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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.

Short Answer

Expert verified

(a)

open chain form of tagatose

(b)

most stable cyclic hemiacetal form of tagatose

Step by step solution

01

Mutarotation

A spontaneous change in the specific rotation of a solution of an optically active compound is known as mutarotation. This implies that the two samples are different but in solution they form an equilibrium mixture. Sugars which are reducing can undergo mutarotation while non-reducing sugars cannot undergo mutarotation.

02

Bromine water oxidation of monosaccharides

Bromine water oxidises aldehyde group (-CHO) of an aldose into carboxylic acid (-COOH). Moreover, bromine water does not oxidize alcohols and ketoses. The product formed from bromine water oxidation is known as an aldonic acid.

03

Tollens test

Tollens reagent, [Ag(NH3)2]OH test is a qualitative test used to distinguish between an aldehyde an ketone. This test is also known as silver mirror test in which aldehydes are readily oxidized to corresponding carboxylic acids whereas ketones are not.

04

Formation of methyl ether

Hydroxy groups of sugar can be converted to methyl ethers by treating with methyl iodide (CH3I) and silver oxide (Ag2O).CH3-I. group can be polarised by silver oxide(Ag2O) , which makes the methyl carbon more strongly electrophilic. Attack by the -OH group of carbohydrate is then followed by deprotonation which gives the ether.

05

Drawing the structures

(a)

open chain form of tagatose

Tagatose reduces tollens reagent to form D-galactonic acid and D-talonic acid.

(b)

most stable cyclic hemiacetal form of tagatose

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Most popular questions from this chapter

Question. 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 fibres. Show the structure of cellulose acetate.

The relative configurations of the stereoisomers of tartaric acid were established by the following synthesis:

(1) D-(+)-glyceraldehydediastereomers A and B (separated)

(2) Hydrolysis of A and B using aqueous Ba(OH)2 gave C and D, respectively.

(3) HNO3 oxidation of C and D gave (-)-tartaric acid and meso-tartaric acid, respectively.

(a) You know the absolute configuration of D-(+)-glyceraldehyde, Use Fischer projections to show the absolute configurations of products A, B, C, and D.

(b) Show the absolute configurations of the three stereoisomers of tartaric acid: (+)-tartaric acid, (-)-tartaric acid, and meso-tartaric acid.

H. G. Khorana won the Nobel Prize in Medicine in 1968 for developing the synthesis of DNA and RNA and for helping to unravel the genetic code. Part of the chemistry he developed was the use of selective protecting groups for the 5โ€ฒ OH group of nucleosides.

The trityl ether derivative of just the 5โ€ฒ OH group is obtained by reaction of the nucleoside with trityl chloride, MMT chloride, or DMT chloride and a base like Et3N. The trityl ether derivative can be removed in dilute aqueous acid. DMT derivatives hydrolyze fastest, followed by MMT derivatives, and trityl derivatives slowest.

(a) Draw the product with the trityl derivative on the 5โ€ฒ oxygen.

(b) Explain why the trityl derivative is selective for the 5โ€ฒ OH group. Why doesnโ€™t it react at 2โ€ฒ or 3โ€ฒ? (c) Why is the DMT group easiest to remove under dilute acid conditions? Why does the solution instantly turn orange when acid is added to a DMT derivative?

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.

(a) Which of the D-aldopentoses will give optically active aldaric acids on oxidation with HNO3 ?

(b) Which of the D-aldotetroses will give optically active aldaric acids on oxidation withHNO3 ?

(c) Sugar X is known to be a D-aldohexose. On oxidation with HNO3 , X gives an optically inactive aldaric acid. When X is degraded to an aldopentose, oxidation of the aldopentose gives an optically active aldaric acid. Determine the structure of X.

(d) Even though sugar X gives an optically inactive aldaric acid, the pentose formed by degradation gives an optically active aldaric acid. Does this finding contradict the principle that optically inactive reagents cannot form optically active products?

(e) Show what products results if the aldopentose formed from degradation of X is further degraded to an aldotetrose. DoesHNO3 oxidize this aldotetrose to an optically active aldaric acid?

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