Problem 38 - Advanced Problems in Organic Chemistry for Competitive Examinations

Question

Williamson's synthesis is applied to prepare ether. Identify the ether from the following which cannot be prepared by it? (a) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{O}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (b) \(\mathrm{PhOCH}_{3}\) (c) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{3}\) (d) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{O}-\mathrm{CH}=\mathrm{CH}_{2}\)

Step-by-Step Solution

Answer

Ether in option (a) cannot be prepared by Williamson's synthesis.

Step 1: Understand Williamson's Ether Synthesis

Williamson's Ether Synthesis is a method to prepare ether by reacting an alkoxide ion with a primary alkyl halide. The reaction occurs via an S_N2 mechanism with the alkoxide ion acting as a nucleophile, substituting the halogen atom in the alkyl halide.

Step 2: Analyze the Structure for Steric Hindrance

In an S_N2 reaction, the presence of bulky groups hinders the nucleophile's attack. The tert-butyl group, which is \((\mathrm{CH}_{3})_{3}\mathrm{C}\), is very bulky and thus resists attack by a nucleophile like an alkoxide ion.

Step 3: Evaluate Each Option for Synthesis Feasibility

Evaluate each ether to see if it can be synthesized:- Option (a) \((\mathrm{CH}_{3})_{3}\mathrm{C}-\mathrm{O}-\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{3}\) cannot be prepared because of the bulky tert-butyl group, which prevents the S_N2 mechanism.- Options (b), (c), and (d) involve less steric hindrance

Key Concepts

SN2 Mechanism

The SN2 mechanism is a type of nucleophilic substitution reaction that is quite straightforward but fascinating. It proceeds in a single step where the nucleophile attacks the electrophile directly from the opposite side of the leaving group. This results in the simultaneous formation of a new bond and the breaking of the old one. One of the greatest things about SN2 reactions is how they invert the configuration at the carbon where the substitution happens. It's like flipping a pancake!

What's really interesting is that the rate of this reaction depends on both the substrate and the nucleophile. This means if either is in high concentration, the reaction speeds up. With an SN2 reaction, you want the nucleophile to be strong. Common examples include small, negatively charged ions like hydroxide. Additionally, for an SN2 reaction to occur smoothly, the substrate should not be too crowded. If it is, the nucleophile cannot get close enough to do its job.

Steric Hindrance in Organic Reactions

Steric hindrance is like trying to fit a large car into a small garage. In chemistry, it means that bulky groups crowd around a reaction site, preventing a nucleophile from easily approaching and reacting. In the case of SN2 reactions, steric hindrance is a big obstacle. Bulky groups around the carbon hinder the nucleophile's attack, making it more challenging for the reaction to occur.

The presence of these bulky groups or hindrances changes the reaction dynamics, often slowing it down a lot. The size and shape of groups attached to a reaction center determine the extent of steric hindrance. For example, tert-butyl groups are particularly large and create a significant barrier. That is precisely why in Williamson's Ether Synthesis, ethers with these large groups can't be formed. The nucleophile simply can't squeeze through to do the substitution.

Preparation of Ethers

Preparing ethers through Williamson's Ether Synthesis is a classic method used in organic chemistry. It's like a construction process where you build the ether using alkoxide ions and primary alkyl halides. This tried-and-true reaction works well for forming simple and symmetrical ethers.

• First, you generate an alkoxide ion, which acts as a strong nucleophile. This can be created by deprotonating an alcohol with a strong base like sodium hydride.
• Next, this alkoxide ion attacks a primary alkyl halide. Because the SN2 mechanism requires a concerted one-step reaction, it is important that the halide is not sterically hindered. This enables a smooth nucleophilic substitution.
• Finally, the result is the formation of an ether where the oxygen atom links two alkyl groups together.

However, remember that not all ethers can be prepared using this method. If there's significant steric hindrance or if secondary or tertiary alkyl halides are involved, the SN2 pathway will struggle, making the ether difficult or impossible to synthesize through this technique.