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Predict the major monobromination product when 3-methylpentane is subjected to a light-induced radical bromination reaction. a) 1-Bromo-3-methylpentane b) 2-Bromo-3-methylpentane c) 1-Bromo-2-methylpentane d) 1-Bromo-4-methylpentane
Predict the major monobromination product when 3-methylpentane is subjected to a light-induced radical bromination reaction. a) 1-Bromo-3-methylpentane b) 2-Bromo-3-methylpentane c) 1-Bromo-2-methylpentane d) 1-Bromo-4-methylpentane
Final answer:When 3-methylpentane undergoes a light-induced monobromination reaction, the major product is 2-bromo-3-methylpentane. This occurs because the bromine radical prefers to attack thetertiary carbon, forming the most stable 3° carbon radical.Explanation:The question asks about the major product of the monobromination reaction of3-methylpentane.Brominationof alkanes via a radical mechanism can follow a mechanism that allows for the bromine atom to be inserted at a tertiary position, if available, due to the stability of tertiary radicals.3-Methylpentane has the structureCH3CH(CH3)CH2CH2CH3. When subjected to light-induced monobromination, the bromine radical will preferentially attack the tertiary carbon to form the most stable3° carbon radical. This results in the formation of the major product: 2-bromo-3-methylpentane. Hence, the answer is (b) 2-Bromo-3-methylpentane.Learn more aboutMonobromination of 3-methylpentanehere:brainly.com/question/32198638#SPJ11...