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प्रश्न
How will you distinguish 1° and 2° hydroxyl groups present in glucose? Explain with reactions.
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उत्तर
On oxidation with nitric acid, glucose as well as gluconic acid both yield a dicarboxylic acid, saccharic acid. This indicates the presence of a primary alcoholic (–OH) group in glucose.
\[\begin{array}{cc}
\phantom{.......}\ce{CHO}\phantom{..............}\ce{COOH}\phantom{............}\ce{COOH}\phantom{....}\\
\phantom{......}|\phantom{...................}|\phantom{..................}|\phantom{........}\\
\phantom{.....}\ce{(CHOH)4 ->[Oxidation] (CHOH)4 <-[Oxidation] (CHOH)4}\\
\phantom{......}|\phantom{...................}|\phantom{..................}|\phantom{........}\\
\phantom{......}\ce{CH2OH}\phantom{............}\ce{\underset{acid}{\underset{Saccharic}{COOH}}\phantom{............}\ce{\underset{acid}{\underset{Gluconic}{CH2OH}}}\phantom{..}}\\
\end{array}\]
Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five –OH groups. Since it exists as a stable compound, five –OH groups should be attached to different carbon atoms.
\[\begin{array}{cc}
\phantom{...}\ce{CHO}\phantom{..................}\ce{CHO}\phantom{......}\ce{O}\phantom{..........}\\
\phantom{........}|\phantom{.......................}|\phantom{..........}||\phantom{...............}\\
\phantom{}\ce{(CHOH)4 ->[Acetic anhydride] (CH - O - C - CH3)4}\\
\phantom{.}|\phantom{.......................}|\phantom{...................}\\
\phantom{.....}\ce{CH2OH}\phantom{................}\ce{CH2 - O - C - CH3}
\phantom{.....}\\
\end{array}\]
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संबंधित प्रश्न
Draw the simple Fisher projection formulae of D - (+) - glucose and D - (-) - fructose
What happens when glucose is treated with hydroxylamine?
Enlist the properties of glucose that can not be explained on the basis of open chain structure of it
Glucose on reaction with HI gives n-hexane. What does it suggest about the structure of glucose?
Answer the following question.
What is the basic structural difference between glucose and fructose?
Choose the appropriate answer(s) for the below representation from the options given
What is the most abundant organic compound on earth?
Glucose does not give Schiff’s test because of the formation of cyclic ____________.
Glucose reacts with acetic anhydride to form ______.
Reduction of glucose by HI suggest that ____________.
Which of the following reactions of glucose can be explained only by its cyclic structure?
Which is the least stable form of glucose?
A solution of D-glucose in water rotates the plane polarised light ____________.
The number of chiral carbon atoms present in cyclic structure α-D(+) glucose:
The number of chiral carbons in ß-D(+) glucose is ____________.
On the basis of which evidences D-glucose was assigned the following structure?
\[\begin{array}{cc}
\ce{CHO}\\
|\phantom{....}\\
\phantom{..}\ce{(CHOH)4}\\
|\phantom{....}\\
\phantom{..}\ce{CH2OH}
\end{array}\]
Match List - I with List - II.
| List I | List II | ||
| (A) | Glucose + HI | (I) | Gluconic acid |
| (B) | Glucose + Br2 water | (II) | Glucose pentacetate |
| (C) | Glucose + acetic anhydride | (III) | Saccharic acid |
| (D) | Glucose + HNO3 | (IV) | Hexane |
Choose the correct answer from the options given below:
When D-glucose reacts with HI, it forms ______.
