JEE Chemistry - MCQ Practice Questions

For first-order: k = 0.693/t₁/₂ = 0.$\frac{693}{30}$ = 0.0231 min⁻¹ ≈ 0.023 min⁻¹

If doubling concentration increases rate by 4 times (2²), reaction is second order

Rate constant k is independent of reactant concentration; it depends on T, nature of reactants, and catalyst

k = A·exp(-Eₐ/RT) = 2 × 10¹³ × exp(-$\frac{50000}{8}$.314×300) = 2 × 10¹³ × exp(-20.03) ≈ 3.4 × 10⁻³ s⁻¹

For elementary reactions, rate law exponents = stoichiometric coefficients. Rate = k[A]²[B]

Catalyst lowers Eₐ (forward and reverse) without changing ΔH, reaction enthalpy change

1/[A] = 2 + 0.4(5) = 2 + 2 = 4; [A] = $\frac{1}{4}$ = 0.25 M

Adding all steps and canceling intermediates (Cl, H): Cl₂ + H₂ → 2HCl

For second-order: t₁/₂ = 1/(k[A]₀); 100 = 1/(k × 0.5); k = 0.04 M⁻¹s⁻¹

Using ln(k₂/k₁) = (Eₐ/R)[(T₂-T₁)/(T₁T₂)]; ln(k₂/k₁) = ($\frac{60000}{8}$.314)[(300)/(600×300)] ≈ 6.2; k₂/k₁ ≈ 500

Understanding:

We need to identify the correct statement about the $\alpha$-helix structure in proteins.

Step 1: Recall structural levels of proteins

Proteins have primary, secondary, tertiary, and quaternary structures. The $\alpha$-helix is a secondary structure.

Step 2: Identify the stabilizing force in $\alpha$-helix

The $\alpha$-helix is stabilized by intramolecular hydrogen bonds. These hydrogen bonds form between the $\text{N-H}$ group of one amino acid residue and the $\text{C=O}$ group of an amino acid four residues ahead in the sequence (not side chains).

Step 3: Eliminate incorrect options

Answer:

The $\alpha$-helix is a secondary structure stabilized by hydrogen bonds between the $\text{N-H}$ of one peptide bond and the $\text{C=O}$ of the peptide bond four residues earlier.

Quick Tip:

Always distinguish: hydrogen bonds → secondary structure ($\alpha$-helix, $\beta$-sheet); disulfide bridges → tertiary structure.

Understanding:

We need to identify the anomeric carbon and the position of the $\text{-OH}$ group in $\alpha$-$\text{D}$-glucose.

Step 1: Define the anomeric carbon

The anomeric carbon is the carbon that becomes a new chiral centre during ring closure (hemiacetal/hemiketal formation). In glucose (a pyranose), this is ${\text{C}}_1$ — the former aldehyde carbon.

Step 2: Distinguish $\alpha$ and $\beta$ forms

In the Haworth projection of $\text{D}$-glucose:

Step 3: Apply to the question

In $\alpha$-$\text{D}$-glucose, the $\text{-OH}$ at ${\text{C}}_1$ (the anomeric carbon) is in the axial position in the chair conformation.

Answer:

The anomeric carbon in $\alpha$-$\text{D}$-glucose is ${\text{C}}_1$ with $\text{-OH}$ in the axial position.

Quick Tip:

"$\alpha$ is axial" — a quick mnemonic to remember that in $\alpha$-$\text{D}$-glucose the $\text{-OH}$ at ${\text{C}}_1$ is axial.

Understanding:

We need to find a disaccharide that:

(i) hydrolyzes to give two glucose molecules, and

(ii) is a reducing sugar.

Step 1: Analyze each option

Step 2: Identify the correct answer

Only Maltose satisfies both conditions: it gives two glucose molecules on hydrolysis and is a reducing sugar.

Answer:

Maltose gives two glucose molecules on hydrolysis and is a reducing sugar due to its free anomeric ${\text{C}}_1$.

Quick Tip:

A disaccharide is a reducing sugar only if at least one anomeric carbon (${\text{C}}_1$ in aldoses) is free (not involved in the glycosidic bond).

Understanding:

We need to identify which biomolecule does NOT reduce Fehling's solution (i.e., is a non-reducing sugar).

Step 1: Principle of Fehling's test

Fehling's solution contains ${\text{Cu}}^{2+}$ ions (as a tartrate complex). Reducing sugars (those with a free aldehyde or ketone group, or a free anomeric carbon) reduce ${\text{Cu}}^{2+}$ to ${\text{Cu}}_2\text{O}$ (brick-red precipitate).

Step 2: Classify each option

Step 3: Conclusion

Sucrose does not reduce Fehling's solution because both anomeric carbons are involved in the $\alpha ,\beta$-1,2-glycosidic linkage.

Answer:

Sucrose is the only non-reducing sugar among the options and will NOT give a positive Fehling's test.

Quick Tip:

Sucrose is the classic example of a non-reducing disaccharide. It does not mutarotate, does not reduce Tollens' or Fehling's reagent.

Understanding:

We need to match each vitamin correctly with the disease caused by its deficiency.

Step 1: Recall vitamin-deficiency disease pairs

Step 2: Check each option

Answer:

Vitamin $\text{C}$ (Ascorbic acid) deficiency causes Scurvy.

Quick Tip:

A useful mnemonic: "${\text{B}}_1$ Beriberi, ${\text{B}}_3$ Pellagra, $\text{C}$ Scurvy, $\text{D}$ Rickets, $\text{A}$ Night-blindness" — memorize in this order.

Understanding:

We need to identify which of the given vitamins is fat-soluble.

Step 1: Classify vitamins by solubility

Vitamins are classified into two groups:

Step 2: Analyze each option

Answer:

Vitamin $\text{K}$ is a fat-soluble vitamin.

Quick Tip:

Remember "ADEK" — the four fat-soluble vitamins. All other common vitamins (B-complex and C) are water-soluble.

Understanding:

We need to calculate the isoelectric point (pI) of glycine.

Formula:

For a simple amino acid with one amino and one carboxyl group, the isoelectric point is:

Step 1: Substitute the values

Step 2: Interpret the result

At $\text{pH}=5.97$, glycine carries equal positive and negative charges (exists as ${\text{H}}_3{\text{N}}^{+}{\text{-CH}}_2{\text{-COO}}^{-}$, the zwitterion), so its net charge is zero.

Answer:

The isoelectric point of glycine is $5.97$.

Quick Tip:

For acidic amino acids (e.g., aspartic acid), use the two acidic $\text{p}{K}_a$ values; for basic amino acids (e.g., lysine), use the two basic $\text{p}{K}_a$ values to calculate pI.

Understanding:

We need to identify the INCORRECT statement about the DNA double helix structure.

Step 1: Verify each statement

Step 2: Analyze option D

The backbone of DNA is made of alternating deoxyribose sugar and phosphate groups. These are connected by $3^{\prime },5^{\prime }$-phosphodiester bonds (not $\text{N}$-glycosidic bonds). The $\text{N}$-glycosidic bond connects the nitrogenous base to the sugar (within a nucleotide), not the backbone linkage between nucleotides.

Step 3: Conclusion

Option D is incorrect because the backbone is held together by phosphodiester bonds, not $\text{N}$-glycosidic bonds.

Answer:

The statement that the backbone of DNA is held by $\text{N}$-glycosidic bonds is incorrect; the backbone uses phosphodiester bonds.

Quick Tip:

N-glycosidic bonds attach bases to sugars (within nucleosides); phosphodiester bonds link nucleotides together in the backbone.

Understanding:

We need to identify the key structural difference between cellulose and starch.

Step 1: Structure of starch

Starch exists in two forms:

Step 2: Structure of cellulose

Cellulose is a linear polymer of $\text{D}$-glucose units connected by $\beta (1\to 4)$ glycosidic linkages. The $\beta$ configuration allows extended straight chains that can align and form strong intermolecular hydrogen bonds, giving cellulose its fibrous, rigid nature.

Step 3: Why the linkage type matters

The $\alpha$ linkage in starch creates a helical structure (digestible by humans). The $\beta$ linkage in cellulose creates straight chains (not digestible by humans, as we lack the enzyme $\beta$-glucosidase/cellulase).

Answer:

Starch contains $\alpha (1\to 4)$ glycosidic linkages while cellulose contains $\beta (1\to 4)$ linkages.

Quick Tip:

"$\alpha$ links make starch (edible); $\beta$ links make cellulose (structural)." This single difference explains all digestibility and physical property contrasts.

Understanding:

We need to identify the correct description of an enzyme as a biochemical catalyst.

Step 1: Fundamental properties of enzymes

Step 2: Eliminate incorrect options

Step 3: Confirm correct option

Option B correctly states that enzymes lower activation energy and are substrate-specific globular proteins.

Answer:

Enzymes lower the activation energy and are specific globular protein catalysts that are not consumed in the reaction.

Quick Tip:

A catalyst (including an enzyme) can never change $\Delta G$ or $K_{eq}$ — it only provides an alternative pathway with lower $E_a$.