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प्रश्न
Give two evidences for presence of formyl group in glucose.
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उत्तर
- Glucose gets oxidized to a six-carbon monocarboxylic acid called gluconic acid on reaction with bromine water which is a mild oxidizing agent. Thus, the carbonyl group in glucose is in the form of formyl (–CHO).
- Hemiacetal group of glucopyranose structure is a potential aldehyde (formyl) group. It imparts reducing properties to glucose. Thus, glucose gives positive Tollen’s test or Fehling test.
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संबंधित प्रश्न
Draw a neat diagram for the following:
Haworth formula of glucopyranose.
Draw a neat diagram for the following:
Haworth formula of maltose
Glucose on oxidation with dilute nitric acid gives _______________
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glucose into gluconic acid
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hydrogen iodide
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Write a chemical reaction to convert glucose into glucose cyanohydrin.
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Which of the following is the basic unit of carbohydrates?
From the following which is a heteropolysaccharide?
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i. Plant cell wall
ii. Exoskeleton of arthropods
iii. Paper from plant pulp
iv. Cotton fibre
Identify the WRONG statement.
By which of the following feature we can identify the relatively small DNA molecules of plasmids?
Identify a non-reducing carbohydrate from the following.
Monosaccharides are ______ in nature.
From the following identify the two types of glucose polymers present in starch.
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All these carbohydrates contain \[\ce{1 -> 4β}\] glycosidic linkage, EXCEPT ____________.
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Identify the number of secondary carbon atoms in glucose.
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On hydrolysis sucrose gives ____________.
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Lactose is made of ______.
Which carbon atoms of fructose are bonded together through oxygen forming fructofuranose?
Match the Column I with Column II and choose the correct answer from options below:
| Column I | Column II |
| A. Purine | 1. Glycogen |
| B. Pyrimidine | 2. Cellulose |
| C. Structural polysaccharide | 3. Glucagon |
| D. Storage polysaccharide | 4. Adenine |
| 5. Cytosine |
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Why carbohydrates are generally optically active?
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