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Chapter 28: Problem 19

Draw the products formed when \(\beta\)-\(\small\mathrm{D}\)-galactose is treated with each reagent. a. Ag\(_2\)O + CH\(_3\)I b. NaH + C\(_6\)H\(_5\)CH\(_2\)CI c. The product in (b), then H\(_3\)O\(^+\) d. Ac\(_2\)O + pyridine e. C\(_6\)H\(_5\)COCI + pyridine f. The product in (c), then C\(_6\)H\(_5\)COCI + pyridine

Short Answer

Expert verified
Products are fully methylated, benzylated, hydroxylated, acetylated, and benzoylated variations of 尾-D-galactose.

Step by step solution

01

Methylation Reaction (a)

Treat 尾-D-galactose with Ag鈧侽 and CH鈧僆. In this methylation reaction, methoxide ions (CH鈧僌鈦�) are generated using Ag鈧侽 that replace the hydroxyl hydrogen atoms, leading to the formation of methyl ethers. All hydroxyl (-OH) groups of galactose are converted to methoxy (-OCH鈧�) groups. Thus, the product is completely methylated 尾-D-galactose.
02

Benzylation Reaction (b)

React 尾-D-galactose with NaH, which deprotonates the hydroxyl groups, forming alkoxide ions. Treat the intermediate with benzyl chloride (C鈧咹鈧匔H鈧侰l), resulting in the conversion of hydroxyl groups into benzyl ethers (-OCH鈧侰鈧咹鈧�). All hydroxyl groups are now benzyl protected.
03

Hydrolysis (c)

Hydrolyze the product formed in step 2 with H鈧僌鈦�. The acidic conditions remove the benzyl protecting groups, reverting them to hydroxyl (-OH) groups while retaining any non-protected hydroxyl groups.
04

Acetylation Reaction (d)

Treat 尾-D-galactose with acetic anhydride (Ac鈧侽) and pyridine. These reagents acetylate the hydroxyl groups, converting them into acetyl groups (-OCOCH鈧�). As a result, all hydroxyl groups become acetyl esters on the galactose molecule.
05

Benzoylation Reaction (e)

React 尾-D-galactose with benzoyl chloride (C鈧咹鈧匔OCl) and pyridine. Each hydroxyl group on the galactose is converted into a benzoyl ester (-OCOR) where R is the phenyl group (C鈧咹鈧�), resulting in fully benzoylated 尾-D-galactose.
06

Benzoylation Post-Hydrolysis (f)

Using the product from Step 3, treat it with benzoyl chloride (C鈧咹鈧匔OCl) and pyridine. Any existing hydroxyl groups are transformed into benzoyl esters. This step benzoylates the hydroxyl groups that are still present after hydrolysis in Step 3.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Methylation Reaction
In the methylation reaction involving 尾-D-galactose, methoxide ions come into play. This process kicks off using silver oxide (Ag鈧侽) in the presence of methyl iodide (CH鈧僆). Here's the block-by-block breakdown:
  • Ag鈧侽 works as a mild base, encouraging the formation of methoxide ions (CH鈧僌鈦�).
  • These methoxide ions target the hydroxyl hydrogen atoms present on the galactose molecule.
  • Consequently, all the hydroxyl (-OH) groups get converted into methoxy (-OCH鈧�) groups.
The outcome is a molecule where all the original hydroxyl groups of galactose are now replaced with methoxy groups. This fully methylated form of 尾-D-galactose is crucial in protecting the hydroxyl groups from undergoing unwanted reactions.
Benzylation Reaction
Benzylation is a protective strategy where 尾-D-galactose is reacted to form benzyl ethers. To initiate, galactose interacts with sodium hydride (NaH). This stepwise reaction functions as follows:
  • NaH acts as a strong base, deprotonating the hydroxyl groups on galactose to form alkoxide anions.
  • The alkoxide ions then react with benzyl chloride (C鈧咹鈧匔H鈧侰l), resulting in the formation of benzyl ether groups (-OCH鈧侰鈧咹鈧�).
Each hydroxyl group on the galactose is now benzyl-protected, preventing it from participating in reactions like esterification until the benzyl groups are removed. This reaction is quite essential in organic synthesis to shield specific sensitive groups.
Acetylation Reaction
Acetylation involves the conversion of hydroxyl groups into acetyl esters. In this reaction with 尾-D-galactose, acetic anhydride (Ac鈧侽) and pyridine are utilized.
  • Pyridine functions as a base, facilitating the stabilization of the reaction intermediates.
  • Acetic anhydride reacts with the hydroxyl groups, forming acetyl groups (-OCOCH鈧�) instead.
By transforming all the hydroxyl groups into acetyl esters, this reaction provides a way to modify galactose chemically. Acetylation is particularly valuable in altering the solubility and reactivity of carbohydrate compounds.
Benzoylation Reaction
In the benzoylation reaction, 尾-D-galactose's hydroxyl groups are converted into benzoyl esters. This process employs benzoyl chloride (C鈧咹鈧匔OCl) in combination with pyridine.
  • The role of pyridine is as a catalyst and base to absorb the generated HCl and smooth the reaction through stabilization.
  • Each hydroxyl group on the carbohydrate is transformed into a benzoyl ester (-OCOR, where R is the phenyl group).
As a result, you obtain fully benzoylated 尾-D-galactose. This transformation is often used to protect alcohols during synthetic procedures, given benzoyl esters' relative stability against some chemical conditions.
Hydrolysis of Ethers
Hydrolysis refers to the process of breaking ether bonds, in this context specifically the benzyl ethers formed during benzylation. Utilizing hydronium ions (H鈧僌鈦�), the reaction proceeds as follows:
  • The acidic environment targets the benzyl ether linkages, cleaving them to restore the original hydroxyl groups.
  • More specifically, the oxygen-benzyl group bonds are broken down by the protonation of the ether oxygen, which facilitates the cleavage process.
Post-hydrolysis, the galactose returns to its form with free hydroxyl groups where benzyl ethers previously stood. This process is significant as it allows the transition from a protected to a reactive state, enabling further chemical modification if needed.

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Most popular questions from this chapter

\(\small\mathrm{D}\)-Arabinose can exist in both pyranose and furanose forms. a. Draw the \(\alpha\) and \(\beta\) anomers of \(\small\mathrm{D}\)-arabinofuranose. b. Draw the \(\alpha\) and \(\beta\) anomers of \(\small\mathrm{D}\)-arabinopyranose.

For \(\small\mathrm{D}\)-arabinose: a. Draw its enantiomer. b. Draw an epimer at C3. c. Draw a diastereomer that is not an epimer. d. Draw a constitutional isomer that still contains a carbonyl group.

Which aldoses are oxidized to optically inactive aldaric acids: (a) \(\small\mathrm{D}\)-erythrose; (b) \(\small\mathrm{D}\)-lyxose; (c) \(\small\mathrm{D}\)-galactose?

Which \(\small\mathrm{D}\)-aldopentoses are reduced to optically inactive alditols using NaBH\(_4\), CH\(_3\)OH?

Draw the structure of (a) a ketotetrose; (b) an aldopentose; (c) an aldotetrose.
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