Polymers

According to question and given monomers in the options, the polymer $X$ is nylon 6,6 as, it contains hydrogen bonding which is due to the presence of amide linkage. Thus, monomer $Y$ and $Z$ are hexamethylene diamine and adipic acid. The reaction is as follows

$ \text { Thus, } Y \text { and } Z \text { are given in } B \text { and } E \text {. } $

$ \begin{aligned} &\text { The reaction involved is as follows }\\ &\begin{aligned} 2 \mathrm{CH}_3 \mathrm{Cl}+\mathrm{Si} & \xrightarrow[573 \mathrm{~K}]{\mathrm{Cu}} \underset{(X)}{\left(\mathrm{CH}_3\right)_2 \mathrm{SiCl}_2} \xrightarrow{\mathrm{H}_2 \mathrm{O}}\left(\mathrm{CH}_3\right)_2 \mathrm{Si}(\mathrm{OH})_2(Y)+2 \mathrm{HCl} \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} &\left(\mathrm{CH}_3\right)_2 \mathrm{Si}(\mathrm{OH})_2(\mathrm{Y}) \xrightarrow{\text { Polymerisation }} \\ &-\left[\mathrm{Si}\left(\mathrm{CH}_3\right)_2-\mathrm{O}\right]- \end{aligned}\\ &\text { Thus, repeating unit in } \mathrm{Z} \text { is - }\\ &\left[\mathrm{Si}\left(\mathrm{CH}_3\right)_2-\mathrm{O}\right]- \end{aligned} $

Dacron and nylon 6,6 are the examples of polyester and polyamides respectively.

The correct match is A-II, B-III, C-IV, D-I.

The monomer of neoprene is chloroprene. It's IUPAC name is, 2-chloro-1,3-butadiene.

Given:

Polymer ‘X’ is mainly used for making unbreakable cups and laminated sheets.

We’re asked: What are the monomers of polymer X?

Step 1: Identify the polymer

• Unbreakable cups and laminated sheets are typically made from melamine-formaldehyde or phenol-formaldehyde type resins.

• However, phenol-formaldehyde resin (commonly known as Bakelite) is specifically known for:

• Being hard and brittle

• Used in electrical switches, unbreakable crockery, laminated sheets, etc.

Step 2: Match with monomers

Bakelite (phenol-formaldehyde resin) is formed by:

$ \text{Phenol} + \text{Formaldehyde} \longrightarrow \text{Phenol-formaldehyde resin (Bakelite)} $

Hence, the monomers are phenol and formaldehyde.

✅ Correct Answer:

Option C: phenol and formaldehyde

• A. Electrodes in batteries:

  • Polyacetylene (II) is a conductive polymer that, when doped, can achieve electrical conductivities similar to metals. This property makes it suitable for use as an electrode material in some types of batteries, particularly in the early development of conducting polymer batteries.

  • Polypropylene (I) is used for battery cases, not electrodes.

  • Oxyacetylene (III) is a gas mixture for welding.

  • Therefore, A matches II.

• B. Welding of metals:

  • Oxyacetylene (III) is a mixture of oxygen and acetylene gas, which when ignited, produces a very hot flame (up to 3200°C). This high-temperature flame is widely used for welding, cutting, and heating metals.

  • Polypropylene (I) and Polyacetylene (II) are polymers and not directly used for welding metals in this context.

  • Therefore, B matches III.

• C. Toys:

  • Polypropylene (I) is a versatile thermoplastic polymer known for its durability, flexibility, chemical resistance, and relatively low cost. It is widely used in the manufacturing of various plastic products, including many types of toys.

  • Polyacetylene (II) is a specialized conductive polymer, not typically used for common toys.

  • Oxyacetylene (III) is a gas mixture, not a material for toys.

  • Therefore, C matches I.

Combining these matches:

A - II

B - III

C - I

Step 1: Find the total mass for each type of molecule

There are 10 molecules, each with a mass of 5,000. So, the total mass for these is:

$ 10 \times 5000 = 50000 $

There are 5 molecules, each with a mass of 50,000. So, the total mass for these is:

$ 5 \times 50000 = 250000 $

Step 2: Add total masses and total number of molecules

Add the total masses together:

$ 50000 + 250000 = 300000 $

Add the total numbers of molecules:

$ 10 + 5 = 15 $

Step 3: Calculate the number average molecular mass

Use the formula:

$ \mathrm{M}_n = \frac{\sum n_i M_i}{\sum n_i} = \frac{300000}{15} = 20000 $

There are three linear polymer

Nylon-6,6 PVC, Novolac

Linear polymers have a chain like structure with monomer linked in single line

There are two cross linked polymer.

Melamine, Bakelite

$ \text { It is a biodegradable polymer. } $

Both statement I and statement II are incorrect. This correct forms are Nylon 6 is a condensation homopolymer, Nylon 6,6 is made from adipic acid and hexamethylene diamine.

$ \text { The complete polymerisation reaction is, } $

$ \text { Thermoplastic polymer (Product } Y \text { ). } $

The incorrect statement about silicon is

It has low dielectric strength.

The correct form is, it has high dielectric strength and due to this it is good insulator.

The correct match is, A-II, B-IV, C- I, D-III

$ \text { Complete reaction is, } $

$ \text { Free radical polymerisation } $

$ \text { (Ziegler - Natta polymerisation) } $

$ \text { The complete polymerisation is, } $

The common monomer for bakelite and melamine is formaldehyde

$ \text { Phenol-formaldehyde resin } $

$ \text { Melamine-formaldehyde resin. } $

High density polythene is prepared using Ziegler-Natta catalyst. Ziegler - Natta catalyst: A transition metal halides + Organoaluminium compound e.g. $\mathrm{TiCl}_4+\mathrm{Al}\left(\mathrm{C}_2 \mathrm{H}_5\right)_3$

Polystyrene is not an example of condensation polymerisation. It is prepared by free radical addition polymerisation.

Condensation polymerisation is a form of a step-growth polymerisation where smaller molecules react with each other to form large structural unit while releasing by product such as water.

Both A and R are correct but R is not the correct explanation of A . The correct explanation is,

The low surface tension of silicones, coupled with the molecular orientation of non-polar methyl groups, create highly hydrophobic barrier that repels water molecules.

When caprolactum is heated with water at a higher temperature, Polymer nylon-6 is obtained. The reaction is as follows,

Methanal is the common monomer of bakelite and melamine polymer.

Bakelite has phenol and methanal as monomer units. Melamine has melamine and methanal as monomer units.

The correct answer is Option D: thermoplastic polymers.

Here's why:

Thermoplastic polymers soften when heated and harden again upon cooling. This property allows them to be remolded repeatedly.

In contrast, thermosetting polymers (such as bakelite) undergo a curing process that permanently locks their structure, so they do not soften upon reheating.

Fibres refer to materials with long, flexible strands and are not classified based on heat-softening properties.

Thus, polymers that can be softened on heating and hardened on cooling are called $\textbf{thermoplastic polymers.}$

Statement given in option (b) is correct, while all other are incorrect.

The correct statements are:

(a) Starch is a polymer of $\alpha-\mathrm{D}$ glucose.

(c) Proteins are made of up to 20 different amino acids, so protein is heteropolymers or copolymers.

(d) Lactose is a disaccharide of $\beta-\mathrm{D}$ glucose and $\beta$-D-galactose.

Set given in option (a) is not

The correct match is A-III, B-I, C-IV, D-II.

A. Elastomer are any rubber material composed of long chain like polymers that are capable of recovering to original shape after stretching $\longrightarrow$ Buna-N Rubber is elastomer. (Inter-particle force is weakest : van der Waals' force)

B. Fiber forming polymers are linear macromolecules that are suitable for synthetic fiber $\longrightarrow$ Terylene is synthetic fiber.

(Inter-particle force is strongest : H-bonding)

C. A thermoplastic polymer are those polymer that can softern on heating and then processed. Polystyrene is an example.

D. Thermosetting polymers are semifluid substance with low molecular mass, when heated in mould undergo a permanent change, in composition. Urea-formaldehyde resin is an example.

Option A: Polystyrene is formed by the polymerization of styrene and is a homopolymer (made from a single type of monomer), not a copolymer. Although it is a thermoplastic, the first part of the match is incorrect.

Option B: Bakelite is a thermosetting resin, but it is produced via a condensation reaction (not addition), so calling it an "addition polymer" is incorrect.

Option C: Nylon 6 is produced from caprolactam by ring-opening polymerization, making it a homopolymer. It is widely used in the textile industry as a fiber.

$\text{Nylon 6 (a homopolymer of caprolactam) → Fiber}$

This option is correctly matched.

Option D: Buna-N, also known as nitrile rubber, is a copolymer (usually made from butadiene and acrylonitrile), not a homopolymer, although it is an elastomer.

Thus, the correctly matched set is:

Option C: Nylon 6 – homopolymer – fiber.

Ethylene glycol is the common monomer for both terylene and glyptal.

Ethylene glycol

Chloroprene is the monomer of neoprene. The complete reaction is as follow.

Among the given, polyethylene terephthalate is a fibre polymer (a polyester).

Its monomers are ethylene glycol and terephthalic acid.

PHBV (IUPAC name) 3-hydroxybutyrate-CO- 3 hydroxy valerate) is a polyhydroxy alkanoate type polymer. The monomers are

(i) 3 hydroxybutyrate ( 3 HB )

$ X=\mathrm{CH}_3-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H} $

(ii) 3 hydroxyvalerate ( 3 HV )

$ Y=\mathrm{C}_2 \mathrm{H}_5 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H} $

Among the given options monomer given in option (d) will give biodegradable polymer i.e., PHBV.

Let's look at each example in List I and try to identify its type:

A. 2-chloro-1,3-butadiene: This is an example of a synthetic rubber, which is a type of polymer.

B. Nylon 2-nylon 6: This is a type of nylon polymer, which is a type of polyester.

C. Polyacrylonitrile: This is an example of a biodegradable polymer, which can be broken down by microorganisms.

D. Dacron: This is an example of an addition polymer, which is a polymer formed by the addition of monomers.

Now, let's look at the types in List II and match them with their corresponding examples:

I. Biodegradable polymer: This matches with example C, which is Polyacrylonitrile.

II. Synthetic Rubber: This matches with example A, which is 2-chloro-1,3-butadiene.

III. Polyester: This matches with example B, which is Nylon 2-nylon 6.

IV. Addition polymer: This matches with example D, which is Dacron.

Therefore, the correct matching of List I with List II is:

A - II (Synthetic rubber)

B - III (Polyester)

C - I (Biodegradable polymer)

D - IV (Addition polymer)

Hence, the correct answer is A-II, B-I, C-IV, D-III.

Statement I is false. Ethene (ethylene) under certain conditions (high pressure, high temperature) in the presence of catalysts does undergo addition polymerization, but it forms polyethylene, not LDP (Low-Density Polyethylene) specifically. The formation of low-density polyethylene (LDPE) vs high-density polyethylene (HDPE) depends on the specific conditions and catalysts used. The catalysts mentioned (AlEt₃ and TiCl₄) are associated with the production of HDPE through the Ziegler-Natta process, not LDPE.

Statement II is true. Caprolactam polymerizes to form Nylon 6 via a ring-opening polymerization mechanism, which is a type of step-growth polymerization. The process is often carried out at high temperatures, and water is a byproduct of the reaction, so the description in Statement II is accurate.

Therefore, Option C (Statement I is false but Statement II is true) is the correct answer.

The polymer X described in the question is synthesized using triethylaluminium and titanium tetrachloride under low pressure. This is a clear indication that the Ziegler-Natta catalyst is used, which is typically employed in the synthesis of high-density polyethylene (HDPE).

High-density polyethylene is made up of linear molecules that pack closely together, giving the material its high density. This is in contrast to low-density polyethylene (LDPE), which has a branched structure that prevents the chains from packing closely together.

So, the polymer X is :

Option A: High-density polyethylene.

• Nylon-2-Nylon-6 is a biodegradable polymer. It is made from two monomers, hexamethylenediamine and adipic acid. The polymer is strong and durable, but it can also be broken down by bacteria and other microorganisms. This makes it a good choice for use in products that are designed to be biodegradable, such as food packaging and medical implants.


• Buna N is a synthetic rubber. It is made from a mixture of 1,3-butadiene and acrylonitrile. The polymer is strong and flexible, and it is resistant to wear and tear. Buna N is used in a variety of applications, including tires, hoses, and belts.


• Urea-formaldehyde is a thermosetting resin. It is made from a mixture of urea and formaldehyde. The polymer is hard and strong, and it is resistant to heat, water, and chemicals. Urea-formaldehyde is used in a variety of applications, including adhesives, coatings, and foams.


• Dacron is a polyester. It is made from a mixture of terephthalic acid and ethylene glycol. The polymer is strong, durable, and lightweight. Dacron is used in a variety of applications, including clothing, fabrics, and bottles.

Caprolactum $\buildrel \Delta \over \longrightarrow $ Nylon $-$ 6

The correct match is A-IV, B-III, C-I, D-II.

• Dacron is a fibre formed by condensation of ethylene glycol and terephthalic acid with the elimination of water molecule.

• Neoprene is a elastomer that are produced by polymerisation of chloroprene. Bakelite is a thermosetting phenol-formaldehyde resin.

• Polystyrene is a thermoplastics polymer made from monomer styrene.

Zeigler Natta catalyst $\left[\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{Al}+\mathrm{TiCl}_4\right]$ is used to prepare H.D.P.E. (high density polythene). For examples toys, grocery bags etc.

Thus, $R$ is the monomers polymer of dacron.

All the given statements are correct i.e. $A, B, C$.

Glyptal is made from monomer ethylene glycol and phthalic acid.

• Bakelite is used in making combs, phonograph records, electrical switches and handles of utensils etc.

• Nylon-2-nylon-6 is an alternating polyamide copolymer of glycine and amino caproic acid.

It is a biodegradable polymer.

$ \begin{aligned} &\text { Weight average molecular mass }\\ &\begin{aligned} & \left(\bar{M}_w\right)=\frac{\Sigma n_i m_i^2}{\Sigma n_i m_i} \\ & =\frac{3 \times\left(10^3\right)^2+3 \times(500)^2+4 \times(200)^2}{3 \times 10^3+3 \times 500+4 \times 200} \\ & =\frac{3,000,000+7,50,000+1,60,000}{5300} \\ & =737.7 \end{aligned} \end{aligned} $