Organic Chemistry

Step 1: Understanding the Problem.
The question asks for the correct organic structure for a compound undergoing an S 2 reaction, which is a bimolecular nucleophilic substitution. In such reactions, the nucleophile attacks the electrophilic carbon while the leaving group departs.
Step 2: Analyzing the Options.
- (A) 1-Bromo-2-chloro-3-butene is the correct structure that fits the S 2 reaction conditions as it provides the right positioning of substituents for substitution at the electrophilic carbon.
- (B), (C), and (D) are incorrect because they involve improper positions for the leaving groups, which would make the S 2 mechanism less favorable.
Step 3: Conclusion.
The correct answer is (A) 1-Bromo-2-chloro-3-butene as it follows the proper structural formation for an S 2 reaction.

Step 1: Understanding the Reaction.
The question asks for the structure of a compound undergoing electrophilic addition with halogens. Electrophilic addition involves an electron-deficient species (like a halogen) attacking a double bond, which opens up the bond and adds the halogen across it.
Step 2: Analyzing the Options.
- (A) (benzene) does not undergo electrophilic addition with halogens under normal conditions because it requires special conditions like halogenation using a catalyst.
- (B) is a simple alkyl halide, but it does not undergo electrophilic addition.
- (C) is a chlorinated benzene, which undergoes substitution reactions, not electrophilic addition.
- (D) represents a chloroalkane that can undergo electrophilic addition, fitting the conditions.
Step 3: Conclusion.
The correct answer is (D) as it fits the conditions for an electrophilic addition reaction with halogens.

Step 1: Understand the meaning of allylic chloride.
An allylic chloride is a compound in which the chlorine atom is attached to an carbon atom that is adjacent to a carbon-carbon double bond. In simple words, chlorine should be on the carbon next to the double bond, not on the double-bonded carbon itself.
Step 2: Differentiate allylic chloride from vinylic chloride.
If chlorine is attached directly to a carbon atom involved in the double bond, then the compound is called a vinylic chloride, not an allylic chloride. This is the key point needed to solve the question.
Step 3: Analyse the given options.

• (A) : Chlorine is on the carbon adjacent to the double bond, so it is an allylic chloride.
  
• (B) : Again, chlorine is on the carbon next to the double bond, so it is allylic chloride.
  
• (C) In this structure, chlorine is directly attached to a double-bond carbon of the ring. Therefore, it is a vinylic chloride, not an allylic chloride.
  
• (D) Here chlorine is attached to the carbon adjacent to the double bond in the ring, so it is allylic chloride.
  

Step 4: Conclusion.
Hence, the compound that is not an allylic chloride is option , because it is a vinylic chloride.
Final Answer:Option (C).

Step 1: Understand the Wacker process.
The Wacker process is an industrial method used for the oxidation of ethene to acetaldehyde. It involves the use of a palladium-based catalyst in aqueous medium.
Step 2: Identify the catalyst.
The main catalyst used in this process is palladium(II) chloride, , along with copper chloride as a co-catalyst.
Step 3: Analyze options.

• (A) : Not used in Wacker process.
• (B) : Not used.
• (C) : Not used.
• (D) : Correct catalyst.

Step 4: Conclusion.
Hence, the catalyst used in the Wacker process is .
Final Answer: .

Step 1: Recall the carbylamine reaction.
The carbylamine reaction is shown by primary amines only. In this reaction, a primary amine reacts with chloroform and alcoholic potassium hydroxide to form an isocyanide or carbylamine. The general reaction is:
Step 2: Identify the given amine.
4-methylaniline is a primary aromatic amine. Its structure is . Since it is a primary amine, it gives the carbylamine test.
Step 3: Determine the product formed.
In the carbylamine reaction, the amino group is converted into the isocyanide group . Therefore, 4-methylaniline forms 4-methylphenyl isocyanide, whose formula is:
This corresponds to option (B).
Step 4: Compare with the other options.

• (A) : This is a nitrile, not the product of carbylamine reaction.
• (B) : Correct. This is the isocyanide formed in the carbylamine reaction.
• (C) : This is a nitro compound, not related to the carbylamine reaction.
• (D) : This is a diazonium salt, formed in diazotisation, not in carbylamine reaction.

Step 5: Conclusion.
Hence, the product formed in the carbylamine reaction of 4-methylaniline is .
Final Answer: .

Step 1: Understanding basic strength of amines.
Amines behave as bases because the nitrogen atom contains a lone pair of electrons, which can accept a proton. Greater the availability of this lone pair, greater will be the basic strength of the amine. Alkyl groups generally increase electron density on nitrogen due to the effect, so substituted amines are usually more basic than ammonia.
Step 2: Effect in aqueous solution.
In aqueous solution, basic strength does not depend only on the effect. It also depends on the stability of the protonated amine formed after accepting a proton. This stability is strongly affected by solvation. Secondary ammonium ions are well stabilised by both electron donation and effective solvation, so secondary amines are generally the strongest bases in water.
Step 3: Compare primary, secondary and tertiary amines.
For ethyl-substituted amines in aqueous medium, the general order is:
This happens because:

• Secondary amine :} It gets strong effect and its conjugate acid is sufficiently solvated, so it is the most basic.
  
• Primary amine : It is also more basic than ammonia, but less basic than secondary amine.
  
• Tertiary amine :} Though it has a strong effect, its conjugate acid is less effectively solvated because of steric hindrance, so its basic strength decreases in water.
  
• Ammonia : It has no alkyl groups, so it is the least basic among these.
  

Step 4: Conclusion.
Therefore, the correct order of basic strength in aqueous solution is:
Final Answer: .