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प्रश्न
On the basis of which evidences D-glucose was assigned the following structure?
\[\begin{array}{cc}
\ce{CHO}\\
|\phantom{....}\\
\phantom{..}\ce{(CHOH)4}\\
|\phantom{....}\\
\phantom{..}\ce{CH2OH}
\end{array}\]
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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.
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संबंधित प्रश्न
How many moles of acetic anhydride will be required to form glucose pentaacetate from 2M of glucose?
(a) 2
(b) 5
(c) 10
(d) 2.5
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?
What do you observe when glucose solution is heated with Tollen’s reagent?
What do you observe when glucose is treated with bromine water?
Write the reactions involved when D-glucose is treated with the following reagent:
H2N-OH
Which of the following statements is incorrect regarding glucose?
Which one of the following compounds is different from the rest?
Glucose reacts with acetic anhydride to form ______.
Which of the following properties of glucose cannot be explained by its open chain structure?
(i) Glucose does not form hydrogen sulphite with NaHSO3.
(ii) On oxidation with HNO3 glucose gives saccharic acid.
(iii) Glucose is found to exist in two different crystalline forms which are named as α and β.
Glucose is found to exist in two different α and β crystalline forms. These forms can be obtained by:
(i) The α form of glucose is obtained by crystallisation from a concentrated solution of glucose at 303 K.
(ii) The β form of glucose is obtained by crystallisation from a concentrated solution of glucose at 303 K.
(iii) The β form is obtained by crystallisation from hot and saturated aqueous solution at 371 K.
(iv) The α form is obtained by crystallisation from hot and saturated aqueous solution at 371 K.
The reaction of glucose with red P + HI is called ____________.
Which is the least stable form of glucose?
The α-D glucose and β-D glucose differ from each other due to difference in carbon atom with respect to its ____________.
In the following reaction, identify A and B:
\[\begin{array}{cc}
\ce{C6H12O6 ->[Acetic anhydride] A}\\
\downarrow \text{Conc. nitric acid}\phantom{...}\\
\ce{B}\phantom{.................}\end{array}\]
Why does compound (A) given below not form an oxime?
(A)
How will you distinguish 1° and 2° hydroxyl groups present in glucose? Explain with reactions.
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:
