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Ribose, \(C_{5} H_{10} O_{5},\) contains an aldehyde group but exists predominantly in the form of the cyclic hemiacetal shown here. We will discuss this cyclic form of ribose in Chapter 20. Which carbon of ribose provides the \(-\) OH group and which provides the CHO group for formation of this cyclic hemiacetal?

Short Answer

Expert verified
The \(-OH\) group comes from \(C_4\) and the \(-CHO\) group from \(C_1\).

Step by step solution

01

Understand Ribose Structure

Ribose is a monosaccharide with the chemical formula \(C_5H_{10}O_5\). It contains five carbon atoms, where each carbon is bonded with the hydroxyl group (\(-OH\)) except one - the carbonyl group (\(-C=O\)) of the aldehyde.
02

Identify the Functional Groups

The aldehyde group in ribose is located at the first carbon (\(C_1\)), which is the carbon with the carbonyl group (\(-CHO\)). The hydroxyl groups (\(-OH\)) are on the other four carbons: \(C_2, C_3, C_4\), and \(C_5\).
03

Formation of Hemiacetal

In the cyclic hemiacetal formation, the aldehyde group at \(C_1\) reacts with a hydroxyl group to form a ring structure. This involves a nucleophilic attack by one of the \(-OH\) groups on the aldehyde carbon.
04

Determine the Reacting Hydroxyl Group

In ribose, the ring is typically formed by a reaction between the \(-OH\) group on \(C_4\) and the aldehyde group on \(C_1\). This creates a five-membered ring (a furanose form).
05

Conclusion

The aldehyde group (\(-CHO\)) is provided by carbon \(C_1\), and the \(-OH\) group involved in forming the cyclic hemiacetal is provided by carbon \(C_4\).

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Key Concepts

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

Monosaccharides
Monosaccharides are the simplest form of carbohydrates. These are single sugar molecules that form the building blocks of more complex carbohydrates like oligosaccharides and polysaccharides. Ribose, a critical sugar in biology, is one such monosaccharide with the chemical formula \( C_5H_{10}O_5 \). It belongs to the aldose family of sugars because it contains an aldehyde group. Unlike larger carbohydrates, monosaccharides are not broken down by hydrolysis into simpler sugars. They typically contain 3 to 7 carbon atoms. Ribose itself has five, which classifies it as a pentose sugar.

Monosaccharides can exist in linear and ring structures. In aqueous solutions, they often prefer the ring form, which is more stable. This change in structure doesn't alter the chemical identity but can affect reactivity and biological roles.
Functional Groups in Carbohydrates
Carbohydrates are diverse, but their basic structure remains simple. They are primarily composed of carbon, hydrogen, and oxygen atoms. Functional groups in these molecules include hydroxyl groups (-OH) and carbonyl groups (C=O). The presence of these groups defines the sugar's properties and reactivity.

For instance, ribose contains multiple hydroxyl groups that make it hydrophilic, meaning it can dissolve easily in water. The carbonyl group determines whether a sugar is an aldose or ketose. In ribose, the carbonyl group is part of the aldehyde (-CHO) found at carbon \( C_1 \), making it an aldose.

The arrangement of the hydroxyl and carbonyl groups also leads to the possibility of forming ring structures. Recognizing these functional groups is vital for understanding chemical reactions involving carbohydrates, such as those leading to hemiacetal formation.
Hemiacetal Formation
Hemiacetal formation is an essential concept in carbohydrate chemistry. It occurs when a molecule containing an aldehyde or ketone reacts with an alcohol. In the case of ribose, the aldehyde group at \( C_1 \) reacts with a hydroxyl group on the same molecule to form a cyclic structure. This is facilitated by a nucleophilic attack where the hydroxyl oxygen attacks the electrophilic carbon in the aldehyde group.

Specifically, in ribose, the hydroxyl group attached to carbon \( C_4 \) typically participates in this reaction, forming a five-membered ring known as a furanose form. This reaction is reversible and happens frequently in solutions, leading to equilibrium between linear and cyclic forms.

The resulting hemiacetal structure is a key feature in the stability and function of many biochemical molecules, and understanding this process is foundational in both organic chemistry and biology.

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Most popular questions from this chapter

Draw a structural formula for the principal organic product formed when each compound is treated with \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} / \mathrm{H}_{2} \mathrm{SO}_{4} .\) If there is no reaction, say so. (a) Butanal (b) Benzaldehyde (c) Cyclohexanone (d) Cyclohexanol

Explain why each name is incorrect. Write the correct IUPAC name for the intended compound. (a) \(3-\) Butanone (b) \(1-\) Butanone (c) \(4-\) Methylbutanal (d) 2,2 -Dimethyl-3-butanone

Answer true or false. (a) The reduction of an aldehyde always gives a primary alcohol. (b) The reduction of a ketone always gives a secondary alcohol. (c) The oxidation of an aldehyde gives a carboxylic acid. (d) The oxidation of a primary alcohol gives a ketone. (e) Tollens' reagent can be used to distinguish between an aldehyde and a ketone. (f) Sodium borohydride, \(\mathrm{NaBH}_{4}\), reduces an aldehyde to a primary alcohol. (g) The addition of one molecule of alcohol to the carbonyl group of a ketone gives a hemiacetal. (h) The reaction of an aldehyde with two molecules of alcohol gives an acetal, plus a molecule of water. (i) The formation of hemiacetals and acetals is reversible. (j) The cyclic hemiacetal formed from 4-hydroxypentanal has two stereocenters and can exist as a mixture of \(2^{2}=4\) stereoisomers.

Acetone is completely soluble in water, but 4 -heptanone is completely insoluble in water. Explain.

2-Methylbutene can be formed by dehydrating two different compounds (A and B) of molecular formula \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O} .\) When \(\mathbf{A}\) is mildly oxidized, compound \(\mathbf{C}\) \(\left(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}\right)\) forms, which can be oxidized further to a carboxylic acid. When compound B undergoes dehydration, 2-methylbutene is formed as the minor product. In addition, compound B cannot be oxidized further. Use this information to determine the identity of compounds \(\mathbf{A}, \mathbf{B},\) and \(\mathbf{C}\).

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