Organic Reactions

(a) Preparation of p-Nitroaniline

Reaction: Nitration of aniline using concentrated nitric acid and sulfuric acid.

Explanation: Aniline reacts with nitrating mixture to give para-nitroaniline as the major product due to the activating nature of the –NH group.

(b) Preparation of Chlorobenzene

Reaction: Chlorination of aniline in the presence of ferric chloride as a Lewis acid catalyst.

Explanation: Chlorine reacts with aniline to give substituted products. The reaction is usually controlled to obtain mono-chlorinated derivative such as chlorobenzene.

(c) Preparation of Phenol

Reaction: Oxidation of aniline with potassium dichromate in acidic medium.

Explanation: The –NH group of aniline is oxidized to a hydroxyl group, yielding phenol.

Conversion of Chlorobenzene to Biphenyl

To convert chlorobenzene to biphenyl, the Wurtz-Fittig reaction is employed.

In this reaction, two molecules of chlorobenzene undergo a coupling reaction in the presence of metallic sodium and dry ether. The sodium facilitates the removal of chlorine atoms and formation of a new carbon-carbon bond between the phenyl groups, resulting in the formation of biphenyl.

Reaction Equation:

Product: Biphenyl

(a) Conversion of Alcohol to Alkyl Chloride using Thionyl Chloride

In this reaction, a hydroxymethyl group ( ) reacts with thionyl chloride ( ) to form the corresponding chloro derivative.

The product is benzyl chloride, and the reaction proceeds via the conversion of the hydroxyl group into a good leaving group, followed by substitution with chlorine.

Product structure:

Quick Tip:
Thionyl chloride ( ) is widely used to convert alcohols into alkyl chlorides with high efficiency.

(b) Anti-Markovnikov Addition of HBr to Styrene

In this reaction, styrene ( ) reacts with in the presence of a peroxide.

Due to the peroxide effect (Kharasch effect), the reaction follows a free radical mechanism, giving the anti-Markovnikov product. This means that the bromine atom attaches to the less substituted carbon of the double bond.

Major Product: 1-bromo-2-phenylethane

This reaction is a classic example of free radical addition to alkenes influenced by the presence of peroxides.

Gabriel Phthalimide synthesis involves the reaction of phthalimide with an alkyl halide to form an amine. This method works well for aliphatic amines but fails for aromatic primary amines because the reaction does not readily occur with aryl halides. Aromatic rings are generally less reactive in nucleophilic substitution reactions due to the resonance stabilization of the ring, which makes it harder for the nucleophile (amide ion) to attack. Hence, Gabriel€™s synthesis does not work efficiently for preparing aromatic primary amines.

(a) Reaction Sequence: 1. The reaction of with results in the formation of ethyl cyanide (A) through nucleophilic substitution, where the chlorine is replaced by the cyanide ion. 2. The hydrogenation of ethyl cyanide ( ) in the presence of and a nickel catalyst leads to the reduction of the nitrile group to an amine, yielding ethylamine (B). (b) Reaction Sequence: 1. The reaction of aniline ( ) with sodium nitrite ( ) and hydrochloric acid ( ) forms a diazonium salt, (A). 2. The treatment of with an acid (e.g., ) causes the diazonium ion to undergo hydrolysis, leading to the formation of phenol (B).

To determine the best reagent for converting propanamide into propanamine, we need to understand the reduction process involved.

Propanamide (CH₃CH₂CONH₂) to Propanamine (CH₃CH₂CH₂NH₂):

This conversion involves reducing the carbonyl group (C=O) of the amide to an amine group (-CH₂NH₂). A common method for achieving this is using reducing agents.

1. Excess H₂: This is used for catalytic hydrogenation, which typically requires a catalyst such as Pd or Pt. While it can reduce double bonds, it is not selective for amides.
2. Br₂ in aqueous NaOH: Used in the Hofmann Bromamide degradation to convert primary amides to primary amines with one fewer carbon atom, not suitable for this conversion.
3. LiAlH₄ in ether: A strong reducing agent, typically used for the reduction of carboxylic acids, esters, and amides to alcohols or amines. While effective, the specific choice here is different.
4. Iodine in the presence of red phosphorus: This reagent is used in the reduction of amides to amines without reducing other functional groups, ideal for this transformation.

Given the options, iodine in the presence of red phosphorus is the best choice for converting propanamide into propanamine. It specifically facilitates the conversion of the carbonyl group in amides to amine groups, maintaining the carbon chain structure intact.

Step 1: Understanding the Reaction

Propyl magnesium bromide is a Grignard reagent that reacts with carbon dioxide (CO₂) to form a carboxylate anion. This anion is then hydrolyzed to form a carboxylic acid. The reaction can be summarized as follows:

RMgBr + CO₂ → RCO⁻ + MgBr⁺

Upon hydrolysis, the carboxylate anion is converted into a carboxylic acid:

RCO⁻ + H₂O → RCOOH

Step 2: Conclusion

Thus, the acid formed is propionic acid (C₃H₇COOH), corresponding to option (A).

Propionic acid is the correct product of this Grignard reaction with CO₂.

When D-glucose reacts with hydroxylamine ( ), the product obtained is D-glucose oxime. The reaction involves the formation of an oxime by the condensation of the aldehyde group (at C-1) of D-glucose with hydroxylamine, eliminating a molecule of water ( ).

Reaction:

Structure of D-Glucose Oxime:
The oxime forms at the C-1 position of D-glucose, replacing the aldehyde group ( ) with the oxime group ( ). The rest of the glucose structure remains unchanged.

Key Points:

• The reaction is specific to the open-chain form of D-glucose, which contains an aldehyde group.
• In aqueous solutions, D-glucose exists predominantly in the cyclic hemiacetal form, but a small equilibrium amount of the open-chain form allows this reaction to proceed.
• Oximes are stable compounds and can be used for characterization or protection of carbonyl groups in organic synthesis.

Final Answer:
The product is D-glucose oxime ( ), formed at the C-1 position.

To solve the problem, we need to determine the product formed when a secondary alcohol is oxidized with chromic anhydride (CrO ).

1. Understanding Secondary Alcohols:
A secondary alcohol has the general structure R R CHOH, where R and R are alkyl groups, meaning the carbon bearing the -OH group is attached to two other carbons.

2. Identifying the Reagent:
Chromic anhydride (CrO ) is a strong oxidizing agent, often used in acidic conditions (e.g., with H SO ) to form chromic acid, a common reagent for alcohol oxidation. It can also be used in other forms, such as in the Jones oxidation (CrO in aqueous sulfuric acid and acetone).

3. Oxidation of Secondary Alcohol:
When a secondary alcohol is oxidized with a strong oxidizing agent like CrO , the -OH group is converted to a carbonyl group (C=O). Since the carbon is attached to two other carbons, the product is a ketone (R R C=O). For example, if the secondary alcohol is 2-propanol (CH CH(OH)CH ), oxidation with CrO yields acetone (CH COCH ), a ketone.

4. Considering Reaction Conditions:
CrO in acidic conditions ensures complete oxidation. Secondary alcohols do not oxidize further to carboxylic acids (unlike primary alcohols), as there is no hydrogen on the carbonyl carbon to allow further oxidation to a carboxyl group.

Final Answer:
Oxidation of a secondary alcohol with chromic anhydride (CrO ) forms a ketone.

Conversion of Chlorobenzene to Biphenyl:

The conversion of chlorobenzene (C₆H₅Cl) to biphenyl (C₆H₅-C₆H₅) involves a reaction known as the **Wurtz-Fittig Reaction**. This reaction is a type of cross-coupling reaction where two aryl halides are coupled in the presence of sodium metal.

Reaction:

2 C₆H₅Cl + Na → C₆H₅-C₆H₅ + 2 NaCl

Mechanism:
1. The reaction is carried out in dry ether and requires heat. Sodium metal donates electrons to the chlorobenzene, generating a phenyl radical (C₆H₅•) and sodium chloride (NaCl). 2. The phenyl radicals couple together to form biphenyl (C₆H₅-C₆H₅).

Conditions:
- Sodium metal (Na) is used as a reducing agent.
- The reaction takes place in the presence of anhydrous ether to prevent moisture from interfering with the reaction.

Conclusion:
By heating chlorobenzene with sodium metal in ether, chlorobenzene undergoes a Wurtz-Fittig reaction to form biphenyl.