Chapter 21: Problem 80
Explain how \(\mathrm{NaBH}_{4}\) in \(\mathrm{CH}_{3} \mathrm{OH}\) can reduce hemiacetal \(\mathbf{A}\) to 1,4 -butanediol \(\left(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\).
Short Answer
Expert verified
NaBH鈧 in CH鈧僌H reduces the hemiacetal to form 1,4-butanediol by breaking C-O bonds and adding hydrogens to form 鈥揙H groups.
Step by step solution
01
Understanding the Reaction Components
We start by recognizing that NaBH鈧 is a common reducing agent whereas CH鈧僌H (methanol) is usually the solvent used in this reaction. The hemiacetal compound A contains an ether 鈥揙鈥 group and an 鈥揙H group, which is a characteristic structure of hemiacetals. Our target compound is 1,4-butanediol, indicating a four-carbon chain with hydroxyl groups at each end.
02
Mechanism of Reduction by NaBH鈧
NaBH鈧 reduces the carbonyl group (C=O) typically found in esters, carboxylic acids, aldehydes, and hemiacetals to alcohols. Hemiacetal A has a carbon where the 鈥揙H group is bonded to the ether oxygen. This is reduced by NaBH鈧 in CH鈧僌H, which donates a hydride (H鈦 ion) to attack the electron-deficient carbon atom, breaking the C-O bond and forming an alcohol.
03
Transformation into 1,4-Butanediol
After the initial attack by hydride on the hemiacetal, the intermediate formed will tend to eject the ether side (oxygen bonded to another carbon chain in A), leaving behind the structure that upon subsequent rearrangement becomes an alcohol on one end and the original alcohol 鈥揙H on the other end, forming 1,4-butanediol.
04
Identifying the Product
The NaBH鈧 facilitated the conversion of hemiacetal A by reducing the hemiacetal carbon to an alcohol, as 魔emiacetal which creates the 1,4-butanediol structure with hydroxyl groups on the first and fourth carbon. This final compound corresponds to HOCH鈧侰H鈧侰H鈧侰H鈧侽H.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
NaBH鈧 Reduction Mechanism
In organic chemistry, sodium borohydride \(\mathrm{NaBH}_{4}\) is a well-known reducing agent, frequently employed to convert carbonyl groups into alcohols. This compound is typically used in the presence of a solvent such as methanol \(\mathrm{CH}_{3}\mathrm{OH}\) which stabilizes the reaction. The core mechanism begins when \(\mathrm{NaBH}_{4}\) releases a hydride ion \(\mathrm{H}^{-}\), a negatively charged hydrogen ion.
- Hydride is attracted to electron-deficient regions, such as carbon atoms in carbonyl groups.
- The 鈥揙H group in a hemiacetal makes the neighboring carbon an electron-deficient site.
Hemiacetal Conversion
Hemiacetals are unique organic compounds that contain an ether linkage \(-O-\) and an alcohol \(-OH\) group bonded to the same carbon. They are generally considered intermediates in organic reactions, especially in acetal formation or reduction. In the context of reduction with \(\mathrm{NaBH}_{4}\), these functional groups play a pivotal role.
- In the reduction process, the ether-like oxygen and hydroxyl group weaken the corresponding carbon鈥檚 bonds.
- The hydride, released by \(\mathrm{NaBH}_{4}\),effectively targets these weakened links.
1,4-Butanediol Formation
The transformation to 1,4-butanediol involves a sequence of molecular conversions starting from a hemiacetal. The aim is to construct a clear pathway that links structural alterations with chemical behavior.
- Initially, the \(\mathrm{NaBH}_{4}\) attacks the carbon center in the hemiacetal, enforcing a conversion to an alcohol.
- This new alcohol intermediate has a hydroxyl group at one end due to the hydride addition.