JEE Chemistry Biomolecules MCQs Set 29

Answer: (d) \( \ce{C20H34O17} \)

Explanation:
1. A pentose is \( \ce{C5H10O5} \).
2. Four pentose units: \( \ce{C20H40O20} \).
3. Each glycosidic linkage removes one \( \ce{H2O} \); a tetrasaccharide has 3 linkages, so subtract \( \ce{H6O3} \).
4. Formula: \( \ce{C20H34O17} \).

Teacher's Note:
1. n monosaccharides joined by (n-1) glycosidic bonds lose (n-1) molecules of water.
2. Subtract water from the sum of monosaccharide formulae.

Answer: (c) originally supplied by the sun

Explanation:
1. Plants make glucose by photosynthesis using solar energy.
2. That energy is stored in the chemical bonds of glucose.
3. Oxidation of glucose in tissues releases the stored solar energy.

Teacher's Note:
1. Photosynthesis converts light energy to chemical energy.
2. Respiration releases this energy for the cell.

Answer: (c) Bromine water

Explanation:
1. Fructose isomerises to glucose-type enediol in basic medium, so it also reduces Tollens' and Fehling's reagents.
2. Bromine water is a mild acidic oxidant that oxidises the aldehyde group of glucose to gluconic acid but does not oxidise a ketose like fructose.
3. So only bromine water differentiates them.

Teacher's Note:
1. Bromine water (acidic) distinguishes aldoses from ketoses.
2. Tollens' and Fehling are alkaline, so ketoses react through enediol rearrangement.

Answer: (d) reacts with neither phenylhydrazine nor Tollens' reagent

Explanation:
1. Five OH groups (including the hemiacetal OH) are acetylated.
2. This shows that glucose exists in the cyclic form, with no free CHO group in the pentaacetate.
3. With no free aldehyde group it neither reacts with phenylhydrazine nor reduces Tollens' reagent.

Teacher's Note:
1. Pentaacetate formation is evidence for five OH groups; the absence of CHO reactions supports the cyclic structure.
2. Acetylation locks the ring so that mutarotation stops.

Answer: (d) meso form

Explanation:
1. D-glucose has four stereocentres and no plane or centre of symmetry.
2. So it can exist as (+), (-) (the L-form) and racemic mixture, but a meso form is impossible.

Teacher's Note:
1. Meso needs internal symmetry.
2. Glucose's chain has different groups at both ends (CHO and CH2OH).

Answer: (b) Anomers are also epimers

Explanation:
1. Epimers are diastereomers that differ in configuration at exactly one stereocentre.
2. Anomers differ only at the anomeric carbon (C-1 in aldoses), so they are a special type of epimer.
3. The reverse is not true: C-2 epimers (glucose and mannose) are not anomers.

Teacher's Note:
1. Anomers are the subset of epimers that differ at the anomeric carbon.
2. Glucose and mannose are epimers but not anomers.

Answer: (c) alpha-L-(-)-glucose

Explanation:
1. An enantiomer has the opposite configuration at every stereocentre.
2. Inverting all centres of alpha-D gives alpha-L (the anomeric configuration relative to the others is retained).
3. The enantiomer rotates plane-polarised light equally in the opposite direction: (-).

Teacher's Note:
1. D/L designation changes with enantiomer; alpha/beta stays the same relative to the ring.
2. D and L are not related to the sign of rotation.

Answer: (b) diastereomers

Explanation:
1. Reduction creates a new stereocentre at the carbonyl carbon, giving two products that differ only at that carbon.
2. One is meso-erythritol and the other is D-threitol.
3. Stereoisomers that are not mirror images are diastereomers.

Teacher's Note:
1. Reduction of a prochiral carbonyl gives two products which are epimers/diastereomers when the molecule already has a stereocentre.
2. A meso compound and a chiral compound are always diastereomers.

Answer: (d) all of the three

Explanation:
1. The two cyanohydrins are stereoisomers that are not mirror images, so they are diastereomers.
2. They differ at exactly one stereocentre, so they are epimers.
3. The differing centre is C-2, so they are C-2 epimers.

Teacher's Note:
1. Epimers are a type of diastereomer.
2. Kiliani-Fischer synthesis gives a pair of C-2 epimers.

Answer: (d) all of the above statements are correct

Explanation:
1. alpha- and beta-D-glucose differ only in configuration at C-1.
2. They are not mirror images, so they are diastereomers.
3. Differing at one centre makes them epimers; differing at the anomeric carbon makes them anomers.
4. All three statements hold.

Teacher's Note:
1. Anomers are epimers; epimers are diastereomers.
2. The two anomers interconvert in solution (mutarotation).

Answer: (c) both enolisation and tautomerisation

Explanation:
1. In base, the alpha-hydrogen of the aldehyde is removed to give an enediol (enolisation).
2. The enediol is the common enol-form for glucose, mannose and fructose.
3. Ketonisation of the enediol (keto-enol tautomerism) gives D-fructose, D-glucose or D-mannose.

Teacher's Note:
1. Lobry de Bruyn-van Ekenstein transformation uses an enediol intermediate.
2. Keto-enol interconversion is tautomerism.

Answer: (d) C-1 in glucose and C-2 in fructose

Explanation:
1. The anomeric carbon is the new stereocentre formed from the carbonyl carbon on ring closure.
2. In glucose (aldose) the carbonyl carbon is C-1.
3. In fructose (ketose) it is C-2.

Teacher's Note:
1. Anomeric carbon = former carbonyl carbon.
2. It carries the hemiacetal/hemiketal OH.

Answer: (c) both diastereomers and epimers

Explanation:
1. Reduction of C-2 carbonyl creates a new stereocentre, giving two hexitols.
2. They differ only at C-2, so they are epimers.
3. They are not mirror images, so they are also diastereomers.

Teacher's Note:
1. Epimers are special diastereomers.
2. Anomers apply only to cyclic forms.

Answer: (d) Z is n-heptanoic acid and Z' is a branched (substituted) hexanoic acid

Explanation:
1. For the aldose: cyanohydrin, then hydrolysis to aldonic acid (7 carbons), then HI/P reduces all OH to give n-heptanoic acid.
2. For the ketose, the cyano group is attached to C-2, which also bears the chain: the product has a branch (methyl-type) at C-2, i.e. 2-methylhexanoic acid.
3. So Z' is a substituted hexanoic acid.

Teacher's Note:
1. The CN group is attached to the former carbonyl carbon; in a ketose that carbon is internal, giving branching.
2. HI/P completely reduces sugar acids to straight (or branched) alkanoic acids.

Answer: (b) \( \ce{CH2OH-CO-CHOH-CH2OH} \) (a ketotetrose)

Explanation:
1. Periodic acid cleaves C-C bonds between adjacent OH/carbonyl groups.
2. In an aldose the terminal CHO becomes formic acid.
3. In a ketose the C=O carbon, which sits between two CHOH/CH2OH carbons, is oxidised to \( \ce{CO2} \).

Teacher's Note:
1. Aldoses give HCOOH; ketoses give CO2 from the ketone carbon.
2. Count oxidised carbons: ends become HCHO.

Answer: (b) Hemiacetals

Explanation:
1. Benedict's reagent reacts with sugars that can ring-open to give a free aldehyde (or alpha-hydroxy ketone).
2. Hemiacetals can open to the free carbonyl form, so they reduce Benedict's reagent.
3. Acetals (glycosides) and acetylated sugars cannot open.

Teacher's Note:
1. Reducing sugars have a free hemiacetal/hemiketal OH.
2. Glycosides are non-reducing.

Answer: (d) No characteristic colour or precipitate will be formed

Explanation:
1. Glucose with methanolic HCl gives methyl glucoside (an acetal).
2. Acetals cannot open to the carbonyl form under the basic conditions of Tollens' test.
3. So no silver mirror or colour appears.

Teacher's Note:
1. Glycosides are non-reducing.
2. Mutarotation is also absent in glycosides.

Answer: (b) Glycosides show the phenomenon of mutarotation

Explanation:
1. Fructose and other ketoses exist in cyclic hemiketal forms which interconvert in solution, so they mutarotate.
2. Glycosides are fixed acetals and cannot interconvert, so they do not mutarotate.
3. Both mutarotation and reduction of Tollens' reagent need a free hemiacetal OH.

Teacher's Note:
1. Glycosides: no mutarotation, no reduction of Tollens' reagent.
2. Mutarotation implies a free anomeric OH.

Answer: (a) The OH groups of sugars are more acidic than those of a typical alcohol

Explanation:
1. Sugars have many electronegative oxygens in the molecule which exert a -I effect and stabilise the alkoxide.
2. The anomeric OH in particular is more acidic (hemiacetal).
3. Glucose has pKa about 12.3 against about 16 for ethanol.

Teacher's Note:
1. Neighbouring electron-withdrawing OH groups stabilise the conjugate base.
2. Glucose dissolves in aqueous NaOH forming alkoxides.

Answer: (c) invert sugar

Explanation:
1. Honey is mainly an equimolar mixture of glucose and fructose (invert sugar).
2. Fructose is the sweetest natural sugar, so the mixture is sweeter than glucose alone.

Teacher's Note:
1. Invert sugar = glucose + fructose from hydrolysis of sucrose.
2. Fructose is sweeter than sucrose.

Answer: (c) Inulin

Explanation:
1. Amylose, amylopectin and glycogen are polymers of alpha-D-glucose.
2. Inulin is a polymer of fructose.

Teacher's Note:
1. Starch = amylose + amylopectin; glycogen is animal starch.
2. Inulin is a fructan found in roots such as chicory.

Answer: (d) Sucrose

Explanation:
1. Amylose, amylopectin and inulin are long polymers of monosaccharides.
2. Sucrose is a disaccharide of glucose and fructose.

Teacher's Note:
1. Polysaccharides: starch, cellulose, glycogen, inulin.
2. Sucrose, maltose, lactose are disaccharides.