Like glucose, galactose mutarotates when it dissolves in water. The specific rotation of α-D-galactopyranoseis +150.70 , and that of the β anomer is +52.80 . When either of the pure anomers dissolves in water, the specific rotation gradually changes to +80.20. Determine the percentages of the two anomers present at equilibrium.

Short Answer

Expert verified

αanomer=28%andβanomer=72%

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.

02

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).

03

Calculation

If the fraction of galactose present as the αanomer is role="math" localid="1664865398062" α=+150.70 is a, the fraction present as the βanomer is α=+52.80, and the specific rotation of the mixture is +80.2, then we can write as:

role="math" localid="1664865861024" a+150.70+b+52.80=+80.20

The fraction present as the α anomer (a) plus the fraction present as β anomer (b) should account for all the galactose.

a+b = 1 or we can write b = 1-a

Now,

a+150.70+b+52.80=+80.20a+150.70+1-a52.80=80.20a×150.70+52.80-a×52.80=80.20a150.70-52.80=80.20-52.80a×97.90=27.40a=27.4097.90a=0.28

So,

b=1-a=1-0.28=0.72

Hence, the amounts of two anomers present at equilibrium are α anomer = 28% and β anomer = 72%.

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Most popular questions from this chapter

Predict the products formed when the following sugars react with excess acetic anhydride and pyridine.

  1. α-D-glucopyranose
  2. β-D-ribofuranose

(a) Figure 23-2 shows that the degradation of D-glucose gives D-arabinose, an aldopentose. Arabinose is most stable in its furanose form. Draw D-arabinofuranose.

(b) Ribose, the C2 epimer of arabinose, is most stable in its furanose form. Draw D-ribofuranose.

(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?

Draw the following sugar derivatives.

  1. ethyl-β-D-glucopyranoside
  2. 2,3,4,6-tetra-O-ethyl-D-mannopyranose
  3. 1,3,6-tri-O-ethyl-D-fructofuranose
  4. Ethyl-2,3,4,6-tetra-O-methyl-β-D-galactopyranoside

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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