Chapter 16, Problem 16.89P

Interpretation Introduction

(a)

Interpretation:

The structure of the alcohol that, on heating with sulfuric acid, gave the product whose IR and NMR spectra are as shown is to be drawn.

Concept introduction:

Alcohols have an electron-rich functional group. The oxygen atom in the OH group has two lone pairs and is highly electronegative. This makes the carbon to which it is bonded an electrophilic atom, while the oxygen can behave as a nucleophile. Alcohols, therefore, undergo nucleophilic substitution as well as elimination reactions.

Although the OH group is a poor leaving group, it can be converted into a good leaving group (${\text{H}}_{\text{2}}{\text{O}}^{\text{+}}-$) under strongly acidic conditions. It can then be replaced by a variety of nucleophiles.

Interpretation Introduction

(b)

Interpretation:

The mechanism for the formation of the reaction from part (a) is to be drawn.

Concept introduction:

Alcohols can undergo nuclophilic substitution as well as elimination reactions. Normally the hydroxyl group is a poor leaving group as the hydroxide ion is a strong base. It is, however, be converted to a good leaving group under strongly acidic conditions. The strong acid protonates the hydroxyl group, converting it to the good leaving group ${\text{H}}_{\text{2}}{\text{O}}^{\text{+}}-$. Heterolysis of this group will form a carbocation. The carbocation can then either lose a proton from an adjacent carbon to form an alkene or add a nucleophile to give a substitution product. The major product formed depends on the reaction conditions. If a sufficiently strong nucleophile is present, substitution is the more likely outcome, particularly at relatively low temperatures. Formation of the elimination product requires the presence of a strong, non-nuclophilic base and higher temperatures.