JEE Chemistry Isomerism MCQs Set 10

Answer: (b) 5

Explanation:
1. Draw the skeletons: n-hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane and 2,3-dimethylbutane.
2. No other arrangement of six carbons is possible, so there are 5 isomers.
3. Answer: (b)

Teacher's Note:
1. Count alkane isomers systematically from the longest chain downwards.
2. C₄H₁₀, C₅H₁₂, C₆H₁₄, C₇H₁₆ have 2, 3, 5, 9 isomers.

Answer: (c) 4

Explanation:
1. The carbon skeleton is either n-butane or isobutane.
2. n-Butane gives 1-chloro and 2-chloro derivatives; isobutane gives 1-chloro-2-methylpropane and 2-chloro-2-methylpropane.
3. Total = 4.
4. Answer: (c)

Teacher's Note:
1. Number of monochloro isomers equals the number of distinct types of hydrogen in each skeleton, added over the skeletons.
2. n-Butane has 2 types, isobutane has 2 types.

Answer: (c) 7

Explanation:
1. Aldehydes: pentanal, 2-methylbutanal, 3-methylbutanal, 2,2-dimethylpropanal = 4.
2. Ketones: pentan-2-one, pentan-3-one, 3-methylbutan-2-one = 3.
3. Total = 4 + 3 = 7.
4. Answer: (c)

Teacher's Note:
1. Aldehyde group must be at a chain end, so build it from the four C₄H₉ groups.
2. Ketones are built from the possible R and R' pairs totalling four carbons.

Answer: (c) 6

Explanation:
1. Acids: butanoic acid and 2-methylpropanoic acid = 2.
2. Esters: methyl propanoate, ethyl ethanoate, propyl methanoate and isopropyl methanoate = 4.
3. Total = 6.
4. Answer: (c)

Teacher's Note:
1. For esters, vary the acid part (1 to 3 carbons) and the alcohol part accordingly.
2. Remember that propyl has two forms, n- and iso-.

Answer: (c) metamerism

Explanation:
1. Both are secondary amines, C₄H₁₁N, with the same functional group.
2. They differ in the alkyl groups on the two sides of the polyvalent nitrogen atom.
3. This is metamerism.
4. Answer: (c)

Teacher's Note:
1. Metamerism needs a polyvalent atom such as O, S or N in the middle.
2. It is shown by ethers, ketones, esters and secondary or tertiary amines.

Answer: (c) Benzaldehyde

Explanation:
1. Keto-enol tautomerism needs at least one hydrogen on the carbon next to the carbonyl group (an alpha hydrogen).
2. Benzaldehyde has no alpha carbon with hydrogen, since the carbonyl carbon is attached directly to the ring.
3. Therefore it cannot form an enol.
4. Answer: (c)

Teacher's Note:
1. Check for alpha hydrogens before deciding on tautomerism.
2. Aromatic ring carbons do not count as alpha carbons.

Answer: (b) functional isomers

Explanation:
1. Both have the formula C₃H₆O.
2. Propanal contains the aldehyde group and propanone contains the ketone group.
3. Different functional groups mean functional isomerism.
4. Answer: (b)

Teacher's Note:
1. Aldehyde and ketone with the same formula are always functional isomers.
2. Alcohols and ethers, acids and esters are other common functional pairs.

Answer: (c) Butan-2-ol

Explanation:
1. Butan-2-ol, CH₃CH(OH)CH₂CH₃, has a carbon bonded to four different groups: H, OH, CH₃ and C₂H₅.
2. The other three have no carbon with four different groups.
3. Answer: (c)

Teacher's Note:
1. Look for a carbon with four different attached groups.
2. Two identical groups on a carbon, such as two methyl groups, rule out chirality there.

Answer: (b) 3

Explanation:
1. The molecule has two stereocentres with identical substituent sets, CH₃, H and Cl.
2. Possible forms: (R,R), (S,S) and the meso form (R,S) = (S,R).
3. The meso form is identical to its mirror image, so only 3 stereoisomers exist.
4. Answer: (b)

Teacher's Note:
1. With two identical stereocentres the number is 2ⁿ minus the meso overlap.
2. Total isomers = 2^n-1 + 2^(n/2)-1 for even n; here 2 + 1 = 3.

Answer: (c) 4

Explanation:
1. The molecule is symmetric about C-3, with C-2 and C-4 stereocentres of identical type.
2. C-3 becomes a pseudo-asymmetric centre when C-2 and C-4 have opposite configurations (R and S).
3. This gives 2 meso forms plus one pair of enantiomers = 4 stereoisomers.
4. Answer: (c)

Teacher's Note:
1. Pseudo-asymmetric centres give two different meso forms.
2. Only the (R,R) and (S,S) isomers are optically active.

Answer: (b) 2

Explanation:
1. The terminal double bond has two H atoms on one carbon, so it has no cis-trans possibility.
2. The C-3 to C-4 double bond has different groups on each carbon, so it gives cis and trans forms.
3. Total = 2.
4. Answer: (b)

Teacher's Note:
1. A double bond shows geometrical isomerism only if each carbon carries two different groups.
2. Count each qualifying double bond separately.

Answer: (d) –CH₂Cl

Explanation:
1. Compare the atoms attached to the first carbon: –CH₂Cl has (Cl, H, H), –CH₂OH has (O, H, H), –C(CH₃)₃ has (C, C, C), and –CH(CH₃)₂ has (C, C, H).
2. Atoms are compared one by one starting with the highest: Cl > O > C.
3. So –CH₂Cl ranks first.
4. Answer: (d)

Teacher's Note:
1. Compare the set of attached atoms in order of atomic number, not the total number of atoms.
2. The first point of difference decides the priority.

Answer: (b) the Cahn-Ingold-Prelog priority rules

Explanation:
1. R and S are assigned by ranking the four groups by CIP priority and viewing with the lowest-priority group away from the observer.
2. Clockwise from highest to lowest gives R and anticlockwise gives S.
3. This has no fixed connection with the (+) or (−) sign of rotation.
4. Answer: (b)

Teacher's Note:
1. R/S is a structural label; (+)/(−) is an experimental observation.
2. d/l and D/L are separate older systems.

Answer: (b) internal compensation

Explanation:
1. Meso-tartaric acid has two stereocentres of opposite configuration.
2. The rotation caused by one half is cancelled by the other half within the same molecule, and the molecule has a plane of symmetry.
3. This is internal compensation.
4. Answer: (b)

Teacher's Note:
1. Internal compensation: the compound is inactive and cannot be resolved.
2. External compensation applies to a racemic mixture, which can be resolved.

Answer: (c) the rotations of the two enantiomers cancel each other

Explanation:
1. A racemic mixture contains equal amounts of the d and l enantiomers.
2. Each rotates plane polarised light by equal amounts in opposite directions, so the net rotation is zero.
3. The individual molecules are still chiral.
4. Answer: (c)

Teacher's Note:
1. Racemic is written (±) or dl.
2. It can be separated into enantiomers by resolution, unlike a meso compound.

Answer: (b) +6°

Explanation:
1. Specific rotation = observed rotation / (length in dm × concentration in g mL⁻¹).
2. = 6 / (2 × 0.5) = +6°.
3. Answer: (b)

Teacher's Note:
1. Length must be in decimetres and concentration in g mL⁻¹.
2. Specific rotation depends on temperature, wavelength and solvent.

Answer: (a) physical properties

Explanation:
1. Diastereomers are stereoisomers that are not mirror images, and they have the same formula and connectivity.
2. Their physical properties such as melting point, boiling point and solubility are different.
3. Answer: (a)

Teacher's Note:
1. Enantiomers share all physical properties except optical rotation.
2. Diastereomers can be separated by ordinary methods like crystallisation.

Answer: (d) fully eclipsed

Explanation:
1. In the fully eclipsed conformation the two methyl groups are on the same side with a dihedral angle of 0°.
2. It has maximum torsional and steric strain.
3. Answer: (d)

Teacher's Note:
1. Order of stability: anti > gauche > partially eclipsed > fully eclipsed.
2. Gauche is about 3.8 kJ mol⁻¹ above anti.

Answer: (b) equatorial, due to the absence of 1,3-diaxial interaction

Explanation:
1. An axial methyl group suffers 1,3-diaxial repulsion with two axial hydrogens.
2. An equatorial methyl group points away from the ring and avoids this strain.
3. About 95% of the molecules exist with methyl equatorial.
4. Answer: (b)

Teacher's Note:
1. Larger substituents prefer the equatorial position.
2. The more bulky the group, the larger the preference.

Answer: (c) CH₃CH=C=CHCH₃

Explanation:
1. An allene is chiral when each terminal carbon carries two different groups.
2. In CH₃CH=C=CHCH₃ each end has H and CH₃, and the two ends lie in perpendicular planes, so the molecule has no plane of symmetry.
3. The other three have two identical groups on at least one terminal carbon.
4. Answer: (c)

Teacher's Note:
1. Chirality here is axial and arises without any stereocentre.
2. Identical groups on one end carbon create a plane of symmetry.

Answer: (d) 6

Explanation:
1. Take the three dibromobenzenes (o-, m-, p-) and place Cl on a free ring position.
2. o-Dibromo gives 2 isomers (3-Cl and 4-Cl); m-dibromo gives 3 isomers (2-, 4- and 5-Cl); p-dibromo gives 1 isomer.
3. Total = 2 + 3 + 1 = 6.
4. Answer: (d)

Teacher's Note:
1. For C₆H₃X₂Y use the number 6; for C₆H₃X₃ (all same) there are 3.
2. Count each distinct set of ring positions only once.

Answer: (a) ring-chain isomers

Explanation:
1. Both have the molecular formula C₄H₈.
2. One is an open-chain alkene and the other has a closed ring.
3. This is ring-chain isomerism.
4. Answer: (a)

Teacher's Note:
1. Alkenes and cycloalkanes with the same carbon count are ring-chain isomers.
2. Their degree of unsaturation is the same.

Answer: (c) 2-Methylbut-2-ene

Explanation:
1. In 2-methylbut-2-ene, (CH₃)₂C=CHCH₃, one double-bonded carbon carries two identical methyl groups.
2. Hence no cis-trans arrangement is possible.
3. The other three each have different or identical-pair arrangements that give cis and trans forms.
4. Answer: (c)

Teacher's Note:
1. Two identical groups on the same alkene carbon rule out geometrical isomerism.
2. Ring compounds with two substituents on different carbons can be cis or trans.

Answer: (b) 2

Explanation:
1. The two methyl groups may be on the same face (cis) or on opposite faces (trans) of the ring.
2. Both forms are achiral because of a plane of symmetry.
3. Total = 2.
4. Answer: (b)

Teacher's Note:
1. Disubstituted rings with two identical substituents give cis and trans.
2. 1,2-Dimethylcyclohexane gives 3 stereoisomers because trans is chiral.

Answer: (c) Pentane-2,4-dione

Explanation:
1. In pentane-2,4-dione the enol form is stabilised by intramolecular hydrogen bonding and conjugation of the C=C with the other C=O.
2. Simple carbonyl compounds have very little enol content.
3. Answer: (c)

Teacher's Note:
1. 1,3-Dicarbonyls enolise much more than ordinary carbonyls.
2. Methylene hydrogens between two C=O groups are acidic.