JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 04

Answer: (c) functional isomers

Explanation:
1. Both have the molecular formula C\(_4\)H\(_8\)O\(_2\).
2. One is a hydroxy-aldehyde and the other a carboxylic acid, so the functional groups differ.

Teacher's Note:
1. Same molecular formula with different functional groups gives functional isomers.
2. Write the molecular formula of each compound first.

Answer: (a) 4

Explanation:
1. Chain isomerism needs a straight-chain and a branched arrangement of the same carbon skeleton.
2. Butane and 2-methylpropane (C\(_4\)H\(_{10}\)) are the smallest such pair.

Teacher's Note:
1. With 3 or fewer carbons only one carbon skeleton is possible.
2. Pentane has three chain isomers.

Answer: (d) 3

Explanation:
1. With 3 carbons the functional group, double bond or substituent can occupy two different positions.
2. For example, propan-1-ol and propan-2-ol, or 1-chloropropane and 2-chloropropane.

Teacher's Note:
1. With only two carbons all positions are equivalent.
2. The source key lists 2 for this question; the correct minimum is 3.

Answer: (b) HOCH\(_2\)CH\(_2\)CHO

Explanation:
1. Propanoic acid is C\(_3\)H\(_6\)O\(_2\).
2. HOCH\(_2\)CH\(_2\)CHO is C\(_3\)H\(_6\)O\(_2\); the others are C\(_4\)H\(_8\)O\(_2\), C\(_3\)H\(_8\)O and C\(_3\)H\(_8\)O\(_2\).

Teacher's Note:
1. Isomers must have identical molecular formulae.
2. Count the carbon, hydrogen and oxygen atoms in each option.

Answer: (b) 5

Explanation:
1. Positional isomers of a ketone differ in the location of the C=O group.
2. Pentan-2-one and pentan-3-one is the smallest pair; butanone has only one position.

Teacher's Note:
1. Ketones need at least three carbons, but the carbonyl position matters only from five.
2. Check the possible positions of C=O in a four-carbon chain.

Answer: (d) CH\(_3\)OCH\(_2\)CH\(_2\)CH\(_3\) and CH\(_3\)CH\(_2\)OCH\(_2\)CH\(_3\)

Explanation:
1. Metamers have the same molecular formula and the same functional group but different alkyl groups on either side of it.
2. Methyl propyl ether and diethyl ether are both C\(_4\)H\(_{10}\)O ethers with different alkyl groups.

Teacher's Note:
1. Metamerism needs a polyvalent functional group such as ether, ketone or secondary amine.
2. The pair in the third option is the same compound written twice.

Answer: (a) the same molecular formula

Explanation:
1. Isomers are compounds with the same molecular formula but different arrangements of atoms.
2. They differ in structure and therefore in physical and often chemical properties.

Teacher's Note:
1. Definition: same molecular formula, different structure.
2. Different structural formulae lead to different properties.

Answer: (c) 5

Explanation:
1. The isomers are hexane, 2-methylpentane, 3-methylpentane, 2,3-dimethylbutane and 2,2-dimethylbutane.
2. These five are all the possible carbon skeletons.

Teacher's Note:
1. Draw systematically: the straight chain first, then one methyl branch, then two methyl branches.
2. C\(_5\)H\(_{12}\) has 3 and C\(_7\)H\(_{16}\) has 9 structural isomers.

Answer: (d) geometrical isomerism

Explanation:
1. The C=N double bond restricts rotation, and the carbon carries H and phenyl while nitrogen carries OH and a lone pair.
2. So syn (Z) and anti (E) forms exist.

Teacher's Note:
1. Oximes show geometrical isomerism about the C=N bond.
2. Each atom of the double bond must carry two different groups.

Answer: (a) butan-2-ol

Explanation:
1. Butan-2-ol, CH\(_3\)CH(OH)CH\(_2\)CH\(_3\), has a carbon bonded to four different groups: H, OH, CH\(_3\) and C\(_2\)H\(_5\).
2. Propan-2-ol and smaller alcohols have no such carbon.

Teacher's Note:
1. Look for a carbon bearing four different groups.
2. Butan-2-ol is also called s-butyl alcohol.

Answer: (c) molecular symmetry (internal compensation)

Explanation:
1. Meso-tartaric acid has a plane of symmetry, so the rotation caused by one half is cancelled by the other half within the molecule.
2. This is internal compensation due to molecular symmetry.

Teacher's Note:
1. A meso compound has chiral centres but is achiral overall.
2. External compensation refers to a racemic mixture.

Answer: (b) CH\(_3\)CH(OH)C\(_2\)H\(_5\)

Explanation:
1. Only butan-2-ol has a carbon attached to four different groups.
2. The other compounds have no chiral carbon.

Teacher's Note:
1. Check each carbon for four different substituents.
2. The CH\(_2\) carbon and the carboxyl carbon cannot be chiral.

Answer: (a) 16

Explanation:
1. Glucose, HOCH\(_2\)(CHOH)\(_4\)CHO, has 4 chiral carbons.
2. Number of stereoisomers = \( 2^4 = 16 \).

Teacher's Note:
1. Use \( 2^n \) for n dissimilar chiral centres.
2. An aldohexose has 4 chiral carbons; a ketohexose has 3.

Answer: (c) only I and II are chiral

Explanation:
1. I (3-methylhexane): the carbon carries H, CH\(_3\), C\(_2\)H\(_5\) and C\(_3\)H\(_7\), all different, so it is chiral.
2. II (3-methylpentan-2-one): the carbon carries H, CH\(_3\), COCH\(_3\) and C\(_2\)H\(_5\), so it is chiral.
3. III (3-methylpentane) has two identical ethyl groups on that carbon, so it is achiral.

Teacher's Note:
1. Check whether any two groups on the stereocentre are the same.
2. Two ethyl groups make the carbon achiral.

Answer: (b) lactic acid, CH\(_3\)CH(OH)COOH

Explanation:
1. Lactic acid has a carbon bonded to H, OH, CH\(_3\) and COOH, so it is chiral.
2. In the other acids the corresponding carbon has two identical groups (H and H, or two CH\(_3\)).

Teacher's Note:
1. Chirality requires four different groups on one carbon.
2. Two methyl groups on the same carbon remove the chirality.

Answer: (d) 180°

Explanation:
1. The fully eclipsed form has the two methyl groups at 0° dihedral angle, and the anti form has them at 180°.
2. So the rotation required is 180°.

Teacher's Note:
1. The dihedral angle between the two methyl groups is 0° in the fully eclipsed form and 180° in the anti form.
2. Rotation by 360° returns the same conformer.

Answer: (c) Anti (staggered with the methyl groups 180° apart)

Explanation:
1. In the anti conformer the bulky methyl groups are farthest apart, minimising steric and torsional strain.
2. The energy order is anti < gauche < partially eclipsed < fully eclipsed.

Teacher's Note:
1. Order of stability: anti > gauche > partially eclipsed > fully eclipsed.
2. Gauche is less stable because the methyl groups are 60° apart.

Answer: (a) enantiomers

Explanation:
1. Rotate the second projection by 180° in the plane of the paper: Br and H are interchanged vertically and F and Cl horizontally.
2. It then has H on top, Br at the bottom, Cl on the left and F on the right.
3. Compared with the first (Cl top, Br bottom, H left, F right), only one pair of groups (H and Cl) is exchanged, so they are mirror images.

Teacher's Note:
1. An odd number of interchanges of groups on a stereocentre gives the enantiomer.
2. A 180° rotation in the plane keeps the same compound.

Answer: (d) zero

Explanation:
1. [A] has the two OH groups on opposite sides, so it is a chiral (2R,3R or 2S,3S) diol.
2. [B] has both OH groups on the same side of the Fischer projection, giving a plane of symmetry, so it is meso.
3. A meso compound is optically inactive.

Teacher's Note:
1. Same-side OH groups in this symmetrical Fischer projection means meso.
2. Meso compounds show no optical rotation.

Answer: (b) 2R, 3S

Explanation:
1. C2: priorities Cl > C3 > CH\(_3\) > H. H is on a horizontal bond, so the apparent direction Cl, C3, CH\(_3\) is anticlockwise and the true configuration is R.
2. C3: priorities Cl > C2 > C\(_2\)H\(_5\) > H. The apparent direction Cl, C2, C\(_2\)H\(_5\) is clockwise, so the true configuration is S.

Teacher's Note:
1. In a Fischer projection, if the lowest priority group is on a horizontal bond, reverse the apparent direction.
2. Assign priorities by atomic number, then by the next atoms.

Answer: (b) a meso compound

Explanation:
1. Both Cl atoms are on the same side and the two ends are identical, so a horizontal plane of symmetry passes through the middle of the C2-C3 bond.
2. So the molecule is meso (R,S) and is optically inactive.

Teacher's Note:
1. A meso form needs identical halves related by a mirror plane.
2. If the two Cl atoms were on opposite sides, the compound would be chiral (R,R or S,S).

Answer: (d) CH\(_3\)CHClCH\(_2\)CH\(_3\) (2-chlorobutane)

Explanation:
1. 2-Chlorobutane has a carbon bonded to H, Cl, CH\(_3\) and C\(_2\)H\(_5\), so it is chiral.
2. Meso-tartaric acid is internally compensated; 1,1-dimethylcyclobutane and 2-methylpropane have no chiral carbon.

Teacher's Note:
1. A single stereocentre without internal symmetry gives optical activity.
2. Meso compounds are inactive.