Question 1Level 2 · ExamSingle correct
The open-chain formula of D-glucose, OHC–(CHOH)4–CH2OH, explains many of its reactions, but some observations require the cyclic (hemiacetal) structure. Which one of the following observations is explained by the open-chain structure alone and does NOT require the cyclic structure? Exactly one option is correct.
AGlucose does not give Schiff's test and does not form the hydrogensulfite addition product with NaHSO3.
BGlucose pentaacetate does not react with hydroxylamine.
CGlucose reacts with hydroxylamine to form an oxime and adds HCN to form a cyanohydrin.
DGlucose exists in two crystalline forms with different melting points and different specific rotations.
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Answer: C
(C) Oxime and cyanohydrin formation are typical carbonyl reactions and follow directly from the free –CHO group of the open-chain formula.
(A) An aldehyde should give Schiff's test and a hydrogensulfite adduct; their absence shows that the –CHO group is largely tied up as a cyclic hemiacetal.
(B) In the pentaacetate the C-1 OH is acetylated, so the ring cannot open to a free –CHO group; no oxime forms. This needs the cyclic structure.
(D) The two crystalline forms are the α- and β-anomers, which exist only because C-1 becomes a new stereocentre on ring closure.
Answer: (C).
Difficulty: HardQuestion 2Level 3 · AdvancedSingle correct
D-Glucose is treated separately with (i) bromine water, giving P; (ii) dilute nitric acid on warming, giving Q; (iii) NaBH4 in water, giving R. Which one of the following statements is correct? Exactly one option is correct.
AP and Q are both dicarboxylic acids.
BQ has molecular formula C6H10O8 and contains four stereocentres.
CR reduces Tollens' reagent because it still contains the C-1 carbon of glucose as a –CH2OH group.
DP is formed by oxidation of the C-6 –CH2OH group of glucose.
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Answer: B
P: bromine water (mild oxidant) oxidises only the –CHO group: gluconic acid, HOOC–(CHOH)4–CH2OH, C6H12O7, a monocarboxylic acid. So (A) and (D) are wrong (C-6 is untouched by Br2 water).
Q: nitric acid oxidises both the –CHO and the primary –CH2OH group: saccharic (glucaric) acid, HOOC–(CHOH)4–COOH. Formula: C6H10O8 (C 6, H 2+4+4=10, O 4+4=8). Its stereocentres are C-2, C-3, C-4, C-5 (each carries H, OH and two different carbon chains): four. (B) is correct.
R: NaBH4 reduces –CHO to –CH2OH: sorbitol (D-glucitol), HOCH2–(CHOH)4–CH2OH, a polyol with no carbonyl group (not even latent), so it does not reduce Tollens' reagent. (C) is wrong.
Answer: (B).
Difficulty: Very hardQuestion 3Level 3 · AdvancedSingle correct
In Fischer projections (carbon chain vertical, C-1 at the top), the positions of the OH groups on the stereocentres are:
D-glucose, OHC–(CHOH)4–CH2OH: C-2 right, C-3 left, C-4 right, C-5 right.
D-mannose, OHC–(CHOH)4–CH2OH: C-2 left, C-3 left, C-4 right, C-5 right.
D-fructose, HOCH2–CO–(CHOH)3–CH2OH: C-3 left, C-4 right, C-5 right.
Reduction of D-glucose with NaBH4 gives D-glucitol (sorbitol) and reduction of D-mannose gives D-mannitol. When D-fructose is reduced with NaBH4, the product is ______. Exactly one option is correct.
AD-glucitol only, because fructose and glucose have the same configuration at C-3, C-4 and C-5
BD-mannitol only, because the C-2 carbonyl carbon of fructose is attacked by hydride from one face only
Ca single optically inactive (meso) hexitol
Da mixture of D-glucitol and D-mannitol
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Answer: D
Reduction converts the C-2 ketone (trigonal, not a stereocentre) into a –CH(OH)– group that is a new stereocentre. Hydride can add to either face of the planar C=O, so both configurations at C-2 are formed.
C-3, C-4, C-5 are untouched (left, right, right), and C-1 and C-6 are both –CH2OH, so the products are HOCH2–(CHOH)4–CH2OH with C-2 right (identical to D-glucitol) and C-2 left (identical to D-mannitol). These are diastereomers, both optically active; neither is meso (the pattern right, left, right, right or left, left, right, right is not symmetric end to end).
(A) ignores the new stereocentre; (B) is unjustified (a planar carbonyl is attacked from both faces); (C) is wrong because both hexitols are chiral.
Answer: (D).
Insight: glucose, mannose and fructose share C-3 to C-5; reduction at C-2 of fructose regenerates a stereocentre, giving the alditols of both aldoses.
Difficulty: Peak challenge