JEE Advanced Mix Test 47

Solution:   

To calculate the gravitational force on the particle inside the spherical shell, we divide the shell into small rings (or differential mass elements) and then sum up the forces contributed by each ring. 

The gravitational force on a small section of the shell (a ring) at angle θ, which subtends a small angle dθ, can be expressed by:

Here:

    G is the gravitational constant.

    m is the mass of the particle.

    ρ  is the mass density of the shell.

    dV is the volume element of the shell.

    r' is the distance from the particle to the differential mass element.

    α  is the angle between the force vector and the line connecting the center of the shell to the particle.

The angle α is the angle between the force vector and the line connecting the center of the shell to the particle. From the geometry, we can express cos α as:

To get the total gravitational force on the particle, we integrate the differential force dF over the angle θ from 0 to \pi , considering the contributions of all the rings in the shell:

CONCEPT:

Semi-Pinacol – Pinacolone Rearrangement

• The reaction shown is a semi-pinacol – pinacolone type rearrangement, where a hydroxyl group (-OH) undergoes a reaction with nitrous acid (HNO₂), followed by heating (Δ), leading to the formation of a carbocation and its rearrangement by ring expansion.
• The reaction is a ring expansion after the formation of a carbocation. The intermediate carbocation rearranges to form a more stable product, which involves the migration of the substituent group and ring expansion.
• The overall process involves the conversion of the starting material into the product by expanding the ring and eliminating a proton (H⁺).

EXPLANATION:

• Step 1: The reaction is initiated by the reaction with nitrous acid (HNO₂), which causes the formation of a carbocation. This intermediate is formed as a result of a nucleophilic attack leading to a positively charged species.
• Step 2: The carbocation undergoes rearrangement, with the ring expanding to stabilize the positive charge. This results in the formation of a new structure, leading to the final product.
• Step 3: A proton (H⁺) is eliminated from the intermediate, forming the final product as shown in the mechanism. This product has a carbonyl group (-C=O) at the site of the original hydroxyl group, resulting in a ketone as the final product.

Therefore, the correct product P is the structure with a ketone group (Option 3) based on the rearranged intermediate.

CONCEPT:

Nucleophilic Substitution and Aromatic Amine Formation

• The reaction involves the nucleophilic substitution of a chlorine atom on a methyl group-substituted phenyl ring by an amine group (NH₂) in the presence of sodium amide (NaNH₂), followed by the addition of ammonia (NH₃).
• The nucleophilic substitution proceeds through an S_NAr mechanism due to the presence of an electron-withdrawing group (-OCH₃) on the aromatic ring, which makes the ring more susceptible to attack by the nucleophile (NH₂).
• In the reaction, the chlorine atom is replaced by the NH₂ group at the position ortho to the methoxy group (-OCH₃) due to the electron-withdrawing nature of the methoxy group, which favors the attack at the ortho position.

• Step 1: The NaNH₂ base deprotonates the amine, generating a strong nucleophile (NH₂⁻), which attacks the aromatic ring where the chlorine atom is present.
• Step 2: The methoxy group (-OCH₃) directs the nucleophilic substitution to the ortho position, resulting in the displacement of chlorine and the attachment of the NH₂ group.
• Step 3: The addition of ammonia (NH₃) in the final step ensures the amine (-NH₂) remains in the product at the ortho position relative to the methoxy group.

Therefore, the correct product is the compound shown in option A,

• The total pressure of the system is the sum of the partial pressures of all gases present taking into account the stoichiometry of the reaction.

Explanation:

Given data:

• t_1/2 = 30 minutes
• The initial pressure of A, P_A = 800 torr
• Time elapsed = 60 minutes

The rate constant (k) for the reaction:

For xA(g) → yB(g) + zC(g) in general equation( at reactant side, there must be single reactant):

Hence the total pressure of the system after 60 minutes is 1400 torr.

Calculation