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प्रश्न
Why carbohydrates are generally optically active?
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उत्तर
Carbohydrates are generally optically active because they have one or more chiral carbon atoms in their molecules. For example, Glucose has four chiral carbons and therefore it is optically active.
\[\begin{array}{cc}
\ce{CHO}\\
|\phantom{....}\\
\ce{^*CHOH}\\
|\phantom{....}\\
\ce{^*CHOH}\\
|\phantom{....}\\
\ce{^*CHOH}\\
|\phantom{....}\\
\ce{^*CHOH}\\
|\phantom{....}\\
\phantom{...}\ce{CH2OH}
\end{array}\]
\[\ce{^*C}\] - Chiral carbon
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संबंधित प्रश्न
Write a commercial method for preparation of glucose.
Which carbon atoms of α- D glucopyranose and β-D-fructofuranose respectively are linked together to form glycosidic linkage in sucrose?
Which one of the following carbohydrates is insoluble in water?
How many optical isomers are possible for a compound having four asymmetric carbon atoms?
The two monosaccharides in a disaccharide are held together by ______ bonds.
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 |
Identify the product obtained in the following conversion.
\[\ce{Glucose ->[(O)][Br2 water] Product}\]
\[\ce{CH2OH - CO - (CHOH)4 - CH2OH}\] is an example of ______.
CH2 OH - CO - (CHOH)4 - CH2 OH is an example of ______.
Why carbohydrates are generally optically active.
