Draw the structures of the compounds named in Problem 23-20 parts (a), (c), and (d). Allose is the C3 epimer of glucose and ribose is the C2 epimer of arabinose.

Short Answer

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Pyranose is a six-membered cyclic hemiacetal while a furanose is a five-membered cyclic hemiacetal.

Step by step solution

01

Pyranose and furanose

Pyranose is a six-membered cyclic hemiacetal while a furanose is a five-membered cyclic hemiacetal.

02

Epimers

Sugars which have the common configuration like the arrangement of H - and - OH around carbon atoms are same except the carbon present at α -position or carbon number 2 are known as epimers. For example, glucose and mannose are epimers of each other.

03

Anomers

The diastereomers resulting from cyclisation are known as anomers. They differ only in the configuration around first carbon (C1) which is referred to as the anomeric carbon (hemiacetal carbon atom).

04

Identifying α and β anomers of pyranoses and furanoses

For α -anomer, hydroxy group (-OH) on the anomeric carbon is down (axial position) while for β -anomer, hydroxy group (-OH) on the anomeric carbon is up (equatorial position). Again, for α -anomer, the anomeric hydroxy (-OH) group is placed trans to the terminal -CH2OH group while for β -anomer, the anomeric hydroxy (-OH) group is placed cis to the terminal -CH2OH group.

05

Structure of the sugars

In chair conformation of methyl α -D-galactopyranoside, the anomeric carbon is C1. The methyl group (-CH3) on the anomeric carbon is down (axial position).

In chair conformation of α -D-galactopyranoside, the anomeric carbon is C1. The hydroxy group (-OH) on the anomeric carbon is down (axial position).

In Haworth projection of β-D-ibofuransoide, the anomeric carbon is C1. The ethyl group (-CH2CH3)on the anomeric carbon is up which is cis to the terminal -CH2CH group.

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Most popular questions from this chapter

Cellulose is converted to cellulose acetateby treatment with acetic anhydride and pyridine. Cellulose acetate is soluble in common organic solvents, and it is easily dissolved and spun into fibers. Show the structure of cellulose acetate.

Treatment of either anomer of fructose with excess ethanol in the presence of a trace of HCI gives a mixture of the α and β anomers of ethyl-D-fructofuranoside. Draw the starting materials, reagents, and products for this reaction. Circle the aglycone in each product.

When the gum of the shrub Sterculia setigera is subjected to acidic hydrolysis, one of the water-soluble components of the hydrolysate is found to be tagatose. The following information is known about tagatose:

(1) Molecular formula C6H12O6

(2) Undergoes mutarotation.

(3) Does not react with bromine water.

(4) Reduces Tollens reagent to give d-galactonic acid and d-talonic acid.

(5) Methylation of tagatose (using excess CH3 I and Ag2O) followed by acidic hydrolysis gives 1,3,4,5-tetra-O-methyltagatose.

(a) Draw a Fischer projection structure for the open-chain form of tagatose.

(b) Draw the most stable conformation of the most stable cyclic hemiacetal form of tagatose.

An important protecting group developed specifically for polyhydroxy compounds like nucleosides is the tetraisopropyl-disiloxanyl group, abbreviated TIPDS, that can protect two alcohol groups in a molecule.

(a) The TIPDS group is somewhat hindered around the Siatoms by the isopropyl groups. Which OHis more likely to react first with TIPDS chloride? Show the product with the TIPDS group on one oxygen.

(b) Once the TIPDS group is attached at the first oxygen, it reaches around to the next closest oxygen. Show the final product with two oxygens protected.

(c) The unprotected hydroxy group can now undergo reactions without affecting the protected oxygens. Show the product after the protected nucleoside from (b) is treated with tosyl chloride and pyridine, followed by NaBr, ending with deprotection with Bu4NF.

All of the rings of the four heterocyclic bases are aromatic. This is more apparent when the polar resonance forms of the amide groups are drawn, as is done for thymine at left. Redraw the hydrogen-bonded guanine-cytosine and adenine-thymine pairs shown in figure 23-24, using the polar resonance forms of the amides. Show how these forms help to explain why the hydrogen bonds involved in these pairings are particularly strong. Remember that a hydrogen bond arises between an electron-deficient hydrogen atom and electron-rich pair of nonbonding electrons.

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