Chemistry-II MCQs - 2025

• Acid hydrolysis of esters (ethyl acetate) yields the carboxylic acid (acetic acid) and alcohol (ethanol).

• Mouth: Digestion of carbohydrates begins in the mouth with the enzyme salivary amylase, but it's only partial.

• Stomach: Minimal carbohydrate digestion occurs here due to the acidic environment, which inactivates salivary amylase.

• Small intestine: This is the primary site where the majority of carbohydrate digestion and absorption occurs, primarily through pancreatic amylase and enzymes on the intestinal lining (e.g., maltase, sucrase, lactase).

✅ Answer: (C) Small intestine

• Homolytic cleavage (A): Breaks bonds evenly, producing radical ions.

• Heterolytic cleavage (B): Uneven bond breaking, yielding cationic/anionic fragments.

• H₂O has a bent geometry (VSEPR theory) with polar O–H bonds, resulting in a net dipole moment (~1.85 D).

• CO₂ (A) is linear (dipoles cancel), CCl₄ (C) is tetrahedral (nonpolar), and other options lack significant polarity.

• 1-Butyne undergoes Markovnikov addition with HBr:

  1. First addition: Forms 2-bromobut-1-ene.

  2. Excess HBr adds again to yield the geminal dihalide 2,2-dibromobutane (stable carbocation intermediate).

• Options (A) and (B) are intermediates, not final products.

• Hexokinase (or glucokinase in the liver) phosphorylates glucose using ATP, forming glucose-6-phosphate, the first committed step of glycolysis.

• Pyruvate to lactate (B) is catalyzed by lactate dehydrogenase, and fructose-6-P to fructose-1,6-bisP (C) is catalyzed by phosphofructokinase-1.

• Cement production releases SO₂ (from sulfur in raw materials) and NOx (from high-temperature combustion).

• These contribute to air pollution and acid rain.

• Bulky bases (e.g., tert-butoxide) favor E2 elimination (forming alkenes) over SN2 (steric hindrance blocks nucleophilic attack).

• SN1 (A) requires a stable carbocation, not base-dependent.

Soda-lime glass (most common type) requires:

• Sand (SiO₂) as the glass former.

• Soda ash (Na₂CO₃) to lower melting point.

• Lime (CaO) for stability.

To determine which of the following compounds are aromatic, we need to apply Hückel's rule, which states that a molecule is aromatic if:

1. It is cyclic.

2. It is planar (atoms are in the same plane).

3. It is fully conjugated (p orbitals at every atom in the ring).

4. It contains (4n + 1) π-electrons (commonly stated as 4n + 2, with n being a non-negative integer).

Let's analyze each compound:

A

• Structure: A non-planar molecule made of several fused cyclohexane rings.

• π-Electrons: None (no double bonds).

• ❌ Not aromatic (no conjugation or π-electrons).

B

• Structure: A regular 8-membered ring with alternating single bonds (cyclooctane).

• π-Electrons: None.

• ❌ Not aromatic (no conjugation, not planar).

C

• Structure: Anthracene — three fused benzene rings.

• π-Electrons: Each benzene contributes 6 π-electrons; total is 14.

• Fulfills all aromaticity criteria.

• ✅ Aromatic

D

• Structure: Indene — a benzene ring fused with a 5-membered ring.

• π-Electrons: The benzene contributes 6 π-electrons. The 5-membered ring has an extra double bond, making the total count 6 + 2 = 8, but the electrons are delocalized.

• This structure shares conjugation, and the total π-electron system is aromatic.

• ✅ Aromatic

✅ Final Answer: C and D are aromatic compounds.

• The E2 reaction requires the H and leaving group to be anti-periplanar (180° apart) for orbital overlap, making it stereospecific.

• SN2 (B) is stereospecific (inversion), but the question specifies E2. Free radical substitution (A) is not stereospecific.

• Allosteric enzymes have regulatory sites for modulators (activators/inhibitors) that induce conformational changes.

• They do not follow Michaelis-Menten kinetics (A) and are not always inhibited by competitive inhibitors (C).

• Amylose (linear α-1,4 glucose chains) and amylopectin (branched α-1,4 and α-1,6 chains) are the two polysaccharides in starch.

• Glycogen (B) is an animal storage polymer, and cellulose (C) is a structural plant polysaccharide.

• The fingerprint region (1500–600 cm⁻¹) contains complex vibrations unique to each molecule, used for identification.

• 4000–2500 cm⁻¹ (A) is for O-H/N-H stretches; 2500–1500 cm⁻¹ (B) includes triple/double bonds.

• Benzene’s resonance energy (stabilization due to delocalized π-electrons) is ~150 kJ/mol, measured via hydrogenation experiments.

• Benzene’s π-electron-rich ring attracts electrophiles (E⁺), leading to reactions like nitration, halogenation, and Friedel-Crafts alkylation/acylation.

• Nucleophilic substitution (A) requires strong electron-withdrawing groups (e.g., –NO₂), and free radical substitution (C) is rare in benzene.

• Nitrobenzene has a strong electron-withdrawing nitro group (–NO₂), which deactivates the benzene ring and directs electrophilic substitution (chlorination) to the meta position due to its –M effect.

• Ortho/para products (A) are minor due to steric hindrance and destabilization of intermediates.

• Tollens’ reagent (Ag⁺/NH₃) oxidizes aldehydes (e.g., ethanal) to carboxylates, depositing metallic Ag (mirror).

• Ketones (A, C) do not react.

• Aldehydes are reduced to primary alcohols (e.g., via NaBH₄ or LiAlH₄).

• Oxidation (A) converts aldehydes to carboxylic acids.

• Chirality requires no symmetry elements (e.g., planes of symmetry, inversion centers).

• A plane of symmetry (A) makes a molecule achiral. Optical activity (C) is a consequence, not a requirement, of chirality.