Which compound in each of the following pairs will react faster in SN2 reaction with –OH?
(i)CH3Br or CH3I
(ii)(CH3)3CCl or CH3Cl
(i) In the SN2 mechanism, the reactivity of halides for the same alkyl group increases in the order.This happens because as the size increases,the halide ion becomes a better-leaving group.R−F<<R−Cl< R−Br<R−I Therefore,CH3I will react faster than CH3Br in SN2 reactions with OH−.
(ii)
The SN2 mechanism involves the attack of the nucleophile at the atom bearing the leaving group. But, in the case of (CH3)3CCl,the attack of the nucleophile at the carbon atom is hindered because of the presence of bulky substituents on that carbon atom bearing the leaving group. On the other hand, there are no bulky substituents on the carbon atom bearing the leaving group in CH3Cl.Hence,CH3Cl reacts faster than(CH3)3CCl in SN2 reaction with OH−.
A compound (A) with molecular formula $C_4H_9I$ which is a primary alkyl halide, reacts with alcoholic KOH to give compound (B). Compound (B) reacts with HI to give (C) which is an isomer of (A). When (A) reacts with Na metal in the presence of dry ether, it gives a compound (D), C8H18, which is different from the compound formed when n-butyl iodide reacts with sodium. Write the structures of A, (B), (C) and (D) when (A) reacts with alcoholic KOH.

A ladder of fixed length \( h \) is to be placed along the wall such that it is free to move along the height of the wall.
Based upon the above information, answer the following questions:
(iii) (b) If the foot of the ladder, whose length is 5 m, is being pulled towards the wall such that the rate of decrease of distance \( y \) is \( 2 \, \text{m/s} \), then at what rate is the height on the wall \( x \) increasing when the foot of the ladder is 3 m away from the wall?