Organic Chemistry

Alkoxide attacks primary alkyl halide by SN2.

Basicity increases with more alkyl groups due to the +I effect in gaseous phase.

Aniline forms diazonium → azo dye.

In SN₁ reactions, the rate-determining step is the departure of the leaving group. In the case of aromatic compounds, a resonance-stabilized carbocation can form if the leaving group is attached to the ring, which makes the reaction faster. Hence, the phenyl bromide (C₆H₅Br) undergoes the reaction most rapidly.

When acetic acid reacts with PCl₅, it forms acyl chloride (CH₃COCl) by replacing the –OH group with a chlorine atom. This reaction is a typical example of chlorination of carboxylic acids.

(a) Phenol + CHCl₃ + NaOH → ortho-hydroxybenzaldehyde
(b) Phenol + CO₂ + NaOH → salicylic acid

The reaction of pivaldehyde (tert-butyraldehyde) [(CH₃)₃C–CHO] with conc. NaOHundergoes the Cannizzaro reaction, where one molecule is reduced to alcohol, and the other is oxidized to a carboxylate ion.

(i) The oxidation of the hydroboration product with hydrogen peroxide (H₂O₂) and hydroxide (OH⁻) leads to the formation of a diol. The main product will be (CH₃–CH₂–CH₂–CH₂OH), 1-propanol.

(ii) Phenol (C₆H₅OH) reacts with aqueous NaOH to form phenoxide ion (C₆H₅O⁻). When phenol reacts with CO₂ in the presence of H⁺, it undergoes carboxylation to form salicylic acid (2-hydroxybenzoic acid).

Main product: Salicylic acid (C₆H₄(OH)COOH)

A colligative property depends only on the number of solute particles and not on their nature. Osmotic pressure is the preferred colligative property for molar mass determination of macromolecules.
Step 1: Definition of Colligative Property Colligative properties are solution properties that depend only on the concentration of solute particles and not on their identity.
Examples of Colligative Properties: - Relative lowering of vapour pressure - Boiling point elevation - Freezing point depression - Osmotic pressure

Step 2: Why Osmotic Pressure is Preferred for Macromolecules? - Osmotic pressure ( ) is highly sensitive to small concentrations, making it ideal for determining the molar mass of macromolecules like proteins and polymers. - Unlike boiling point elevation or freezing point depression, osmotic pressure is measurable at room temperature, preventing thermal degradation of macromolecules.

(a) A peptide linkage is a covalent bond that joins two amino acids through the carboxyl group ( ) of one amino acid and the amino group ( ) of another, forming a bond. This reaction results in the elimination of water ( ) and is known as a peptide bond.

(b) The DNA double helix is held together primarily by hydrogen bonding between the complementary nitrogenous bases. The adenine (A) base forms two hydrogen bonds with thymine (T), and guanine (G) forms three hydrogen bonds with cytosine (C). These hydrogen bonds are essential for the stability of the double helix structure of DNA.

(c) The correct answer is starch. Starch is a polysaccharide composed of long chains of glucose molecules. It serves as a storage carbohydrate in plants. Sucrose and fructose are monosaccharides or disaccharides, and glucose is a monosaccharide.

(d) - Water-soluble vitamins: Vitamin B and Vitamin C. - Fat-soluble vitamins: Vitamin A, D, E, K. These vitamins are classified based on their solubility, with water-soluble vitamins dissolving in water and fat-soluble vitamins dissolving in fats and oils.

The isomer of C₄H₉Cl with the lowest boiling point is tert-butyl chloride (C₄H₉Cl), also known as 2-chloro-2-methylpropane (CH₃)₃C-Cl.

This is because the high degree of branching in the tert-butyl chloride molecule reduces the surface area available for intermolecular interactions, such as Van der Waals forces. With fewer opportunities for these interactions, the molecule is easier to vaporize, leading to a lower boiling point compared to less branched isomers, such as n-butyl chloride or sec-butyl chloride.

An allylic halide is a compound in which the halogen is attached to a carbon that is adjacent to a double bond.

From the given options:

• Option (i): is not an allylic halide because the halogen (Br) is not adjacent to the double bond.
• Option (ii): is an allylic halide, as the halogen (Br) is attached to the carbon that is adjacent to the double bond.

Thus, the correct allylic halide is option (ii).

(a) (i) The major product of the reaction between ethene and HBr is 1-bromoethane (CH₃CH₂Br).

(ii) The product of the reaction between benzene and Br₂ in UV light is bromobenzene (C₆H₅Br).

(b) (i) Grignard reagents are highly reactive and react with water to form hydrocarbons, so they should be prepared under anhydrous conditions to prevent hydrolysis.

(ii) Aqueous KOH leads to nucleophilic substitution, forming alcohols, while alcoholic KOH leads to elimination reactions, producing alkenes.

Due to salt formation with AlCl , aniline becomes a Lewis base, deactivating the benzene ring.
Step 1: Role of Lewis Acid Catalyst in Friedel-Crafts Reaction - Friedel-Crafts alkylation and acylation require a Lewis acid catalyst like AlCl to generate the electrophile.

Step 2: Interaction Between Aniline and AlCl - Aniline ( ) has a lone pair on nitrogen, which interacts with AlCl , forming a salt. - This deactivates the benzene ring, making it less reactive toward electrophilic substitution.

Step 3: Conclusion Due to this salt formation, aniline does not undergo Friedel-Crafts reaction.

N,N-diethylbenzenesulphonamide does not contain any hydrogen atom attached to nitrogen, making it non-acidic and thus insoluble in alkali.

Step 1: Understanding the Nature of N,N-Diethylbenzenesulphonamide - In order for a compound to dissolve in alkali, it must have an acidic hydrogen atom that can react with OH .

Step 2: Why It Is Insoluble - Sulphonamides that contain a hydrogen attached to nitrogen are acidic and dissolve in alkali. - However, in N,N-diethylbenzenesulphonamide, both hydrogen atoms on nitrogen are replaced by ethyl groups, making it non-acidic and hence, insoluble in alkali.

1. Identify A and B in the reaction sequence:

(a) Reaction:
CH₃CH₂Cl + NaCN → A → H₂/Ni → B

Solution:
In the first step, NaCN reacts with CH₃CH₂Cl (ethyl chloride) in a nucleophilic substitution reaction to form a nitrile (A). The reaction proceeds as follows:

CH₃CH₂Cl + NaCN → CH₃CH₂CN (ethyl cyanide, A)

In the second step, the nitrile group is hydrogenated using H₂ in the presence of a nickel catalyst to form an amine (B). The reaction proceeds as:

CH₃CH₂CN + H₂/Ni → CH₃CH₂NH₂ (ethylamine, B)

Identified compounds:
A = CH₃CH₂CN (ethyl cyanide), B = CH₃CH₂NH₂ (ethylamine)

(b) Reaction:
C₆H₅NH₂ + NaNO₂/HCl (0–5°C) → A → C₆H₅NH₂ + H⁺ → B

Solution:
In the first step, aniline (C₆H₅NH₂) reacts with sodium nitrite (NaNO₂) in the presence of hydrochloric acid (HCl) at 0-5°C to form a diazonium salt (A). The reaction proceeds as follows:

C₆H₅NH₂ + NaNO₂/HCl → C₆H₅N₂⁺Cl⁻ (benzenediazonium chloride, A)

In the second step, the diazonium salt is reduced by adding a reducing agent like H⁺ to form aniline (C₆H₅NH₂) again, which is compound B. The reaction proceeds as:

C₆H₅N₂⁺Cl⁻ + H⁺ → C₆H₅NH₂ (aniline, B)

Identified compounds:
A = C₆H₅N₂⁺Cl⁻ (benzenediazonium chloride), B = C₆H₅NH₂ (aniline)

Chlorobenzene is resistant to nucleophilic substitution reactions due to the following reasons:

1. Resonance Stabilization:
The lone pair of electrons on the chlorine atom participates in resonance with the π-electrons of the benzene ring, leading to delocalization of the electron density. This resonance stabilization strengthens the C-Cl bond and makes it less susceptible to nucleophilic attack.

2. Partial Double Bond Character:
Due to resonance, the C-Cl bond in chlorobenzene acquires partial double bond character, which increases its bond strength and reduces its reactivity towards nucleophiles.

3. sp² Hybridization of the Carbon Atom:
The carbon atom bonded to chlorine in chlorobenzene is sp² hybridized (part of the aromatic ring), which makes it more electronegative and less accessible for nucleophilic substitution compared to sp³ hybridized carbons in alkyl halides.

4. High Electron Density on the Ring:
The benzene ring has a high electron density, which repels nucleophiles and further hinders the substitution reaction.

5. Stability of the Aromatic System:
Nucleophilic substitution would disrupt the aromaticity of the benzene ring, which is energetically unfavorable. The aromatic system is highly stable, and reactions that compromise this stability are less likely to occur.

Conclusion:
The combined effects of resonance stabilization, partial double bond character, sp² hybridization, high electron density, and aromatic stability make chlorobenzene resistant to nucleophilic substitution reactions.

To solve the problem, we need to identify the reaction that converts phenol to salicylic acid.

1. Understanding the Conversion:
Phenol has the formula C H OH, and salicylic acid is C H (OH)(COOH), a benzene ring with both a hydroxyl (-OH) and a carboxyl (-COOH) group. The conversion involves introducing a carboxyl group to the phenol ring, specifically at the ortho position relative to the hydroxyl group.

2. Evaluating Option C (Reimer-Tiemann Reaction):
The Reimer-Tiemann reaction involves treating phenol with chloroform (CHCl ) and a strong base (e.g., KOH) to introduce an aldehyde group (-CHO) ortho to the hydroxyl group, forming salicylaldehyde (C H (OH)(CHO)). This does not directly yield salicylic acid, as the product is an aldehyde, not a carboxylic acid. Further oxidation would be needed, which is not part of the standard reaction. Thus, this option is incorrect.

3. Evaluating Option B (Friedel-Crafts Reaction):
The Friedel-Crafts reaction typically involves alkylation or acylation of an aromatic ring using a halogenated compound and a Lewis acid catalyst (e.g., AlCl ). For phenol, the -OH group is strongly activating and ortho-para directing, but Friedel-Crafts acylation (e.g., with CO and HCl) to introduce a carboxyl group directly is not a standard method for salicylic acid synthesis. Moreover, the -OH group can react with the catalyst, complicating the reaction. This option is incorrect.

4. Evaluating Option A (Kolbe Reaction):
The Kolbe-Schmitt reaction (often referred to as the Kolbe reaction in this context) involves heating sodium phenoxide (C H ONa, formed from phenol and NaOH) with carbon dioxide (CO ) under pressure at around 125°C, followed by acidification. This introduces a carboxyl group ortho to the phenoxide group, forming salicylic acid:


This is the standard industrial method for synthesizing salicylic acid from phenol, making this option correct.

5. Evaluating Option D (Coupling Reaction):
A coupling reaction typically refers to reactions like diazo coupling, where a diazonium salt reacts with an aromatic compound (e.g., phenol) to form azo dyes. This does not involve the formation of a carboxyl group or salicylic acid. Thus, this option is incorrect.

Final Answer:
The conversion of phenol to salicylic acid can be accomplished by the Kolbe reaction (option A).

Step 1: Understanding nucleophilic bimolecular substitution (SN2). In an SN2 reaction, the nucleophile attacks the carbon opposite to the leaving group, leading to the inversion of configuration.

Step 2: Conclusion. Thus, SN2 reactions involve inversion of configuration, corresponding to option (C).