Problem 52 - A Complete Resource Book in Chemistry for JEE Main

Question

Tert-butyl methyl ether on heating with \(1 \mathrm{M}\) HI gives (a) \(\mathrm{CH}_{3} \mathrm{I}+\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}\) (b) \(\mathrm{CH}_{3} \mathrm{OH}+\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Cl}\) (c) \(\mathrm{CH}_{3} \mathrm{I}+\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Cl}\) (d) none of these

Step-by-Step Solution

Answer

The answer is (a): \( \text{CH}_{3}\text{I} + \left(\text{CH}_{3}\right)_{3} \text{COH} \).

Step 1: Understanding the Reaction

Tert-butyl methyl ether, when heated with hydrogen iodide (HI), undergoes a nucleophilic substitution reaction. HI is a strong acid that can cleave the ether into an alcohol and an alkyl halide.

Step 2: Analyzing Possible Products

The ether bond in tert-butyl methyl ether is between the tert-butyl group and the methyl group. Upon reaction with HI, this bond can be cleaved, leading to the formation of methanol (\( \text{CH}_3\text{OH} \)) from the methyl group and tert-butyl iodide (\( (\text{CH}_3)_3\text{CI} \)) from the tert-butyl group. However, because iodine is a better nucleophile than chloride, the expected product is tert-butyl alcohol and methyl iodide.

Step 3: Identifying the Correct Products

The cleavage of the ether bond by HI would result in one product being iodide and the other typically an alcohol. The correct separation results in the formation of \( \text{CH}_{3}\text{I} \) and \( (\text{CH}_3)_3\text{COH} \).

Step 4: Matching with Given Options

Reviewing the options provided: (a) \( \text{CH}_{3}\text{I} + \left(\text{CH}_{3}\right)_{3} \text{COH} \) This matches the products from the analysis, confirming option (a) as correct.Options (b) and (c) do not match the expected outcomes of the reaction. Option (d) is incorrect because we found a matching option.

Key Concepts

Ether Cleavage

Ether cleavage is an important chemical reaction where the ether linkage, characterized by an oxygen atom connecting two alkyl groups, is broken. In the context of nucleophilic substitution reactions, this often involves an attacking nucleophile and an acid, typically hydrogen halides like HI or HCl.
In these reactions, the strong acid protonates the oxygen in the ether. This step makes the ether bond more susceptible to being broken as it forms an oxonium ion, which is an activated version of the ether.
The nucleophile, in our example iodine from HI, then attacks the protonated ether, leading to the cleavage of the bond between the oxygen and one of the carbon groups:

• The leaving group, generally the more substituted alkyl group, exits as a carbocation if stable and then typically forms an alkyl halide.
• The other alkyl component gains the oxygen, resulting in the formation of an alcohol.

This specific mechanism ensures a well-defined pathway for products, making ether cleavage a significant topic in organic chemistry.

Alkyl Halide Formation

Alkyl halides are organic compounds where one hydrogen atom in an alkane has been replaced by a halogen atom. In the reaction of tert-butyl methyl ether with HI, one of the main products is methyl iodide (\(\text{CH}_3\text{I}\)), an alkyl halide.
During the reaction, the nucleophile, iodine in this case, effectively 'captures' the smaller, more straightforward methyl group. This is due to iodide's strong nucleophilic properties, which prefer smaller, less sterically hindered groups. As the methyl group detaches, it forms a bond with the iodide, resulting in methyl iodide:

• Methyl iodide is volatile and commonly used in chemical synthesis.
• It showcases the ability of iodide to participate in substitution reactions smoothly.

Understanding the formation of alkyl halides is crucial for predicting the products of ether cleavage reactions. They demonstrate how nucleophiles can transform simple alkanes into more reactive halide compounds.

Alcohol Formation

Alcohol formation in ether cleavage reactions occurs when one of the cleavage products retains the oxygen atom from the original ether. Upon heating tert-butyl methyl ether with HI, tert-butyl alcohol (\((\text{CH}_3)_3\text{COH}\)) is formed.
Tert-butyl alcohol results from the attachment of the oxygen to the more substituted carbon group, often due to the stability provided by its tertiary structure. Consequently, the more stable carbocation intermediate forms, favoring the production of alcohol.

• In tertiary alcohol formations, the oxygen remains bonded to what was originally the more substituted portion of the ether.
• This yields tert-butyl alcohol, a stabilizing secondary product in the reaction sequence.

This process illustrates one of the methods by which ethers can be strategically decomposed into useful alcohols, broadening the scope of synthetic applications and understanding of chemical reactivity.