JEE Chemistry Biomolecules MCQs Set 08

Answer: (a) an acetal linkage

Explanation:
1. The anomeric carbon of a sugar is a hemiacetal carbon
2. Its OH group condenses with the OH of another sugar unit and loses water
3. The resulting C-O-C bond between two sugar units is an acetal (glycosidic) linkage

Teacher's Note:
1. Disaccharides and polysaccharides are held together by glycosidic bonds. They are acetals, which is why they hydrolyse in dilute acid.

Answer: (c) laevorotatory

Explanation:
1. Sucrose is dextrorotatory, \([\alpha]_D=+66.5^\circ\)
2. Hydrolysis gives D-(+)-glucose (\(+52.5^\circ\)) and D-(-)-fructose (\(-92.4^\circ\))
3. Fructose rotates more strongly to the left, so the mixture is laevorotatory (invert sugar)

Teacher's Note:
1. The change of rotation from positive to negative is called inversion, so the mixture is called invert sugar. It is not racemic because the two products are different compounds.

Answer: (d) 8

Explanation:
1. Open-chain fructose has three chiral carbons: C3, C4 and C5
2. Number of optical isomers \(=2^n=2^3=8\)

Teacher's Note:
1. Use \(2^n\) for n different stereocentres. Glucose has four, giving 16; fructose has three, giving 8.

Answer: (c) hemiacetal

Explanation:
1. Glucose is an aldehyde sugar; ring closure joins C1 (the aldehyde carbon) to the OH on C5
2. A carbon bearing one OH and one OR group is a hemiacetal
3. So the cyclic form is a hemiacetal

Teacher's Note:
1. Ring closure of an aldose gives a hemiacetal; for a ketose such as fructose it gives a hemiketal. The anomeric carbon is the new stereocentre.

Answer: (a) Glucose and fructose

Explanation:
1. Phenylhydrazine reacts at C1 and C2 only, so any difference at C1 and C2 disappears
2. Glucose and fructose differ only at C1 and C2 and give the same osazone
3. Galactose differs from glucose at C4, so its osazone is different

Teacher's Note:
1. Glucose, mannose and fructose all give the same osazone because their C3 to C5 configurations are identical. Osazone formation destroys the stereochemistry at C1 and C2.

Answer: (b) Glycine

Explanation:
1. An optically active molecule needs a chiral carbon
2. The alpha carbon of glycine carries two hydrogen atoms (\(\ce{NH2CH2COOH}\))
3. So glycine has no chiral centre and is optically inactive

Teacher's Note:
1. Glycine is the only common amino acid with no chiral carbon. All other standard amino acids have an alpha carbon with four different groups.

Answer: (d) Alanylglycine

Explanation:
1. The N-terminal residue carries the free \(\ce{NH2}\) group and is \(\ce{H2N-CH(CH3)-CO-}\), which is alanine
2. The C-terminal residue, with free COOH, is \(\ce{-NH-CH2-COOH}\), which is glycine
3. The N-terminal acid is named first with the ending -yl: alanylglycine

Teacher's Note:
1. In peptide names the N-terminal amino acid is written first and gets the -yl ending. Reversing the order gives a different dipeptide.

Answer: (c) hydrogen bond

Explanation:
1. The peptide link has a C=O group and an N-H group
2. The N-H of one chain forms a hydrogen bond with the C=O of the neighbouring chain
3. This holds the chains side by side in the secondary structure

Teacher's Note:
1. Hydrogen bonds between C=O and N-H groups stabilise both the alpha helix and the beta pleated sheet. The peptide linkage itself joins amino acids within one chain.

Answer: (a) \(\ce{H2N-C(=NH)-NH-(CH2)3-CH(NH2)-COOH}\)

Explanation:
1. A basic amino acid has more basic groups than acidic groups
2. Option 1 has a guanidino group plus the alpha amino group and only one COOH (this is arginine)
3. Options 3 and 4 have two COOH groups (acidic amino acids) and option 2 has one each of amino, carboxyl and alcohol groups (neutral)

Teacher's Note:
1. Count the basic groups (amino, guanidino, imidazole) against the acidic groups (COOH). Arginine, lysine and histidine are the basic amino acids.

Answer: (b) C6H5-CH(NH2)-COOH

Explanation:
1. In an alpha-amino acid the amino group is on the carbon directly next to COOH
2. Option 2: the carbon carrying COOH also carries the \(\ce{NH2}\) group, so it is an alpha-amino acid
3. In the other options the nitrogen is one carbon further away (beta position)

Teacher's Note:
1. Locate the carbon attached to COOH (alpha carbon) and see whether the amino group is on it. If it is on the next carbon, it is a beta-amino acid.