Identify the product of the following reaction |
$CH_3CH_2Br$ |
The correct answer is Option (3) →
Tertiary alkyl halides react with strong bases like ethoxide ion via elimination (E2) rather than substitution due to steric hindrance. Detailed Explanation Option 1: $CH_3CH_2O-C(CH_3)_3$ Ether formation would require SN2 substitution. However, tert-butyl bromide is highly hindered, making SN2 substitution difficult. Option 2: $H_2C=CH_2$ Formation of ethene would require elimination from ethyl group, which is not present as substrate. Option 3: $(CH_3)_2C=CH_2$ Ethoxide ion acts as a strong base and removes a ẞ-hydrogen from tert-butyl bromide leading to elimination via E2 mechanism. Product formed is isobutene (2-methylpropene). Option 4: $CH_3CH_2Br$ This would require substitution at ethoxide ion, which does not occur. |