[\text{propanal}\!-\!\text{H}^+][\text{MeOH}]_{\text{stoich}},$$

and clearly $$[\text{MeOH}]_{\text{excess}} > [\text{MeOH}]_{\text{stoich}},$$ so $$\text{Rate}_{\text{B}} > \text{Rate}_{\text{C}}.$$

Any rate involving acetone will already be slower due to the less electrophilic carbonyl, cementing the conclusion that

$$\text{Rate}_{\text{B}} > \text{Rate}_{\text{D}} \gt \text{Rate}_{\text{C}} \gt \text{Rate}_{\text{A}}.$$

Hence, the correct answer is Option B.