Question 27.6: Predicting the Products of a Reaction Involving an Alcohol. ......
Predicting the Products of a Reaction Involving an Alcohol
Predict the products of the following reactions. If appropriate, suggest a mechanism by which the reaction occurs by using arrows to show the movement of electrons.
(a) propan-1-ol + sodium hydroxide \xrightarrow[]{H_{2}O}
(b) (R)-2-bromo-3-methylbutane + ethanol \xrightarrow[]{ethanol}
Analyze
First, we write condensed structural formulas for the reactants, and then we consider the role of the alcohol in each case. Alcohols can be oxidized or deprotonated. They can act as either a nucleophile or an electrophile in a substitution reaction or they can be dehydrated. To make a decision, we must consider the carbon atoms in the reactants and classify them as primary, secondary, or tertiary. We must also consider the other reactants, the solvent, and the reaction conditions.
Solve
(a) The condensed structural formula for propan-1-ol is CH_{3}CH_{2}CH_{2}OH. It is a primary alcohol. Neither substitution nor elimination (dehydration) is possible under the conditions specified. We know that sodium hydroxide is a relatively strong base, and so we should consider the feasibility of the following acid–base reaction:
CH_{3}CH_{2}CH_{2}OH + NaOH → CH_{3}CH_{2}CH_{2}O^{−}Na^{+} + H_{2}O
We must compare the strengths of the acids (or the bases) on either side of the equation. We know that OH^{−} and CH_{3}CH_{2}CH_{2}O^{−} are both considered strong bases, but which is stronger? Because both bases have the negative charge localized on oxygen, we expect them to be fairly similar in strength.
Thus, we predict that the reaction will not go to completion. At equilibrium, significant amounts of all species will be present.
(b) The condensed structural formulas for 2-bromo-3-methylbutane and ethanol are CH_{3}CHBrCH(CH_{3})_{2} and CH_{3}CH_{2}OH. The carbon bonded to bromine in the haloalkane is a 2° carbon atom and Br^{−} is a very weak base and a good leaving group. CH_{3}CH_{2}OH is a weak base and also a weak nucleophile. The solvent is polar protic. On the basis of this information, we conclude that S_{N}1 and E1 reactions will occur (see Figure 27-13). The first step in both the S_{N}1 and E1 mechanisms is the formation of the carbocation:
Once the carbocation is formed, either substitution or elimination can occur. The steps involved in the substitution are as follows:
In the diagram above, the nucleophile is shown attacking the carbocation from above, but it can attack from either above or below. Consequently, both (R) and (S) stereoisomers are obtained.
The steps involved in the elimination are shown below. The elimination reaction produces two alkenes, depending on which H atom (red or blue) is abstracted by the CH_{3}CH_{2}OH molecule. The major product is the most highly substituted alkene.
Assess
As shown in (a), the deprotonation of CH_{3}CH_{2}OH cannot be accomplished effectively by using NaOH. However,
the deprotonation can be accomplished by using NaNH_{2}. The pK_{a} \text{for} NH_{3} is about 34, and so pK_{b} \text{for} NH_{2}^{−} is about 14 − 34 = −20. For part (b), keep in mind that a carbocation can undergo a rearrangement (see Are You Wondering 27-2). Consequently, a variety of products are obtained in substitution and elimination reactions involving secondary and tertiary alcohols. We will not explore this complication.
