JEE Chemistry Isomerism MCQs Set 08

Answer: (b) chain isomers

Explanation:
1. Both have the formula C₄H₁₀ but differ in the arrangement of the carbon skeleton (straight chain versus branched chain).
2. Such isomers are chain (nuclear) isomers.
3. Answer: (b)

Teacher's Note:
1. Different carbon skeleton: chain isomerism.
2. Different location of a group: position isomerism.

Answer: (b) ring-chain isomerism

Explanation:
1. Both have the formula C₃H₆, one is an open chain alkene and the other a ring.
2. Isomers differing in having an open chain or a ring structure are ring-chain isomers.
3. Answer: (b)

Teacher's Note:
1. Same formula, one with a ring.
2. Degrees of unsaturation are equal.

Answer: (c) CH₃COCH₃

Explanation:
1. Keto–enol tautomerism requires a carbonyl compound with at least one α-hydrogen.
2. Acetone has α-hydrogens, whereas formaldehyde, benzophenone and pivaldehyde ((CH₃)₃CCHO) have none.
3. Answer: (c)

Teacher's Note:
1. Needs an α-hydrogen.
2. The enol form has a C=C and –OH.

Answer: (c) CH₃COCH₂COCH₃

Explanation:
1. The enol form of pentane-2,4-dione is stabilised by intramolecular hydrogen bonding and conjugation of C=C with C=O.
2. So it has a large enol content (about 80%), compared with a tiny fraction for simple ketones and aldehydes.
3. Answer: (c)

Teacher's Note:
1. The more stabilised the enol, the larger its fraction.
2. 1,3-Dicarbonyls are strongly enolised.

Answer: (c) 5

Explanation:
1. Benzyl alcohol, methyl phenyl ether (anisole), and o-, m- and p-cresol.
2. That makes 5 isomers.
3. Answer: (c)

Teacher's Note:
1. Alcohol, ether, and three phenols.
2. All contain C₆ ring plus one carbon.

Answer: (b) functional isomers

Explanation:
1. Both have the formula C₃H₆O, but one is an aldehyde and the other a ketone.
2. Different functional groups mean functional isomerism.
3. Answer: (b)

Teacher's Note:
1. Aldehyde and ketone: functional isomers.
2. Different functional group.

Answer: (b) 3

Explanation:
1. Propan-1-ol, propan-2-ol and methoxyethane.
2. Total 3 structural isomers.
3. Answer: (b)

Teacher's Note:
1. Two alcohols and one ether.
2. No stereoisomers here.

Answer: (c) 4

Explanation:
1. The two butyl skeletons give 1-bromobutane, 2-bromobutane, and with the branched skeleton 1-bromo-2-methylpropane and 2-bromo-2-methylpropane.
2. Total = 4.
3. Answer: (c)

Teacher's Note:
1. Four butyl groups.
2. Position and chain isomerism together.

Answer: (b) 4

Explanation:
1. 1,1-Dichloropropane, 1,2-dichloropropane, 1,3-dichloropropane and 2,2-dichloropropane.
2. That makes 4 (1,2-dichloropropane is also chiral, but stereoisomers are not counted here).
3. Answer: (b)

Teacher's Note:
1. Systematically place two Cl on a 3-carbon chain.
2. Count each distinct pattern once.

Answer: (c) 5

Explanation:
1. Hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane and 2,3-dimethylbutane.
2. That makes five.
3. Answer: (c)

Teacher's Note:
1. Alkane isomers: C₄ 2, C₅ 3, C₆ 5, C₇ 9.
2. Check by longest chain.

Answer: (b) cis isomer is polar and has the higher boiling point

Explanation:
1. In the trans isomer the two C–Cl bond moments cancel, so its dipole moment is zero.
2. The cis isomer is polar (μ ≈ 1.9 D), with stronger dipole–dipole forces and a higher boiling point (60 °C against 48 °C).
3. Answer: (b)

Teacher's Note:
1. Symmetric trans isomer has zero dipole.
2. Polar molecules usually boil higher.

Answer: (b) fumaric acid has the higher melting point and maleic acid forms an anhydride on heating

Explanation:
1. Maleic acid is the cis isomer and fumaric acid is the trans isomer.
2. Fumaric acid has a much higher melting point (about 287 °C) because of its symmetry and packing, while maleic acid (about 135 °C) is more soluble and forms the cyclic anhydride on heating as the two –COOH groups are close.
3. Answer: (b)

Teacher's Note:
1. Cis dicarboxylic acid forms a cyclic anhydride.
2. Symmetrical trans isomer packs better in crystals.

Answer: (b) Z

Explanation:
1. On each carbon compare the two atoms: Br > H on one carbon and Cl > H on the other.
2. The higher priority groups, Br and Cl, are on the same side, so the configuration is Z.
3. Answer: (b)

Teacher's Note:
1. Same side: Z (zusammen).
2. Opposite sides: E (entgegen).

Answer: (b) has chiral centres but is optically inactive due to internal compensation

Explanation:
1. A meso compound contains stereocentres but also an internal plane of symmetry, so the rotation of one half is cancelled by the other half.
2. It is therefore optically inactive and cannot be resolved, unlike a racemic mixture, which is separable.
3. Answer: (b)

Teacher's Note:
1. Internal compensation versus external compensation.
2. Meso compound is a single compound.

Answer: (b) the rotation of one enantiomer is cancelled by the equal and opposite rotation of the other (external compensation)

Explanation:
1. A racemic mixture contains equal amounts of the d and l forms.
2. The two rotate plane-polarised light by equal angles in opposite directions, so the net rotation is zero; it can be resolved into the two enantiomers.
3. Answer: (b)

Teacher's Note:
1. Racemic: 50:50 mixture of enantiomers.
2. Separation is called resolution.

Answer: (b) are stereoisomers that are not mirror images and have different physical properties

Explanation:
1. Diastereomers are stereoisomers that are not mirror images of each other, so they have different melting points, boiling points and solubilities.
2. This is why they can be separated by ordinary methods such as crystallisation.
3. Answer: (b)

Teacher's Note:
1. Enantiomers: mirror images.
2. Diastereomers: all other stereoisomers.

Answer: (c) 4

Explanation:
1. It has one stereocentre (C-4) giving two enantiomers, and a double bond that can be E or Z.
2. Total = 2 × 2 = 4.
3. Answer: (c)

Teacher's Note:
1. Combine stereocentre and double bond.
2. Each independent element doubles the count.

Answer: (c) 16

Explanation:
1. An aldohexose, CH₂OH–(CHOH)₄–CHO, has four different stereocentres.
2. The number of stereoisomers = 2⁴ = 16.
3. Answer: (c)

Teacher's Note:
1. Four chiral carbons: C2 to C5.
2. 2ⁿ with n = 4.

Answer: (b) R

Explanation:
1. With the lowest priority group pointing away, a clockwise sense of 1 → 2 → 3 is assigned R (rectus).
2. An anticlockwise sense would be S (sinister).
3. Answer: (b)

Teacher's Note:
1. Clockwise: R.
2. Anticlockwise: S.

Answer: (a) it avoids 1,3-diaxial interactions with axial hydrogens

Explanation:
1. An axial methyl group is close to the two axial hydrogens on the same side of the ring (1,3-diaxial interaction), which causes steric strain of about 7 kJ mol⁻¹.
2. In the equatorial position it points away from the ring and avoids this strain.
3. Answer: (a)

Teacher's Note:
1. Axial substituents have 1,3-diaxial strain.
2. Large groups prefer the equatorial position.

Answer: (b) CH₃CH=C=CHCH₃

Explanation:
1. An allene is chiral when each terminal carbon carries two different groups, as in penta-2,3-diene (CH₃CH=C=CHCH₃): the two ends lie in perpendicular planes.
2. In the others at least one end has two identical groups, so there is a plane of symmetry.
3. Answer: (b)

Teacher's Note:
1. Chiral axis in allenes.
2. Each end must have two different groups.

Answer: (b) rotation about the single bond between the rings is restricted, making the two rings non-coplanar and the molecule chiral

Explanation:
1. Bulky ortho substituents prevent free rotation about the biphenyl C–C single bond (atropisomerism).
2. The rings stay in perpendicular planes, and the molecule has no plane or centre of symmetry, so it is chiral although it has no stereocentre.
3. Answer: (b)

Teacher's Note:
1. Restricted rotation: atropisomers.
2. Chirality needs no stereocentre.

Answer: (a) CH₃NO₂

Explanation:
1. Nitromethane has an α-hydrogen and can shift it to oxygen: CH₃NO₂ ⇌ CH₂=N(O)OH (aci-nitro form).
2. The other compounds have no such tautomerisable hydrogen.
3. Answer: (a)

Teacher's Note:
1. Primary and secondary nitroalkanes show nitro–aci tautomerism.
2. Needs an α-hydrogen.

Answer: (b) the enantiomer (opposite configuration)

Explanation:
1. A 180° rotation in the plane keeps the same configuration, but a 90° rotation interchanges the sense of the horizontal and vertical bonds.
2. The result represents the enantiomer, so such a rotation is not allowed without changing the meaning.
3. Answer: (b)

Teacher's Note:
1. Allowed: 180° in-plane rotation.
2. Odd number of group swaps inverts the configuration.