GSEB Class 12 Chemistry Solutions Chapter 15 Polymers

Gujarat Board Textbook Solutions Class 12 Chemistry Chapter 15 Polymers

GSEB Class 12 Chemistry Polymers InText Questions And Answers

Question 1. What are polymers?
Answer: Polymers are large molecular weight compounds, typically ranging from \(10^3\) to \(10^7\) atomic mass units (u), created by linking together numerous smaller, simpler molecules. These substances are also known as macromolecules. Examples include polyethene, bakelite, polyvinyl chloride (PVC), and Teflon.
In simple words: Polymers are big molecules made from many small units joined together, like a long chain of beads.

🎯 Exam Tip: Focus on understanding the definition of a polymer and being able to provide a few common examples.

Question 2. How are polymers classified on the basis of structure?
Answer: Based on their structural arrangement, polymers are categorized into three main types:
(i) Linear polymers: These consist of straight chains, exemplified by polythene and polyvinyl chloride.
(ii) Branched-chain polymers: These polymers feature chains with side branches, such as low-density polythene.
(iii) Cross-linked polymers: These have a complex network structure formed by strong covalent bonds, seen in materials like bakelite and melamine.
In simple words: Polymers are grouped by how their parts connect: in straight lines, with branches, or in a strong 3D mesh.

🎯 Exam Tip: Remember the three structural classifications of polymers and be ready to provide one example for each type.

Question 3. Write the names of monomers of the following polymers:
(i) \( \[ -NH-(CH_2)_6-NH-CO-(CH_2)_4-CO- \]_n \)
(ii) \( \[ -CO-(CH_2)_5-NH- \]_n \)
(iii) \( \[ -CF_2-CF_2- \]_n \)
Answer: The constituent monomers for the given polymers are:
(i) The polymer represented by \( \[ -NH-(CH_2)_6-NH-CO-(CH_2)_4-CO- \]_n \) is formed from hexamethylene diamine and adipic acid.
(ii) The polymer \( \[ -CO-(CH_2)_5-NH- \]_n \) is derived from caprolactam.
(iii) The polymer \( \[ -CF_2-CF_2- \]_n \) is synthesized from tetrafluoroethene.
In simple words: Monomers are the small building blocks that link up to make a polymer.

🎯 Exam Tip: Practice identifying common monomers from their polymer structures, especially for frequently encountered polymers like Nylon-6,6, Nylon-6, and Teflon.

Question 4. Classify the following as addition and condensation polymers: Terylene, Bakelite, Polyvinyl chloride, Polythene.
Answer: The classification of the listed polymers into addition or condensation types is as follows:
(i) Terylene: This is a condensation polymer.
(ii) Bakelite: This is also a condensation polymer.
(iii) Polyvinyl chloride: This falls under addition polymers.
(iv) Polyethylene: This is categorized as an addition polymer.
In simple words: Addition polymers join without losing atoms; condensation polymers join and release small molecules like water.

🎯 Exam Tip: Understand the fundamental difference between addition and condensation polymerization mechanisms to correctly classify polymers.

Question 5. Explain the difference between Buna-N, and Buna-S.
Answer: The key distinction between Buna-N and Buna-S lies in their monomeric composition. Buna-N is a copolymer formed from the combination of 1,3-butadiene and acrylonitrile. In contrast, Buna-S is a copolymer synthesized from 1,3-butadiene and styrene.
In simple words: Buna-N and Buna-S are both synthetic rubbers, but they use different second monomers alongside 1,3-butadiene.

🎯 Exam Tip: Memorize the specific monomers for common synthetic rubbers like Buna-N and Buna-S, as this is a frequent point of comparison.

Question 6. Arrange the following polymers in increasing order of their intermolecular forces.
(i) Buna-S, Polyethene, Nylon-6,6.
(ii) Nylon-6, Neoprene, Polyvinylchloride.
Answer: To arrange the polymers based on increasing intermolecular forces:
(i) For Buna-S, Polyethene, and Nylon-6,6: The order is Buna-S < Polyethene < Nylon-6,6.
(ii) For Neoprene, Polyvinyl chloride, and Nylon-6: The order is Neoprene < Polyvinyl chloride < Nylon-6.
In simple words: Intermolecular forces determine a polymer's strength and properties; stronger forces mean a more rigid material.

🎯 Exam Tip: Understanding the types of intermolecular forces (van der Waals, dipole-dipole, hydrogen bonding) is crucial for predicting the relative strengths and properties of polymers.

Question 7. Is \( \[CH_2-CH(C_6H_5)\]_n \) a homopolymer or a copolymer?
Answer: The polymer represented by \( \[CH_2-CH(C_6H_5)\]_n \) is classified as a homopolymer. Its single monomer unit is styrene, with the chemical formula \( C_6H_5CH=CH_2 \).
In simple words: A homopolymer is made from only one type of building block, while a copolymer uses two or more different types.

🎯 Exam Tip: Distinguishing between homopolymers and copolymers requires identifying whether the repeating unit comes from one or multiple monomer species.

GSEB Class 12 Chemistry Polymers Text Book Questions And Answers

Question 1. Explain the terms polymer and monomer.
Answer:
(i) A polymer is a large molecule (macromolecule) characterized by its high molecular weight, composed of numerous identical or similar repeating structural units that originate from smaller molecules called monomers.
(ii) A monomer is a simple, reactive molecule that possesses the ability to chemically combine with other monomers through a process called polymerization, thereby forming a polymer.
In simple words: Polymers are long chains, and monomers are the individual links that make up those chains.

🎯 Exam Tip: Provide clear, concise definitions for both polymer and monomer, emphasizing their relationship and roles in polymerization.

Question 2. What are natural and synthetic polymers? Give two examples of each type.
Answer:
(i) Natural polymers are macromolecules with high molecular masses that occur organically within living organisms, specifically in plants and animals. Prominent examples include proteins and nucleic acids.
(ii) Synthetic polymers, conversely, are high molecular mass macromolecules produced artificially by humans. This category encompasses various synthetic plastics, fibers, and rubbers. Polythene and dacron are two illustrative examples of synthetic polymers.
In simple words: Natural polymers come from living things; synthetic polymers are made in labs by people.

🎯 Exam Tip: Be ready to classify polymers as natural or synthetic and provide specific examples for each category.

Question 3. Distinguish between the terms homopolymer and copolymer and give an example of each.
Answer: Here's a distinction between homopolymers and copolymers, with examples:
(i) A homopolymer is a polymer derived from a single type of monomer molecule. For instance, polyethene is a homopolymer formed from the polymerization of ethene: \[ nCH_2=CH_2 \xrightarrow{\text{Polymerisation}} \[CH_2-CH_2\]_n \]
(ii) A copolymer is a polymer synthesized from two or more distinct types of monomer molecules. A common example is Buna-S, which is a copolymer of 1,3-butadiene and styrene: \[ nCH_2=CH-CH=CH_2 + nC_6H_5CH=CH_2 \xrightarrow{\text{Na}} \[CH_2-CH=CH-CH_2-CH(C_6H_5)-CH_2\]_n \]
In simple words: Homopolymers are chains of identical units, while copolymers are chains built from different kinds of units alternating or arranged in blocks.

🎯 Exam Tip: Understand the structural difference and reaction types for homopolymerization and copolymerization, and be able to draw or name the monomers for common examples.

Question 4. How do you explain the functionality of a monomer?
Answer: The functionality of a monomer refers to the total number of reactive sites or bonding points available within that monomer molecule. These sites are where the monomer can form covalent bonds with other monomers during the polymerization process.
In simple words: A monomer's functionality is how many places it can connect with other monomers to build a polymer chain.

🎯 Exam Tip: Recognize that functionality dictates the type of polymer formed (e.g., linear vs. cross-linked) and is key to understanding polymer architecture.

Question 5. Define the term polymerisation.
Answer: Polymerization is a chemical process where multiple monomer molecules combine chemically to form a high molecular mass polymer. This involves the linking of repeating structural units through covalent bonds.
In simple words: Polymerization is the chemical reaction where many small molecules join to make one very large molecule.

🎯 Exam Tip: A clear definition of polymerization, highlighting the role of monomers and covalent bonds, is essential.

Question 6. Is \( \[NH-CHR-CO\]_n \), a homopolymer or copolymer?
Answer: The polymer with the repeating unit \( \[NH-CHR-CO\]_n \) is classified as a homopolymer. This is because its structure indicates that it is formed from a single type of monomer, specifically an amino acid, which provides the repeating unit upon polymerization.
In simple words: If a polymer's repeating pattern comes from only one kind of starting molecule, it's a homopolymer.

🎯 Exam Tip: The ability to visually identify repeating units from a polymer's chemical formula helps determine if it's a homopolymer or copolymer.

Question 7. In which classes, the polymers are classified on the basis of molecular forces?
Answer: Based on the types and strengths of intermolecular forces existing between their polymer chains, polymers are categorized into the following classes:
(i) Elastomers
(ii) Fibres
(iii) Thermoplastics
(iv) Thermosetting plastics
In simple words: Polymers are grouped by how strongly their chains attract each other, which affects their properties like flexibility or strength.

🎯 Exam Tip: Know the four main classifications based on intermolecular forces and be able to associate general properties with each type.

Question 8. How can you differentiate between addition and condensation polymerisation?
Answer: The primary distinction between addition and condensation polymerization lies in their reaction mechanisms:
In addition polymerization, monomer molecules (either identical or different) combine directly without the loss of any atoms, resulting in a large polymer molecule. The molecular formula of the repeating unit is identical to that of the monomer.
Conversely, condensation polymerization involves the reaction of two or more bifunctional or polyfunctional monomer molecules, typically accompanied by the elimination of small molecules such as water, alcohol, or hydrogen chloride. This process leads to the formation of polymers.
In simple words: Addition polymerization just links monomers; condensation polymerization links monomers *and* removes small molecules.

🎯 Exam Tip: Clearly stating the presence or absence of byproduct formation is the key differentiator between addition and condensation polymerization.

Question 9. Explain the term copolymerisation and give two examples.
Answer: Copolymerization is a polymerization process where two or more different types of monomers are reacted together to form a polymer chain. The resulting polymer, known as a copolymer, incorporates repeating units from each of the distinct monomers within its structure.
Two examples of copolymerization include the formation of:
(i) A copolymer from 1,3-butadiene and styrene (e.g., Buna-S).
(ii) A copolymer from 1,3-butadiene and acrylonitrile (e.g., Buna-N).
In simple words: Copolymerization is like building a wall with two different types of bricks mixed together.

🎯 Exam Tip: Define copolymerization precisely and be ready to provide examples of specific monomer pairs that form copolymers.

Question 10. Write the free radical mechanism for the polymerisation of ethene.
Answer: The free radical polymerization of ethene typically proceeds through three main steps: initiation, propagation, and termination.
Initiation: This step begins with a free radical initiator, often benzoyl peroxide, which decomposes to form radicals. These radicals then attack the ethene monomer, creating a new radical species: \[ C_6H_5-C(=O)-O-O-C(=O)-C_6H_5 \xrightarrow{\text{heat}} 2 C_6H_5-C(=O)-O^\bullet \]
\( C_6H_5-C(=O)-O^\bullet \implies C_6H_5^\bullet + CO_2 \)
\( C_6H_5^\bullet + CH_2=CH_2 \implies C_6H_5-CH_2-CH_2^\bullet \]
Propagation: The newly formed radical adds to successive ethene molecules, leading to the growth of the polymer chain: \[ C_6H_5-CH_2-CH_2^\bullet + nCH_2=CH_2 \implies C_6H_5-(CH_2-CH_2)_n-CH_2-CH_2^\bullet \]
Termination: The growing polymer chains stop growing when two radicals combine or disproportionate, forming a stable polymer molecule:
\( C_6H_5-(CH_2-CH_2)_n-CH_2-CH_2^\bullet + C_6H_5-(CH_2-CH_2)_m-CH_2-CH_2^\bullet \implies \)
\( C_6H_5-(CH_2-CH_2)_n-(CH_2-CH_2)-(CH_2-CH_2)_m-C_6H_5 \) The final product is polythene.
In simple words: Free radical polymerization of ethene uses a 'starter' molecule to create active sites that rapidly add more ethene units, forming long chains, until these active sites combine and stop the growth.

🎯 Exam Tip: Memorize the three steps of free radical polymerization – initiation, propagation, and termination – and be able to illustrate them with ethene as an example.

Question 11. Define thermoplastics and thermosetting polymers with two examples of each.
Answer: The distinctions between thermoplastic and thermosetting polymers are summarized as follows:
(i) Thermoplastics:
* Characterized by linear or slightly branched polymer chains.
* Intermolecular forces, primarily weak van der Waals forces, allow them to soften and melt when heated.
* They can be repeatedly softened, reshaped, and hardened upon cooling, making them recyclable.
* Examples include polystyrene, polyvinyl chloride (PVC), and terylene.
(ii) Thermosetting Polymers:
* Possess extensive cross-linked polymer networks.
* The strong covalent cross-links prevent them from melting or softening upon heating; instead, they degrade at high temperatures. They are non-fusible.
* Once molded and cured, they become permanently hard and rigid and cannot be re-molded or recycled.
* Examples include formaldehyde resins (like Bakelite) and glyptal.
In simple words: Thermoplastics can be melted and reformed many times because their chains are not strongly linked, while thermosets form permanent, rigid networks when heated and cannot be reshaped.

🎯 Exam Tip: Create a comparative table in your mind highlighting recyclability, structure (linear vs. cross-linked), and thermal behavior (melt vs. degrade) for thermoplastics and thermosets, along with key examples.

Question 12. Write the monomers used for getting the following polymers,
(i) Polyvinyl chloride
(ii) Teflon
(iii) Bakelite
Answer: The monomers required to form the specified polymers are:
(i) For polyvinyl chloride, the monomer is vinyl chloride (\( CH_2=CHCl \)).
(ii) For Teflon, the monomer is tetrafluoroethene (\( CF_2=CF_2 \)).
(iii) For Bakelite, the monomers are formaldehyde (HCHO) and phenol (\( C_6H_5OH \)).
In simple words: To build a polymer, you need specific monomer 'bricks'; knowing the polymer helps identify its unique monomer.

🎯 Exam Tip: It's essential to recall the correct chemical names and structures of monomers for important commercial polymers.

Question 13. Write the name and structure of one of the common initiators used in free radical addition polymerisation.
Answer: Benzoyl peroxide is a frequently used initiator in free radical addition polymerization. Its chemical structure is: \[ C_6H_5-C(=O)-O-O-C(=O)-C_6H_5 \] This molecule readily cleaves to form free radicals, initiating the polymerization process.
In simple words: Benzoyl peroxide is a 'starter' chemical that breaks apart to make free radicals, which then begin a polymerization reaction.

🎯 Exam Tip: Remember benzoyl peroxide as a common free radical initiator and its general structure.

Question 14. How does the presence of double bonds in rubber molecules influence their structure and reactivity?
Answer: The double bonds within natural rubber molecules (cis-1,4-polyisoprene) play a crucial role in determining its structure and reactivity. Structurally, these double bonds, specifically between the C2 and C3 atoms of each isoprene unit, impose a cis-configuration. This geometry prevents the polymer chains from packing closely, resulting in weak intermolecular attractions. Consequently, natural rubber adopts a coiled or helical structure, which is responsible for its characteristic elasticity. In terms of reactivity, the presence of double bonds makes natural rubber susceptible to addition reactions, such as vulcanization, which modifies its properties.
In simple words: The double bonds in natural rubber, arranged in a 'cis' way, keep the chains from getting too close, making the rubber stretchy and reactive to chemical changes like vulcanization.

🎯 Exam Tip: Connect the specific cis-configuration around double bonds in natural rubber to its physical properties (coiled structure, elasticity) and chemical reactivity (addition reactions).

Question 15. Discuss the main purpose of vulcanisation of rubber.
Answer: The primary objectives of vulcanization are to significantly enhance the properties of natural rubber. This process improves its elasticity, reduces its tendency to absorb water, and increases its resistance to both oxidation and various organic solvents. During vulcanization, sulfur atoms create cross-links between the individual polymer chains, which leads to a more rigid and durable rubber material.
In simple words: Vulcanization is a process that makes rubber stronger, less sticky, more elastic, and more resistant to wear by forming sulfur bridges between polymer chains.

🎯 Exam Tip: List at least three key improvements that vulcanization brings to rubber and mention the role of sulfur in cross-linking.

Question 16. What are the monomeric repeating units of Nylon-6 and Nylon-6,6?
Answer: The monomeric repeating unit for Nylon-6 is \( \[ -NH-(CH_2)_5-CO- \] \).
For Nylon-6,6, the monomeric repeating unit is derived from two distinct monomers: hexamethylene diamine (\( H_2N(CH_2)_6NH_2 \)) and adipic acid (\( HOOC(CH_2)_4COOH \)). The repeating unit in the polymer chain is \( \[ -NH-(CH_2)_6-NH-CO-(CH_2)_4-CO- \] \).
In simple words: Nylon-6 comes from a single monomer, while Nylon-6,6 is made from two different monomers that link up.

🎯 Exam Tip: Distinguish between Nylon-6 and Nylon-6,6 by identifying their respective monomer(s) and their repeating unit structures.

Question 17. Write the names and structures of the monomers of the following polymers:
Answer: The names and corresponding structures of the monomers for the specified polymers are detailed below:

🎯 Exam Tip: A thorough understanding of common synthetic polymers requires memorizing their constituent monomers' names and drawing their chemical structures accurately.

Question 18. Identify the monomer in the following polymeric structures.
(i) \( \[ -C(=O)-(CH_2)_8-C(=O)-NH-(CH_2)_6-NH- \]_n \)
(ii)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख मेलामाइन-फॉर्मेल्डीहाइड रेजिन की बहुलक संरचना को दर्शाता है। इसमें मेलामाइन इकाइयों को फॉर्मेल्डीहाइड से प्राप्त मेथिलोल समूहों के माध्यम से आपस में जोड़ा गया है, जिससे एक क्रॉस-लिंक्ड नेटवर्क बनता है।
Answer: The monomers that constitute the given polymeric structures are:
(i) For the polymer \( \[ -C(=O)-(CH_2)_8-C(=O)-NH-(CH_2)_6-NH- \]_n \), the monomers are decanoic acid (\( HOOC-(CH_2)_8-COOH \)) and hexamethylene diamine (\( H_2N(CH_2)_6NH_2 \)).
(ii) For the melamine-formaldehyde resin structure, the monomers are melamine and formaldehyde.
In simple words: You can break down complex polymer structures to find the smaller, simpler molecules (monomers) they were built from.

🎯 Exam Tip: Being able to deduce the monomer(s) from a given polymer's repeating unit structure is a critical skill in polymer chemistry.

Question 19. How is dacron obtained from ethylene glycol and terephthalic acid?
Answer: Dacron, also known as Terylene, is synthesized through the condensation polymerization of two monomers: ethylene glycol and terephthalic acid. This reaction typically occurs at temperatures ranging from 420K to 460K, in the presence of a catalyst system comprising a mixture of zinc acetate and antimony trioxide. The process involves the elimination of water molecules as the monomers link together to form the polyester chain: \[ nHOCH_2CH_2OH + nHOOC-C_6H_4-COOH \implies \[ -O-CH_2CH_2O-C(=O)-C_6H_4-C(=O)- \]_n + 2nH_2O \]
In simple words: Dacron is made by joining ethylene glycol and terephthalic acid, releasing water in the process, at high temperatures with a catalyst.

🎯 Exam Tip: Understand Dacron's formation as a condensation polymerization, identify its monomers, and recall the reaction conditions.

Question 20. What is a biodegradable polymer ? Give an example of a biodegradable aliphatic polyester.
Answer: A biodegradable polymer is a type of polymer that can be naturally broken down into simpler substances by microorganisms within biological systems. This degradation typically occurs through processes like enzymatic hydrolysis or oxidation.
An illustrative example of a biodegradable aliphatic polyester is PHBV (polyhydroxybutyrate-co-hydroxyvalerate). It is a copolymer formed from 3-hydroxybutanoic acid and 3-hydroxypentanoic acid: \[ n \text{ HO-CH(CH}_3\text{)-CH}_2\text{-COOH} + n \text{ HO-CH(C}_2\text{H}_5\text{)-CH}_2\text{-COOH} \implies \] \[ \[ -O-\text{CH(CH}_3\text{)-CH}_2\text{-C(=O)-O-CH(C}_2\text{H}_5\text{)-CH}_2\text{-C(=O)-} \]_n + 2nH_2O \] Here, the 'R' groups in the hydroxy acids correspond to \( -CH_3 \) or \( -C_2H_5 \).
In simple words: Biodegradable polymers can be broken down naturally by living organisms, helping reduce pollution, and PHBV is a good example.

🎯 Exam Tip: Define biodegradable polymers clearly and remember PHBV as a prominent example, including its monomers.

GSEB Class 12 Chemistry Polymers Additional Important Questions And Answers

Question 1. Natural rubber is an elastomer.
(i) Briefly explain the properties of elastomers.
(ii) Write the monomer present in natural rubber.
(iii) If natural rubber is not hard, how will you make it hard? Explain the chemistry behind it.
Answer:
(i) Elastomers are a class of polymers known for their exceptional elasticity, meaning they can be significantly stretched under stress and then return to their original shape once the stress is removed. Their polymeric chains are held together by weak intermolecular forces, which allow for considerable chain movement and flexibility.
(ii) The monomer present in natural rubber is isoprene (2-methyl-1,3-butadiene).
(iii) Natural rubber can be made harder and more rigid through a process called vulcanization. This involves heating the rubber with sulfur. During vulcanization, sulfur atoms form cross-links between the individual polymer chains. These covalent cross-links restrict chain movement, making the material less pliable and increasing its stiffness and strength.
In simple words: Elastomers are very stretchy materials that can bounce back to their original form after being pulled. Isoprene is the building block of natural rubber. Natural rubber can be hardened by adding sulfur, which creates strong links between its polymer chains, making it stiffer.

🎯 Exam Tip: Focus on elasticity, weak intermolecular forces, and reversibility as key properties of elastomers. Identify isoprene as the monomer for natural rubber, including its chemical name and structure. Explain vulcanization as the method for hardening natural rubber, emphasizing the role of sulfur in forming cross-links.

Question 2. In a factory the supervisor told the technical assistant that they got a bulk order for polythene. Then the technical assistant asked whether it was LDPE or HDPE.
(i) What do the abbreviations stand for?
(ii) How are they prepared? Give equation?
(iii) List out their applications.
Answer:
(i) The abbreviations stand for:
* LDPE: Low-Density Polyethylene
* HDPE: High-Density Polyethylene
(ii) Their preparation methods and equations are as follows:
a. Low-Density Polyethylene (LDPE): LDPE is formed when ethene undergoes polymerization under very high pressure (around 2000 atmospheres) and high temperatures (500-600 K) in the presence of oxygen or a peroxide initiator. This process results in a highly branched polymer structure. \[ nCH_2=CH_2 \xrightarrow{\text{2000 atm, O}_2, \text{ 500-600 K}} \[CH_2-CH_2\]_n \]
b. High-Density Polyethylene (HDPE): HDPE is produced by polymerizing ethene at lower temperatures (333-343 K) and moderate pressures, using a Ziegler-Natta catalyst (e.g., \( (C_2H_5)_3Al \) and \( TiCl_4 \)). This method yields a linear polymer with minimal branching. \[ nCH_2=CH_2 \xrightarrow{\text{333-343 K, Ziegler-Natta catalyst}} \[CH_2-CH_2\]_n \]
(iii) The applications for each type are:
* LDPE: Due to its flexibility and translucency, LDPE is used for packaging films, squeeze bottles, and pipes.
* HDPE: Being hard, tough, and chemically inert, HDPE is employed in the manufacture of toys, rigid pipes, and various household articles.
In simple words: LDPE and HDPE are both polythene, but made differently: LDPE is soft and branched (for films), while HDPE is hard and linear (for rigid items).

🎯 Exam Tip: Understand the synthesis conditions (pressure, temperature, catalyst) that differentiate LDPE and HDPE, how these conditions affect their structure (branched vs. linear), and their corresponding applications.

Question 3. How does thermoplastic polymer differ from thermosetting polymer?
Answer: The distinctions between thermoplastic and thermosetting polymers are summarized as follows:
Thermoplastics:
* Characterized by linear or slightly branched polymer chains.
* Intermolecular forces, primarily weak van der Waals forces, allow them to soften and melt when heated.
* They can be repeatedly softened, reshaped, and hardened upon cooling, making them recyclable.
* Examples include polystyrene, polyvinyl chloride (PVC), and terylene.
Thermosetting Polymers:
* Possess extensive cross-linked polymer networks.
* The strong covalent cross-links prevent them from melting or softening upon heating; instead, they degrade at high temperatures. They are non-fusible.
* Once molded and cured, they become permanently hard and rigid and cannot be re-molded or recycled.
* Examples include formaldehyde resins (like Bakelite) and glyptal.
In simple words: Thermoplastics can be melted and reformed many times because their chains are not strongly linked, while thermosets form permanent, rigid networks when heated and cannot be reshaped.

🎯 Exam Tip: Create a comparative table in your mind highlighting recyclability, structure (linear vs. cross-linked), and thermal behavior (melt vs. degrade) for thermoplastics and thermosets, along with key examples.

Question 4. Identify me.
(i) I am a polymer resistant to heat and chemicals. People use me to make non-sticky frying pans.
(ii) I am a polymer formed from ethylene glycol and terephthalic acid, used for making heart valves.
(iii) I am a polymer used for making unbreakable crockery items.
Answer: Identifying the polymers based on their descriptions:
(i) Teflon: This polymer is highly resistant to heat and chemicals, making it ideal for non-stick coatings on frying pans.
(ii) Terylene (or Dacron): Formed from ethylene glycol and terephthalic acid, this polymer finds application in materials such as heart valves and synthetic fibers.
(iii) Melamine-formaldehyde resin: This polymer is known for its hardness and durability, making it suitable for manufacturing unbreakable crockery and decorative laminates.
In simple words: These are common polymers identified by their uses: Teflon for non-stick, Terylene for strong fibers and medical uses, and Melamine resin for tough dishes.

🎯 Exam Tip: Be familiar with the applications of common polymers, as this helps in identifying them and understanding their properties.

Question 5. List some important differences between Natural rubber and the Vulcanization of rubber.
Answer: Here are the key differences between natural rubber and vulcanized rubber:
Natural Rubber:
* Characteristically soft and exhibits stickiness.
* Possesses low tensile strength, meaning it breaks easily under tension.
* Shows limited elasticity, deforming permanently after significant stretching.
* Its functional temperature range is narrow, typically from 10°C to 60°C.
Vulcanized Rubber:
* Becomes hard and loses its sticky nature.
* Exhibits high tensile strength, making it much more durable.
* Demonstrates significantly improved elasticity, returning to its original shape effectively after stretching.
* Can be utilized over a wider temperature range, generally from 40°C to 100°C.
In simple words: Natural rubber is soft and weak, but vulcanized rubber is strong, hard, and much more useful because of cross-links.

🎯 Exam Tip: Focus on how vulcanization enhances the mechanical properties (hardness, tensile strength, elasticity) and thermal stability of natural rubber.

Question 6. Write a note on
(i) polyamides
(ii) polyesters
Answer: Here are notes on polyamides and polyesters:
(i) Polyamides: These are polymers characterized by the presence of amide linkages (\( -CO-NH- \)) within their main chain. Examples include Nylon-66 and Nylon-6. Nylon-66, specifically, is a polyamide formed by the condensation polymerization of hexamethylene diamine (\( H_2N(CH_2)_6NH_2 \)) and adipic acid (\( HOOC(CH_2)_4COOH \)), with the elimination of water: \[ nH_2N(CH_2)_6NH_2 + nHOOC(CH_2)_4COOH \implies \[ -HN(CH_2)_6NHCO(CH_2)_4CO- \]_n + 2nH_2O \]
(ii) Polyesters: These polymers contain ester linkages (\( -CO-O- \)) within their backbone. Common examples include Terylene (Dacron) and glyptal. Terylene is a polyester synthesized from ethylene glycol (\( HOCH_2CH_2OH \)) and terephthalic acid (\( HOOC-C_6H_4-COOH \)), also through condensation polymerization: \[ nHOOC-C_6H_4-COOH + nHOCH_2CH_2OH \implies \[ -OC-C_6H_4-COOCH_2CH_2O- \]_n + 2nH_2O \]
In simple words: Polyamides are polymers like nylon, made by linking units with amide bonds, often from diamines and diacids. Polyesters are polymers like Terylene, made by linking units with ester bonds, typically from diols and dicarboxylic acids.

🎯 Exam Tip: Define polyamides by their characteristic amide linkage and use Nylon-6,6 as a key example, identifying its monomers. Define polyesters by their characteristic ester linkage and use Terylene (Dacron) as a key example, identifying its monomers.