Q.Give the plausible explanation for the following:

Concept: Structural Chemistry of Biomolecules

These questions probe the relationship between molecular structure and chemical behavior in carbohydrates and nucleic acids.

1. Glucose and the 2,4-DNP test The 2,4-dinitrophenylhydrazine (2,4-DNP) test detects free carbonyl groups (aldehydes or ketones). In aqueous solution, glucose exists predominantly (~99%) in the cyclic hemiacetal form (pyranose ring), where the aldehyde group at C-1 has reacted intramolecularly with the C-5 hydroxyl. Only a trace amount exists in the open-chain form with a free $\ce{-CHO}$ group. This equilibrium concentration is too low to give a visible precipitate with 2,4-DNP under standard test conditions, so glucose appears negative.
2. DNA strand complementarity DNA's two strands run antiparallel (one 5'→3', the other 3'→5') and are held together by specific base pairing: adenine pairs with thymine (A–T, two H-bonds) and guanine pairs with cytosine (G–C, three H-bonds). If one strand has sequence 5'-ATGC-3', the complementary strand must be 3'-TACG-5'. The strands are not identical because they carry different base sequences, but they are complementary because each base on one strand dictates its partner on the other, enabling faithful replication and information storage.
3. Starch vs. cellulose structure …

(a) Why glucose doesn't give the 2,4-DNP test

The 2,4-dinitrophenylhydrazine (2,4-DNP) test is a classic carbonyl test. It detects free aldehyde or ketone groups by forming a yellow-to-orange precipitate (a hydrazone). The question is: glucose is an aldehyde sugar (an aldohexose), so why doesn't it respond?

The answer lies in structure, not classification. In aqueous solution, glucose does not exist as the open-chain aldehyde form. Instead, it undergoes intramolecular cyclization: the hydroxyl group on C-5 attacks the carbonyl carbon (C-1), forming a six-membered ring called a pyranose (or occasionally a five-membered furanose). This cyclic form is a hemiacetal, not an aldehyde.

At equilibrium in water, more than 99% of glucose molecules are in the cyclic form. The tiny fraction of open-chain aldehyde present at any moment is insufficient to give a positive 2,4-DNP test under normal conditions. The reagent requires a free carbonyl group to attack, and the hemiacetal oxygen occupies that reactive site.

(b) Why the two DNA strands are complementary, not identical

DNA's double helix consists of two polynucleotide strands running antiparallel (one 5'→3', the other 3'→5'). If the strands were identical, they would have the same sequence of bases read in the same direction—but that would violate the structural rules of base pairing.

The key concept is complementary base pairing through hydrogen bonds:

These pairs are dictated by:

1. Geometry: A purine (A or G, two-ring) must pair with a pyrimidine (T or C, one-ring) to maintain constant width across the helix (about 2 nm).
2. Hydrogen bonding: A and T form two H-bonds; G and C form three. No other pairing (A–C, G–T) provides the correct number or geometry of hydrogen bonds.

If one strand reads 5'-ATGC-3', the complementary strand must read 3'-TACG-5' (or equivalently 5'-GCAT-3' when flipped). The sequences are not the same—they are determined by each other. This complementarity is the basis of DNA replication: each strand serves as a template for synthesizing its partner.

(c) Why starch and cellulose differ despite both being glucose polymers

Both starch and cellulose are polysaccharides made exclusively of D-glucose monomers, yet their properties could not be more different: starch is digestible by humans and forms granules; cellulose is indigestible and forms tough plant fibers. The difference is entirely due to the type of glycosidic linkage.

Starch: α-glycosidic linkages

Starch consists of two components:

• Amylose: unbranched chains of glucose linked by α(1→4) glycosidic bonds.
• Amylopectin: branched chains with α(1→4) bonds in the linear segments and α(1→6) bonds at branch points. …