JEE Aldehydes & Ketones PYQs With Video Solutions PDF

1. Identify the Nature of the Chemical Transformation:
The given reaction indicates the complete reduction of an aliphatic ketone ($$\text{R-CO-R'}$$) into its corresponding hydrocarbon alkane ($$\text{R-CH}_2\text{-R'}$$). The carbonyl functional group ($$\text{C}=\text{O}$$) is selectively converted entirely into a methylene group ($$\text{CH}_2$$).

2. Evaluate the Given Reagent Options:

• Option A ($$\text{NaBH}_4 / \text{Ethanol}$$): Sodium borohydride is a mild reducing agent. It reduces aldehydes and ketones to primary and secondary alcohols, respectively, rather than reducing them completely to alkanes.
• Option B ($$\text{NH}_2\text{NH}_2 / \text{KOH, glycol, }\Delta$$): This combination represents the Wolff-Kishner reduction . Hydrazine ($$\text{NH}_2\text{NH}_2$$) reacts with the carbonyl compound to form a hydrazone intermediate. Under strongly basic conditions ($$\text{KOH}$$) and high temperature in a high-boiling solvent like ethylene glycol, the hydrazone decomposes with the evolution of nitrogen gas ($$\text{N}_2$$) to produce the corresponding alkane. This is completely correct.
• Option C ($$\text{LiAlH}_4 / \text{Dry ether}$$): Lithium aluminum hydride is a powerful reducing agent, but it reduces ketones to secondary alcohols, not to alkanes.
• Option D ($$\text{H}_2 / \text{Pd-BaSO}_4, \text{ quinoline}$$): The Rosenmund reduction specifically converts acyl chlorides ($$\text{R-COCl}$$) into aldehydes ($$\text{R-CHO}$$). It does not reduce ketones to alkanes.

Statement A is correct.

Benzaldehyde is an aromatic aldehyde. It readily reduces Tollens' reagent to produce a silver mirror because Tollens' reagent oxidizes aldehydes to carboxylates. However, Benzaldehyde does not reduce Fehling's solution under ordinary conditions. Fehling's test is generally given by aliphatic aldehydes, while aromatic aldehydes such as Benzaldehyde are negative.

Statement B is incorrect.

Acetaldehyde contains the $$CH_3CO-$$ unit and therefore gives a positive Iodoform test. However, it possesses $$\alpha$$-hydrogen atoms and consequently undergoes Aldol condensation in the presence of alkali. Cannizzaro reaction is shown only by aldehydes that do not contain any $$\alpha$$-hydrogen atoms. Hence Acetaldehyde does not undergo the Cannizzaro reaction.

Statement C is correct.

Hydrogen cyanide is a weak acid and undergoes only limited ionization in solution:

$$HCN \rightleftharpoons H^+ + CN^-$$

Addition of a small amount of base increases the concentration of the nucleophile $$CN^-$$ through the equilibrium:

$$HCN + OH^- \rightleftharpoons CN^- + H_2O$$

The increased concentration of $$CN^-$$ accelerates nucleophilic attack on the carbonyl carbon and therefore increases the rate of cyanohydrin formation.

Statement D is incorrect.

Nucleophilic addition reactions occur more readily with aldehydes than with ketones because aldehydes possess lower steric hindrance and a more electrophilic carbonyl carbon.

Benzaldehyde is an aldehyde whereas Acetophenone is a ketone. In Acetophenone, the additional methyl group exerts a $$+I$$ effect and increases steric hindrance, reducing the susceptibility of the carbonyl carbon towards nucleophilic attack.

Therefore:

$$\text{Benzaldehyde} > \text{Acetophenone}$$

in nucleophilic addition reactivity.

Hence, only Statements A and C are correct.

$$(B)\ \text{A and C only}$$

$$\text{}$$

Step 1: Halogenation of the Ketone (Formation of Intermediate X)
When 3-methylbutan-2-one ($$\text{CH}_3-\text{CH(CH}_3)-\text{CO}-\text{CH}_3$$) is treated with bromine ($$\text{Br}_2$$) in a non-polar solvent like chloroform ($$\text{CHCl}_3$$), it undergoes electrophilic $$\alpha$$-halogenation:

• The reaction proceeds via an enol intermediate. Under neutral or weakly acidic conditions, bromination occurs preferentially at the more substituted $$\alpha$$-carbon due to the formation of a more stable, highly substituted enol intermediate ($$\text{(CH}_3)_2\text{C}=\text{C(OH)-CH}_3$$).
• Thus, substitution takes place at the $$C-3$$ position, yielding a vicinal-like structural framework for the next step:
$$\text{Intermediate X} = \text{3-bromo-3-methylbutan-2-one } [(\text{CH}_3)_2\text{C(Br)}-\text{CO}-\text{CH}_3]$$

Step 2: Elimination and Rearrangement via Strong Base (Formation of Final Product Y)
When 3-bromo-3-methylbutan-2-one is treated with an excess of sodamide ($$\text{NaNH}_2$$), a remarkably strong base, it initiates a series of dehydrohalogenation and base-catalyzed elimination events:

• The strong amide ion ($$\text{NH}_2^-$$) abstracts a proton from the less hindered methyl group adjacent to the carbonyl ($$C-1$$), leading to intermediate enolization and subsequent loss of the bromide ion ($$\text{Br}^-$$) from the $$C-3$$ position.
• This successive loss of water and hydrogen bromide elements transforms the $\alpha$-halo-ketone structure into an alkyne network. Because an excess of the highly powerful base $$\text{NaNH}_2$$ is present, it drives the isomerization of any internal alkyne or allene intermediate toward the thermodynamically stable, terminal sodium acetylide salt by abstracting the acidic terminal acetylenic proton.
• Finally, adding water ($$\text{H}_2\text{O}$$) during the protonation workup protonates the terminal acetylide anion to yield the terminal alkyne: 3-methylbut-1-yne .

$$\text{Final Product Y}$$ is $$\text{CH}_3\text{CH(CH}_3)\text{C}\equiv\text{CH}$$

Conclusion:
The combination matches the structural formula written in Option A.

Answer:
The correct organic structural formula is given in Option A .