Advertisements
Advertisements
प्रश्न
On the basis of which evidences D-glucose was assigned the following structure?
\[\begin{array}{cc}
\ce{CHO}\\
|\phantom{....}\\
\phantom{..}\ce{(CHOH)4}\\
|\phantom{....}\\
\phantom{..}\ce{CH2OH}
\end{array}\]
Advertisements
उत्तर
This structure was assigned on the basis of the following evidences:
1. Molecular formula: The molecular formula of glucose has been found to be \[\ce{C6H12O6}\].
2. Straight chain structure:
(i) When aqueous solution of glucose is treated with sodium amalgam (Na/Hg) or sodium borohydride, it is reduced to sorbitol (or glucitol) a hexahedric alcohol.
\[\begin{array}{cc}
\phantom{.}\ce{CHO}\phantom{.......................}\ce{CH2OH}\phantom{..}\\
\phantom{}|\phantom{...........................}|\phantom{........}\\
\ce{(CHOH)4 + 2[H] ->[Na amalgam] (CHOH)4}\\
\phantom{}|\phantom{...........................}|\phantom{........}\\\
\phantom{..}\ce{CH2OH}\phantom{....................}\ce{\underset{Sorbitol}{CH2OH}\phantom{....}}
\end{array}\]
(ii) Prolonged heating with hydriodic acid and red phosphorus at 100°C gives a mixture of n-hexane and 2-iodohexane.
\[\begin{array}{cc}
\ce{\underset{Glucose}{CH2OH(CHOH)4CHO} ->[Hl][red P, 100°C] \underset{n-hexane}{CH3(CH2)4CH3} + CH3CH(CH2)3CH3}\\
\phantom{............................................}|\\
\phantom{............................................}\ce{\underset{2-Iodohexane}{I}}
\end{array}\]
The formation of n-hexane suggests that all the six carbon atoms in glucose are arranged in a straight chain structure of glucose.
3. Presence of five hydroxyl (-OH) groups: On acetylation with acetic anhydride, glucose gives a pentaacetate. This confirms that glucose contains five –OH groups. We know that the presence of two or more –OH groups on the same carbon atom makes the molecules unstable. Now since glucose is a stable compound, therefore, the five -OH groups must present on different carbon atoms.
4. Presence of one primary alcoholic group: On oxidation with cone, nitric acid, both glucose and gluconic acid give the same dicarboxylic acid, saccharic acid or glucaric acid. The primary alcoholic group \[\ce{(CH2OH)}\] is always present at the end of the carbon chain.
5. Presence of an aldehyde (-CHO) group: Glucose reacts with hydroxylamine, \[\ce{NH2OH}\] to form glucose CHO oxime. Which suggest that glucose contains a carbonyl \[\ce{(CHOH)4}\] (>C = O) groups.
APPEARS IN
संबंधित प्रश्न
Draw the simple Fisher projection formulae of D - (+) - glucose and D - (-) - fructose
Maltose is a
(a) Polysaccharide
(b) Disaccharide
(c) Trisaccharide
(d) Monosaccharide
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?
The following compound can be called as:
Acetylation of glucose yields ____________.
Which of the following statements is incorrect regarding glucose?
Which one of the following compounds is different from the rest?
When glucose reacts with bromine water, the main product is ____________.
Glucose reacts with acetic anhydride to form ______.
The number of chiral carbon atoms present in cyclic structure α-D(+) glucose:
The α-D glucose and β-D glucose differ from each other due to difference in carbon atom with respect to its ____________.
The two forms of D-glucopyranose obtained from the solution of D-glucose are called ____________.
Which one of the following reactions is not explained by the open chain Structure of glucose?
In the following reaction, identify A and B:
\[\begin{array}{cc}
\ce{C6H12O6 ->[Acetic anhydride] A}\\
\downarrow \text{Conc. nitric acid}\phantom{...}\\
\ce{B}\phantom{.................}\end{array}\]
Write the reactions of D-glucose which can’t be explained by its open-chain structure. How can cyclic structure of glucose explain these reactions?
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:
Give the reaction of glucose with hydrogen cyanide. Presence of which group is confirmed by this reaction?
