JEE Chemistry Biomolecules MCQs Set 09

Answer: (c) Sucrose

Explanation:
1. A disaccharide is non-reducing only if the anomeric carbons of both units are tied up in the glycosidic bond.
2. In maltose, lactose and cellobiose the link is C1 to C4, so one unit keeps a free hemiacetal group and the sugar reduces Tollens' and Fehling's reagents.
3. In sucrose the link is between C1 of glucose and C2 of fructose, so no free hemiacetal or hemiketal group remains.
4. Hence sucrose is the non-reducing disaccharide.

Teacher's Note:
1. A free anomeric (hemiacetal) carbon is what makes a sugar reducing.
2. Sucrose has its two anomeric carbons joined to each other, which is why it does not reduce Tollens' or Fehling's reagent.

Answer: (d) a, b

Explanation:
1. The -NH2 group on the carbon next to -COOH makes lysine an α-amino acid, so (a) is true and (d) is false.
2. It has two amino groups and one carboxyl group, so it is a basic amino acid: (b) is true.
3. Lysine is an essential amino acid that the body cannot make, so (c) is false.
4. Correct statements: a and b.

Teacher's Note:
1. Count the amino and carboxyl groups to decide acidic, basic or neutral.
2. Lysine, valine, leucine and a few others are essential amino acids that must come from the diet.

Answer: (c) The pentaacetate of glucose does not react with hydroxylamine

Explanation:
1. Pentaacetate formation shows five -OH groups and the oxime formation and bromine water oxidation show a free -CHO group; all of these fit the open-chain structure.
2. If glucose were purely open-chain, its pentaacetate would still carry the free -CHO group and should react with hydroxylamine.
3. That the pentaacetate does not react shows the -CHO group is not free but tied up as a cyclic hemiacetal.
4. So only option (c) needs the cyclic structure.

Teacher's Note:
1. The open-chain structure explains the aldehyde reactions, while the cyclic hemiacetal explains why some aldehyde tests fail.
2. Failure to give the oxime, Schiff's test and NaHSO3 addition points to a ring form.

Answer: (b) Anomers

Explanation:
1. Epimers differ in configuration at any one stereocentre other than the anomeric carbon.
2. Cyclic forms that differ only in configuration at the new stereocentre C-1 (formed on ring closure) are anomers.
3. Hence the pair is anomers.

Teacher's Note:
1. Anomers are a special kind of epimer that differ at the hemiacetal carbon.
2. The α and β forms of glucose are the standard example.

Answer: (b) \( 5\,HCOOH + HCHO \)

Explanation:
1. Periodic acid cleaves each C-C bond that carries an -OH (or C=O) on adjacent carbons.
2. Glucose has 6 carbons and 5 such bonds. The terminal CHO carbon becomes HCOOH, the terminal CH2OH becomes HCHO, and each of the four middle CHOH carbons becomes HCOOH.
3. Product: 5 HCOOH + 1 HCHO.

Teacher's Note:
1. Each secondary alcohol carbon in the chain ends up as formic acid.
2. Only the primary alcohol end gives formaldehyde.

Answer: (d) Caprolactam : Nylon-6,10

Explanation:
1. Orlon is polyacrylonitrile, Nylon-6,6 comes from hexamethylenediamine and adipic acid, and Dacron from ethylene glycol and terephthalic acid, so (a), (b) and (c) are correct.
2. Caprolactam is the monomer of nylon-6, not nylon-6,10 (which is made from hexamethylenediamine and sebacic acid).
3. So (d) is the wrong match.

Teacher's Note:
1. Nylon-x,y names tell the carbon counts of the diamine and the diacid.
2. Nylon-6 alone is made from one monomer, caprolactam.

Answer: (b) Buta-1,3-diene and styrene

Explanation:
1. Buna-S is a copolymer of buta-1,3-diene and styrene (Bu = butadiene, na = sodium catalyst, S = styrene).
2. Butadiene with acrylonitrile gives Buna-N.

Teacher's Note:
1. The last letter in Buna-S and Buna-N names the comonomer.
2. S stands for styrene and N for acrylonitrile.

Answer: (b) c, d

Explanation:
1. Nylon-6,6 and nylon-6,10 contain amide (-CONH-) links, so they are polyamides.
2. Glyptal (glycerol with phthalic acid) and Dacron (ethylene glycol with terephthalic acid) are linked by ester groups, so they are polyesters.
3. The non-polyamides are c and d.

Teacher's Note:
1. Polyamides have -CONH- links and polyesters have -COO- links.
2. Check the functional group in the repeating unit, not the trade name.

Answer: (c) \( (-O-CH(CH_3)-CH_2-CO-O-CH(CH_2CH_3)-CH_2-CO-)_n \)

Explanation:
1. Option (a) is neoprene, (b) is Buna-N and (d) is nylon-6,6; these synthetic polymers are not readily broken down by microbes.
2. Option (c) is the aliphatic polyester PHBV (poly-3-hydroxybutyrate-co-3-hydroxyvalerate), which bacteria degrade.
3. Hence (c) is biodegradable.

Teacher's Note:
1. Aliphatic polyesters such as PHBV are used for biodegradable packaging.
2. PHBV is made from 3-hydroxybutanoic acid and 3-hydroxypentanoic acid.

Answer: (c) a, c, d

Explanation:
1. Mutarotation needs a free anomeric hemiacetal group that can open and re-close, interconverting the α and β forms.
2. Glucose, maltose and galactose all have a free hemiacetal group.
3. Sucrose has no free anomeric carbon, so it does not show mutarotation.
4. Hence a, c, d.

Teacher's Note:
1. Only sugars that can ring-open show mutarotation.
2. Sucrose is locked at both anomeric carbons.

Answer: (d) Epimers

Explanation:
1. Glucose and mannose differ only in the configuration at C-2.
2. Diastereomers that differ at a single stereocentre other than the anomeric carbon are called epimers.
3. Hence they are epimers.

Teacher's Note:
1. Glucose and galactose are C-4 epimers, glucose and mannose are C-2 epimers.
2. Do not confuse epimers with anomers, which differ at C-1.

Answer: (a) Sucrose

Explanation:
1. Sucrose is dextrorotatory (+66.5°). On hydrolysis it gives an equimolar mixture of D-glucose (+52.7°) and D-fructose (-92.4°).
2. The mixture has a net rotation of about -19.95°, so the sign of rotation inverts from dextro to laevo.
3. This mixture is called invert sugar, so it comes from sucrose.

Teacher's Note:
1. The change in sign of rotation on hydrolysis of sucrose is called inversion.
2. Invert sugar is an equimolar mixture of glucose and fructose.

Answer: (d) their configurations differ only at C-1 and C-2

Explanation:
1. Osazone formation with phenylhydrazine involves only C-1 and C-2 (the carbonyl carbon and the carbon next to it), which become -CH=N-NHPh and -C=N-NHPh.
2. In glucose and fructose the configurations at C-3, C-4 and C-5 are the same; they differ only at C-1 and C-2.
3. Once C-1 and C-2 are converted, both give the same osazone.

Teacher's Note:
1. Osazone formation destroys the stereo-difference at C-1 and C-2.
2. Glucose, mannose and fructose all give the same osazone.

Answer: (c) 9

Explanation:
1. In nucleosides the sugar C-1' is bonded to N-9 of purine bases (adenine, guanine).
2. In pyrimidine nucleosides the sugar is attached at N-1 of the base.
3. So the purine position is 9.

Teacher's Note:
1. Purine: N-9. Pyrimidine: N-1.
2. This is a N-glycosidic bond.

Answer: (b) a, b

Explanation:
1. Methionine has the side chain \( CH_3-S-CH_2CH_2- \) and cysteine has \( HS-CH_2- \), so both contain sulphur.
2. Leucine and glutamic acid contain only C, H, O and N.
3. Hence a and b.

Teacher's Note:
1. Cysteine units join through -S-S- bridges in proteins.
2. Methionine and cysteine are the two sulphur-containing standard amino acids.

Answer: (d) Cobalt

Explanation:
1. Vitamin B-12 (cyanocobalamin) is a coordination compound with cobalt at the centre of a corrin ring.
2. So the metal is cobalt.

Teacher's Note:
1. Cobalt is needed only in tiny amounts, as part of vitamin B-12.
2. Haemoglobin has iron and chlorophyll has magnesium, so do not confuse them.

Answer: (b) Anomers

Explanation:
1. The two forms differ only in the configuration at the hemiacetal carbon C-1 formed on ring closure.
2. Such cyclic forms are called anomers.
3. They are not mirror images (both are D) so they are not enantiomers.

Teacher's Note:
1. The α and β forms interconvert in water through the open-chain form.
2. This interconversion is mutarotation.

Answer: (a) 2,4-DNP

Explanation:
1. Glucose has a free aldehyde group only in a very small fraction of molecules in solution, since it is mostly in the cyclic hemiacetal form.
2. That small amount keeps reducing Tollens' and Fehling's reagents as the equilibrium shifts, so both tests are positive.
3. It does not give the 2,4-DNP test, so the answer is (a).

Teacher's Note:
1. Glucose does not give Schiff's test or the NaHSO3 addition either.
2. Reducing tests work because they pull the ring-opening equilibrium.

Answer: (d) All of these

Explanation:
1. α-D-glucose has a free hemiacetal group, so it ring-opens and shows mutarotation: (a) is correct.
2. Being cyclic, it does not give the 2,4-DNP test: (b) is correct.
3. Through the open-chain form it reduces Tollens' reagent: (c) is correct.
4. Hence all of these.

Teacher's Note:
1. Cyclic hemiacetal forms are in equilibrium with a small amount of free aldehyde.
2. That aldehyde accounts for reduction but not for 2,4-DNP.

Answer: (c) Cellulose

Explanation:
1. Glucose, fructose and D-ribose cannot be hydrolysed further into simpler sugars.
2. Cellulose is a polysaccharide made up of many β-glucose units.
3. So cellulose is the answer.

Teacher's Note:
1. Monosaccharides do not hydrolyse into smaller sugars.
2. Starch, cellulose and glycogen are polysaccharides.

Answer: (a) Mercurous and mercuric nitrate in \( HNO_3 \)

Explanation:
1. Millon's reagent is a solution of mercurous and mercuric nitrates in nitric acid.
2. It gives a white precipitate with proteins that turns red on heating, due to the phenolic group of tyrosine.

Teacher's Note:
1. Millon's test detects tyrosine, a phenolic amino acid.
2. The reagent is made from mercury dissolved in nitric acid.

Answer: (d) α-Amino acids

Explanation:
1. During digestion enzymes break proteins into peptones and peptides.
2. Further hydrolysis of the peptide bonds gives the free α-amino acids, which are absorbed.

Teacher's Note:
1. Pepsin, trypsin and peptidases break down proteins step by step.
2. The final products are the building blocks, the α-amino acids.

Answer: (d) Glucose

Explanation:
1. Glucose is a reducing sugar and reduces silver nitrate (Tollens' reagent) to metallic silver.
2. This silver deposit forms the mirror coating; sucrose, cellulose and starch are non-reducing.

Teacher's Note:
1. Glucose is the reducing agent in the Tollens' silver mirror reaction.
2. Sucrose can only be used after hydrolysis.