Polymers and Biomolecules are very important chapters in JEE Main Chemistry. Generally, 1 question is asked from each chapter, Polymers and Biomolecules, which bears about 8 marks in total. Therefore, their overall weightage in JEE Main is 2-3%. These chapters cannot be skipped. Most of the questions asked from Polymers and Biomolecules are quite easy to score. 

Some of the topics tested under Polymers include - Different Types of Polymers and Polymerization. Lipids, Carbohydrates, and Proteins are some of the topics tested under Biomolecules. Read the article below to find study notes for polymers and biomolecules to help you cover JEE Main Chemistry Syllabus.

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Polymers

• Polymers are known as macromolecules. Macromolecules are known to be a grouping of small molecules to produce a big molecule. They can be man-made as well. 
• The first syntheses had the goal of making substitutes for natural macromolecules, like rubber and silk. However, with the growth of technology, there can be a production of several substances which have no natural counterparts.
• Some of the synthetic macromolecular compounds are - elastomers, that have a specific kind of elasticity of rubber; fibers, which are long, thin, and thread-like - the strength along the fiber is characteristic of wool, cotton, and silk; and plastics, that can be extruded as pipes or sheets and painted on surfaces, or molded to make numerous objects.
• Therefore, a polymer is a very huge molecule which consists of several repeating small molecular units.
• The small molecular units that make up the polymer are known as monomers.
• The chemical reaction that binds the monomers together is known as polymerization

Also Check

• JEE Main Study Notes for Equilibrium
• JEE Main Study Notes for s block Elements
• JEE Main Study Notes for States of Matter

Classification of Polymers According to Source

Natural Polymers 

• Natural Polymers are derived from animals or plants.
• They are found in nature. 
• Some examples include, DNA, RNA, Natural Rubber, Cellulose, and Protein.
• Natural rubber is a polymer of isoprene.
• It is derived from the latex of rubber trees. In chemical terms, it is known to be the polymer of isoprene units. It is soft, sticky, and possesses very limited capacity to retain its shape. Such properties exist due to the absence of cross-linkage among the polymeric chains. Natural Rubber turns soft at high temperatures (T> 335 K) and becomes brittle at low temperatures (T<283 K). It also shows a high water absorption capacity because of which it is sticky and a semi-fluid-like substance. Another property is that it is soluble in non–polar solvents and gets radially attacked by oxidizing agents.

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Semisynthetic Polymers

• These polymers are made in the laboratory by introducing some modifications in natural polymers to enhance their quality. 
• Some examples include, Rayon, Cellulose nitrate, and Cellulose acetate

Synthetic Polymers

• These get synthesized in the laboratory from chemicals.
• Some examples include, nylon, backelite, polythene, P.V.C., and Teflon.

Classification of Polymers According to Structure 

Linear Polymers 

• These get formed when monomers are linked together to form long and straight chains of polymer molecules.
• They do not have any branching.
• Some examples include, nylons and polyethylene.

Branched Chain Polymers 

• The monomeric units in these polymers are part of a branched chain.
• They have low densities, low melting points, and low tensile strengths when compared to linear polymers.
• Some examples include, glycogen, amylopectin, and low density polyethylene. 

Cross Linked or Three Dimensional Network Polymers 

• These polymers are formed when linear polymeric chains are connected together to make up a three dimensional network structure.
• They are very rigid, hard, and brittle.
• Some examples include, bakelite and melamine formaldehyde resin.

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Classification of Polymers Based on Intermolecular Forces

Elastomers

• They have a very weak intermolecular force of attraction among the polymer chains.
• They also have very weak Van der Waals’ forces holding the various chains of elastomers together.
• They have a high degree of elasticity and can thus be stretched and reverted back to their initial shape when force is eliminated.
• These polymers have a few cross-links to prevent the chains from slipping over each other.
• An example is vulcanized rubber.

• These polymers are formed when linear polymeric chains are connected together to make up a three dimensional network structure.
• They are very rigid, hard, and brittle.
• Some examples include, bakelite and melamine formaldehyde resin.

Fibers

• They have a very strong intermolecular force.
• They are held together through hydrogen bonds and/or dipole- dipole attraction..
• These polymers have tensile strength, less elasticity, high melting point, and low solubility.
• Some examples include, nylon, cellulose, wool, terylene, and silk.

Thermoplastic Polymers

• The intermolecular forces between these polymers are higher than elastomers but lower than fibers.
• They do not contain any cross linkage.
• These polymers are hard at room temperature. When heated, the individual chains slip past each other and the polymers become soft.
• As they readily become soft on heating, they can be shaped into anything. This is why they are used for manufacturing toys and chairs. 
• Some examples include, polystyrene, polyethene, and polyvinyl chloride. 

Thermosetting Polymers

• They are cross linked polymers.
• Cross linking becomes more on heating.
• They turn hard and infusable on heating. Also, they cannot be softened further due to the huge number of cross links.
• Some examples include, phenol formaldehyde resin and melamine-formaldehyde resin.

Also Check

• JEE Main Study Notes for Surface Chemistry
• JEE Main Study Notes for States of Matter

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Classification of Polymers According to Mode of Polymerization

Addition Polymers

These polymers are formed through the repeated addition of monomer molecules that contain double or triple bonds. Some examples include, polythene (from ethene) and polypropene (from propene). 

Homopolymers

These are addition polymers that are formed through the polymerisation of a single monomeric type. 

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Copolymers

These are addition polymers that are formed by more than one type of monomer.

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Types of Copolymers

Random Copolymers: In these polymers, the monomer units consist of random distribution in the chain segment. 

nA + nB → -A-A-B-A-B-B-A-B-A-A-A-B-

Alternating Copolymers: In these polymers, the monomer units occur in an alternative fashion in the polymer chain segment.

nA + nB → -A-B-A-B-A-B-A-B-

Block Copolymers: In these polymers, various blocks of identical monomer units are alternative with one another. 

nA + nB → -A-A-B-B-B-B-B-A-A-A-A-A-B-B-B-B-B-A-

Graft Copolymers: In these polymers, the homopolymer branches of one polymer unit get grafted onto the homopolymer chain of another monomer unit.

-A-A-A-A-A-A-A-A-

| |

B B

| |

B B

| |

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Condensation Polymers

• They are derived through a repeated condensation reaction among two distinct bi-functional or tri-functional monomeric units.
• Their formation involves the removal of small molecules like HCl, water, and alcohol.
• Some examples include, Nylon 6, 6, and terylene (dacron).
• Nylon 6, 6 is formed through the condensation of hexamethylenediamine with adipic acid.

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Biomolecules

Any molecule that exists in living organisms is known as a biomolecule. Biomolecules include macromolecules like lipids, carbohydrates, nucleic acids, and proteins, and small molecules like primary and secondary metabolites, and natural products.

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Proteins

Proteins are biomolecules which are crucial for the existence of living organisms. Amino acids are known to be the building blocks of proteins. Naturally occurring amino acids are 22 in number. They consist of the carboxyl group, amino group, hydrogen atom, and R group, which is the alkyl group. The R group is variable. This means that it varies with different amino acids. The 4 groups mentioned are attached to a single carbon atom that is called the Alpha-Carbon. 

Fig.1. Structure of Amino Acid

The simplest form of amino acid is glycine. In glycine, the R group undergoes replacement with the hydrogen atom. The bond formed among the two amino acids is known as the peptide bond.

Fig.2. Amino Acid Sequence-Protein

Structure of Proteins 

Primary Structure: This is the sequence of amino acids in a protein. The left end is shown by the first amino acid and the right end is shown by the last amino acid. The first amino acid is known as the N-terminal amino acid. The last amino acid is known as the C-terminal amino acid.

Secondary Structure: The protein is not a linear chain of amino acids. Instead, the chain bends at some points and also forms helices. By repeating the local structures regularly, the secondary structure to protein gets formed.

Tertiary Structure: This is the general shape of a protein molecule and also the spatial relationship between the secondary structures. In simpler words, the different folds that provide three-dimensional looks to a protein make up its tertiary structure.

Quaternary Structure: This is the way in which the individually folded polypeptides are arranged with regard to one another.

Fig.3. Structure of Protein

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Types of Biomolecules Which are Enzymes

 Lipids

Lipids are generally insoluble in water. They are known to be fatty acid esters. Importantly, lipids are the primary element of cell membranes. They can either be simple fatty acids or contain phosphorus and phosphorylated organic compounds. Those which contain phosphorus are known as phospholipids. 

Fig.4. Structure of Phospholipids

A fatty acid consists of a carboxyl group which is attached to an R group. The R group could be a methyl, ethyl or a higher number of the CH2 group (1 carbon to 19 carbons).

Primarily, there are two kinds of fatty acids - saturated and unsaturated fatty acids.

Saturated fatty acids do not have any double bond among the carbon atoms. An example is butyric acid.

Unsaturated fatty acids have double bonds among the carbon atoms. An example is linoleic acid.

Carbohydrates

Carbohydrates consist of atoms of carbon, hydrogen, and oxygen. They include sugars, starch, and cellulose. The simplest form of carbohydrates is glucose. The bond among two sugar molecules is called the glycosidic bond. 

Fig.5. Structure of Glucose

Carbohydrates which are made up of a single sugar molecule are called monosaccharide. An example is glucose. Monosaccharides that consist of more than one unit of sugar molecule are called oligosaccharides. An example is fructose. Lengthy chains of sugars are known as polysaccharides. When a polysaccharide consists of similar monosaccharides, it is known as a homopolymer. An example is cellulose. When a polysaccharide consists of different monosaccharides, it is known as a heteropolymer.

The right end of a polysaccharide chain is known as the reducing end and the left end is known as the non-Reducing end.

Fig.6. Reducing and Non-Reducing ends in Maltose

Note that starch is a homopolymer of glucose. It is the primary storage sugar in plants. In a similar manner, the storage form of sugar is glycogen in animals. This is stored in their livers. When the body requires glucose during starvation or fasting, glycogen breaks down into glucose in order to fulfil the energy needs of the body. The sugar existing in DNA is deoxyribose and in RNA, the sugar is ribose.

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Nucleic Acids

Nucleic Acids are organic molecules that exist in living cells. They are polymers of nucleotides. The three chemically significant components of a nucleotide are - phosphate group, sugar called deoxyribose, and nitrogenous bases. Nitrogenous bases are of two types - purines and pyrimidines. Purines include adenine and guanine. Pyrimidines include thymine and cytosine. DNA contains all the four bases - adenine, thymine, guanine, and cytosine. However, in RNA, thymine gets replaced by uracil.

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Types of Biomolecules is DNA

DNA is a nucleic acid biomolecule.

Fig.6. Structure of Nitrogenous Bases

Formation of Polymers

a) In chain reaction polymerization

Rad. + CH2 = CH2 → RadCH2CH­2→ RadCH2CH2CH2CH­2× → etc.

b) In step reaction polymerization

c) Free-radical vinyl polymerization

d) Copolymerization 

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Important Polymers

a) Natural Rubber

This is an addition polymer of isoprene (2-methyl-1,3-butadiene). On an average, it has a chain length of 5000 monomer units of isoprene.

b) Synthetic rubber ((Polychloroprene) or Neoprene)

It is formed through the free radical polymerisation of chloroprene in 

A thermoplastic and need not to be vulcanised. Synthetic rubber is superior to natural rubber because of its resistance to the reactions of air, light chemicals, heat, alkalis, and acids that are lower than 50% in strength.

c) Buna rubbers

i) Buna - N or GRA: A synthetic rubber that is formed through copolymerisation of one part of aryl nitrile and two parts of butadiene. 

It is more rigid and has low response to heat. It is very resistant to the swelling action of petrol, oils, and other organic solvents.

ii) Buna -S or GRS (General Purpose Styrene Rubber): This is a copolymer made from three moles of butadiene and one mole of styrene. It is an elastomer.

It is typically vulcanised with sulphur and compounded with carbon black. It is very resistant to wear and tear. 

d) Teflon: It is a polymer of tetrafluoroethylene (F2C=CF2) which when polymerized produces Telfon.

It is a thermoplastic polymer that has a high softening point (600K). It is inert to most chemicals with the exception of fluorine and molten alkali metals. Also, it withstands high temperatures. 

e) Nylon - 66: This is a condensation polymer that is formed from a reaction between adipic acid and hexamethylenediamine. It is a thermoplastic polymer.

f) Nylon 6 or Perelon: This is a polyamide that is made from the heating of caprolactum at 530-540 K for a long period of time. 

g) Dacron

h) Phenol-formaldehyde polymer

An example is bakelite novolac. 

Points to Remember for Biomolecules

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Carbohydrates

Classification of carbohydrates

• Monosaccharides: These are polyhydroxy aldehydes or polyhydroxy ketones that cannot be decomposed by hydrolysis to form simpler forms of carbohydrates. Examples: glucose, ribose, fructose, mannose, arabinose, and gulose.
• Oligosaccharides: These produce a definite number (2-9) of monosaccharide molecules by hydrolysis.
• Trisaccharides: They produce three monosaccharide molecules by hydrolysis. Example: raffinose, which has the molecular formula of C18H32O16.
• Polysaccharides: They have a high molecular weight and produce several monosaccharide molecules by hydrolysis. Examples are starch and cellulose, Both of them have the molecular formula of (C6H10O5)n.

Monosaccharides and oligosaccharides are crystalline solids. They are soluble in water and are sweet in taste. Collectively, they are called sugars. Polysaccharides are amorphous, insoluble in water, and have no taste. They are known as non-sugars. 

Monosaccharides

• The Aldoses that consist of an aldehyde group (-CHO).
• The Ketoses that consist of a ketone group (>C=O).

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Stereo Isomerism in Carbohydrates 

• Suppose the hydroxyl group on the asymmetric carbon atom far from the aldehyde or ketone group displays to the right, the compound is considered to be a member of the D-family.
• Suppose the hydroxyl group on the far asymmetric carbon displays to the left, the compound is considered to be a member of the L-family. 

• Maximum Number of Optical Isomers = 2n, where n = the number of asymmetric carbon atoms.

• Epimer: A pair of diastereomers that differs only in the configuration of one carbon atom. 

Cyclic Form of Monosaccharide

• Pyranose and Furanose Forms:

• Anomers: A pair of stereoisomers that differs in configuration when it is around C1. The C1 carbon is known as the anomeric carbon.
• In the a-anomer, the OH group at C1 is to the right and in the b-anomer, the OH group at C1 is to left.

Mutarotation

The change with time in the rotation of an optically active sugar in solution to an equilibrium value. During this process, the ring opens and later recloses in the inverted position or in the initial position producing a mix of a-and-b-forms.

Reactions of Glucose

a) With HI/P: It goes through reduction to produce n-hexane and with sodium amalgam, it produces sorbitol. 

e) Oxidation: The glucose when oxidized with Br2 produces gluconic acid which on further oxidation with HNO3 forms glucaric acid.

f) With Tollen reagent and Fehling solution: Glucose gives silver mirror and red ppt. of Cu2O respectively.

g) With acetic anhydride: In the presence of pyridine, glucose gives pentaacetate.

j) Glycoside formation: When a small quantity of gaseous HCl gets passed into a solution of D (+) glucose in methanol, the reaction results in the production of amomeric methyl acetals.

Carbohydrate acetals are generally known as glycosides and an acetal of glucose is known as glucoside.

k) Kiliani - Fischer Synthesis: A process of lengthening the carbon chain of an aldose.

l) Ruff Degradation: A contrast to Kiliani Fischer synthesis which can shorten the chain by a similar unit.

a) Sucrose: A non-reducing sugar that is made through the condensation of one molecule of glucose and one molecule of fructose.

Hydrolysis: The hydrolysis of sucrose with hot dilute acid produces D-glucose and D-fructose.

Polysaccharides

a. Starch:

i. It is a polymer of glucose.

ii. It is a mixture of two components, a water soluble component known as amylose (20%) and a water insoluble component known as amylopectin (80%).

iii. Amylose and amylopectin are the polymers of a-D-glucose.

iv. Amylose is a linear polymer of a-D-glucose.

vi Amylopectin is a highly branched polymer

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Amino Acids

Amino acids are molecules that consist of two functional groups, a carboxylic group and an amino group. 

H2N CH2 COOH : Amino acetic acid, or Glycine

CH3 CH (NH­2) COOH : a - Amino propionic acid or Alanine

H2N CH2 CH2COOH : b - Amino propionic acid

Acidic Amino Acid: Have a second carboxyl group or a potential carboxyl group in the form of carboxamide.

Basic Amino Acids: Have a second basic group that could be an amino group.

IsoElectric Point:

It is the hydrogen ion concentration of the solution in which a specific amino acid does not migrate due to the influence of an electric field. 

Peptides 

Peptides are the amides that are produced through an interaction among amino groups and carboxyl groups of amino acids.

They are called dipeptides, tripeptides, or polypeptides depending on the number of amino acid residues per molecule. 

Two amino acids are connected through a –CO-NH group. This is known as the peptide bond.

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Previous Year Solved Questions on Polymers and Biomolecules 

Q1: What is the correct functional group X and the reagent/reaction conditions Y in the scheme given below? 

(A) X = COOCH3, Y = H2/Ni/heat

(B) X = CONH2, Y = H2/Ni/heat

(C) X = CONH2, Y = Br2/NaOH

(D) X = CN, Y = H2/Ni/heat

Answer: A,B,C,D

Solution:

Condensation polymers are produced through the condensation of diamine or diols with dicarboxylic acids.

Therefore, X may be -COR, -CONH2, -CN

Q2: 

a) What is the structure of nylon-6 which is made through alkaline polymerisation of caprolactam?

b) Suggest a mechanism for this process. Is the polymerisation of the chain-reaction or step-reaction type?

Solution: 

The reaction is an anionic chain-reaction polymerization. It involves nucleophilic substitution in the acyl group of the cyclic amide. The base can be OH–or the anion made through abstraction of the –NH proton from one molecule of lactam.

Q3: Identify the compound C. 

Solution:

Solved Examples of Biomolecules 

Q1: 

a) Give structures for H through K. Given,

b) Explain the last step.

(c) What is net structural change?

(d) Name this overall method.

(e) What is the possibility of epimer formation?

Solution:

a) H is an oxime HOCH2(CHOH)4CH = NOH; I is the fully acetylated oxime, AcOCH2(CHOAc)4CH = NOAc which loses 1 mole of HOAc to give J, AcOCH2(CHOAc)4 CºN; K is an aldopentose, HOCH2(CHOH)3CHO.

b) The acetates go through transesterification to produce methyl acetate. This frees up all the sugar OH’s. After this, the reversal of HCN addition happens. 

c) From the carbon chain, there is a loss of one C.

d) Wohl degradation.

e) The a-CHOH turns into –CH = O with no configurational changes of the other chiral carbons. Therefore, no epimers get formed.

Q2: 

a) How can aldohexose be used to synthesize 2-ketohexose?

(b) As glucose gets converted to fructose through this process, what can be said about the configurations of C3, C4 , and C5 in the sugars?

Solution:

 a)

Aldohexose reacts with one molecule of phenylhydrazine that condenses with the aldehyde group to produce phenylhydrazone. When warmed with a big quantity of phenylhydrazine, the secondary alcoholic group which is adjacent to the aldehyde group gets oxidised to a ketonic group by another molecule of phenylhydrazine. The third molecule of phenylhydrazine condenses to produce osazone with this ketonic group. The phenylhydrazine group gets shifted from osazone to C6H5CHO and gives C6H5CH = N×NHC6H5 and a dicarbonyl compound known as an osone. The more reactive aldehyde group of the osone gets reduced and produces the 2-ketohexose. 

b) The configurations of the carbons that do not change in the reactions should be identical to be able to get the same osazone. 

Q3: Why should the isoelectric point for Aspartic acid (2.98) be much lower than that of leucine?

Solution: This could be explained by considering the ion equilibria given below - 

It is evident that the ions (A) and (B) are neutral, while (C) is a cation and (D) is dianion. In the species (D), the anion is found from the second —COOH group that exists in aspartic acid and is not possible in leucine. At neutral pH, a sizeable concentration of (D) will exist in the aqueous solution. It will hence be necessary to reduce the pH of such a solution when the formation of (D) has to be suppressed to a stage where the anions and cations exist in equal concentration (the isoelectric point).

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Tricks to Solve Questions from Polymers and Biomolecules 

1. As these are mostly theory-based, it is highly recommended that you draw all the diagrams and equations of every concept multiple times. 

2. Attempt questions from within a particular text and those after the chapters.

3. Focus primarily on preparing from NCERT books because most of the questions asked in JEE Main Chemistry are from them.

4. While attempting MCQs, take a moment to mentally visualize the concepts, diagrams, and equations you have learnt that pertain to a specific question. Then, choose the correct answer. 

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