Problem 75 - General Chemistry: Principles and Modern Applications

Question

Predict and name the product(s) obtained from the following reaction. Write out the mechanism for the reaction and use curved arrows to show the movement of electrons. $$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}=\mathrm{CH}_{2} \stackrel{\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{H}_{2} \mathrm{O}}{\longrightarrow}$$

Step-by-Step Solution

Answer

The reaction of 2-methyl-1-butene with sulfuric acid and water yields 2-methyl-2-butanol. This involves a protonation step to form a carbocation, a nucleophilic attack by water, and a deprotonation step to form the alcohol.

Step 1: Protonation

Protonate the alkene in 2-methyl-1-butene using a proton (H+) from sulfuric acid (H2SO4) makes a positive carbocation. As a general rule, the proton will attach to the carbon atom that results in the most stable carbocation. In this case, that will be the secondary carbon atom, not part of the methyl group. After protonation, you get (CH3)2C+(CH2)CH3.

Step 2: Nucleophilic attack

The carbocation will undergo a nucleophilic attack by a water molecule. Water, being a stronger nucleophile, donates an electron pair to form a bond with the positively charged carbon atom. The resulting species is (CH3)2C(OH)(CH2)CH3 positively charged since it retains one proton from water.

Step 3: Deprotonation

A second water molecule will now act as a base, it removes the extra proton (H+) from the carbon bearing the hydroxyl group (OH). The result is a neutral alcohol - 2-methyl-2-butanol.

Key Concepts

Alkene Protonation

Alkene protonation is the initial step in many organic reactions involving an alkene. To understand this concept, it's vital to know that alkenes have a carbon-carbon double bond, which is a region of high electron density. This makes them reactive towards electron-seeking species, such as protons (H⁺). In the reaction, 2-methyl-1-butene undergoes protonation, where a proton from sulfuric acid ( H_2SO_4) attacks the double bond. This step adds a proton to one of the carbons in the double bond, resulting in the formation of a carbocation. Not just any protonation will suffice — the goal is to create the most stable carbocation. This happens when the proton attaches to the less substituted carbon, forming a more stable secondary carbocation structure. Therefore, choosing the correct carbon for protonation to enhance stability is critical.

Carbocation Intermediate

Forming a carbocation intermediate is a hallmark of reactions involving alkenes. After the protonation step, the splitting of the double bond leaves behind a positively charged carbon atom, known as a carbocation. This carbocation is a high-energy, unstable species that seeks stability by gaining electrons. The stability of this intermediate plays a crucial role in determining the reaction pathway and product formation. Key things to keep in mind about carbocations are:

• They are stabilized by surrounding carbon atoms – more neighboring carbons mean greater stability.
• Resonance also contributes to carbocation stability, but here we focus on the inductive effect.

In our example, the most stable intermediate is a secondary carbocation, denoted as (CH₃)₂C⁺(CH₂)CH₃ , which plays a pivotal part in the subsequent reaction steps.

Nucleophilic Attack

Nucleophilic attack is the next exciting phase after the carbocation formation. This happens when an electron-rich species, known as the nucleophile, donates an electron pair to a positively charged carbon, forming a bond. In the described mechanism, water acts as the nucleophile due to its lone pairs of electrons. The water molecule attacks the carbocation at the positively charged carbon atom. This step transforms the carbocation into an alcohol intermediate. Be mindful of why water is such a potent nucleophile here:

• Water's oxygen atom has lone electron pairs that are ready to donate.
• This donation satisfies the electron deficiency of the carbocation, leading to an alcohol formation.

The intermediate formed is (CH₃)₂C(OH)(CH₂)CH₃ , now ready for refinement through deprotonation.

Deprotonation

Deprotonation is the step where an entity like water removes an extra proton from the intermediate. Here, after the initial nucleophilic attack, the alcohol-like intermediate holds an additional proton, making it a charged species. A second water molecule acts as a base to snatch away this proton, primarily from the hydroxyl group (OH), stabilizing the intermediate back to a neutral molecule. This step cleverly converts the intermediate (CH₃)₂C(OH)(CH₂)CH₃ into the final alcohol form. The process of deprotonation is essential in completing the reaction cycle and ensuring the molecule becomes stable and neutral. Understanding this process helps recognize the intricate dance of electrons in organic reactions.

Alcohol Formation

Alcohol formation marks the successful culmination of these mechanistic steps. After protonation, nucleophilic attack, and deprotonation, the final stable product is an alcohol. In this reaction, we see the transformation of 2-methyl-1-butene to 2-methyl-2-butanol. This entire process illustrates how organic synthesis effectively converts a simple alkene into an alcohol by orchestrating a series of precise electron-moving steps:

• Alkene double bonds offer electron density for initial reactivity.
• Carbocation intermediates harness reactivity potential for nucleophilic capture.
• Deprotonation solidifies the alcohol's structural stability.

Overall, understanding these steps not only clarifies the specific reaction but also broadens knowledge of organic reaction mechanisms.