JEE Chemistry Isomerism MCQs Set 01

Answer: (a) 5

Explanation:
1. C₅H₁₀ fits CnH2n, so it is either an alkene or a cycloalkane. We count only rings and ignore stereoisomers.
2. Ring of 5: cyclopentane (1). Ring of 4 plus CH₃: methylcyclobutane (1). Ring of 3: ethylcyclopropane, 1,1-dimethylcyclopropane and 1,2-dimethylcyclopropane (3).
3. Total = 1 + 1 + 3 = 5 cyclic structural isomers (cis and trans 1,2-dimethylcyclopropane are stereoisomers, not extra structural isomers).

Teacher's Note:
1. List isomers by ring size: 5-ring, 4-ring, 3-ring. This avoids missing or repeating any structure.

Answer: (a) II only

Explanation:
1. Tautomers are constitutional isomers that interconvert by shifting a hydrogen and a double bond, for example keto and enol forms.
2. Compound II (cyclohex-1-en-1-ol) is the enol form of cyclohexanone, so it is a tautomer.
3. Compound III (cyclohex-2-en-1-ol) is an allylic alcohol. It has the same formula but cannot be reached by a simple H shift from the ketone, so it is only an isomer.

Teacher's Note:
1. Tautomerism needs the OH to sit on the carbon that carries the C=C (a vinylic OH). An OH next to a C=C (allylic alcohol) is not an enol.

Answer: (c) 9

Explanation:
1. C₃H₆O has one degree of unsaturation (a C=C, a C=O or a ring).
2. With C=O: propanal and propanone (2). With C=C: prop-2-en-1-ol, prop-1-en-1-ol, prop-1-en-2-ol and methoxyethene (4). With a ring: oxetane, methyloxirane and cyclopropanol (3).
3. Total = 2 + 4 + 3 = 9.

Teacher's Note:
1. Count by type of unsaturation: carbonyl, alkene, ring. Counting stereoisomers separately (such as cis and trans propen-1-ol) is not part of structural isomers.

Answer: (c) 9

Explanation:
1. C₄H₆ has two degrees of unsaturation: two double bonds, one triple bond, one ring plus one double bond, or two rings.
2. Alkynes: but-1-yne and but-2-yne (2). Dienes: buta-1,2-diene and buta-1,3-diene (2).
3. Rings: cyclobutene, 1-methylcyclopropene, 3-methylcyclopropene, methylenecyclopropane (4) and bicyclo[1.1.0]butane (1). Total = 2 + 2 + 5 = 9.

Teacher's Note:
1. Use the degrees of unsaturation to decide which skeleton types are possible, then count each type separately.

Answer: (d) 4

Explanation:
1. An ester R–COO–R' has a total of 4 carbons including the carbonyl carbon.
2. Formate esters (acid part has 1 C, alkyl has 3 C): propyl methanoate and isopropyl methanoate (2).
3. Acetate: ethyl ethanoate (1). Propanoate: methyl propanoate (1). Total = 4.

Teacher's Note:
1. Vary the split of carbons between the acid part and the alcohol part: 1+3, 2+2, 3+1. A 3-carbon alkyl group has two forms, propyl and isopropyl.

Answer: (c) H_γ

Explanation:
1. The carbons next to C=O (the alpha carbons) are sp² carbons of C=C, and their hydrogens are vinylic, so they cannot be removed to give an enol.
2. The CH₂ carbon opposite the C=O (the gamma position) has hydrogens that are conjugated with the C=O through both double bonds.
3. Loss of H_γ and shifting of the double bonds gives the aromatic phenol, which is very stable. So enolisation involves H_γ.

Teacher's Note:
1. Cyclohexa-2,5-dienone converts completely into phenol because aromatic stabilisation drives the shift. This is the reason phenol exists almost entirely in its enol form.

Answer: (d)

Explanation:
1. Option 1 (a bicycloheptanone with the C=O on the one-carbon bridge): the alpha carbons are bridgeheads, and a double bond at a bridgehead of this small ring is too strained, so no enol forms.
2. Option 2: one alpha carbon is a bridgehead (same problem) and the other carries two phenyl groups and no hydrogen.
3. Option 3: the alpha carbons are sp² and the gamma carbon is quaternary (two methyls), so there is no hydrogen to move.
4. Option 4: the extra methyl on the beta carbon is a vinylogous (extended) alpha position. Moving one of its hydrogens gives a conjugated dienol, so tautomerism is possible.

Teacher's Note:
1. Look for hydrogens on the alpha carbon, or on a carbon conjugated to the C=O through C=C (vinylogous). Bridgehead C=C in small bicyclic rings is not allowed (Bredt's rule).

Answer: (d) (a) and (c)

Explanation:
1. (a) 4-Pyridone (N–H and C=O): it is in equilibrium with 4-hydroxypyridine, an aromatic hydroxy form.
2. (b) N-Methyl-4-pyridone has no N–H hydrogen to shift, so it is locked in the keto form.
3. (c) 2-Pyridone (N–H and C=O): it is in equilibrium with 2-hydroxypyridine (lactam–lactim tautomerism).
4. So (a) and (c) show tautomerism.

Teacher's Note:
1. A mobile hydrogen on nitrogen next to or conjugated to a C=O gives lactam–lactim tautomerism. Replacing the N–H by N–CH₃ blocks it.

Answer: (d) II > III > I

Explanation:
1. II is the keto form. For simple ketones the keto form is much more stable than the enol (the C=O bond is stronger than C=C plus O–H).
2. Between the two enols, III has a tetra-substituted C=C (both carbons carry two alkyl groups or OH/alkyl), while I has a tri-substituted C=C. More substituted alkenes are more stable, so III > I.
3. Order: II > III > I.

Teacher's Note:
1. Rule of thumb: keto beats enol for simple ketones, and among enols the more substituted double bond is more stable.

Answer: (c) The eclipsed and the anti form of Cl–CH₂–CH₂–Cl

Explanation:
1. Conformers differ only by rotation about a single bond, and they interconvert at room temperature.
2. Option 1: the double bond is in a different position, so these are positional isomers.
3. Option 2: the two Fischer projections are mirror images of 2-chlorobutane, so they are enantiomers.
4. Option 4: cis and trans-1,2-dichloroethene differ in arrangement about a double bond (geometrical isomers).
5. Option 3: the syn (eclipsed) and anti forms of 1,2-dichloroethane differ only by rotation about the C–C single bond, so they are conformers.

Teacher's Note:
1. Conformers need only single-bond rotation. If making the change requires breaking a bond, the pair are configurational or structural isomers.

Answer: (c) Configurational isomers

Explanation:
1. Geometrical (cis–trans) isomers differ in the spatial arrangement of groups about a double bond or ring, which cannot rotate freely.
2. Isomers that cannot be interconverted by simple rotation about single bonds, and have the same connectivity, are configurational isomers. Geometrical and optical isomers are the two kinds.

Teacher's Note:
1. Stereoisomers are divided into configurational (geometrical and optical) and conformational. Only the second type interconverts by bond rotation.

Answer: (d) (I) and (III)

Explanation:
1. C=N compounds show syn–anti (geometrical) isomerism when the carbon has two different groups; the nitrogen has its lone pair as the second 'group'.
2. (I) CH₂=CH–CH=N–CH₃: the carbon has H and vinyl (different), so isomerism is possible.
3. (II) (CH₃)₂C=N–CH₃: the carbon has two identical methyl groups, so no isomerism.
4. (III) CH₃–C(C₂H₅)=N–OH (butanone oxime): the carbon has methyl and ethyl, so syn and anti forms exist.
5. So (I) and (III).

Teacher's Note:
1. Oximes and imines behave like alkenes: they show isomerism only if the carbon of C=N carries two different groups.

Answer: (b) (II)

Explanation:
1. Geometrical isomerism in a ring (cis–trans) needs two ring carbons that each carry two different groups (a substituent and a hydrogen, or two different substituents).
2. (I) 1,1,2-trimethylcyclohexane: C1 carries two identical methyl groups, so no cis–trans isomers.
3. (II) 1-ethyl-1,2-dimethylcyclohexane: C1 carries ethyl and methyl (different) and C2 carries methyl and H, so cis and trans forms exist.
4. (III) 1,1,4-trimethylcyclohexane: C1 carries two identical methyl groups, so none.

Teacher's Note:
1. For ring cis–trans isomerism, if either substituted ring carbon has two identical substituents, the isomerism disappears.

Answer: (b) 2

Explanation:
1. The ring carbons C1 and C2 each carry a methyl and a hydrogen (two different groups), so cis–trans isomerism exists.
2. Isomers: cis (methyls on the same face) and trans (methyls on opposite faces). The trans form is chiral but it is still one geometrical isomer.
3. Number of geometrical isomers = 2.

Teacher's Note:
1. Geometrical isomers of a ring are counted as the cis and trans arrangements. Their optical forms are counted separately.

Answer: (b) 3

Explanation:
1. C1 and C2 are both stereocentres, but the two ends are identical, so the formula 2ⁿ overcounts.
2. cis-1,2-Dibromocyclohexane is meso (a mirror plane), so it is one isomer. trans-1,2-Dibromocyclohexane exists as a pair of enantiomers.
3. Total = 1 + 2 = 3 stereoisomers.

Teacher's Note:
1. With two identical stereocentres, expect a meso form. The count is 2 enantiomers plus 1 meso = 3.

Answer: (b) (2)

Explanation:
1. A biphenyl is chiral only if rotation about the ring–ring bond is blocked by bulky ortho groups, and each ring has two different ortho groups.
2. (1) 3-Nitro-3'-iodobiphenyl has no ortho groups, so it rotates freely and is not chiral.
3. (2) 2,2'-Dibromo-6,6'-diiodobiphenyl: each ring has Br on one ortho position and I on the other, so each ring is unsymmetrical and the molecule is chiral.
4. (3) One ring has two I, the other two Br. Each ring has identical ortho groups, so each ring has a symmetry plane and the molecule is achiral.

Teacher's Note:
1. Chiral biphenyls (atropisomers) need both rings to be unsymmetrical. If either ring has identical ortho groups, the molecule is achiral.

Answer: (d) All of these

Explanation:
1. (1) 1-Ethylidene-2,4-dimethylcyclohexane has stereocentres at C2 and C4 (each with four different paths or groups), so it is chiral.
2. (2) 1-Ethylidene-3-methylcyclohexane: C3 has H, CH₃ and two different ring paths, so it is chiral.
3. (3) 1-Ethyl-2-methylcyclohexane has two stereocentres with different substituents, so it is chiral.
4. All three are optically active.

Teacher's Note:
1. In a ring, a carbon with two different substituents (one of them H) is a stereocentre if the two ring paths around it are different.

Answer: (b) CH₂(OH)CH₂OH

Explanation:
1. Conformational isomerism needs at least one single bond between two atoms that each carry other groups, so rotation gives different arrangements.
2. CD₃–H (methane type) and NH₃ have no bond with substituents on both ends, so no distinct conformers.
3. CH₂(OH)CH₂OH (ethane-1,2-diol) has a C–C bond with groups on both carbons, so staggered and eclipsed (and gauche/anti) forms exist.

Teacher's Note:
1. Look for a C–C (or similar) single bond with substituents on both atoms. Simple molecules with only one central atom cannot have conformers.

Answer: (a) 4

Explanation:
1. An aldehyde has the –CHO group, so the rest is a C₄H₉ alkyl group attached to it.
2. C₄H₉ has 4 isomers: n-butyl, sec-butyl, isobutyl and tert-butyl.
3. The aldehydes are pentanal, 2-methylbutanal, 3-methylbutanal and 2,2-dimethylpropanal, so 4.

Teacher's Note:
1. The number of aldehydes with formula C_nH_2nO equals the number of alkyl groups with n−1 carbons. A butyl group has four isomers.

Answer: (b) (2)

Explanation:
1. (1) Butan-2-ol is chiral but an alcohol with no C=O has no keto–enol tautomerism.
2. (2) 4-Methylcyclohex-2-enone has alpha hydrogens on the CH₂ next to C=O (so enolisation is possible) and C4 carries H, CH₃ and two different ring paths (a stereocentre). It shows both.
3. (3) Butan-2-one shows tautomerism but is achiral.
4. (4) Pentane-2,4-dione shows tautomerism but is achiral.

Teacher's Note:
1. Check two things separately: an alpha hydrogen next to a C=O (for tautomerism) and a carbon with four different groups (for optical activity).

Answer: (b) (2)

Explanation:
1. (1) 1,2-Dimethylcyclohexane: the trans isomer has two stereocentres with no mirror plane, so it is optically active as a compound type.
2. (2) 1,4-Dimethylcyclohexane: each ring carbon has two identical ring paths, so there is no stereocentre. It has only cis–trans forms, both achiral.
3. (3) 1,4-Dimethylcyclohexene: C4 carries H, CH₃ and two different ring paths (one leading to the C=C), so it is chiral.
4. So only (2) does not show optical isomerism.

Teacher's Note:
1. In a ring, a carbon is a stereocentre only if the two paths around the ring from it are different. In a symmetric 1,4-disubstituted ring they are not.

Answer: (a) Positional isomers

Explanation:
1. The first is 3-cyanopentanal and the second is 4-cyanopentanal.
2. Both have the same five-carbon chain and the same groups (–CHO and –CN). Only the position of the CN group has changed.
3. So they are positional isomers.

Teacher's Note:
1. Same carbon skeleton and same functional groups, but a different position: positional isomerism.

Answer: (b) (2)

Explanation:
1. Geometrical isomerism needs a C=C with two different groups on each carbon.
2. (1) The CH=CH–CH₃ unit has H and a large group on one carbon, and H and CH₃ on the other, so cis–trans forms exist.
3. (2) The only double bond is the terminal CH=CH₂, whose =CH₂ carbon has two H, so no geometrical isomerism.
4. (3) The CH=CHCl unit has two different groups on each carbon, so it shows isomerism.
5. (4) CH₃CH=CHCH₂CH₃ (pent-2-ene) shows cis–trans isomerism.

Teacher's Note:
1. A terminal vinyl group (–CH=CH₂) never gives geometrical isomers.

Answer: (c) Internal compensation

Explanation:
1. Meso-tartaric acid has two stereocentres of opposite configuration (R and S) in the same molecule.
2. The rotation of plane-polarised light caused by one centre is cancelled by the equal and opposite rotation of the other. This is internal compensation, and the molecule has a plane of symmetry.
3. External compensation is the cancelling in a racemic mixture, which involves two different kinds of molecules.

Teacher's Note:
1. Meso compound: internal compensation within one molecule. Racemic mixture: external compensation between two enantiomers.

Answer: (d) 8

Explanation:
1. There are two C=C bonds. In each, the two carbons carry different groups, so each gives E/Z isomerism: 2 × 2 = 4.
2. The carbon bearing the CH₃ is bonded to H, CH₃, propenyl and pentenyl groups. The two alkenyl arms are different, so it is a stereocentre: 2 forms.
3. Total = 2 × 2 × 2 = 8.

Teacher's Note:
1. Because the two arms differ (CH₃ end and C₂H₅ end), there is no symmetry and no meso form. The simple 2ⁿ rule applies.

Answer: (c) (3)

Explanation:
1. A ring carbon is a stereocentre only if its four attachments are all different, counting the two ring paths.
2. (1) The carbon carries two OH groups, so it is not a stereocentre.
3. (2) The carbon carries SH and Br, but both ring paths are identical (CH₂–CH₂), so achiral.
4. (3) The carbon carries OH and Br and its two ring paths differ (one leads to the C=C, the other to CH₂), so it is a stereocentre: optically active.
5. (4) The carbon carries CH₃ and H, and the ring paths are identical, so achiral.

Teacher's Note:
1. In a monosubstituted or symmetrical ring, the ring carbon is not chiral because the two ring paths are the same. Unsaturation on one side makes them different.

Answer: (d) Dicyclohexyl ketone

Explanation:
1. The oxime C=N–OH shows syn–anti isomerism only if the two groups on carbon are different.
2. Acetophenone (Ph and CH₃), benzaldehyde (Ph and H) and cyclohexyl methyl ketone (cyclohexyl and CH₃) all have two different groups, so their oximes show isomerism.
3. Dicyclohexyl ketone has two identical cyclohexyl groups, so its oxime has no syn–anti isomers.

Teacher's Note:
1. The rule is the same as for C=C: two identical groups on the carbon of C=N remove the isomerism.

Answer: (a) (2R,3E)-Pent-3-en-2-ol

Explanation:
1. The chain is CH₃–CH=CH–CH(OH)–CH₃. OH is on C2 (lowest locant), the double bond is at C3: pent-3-en-2-ol.
2. The double bond has the chain ends on opposite sides in the zigzag drawing, so it is E.
3. At C2 the priorities are OH > CH=CH (C3) > CH₃ > H. With H pointing away (hashed), OH to C3 to CH₃ runs clockwise, so C2 is R.
4. Name: (2R,3E)-pent-3-en-2-ol.

Teacher's Note:
1. Write the stereodescriptors in locant order: (2R,3E). The OH gets the lowest locant, so the name pent-2-en-3-ol is wrong.