17.4 Alcohols from Carbonyl Compounds: Reduction
- Page ID
- 44254
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After completing this lesson, you should be able to
- determine whether a given reaction should be classified as an oxidation or a reduction.
- write an equation to represent the reduction of an aldehyde or ketone using sodium borohydride or lithium aluminum hydride.
- discuss the relative advantages and disadvantages of using sodium borohydride or lithium aluminum hydride to reduce aldehydes or ketones to alcohols.
- identify the product formed from the reduction of a given aldehyde or ketone.
- identify the aldehyde or ketone that should be used to produce a given alcohol in a reduction reaction.
- identify the best reagent to carry out the reduction of a given aldehyde or ketone.
- write an equation to represent the reduction of an ester or a carboxylic acid to an alcohol.
- identify the product formed from the reduction of a given ester or carboxylic acid.
- identify the esters or carboxylic acids that could be reduced to form a given alcohol.
- determine the most appropriate reducing agent for bringing about the reduction of one specific functional group in a compound containing two or more functional groups that are capable of being reduced; for example, the reduction of a keto ester to a hydroxy ester.
Key Terms
- (organic) oxidation
- (organic) reduction
Study Notes
In your course in first-year general chemistry, you probably discussed oxidation-reduction reactions in terms of the transfer of electrons and changes in oxidation numbers (oxidation states). In organic chemistry, it is often more convenient to regard reduction as the gain of hydrogen or loss of oxygen, and oxidation as the gain of oxygen or the loss of hydrogen. There is no contradiction in using these various definitions. For example, when hydrogen is added across the double bond of ethene to reduce it to ethane, the oxidation number of the doubly bonded carbon atoms decreases from −II to −III. Similarly, when 2-propanol
is oxidized to acetone
hydrogen is removed from the compound and the oxidation number of the central carbon atom increases from 0 to +II. If necessary, you should review the concept of oxidation number.
Reduction of Aldehydes and Ketones
The most common sources of the hydride Nucleophile are lithium aluminum hydride (LiAlH4) and sodium borohydride (NaBH4). Note! The hydride anion is not present during this reaction; rather, these reagents serve as a source of hydride due to the presence of a polar metal-hydrogen bond. Because aluminum is less electronegative than boron, the Al-H bond in LiAlH4 is more polar, thereby, making LiAlH4 a stronger reducing agent.
Addition of a hydride anion (H:-) to an aldehyde or ketone gives an alkoxide anion, which on protonation yields the corresponding alcohol. Aldehydes produce 1º-alcohols and ketones produce 2º-alcohols.
In metal hydrides reductions the resulting alkoxide salts are insoluble and need to be hydrolyzed (with care) before the alcohol product can be isolated. In the sodium borohydride reduction the methanol solvent system achieves this hydrolysis automatically. In the lithium aluminum hydride reduction water is usually added in a second step. The lithium, sodium, boron and aluminum end up as soluble inorganic salts at the end of either reaction. Note! LiAlH4 and NaBH4 are both capable of reducing aldehydes and ketones to the corresponding alcohol.
Example 17.4.1
Mechanism
This mechanism is for a LiAlH4 reduction. The mechanism for a NaBH4 reduction is the same except methanol is the proton source used in the second step.
1) Nucleopilic attack by the hydride anion
2) The alkoxide is protonated
Reduction of Carboxylic Acids and Esters
Carboxylic acids can be converted to 1o alcohols using Lithium aluminum hydride (LiAlH4). Note that NaBH4 is not strong enough to convert carboxylic acids or esters to alcohols. An aldehyde is produced as an intermediate during this reaction, but it cannot be isolated because it is more reactive than the original carboxylic acid.
Esters can be converted to 1o alcohols using LiAlH4, while sodium borohydride (\(NaBH_4\)) is not a strong enough reducing agent to perform this reaction.
Contributors
Dr. Dietmar Kennepohl FCIC (Professor of Chemistry, Athabasca University)
Prof. Steven Farmer (Sonoma State University)