Question

The reaction of t-buty1 chloride with water is strongly accelerated by sodium hydroxide. How would the ratio of elimination to substitution products be effected thereby?

Short answer

The presence of sodium hydroxide (NaOH) in the reaction of t-butyl chloride with water accelerates the substitution reaction due to its strong nucleophilic nature, resulting in the formation of t-butyl alcohol. However, no elimination products are formed in this reaction. Therefore, the ratio of elimination to substitution products would be 0:1, indicating that the addition of NaOH increases the substitution product while having no effect on elimination products.

Step-by-step solution

Write the reaction of t-butyl chloride with water

The reaction of t-butyl chloride (t-C4H9Cl) with water (H2O) can be represented as follows: t-C4H9Cl + H2O → t-C4H9OH + HCl In this reaction, t-butyl chloride undergoes nucleophilic substitution and forms t-butyl alcohol (t-C4H9OH) and hydrochloric acid (HCl).

Identify elimination and substitution products

For this reaction, the substitution product is the newly formed t-butyl alcohol, as it is formed due to the replacement of Cl- ion in t-butyl chloride. An elimination reaction could occur instead of the substitution if a proton would be removed from the β-carbon of t-butyl chloride, but the structure of t-butyl chloride makes this impossible. Therefore, in this case, there are no elimination products formed.

Analyze the effect of sodium hydroxide

Sodium hydroxide (NaOH), being a strong base, will accelerate the reaction of t-butyl chloride with water. This is because, in the presence of a strong base like NaOH, it will act as a nucleophile, attacking the carbon with the leaving group (Cl-) and facilitating the substitution reaction. The reaction will be faster and more complete in the presence of NaOH due to the increased nucleophile strength.

Determine the effect on the ratio of elimination to substitution products

Since there is no elimination product formed in this reaction and the substitution product is t-butyl alcohol, the presence of sodium hydroxide will accelerate the formation of the substitution product. The ratio of elimination to substitution products would therefore be 0 to 1 (0:1). In other words, with the addition of sodium hydroxide, there will be an increase in the substitution product (t-butyl alcohol) as the reaction is accelerated, while no elimination products are formed.

Key concepts

t-butyl Chloride Reaction

The reaction involving t-butyl chloride (t-C4H9Cl) and water is a classic example of nucleophilic substitution, specifically known as the SN1 reaction. In this process, the chlorine atom in t-butyl chloride is replaced by a hydroxyl group from water to form t-butyl alcohol (t-C4H9OH), with hydrochloric acid (HCl) as the byproduct. The 't' in t-butyl stands for tertiary, which indicates that the central carbon atom is attached to three other carbon atoms.

Due to its highly hindered tertiary structure, t-butyl chloride is an excellent candidate for SN1 reactions. Sn1 reactions involve two steps; first, the formation of a carbocation intermediate when the chlorine leaves, and second, the attack of the nucleophile. Since elimination reactions generally require the presence of a β-hydrogen, which t-butyl chloride lacks due to its quaternary carbon, they are not favored in this scenario. The presence of sodium hydroxide in the mixture, which we will discuss later, significantly affects the rate but doesn't change the reaction pathway from substitution to elimination.

Substitution vs Elimination

Understanding the difference between substitution and elimination reactions is crucial for predicting the outcome of a reaction. Substitution reactions involve the exchange of one atom or group for another on a specific carbon atom in the molecule. In the case of secondary and tertiary alkyl halides like t-butyl chloride, these reactions often proceed through the SN1 mechanism, where the rate is dependent on the step involving the formation of the carbocation.

Elimination reactions, on the other hand, result in the removal of two atoms or groups from adjacent carbons to form a double bond, following either the E1 or E2 mechanism. These reactions typically require a β-hydrogen, which is absent in t-butyl chloride. Therefore, in the exercise's context, the t-butyl chloride will not undergo an elimination reaction; instead, it will always favor substitution in the presence of both water and sodium hydroxide, reflecting a 0:1 ratio of elimination to substitution products.

Sodium Hydroxide as Nucleophile

Sodium hydroxide (NaOH) plays a pivotal role when it's introduced into the reaction mixture of t-butyl chloride and water. As a strong base and nucleophile, NaOH enhances the reactivity of the water molecule by abstracting a proton, thereby increasing the concentration of hydroxide ions (OH-). These hydroxide ions are more nucleophilic than water, making them more efficient in attacking the carbocation intermediate that forms when t-butyl chloride reacts.

In the provided exercise, NaOH doesn't act directly as the nucleophile in the reaction because it serves to accelerate the conversion of t-butyl chloride to t-butyl alcohol by facilitating the formation of the more nucleophilic hydroxide ions from water. As a strong base, intuition might suggest NaOH could induce an elimination reaction, but in the case of t-butyl chloride, such elimination is structurally impossible. Thus, NaOH's role in the reaction is to reinforce the nucleophilic substitution pathway, ensuring a more rapid and complete transition to the substitution product, t-butyl alcohol.