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प्रश्न
Monosaccharides contain carbonyl group hence are classified, as aldose or ketose. The number of carbon atoms present in the monosaccharide molecule are also considered for classification. In which class of monosaccharide will you place fructose?
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उत्तर
Monosaccharides contain carbonyl group. Hence, are classified as aldose or ketose. When aldehyde group is present, the monosaccharides are known as aldose. When ketone group is present, the monosaccharides are known as ketose. Fructose has molecular formula \[\ce{C6H12O6}\] containing six carbon and keto group and is classified as ketohexose.
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संबंधित प्रश्न
Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain.
Write two main functions of carbohydrates in plants.
What are reducing sugars?
Classify the following into monosaccharides and disaccharides.
Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose
The number of primary and secondary hydroxyl groups in ribose are -
(a) 1, 3
(b) 2, 3
(c) 3, 1
(d) 3, 2
The following molecule is called as:
\[\begin{array}{cc}
\ce{CHO}\phantom{..}\\
|\phantom{......}\\
\ce{(CHOH)4}\\
|\phantom{......}\\
\ce{CH2OH}
\end{array}\]
Which of the following gives a positive Fehling solution test?
Which of the following monosaccharide is pentose?
Monosaccharides usually contains ____________ carbon atoms.
A carbohydrate that cannot be hydrolysed into simpler units is called ____________.
Glucose can’t be classified as ____________.
Which of the following reagent cannot distinguish between glucose and fructose?
Three cyclic structures of monosaccharides are given below which of these are anomers.
| (I) |  |
| (II) |  |
| (III) |  |
Structure of a disaccharide formed by glucose and fructose is given below. Identify anomeric carbon atoms in monosaccharide units.
Assertion: Deoxyribose, \[\ce{C5H10O4}\] is not a carbohydrate.
Reason: Carbohydrates are hydrates of carbon so compounds which follow \[\ce{C_x(H2O)_y}\] formula are carbohydrates.
Reagent X is:
