Question

Sodium metal was added to tert-butyl alcohol and allowed to react. When the metal was consumed, ethyl bromide was added to the resulting mixture. Work-up of the reaction mixture yielded a compound of formula \(\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{O}\). In a similar experiment, sodium metal was allowed to react with ethanol. When tert-butyl bromide was added, a gas was evolved, and work-up of the remaining mixture gave ethanol as the only organic material. (a) Write equations for all reactions, (b) What familiar reaction type is involved in each case? (c) Why did the reactions take different courses?

Short answer

The reactions are \(\mathrm{S}_{\mathrm{N}}2\) nucleophilic substitution reactions. In the first reaction, sodium tert-butoxide, a bulky nucleophile, reacts with ethyl bromide to form ethyl tert-butyl ether. In the second reaction, sodium ethoxide reacts with tert-butyl bromide, undergoing an elimination reaction due to steric hindrance, producing methane gas and leaving ethanol as the only organic product.

Step-by-step solution

a) Write equations for all reactions

First, let's analyze each reaction separately and write their balanced equations. 1. Sodium metal reacts with tert-butyl alcohol: Na + (CH3)3COH → (CH3)3CO-Na+ + 1/2 H2 2. Sodium alkoxide reacts with ethyl bromide: (CH3)3CO-Na+ + CH3CH2Br → (CH3)3COCH2CH3 + NaBr In the second experiment, 1. Sodium metal reacts with ethanol: 2 Na + 2 CH3CH2OH → 2 CH3CH2O-Na+ + H2 2. Sodium ethoxide reacts with tert-butyl bromide: 2 CH3CH2O-Na+ + (CH3)3CBr → 2 CH3CH2OH + NaBr + CH4 Note: As tert-butyl bromide reacts with sodium ethoxide, a gas (methane) is evolved.

b) What familiar reaction type is involved in each case?

The reactions involved in both cases are nucleophilic substitution reactions, specifically, \(\mathrm{S}_{\mathrm{N}}2\) reactions.

c) Why did the reactions take different courses?

The two reactions took different courses due to differences in the substrates and the nucleophiles involved: 1. In the first reaction, tert-butyl alcohol is reacting with sodium metal leading to the formation of sodium tert-butoxide, which is a strong and bulky nucleophile. Due to its bulky nature, it prefers to react with primary alkyl halides such as ethyl bromide, resulting in the formation of a new compound (ethyl tert-butyl ether). 2. In the second reaction, ethanol reacts with sodium metal to form sodium ethoxide, which is a less bulky and more reactive nucleophile compared to sodium tert-butoxide. When tert-butyl bromide is added, sodium ethoxide undergoes an elimination reaction due to the high steric hindrance of the tert-butoxide anion. As a result, methane gas is formed and the only organic product isolated from the reaction mixture is ethanol itself.

Key concepts

Sodium Metal Reactivity

Sodium, a highly reactive alkali metal, eagerly donates its electron to elements or molecules that accept electrons, such as alcohols. In contact with tert-butyl alcohol or ethanol, sodium forms the corresponding alkoxide ions and hydrogen gas. This vigorous reaction sets the stage for subsequent chemical transformations.

In educational terms, think of sodium as a generous friend, stretching out its single valence electron to bond with others. The resulting 'alkoxide' species are full of potential, ready to engage in even more complex chemical dances, like those we see in nucleophilic substitution reactions.

SN2 Mechanism

The SN2 mechanism is a bimolecular nucleophilic substitution where a nucleophile attacks the electrophile and a leaving group departs all in a single, concerted step. The '2' in SN2 stands for this two-player dance. It requires the nucleophile to approach from the backside, flipping the configuration at the carbon atom involved.

Imagine the carbon atom as a crowded roundabout. The nucleophile must navigate to the opposite side for a successful maneuver, kicking out the leaving group and taking its place. The SN2 mechanism thus depends on the accessibility of this 'roundabout', which brings us to the next critical concept—steric hindrance.

Steric Hindrance

Steric hindrance refers to the physical obstruction caused by the size of groups attached to the carbon atoms participating in a chemical reaction. It's like a molecular traffic jam preventing the nucleophile from reaching the reactive center during an SN2 reaction.

In our analogy, if the central carbon is surrounded by large groups, it's like having oversized vehicles blocking the roundabout. For tert-butyl alcohol, the bulky tert-butyl group makes an SN2 reaction nearly impossible. This is why in the exercise, tert-butyl alcohol and its derivatives follow a different reaction path than the more accessible, less hindered ethanol. Steric bulk doesn't just slow things down; it can change the course of a chemical reaction entirely!

Alkoxide Formation

Alkoxide ions form when an alcohol deprotonates, usually through reaction with a metal such as sodium. These alkoxide ions are like supercharged alcohol molecules, missing their 'peacekeeping' hydrogen and ready to react with other compounds.

For students, think of an alkoxide as a pivotal intermediary, much like a referee who’s ready to blow the whistle and trigger the next play. Their formation from alcohols like tert-butyl alcohol and ethanol is essential to the continuation of reactions, particularly in the creation of ethers or the shift from a simple alcohol to a more complex molecule in the SN2 mechanism. The exact nature of these subsequent reactions, however, can greatly depend on factors such as steric hindrance, as previously discussed.