Chemistry Class 12 Biomolecules Notes for IIT JEE & NEET

Biomolecules is Unit 10 of 10 in the current NCERT Class 12 Chemistry book (it was Chapter 14 in older books), and it carries 7 marks in the CBSE theory paper. Most questions come directly from NCERT definitions, structures and tables. These Chemistry Class 12 Biomolecules Notes put the whole chapter in one place, with the reason behind each fact, and they show you the topics that overlap with Biology for NEET.

Chemistry Class 12 Biomolecules Notes: Chapter Overview and Syllabus

The chapter covers:

• Carbohydrates: classification (aldoses and ketoses), monosaccharides (glucose and fructose), D and L configuration, oligosaccharides (sucrose, lactose and maltose) and polysaccharides (starch, cellulose and glycogen)
• Proteins: α-amino acids, the peptide bond, polypeptides, primary, secondary, tertiary and quaternary structures, and denaturation
• Enzymes
• Vitamins: classification and functions
• Nucleic acids: chemical constituents of DNA and RNA, and their biological functions

Hormones are not part of the current NCERT chapter. Check the latest syllabi of CBSE, NTA (JEE Main) and NMC (NEET) for the exact topic lists.

NEET note: NEET asks questions on biomolecules from two textbooks, Class 12 Chemistry (this chapter) and Class 11 Biology (Biomolecules). The Biology chapter adds topics such as enzyme kinetics, so study both.

Tip from the eSaral Chemistry team: Make four tables from memory (carbohydrates, protein structure, DNA vs RNA, and vitamins with deficiency diseases), and revise them often. Most questions in this chapter can be answered from these four tables.

Chemistry Class 12 Biomolecules Notes

India's Best Exam Preparation for Class 12th - Download Now

India's Best Exam Preparation for Class 12th - Download Now

India's Best Exam Preparation for Class 12th - Download Now

India's Best Exam Preparation for Class 12th - Download Now

Carbohydrates: Classification and Structure of Glucose in Chemistry Class 12 Biomolecules Notes

Carbohydrates are optically active polyhydroxy aldehydes or ketones, or compounds that give them on hydrolysis.

Classification of Carbohydrates

Structure of Glucose in Chemistry Class 12 Biomolecules Notes

The molecular formula of glucose is $\mathrm{C_6H_{12}O_6}$. Its open-chain structure is $\mathrm{HOCH_2{-}(CHOH)_4{-}CHO}$, a pentahydroxy aldehyde. The evidence for this structure is:

The cyclic structure: glucose does not give Schiff’s test or the sodium hydrogen sulphite test, and its pentaacetate does not react with $\mathrm{NH_2OH}$. These facts show that the free aldehyde group is absent in the ring form. The $-\mathrm{OH}$ group at C-5 adds to the aldehyde group at C-1 and forms a six-membered pyranose ring (a hemiacetal). Fructose forms a five-membered furanose ring.

Anomers and mutarotation: ring closure makes C-1 a new chiral centre (the anomeric carbon), which gives two forms:

The spontaneous change in the specific rotation of a freshly prepared solution until it reaches $+52.5^\circ$ is called mutarotation. For the D and L configuration, the reference is glyceraldehyde: in D-glucose, the $-\mathrm{OH}$ on the last chiral carbon (C-5) is on the right side in the Fischer projection.

Reducing and Non-Reducing Sugars in Chemistry Class 12 Biomolecules Notes

A reducing sugar has a free aldehyde or a free hemiacetal (anomeric) group, so it reduces Fehling’s solution and Tollens’ reagent. Glucose, fructose, maltose, lactose and galactose are reducing. Sucrose is non-reducing, because the anomeric carbons of both glucose (C-1) and fructose (C-2) are used in the glycosidic bond.

Inversion of sucrose: sucrose is dextrorotatory ($+66.5^\circ$). On hydrolysis, it gives a mixture of glucose ($+52.5^\circ$) and fructose ($-92.4^\circ$) that is laevorotatory, so the process is called inversion, and the mixture is invert sugar:

$$\mathrm{C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^{+}} C_6H_{12}O_6\ (\text{glucose}) + C_6H_{12}O_6\ (\text{fructose})}$$

Polysaccharides

Humans cannot digest cellulose, because they lack the enzyme cellulase, which breaks the $\beta$-glycosidic links.

Proteins: Amino Acids, Structure and Denaturation in Chemistry Class 12 Biomolecules Notes

α-Amino Acids

An α-amino acid has an amino group and a carboxyl group on the same carbon:

$$\mathrm{H_2N{-}CHR{-}COOH}$$

Essential amino acids cannot be made by the body, so they must come from food. NCERT lists ten (arginine, histidine, isoleucine, leucine, lysine, methionine, phenylalanine, threonine, tryptophan and valine), while many references list eight for adults, because arginine and histidine are only conditionally essential. All natural α-amino acids except glycine are optically active, and the L form is the common one.

Zwitterion: in water, the proton moves from $-\mathrm{COOH}$ to $-\mathrm{NH_2}$ and gives a dipolar ion:

$$\mathrm{{}^{+}H_3N{-}CHR{-}COO^{-}}$$

Amino acids are amphoteric. At the isoelectric point (the pH at which the net charge is zero), the zwitterion predominates. For a neutral amino acid, $\mathrm{pI} = \dfrac{\mathrm{p}K_{a1} + \mathrm{p}K_{a2}}{2}$.

Peptide Bond and Structure of Proteins in Chemistry Class 12 Biomolecules Notes

A peptide bond ($-\mathrm{CO{-}NH}-$) forms by the loss of water between the $-\mathrm{COOH}$ of one amino acid and the $-\mathrm{NH_2}$ of the next. A chain of $n$ amino acids has $n - 1$ peptide bonds.

Fibrous proteins (keratin, myosin) are insoluble in water, and globular proteins (insulin, albumin) are soluble in water.

Denaturation is the loss of the secondary, tertiary and quaternary structures, with the peptide bonds and the primary structure intact. Heat, a change in pH, heavy metal salts and urea cause it. Boiling an egg is an example, and it is irreversible.

Enzymes in Chemistry Class 12 Biomolecules Notes

Enzymes are biocatalysts, and most are globular proteins. They lower the activation energy of a reaction. The general scheme is:

$$\mathrm{E + S \rightleftharpoons ES \longrightarrow E + P}$$

where E is the enzyme, S the substrate, ES the enzyme–substrate complex and P the product.

Properties: enzymes are highly specific, they work at an optimum temperature and pH (pepsin is at about pH 2 and trypsin at about pH 8), and they are not used up in the reaction. Inhibitors reduce their activity. A competitive inhibitor resembles the substrate and competes for the active site.

Vitamins in Chemistry Class 12 Biomolecules Notes

Vitamins are organic compounds that the body needs in small amounts. Fat-soluble vitamins are A, D, E and K, and they are stored in the body. Water-soluble vitamins are the B group and C, and they must be supplied regularly.

Nucleic Acids: DNA and RNA in Chemistry Class 12 Biomolecules Notes

A nucleoside is a base joined to a sugar, and a nucleotide is a nucleoside joined to a phosphate group. Nucleotides join through phosphodiester linkages to form the nucleic acid chain.

The purine bases are adenine and guanine (two rings), and the pyrimidine bases are cytosine, thymine and uracil (one ring). Chargaff’s rules for DNA: $[\mathrm{A}] = [\mathrm{T}]$ and $[\mathrm{G}] = [\mathrm{C}]$, so purines equal pyrimidines.

Key Facts in Chemistry Class 12 Biomolecules Notes

Solved Examples from Chemistry Class 12 Biomolecules Notes

Example 1 (invert sugar): The specific rotations of glucose and fructose are $+52.5^\circ$ and $-92.4^\circ$. Find the rotation of an equimolar mixture of the two.

$\dfrac{52.5 + (-92.4)}{2} \approx -20^\circ$, so the mixture is laevorotatory, as in invert sugar.

Example 2 (isoelectric point): For glycine, $\mathrm{p}K_{a1} = 2.34$ and $\mathrm{p}K_{a2} = 9.60$. Find the isoelectric point.

$\mathrm{pI} = \dfrac{2.34 + 9.60}{2} = 5.97$.

Example 3 (Chargaff’s rules): A DNA sample has 30% adenine. Find the percentage of the other bases.

$\mathrm{T} = 30\%$, so $\mathrm{G} + \mathrm{C} = 40\%$, which gives $\mathrm{G} = \mathrm{C} = 20\%$.

Example 4 (hydrogen bonds): A DNA fragment has 100 base pairs, of which 40 are G and C pairs. Find the number of hydrogen bonds.

$40 \times 3 + 60 \times 2 = 120 + 120 = 240$.

Example 5 (peptide bonds): A protein has 4 polypeptide chains with a total of 100 amino acids. How many peptide bonds are there?

Number of peptide bonds = $100 - 4 = 96$.

Example 6 (reducing sugars): Which of sucrose, maltose and lactose does not reduce Tollens’ reagent? Name the products of hydrolysis of each.

Sucrose does not, because it has no free anomeric group. Sucrose gives glucose and fructose, maltose gives two glucose molecules, and lactose gives glucose and galactose.

Example 7 (stereoisomers): How many stereoisomers does open-chain glucose have?

Glucose has 4 chiral carbons, so there are $2^4 = 16$ stereoisomers.

Common Mistakes in Chemistry Class 12 Biomolecules Notes Problems

How to Study Chemistry Class 12 Biomolecules Notes

1. Read the NCERT chapter first, and download the textbook from eSaral’s NCERT books for Class 12 page.
2. Make the four tables (carbohydrates, proteins, DNA and RNA, vitamins) from memory, and add the evidence table for glucose.
3. Learn the structure of glucose with eSaral’s chemical formula of glucose page, and solve the numericals on this page.
4. Revise with eSaral’s Class 12 Chemistry notes, and practise with eSaral’s Biomolecules NEET previous year questions and the NEET chapter-wise PYQs.