Class 12 Chemistry Chapter 10: Biomolecules MCQs with answers
10 multiple-choice questions on Biomolecules, each with its answer and the reason, written from the NCERT chapter. For the CBSE board, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.
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Answer: (d) 2-deoxyribose, C₅H₁₀O₄
C₅H₁₀O₄ would need five oxygens to be C₅(H₂O)₅, yet 2-deoxy-D-ribose is an aldopentose present in nucleic acids. Lactic acid, C₃(H₂O)₃, and methanal, C(H₂O), fit the formula but are not carbohydrates; fructose fits and is a carbohydrate, so it is not the exception asked for.
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Answer: (d) sucrose
All monosaccharides, aldose or ketose, are reducing sugars, and maltose has a free aldehyde group available at C1 of its second unit. In sucrose the glycosidic linkage uses the reducing groups of both glucose and fructose, so it is non-reducing.
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Answer: (c) Bromine water oxidises it to a monocarboxylic acid with the same six carbon atoms.
Oxime and cyanohydrin formation show only that a >C=O group is present; ketones give them too. HI shows the straight chain. Oxidation by a mild oxidising agent to gluconic acid without loss of carbon is possible only for –CHO, so the carbonyl is an aldehyde group.
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Answer: (c) a primary alcoholic (–CH₂OH) group
Gluconic acid already has COOH at C-1; a second COOH can appear at C-6 only if C-6 carries a primary alcoholic –CH₂OH group, which nitric acid oxidises.
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Answer: (b) the –OH on C-5, the lowest asymmetric carbon, is on the right in the Fischer projection, as in (+)-glyceraldehyde
D gives the configuration relative to glyceraldehyde, judged at the lowest asymmetric carbon (C-5) with –CHO at the top; the sign of rotation is shown separately by (+). In D-glucose the –OH on C-3 is on the left, so not every –OH is on the right.
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Answer: (b) 4 and 5
Open chain: C-2, C-3, C-4 and C-5 each carry four different groups; C-1 (CHO) and C-6 (CH₂OH) do not, so 4. When the C-5 –OH adds to the –CHO, C-1 becomes a new asymmetric centre, the anomeric carbon, which is why α and β forms exist: 5 in the ring.
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Answer: (d) up and down
The α and β anomers differ only at C1. In the Haworth ring the β form has the C1 –OH up (α has it down), while the C4 –OH points down in both. Down and down is the α form at C1; the other options misplace C4.
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Answer: (d) C-1
The two cyclic forms are anomers: they differ only in the position of the –OH at C-1, the anomeric carbon, which was the aldehyde carbon before cyclisation.
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Answer: (c) the –OH at C-5 adds to the keto group at C-2
Addition of the C-5 –OH to the >C=O at C-2 closes a ring of one oxygen and four carbons (C-2 to C-5), like furan. C-1 carries no carbonyl; a C-4 –OH would give a four-membered ring and a C-6 –OH a six-membered one.
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Answer: (a) −19.95°
Hydrolysis of sucrose gives equal amounts of the two: (+52.5° − 92.4°)/2 = −19.95°. The sign is negative because fructose's laevorotation is larger. −39.9° forgets to take the mean; −72.45° adds the magnitudes.