Chapter 10 #38

What follows are answers for problem 38 from Chapter 10, Bruice. I am including notes that indicate where special attention must be paid to each reaction.

a) This is the cleavage of an ether into two compounds, an alcohol and an alkyl halide. The formation of the major product depends on the stability of the leaving group. A tertiary carbocation is capable of forming, so the halide ion combines with the carbocation. The hydroxide combines with the methyl group.

b) This situation is also an ether cleavage, except in this case we cannot form a tertiary carbocation. The halide attaches to the less sterically hindered group.

c) This case is a dehydration of an alcohol. The reaction forms a carbocation, then undergoes a 1,2 Methyl Shift. The final result is an alkene.

d) This reaction is the oxidation of a primary alcohol. The clue for this reaction is the presence of Chromatic Acid. Because the alcohol is primary the reaction proceeds to the formation of a carboxylic acid.

e) This reaction is the nucleophilic substitution of an epoxide, occurring under acidic conditions (as indicated by the presence of HCl). Under acidic conditions the nucleophile attack occurs on the more substituted carbon. The hydroxide attaches to the less substituted carbon.

f) This reaction is also a nucleophilic substitution of an epoxide, except this reaction occurs under basic conditions. In this case the nucleophile attacks the less substituted carbon, based on lesser steric resistance from the lower degree of substitution.

g) This reaction is the activation of an alcohol and subsequent synthesis. TsCl ultimately becomes the leaving group, and is replaced via SN2. This means that the stereochemistry changes.

h) This reaction is the oxidation of a secondary alcohol. Because the reactant is a secondary compound the reaction stops with the formation of a ketone.

i) This reaction is the elimination reaction of a quaternary ammonium hydride. This reaction is an anti-Zaitsev reaction; the hydrogen is removed from the least substituted carbon. Ammonium ions function very similarly to akyl flourides. The leaving groups are more basic, and therefore create a partial negative charge on the carbon in question. This creates a "carbanion" like effect.

Edit (2/2/13): Removed the previous notes and replaced with easier-to-read copy in pen.