Problem 120 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following is correct? (a) Reduction of any aldehyde gives secondary alcohols (b) Reduction of vegetable oil with \(\mathrm{H}_{2} \mathrm{SO}_{4}\) gives glycerine (c) Reaction of ethanolic iodine with \(\mathrm{NaOH}\) gives iodoform (d) Sucrose on reaction with \(\mathrm{NaOH}\) gives invert sugar

Step-by-Step Solution

Answer

Option (c) is correct: Reaction of ethanolic iodine with \( \mathrm{NaOH} \) gives iodoform.

Step 1: Analyze Option (a)

Option (a) states that the reduction of any aldehyde gives secondary alcohols. This is incorrect because the reduction of an aldehyde typically yields a primary alcohol. When an aldehyde is reduced, a hydrogen atom is added to the carbon-oxygen double bond (carbonyl group), resulting in a primary alcohol.

Step 2: Analyze Option (b)

Option (b) suggests that the reduction of vegetable oil with \( \mathrm{H}_{2} \mathrm{SO}_{4} \) gives glycerine. This statement is inaccurate. Sulfuric acid (\( \mathrm{H}_{2} \mathrm{SO}_{4} \)) is not used for the reduction of vegetable oils to glycerine. Instead, reduction typically involves hydrogenation processes or other agents, not \( \mathrm{H}_{2} \mathrm{SO}_{4} \).

Step 3: Analyze Option (c)

Option (c) involves the reaction of ethanolic iodine with \( \mathrm{NaOH} \) to give iodoform. This statement is correct. The iodoform test involves the formation of iodoform (\( \mathrm{CHI}_3 \)) when iodine and a base like \( \mathrm{NaOH} \) react with a methyl ketone compound or ethanol.

Step 4: Analyze Option (d)

Option (d) claims that sucrose on reaction with \( \mathrm{NaOH} \) gives invert sugar. This is incorrect. Invert sugar is typically produced by the hydrolysis of sucrose into glucose and fructose using an acid or enzymatic catalyst, not \( \mathrm{NaOH} \).

Key Concepts

Aldehydes

Aldehydes are a fundamental class of organic compounds characterized by the presence of a carbonyl group (C=O) bonded to a hydrogen atom and an R group, which can be a hydrocarbon chain or an aromatic group. The general formula for an aldehyde is RCHO, where R is the hydrocarbon portion. Since aldehydes contain this carbonyl group at the end of the carbon chain, they are prone to specific types of reactions.
Aldehydes are quite reactive due to their polar carbonyl group. This polarity makes them susceptible to nucleophilic addition reactions, which is the most common type of reaction they undergo. In these reactions, a nucleophile, which is a molecule or ion with a pair of electrons to donate, attacks the electrophilic carbonyl carbon, leading to the addition of new groups to the molecule.

• Aldehydes are commonly found in nature and serve important roles as flavors and fragrances.
• They are often used in the production of perfumes, as preservatives, and as intermediates in the synthesis of various chemical products.
• Their simplest member, formaldehyde, is widely used in the polymer industry for the production of plastics and other materials.

Reduction Reactions

Reduction reactions in organic chemistry involve the gain of hydrogen atoms or the loss of oxygen atoms. These reactions are essential in converting carbonyl compounds like aldehydes and ketones into alcohols. Specifically, when aldehydes are reduced, they yield primary alcohols.
The reduction of aldehydes to alcohols typically involves the use of reducing agents. One common reducing agent is sodium borohydride (NaBH₄), which selectively reduces aldehydes and ketones to alcohols without affecting other functional groups. Another highly effective reducing agent is lithium aluminum hydride (LiAlH₄), which can reduce a broad range of functional groups but requires more stringent conditions.

• The process involves adding hydrogen to the carbonyl carbon and oxygen, converting the double bond into a single bond, and rendering the compound into an alcohol.
• The hydrogen is typically sourced from a reducing agent rather than free hydrogen molecules.
• Reduction is a crucial step in various organic synthesis processes, enabling the transformation of more oxidized compounds into more reduced forms.

Iodoform Test

The iodoform test is a chemical reaction used to identify the presence of methyl ketones and some secondary alcohols. When a compound containing a methyl group adjacent to a carbonyl group is treated with iodine and a base such as sodium hydroxide (NaOH), yellow crystals of iodoform ( CHI₃ ) are formed.
The test is specific because very few types of compounds give a positive result. Along with methyl ketones, some secondary alcohols, which can be oxidized to give a methyl ketone, and ethanol also do. This reaction involves the halogenation of the methyl group followed by the cleavage of the carbon-carbon bond, resulting in the formation of iodoform.

• The characteristic yellow precipitate of iodoform signifies a positive test.
• The iodoform test is a handy tool in organic chemistry for identifying certain functional groups in a compound.
• Due to its specificity, it is frequently used in educational laboratories to illustrate practical aspects of organic functional group identification.