Question

-Chloro-1-butene reacts with sodium ethoxide in ethanol to produce 3 -ethoxy-1butene. The reaction is second order, first order in 3-chloro-1-butene, and first order in sodium ethoxide. In the absence of sodium ethoxide, 3 -chloro-1-butene reacts with ethanol to produce both 3 -ethoxy-1-butene and 1 -ethoxy-2-butene. Explain these results.

Short answer

Question: Explain how the reaction of 3-chloro-1-butene with sodium ethoxide in ethanol produces 3-ethoxy-1-butene and discuss the reaction kinetics and the side reaction that occurs in the absence of sodium ethoxide. Answer: The reaction between 3-chloro-1-butene and sodium ethoxide in ethanol produces 3-ethoxy-1-butene via an SN2 mechanism, in which sodium ethoxide acts as a strong nucleophile and attacks the carbon of 3-chloro-1-butene. The reaction is second order overall and first order in both 3-chloro-1-butene and sodium ethoxide. When sodium ethoxide is absent and only ethanol is present, the reaction proceeds through an SN1 mechanism involving a carbocation intermediate. This can lead to the formation of two possible products: 1) ethanol attacking the positively charged carbon at the 3rd position, yielding 3-ethoxy-1-butene and 2) ethanol attacking the positively charged carbon at the 2nd position after a hydride shift, yielding 1-ethoxy-2-butene. The formation of two products occurs due to the carbocation intermediate and the SN1 mechanism, which allows for multiple nucleophilic attack possibilities.

Step-by-step solution

Understanding the reaction and its kinetics

The main reaction is between 3-chloro-1-butene and sodium ethoxide (NaOEt) in ethanol (EtOH) to produce 3-ethoxy-1-butene. The reaction is second order overall, which means that the rate of the reaction depends on the concentration of both reactants (3-chloro-1-butene and sodium ethoxide). Since it is first order in both reactants, we can write the rate law as: Rate = k [3-chloro-1-butene] [NaOEt]

Reaction of 3-chloro-1-butene with ethanol

In the absence of sodium ethoxide, 3-chloro-1-butene reacts with ethanol to produce both 3-ethoxy-1-butene and 1-ethoxy-2-butene. This indicates that the reaction has more than one pathway when sodium ethoxide is not present.

Explaining the formation of two products

To understand the formation of the two products, we need to analyze the mechanism of the reaction. Sodium ethoxide is a strong nucleophile, which means that it can react with 3-chloro-1-butene in a nucleophilic substitution reaction (SN2). In this case, the ethoxide ion attacks the carbon atom of 3-chloro-1-butene from the opposite side of the leaving group (chloride ion). This leads to the inversion of the configuration of the carbon and the formation of 3-ethoxy-1-butene. When sodium ethoxide is absent, and only ethanol is present, the reaction goes through an SN1 mechanism, which involves the formation of a carbocation intermediate. In this case, the chlorine leaves the 1-butene first, forming a carbon with a positive charge at the 3rd position, which results in a carbocation. The ethanol molecules can now act as a nucleophile and attack the carbocation, which can lead to the formation of two possible products: 1. Ethanol attacking the positively charged carbon at the 3rd position - resulting in 3-ethoxy-1-butene. 2. Ethanol attacking the positively charged carbon at the 2nd position (hydride shift) - resulting in 1-ethoxy-2-butene. The fact that two products are formed in the absence of sodium ethoxide is due to the carbocation intermediate and the SN1 mechanism, which allows for multiple nucleophilic attack possibilities.