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प्रश्न
Why does compound (A) given below not form an oxime?
(A)
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उत्तर
Glucose pentaacetate (structure A) does not have a free –OH group at C, and therefore, cannot be converted to the open chain form to give a free –CHO group and hence it does not form the oxime.
(Structure A)
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संबंधित प्रश्न
Write the reaction that indicates the presence of -CHO group in glucose
Draw the simple Fisher projection formulae of D - (+) - glucose and D - (-) - fructose
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
Glucose on reaction with HI gives n-hexane. What does it suggest about the structure of glucose?
Write the reaction involved when D-glucose is treated with the following reagent:
(CH3CO)2O
The number of asymmetric carbon atom(s) below the figure is/are
What is the most abundant organic compound on earth?
Oxime is formed by treating glucose with ____________.
When glucose reacts with bromine water, the main product is ____________.
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?
Choose the correct relationship for glucose and fructose:
The number of chiral carbons in ß-D(+) glucose is ____________.
Which one is correct?
In the following reaction, identify A and B:
\[\begin{array}{cc}
\ce{C6H12O6 ->[Acetic anhydride] A}\\
\downarrow \text{Conc. nitric acid}\phantom{...}\\
\ce{B}\phantom{.................}\end{array}\]
Account for the following:
There are 5 OH groups in glucose
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:
