MCAT Organic Chemistry Question 70: Answer and Explanation

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Question: 70

10. Treating 2-methyl-1-propanol with methylsulfonyl chloride in base, followed by reaction with pyridinium chlorochromate, and a final step in strong acid, will give an end product of:

  • A. 2-methyl-1-propanol.
  • B. 2-methylpropanal.
  • C. 2-methylpropanoic acid.
  • D. 2-methyl-1-propane.

Correct Answer: A

Explanation:

Methylsulfonyl chloride serves as a protecting group for alcohols, which are converted into mesylates. Reacting with this reagent before continuing with what would normally be an oxidation reaction keeps the alcohol from reacting; when the protecting group is then removed using strong acid, the resultant product is the same as the initial reactant. Neither of the oxidation products in choices (B) or (C), nor the reduction product in choice (D), will be formed.

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