SOLUTIONS
2024 BOARD EXAM
CBSE CLASS 12-PCB CHEMISTRY Board Paper 2024 — Set 5
2024
CHEMISTRY
CLASS 12-PCB
CBSE EXAMINATION PAPER-2024
CHEMISTRY
(Solved)
General Instructions :
Read the following instructions carefully and follow them :
- This question paper contains 16 questions. All questions are compulsory.
- This question paper is divided into 4 sections.
- Section A – questions number 1 to 1 are case based questions
- Section B – questions number 2 to 13 are multiple choice questions
- Section C – questions number 14 to 15 are very short answer
- Section D – questions number 16 to 16 are short answer
- There is no overall choice given in the question paper. However, an internal choice has been provided in few questions.
- Use of calculator is NOT allowed.
Section A
Section B
The specific sequence in which amino acids are arranged in a protein is called its
The correct answer is 'Primary structure' because the primary structure of a protein refers to the specific sequence of amino acids that make up the protein chain. This sequence is crucial as it determines the protein's unique characteristics and function.
Out of the following alkenes, the one which will produce tertiary butyl alcohol on acid catalysed hydration is
The correct option is (CH₃)₂C=CH₂. This alkene, also known as isobutylene, upon acid-catalyzed hydration follows Markovnikov's rule. The hydrogen attaches to the less substituted carbon (the double bond), and the hydroxyl group attaches to the more substituted carbon, resulting in the formation of tertiary butyl alcohol.
Auto-oxidation of chloroform in air and light produces a poisonous gas known as
The correct option is Phosgene. When chloroform undergoes auto-oxidation in the presence of air and light, it can produce phosgene, which is a highly toxic gas. Other options like tear gas, mustard gas, and phosphine are not products of chloroform's auto-oxidation.
Transition metals are known to make interstitial compounds. Formation of interstitial compounds makes the transition metal
The correct option is 'more hard'. Interstitial compounds are formed when small atoms fill the interstitial spaces between the larger metal atoms. This can lead to an increase in hardness and strength of the material, making it more difficult to deform compared to the pure metal.
Isotonic solutions have the same
The correct option is 'osmotic pressure' because isotonic solutions are defined as solutions that have the same osmotic pressure, which means they exert equal pressure across a semipermeable membrane, preventing water movement in or out of the cells.
Which of the following cell was used in Apollo space programme?
The correct answer is H₂-O₂ fuel cell. The Apollo space program utilized hydrogen-oxygen fuel cells to produce electricity and water for the spacecraft, making them essential for the missions.
The rate of a reaction increases sixteen times when the concentration of the reactant increases four times. The order of the reaction is
The order of the reaction can be calculated using the rate law. If the concentration is increased by a factor of 4 and the rate increases by a factor of 16, we can use the equation Rate ∝ [A]^n. Here, 4^n = 16. Solving for n gives us n = 2, meaning the order of the reaction is 2.0.
Dilution affects both conductivity as well as molar conductivity. Effect of dilution on both is as follows
The correct option is 'conductivity decreases whereas molar conductivity increases on dilution.' When a solution is diluted, the concentration of ions decreases, which reduces conductivity. However, molar conductivity, which is the conductivity per unit concentration, increases because the ability of ions to move freely increases in a more dilute solution.
Assertion (A): Zr and Hf are of almost similar atomic radii.
Reason (R): This is due to Lanthanoid contraction.
Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of the Assertion (A). The atomic radii of zirconium (Zr) and hafnium (Hf) are similar due to the lanthanoid contraction, which decreases the size of the lanthanide ions and subsequently affects the size of the elements in the same group of the periodic table.
Assertion (A): The units of rate constant of a zero order reaction and rate of reaction are the same
Reason (R): In a zero order reaction, the rate of reaction is independent of the concentration of reactants.
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). In a zero order reaction, the rate is constant and the units of the rate constant (k) are the same as the units of rate of reaction (e.g., mol/L/s), but the independence of the rate from reactant concentration does not explain the equality of units.
Assertion (A): Inversion of configuration is observed in SN₂ reaction.
Reason (R): The reaction proceeds with the formation of a carbocation.
Assertion (A) is true, but Reason (R) is false. The S₂ reaction (bimolecular nucleophilic substitution) indeed leads to an inversion of configuration; however, it does not involve the formation of a carbocation, which is characteristic of S₁ reactions. Therefore, while the assertion about the S₂ reaction is correct, the reasoning does not support it.
Assertion (A): p-methoxyphenol is a stronger acid than p-nitrophenol.
Reason (R): Methoxy group shows +I effect whereas nitro group shows -I effect.
Assertion (A) is false, but Reason (R) is true. p-Nitrophenol is actually a stronger acid than p-methoxyphenol due to the electron-withdrawing -I effect of the nitro group, which stabilizes the negative charge of the phenoxide ion more effectively than the +I effect of the methoxy group destabilizes it.
Section C
Carry out the following conversions:
(i) Nitrobenzene to Aniline
(ii) Aniline to Phenol.
(ii) Aniline to Phenol: Aniline is first diazotized by reacting with NaNO2 and HCl at 0-5 deg C to form benzenediazonium chloride. Then, on warming with water, diazonium group is replaced by hydroxyl group, producing Phenol (C6H5OH).
Show that in case of a first order reaction, the time taken for completion of 99% reaction is twice the time required for 90% completion of the reaction. (log 10 = 1)
For a first order reaction, the integrated rate law is given as:
ln([R]_0/[R]) = kt
where [R]_0 is the initial concentration and [R] is the concentration at time t.
Let t_90 be the time for 90% completion, so 10% reactant remains:
[R] = 0.1 [R]_0
ln([R]_0/0.1 [R]_0) = k t_90
ln(10) = k t_90
Using log 10 = 1, ln(10) = 2.303 * 1 = 2.303
So, k t_90 = 2.303
Let t_99 be the time for 99% completion, so 1% reactant remains:
[R] = 0.01 [R]_0
ln([R]_0/0.01 [R]_0) = k t_99
ln(100) = k t_99
ln(100) = 2.303 * 2 = 4.606
So, k t_99 = 4.606
Dividing the two equations:
(k t_99) / (k t_90) = 4.606 / 2.303 = 2
Therefore, t_99 / t_90 = 2
Hence, the time taken for 99% completion is twice the time taken for 90% completion for a first order reaction.
Section D
Draw the structures of major product(s) in each of the following reactions
(b) Dinitration: When dinitration is carried out on 3-methylphenol under harsher conditions, two nitro groups attach preferentially at positions ortho and para to the -OH group since it is a strong activating group. Positions 2 and 4 (with respect to -OH) get nitrated leading to 2,4-dinitro-3-methylphenol as the major product. The methyl group at position 3 influences substitution pattern but the strong activation by -OH dominates.
In summary, mononitration mainly gives 2-nitro-3-methylphenol whereas dinitration forms 2,4-dinitro-3-methylphenol as the major products. The directing effect of -OH group determines substitution positions in these electrophilic aromatic substitution reactions. Structures can be drawn showing benzene ring with OH at position 1, methyl at position 3 and nitro groups at the substituted positions.
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