Question

List-I contains reaction and List-II contains major products.
Match each reaction in List-I with one or more products in List-II and choose the correct option.

(a)- P$\to $ 1, 5; Q $\to $ 2; R $\to $ 3; S $\to $ 4
(b)- P$\to $ 1, 4; Q $\to $ 2; R $\to $ 4; S $\to $ 3
(c)- P$\to $ 1, 4; Q $\to $ 1, 2; R $\to $ 3; S $\to $ 4
(d)- P$\to $ 4, 5; Q $\to $ 4; R $\to $ 4; S $\to $ 3, 4

Answer 1

Hint: If the reactants contain bulky bases, then the reaction will be ${{E}_{2}}$ an elimination reaction in which an alkene will be formed and if the base is not bulky, then there will be a substitution reaction.

Complete answer: In reaction P, the two reactants are sodium tertiary butoxide and tert butyl bromide. The sodium tertiary butoxide is a bulky base, so it will attack the carbon atom that has the least steric hindrance. So, it will attack the acidic hydrogen from tertiary butyl bromide and ${{E}_{2}}$ a reaction will occur. Now, there will be a negative charge on the carbon atom, and this will release the bromine and will form alkene as one product and alcohol. The reaction is given below:

In Q, the reactants are ether and HBr, the hydrogen ion of the HBr will attack the ether. Now, the tertiary group will be released forming a carbocation, on this carbocation, the bromide ion will attack, and for tertiary butyl bromide. The reaction is given below:

In R, the base is methoxide, so it will attack the acidic hydrogen from tertiary butyl bromide and ${{E}_{2}}$ reaction will occur. Now, there will be a negative charge on the carbon atom, and this will release the bromine and will form an alkene. The reaction is given below:

In S, the base is bulky but the MeBr is a small group due to which elimination will not occur instead the Sodium will be substituted with the methyl group. The reaction is given below:

Therefore, the correct answer is option (b)- P$\to $ 1, 4; Q $\to $ 2; R $\to $ 4; S $\to $ 3.

Note: In Q, the bond of Carbon and tertiary carbocation breaks instead of the bond between carbon and methyl because tertiary carbocation is more stable than primary carbocation.