The two stereoisomeric forms of glucose, i.e., α-D-glucose and β-D-glucose exist in separate crystalline forms and thus have different melting points and specific roations. For example α-D-glucose has a m.p. of 419 K with a specific rotation of +112° while  β-D-glucose has a m.p. of 424 K and has a specific rotation of +19°. However, when either of these two forms is dissolved in water and allowed to stand, it gets converted into an equilibrium mixture of α-and β-forms through a small amount of the open chain form.


As a result of this equilibrium, the specific rotation of a freshly prepared solution of α-D-glucose gradually decreases from of +112° to +52.7° and that of β-D-glucose gradually increases from +19° to +52.7°.


This change in specific rotation of an optically active compound in solution with time, to an equilibrium value, is called mutarotation. During mutarotation, the ring opens and then recloses either in the inverted position or in the original position giving a mixture of α-and-β-forms. All reducing carbohydrates, i.e., monosaccharides and disacchardies (maltose, lactose etc.) undergo mutarotation in aqueous solution.
    Glucose forms an oxime but glucose pentaacetate does not. Explain

(a)    With HI/P: It undergoes reduction to form n-hexane while with sodium amalgam it forms sorbitol.


(b)    With H2O: It forms neutral solution
(c)    With Hydroxylamine (NH2OH)


(d)    With HCN: It forms addition product cyanohydrin


(e)    Oxidation: Glucose on oxidation with Br2 gives gluconic acid which on further oxidation with HNO3 gives glucaric acid


(f)    With Tollen reagent and Fehling solution. Glucose forms silver mirror and red ppt. of Cu2O respectively.
(g)    With acetic anhydride. In presence of pyridine glucose forms pentaacetate.


(h)    With phenylhydrazine: it forms glucosazone


(j)     Glycoside formation: When a small amount of gaseous HCl is passed into a solution of  D (+) glucose in methanol , a reaction takes place that results in the formation of anomeric methyl acetals.

(i)    With conc. HCl acid: Glucose gives laevulinic acid


(j)     Glycoside formation: When a small amount of gaseous HCl is passed into a solution of  D (+) glucose in methanol , a reaction takes place that results in the formation of anomeric methyl acetals.


Carbohydrate acetals, generally are called glycosides and an acetal of glucose is called glucoside.
Solved example.     Glucose and fructose give the same osazone. One may therefore conclude that 
            (A)    glucose and fructose have identical structures
            (B)    glucose and fructose are anomers 
            (C)    the structures of glucose and fructose have mirror – image relationship
            (D)    the structure of glucose and fructose differ only in those carbon atoms which take part in asazone formation.
Solution:        (D) Glucose and fructose give the same osazone, but differ from each other only in configuration at C1 and C2. 
Other reactions
(a)    Kiliani - Fischer Synthesis: - This is a method of lengthening the carbon chain of an aldose. To illustrate, we take synthesis of D-threose and D-erythrose (Aldotetroses) from D-glyceraldehyde (an aldotriose).
    Addition of HCN to glyceraldehyde produces two epimeric cyanohydrins because reaction creates a new stereocenter. The cyanohydrins can be separated easily (since they are diastereomers) and each can be converted to an aldose through hydrolysis, acidification and lactonisation, and reduction with Na—Hg in presence of H2SO4. One cyanohydrin ultimately yields D-erythrose and D-threose. 
    Here we can see that both sugars are D-sugars because starting compound is 
D-glyceraldehyde and its stereocentrer is unaffected by its synthesis.
(b)    Ruff Degradation: It is opposite to Kiliani Fischer synthesis that can be used to shorten the chain by a similar unit. The ruff degradation involves (i) Oxidation of the aldose to an aldonic acid using Bromine water. (ii)Oxidative decarboxylation of the aldonic acid to the next lower aldose using H2O2 and Fe2(SO4)3. D-ribose for example can be reduced to D-erythrose.


Solved example.     The open – chain glucose on oxidation with HIO4 gives 
            (A)    5 HCOOH + H2C = O    (B)    4 HCOOH + 2H2C = O 
            (C)    3 HCOOH + 3H2C = O    (D)    2HCOOH + 4H2C = O
Solution:        (A) C = O both CHO and CHOH form HCOOH. While HOCH2 forms H2C = O giving 5 HCOCH + H2C = O. 

Exercise .
    Fructose reduces Tollen’s reagent due to 
    (A)    presence of ketonic group     
    (B) presence of NH4OH in Tollen’s reagent 
    (C)    rearrangement of fructose into mixture of glucose, glactose and mannose 
    (D)    both (B) and (C)   

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