When n-butyl magnesium iodide is treated with water, the product is:
Correct answer: A. n-butane
- A. n-butane
- B. Propane
- C. Iso-butane
- D. Alcohol
Explanation
Explanation:Alkanes are produced when grignard reacts with:Water Ammonia AlcoholAs a result, n-butane will be produced in the above reaction. This option is not correct because propane is not formed by the reaction of n-butyl magnesium iodide with water. This option is also incorrect because iso-butane is not formed by the reaction of n-butyl magnesium iodide with water. This option is also incorrect because alcohol is not formed by the reaction of n-butyl magnesium iodide with water.
This question appeared on the UHS MDCAT 2009 paper, which you can sit online with every answer explained.
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About Nucleophilic Substitution
Nucleophilic substitution replaces a leaving group in an alkyl halide when an electron-rich nucleophile attacks the carbon atom. The topic compares SN1 and SN2 mechanisms, including reaction conditions, substrate structure, solvent effects, nucleophile strength, carbocation formation and stereochemical changes, and distinguishes substitution from elimination.
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