Problem 12
Question
Which of the following compounds will not show mutarotation? (a) \(\beta-\mathrm{D}(+)\) glucopyranose (b) \(\alpha-\mathrm{D}(+)\) glucospyranose (c) \(\beta-\mathrm{D}(+)\) galactopyranose (d) methyl- \(\alpha\)-D-glucopyranoiside
Step-by-Step Solution
Verified Answer
Methyl-\(\alpha\)-D-glucopyranoiside will not show mutarotation.
1Step 1: Understanding Mutarotation
Mutarotation is the change in optical rotation that occurs by the conversion of one anomer to another. It necessitates a free hydroxyl group at the anomeric carbon that can open to form an open-chain structure and is capable of converting from one anomer to another.
2Step 2: Evaluate Each Compound
(a) \( \beta-\mathrm{D}(+) \) glucopyranose is able to undergo mutarotation because it has a free anomeric hydroxyl group. (b) \( \alpha-\mathrm{D}(+) \) glucopyranose can also undergo mutarotation for the same reason, the free anomeric hydroxyl group. (c) \( \beta-\mathrm{D}(+) \) galactopyranose also has a free anomeric hydroxyl group, thus it can undergo mutarotation. (d) Methyl- \( \alpha \)-D-glucopyranoiside contains a methyl group bonded to the anomeric carbon, preventing the opening of the ring and thus cannot undergo mutarotation.
3Step 3: Conclusion
Since mutarotation requires a free hydroxyl group at the anomeric carbon to allow the ring to open, compounds with an anomeric carbon bonded to a different atom rather than hydroxyl like a methyl group won’t undergo mutarotation. Hence, methyl-\( \alpha \)-D-glucopyranoiside will not undergo mutarotation.
Key Concepts
Anomeric CarbonOptical RotationGlucopyranoseGalactopyranose
Anomeric Carbon
The anomeric carbon is a pivotal player in the phenomenon of mutarotation. It is specifically the carbon atom that becomes a new chiral center when a sugar forms a cyclic structure, such as in glucopyranose or galactopyranose. This transformation occurs when the hydroxyl group at the penultimate carbon attacks the carbonyl carbon, creating a hemiacetal or hemiketal.
- For glucopyranose and galactopyranose, this carbon is originally part of an aldehyde or ketone group in the open-chain form.
- The anomeric carbon is central to the ring closure that leads to two anomers, often denoted as alpha (α) and beta (β).
- A free hydroxyl group attached to this carbon makes the ring dynamic, allowing it to open and close, thereby facilitating the conversion between anomers.
Optical Rotation
Optical rotation refers to the rotation of polarized light by a chiral substance in solution. When you dissolve an optically active compound, like a sugar, in solution, and pass polarized light through it, the orientation of the light is rotated.
The extent of this rotation is dependent on:
- The concentration of the substance.
- The path length of the light through the solution.
- The specific rotation of the compound, a characteristic property of each chiral molecule.
Glucopyranose
Glucopyranose is a term used for the cyclic form of glucose. In water, glucose predominantly exists in a six-membered ring known as pyranose. There are two anomers of glucopyranose, namely, alpha (α) and beta (β), which differ in the configuration around the anomeric carbon.
- The α-anomer has the hydroxyl group on the anomeric carbon trans (opposite side) to the CH₂OH group.
- The β-anomer positions the hydroxyl group cis (same side) to the CH₂OH group.
Galactopyranose
Galactopyranose is the cyclic form of galactose, a monosaccharide similar to glucose. It adopts a six-membered ring structure, or pyranose, in solution. Like glucopyranose, galactopyranose possesses two anomeric forms, α and β, influenced by the orientation of the hydroxyl group on the anomeric carbon.
- The α-galactopyranose has the hydroxyl group on the anomeric carbon positioned opposite to the CH₂OH group, while the β form has it on the same side.
- It plays a significant role in biological systems and is a component in glycolipids and glycoproteins.
Other exercises in this chapter
Problem 10
The aldehyde which is formed during photosynthesis of plants is (a) propanal (b) methanal (c) acetaldehyde (d) phenylmethanal
View solution Problem 11
Which one of the following on reduction with \(\mathrm{NaBH}_{4}\) gives an equimolar mixture of sorbitol and mannitol? (a) mannose (b) glucose (c) xylose (d) f
View solution Problem 13
The conversion of maltose to glucose is possible by the enzyme (a) zymase (b) lactase (c) maltase (d) diastase
View solution Problem 14
Oxidation of fructose with bromine water gives (a) gluconic acid (b) oxalic acid (c) trimethyl glutaric acid (d) tartaric acid
View solution