Question

p-cresol \(\stackrel{\mathrm{CHCl}_{3} \mathrm{NaOH}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{HCN}}{\longrightarrow} \mathrm{B} \stackrel{\mathrm{H}, \mathrm{O}^{+}}{\longrightarrow} \mathrm{C}\) Which of the following statements is not true about the product (C) in the above sequence of reactions? (a) It gives violet colour with neutral ferric chloride solution. (b) It is optically active. (c) There are six \(\mathrm{sp}^{2}\) hybridised carbon atoms in one molecule of it. (d) It liberates \(\mathrm{CO}_{2}\) when treated with \(\mathrm{NaHCO}_{3}\) solution.

Short answer

Answer: (b) It is optically active.

Step-by-step solution

Reaction of p-cresol with CHCl3 and NaOH to form Product A

This step is a haloform reaction. p-cresol reacts with CHCl3 and NaOH to give a carboxylic acid as Product A. The carboxyl group will be present in the ortho position to the hydroxyl group. The reaction is as follows: p-cresol + CHCl3 + 3 NaOH → Ortho-chloro-para-hydroxybenzoic acid + CHCl3

Reaction of Product A with HCN to form Product B

In this step, Product A reacts with HCN, which is a nucleophilic addition reaction. The nitrile group replaces the chlorine atom in the ortho position to the hydroxyl group, forming Product B. The reaction is as follows: Ortho-chloro-para-hydroxybenzoic acid + HCN → Ortho-cyano-para-hydroxybenzoic acid

Reaction of Product B with H and O+ to form Product C

In this step, Product B undergoes hydrolysis in acidic conditions to form the final product C. The nitrile group is hydrolyzed to form a carboxylic acid group. The reaction is as follows: Ortho-cyano-para-hydroxybenzoic acid + H (acids) → Ortho-carboxy-para-hydroxybenzoic acid Now that we have the structure of Product C, we can verify the given statements. (a) It gives a violet color with neutral ferric chloride solution: Product C is an ortho-hydroxybenzoic acid, and these types of compounds form coordinate complexes with FeCl3, resulting in a violet color. So, this statement is true. (b) It is optically active: There is no chiral center in the molecule of Product C. Therefore, this statement is false. (c) There are six sp2 hybridized carbon atoms in one molecule of it: In the final product C, there are six carbon atoms in the benzene ring which are sp2 hybridized. So, this statement is true. (d) It liberates CO2 when treated with NaHCO3 solution: Ortho-carboxy-para-hydroxybenzoic acid will liberate CO2 when treated with NaHCO3 solution, as it forms water and sodium salt of the carboxylic acid. So, this statement is true. Hence, the statement (b) "It is optically active" is not true about the product C in the above sequence of reactions.

Key concepts

Haloform Reaction

The haloform reaction is a transformation that holds a special place in organic chemistry due to its ability to create important building blocks for various compounds.

In simple terms, this reaction involves the conversion of a methyl ketone into a carboxylic acid halide by the action of a halogen in the presence of a base. It's a beautifully orchestrated, multi-step process where initially, the halogen atom is added to the methyl group adjacent to the carbonyl group. After multiple halogenation steps, the trihalomethyl group becomes a good leaving group, which then leaves with the help of the base, forming a carboxylate ion and a haloform (CHX3), where 'X' is a halogen.

Typical reagents used in a lab setting for the haloform reaction would include chloroform (CHCl3) and sodium hydroxide (NaOH), just as in the textbook exercise concerning p-cresol.

Nucleophilic Addition Reaction

Nucleophilic addition reactions are ubiquitous in the world of synthetic organic chemistry. These reactions occur when a nucleophile, an atom or molecule with a free pair of electrons, adds to an electrophilic, or electron-poor, carbon. This type of reaction is essential for the formation of many functional groups, including alcohols, amines, and acids.

In the context of the provided exercise, the nucleophilic addition reaction comes into play when hydrocyanic acid (HCN) attacks the carbon atom of the carbonyl group in the ortho-chloro-para-hydroxybenzoic acid. This results in the substitution of the chlorine atom and forms a product that incorporates a nitrile group (-CN), further transforming into a carboxylic acid upon hydrolysis.

Optical Activity

Optical activity is a fascinating property of certain molecules that have the ability to rotate the plane of polarized light as it passes through them. This rotational phenomenon is a hallmark of molecular asymmetry, usually due to the presence of chiral centers in the compound. A molecule is considered chiral if it is not superimposable on its mirror image, much like left and right hands.

It’s of critical importance for students to understand that not all molecules exhibit this property – it’s exclusive to those with chiral centers that lack a plane of symmetry. As brought out in the problem explanation, Product C does not possess any chiral centers, disqualifying it from being optically active despite its complex structure.

sp2 Hybridization

Understanding the concept of hybridization is key to deciphering the shape and reactivity of molecules in organic chemistry. sp2 hybridization occurs when one s-orbital and two p-orbitals from an atom mix to form three equivalent hybrid orbitals. These hybrid orbitals have a trigonal planar arrangement, with 120-degree angles between them, which directly impacts a molecule's geometry and reactivity.

In aromatic compounds like benzene, the carbon atoms are sp2 hybridized, giving the molecule a flat, hexagonal shape. Each carbon atom forms three sigma bonds, two with neighboring carbon atoms and one with a hydrogen atom, while the remaining p-orbital is used to form the delocalized π-bonds that give aromatic compounds their unique stability and reactivity. For the molecule in our exercise, six carbons within the benzene ring exhibit sp2 hybridization, which contributes to the molecule's stability and its ability to undergo certain chemical reactions.