Advertisements
Advertisements
प्रश्न
Why carbohydrates are generally optically active?
Advertisements
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
APPEARS IN
संबंधित प्रश्न
Draw a neat diagram for the following:
Haworth formula of maltose
From the following identify an example of disaccharides.
Which carbon atoms of α- D glucopyranose and β-D-fructofuranose respectively are linked together to form glycosidic linkage in sucrose?
Raffinose, sucrose and stachyose are respectively ____________.
Identify the INCORRECT statement regarding glucose.
Identify the number of secondary carbon atoms in glucose.
Which among the following is a product of hydrolysis of one mole raffinose?
α-D (+) Glucose and β-D (+) glucose are ____________.
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 |
\[\ce{CH2OH - CO - (CHOH)4 - CH2 OH}\] is an example of ______.
