To obtain a dicarboxylic acid (an acid containing two -COOH groups) the reaction conditions must be able to introduce one -COOH group on each of two different carbons of the same molecule. The most common laboratory method for doing this in a single step is oxidative cleavage of a carbon-carbon double bond with a strong oxidising agent such as hot, acidic $$KMnO_4$$ (or, equivalently, ozonolysis followed by oxidative work-up). Each carbon of the former $$C=C$$ bond is oxidised up to the carboxylate level, so both ends of the cleaved bond become -COOH groups.

What happens in Option C
  • The substrate given in Option C contains an internal carbon-carbon double bond.
  • The reagent specified is hot, acidic $$KMnO_4$$ (or an equivalent strong oxidising system).
  • Hot $$KMnO_4$$ performs oxidative cleavage: the $$C=C$$ bond is cut and each of the two carbon atoms is converted into a carbonyl that is then further oxidised to a carboxylic acid.
  • Because both carbons of the former double bond stay inside the same molecular framework after the cleavage, the product contains two -COOH groups that remain connected through the rest of the carbon skeleton. Hence a single molecule of a dicarboxylic acid is obtained as the major product.

Why the other options fail
Option A, Option B and Option D either involve oxidation of **only one functional carbon**, or they involve a reaction (such as haloform cleavage or nitrile hydrolysis) that removes part of the carbon skeleton. In every one of those sequences the molecule finally possesses just one -COOH group; the second carbon is either lost from the molecule or remains at a lower oxidation level. Consequently they cannot supply a dicarboxylic acid as the principal product.

Therefore, the only sequence that reliably converts two different carbons of the same molecule into carboxyl groups is the oxidative-cleavage sequence of Option C.

Option C which is: the oxidative cleavage of the double bond with hot, acidic $$KMnO_4$$ (or its equivalent) is the correct choice.

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.

Nucleophilic addition to a carbonyl group occurs at the electrophilic carbon of the $$C = O$$ bond. The rate depends on two main factors:
  • Electronic effects: groups that withdraw electrons ( $$-I$$ and/or $$-R$$) increase the partial positive charge on the carbonyl carbon and enhance reactivity. Groups that donate electrons ( $$+I$$ and/or $$+R$$) decrease the charge and slow the reaction.
  • Steric effects: less‐hindered carbonyls are more accessible to the nucleophile.

First compare aldehydes and ketones attached to an aromatic ring.
  • Aldehydes have only one alkyl/aryl group attached to the carbonyl carbon, so they are less sterically hindered and receive less $$+I$$ donation than ketones. Hence aromatic aldehydes are more reactive than aromatic ketones.
  • Therefore $$\text{acetophenone (Ar-CO-CH}_3)$$ is expected to be the least reactive of the four compounds.

Now compare the three aromatic aldehydes:
  1. $$\text{p-nitrobenzaldehyde}$$ contains a $$NO_2$$ group. $$NO_2$$ is strongly electron-withdrawing by both $$-I$$ (inductive) and $$-R$$ (resonance) effects, so it makes the carbonyl carbon highly electrophilic. This gives the HIGHEST reactivity.
  2. $$\text{benzaldehyde}$$ has no extra substituent on the ring, so its reactivity is taken as the reference among aldehydes.
  3. $$\text{p-tolualdehyde}$$ (para-methylbenzaldehyde) contains a $$CH_3$$ group. $$CH_3$$ is electron-donating by $$+I$$ and weakly $$+R$$, so it reduces the electrophilicity of the carbonyl carbon, lowering the reactivity below that of unsubstituted benzaldehyde.

Putting the steric and electronic arguments together:

$$\text{Least reactive} \;\; \longrightarrow \;\; \text{acetophenone} \lt \text{p-tolualdehyde} \lt \text{benzaldehyde} \lt \text{p-nitrobenzaldehyde} \;\; \longrightarrow \;\; \text{Most reactive}$$

This sequence matches Option D.

Answer: Option D  $$\left(\text{acetophenone} \lt \text{p-tolualdehyde} \lt \text{benzaldehyde} \lt \text{p-nitrobenzaldehyde}\right)$$

A) Benzaldehyde: Gives a negative result. Although it is an aldehyde, aromatic aldehydes are generally not oxidized by the relatively weak Fehling's solution.

B) Acetophenone: Gives a negative result. It is a ketone, and ketones lack the oxidizable hydrogen atom attached to the carbonyl carbon necessary to reduce Fehling's solution.

C) Fructose: Gives a positive result. Despite being a keto-hexose, it undergoes tautomerization under the alkaline conditions of the test to form oxidizable aldoses.
$$ HOCH_{2} - CO - (CHOH)_{3}-CH_{2}-OH $$

D) Acetaldehyde: Gives a positive result. As an aliphatic aldehyde, it is easily oxidized by the $$Cu^{2+}$$ ions in Fehling's solution to form a carboxylic acid and a red precipitate of $$Cu_2O$$.

E) Phenylacetaldehyde: Gives a positive result. Even though it contains an aromatic ring, the aldehyde group is attached to an aliphatic side chain, allowing it to behave like a typical aliphatic aldehyde in this test.

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%.

We need to find the ratio $$K_P/K_C$$ for the reaction $$CO_{(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons CO_{2(g)}$$. The relationship between $$K_P$$ and $$K_C$$ is given by $$K_P = K_C(RT)^{\Delta n_g}$$ where $$\Delta n_g$$ is the change in the number of moles of gaseous species (moles of gaseous products minus moles of gaseous reactants).

Moles of gaseous products = 1 (CO$$_2$$) and moles of gaseous reactants = 1 (CO) + 1/2 (O$$_2$$) = 3/2, so $$\Delta n_g = 1 - \frac{3}{2} = -\frac{1}{2}$$.

Substituting into the formula gives $$\frac{K_P}{K_C} = (RT)^{\Delta n_g} = (RT)^{-1/2} = \frac{1}{\sqrt{RT}}$$, therefore the correct answer is Option (1): $$\frac{1}{\sqrt{RT}}$$.

We need to evaluate two statements about buffer solutions.

Analysis of Statement (I): "A Buffer solution is the mixture of a salt and an acid or a base mixed in any particular quantities."

This statement is FALSE. While a buffer solution is indeed a mixture of a weak acid and its conjugate base (or a weak base and its conjugate acid), it does not work with just "any" quantities. For a buffer to function effectively, the components must be mixed in appropriate proportions. Specifically:

- The buffer capacity depends on the ratio of the concentrations of the conjugate pair.

- An effective buffer typically has a pH within one unit of the $$pK_a$$ of the weak acid, which requires the ratio $$[salt]/[acid]$$ to be between 0.1 and 10.

- If the ratio is too extreme, the solution will not resist pH changes effectively.

Analysis of Statement (II): "Blood is naturally occurring buffer solution whose pH is maintained by $$H_2CO_3/HCO_3^-$$ concentrations."

This statement is TRUE. Blood is a naturally occurring buffer that maintains its pH around 7.4. The primary buffer system in blood is the carbonic acid-bicarbonate buffer system:

$$H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$$

When excess acid ($$H^+$$) enters the blood, bicarbonate ions ($$HCO_3^-$$) neutralise it. When excess base ($$OH^-$$) enters, carbonic acid ($$H_2CO_3$$) neutralises it. This maintains the blood pH within a narrow, life-sustaining range.

The correct answer is Option (3): Statement I is false but Statement II is true.

Find the equilibrium constant for $$X \rightleftharpoons W$$ given intermediate steps.

$$X \rightleftharpoons Y$$, $$K_1 = 1.0$$

$$Y \rightleftharpoons Z$$, $$K_2 = 2.0$$

$$Z \rightleftharpoons W$$, $$K_3 = 4.0$$

When reactions are added, their equilibrium constants are multiplied. Adding all three:

$$X \rightleftharpoons Y \rightleftharpoons Z \rightleftharpoons W$$

$$K = K_1 \times K_2 \times K_3 = 1.0 \times 2.0 \times 4.0 = 8.0$$

The correct answer is Option (4): 8.0.

The reaction: $$Fe_2O_{3(s)} + 3CO_{(g)} \rightleftharpoons Fe_{(l)} + 3CO_{2(g)}$$

Le Chatelier's Principle: The equilibrium shifts in response to changes in concentration, pressure, or temperature.

Key point: Pure solids and pure liquids do NOT affect the equilibrium position because their concentrations (activities) are constant. They do not appear in the equilibrium expression.

$$Fe_2O_3$$ is a solid. Adding more $$Fe_2O_3(s)$$ will not disturb the equilibrium because its activity remains 1.

Adding/removing $$CO_2$$ or removing $$CO$$ would shift the equilibrium as they are gaseous species that appear in the equilibrium expression.

The correct answer is Option (3): Addition of $$Fe_2O_3$$.

The reaction is $$H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$$.

The cylinder is filled with equal number of molecules of $$H_2$$, $$I_2$$, and $$HI$$ at $$-20°C$$ (253 K) and 1 atm.

Since the number of moles of gaseous reactants equals the number of moles of gaseous products ($$\Delta n_g = 0$$), $$K_p = K_c$$ for this reaction.

Let the number of moles of each gas be $$n$$. The partial pressures are all equal (since equal moles at same T, V): each gas has partial pressure $$P/3$$ where $$P = 1$$ atm.

$$K_p = \frac{(P_{HI})^2}{P_{H_2} \cdot P_{I_2}} = \frac{(P/3)^2}{(P/3)(P/3)} = \frac{P^2/9}{P^2/9} = 1$$

So $$K_p = 1 = x \times 10^{-1}$$.

This gives $$x = 10$$.

The correct answer is Option B: 10.

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.

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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.

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:

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:

The available 87 g of acetone corresponds to:

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

Thus, the mass of benzaldehyde needed is:

Therefore, 318 g of benzaldehyde is required.

Formation of Product A (Elimination)

• Reaction: Bromoethane ($$\text{CH}_3\text{CH}_2\text{Br}$$) reacting with $$\text{NaOH}$$ in alcoholic medium ($$\text{C}_2\text{H}_5\text{OH}$$).
• Mechanism: Alcoholic $$\text{NaOH}$$ acts as a strong base, favoring $$\text{E}2$$ dehydrohalogenation.
• Product A: Ethene ($$\text{CH}_2=\text{CH}_2$$)
• Hydrogen Count in A: 4

2. Formation of Product B (Substitution)

• Reaction: Bromoethane ($$\text{CH}_3\text{CH}_2\text{Br}$$) reacting with $$\text{NaOH}$$ in aqueous medium ($$\text{H}_2\text{O}$$).
• Mechanism: Aqueous $$\text{NaOH}$$ provides strong nucleophilic hydroxyl ions ($$\text{OH}^-$$), favoring $$\text{S}_\text{N}2$$ substitution.
• Product B: Ethanol ($$\text{CH}_3\text{CH}_2\text{OH}$$)
• Hydrogen Count in B: 6

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 $$\boxed{3}$$.

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.

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: $$\boxed{\text{A is true but R is false}}$$.

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 (-CH(OH)-_)

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).

The iodoform test is positive for compounds containing either a $$CH_3CO-$$ group (methyl ketone) or a $$CH_3CH(OH)-$$ group (secondary alcohol with methyl group adjacent to OH).

Analyzing each compound:

(a) 1-Phenylbutan-2-one (C₆H₅CH₂COCH₂CH₃): The carbonyl has CH₂ groups on both sides, not a methyl group. Negative ✗

(b) 2-Methylbutan-2-ol ((CH₃)₂C(OH)CH₂CH₃): This is a tertiary alcohol with no CH₃CHOH pattern. Negative ✗

(c) 3-Methylbutan-2-ol (CH₃CH(OH)CH(CH₃)₂): Has a CH₃CHOH group. Positive ✓

(d) 1-Phenylethanol (C₆H₅CH(OH)CH₃): Has a CH₃CHOH group. Positive ✓

(e) 3,3-Dimethylbutan-2-one (CH₃COC(CH₃)₃): Has a CH₃CO group. Positive ✓

(f) 1-Phenylpropan-2-ol (C₆H₅CH₂CH(OH)CH₃): Has a CH₃CHOH group. Positive ✓

Number of compounds giving positive iodoform test = 4 (c, d, e, f).

Glyoxylic acid has the structure: OHC-COOH (an aldehyde and a carboxylic acid group).

When glyoxylic acid reacts with excess MeMgBr (Grignard reagent):

The first equivalent of MeMgBr reacts with the acidic -COOH group (acid-base reaction):

$$-\text{COOH} + \text{CH}_3\text{MgBr} \rightarrow -\text{COO}^-\text{MgBr}^+ + \text{CH}_4$$

The second equivalent of MeMgBr performs nucleophilic addition to the -CHO group:

$$-\text{CHO} + \text{CH}_3\text{MgBr} \rightarrow -\text{CH(OMgBr)CH}_3$$

After hydrolysis with H$$_3$$O$$^+$$, the product is 2-hydroxypropionic acid (lactic acid type).

So for $$y$$ moles of glyoxylic acid, $$x = 2y$$ moles of MeMgBr are needed.

$$\frac{x}{y} = 2$$

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.

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.

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).

We need to identify the products A and B formed in the given reaction.

Based on the JEE 2022 context, this question involves ozonolysis of a suitable alkene. Ozonolysis cleaves the carbon-carbon double bond and produces aldehydes and/or ketones depending on the substituents.

When an alkene undergoes ozonolysis (treatment with $$O_3$$ followed by reductive workup with Zn/$$H_2O$$), the double bond is cleaved, and each carbon of the double bond gets an oxygen (=O) to form carbonyl compounds.

For propene ($$CH_3CH=CH_2$$):

$$CH_3CH=CH_2 \xrightarrow{O_3/Zn, H_2O} CH_3CHO + HCHO$$

The products are:

$$HCHO$$ (formaldehyde) — from the $$=CH_2$$ end

$$CH_3CHO$$ (acetaldehyde) — from the $$CH_3CH=$$ end

Based on the given options and the convention in the question:

$$A = HCHO$$ (Formaldehyde)

$$B = CH_3CHO$$ (Acetaldehyde)

Option A: Acrolein and formaldehyde — Incorrect

Option B: Acetaldehyde and acrolein — Incorrect

Option C: Formaldehyde and acetaldehyde — Correct

Option D: Acrolein and acetaldehyde — Incorrect

Hence, the correct answer is Option C: Formaldehyde and acetaldehyde.

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.

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).

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.

First we write the full structure of the given ester. Pent-4-ynoic acid contains five carbon atoms in the main chain and the triple bond starts at carbon-4. Converting the -COOH group into an ethyl ester gives

$$\mathrm{CH_3CH_2C\!\equiv\!C-CO_2CH_2CH_3}$$

Now we examine the nature of each functional group. The molecule has

(i) an ester carbonyl $$-CO_2-$$, and

(ii) a terminal alkyne hydrogen $$\mathrm{\;-C\!\equiv\!C-H}$$ at carbon-5. A terminal alkyne hydrogen is appreciably acidic (pK$$_a\approx25$$) and is removed instantaneously by a Grignard reagent, because every Grignard reagent behaves as a very strong base as well as a nucleophile.

We next recall the two standard facts we shall need.

• Acid-base reaction with a terminal alkyne $$\mathrm{R-C\!\equiv\!C-H + R'MgX \;\longrightarrow\; R-C\!\equiv\!C^{\;-}MgX^{+} + R'H}$$

• Nucleophilic addition of a Grignard reagent to an ester For every mole of an ester, two moles of a Grignard reagent are required to reach the tertiary-alcohol stage:

$$\mathrm{RCOOR' + RMgX \;\xrightarrow{\;(1)\;}\; R-CO-R \;(\text{ketone})}$$

and then

$$\mathrm{R-CO-R + RMgX \;\xrightarrow{\;(2)\;}\; R-C(OMgX)R_2 \;\xrightarrow{\;H_2O\;/\;H^+\;} R-C(OH)R_2}$$

Keeping these two rules in mind, we follow the reaction of ethyl pent-4-yn-oate with methylmagnesium bromide step by step.

Step 1 - deprotonation of the terminal alkyne

$$\mathrm{CH_3CH_2C\!\equiv\!C-H + CH_3MgBr \;\longrightarrow\; CH_3CH_2C\!\equiv\!C^{\;-}MgBr^{+} + CH_4\uparrow}$$

This consumes one mole of $$\mathrm{CH_3MgBr}$$ and produces methane gas. The alkyne is now converted into its magnesium acetylide; the ester function is still intact.

Step 2 - first nucleophilic attack on the ester carbonyl

$$\mathrm{CH_3CH_2C\!\equiv\!C^{\;-}MgBr^{+}\;-CO_2CH_2CH_3 + CH_3MgBr \;\longrightarrow\; CH_3CH_2C\!\equiv\!C-COCH_3 + CH_3CH_2OMgBr}$$

After the first nucleophilic addition and expulsion of the ethoxide ion, a ketone $$\mathrm{RC(=O)CH_3}$$ is formed. A second mole of the Grignard reagent has therefore been used.

Step 3 - second nucleophilic attack on the ketone

$$\mathrm{CH_3CH_2C\!\equiv\!C-COCH_3 + CH_3MgBr \;\longrightarrow\; CH_3CH_2C\!\equiv\!C-C(OMgBr)(CH_3)_2}$$

Hydrolysing the adduct finally gives the alcohol:

$$\mathrm{CH_3CH_2C\!\equiv\!C-C(OH)(CH_3)_2}$$

The carbon bearing the -OH group now carries three alkyl substituents (two $$\mathrm{CH_3}$$ groups and one $$\mathrm{CH_2C\!\equiv\!CCH_2CH_3}$$ group). By definition, such an alcohol is a tertiary (3°) alcohol. Hence Statement I is perfectly correct.

Let us count the number of moles of $$\mathrm{CH_3MgBr}$$ that were consumed overall:

• one mole in Step 1 (acid-base reaction), • one mole in Step 2 (first nucleophilic addition), • one mole in Step 3 (second nucleophilic addition).

Total $$= 3$$ moles of $$\mathrm{CH_3MgBr}$$ per mole of the ester. Statement II, which claims that only two moles are utilised, is therefore incorrect.

We thus conclude that Statement I is true whereas Statement II is false.

Hence, the correct answer is Option B.

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

To decide which reagent is incapable of reducing an organic functional group in the laboratory, we recall the basic principle of reduction in organic chemistry: reduction involves the addition of hydrogen (or the removal of oxygen) and is usually accomplished by either (i) nascent hydrogen produced in situ or (ii) catalytic hydrogenation in which molecular hydrogen is added across a multiple bond in the presence of a metal catalyst.

We now examine each option in the light of this principle.

Option A: $$\text{Zn / H}_2\text{O}$$ — Metallic zinc reacts with water (especially in the presence of acid) to generate nascent hydrogen according to the well-known reaction

The freshly produced $$[\text{H}]$$ (nascent hydrogen) is able to reduce several functional groups (for example, the Clemmensen reduction formally uses Zn/Hg in acid to reduce carbonyl groups). Hence Zn/H$$_2$$O can act as a reducing system.

Option B: $$\text{Pt-C / H}_2$$ — In catalytic hydrogenation, finely divided platinum supported on carbon adsorbs molecular hydrogen according to

These adsorbed hydrogen atoms readily add to C=C, C≡C and several other reducible functional groups. Therefore Pt-C/H$$_2$$ is a powerful laboratory reducing agent.

Option C: $$\text{Pd-C / H}_2$$ — Palladium on carbon functions in exactly the same catalytic manner as platinum. It is one of the most frequently used reagents for hydrogenation of alkenes, alkynes, nitro groups, etc. Hence Pd-C/H$$_2$$ also reduces functional groups effectively.

Option D: $$\text{Na / H}_2$$ — Here we merely have metallic sodium physically mixed with molecular hydrogen. Sodium metal does not adsorb hydrogen in a way that produces active $$[\text{H}]$$ atoms, nor does it catalyse the dissociation of H$$_2$$. Consequently, no nascent or catalytic hydrogen is available and the mixture fails to reduce ordinary organic functional groups under experimental conditions.

Thus, among the four reagents listed, the only one that cannot serve as a practical reducing system is $$\text{Na / H}_2$$.

Hence, the correct answer is Option 4.

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

The reaction shown yields a mixture of two products:

$$1,4-benzenedimethanol$$ and $$4-(hydroxymethyl)benzoic acid$$.

The reaction shown is a classic example of the Cannizzaro reaction. The starting material, 4-(hydroxymethyl)benzaldehyde, contains an aldehyde group attached directly to a benzene ring. Because the carbon atom adjacent to the carbonyl group is part of the aromatic ring, it does not possess any alpha-hydrogens (hydrogens attached to the alpha-carbon).

When an aldehyde lacking alpha-hydrogens is heated in the presence of a strong base like sodium hydroxide ($$NaOH$$), it undergoes a base-induced disproportionation reaction. In this redox process, two molecules of the reactant participate simultaneously. One molecule is oxidized to form a carboxylic acid salt, while the other molecule is reduced to form a primary alcohol. The existing hydroxymethyl group ($$-CH_2OH$$) at the para position is stable under these conditions and does not actively participate in the disproportionation; it may transiently form an alkoxide in the strong base, but it remains structurally unchanged.

Following the basic reaction, the acidic workup ($$H_3O^+$$) is introduced. This step protonates the intermediate carboxylate salt to form the final neutral carboxylic acid. Therefore, the reaction yields two distinct organic products from the original starting material: the reduced product, where the aldehyde is converted to an alcohol ($$1,4-benzenedimethanol$$), and the oxidized product, where the aldehyde is converted to a carboxylic acid $$4-(hydroxymethyl)benzoic acid$$ that is the final product. It is a monocarboxylic acid with a hydroxy group to it.

Thus the correct option is C.

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.

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 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.

Here is the explanation in a clear, pastable paragraph format:

The starting material in this reaction sequence is a cyanohydrin (specifically, acetone cyanohydrin). To form major product A, the reactant is treated with lithium aluminum hydride (LiAlH₄) followed by acidic workup. LiAlH₄ is a strong reducing agent that specifically targets the nitrile group (-CN) and reduces it to a primary amine (-CH₂NH₂). Because it does not react with tertiary alcohols, the hydroxyl group (-OH) remains completely intact, resulting in an amino alcohol for product A. To form major product B, the reactant is subjected to strongly acidic conditions using hydronium ions and sulfuric acid (H₃O⁺ / H₂SO₄). This triggers two consecutive reactions. First, the acid completely hydrolyzes the nitrile group (-CN) into a carboxylic acid group (-COOH). Immediately following this, the strong dehydrating nature of sulfuric acid causes the elimination of the tertiary hydroxyl group (-OH) along with a hydrogen from an adjacent methyl group. This dehydration forms a stable, conjugated carbon-carbon double bond, yielding an alpha, beta-unsaturated carboxylic acid as product B.

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.

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.

The target molecule is 2-methylpropan-2-ol, which is (CH₃)₃COH — a tertiary alcohol.

The starting material is ethyl ethanoate (ethyl acetate): CH₃COOC₂H₅.

The Grignard reagent is CH₃MgBr (methylmagnesium bromide).

In the first step, CH₃MgBr attacks the carbonyl carbon of ethyl ethanoate. One equivalent of Grignard adds to give a tetrahedral intermediate (a magnesium alkoxide): CH₃-C(CH₃)(OMgBr)-OC₂H₅. This intermediate is a hemiketal-like species that collapses: the ethoxide (C₂H₅O⁻) is expelled as a leaving group, generating a ketone intermediate — in this case acetone (propan-2-one): CH₃COCH₃.

In the second step, a second equivalent of CH₃MgBr adds to the acetone formed. The Grignard adds to the carbonyl of acetone to give the magnesium alkoxide: (CH₃)₃COMgBr. Aqueous workup gives the tertiary alcohol: (CH₃)₃COH = 2-methylpropan-2-ol.

Thus, 2 equivalents of CH₃MgBr are required in total: 1 equivalent to react with ethyl ethanoate to produce acetone, and 1 equivalent to add to the acetone to give 2-methylpropan-2-ol.

Therefore, the answer is $$\boxed{2}$$ equivalents.

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 $$\boxed{3}$$.

We begin by recalling that to convert a given aromatic starting material into an aldehyde group $$\left(-CHO\right)$$, different classical name reactions are employed. The reagent sets themselves often reveal which reaction is intended. Hence, for every combination we shall first identify the reaction indicated by the reagent and then verify that the starting material is suitable for that reaction.

Let us look at Item I where the starting material is benzene $$\left(\text C_6\text H_6\right)$$. To introduce a formyl group directly on the benzene ring, the famous Gattermann-Koch formylation is used. The reagent for this reaction is stated, before any calculation, as

$$\text{CO} + \text{HCl} + \text{AlCl} _3\;($$ and sometimes CuCl $$).$$

This exactly matches reagent $$R : \text{CO},\;\text{HCl}\text{ and }\text{AlCl}_3.$$

Therefore, Item I must be paired with $$R.$$ After the formylation, the product obtained is benzaldehyde $$\left(\text C_6\text H_5\text{CHO}\right).$$

Next, consider Item II where the starting material is benzonitrile $$\left(\text C_6\text H_5\text{CN}\right).$$ Converting a nitrile into an aldehyde is carried out by the Stephen reduction. Before doing any algebra, the standard reagent for Stephen reduction is stated as

$$\text{SnCl} _2 + 2\, \text{HCl} \quad$$ followed by $$\quad \text{H} _3 \text{O} ^+.$$

This reagent set appears verbatim as $$P : \text{HCl and }\text{SnCl}_2,\;\text{H}_3\text O^+.$$

So, Item II must be paired with $$P.$$ The nitrile is thus reduced to the imidoyl chloride intermediate and on hydrolysis gives benzaldehyde.

Finally, Item III has benzoyl chloride $$\left(\text C_6\text H_5\text{COCl}\right)$$ as the starting material. Converting an acyl chloride to an aldehyde is the well-known Rosenmund reduction. The textbook reagent for Rosenmund reduction is stated as

$$\text H_2,\;\text{Pd-BaSO}_4$$ poisoned with sulfur or quinoline.

That is precisely reagent $$Q : \text H_2,\; \text{Pd-BaSO} _4,\; \text S\$$ and quinoline $$.$$

Hence, Item III must be paired with $$Q.$$ The catalytic hydrogenation stops at the aldehyde stage, giving benzaldehyde.

Summarising the matches:

$$\begin{aligned} \text{(I) Benzene} &\longrightarrow R \;(\text{Gattermann-Koch})\\ \text{(II) Benzonitrile} &\longrightarrow P \;(\text{Stephen reduction})\\ \text{(III) Benzoyl chloride} &\longrightarrow Q \;(\text{Rosenmund reduction}) \end{aligned}$$

The correspondence obtained is exactly

(I)-(R), (II)-(P) and (III)-(Q).

Looking at the options given, this set is listed as Option C.

Hence, the correct answer is Option C.

We begin by noting that both compounds A and B have the molecular formula $$C_3H_6O$$. Such a formula fits carbonyl compounds having three carbon atoms, namely

$$\text{propanal : }CH_3CH_2CHO\quad\text{(an aldehyde)}$$

$$\text{acetone : }CH_3COCH_3\quad\text{(a ketone)}$$

When a carbonyl compound is treated with a Grignard reagent, an alkyl group is added to the carbonyl carbon and the oxygen is converted into an alcohol after acidic work-up. Stating the general result first,

$$\text{R-C(=O)-R'} + CH_3MgBr \overset{H_3O^+}{\rightarrow} \text{R-C(OH)(CH}_3\text{)-R'}$$

So each carbonyl compound of three carbons will give an alcohol of four carbons, because one extra $$CH_3$$ group is introduced.

1. Reaction of propanal with $$CH_3MgBr$$

Starting carbonyl:

$$CH_3CH_2CHO$$

Addition of $$CH_3^-$$ from the Grignard reagent at the carbonyl carbon followed by protonation gives

$$CH_3CH_2C(OH)(CH_3)H \;=\;CH_3CH_2CH(OH)CH_3$$

This product is butan-2-ol, a secondary alcohol. Let us call it product C.

2. Reaction of acetone with $$CH_3MgBr$$

Starting carbonyl:

$$CH_3COCH_3$$

Addition of $$CH_3^-$$ and protonation gives

$$CH_3C(OH)(CH_3)CH_3 \;=\;(CH_3)_3C\,OH$$

This product is 2-methyl-2-propanol (tert-butyl alcohol), a tertiary alcohol. Let us call it product D.

Now we match C and D with the chemical tests given.

(i) Ceric ammonium nitrate test - any alcohol answers it, so both C and D should be positive, exactly as observed.

(ii) Lucas test - the rate of turbidity distinguishes the class of alcohol.

• Secondary alcohols turn turbid in about five minutes.
• Tertiary alcohols turn turbid immediately.

Product C is a secondary alcohol (butan-2-ol) → turbidity after five minutes.
Product D is a tertiary alcohol (tert-butyl alcohol) → instantaneous turbidity.

(iii) Iodoform test - positive for ethanol or any secondary alcohol that contains the fragment $$CH_3-CH(OH)-$$.

• Butan-2-ol possesses the $$CH_3-CH(OH)-$$ group, so C gives a positive iodoform test.
• tert-Butyl alcohol lacks that fragment, so D gives a negative iodoform test.

The behaviour predicted for butan-2-ol and tert-butyl alcohol coincides perfectly with the observations. Therefore

$$C = CH_3CH_2CH(OH)CH_3\quad(\text{butan-2-ol})$$

$$D = (CH_3)_3C\,OH\quad(\text{tert-butyl alcohol})$$

Looking at the options, these structures are listed together only in Option A:

$$C = H_3C-CH_2-CH(OH)-CH_3;\; D = H_3C-C(CH_3)_2-OH$$

Hence, the correct answer is Option A.

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.

Since, the para position is blocked the Br atom will come and attach at ortho position due to +M effect of -OH because this is EAS Reaction.

• 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.

We are given the substrate cyclopentanone, which is a five-membered cyclic ketone having the carbonyl group $$\gt \! C = O$$ at one of the ring carbons. The reagent set is sodium borohydride in ethanol, written as $$NaBH_4/EtOH.$$

First, we recall the general reduction formula for a ketone by sodium borohydride. The well-known relationship is

$$R_2C=O \;+\; NaBH_4 \;\overset{\text{protic solvent}}{\rightarrow} \; R_2C(OH)H.$$

Stated in words, NaBH4 delivers a hydride ion $$H^-$$ to the electrophilic carbonyl carbon, converting the $$C=O$$ into an alkoxide $$C-O^-. $$ The protic solvent (here ethanol) then protonates that alkoxide to give the corresponding alcohol.

Applying this step by step to cyclopentanone, we have

1. Hydride addition:

$$ \underset{\text{cyclopentanone}}{C_5H_8O} \;+\; BH_4^{-} \;\longrightarrow\; \underset{\text{alkoxide}}{C_5H_9O^{-}} \;+\; BH_3$$

2. Protonation by ethanol (the solvent is an available proton donor):

$$ C_5H_9O^- \;+\; EtOH \;\longrightarrow\; C_5H_{10O \;+\; EtO^- }$$

Combining the two stages, the net result is conversion of the ketone functional group into a single hydroxyl group without adding any other substituent. Therefore the ring now bears an $$-OH$$ at the same carbon that formerly possessed the carbonyl.

So we obtain cyclopentanol, a secondary alcohol in a five-membered ring, and nothing else such as an $$-OEt$$ group or a double bond is introduced. Among the listed choices, the structure that matches is simply “Cyclopentanol.”

Hence, the correct answer is Option C.

The reaction in question is the addition of a single equivalent of a Grignard reagent, written in the general form $$\mathrm{R{-}MgX},$$ to the carbonyl group of an aldehyde. A necessary condition for this nucleophilic addition is that the Grignard reagent should reach the carbonyl carbon without being destroyed en-route. The reagent is both a very strong base and a very strong nucleophile; therefore any group present in the substrate that is (i) proton-donating (acidic) or (ii) itself an electrophilic carbonyl capable of reacting faster than the aldehyde will consume the Grignard reagent before it can add to the aldehyde carbonyl.

We first recall two key facts.

1. A Grignard reagent instantly abstracts an acidic proton. Symbolically,

$$\mathrm{R{-}MgX \;+\; H{-}Z \;\longrightarrow\; R{-}H \;+\; Z^{-}MgX^{+}},$$

where $$\mathrm{H{-}Z}$$ is any group with an O-H, N-H, S-H, or similarly acidic bond.

2. A Grignard reagent will add to a carbonyl group provided it has not already been destroyed by acid-base reaction.

Now we examine each option.

(A) Benzaldehyde, $$\mathrm{C_6H_5{-}CHO}.$$

No acidic proton other than the aldehydic one, and that proton is attached to carbon, not to a hetero-atom. Hence the Grignard reagent reaches the carbonyl safely and adds:

$$\mathrm{R{-}MgX + C_6H_5{-}CHO \xrightarrow{Et_2O} C_6H_5{-}CH(OMgX)R \xrightarrow{H_2O} C_6H_5{-}CH(OH)R}$$

So benzaldehyde does give the expected Grignard product.

(B) An aldehyde that also contains a $$\mathrm{HO_2C{-}}$$ (carboxylic acid) group.

The -COOH unit possesses an acidic O-H proton. Using the acid-base equation stated above, one equivalent of the Grignard reagent is consumed immediately:

$$\mathrm{R{-}MgX + HO_2C{-}R' \longrightarrow R{-}H + ^{-}O_2C{-}R' \,MgX^{+}}$$

With the reagent destroyed, nothing is left to attack the aldehyde carbonyl. No addition product appears.

(C) An aldehyde that also contains a $$\mathrm{H_3CO{-}}$$ (methoxy) group.

The methoxy group has no acidic proton; it is an ether. Therefore the Grignard reagent is not quenched and can still add to the aldehyde carbonyl exactly as in (A). Product formation is unimpeded.

(D) An aldehyde that also contains a $$\mathrm{HOCH_2{-}}$$ (alcohol, -CH2OH) group.

The alcohol O-H is acidic. One equivalent of Grignard reagent is again lost through proton abstraction:

$$\mathrm{R{-}MgX + HOCH_2{-}R' \longrightarrow R{-}H + ^{-}OCH_2{-}R' \,MgX^{+}}$$

Once more, no nucleophile remains to attack the aldehydic carbonyl, and no Grignard addition product is obtained.

Summarising, aldehydes (B) and (D) fail to furnish Grignard addition products with one equivalent of Grignard reagent, while (A) and (C) succeed.

Hence, the correct answer is Option D.

First, the NaBH4 reduces the ketone to a secondary alcohol. This creates a halohydrin intermediate (a molecule with an -OH and a -Br on adjacent carbons).

In the presence of methanol or the slightly basic conditions generated during a borohydride reduction, the alcohol group can become deprotonated to form an alkoxide. This oxygen then performs an internal nucleophilic attack on the carbon holding the bromine.

We are given:

Compound A has molecular formula $$C_9H_{10}O$$ and gives a positive iodoform test.

Oxidation of A with $$KMnO_4/KOH$$ gives acid B with molecular formula $$C_8H_6O_4$$.

The anhydride of B is used for the preparation of phenolphthalein.

Step 1: Identify acid B.

Phenolphthalein is prepared by heating phthalic anhydride with phenol in the presence of a dehydrating agent (like concentrated $$H_2SO_4$$ or $$ZnCl_2$$).

Phthalic anhydride is the anhydride of phthalic acid (benzene-1,2-dicarboxylic acid), which has the molecular formula $$C_8H_6O_4$$.

So acid B is phthalic acid (ortho-benzenedicarboxylic acid).

Step 2: Identify compound A.

Since A gives a positive iodoform test, it must contain a $$-COCH_3$$ (methyl ketone) group. The iodoform test is positive for compounds containing the $$CH_3CO-$$ group.

The molecular formula of A is $$C_9H_{10}O$$. The degree of unsaturation is:

$$\text{DoU} = \frac{2(9) + 2 - 10}{2} = \frac{10}{2} = 5$$

A benzene ring accounts for 4 degrees of unsaturation (3 double bonds + 1 ring), and the $$C=O$$ of the ketone accounts for 1 more. Total = 5. So A is a substituted acetophenone (phenyl methyl ketone) with an additional methyl group on the ring.

Compound A = methyl acetophenone = $$CH_3-C_6H_4-COCH_3$$.

Step 3: Determine the position of the methyl group.

When A is oxidized with $$KMnO_4/KOH$$, both the $$-CH_3$$ group on the ring and the $$-COCH_3$$ group are oxidized to $$-COOH$$ groups.

The product B is phthalic acid, which has the two $$-COOH$$ groups in the ortho (1,2-) position.

This means the $$-CH_3$$ and $$-COCH_3$$ groups on A must be at ortho positions relative to each other.

Therefore, compound A is ortho-methyl acetophenone (2-methylacetophenone or 1-(2-methylphenyl)ethanone).

The answer is Option B: ortho-methyl acetophenone.

Here an internal Aldol reaction will take place where in a carbanion will be formed at the carbon on the left of ketone rather than right because the formation of a 5 membered ring is more preferrable compared to the formation of a 7 membered ring due to steric factors. This carbanion will in turn attack on the aldehyde group and reduce it to alcohol. Upon heating -OH group will leave with a hydrogen and a double bond will be formed.

We recall the general acid-catalysed acetal formation:

$$\text{Aldehyde}(RCHO) + 2\,\text{Alcohol}(R'OH) \xrightarrow{HCl} RCH(OR')_2 + H_2O$$

The mechanism involves these key steps:

1. Protonation of the carbonyl oxygen to give a more electrophilic $$RCH^+=O$$ species.
2. Nucleophilic attack of the alcohol oxygen on the carbonyl carbon to form a hemiacetal.
3. Proton transfers followed by loss of water and a second alcohol attack to give the acetal.

The slow, rate-determining step is the nucleophilic attack of the alcohol on the protonated carbonyl carbon. Hence:

• The electrophile should be as reactive and as unhindered as possible.
• The nucleophile (alcohol) should be as unhindered and as nucleophilic as possible.

Reactivity order of carbonyl compounds toward nucleophilic addition is

$$\text{HCHO} \; \lt \; CH_3CHO \; \lt \; R_2C=O$$

but when we talk of ease of attack, “<” actually reads “less hindered & more reactive → reacts faster,” so

Formaldehyde (HCHO) > Acetaldehyde ($$\text{CH}_3\text{CHO}$$) > Ketones

Among alcohols, steric bulk decreases nucleophilicity:

MeOH (1°) > EtOH (1° larger) > 2° alcohols > t-BuOH (3°)

Now we examine each option.

A. Aldehyde = $$\text{HCHO}$$ (least hindered, most electrophilic); Alcohol = $$\text{MeOH}$$ (least hindered, most nucleophilic).
B. Aldehyde = $$CH_3CHO$$ (more hindered than HCHO); Alcohol = $$\text{t-BuOH}$$ (very hindered).
C. Aldehyde = $$CH_3CHO$$; Alcohol = $$\text{MeOH}$$ (good alcohol but aldehyde slightly less reactive).
D. Aldehyde = $$\text{HCHO}$$; Alcohol = $$\text{t-BuOH}$$ (excellent aldehyde but very poor, hindered alcohol).

Clearly, Option A combines the most electrophilic carbonyl with the most nucleophilic alcohol, minimising steric hindrance in both partners. This pair will give the fastest and highest-yielding acetal formation under $$HCl$$ catalysis.

Hence, the correct answer is Option A.