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).
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :