Predict the major product $ P $ in the following sequence of reactions:
(i) HBr, benzoyl peroxide
(ii) KCN
(iii) Na(Hg), $C_{2}H_{5}OH$
The sequence of reactions involves the following steps:
Step (i) HBr, benzoyl peroxide: This is a free radical halogenation reaction, where the alkyl group on the benzene ring undergoes bromination at the benzylic position due to the formation of free radicals, resulting in the product \( C_6H_5-CH_2Br \) (benzyl bromide).
Step (ii) KCN: The next step is the nucleophilic substitution of the bromine atom by the cyanide ion (CN-), leading to the formation of \( C_6H_5-CH_2-CN \) (benzyl cyanide).
Step (iii) Na(Hg), C2H5OH: This is a reduction reaction (Clemmensen reduction), which typically reduces a carbonyl group (in this case, the nitrile group) to a methylene group (–CH2–). However, since there is no carbonyl group here, this reaction doesn't affect the cyanide group. Hence, the major product remains as \(C_6H_5-CH_2-CN \).
Thus, the major product is \( C_6H_5-CH_2-CN \), and the correct answer is (3).
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).


Consider a water tank shown in the figure. It has one wall at \(x = L\) and can be taken to be very wide in the z direction. When filled with a liquid of surface tension \(S\) and density \( \rho \), the liquid surface makes angle \( \theta_0 \) (\( \theta_0 < < 1 \)) with the x-axis at \(x = L\). If \(y(x)\) is the height of the surface then the equation for \(y(x)\) is: (take \(g\) as the acceleration due to gravity) 
A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is : 