Problem 53 - A Complete Resource Book in Chemistry for JEE Main

Question

Tert-butyl methyl ether on heating with \(1 \mathrm{M} \mathrm{HI}\) 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

Option (a): \\(\text{CH}_3\text{I} + (\text{CH}_3)_3\text{COH}\\) is correct.

Step 1: Understand the Reaction

Tert-butyl methyl ether reacts with hydroiodic acid (HI) when heated. We need to determine the products formed from this reaction. HI is a strong acid and can cleave ethers through a process called ether cleavage, producing an alcohol and an alkyl halide.

Step 2: Consider the Nature of the Ether

Tert-butyl methyl ether has two different alkyl groups attached to the oxygen: a tert-butyl group \((\text{C}_4\text{H}_9^-)\) and a methyl group \(\text{CH}_3^-)\). The mechanism of cleavage often produces more stable carbocation, so it typically involves the more hindered or bulky group as the leaving alcohol.

Step 3: Predict Products Based on Stability

The tert-butyl group forms a more stable carbocation than the methyl group because it is tertiary. Therefore, the \(\text{C-O}\) bond between the tert-butyl group and the oxygen is more likely to be cleaved. This results in tert-butyl alcohol \(\text{(CH}_3\text{)}_3\text{COH}\) and methyl iodide \(\text{CH}_3\text{I}\) as products.

Step 4: Analyze Given Options

From the given options, (a) \(\text{CH}_3\text{I} + (\text{CH}_3)_3\text{COH}\) matches our predicted reaction products: methyl iodide and tert-butyl alcohol.

Key Concepts

Tert-butyl methyl ether

Tert-butyl methyl ether is an organic compound that is part of the ether family. Ethers have the general formula R-O-R', where R and R' are carbon chains or alkyl groups. In the case of tert-butyl methyl ether, one side of the oxygen bond is connected to a tert-butyl group \((\text{(CH}_3\text{)}_3\text{C}-)\) and the other to a methyl group \((\text{CH}_3-)\).
This mix of groups makes tert-butyl methyl ether interesting, as it combines a bulky tertiary and a simple primary carbon chain. An ether group introduces a level of stability due to the lack of acidic hydrogens, making them less reactive than alcohols or amines in nucleophilic substitution reactions.
Ethers, however, aren't completely inert; they can undergo cleavage, especially with strong acids like hydroiodic acid. This feature is crucial when considering reactions that target the ether linkage, like ether cleavage.

Hydroiodic acid reaction

Hydroiodic acid (HI) is a strong acid and a powerful halogen. It plays a special role in reactions with ethers, specifically in ether cleavage reactions. When tert-butyl methyl ether is heated with HI, the process begins with the protonation of the ether oxygen. This makes the ether group a good leaving group, simplifying the cleavage of the C-O bond.

• The protonation step increases the positive charge on oxygen, destabilizing the ether bond.
• This leads to the cleavage in which one of the two alkyl groups connected to the oxygen is removed.

In this particular reaction, HI acts as both a proton donor and a source of iodide ions. The iodide ions can then displace part of the molecule to form an alkyl iodide. This shows the dual role of HI as both an acid and a nucleophile, crucial for breaking the ether.

Carbocation stability

Carbocation stability is a key concept when predicting the products of ether cleavage reactions. This is because the stability of a potential carbocation influences which bond is cleaved. A carbocation is a positively charged carbon atom with only three bonds instead of the usual four.
In our reaction, we compare the tertiary carbocation from tert-butyl \((\text{C}_3\text{H}_9^{+})\) with the primary carbocation from the methyl group \((\text{CH}_3^{+})\). Tertiary carbocations are more stable than primary ones due to the following factors:

• The positive charge is better stabilized by the three surrounding carbon atoms.
• More opportunities for hyperconjugation, which disperses charge.

Therefore, the bond with the more stable tert-butyl carbocation is more likely to break, leading to the formation of tert-butyl alcohol.

Alkyl halide formation

Alkyl halides are organic compounds where a halogen atom is bonded to an alkyl group. In the reaction of tert-butyl methyl ether with HI, the bond between the methyl group and oxygen breaks to form methyl iodide \(\text{CH}_3\text{I}\). This occurs because the iodide ion replaces the ether's bond with methyl.
Formation of alkyl halides via ether cleavage encompasses a broader understanding of substitution reactions. Features that favor alkyl halide formation include:

• Strong acids providing protons to facilitate the breakdown of ethers.
• Nucleophilic halides like iodide promoting substitution reactions.

Through this reaction, the methyl group, originally part of the ether, now forms a stable methyl iodide compound. This is a classic example of how ether cleavage operates to produce different types of compounds, including useful alkyl halides.