Chapter 20: Problem 7
Draw the products formed when \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2}\) is treated with each reagent: (a) LiAlH \(_{4}\), then \(\mathrm{H}_{2} \mathrm{O}\); (b) \(\mathrm{NaBH}_{4}\) in \(\mathrm{CH}_{3} \mathrm{OH}\); (c) \(\mathrm{H}_{2}\left(1\right.\) equiv), \(\mathrm{Pd}-\mathrm{C} ;\) (d) \(\mathrm{H}_{2}\) (excess), \(\mathrm{Pd}-\mathrm{C}\); (e) \(\mathrm{NaBH}_{4}\) (excess) in \(\mathrm{CH}_{3} \mathrm{OH} ;\) (f) \(\mathrm{NaBD}_{4}\) in \(\mathrm{CH}_{3} \mathrm{OH}\).
Short Answer
Step by step solution
Understanding the Reactants
Reduction with LiAlH4 followed by H2O
Reduction with NaBH4 in CH3OH
Partial Hydrogenation with H2 (1 equiv), Pd-C
Complete Hydrogenation with H2 (excess), Pd-C
Excess Reduction with NaBH4 in CH3OH
Deuterium Labeling with NaBD4 in CH3OH
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ketone Reduction
A ketone, characterized by a carbonyl group (\(C=O\)) bonded to two carbon atoms, can be reduced using various reagents. In the given exercise, lithium aluminum hydride (\( ext{LiAlH}_4\)) and sodium borohydride (\( ext{NaBH}_4\)) are used to perform this reduction.
The general outcome of reducing a ketone is the conversion of the carbonyl group into an alcohol group (\(C-OH\)). With \( ext{LiAlH}_4\) and \( ext{NaBH}_4\), only the ketone in the molecule \( ext{CH}_3 ext{COCH}_2 ext{CH}_2 ext{CH}= ext{CH}_2\) is reduced while any present alkenes remain unchanged. Both reagents selectively reduce the ketone to yield the alcohol \( ext{CH}_3 ext{CH(OH)CH}_2 ext{CH}_2 ext{CH}= ext{CH}_2\).
This conversion allows organic chemists to strategically alter molecular structures in complex synthesis pathways.
Alkene Hydrogenation
This process is often catalyzed by metal catalysts like palladium on carbon (Pd-C) in both complete and partial transformations. In partial hydrogenation, using an equivalent amount of \( ext{H}_2\), the double bond in \( ext{CH}_3 ext{COCH}_2 ext{CH}_2 ext{CH}= ext{CH}_2\) can be reduced to a single bond, forming \( ext{CH}_3 ext{COCH}_2 ext{CH}_2 ext{CH}_2 ext{CH}_3\).
- The ketone remains unaffected in partial hydrogenation.
- Complete hydrogenation uses excess \( ext{H}_2\), leading to the reduction of both alkene and ketone groups.
LiAlH4 Reduction
In the exercise:
- \( ext{LiAlH}_4\) targets the ketone group in \( ext{CH}_3 ext{COCH}_2 ext{CH}_2 ext{CH}= ext{CH}_2\), converting it to \( ext{CH}_3 ext{CH(OH)CH}_2 ext{CH}_2 ext{CH}= ext{CH}_2\).
- It doesn't reduce alkenes, which only require weaker reducing conditions.
NaBH4 Reaction
In the context of the problem:
- \( ext{NaBH}_4\) reduces the ketone in \( ext{CH}_3 ext{COCH}_2 ext{CH}_2 ext{CH}= ext{CH}_2\) to yield a secondary alcohol \( ext{CH}_3 ext{CH(OH)CH}_2 ext{CH}_2 ext{CH}= ext{CH}_2\).
- It uses methanol (\( ext{CH}_3 ext{OH}\)) as a solvent, which may also act as a proton source.
Pd-C Catalysis
In the exercise:
- When paired with 1 equivalent of \( ext{H}_2\), it selectively reduces the alkene in \( ext{CH}_3 ext{COCH}_2 ext{CH}_2 ext{CH}= ext{CH}_2\) without touching the ketone, producing \( ext{CH}_3 ext{COCH}_2 ext{CH}_2 ext{CH}_2 ext{CH}_3\).
- Excess \( ext{H}_2\) allows Pd-C to reduce both the alkene and carbonyl group, leading to a fully saturated product.