/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 80 Explain how \(\mathrm{NaBH}_{4}\... [FREE SOLUTION] | 91影视

91影视

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

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.
When the hydride ion attacks, it adds itself to this carbon, which leads to the breaking of the carbon-oxygen bond. This bond-breaking is crucial because it removes the oxygen's influence on the carbon center, allowing for the formation of a new alcohol group. The reaction is mild yet highly effective, highlighting its widespread use in transforming various functional groups, including those found in hemiacetals, into alcohol forms.
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.
In hemiacetal conversion, once a hydride ion from \(\mathrm{NaBH}_{4}\) attacks the carbon, it forces a change in the molecular configuration. This results in the hybrid molecule ejecting the ether component as a leaving group. Thus, the carbon atom goes from being part of a complex structure to a simpler alcohol form. This mechanism efficiently turns hemiacetal structures into simple, alcohol-based molecules, transforming them along the reaction pathway to form products such as 1,4-butanediol.
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.
Upon subsequent structural rearrangements, the molecule's carbon backbone adjusts, positioning another hydroxyl group on the opposite end. Enabling the creation of 1,4-butanediol, characterized by a four-carbon chain with hydroxyls (\(-OH\)) at both ends. This highlights the efficiency of sodium borohydride in forming alcohol diols from complex starting structures, illustrating the power of simple reagents in synthetic organic chemistry.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What reagents are needed to convert each compound to benzaldehyde \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}\right) ?\) More than one step may be required. a. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}\) b. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCl}\) c. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOCH}_{3}\) d. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) e. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}\) f. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}=\mathrm{CH}_{2}\) g. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}=\mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) h. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}\left(\mathrm{OCH}_{2} \mathrm{CH}_{3}\right)_{2}\)

Give the structure corresponding to each name. a. 2-methyl-3-phenylbutanal b. dipropyl ketone c. 3,3 -dimethylcyclohexanecarbaldehyde d. \(\alpha\) -methoxypropionaldehyde e. 3-benzoylcyclopentanone f. 2-formylcyclopentanone g. \((R)-3\) -methyl-2-heptanone h. \(m\) -acetylbenzaldehyde i. 2-sec-butyl-3-cyclopentenone j. 5,6 -dimethyl-1-cyclohexenecarbaldehyde

Rank the compounds in each group in order of increasing reactivity towards nucleophilic attack. a. \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{O} \quad \mathrm{CH}_{2}=\mathrm{O} \quad\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{O}\) b.

Give the structure corresponding to each IUPAC name. a. 2-isobutyl-3-isopropylhexanal b. trans-3-methylcyclopentanecarbaldehyde c. 1-methylcyclopropanecarbaldehyde d. 3,6-diethylnonanal

Which compound in each pair forms the higher percentage of gem-diol at equilibrium: (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{3}\); (b) \(\mathrm{CH}_{3} \mathrm{CF}_{2} \mathrm{CHO}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO} ?\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.