Question:
R–OH + LiAlH\(_4\) \( \rightarrow \) ?
R–OH + LiAlH\(_4\) \( \rightarrow \) ?
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LiAlH\(_4\) reduces alcohols to primary alcohols in most cases, and it is a strong reducing agent.
Updated On: Jan 6, 2026
- RCH\(_2\)OH
- RCHO
- RCOR
- RCH\(_2\)OH
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The Correct Option is D
Solution and Explanation
Step 1: Understand the reaction with LiAlH\(_4\).
Lithium aluminum hydride (LiAlH\(_4\)) is a strong reducing agent. When alcohols react with LiAlH\(_4\), they are reduced to the corresponding aldehydes or alcohols depending on the condition.
Step 2: Conclusion. Thus, R–OH reacts with LiAlH\(_4\) to form a primary alcohol, RCH\(_2\)OH.
Step 2: Conclusion. Thus, R–OH reacts with LiAlH\(_4\) to form a primary alcohol, RCH\(_2\)OH.
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