JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 27

Answer: (b) 5

Explanation:
1. The carbon skeletons possible for six carbons are: hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane and 2,3-dimethylbutane.
2. No other skeleton exists, so there are 5 isomers.

Teacher's Note:
1. Systematically shorten the main chain: 6, 5, 4 carbons.
2. Alkanes with 4, 5, 6, 7 carbons have 2, 3, 5, 9 isomers.

Answer: (a) functional isomerism

Explanation:
1. Both have the molecular formula C2H6O.
2. One has an alcohol group and the other an ether group, so they are functional isomers.

Teacher's Note:
1. Different functional groups with the same molecular formula: functional isomerism.
2. Alcohols and ethers, acids and esters, aldehydes and ketones are common functional pairs.

Answer: (d) position isomers

Explanation:
1. The carbon skeleton (four carbons in a row) and the functional group (OH) are the same.
2. Only the position of OH differs (C1 and C2), so they are positional isomers.

Teacher's Note:
1. Same skeleton and functional group but different locant: position isomers.
2. Chain isomers have different carbon skeletons.

Answer: (c) metamerism

Explanation:
1. Both are ethers of formula C4H10O.
2. They differ in the alkyl groups on either side of the same divalent atom (O), which is metamerism.

Teacher's Note:
1. Metamerism occurs with a polyvalent functional atom such as O, S or N.
2. Same functional group, different distribution of carbon atoms on the two sides.

Answer: (b) 7

Explanation:
1. Alcohols: butan-1-ol, butan-2-ol, 2-methylpropan-1-ol, 2-methylpropan-2-ol (4).
2. Ethers: methoxypropane, ethoxyethane, 2-methoxypropane (3).
3. Total = 4 + 3 = 7.

Teacher's Note:
1. Count alcohols and ethers separately, then add.
2. All seven are structural isomers of the same formula.

Answer: (a) 4

Explanation:
1. Butan-1-amine and butan-2-amine come from the straight chain.
2. 2-Methylpropan-1-amine and 2-methylpropan-2-amine come from the branched chain.
3. Total = 4.

Teacher's Note:
1. Two carbon skeletons, each with two different NH2 positions, give four primary amines.
2. Chain isomerism and position isomerism are both involved.

Answer: (d) \( \ce{CH3-CH=CH-CH3} \)

Explanation:
1. Cis–trans isomerism needs two different groups on each doubly bonded carbon.
2. But-2-ene has H and CH3 on each carbon, so it has cis and trans forms.
3. The other alkenes have two identical groups on at least one doubly bonded carbon.

Teacher's Note:
1. Check each sp2 carbon for two different substituents.
2. A =CH2 or =C(CH3)2 end rules out geometrical isomerism.

Answer: (c) 3

Explanation:
1. The molecule is symmetrical, with two stereogenic double bonds.
2. The combinations are (E,E), (Z,Z) and (E,Z); (E,Z) and (Z,E) are the same compound.
3. Total = 3.

Teacher's Note:
1. Unsymmetrical polyenes with n stereogenic double bonds have 2^n isomers; symmetrical ones have fewer.
2. Symmetry makes (E,Z) identical to (Z,E).

Answer: (b) Z

Explanation:
1. On each carbon the substituent of higher priority is the halogen (Cl above H, Br above H).
2. Both higher-priority groups lie on the same side, so the configuration is Z (zusammen).

Teacher's Note:
1. E/Z is decided by the positions of the higher-priority groups on each doubly bonded carbon.
2. R and S refer to chiral centres, not to double bonds.

Answer: (a) cis-1,2-dichloroethene

Explanation:
1. In the cis isomer the two C–Cl bond dipoles add up to a net dipole.
2. In the trans isomer they cancel, so the dipole moment is nearly zero.

Teacher's Note:
1. Symmetrical trans isomers usually have lower dipole moment than cis isomers.
2. Higher polarity gives cis isomers generally higher boiling points.

Answer: (d) 2

Explanation:
1. The molecule has one chiral carbon (C2) bonded to H, OH, CH3 and COOH.
2. With one stereocentre there are 2 stereoisomers: the d and l enantiomers.

Teacher's Note:
1. n dissimilar stereocentres give 2^n stereoisomers.
2. Lactic acid has one stereocentre.

Answer: (c) 3

Explanation:
1. There are two chiral carbons, but they carry identical sets of groups (the molecule is symmetrical).
2. The isomers are the (R,R) and (S,S) enantiomers and one achiral meso form (R,S).
3. Total = 3.

Teacher's Note:
1. For two identical chiral centres: 2 enantiomers + 1 meso = 3.
2. For two different chiral centres: 4 stereoisomers.

Answer: (b) 4

Explanation:
1. The two chiral carbons (C2 and C3) are not equivalent because the ends of the chain differ (CH3 and CH2CH3).
2. Hence there is no meso form and 2² = 4 stereoisomers.

Teacher's Note:
1. Formula 2^n applies when the chiral centres are not mirror-related by symmetry.
2. The molecule is unsymmetrical, so no meso form exists.

Answer: (a) 3

Explanation:
1. Tartaric acid has two identical stereocentres.
2. The isomers are (+)-tartaric acid, (−)-tartaric acid and meso-tartaric acid.

Teacher's Note:
1. Symmetrical molecule with two chiral centres: 3 stereoisomers.
2. The meso form is optically inactive because of internal compensation.

Answer: (d) meso-tartaric acid

Explanation:
1. meso-Tartaric acid has a plane of symmetry between the two chiral carbons.
2. The rotations of the two halves cancel (internal compensation), so it is optically inactive.
3. The other three compounds are optically active.

Teacher's Note:
1. A plane of symmetry removes optical activity even with chiral carbons.
2. Meso forms can be recognised by the internal mirror plane.

Answer: (c) 8

Explanation:
1. The chain has three different stereocentres (C2, C3 and C4).
2. Number of stereoisomers = 2³ = 8.

Teacher's Note:
1. Count the carbons bonded to four different groups.
2. Open-chain glucose has four, giving 2⁴ = 16.

Answer: (b) 10

Explanation:
1. The molecule is symmetrical with n = 4 chiral carbons (n even).
2. Total stereoisomers = 2^(n−1) + 2^((n/2)−1) = 8 + 2 = 10.
3. Of these, 2 are meso.

Teacher's Note:
1. For symmetrical molecules with an even number n of identical centres: 2^(n−1) + 2^(n/2 − 1).
2. Odd n gives 2^(n−1).

Answer: (a) \( \ce{CH3-CH2-CHCl-CH3} \)

Explanation:
1. 2-Chlorobutane has C2 bonded to H, Cl, CH3 and C2H5: four different groups.
2. In the other molecules the stereocentre candidate has two identical groups.

Teacher's Note:
1. A chiral carbon needs four different substituents.
2. Check for two identical groups first.

Answer: (d) +35°

Explanation:
1. \( [\alpha]=\frac{\alpha}{l\times c}=\frac{+70}{1\times 2}=+35^\circ \)
2. The specific rotation does not change with concentration.

Teacher's Note:
1. Specific rotation is the rotation for 1 dm path length and 1 g/mL concentration.
2. Doubling the concentration doubles the observed rotation but not [α].

Answer: (c) the rotations due to the two enantiomers cancel (external compensation)

Explanation:
1. A racemic mixture contains equal amounts of the d and l forms.
2. Equal and opposite rotations cancel; this is external compensation.
3. The individual molecules are still chiral.

Teacher's Note:
1. Racemic mixture: external compensation. Meso compound: internal compensation.
2. A racemate can in principle be resolved into its enantiomers.

Answer: (b) 75%

Explanation:
1. Enantiomeric excess = 30/60 × 100 = 50%.
2. The (+) form exceeds the (−) form by 50%, and the rest (50%) is racemic.
3. So (+) = 50 + 25 = 75% and (−) = 25%.

Teacher's Note:
1. % ee = (observed [α] / [α] of pure enantiomer) × 100.
2. The racemic part is shared equally by the two forms.

Answer: (a) the enantiomer

Explanation:
1. One swap of two groups at a stereocentre inverts the configuration.
2. For a molecule with one chiral carbon this gives the mirror image, the enantiomer.

Teacher's Note:
1. Two swaps restore the original configuration.
2. Rotating a Fischer projection by 180° in the plane of the paper gives the same molecule; by 90° gives the enantiomer.

Answer: (d) OH > CHO > CH2OH > H

Explanation:
1. Atomic number first: O (OH) is highest and H is lowest.
2. Between CHO and CH2OH: the carbon of CHO has (O, O, H) because the double bond is counted twice, while CH2OH has (O, H, H).
3. So CHO > CH2OH.

Teacher's Note:
1. Compare the atoms attached to the first atom; duplicate atoms for double bonds.
2. The lowest priority group (H) points away when assigning R or S.

Answer: (c) anti (staggered, dihedral angle 180° between the methyl groups)

Explanation:
1. In the anti form the two bulky methyl groups are as far apart as possible, so van der Waals strain is minimum.
2. Stability order: anti > gauche > partially eclipsed > fully eclipsed.

Teacher's Note:
1. Staggered forms are free from torsional strain; anti also has least steric strain.
2. Fully eclipsed has the highest energy.

Answer: (b) 60°

Explanation:
1. In the staggered gauche conformer the methyl groups are 60° apart.
2. 0° is fully eclipsed, 120° is partially eclipsed and 180° is anti.

Teacher's Note:
1. Staggered conformations have angles 60°, 180° and 300°; eclipsed ones 0°, 120° and 240°.
2. Gauche = staggered with bulky groups adjacent.