Polymers

Step 1: Understand the Ziegler-Natta Catalyst:
The Ziegler-Natta catalyst is a compound used in the polymerization process, primarily in the manufacture of high-density polyethylene (HDPE) and low-density polyethylene (LDPE). It facilitates the polymerization of alkenes and can also be used for producing other types of plastics.

Step 2: Identify the Correct Product of Ziegler-Natta Catalyst:
The Ziegler-Natta catalyst is particularly used in the polymerization of ethylene to produce polyethylene, which can be either high-density or low-density polyethylene depending on the polymerization conditions.

Step 3: Analyze the Options:

• Polyacrylonitrile: This is a synthetic polymer, but it is not produced by the Ziegler-Natta catalyst. It is produced through a different process, so this option is incorrect.
• Terylene: Terylene (also known as polyester) is a synthetic polymer made from terephthalic acid and ethylene glycol. It is not made using the Ziegler-Natta catalyst, so this option is incorrect.
• High density polyethylene (HDPE): This is produced using the Ziegler-Natta catalyst, making this the correct option.
• Low density polyethylene (LDPE): This is also produced using the Ziegler-Natta catalyst under different conditions, so this option is incorrect.

Final Answer:
The correct product made using the Ziegler-Natta catalyst is High density polyethylene (HDPE).

Step 1: Understand the Monomers of Bakelite and Melamine - Bakelite and Melamine are both thermosetting polymers, and both are made by the polymerization of formaldehyde (methanal). - Bakelite is formed by the reaction of phenol with formaldehyde, while Melamine is made by polymerizing melamine (a nitrogen-rich compound) with formaldehyde.

Step 2: Identify the Common Monomer - The common monomer for both Bakelite and Melamine polymers is formaldehyde (methanal), which reacts with different monomers (phenol for Bakelite, melamine for Melamine) to form the respective polymers. Hence, the correct answer is methanal.

The reaction sequence starts with an aldehyde group, which reacts with cyanide to form a cyanohydrin intermediate. Upon hydrolysis, the cyano group is converted into a carboxyl group, resulting in hexahydroxyheptanoic acid as the product Y.

Step 1: Identifying the Reaction Pathway
1. The given transformation involves polymerization reactions:
- The first step converts a polymer used in making safety helmets into ethylene glycol ( ).
- The second step converts ethylene glycol into a polymer used in paint manufacturing.
2. Identifying the polymers:
- The polymer used in safety helmets is polycarbonate or poly(terephthalic acid-co-bisphenol A).
- The polymer used in paint manufacturing is polyester, commonly synthesized from terephthalic acid and ethylene glycol.
Step 2: Determining X and Y
- Y (Reactant for first step): Terephthalic acid derivative leading to ethylene glycol formation.
- X (Reactant for second step): Terephthalic acid, which reacts with ethylene glycol to form polyester.
Step 3: Evaluating the Given Options
- Option (1): Correct, as it correctly identifies the structure of terephthalic acid derivatives leading to polymer formation.
- Option (2): Incorrect, as it swaps the structures incorrectly.
- Option (3): Incorrect, as it does not correctly represent the intermediate.
- Option (4): Incorrect, as it provides incorrect reactants.
Thus, the correct answer is

Option (1).

Step 1: Understanding the Reaction
- The given alkene X ( ) reacts with HBr to form Y ( ).
- The formed Y undergoes hydrolysis via the mechanism.

Step 2: Identifying the Mechanism
- The mechanism involves the formation of a stable carbocation intermediate.
- Tertiary carbocations are more stable than secondary and primary carbocations.

Step 3: Analyzing the Given Alkenes
- Option (1): Forms a primary carbocation upon HBr addition (Least stable)
- Option (2): Forms a secondary carbocation (Less stable)
- Option (3): Forms a primary carbocation, unfavorable for
- Option (4): Forms a tertiary carbocation, which is the most stable

Step 4: Conclusion
- Since the most stable tertiary carbocation is formed in option (4), this is the correct alkene .