Biomolecules
Carbohydrates, Proteins, Nucleic Acids, Structural Chemistry & Solved PYQs
1. Carbohydrates and their Classification
Biomolecules are lifeless chemical molecules which combine in specific ways to produce life and control biological reactions. Important biomolecules include carbohydrates, proteins, nucleic acids, lipids, and enzymes.
Modern Definition of Carbohydrates: They are defined as optically active polyhydroxy aldehydes or polyhydroxy ketones, or the compounds which produce such units on hydrolysis.
Classification on the basis of Hydrolysis
- Monosaccharides: Simplest carbohydrates that cannot be hydrolyzed further into simpler units. (e.g., Glucose, Fructose, Ribose). They consist of 3 to 7 carbon atoms.
- Oligosaccharides: Carbohydrates that yield 2 to 10 monosaccharide units on hydrolysis.
- Disaccharides: Yield 2 units (e.g., Sucrose $\rightarrow$ Glucose + Fructose; Maltose $\rightarrow$ 2 Glucose).
- Trisaccharides: Yield 3 units (e.g., Raffinose).
- Polysaccharides: Yield a large number of monosaccharide units on hydrolysis. They are not sweet in taste and are hence called non-sugars. (e.g., Starch, Cellulose, Glycogen).
Reducing and Non-reducing Sugars
- Reducing Sugars: Carbohydrates that can reduce Tollens' reagent or Fehling's solution. They have a free (uncombined) aldehyde or ketone group. All monosaccharides and most disaccharides (like maltose and lactose) are reducing.
- Non-reducing Sugars: Do not reduce Tollens' or Fehling's reagents because their functional (reducing) groups are bonded together in a glycosidic linkage. (e.g., Sucrose).
2. Structure of Glucose (Aldohexose)
Glucose ($C_6H_{12}O_6$) is the most abundant organic compound on earth. The elucidation of its open-chain structure is based on the following chemical evidence:
1. Straight Chain of 6 Carbon Atoms
Prolonged heating of glucose with Hydrogen Iodide (HI) and red phosphorus yields n-hexane. This proves that all six carbon atoms are linked in a straight unbranched chain.
$CHO-(CHOH)_4-CH_2OH \xrightarrow{HI, \Delta} CH_3-CH_2-CH_2-CH_2-CH_2-CH_3$
2. Presence of a Carbonyl Group ($>C=O$)
Glucose reacts with hydroxylamine ($NH_2OH$) to form an oxime and adds a molecule of hydrogen cyanide ($HCN$) to form a cyanohydrin. These reactions confirm the presence of a carbonyl group.
3. Carbonyl Group is an Aldehyde ($-CHO$)
Glucose gets oxidized to a six-carbon carboxylic acid (Gluconic acid) on reaction with a mild oxidizing agent like bromine water. This proves that the carbonyl group is an aldehyde group present at the end of the chain.
$CHO-(CHOH)_4-CH_2OH \xrightarrow{Br_2 \text{ water}} COOH-(CHOH)_4-CH_2OH$
4. Presence of 5 Hydroxyl ($-OH$) Groups
Acetylation of glucose with acetic anhydride gives glucose pentaacetate, confirming the presence of five $-OH$ groups. Since glucose is stable, these five $-OH$ groups must be attached to different carbon atoms.
5. Presence of one Primary Alcoholic ($-CH_2OH$) Group
Oxidation of glucose or gluconic acid with strong oxidizing agent like dilute Nitric acid ($HNO_3$) yields a dicarboxylic acid called Saccharic acid. This indicates the presence of a primary alcoholic group.
Cyclic Structure of Glucose
The open-chain structure could not explain certain properties (e.g., glucose does not give Schiff's test or form addition product with $NaHSO_3$). It was found that glucose exists in two cyclic hemiacetal forms ($\alpha$-D-glucose and $\beta$-D-glucose) formed by intramolecular reaction between the $-OH$ group at C-5 and the $-CHO$ group at C-1. The cyclic structure is a 6-membered ring called Pyranose.
3. Disaccharides and Polysaccharides
Glycosidic Linkage
The two monosaccharide units in a disaccharide are joined together by an oxide linkage formed by the loss of a water molecule. Such a linkage between two monosaccharide units through an oxygen atom is called a glycosidic linkage.
| Disaccharide | Monosaccharide Units | Glycosidic Linkage Type | Reducing Nature |
|---|---|---|---|
| Sucrose (Cane sugar) | $\alpha$-D-Glucose + $\beta$-D-Fructose | C1 of Glucose and C2 of Fructose | Non-reducing (both reducing groups are involved in bond) |
| Maltose (Malt sugar) | $\alpha$-D-Glucose + $\alpha$-D-Glucose | C1 of one glucose and C4 of another | Reducing (free aldehyde at C1 of second unit) |
| Lactose (Milk sugar) | $\beta$-D-Galactose + $\beta$-D-Glucose | C1 of galactose and C4 of glucose | Reducing |
Polysaccharides
- Starch: Main storage polysaccharide in plants. It is a polymer of $\alpha$-glucose and consists of two components:
- Amylose: Water-soluble, linear unbranched polymer (15-20%). Linked by $\alpha$-(1$\rightarrow$4) glycosidic bonds.
- Amylopectin: Water-insoluble, highly branched polymer (80-85%). Has $\alpha$-(1$\rightarrow$4) linear links and $\alpha$-(1$\rightarrow$6) branching links.
- Cellulose: Predominant constituent of plant cell walls. It is a straight-chain polysaccharide composed only of $\beta$-D-glucose units joined by $\beta$-(1$\rightarrow$4) glycosidic linkages.
- Glycogen: "Animal starch", structurally similar to amylopectin but much more highly branched. Stored in liver, muscles, and brain.
4. Proteins and Amino Acids
Proteins are high molecular mass complex polyamides. They are polymers of $\alpha$-amino acids. All proteins contain C, H, O, N, and sometimes S.
$\alpha$-Amino Acids
These contain an amino group ($-NH_2$) and a carboxyl group ($-COOH$) attached to the same ($\alpha$) carbon atom. There are 20 naturally occurring amino acids (e.g., Glycine, Alanine).
- Essential Amino acids: Cannot be synthesized in the human body and must be supplied through diet (e.g., Valine, Leucine).
- Non-essential Amino acids: Can be synthesized by the human body (e.g., Glycine, Alanine).
Zwitterion (Dipolar Ion)
In aqueous solution, the carboxyl group ($-COOH$) of an amino acid loses a proton ($H^+$) and the amino group ($-NH_2$) accepts it. This internal acid-base reaction results in the formation of a dipolar ion called a Zwitterion.
$H_2N-CH(R)-COOH \rightleftharpoons H_3N^+-CH(R)-COO^-$
It is electrically neutral overall but contains both positive and negative charges. Amino acids show amphoteric behavior due to zwitterion formation.
Peptide Bond (Peptide Linkage)
Proteins are formed by joining amino acids together. The $-COOH$ group of one amino acid reacts with the $-NH_2$ group of another molecule, eliminating a water molecule to form an amide linkage. This $-CO-NH-$ linkage connecting two amino acids is called a peptide bond.
Structure of Proteins
- Primary Structure: The exact linear sequence of specific amino acids linked by peptide bonds in a polypeptide chain.
- Secondary Structure: The shape in which the long polypeptide chain exists due to hydrogen bonding. Common forms are $\alpha$-helix (right-handed coil) and $\beta$-pleated sheet.
- Tertiary Structure: The overall 3D folding of the polypeptide chain, stabilized by H-bonds, disulfide linkages ($-S-S-$), van der Waals forces, and electrostatic forces. Classifies proteins into fibrous and globular.
- Quaternary Structure: The spatial arrangement of two or more polypeptide chains (subunits) with respect to each other (e.g., Hemoglobin).
Denaturation of Proteins
When a native protein is subjected to a physical change (like change in temperature/heating) or a chemical change (like change in pH), the hydrogen bonds are disturbed. The globules unfold, helices get uncoiled, and the protein loses its biological activity. This is called denaturation. During denaturation, secondary and tertiary structures are destroyed, but the primary structure remains intact. (Example: Coagulation of egg white on boiling, curdling of milk).
5. Enzymes
Enzymes are biocatalysts that catalyze highly specific biochemical reactions in living organisms. Almost all enzymes are globular proteins.
- Specificity: Each enzyme catalyzes only one specific reaction. E.g., Maltase hydrolyzes only maltose.
- Mechanism (Lock and Key Model): The substrate molecule (key) fits exactly into the active site (lock) of the enzyme to form an enzyme-substrate complex. This lowers activation energy. The complex then breaks down to yield products and release the free enzyme.
6. Nucleic Acids (DNA and RNA)
Nucleic acids are long-chain polymers of nucleotides, responsible for the transmission of inherent characters from parents to offspring (heredity) and protein synthesis.
Composition of Nucleotides
A nucleotide is composed of three units:
- Pentose Sugar: $\beta$-D-ribose in RNA, and $\beta$-D-2-deoxyribose in DNA.
- Nitrogenous Base:
- Purines (double ring): Adenine (A), Guanine (G).
- Pyrimidines (single ring): Cytosine (C), Thymine (T, only in DNA), Uracil (U, only in RNA).
- Phosphate Group: Forms phosphodiester linkages connecting the 5'-carbon of one sugar to the 3'-carbon of the next.
Note: A molecule containing only the sugar and base is called a Nucleoside.
DNA Double Helix Model (Watson and Crick)
DNA consists of two right-handed antiparallel polynucleotide strands coiled around a common axis. The sugar-phosphate backbone lies on the outside, and bases project inward. The two strands are held together by specific hydrogen bonds between complementary base pairs:
- Adenine pairs with Thymine with 2 H-bonds (A = T).
- Guanine pairs with Cytosine with 3 H-bonds (G $\equiv$ C).
7. Solved Textbook Reasoning Questions
Question 1: Inversion of Sugar
Why is sucrose called invert sugar?
Answer:
Sucrose is dextrorotatory ($+66.5^\circ$) in nature. Upon hydrolysis with dilute acid or the enzyme invertase, it yields an equimolar mixture of D-(+)-glucose ($+52.5^\circ$) and D-(-)-fructose ($-92.4^\circ$). Because the laevorotation of fructose is greater than the dextrorotation of glucose, the resulting mixture becomes laevorotatory. The process involving the change in the sign of optical rotation from dextro to laevo during hydrolysis is called inversion, and the equimolar mixture formed is known as invert sugar.
Question 2: Amino Acid Solubility
Why do amino acids have high melting points and are soluble in water unlike typical haloalkanes or amines?
Answer:
In solid state and aqueous solution, amino acids exist predominantly as zwitterions (dipolar ions) formed by the transfer of a proton from the carboxyl group to the amino group. Because they exist as dipolar ions, strong electrostatic interactions (ionic bonding) exist between the molecules, making their crystal lattice very strong. Hence, they behave like ionic salts, exhibiting high melting points and high solubility in polar solvents like water.
8. Board PYQs with Complete Answers
Verified previous year questions from the Maharashtra State Board HSC Chemistry exams.
1 Mark Questions (VSA)
Q1. Define: Peptide bond. (March 2013, Oct 2018, March 2022)
Answer: The amide linkage ($-CO-NH-$) formed between the carboxyl group of one amino acid and the amino group of another amino acid with the elimination of a water molecule is called a peptide bond.
Q2. Name the base present in RNA but not in DNA. (March 2016, Oct 2021)
Answer: Uracil.
Q3. Give one example of a naturally occurring reducing sugar. (March 2017)
Answer: Glucose (or Maltose, Lactose).
2 Mark Questions (SA-I)
Q4. What is a zwitterion? Explain with an example. (March 2014, March 2019)
In aqueous solution, the acidic carboxyl group ($-COOH$) of an $\alpha$-amino acid can lose a proton, and the basic amino group ($-NH_2$) can accept a proton. This transfer forms a dipolar ion which is electrically neutral overall but contains both a positive and a negative charge. This is known as a zwitterion.
Example for Glycine: $H_2N-CH_2-COOH \rightleftharpoons H_3N^+-CH_2-COO^-$
Q5. Write a short note on denaturation of proteins. (Oct 2015, March 2020)
When a native protein is subjected to a physical change (like heating) or a chemical change (like adding acid/base to change pH), the hydrogen bonds stabilizing it are disturbed. Due to this, the globules unfold and helices uncoil, causing the protein to lose its biological activity. This phenomenon is called denaturation. During this process, secondary and tertiary structures are destroyed, but the primary structure remains intact. (Example: Coagulation of egg white).
3/4 Mark Questions (LA)
Q6. State the reactions of D-glucose with: a) Hydroxylamine ($NH_2OH$) b) Bromine water c) Hydrogen Iodide ($HI$) and Red Phosphorus. What does each reaction prove about the structure of glucose? (Oct 2014, March 2018, March 2023)
a) Reaction with Hydroxylamine ($NH_2OH$):
Glucose reacts to form glucose oxime.
Proves: The presence of a carbonyl group ($>C=O$).
b) Reaction with Bromine water:
Glucose is oxidized to a 6-carbon carboxylic acid, Gluconic acid.
Proves: The carbonyl group is an aldehyde ($-CHO$) present at the end of the chain.
c) Reaction with HI and Red P:
Prolonged heating of glucose gives n-hexane.
Proves: All six carbon atoms are linked in a straight, unbranched chain.
Q7. Distinguish between DNA and RNA. (Write any 3 points). (March 2015, Oct 2021)
| Feature | DNA (Deoxyribonucleic acid) | RNA (Ribonucleic acid) |
|---|---|---|
| 1. Sugar present | $\beta$-D-2-deoxyribose | $\beta$-D-ribose |
| 2. Nitrogenous Bases | Adenine, Guanine, Cytosine, Thymine | Adenine, Guanine, Cytosine, Uracil |
| 3. Structure | Double-stranded $\alpha$-helix | Single-stranded (may fold back on itself) |
| 4. Function | Controls heredity and transmits genetic information. | Controls protein synthesis. |
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