A.The slow step in the conversion of alcohol to alkyl halide is theloss of water

Question

A.The slow step in the conversion of alcohol to alkyl halide is theloss of water from the protonated alcohol to

form the carbocation intermediate. Explain the order ofreactivity for the following alcohols in this reaction:

1-methylcyclohexanol > cyclohexanol >> 1-hexanol

B.The starting material in this experiment is chiral, but it isoptically inactive because it is a racemic mixture.

Wecould purchase and use this material in optically active (singleenantiomer) form. Even if we started with

asingle enantiomer of the alcohol, we would obtain a product withlittle to no optical activity. Explain.

C. The McMurry lecture text gives other reagents that can be usedto convert alcohols to alkyl halides

(particularly for 1 and 2 alcohols). What are they?

Explanation / Answer

A) The alcohol that forms the most stablecarbocation will react fastest 1-methylcyclohexanol (3y) > cyclohexanol(2y) >1-hexanol (1y) B) The carbocation intermediate is planar, and the halide may beintroduced above or below the plane ofthe carbocation. Regardless of whether westart with a racemic mixture or a pure enatiomer, the end productwill always yield a racemicmix. To avoid this, you must use reagents that do not undergoSn1. C) PBr3 or PCl3, SOCl2/Pyridine,HBr/H2SO4/ (mainly for 1yalcohols), HCl/ZnCl2/ (Lucas Reagent; mainly for1y alcohols)

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