Question

2-Methyl propyl bromide reacts with C₂H₅O^- and gives 'A' whereas on reaction with C₂H₅OH it gives 'B'. The mechanism followed in these reactions and the products 'A' and 'B' respectively are

• A. S_N2. A iso-butyl ethyl ether; S_N1, B = tert- butyl ethyl ether
• B. S_N1, A tert-butyl ethyl ether; S_N2, B = iso- butyl ethyl ether
• C. S_N1, A = tert-butyl ethyl ether; S_N1, B = 2- butyl ethyl ether
• D. S_N2, A = 2-butyl ethyl ether; S_N2, B = iso- butyl ethyl ether

Answer 1

The Correct Option is A

The given question asks about the products formed from the reaction of 2-methyl propyl bromide with C₂H₅O^- (ethoxide ion) and C₂H₅OH (ethanol), along with the mechanisms involved—S_N1 or S_N2.

1. Let's first discuss the reactions:
   1. Reaction with C₂H₅O^- (ethoxide ion):
      • 2-Methyl propyl bromide is a primary alkyl halide.
      • When it reacts with the ethoxide ion, a strong nucleophile, it undergoes an S_N2 mechanism. The reason is that S_N2 reactions are favored by primary alkyl halides and strong nucleophiles.
      • The nucleophile attacks the carbon atom bonded to the bromine atom, displacing the bromide ion and forming an ether. The product is iso-butyl ethyl ether.
   2. Reaction with C₂H₅OH (ethanol):
      • In the presence of ethanol, which is a weak nucleophile and polar protic solvent, the reaction follows an S_N1 mechanism. This mechanism is favored by conditions where ionization of the leaving group (Br^-) is facilitated.
      • The slow step involves the formation of a carbocation, and in this case, there is a possibility for carbocation rearrangement.
      • However, due to the stability offered by branching, the most stable carbocation formed is the tertiary structure, leading to the formation of tert-butyl ethyl ether.
2. Now, based on the above step-by-step analysis:
   • The reaction of 2-methyl propyl bromide with C₂H₅O^- follows an S_N2 mechanism producing iso-butyl ethyl ether as product 'A'.
   • The reaction with C₂H₅OH follows an S_N1 mechanism leading to tert-butyl ethyl ether as product 'B'.

Thus, the correct answer is: S_N2. A = iso-butyl ethyl ether; S_N1, B = tert-butyl ethyl ether.