Question:
CH3I cannot be prepared by direct iodination of methane as the reaction is slow and reversible. However, it can be prepared by carrying out the reaction in the presence of
CH3I cannot be prepared by direct iodination of methane as the reaction is slow and reversible. However, it can be prepared by carrying out the reaction in the presence of
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Remember: Iodination requires an oxidizing agent (\(HIO_3\)/\(HNO_3\)), while bromination and chlorination do not, as \(HBr\) and \(HCl\) are not strong enough reducing agents to reverse the reaction easily.
Updated On: Apr 29, 2026
- \(H_2SO_4\)
- \(NaOH\)
- \(HCl\)
- \(H_3PO_4\)
- \(HIO_3\)
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Solution and Explanation
Step 1: Understanding the Concept:
The iodination of alkanes is a reversible and extremely slow reaction.
The byproduct hydrogen iodide (\(HI\)) is a strong reducing agent that reacts with the formed alkyl iodide to regenerate the original alkane.
Step 3: Detailed Explanation:
The reaction is:
\[ CH_4 + I_2 \rightleftharpoons CH_3I + HI \]
To make the reaction proceed in the forward direction and obtain a good yield of methyl iodide (\(CH_3I\)), the \(HI\) produced must be removed immediately as it forms.
This is achieved by adding a strong oxidizing agent like iodic acid (\(HIO_3\)) or nitric acid (\(HNO_3\)).
The oxidizing agent reacts with \(HI\) to convert it back into iodine (\(I_2\)):
\[ 5HI + HIO_3 \to 3I_2 + 3H_2O \]
This shift in equilibrium (per Le Chatelier's principle) allows the preparation of the alkyl iodide.
Step 4: Final Answer:
The reaction is carried out in the presence of \(HIO_3\).
The iodination of alkanes is a reversible and extremely slow reaction.
The byproduct hydrogen iodide (\(HI\)) is a strong reducing agent that reacts with the formed alkyl iodide to regenerate the original alkane.
Step 3: Detailed Explanation:
The reaction is:
\[ CH_4 + I_2 \rightleftharpoons CH_3I + HI \]
To make the reaction proceed in the forward direction and obtain a good yield of methyl iodide (\(CH_3I\)), the \(HI\) produced must be removed immediately as it forms.
This is achieved by adding a strong oxidizing agent like iodic acid (\(HIO_3\)) or nitric acid (\(HNO_3\)).
The oxidizing agent reacts with \(HI\) to convert it back into iodine (\(I_2\)):
\[ 5HI + HIO_3 \to 3I_2 + 3H_2O \]
This shift in equilibrium (per Le Chatelier's principle) allows the preparation of the alkyl iodide.
Step 4: Final Answer:
The reaction is carried out in the presence of \(HIO_3\).
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