JEE Chemistry Isomerism MCQs Set 07

Answer: (c) \( \ce{ClCH=CHCl} \)

Explanation:
1. Geometrical isomerism needs restricted rotation (a double bond or ring) and two different groups on each atom of the restricted bond.
2. \( \ce{ClCH2CH2Cl} \) has free rotation; in \( \ce{Cl2C=CH2} \) and \( \ce{Cl2C=CCl2} \) one or both ends carry two identical groups.
3. Only \( \ce{ClCH=CHCl} \) has a different pair (H, Cl) on each carbon.

Teacher's Note:
1. Condition: restricted rotation plus a different pair of groups at each end of the bond.
2. Two identical groups on one carbon of C=C rule out geometrical isomerism.

Answer: (d)

Explanation:
1. 1,1-Dibromo-3,3-dichlorocyclobutane has two identical groups on each substituted ring carbon, so there is only one arrangement.
2. 1,2-Dimethylcyclohexane, 1,2,3,4-tetramethylcyclobutane and 1,2,3-trimethylcyclopropane all have a different pair of groups (H and CH3) on the substituted ring carbons and show cis-trans isomerism.

Teacher's Note:
1. Ring geometrical isomerism needs two different groups on at least two ring carbons.
2. A carbon carrying two identical groups cannot give cis-trans isomers.

Answer: (b)

Explanation:
1. In the structure with D, H on one carbon and CHO, COOCH3 on the other: D > H on the first carbon; \( \ce{COOCH3} \) (O,O,O) outranks CHO (O,O,H) on the second.
2. These two senior groups lie on opposite sides, so the configuration is E.
3. In the other three structures the senior groups (Cl and Br; the two methyls; methyl and isopropyl) lie on the same side, so they are Z.

Teacher's Note:
1. E: senior groups on opposite sides. Z: on the same side.
2. Compare the first point of difference by atomic number, going outwards.

Answer: (a) 3

Explanation:
1. C-2 carries H, OH, \( \ce{CH3} \) and \( \ce{CH2-} \): chiral.
2. C-4 carries H, Br, \( \ce{CH2-} \) and \( \ce{CH(C2H5)CH3} \): chiral.
3. C-5 carries H, ethyl, methyl and the \( \ce{CHBr-} \) chain: chiral.
4. C-1, C-3 and the ethyl and methyl carbons each have two or three identical groups, so there are 3 chiral carbons.

Teacher's Note:
1. A chiral carbon has four different groups.
2. CH2 and CH3 carbons can never be chiral.

Answer: (d) 4

Explanation:
1. C-2=C-3 and C-4=C-5 are stereogenic double bonds.
2. The terminal \( \ce{CH=CH2} \) has two hydrogens on one carbon and shows no isomerism.
3. So \( 2^2 = 4 \).

Teacher's Note:
1. A terminal double bond =CH2 is never stereogenic.

Answer: (a) 4

Explanation:
1. There are two chiral carbons (C-2 and C-3) with different groups, so \( 2^2 = 4 \) stereoisomers.
2. All four are optically active (two enantiomeric pairs).

Teacher's Note:
1. Unsymmetrical molecule with n chiral centres: \( 2^n \) stereoisomers.

Answer: (c) 2

Explanation:
1. C-2 and C-4 are equivalent chiral centres and C-3 is a pseudo-asymmetric centre, so there are only 4 stereoisomers.
2. Two of them (r and s forms) are meso with a plane of symmetry; the other two are an enantiomeric pair, so 2 are optically active.

Teacher's Note:
1. Symmetric chain: count stereoisomers carefully; meso forms are inactive.

Answer: (d) 16

Explanation:
1. The molecule has a stereogenic centre (C-Cl), a chiral axis along the C=C=C unit (different groups at each end of it) and two stereogenic double bonds (C=C at each end).
2. Number of stereoisomers = \( 2^4 = 16 \), and with no symmetry all of them are optically active.

Teacher's Note:
1. A chiral C=C=C axis counts as a stereogenic element.
2. No internal symmetry means no meso forms.

Answer: (b) 20% and +2.7 units

Explanation:
1. Enantiomeric excess = \( (6-4)/(6+4)\times 100 = 20\% \).
2. Specific rotation of the mixture = 20% x 13.5 = +2.7 units, with the (+) isomer in excess.

Teacher's Note:
1. ee = |major - minor| / total x 100.
2. Observed [alpha] = ee x [alpha] of pure enantiomer.

Answer: (a) -30

Explanation:
1. ee = 60 - 40 = 20%.
2. \( [\alpha] = 0.20\times(-150) = -30 \).

Teacher's Note:
1. Only the excess enantiomer contributes to the net rotation.

Answer: (d) a single stereoisomer (optically active)

Explanation:
1. The acid has one stereocentre, with configuration retained, and the alcohol has one stereocentre, with configuration retained, since the C-O bond of the alcohol is not broken.
2. Both are pure single enantiomers, so the ester is one stereoisomer, which is optically active.

Teacher's Note:
1. Esterification does not affect either stereocentre.
2. Racemic reactants would give diastereomers.

Answer: (a) R

Explanation:
1. Priorities: Br > Cl > \( \ce{CH3} \) > H.
2. Viewing with H pointing away from the observer, the sequence Br, Cl, CH3 runs clockwise for this drawing, so the configuration is R.
3. E and Z describe double bonds, so they do not apply to a stereocentre.

Teacher's Note:
1. Assign priorities by atomic number, orient the lowest priority away, then trace 1, 2, 3.

Answer: (c) S

Explanation:
1. Priorities: \( \ce{NH2} \) > \( \ce{COOH} \) > \( \ce{CH3} \) > H.
2. In this drawing the sequence \( \ce{NH2} \), COOH, \( \ce{CH3} \) is anticlockwise with H pointing away: configuration S.
3. E/Z are for double bonds.

Teacher's Note:
1. COOH outranks CH3 because (O,O,O) beats (H,H,H).
2. Natural L-alanine is S.

Answer: (d) diastereomers

Explanation:
1. Only one of the two stereocentres is inverted, so the compounds are stereoisomers that are not mirror images: diastereomers.
2. The (2R,3S) form is meso, the (2R,3R) form is chiral.

Teacher's Note:
1. Diastereomers differ in physical properties and can be separated by fractional distillation or crystallisation.

Answer: (b) A compound can be chiral without having any chiral centre

Explanation:
1. Compounds with a C=C=C unit, biphenyls with bulky ortho groups and spiro compounds can be chiral through an axis of chirality with no chiral carbon.
2. Not all stereogenic centres are chiral centres, for example the carbons of a cis-trans double bond; and a plane of symmetry makes a molecule achiral.

Teacher's Note:
1. Chirality means non-superimposability on the mirror image.
2. Elements of chirality: centre, axis, plane.

Answer: (a) Geometrical isomers are not mirror images of each other, and cyclodecene can exist in both cis and trans forms

Explanation:
1. Geometrical isomers are diastereomers, not enantiomers.
2. Rings of eight or more carbons are large enough to hold a trans double bond, so cyclodecene exists as cis and trans.
3. A double bond gives geometrical isomerism only if each carbon carries two different groups; and single bonds rotate freely.

Teacher's Note:
1. The smallest isolable trans cycloalkene is trans-cyclooctene.

Answer: (d) free rotation about the C-C single bond

Explanation:
1. The sigma bond allows rotation with a small barrier (about 12 kJ/mol), so the molecule continuously passes through many arrangements (conformations), of which staggered is the most stable and eclipsed the least stable.

Teacher's Note:
1. Conformations differ by rotation about single bonds and are interconvertible at room temperature.

Answer: (a) cyclopropane

Explanation:
1. The C-C-C angle of a planar ring is 60 degrees in cyclopropane, a deviation of 49.5 degrees from 109.5 degrees, the largest of all.
2. The deviation is smaller for cyclobutane (90 degrees) and cyclopentane (108 degrees).

Teacher's Note:
1. Smaller ring means a larger deviation from the tetrahedral angle in the planar model.

Answer: (c) anti > gauche > partially eclipsed > fully eclipsed

Explanation:
1. Anti has the two methyls far apart (lowest energy, 0 kJ/mol); gauche has them 60 degrees apart (about 3.8 kJ/mol higher); partially eclipsed is about 16 kJ/mol; fully eclipsed, with the methyls overlapping, is the highest (about 19 kJ/mol).

Teacher's Note:
1. Steric repulsion between methyls and eclipsing strain both raise the energy.

Answer: (d) half-chair

Explanation:
1. The stability order is chair > twist boat > boat > half-chair; the half-chair (a transition-state structure on the way from chair to twist boat) has the greatest angle and torsional strain, about 45 kJ/mol above the chair.

Teacher's Note:
1. Chair 0 kJ/mol, twist boat about 23, boat about 29, half-chair about 45.

Answer: (b) in the chair all C-H bonds are staggered, while in the boat some are eclipsed (and flagpole hydrogens clash)

Explanation:
1. The chair has all adjacent C-H bonds staggered and no flagpole interaction, while the boat has four eclipsed C-C bonds along the sides and flagpole repulsion.

Teacher's Note:
1. Chair: staggered, strain-free (angles about 111 degrees, practically strain free).

Answer: (d) the chair with both methyl groups equatorial

Explanation:
1. In the trans-1,4 isomer the methyl groups are either both equatorial (ee) or both axial (aa); a ring flip interconverts them.
2. The diequatorial form avoids 1,3-diaxial interactions and is more stable.
3. One axial and one equatorial (ae) is the cis-1,4 isomer.

Teacher's Note:
1. Bulky groups prefer the equatorial position.
2. trans-1,4: ee or aa; cis-1,4: ae.

Answer: (a) infinite

Explanation:
1. Rotation about the C-C bond is continuous, so an infinite number of arrangements (conformations) exist between the staggered and eclipsed extremes.

Teacher's Note:
1. Only two special conformations are named: staggered and eclipsed.