JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 05

Answer: (b) 2,3-pentadiene

Explanation:
1. In CH\(_3\)CH=C=CHCH\(_3\) the two end carbons each carry H and CH\(_3\), and the two double-bond planes are perpendicular.
2. The molecule is chiral (axial chirality) without any stereogenic carbon.

Teacher's Note:
1. Allenes with two different groups at each end are chiral.
2. Lactic acid has a chiral carbon, and 1,3-pentadiene has none and is not chiral.

Answer: (b) 4

Explanation:
1. The primary alcohols are pentan-1-ol, 2-methylbutan-1-ol, 3-methylbutan-1-ol and 2,2-dimethylpropan-1-ol.
2. Total = 4.

Teacher's Note:
1. Primary alcohols have the OH on a CH\(_2\) group.
2. Count for each carbon skeleton separately.

Answer: (d) 5

Explanation:
1. Pentan-2-one and pentan-3-one are metamers (different alkyl groups on either side of C=O).
2. Butanone has no ketone metamer, so five carbons are required.

Teacher's Note:
1. Metamers differ in the alkyl groups attached to the same functional group.
2. Pentan-2-one and pentan-3-one are also positional isomers; both labels apply.

Answer: (a) 8

Explanation:
1. The double bond gives 2 geometrical forms.
2. The two chiral carbons are different, giving \( 2^2 = 4 \) optical forms.
3. Total = 2 × 4 = 8.

Teacher's Note:
1. Multiply the number of geometrical and optical isomers.
2. The two chiral centres are not identical, so no meso form is expected.

Answer: (c) isobutyl alcohol

Explanation:
1. Diethyl ether, methyl n-propyl ether and 2-methoxypropane are all ethers with different alkyl groups and are metamers of each other.
2. Isobutyl alcohol is an alcohol, so it is a functional isomer, not a metamer.

Teacher's Note:
1. Metamers belong to the same functional class.
2. An alcohol and an ether are functional isomers.

Answer: (c) Bond angles and bond lengths remain the same

Explanation:
1. Conformers differ only by rotation about the C-C single bond.
2. Bond lengths and bond angles are unchanged; only dihedral angles change.

Teacher's Note:
1. Rotation about a single bond changes dihedral angle only.
2. Bond angles and lengths define the configuration.

Answer: (a) 4

Explanation:
1. The double bond gives E and Z forms.
2. C4 is chiral (R and S).
3. Total = 2 × 2 = 4.

Teacher's Note:
1. Multiply the number of geometrical and optical possibilities.
2. Check that the chiral carbon has four different groups: H, CH\(_3\), C\(_2\)H\(_5\) and the propenyl group.

Answer: (d) MP of trans is higher but dipole moment of trans is lower than cis

Explanation:
1. The trans isomer is symmetric and packs better in the crystal, so it melts at a higher temperature.
2. The bond dipoles of trans cancel (dipole moment close to zero), whereas the cis isomer has a net dipole moment.

Teacher's Note:
1. Trans isomers usually have higher melting points and lower dipole moments.
2. Cis isomers are generally more polar.

Answer: (d) A is false, but R is true.

Explanation:
1. The two CH\(_2\) planes are perpendicular to each other, so the hydrogens are not coplanar.
2. The central carbon is sp hybridised, so the reason is also false.

Teacher's Note:
1. Allene has sp2-sp-sp2 carbons.
2. Its end groups lie in perpendicular planes.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The compound has a double bond (E/Z) and a chiral carbon at C4 (R/S), so stereoisomerism occurs.
2. The reason is true but incomplete, since the stereoisomerism also includes optical isomerism; it is therefore not the full explanation.

Teacher's Note:
1. Count both geometrical and optical isomerism.
2. Reason is true but not the complete explanation.

Answer: (d) 60°

Explanation:
1. In the staggered form each C-H bond on one carbon bisects the angle between two C-H bonds on the other.
2. The dihedral angle is therefore 60°.

Teacher's Note:
1. Eclipsed ethane has a dihedral angle of 0°.
2. Staggered ethane has 60°.

Answer: (a) 6

Explanation:
1. Open-chain: but-1-ene, cis-but-2-ene, trans-but-2-ene and 2-methylpropene.
2. Cyclic: cyclobutane and methylcyclopropane.
3. Total = 6.

Teacher's Note:
1. Include geometrical isomers and rings.
2. The degree of unsaturation is 1.

Answer: (c) 3

Explanation:
1. The two chiral centres are identical, so a meso form exists.
2. There are two enantiomers (R,R and S,S) and one meso form (R,S).

Teacher's Note:
1. Identical centres give 2n-1 type reduction: 3 instead of 4.
2. Meso forms are optically inactive.

Answer: (b) 3

Explanation:
1. The molecule is symmetrical with two double bonds.
2. The isomers are (E,E), (Z,Z) and (E,Z); the (Z,E) form is identical to (E,Z).

Teacher's Note:
1. For a symmetrical polyene, the unsymmetrical isomer is counted only once.
2. An unsymmetrical polyene would give \( 2^2 = 4 \).

Answer: (a) 8

Explanation:
1. For n different chiral centres, the number is \( 2^n \).
2. With n = 3, \( 2^3 = 8 \).

Teacher's Note:
1. Use \( 2^n \) when no centre is repeated.
2. Meso forms appear only when centres are identical.

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

Explanation:
1. A racemic mixture contains equal amounts of the (+) and (-) enantiomers.
2. Their equal and opposite rotations cancel; this is external compensation.

Teacher's Note:
1. Racemic mixtures are separable into enantiomers (resolution), unlike meso compounds.
2. Each molecule in it is still chiral.

Answer: (b) CH\(_3\)CH=NOH

Explanation:
1. For geometrical isomerism the carbon of C=N must have two different groups (here CH\(_3\) and H).
2. In (CH\(_3\))\(_2\)C=NOH the carbon bears two identical methyls, and in CH\(_2\)=NOH two hydrogens.

Teacher's Note:
1. The nitrogen always has two different groups (OH and the lone pair).
2. Check the carbon end of the double bond.

Answer: (d) the direction of rotation of plane-polarised light

Explanation:
1. Enantiomers have identical physical properties except that they rotate plane-polarised light by equal angles in opposite directions.
2. They also react differently with chiral reagents.

Teacher's Note:
1. Equal magnitude, opposite sign of specific rotation.
2. Diastereomers differ in all physical properties.

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

Explanation:
1. The anti conformer has least strain as the methyl groups are 180° apart.
2. The gauche form has some steric strain (methyls at 60°), the partially eclipsed has torsional strain and the fully eclipsed has both torsional and steric strain.

Teacher's Note:
1. Stability decreases as torsional and steric strain increase.
2. Gauche is staggered, so more stable than any eclipsed form.

Answer: (a) 2

Explanation:
1. The two carbons next to the COOH groups (C2 and C4) carry H, OH, COOH and a CHOH group, which are different at each end, so they are true asymmetric centres.
2. The central carbon carries two identical-constitution but opposite-handed groups, so it is pseudo-asymmetric.

Teacher's Note:
1. A pseudo-asymmetric centre has two enantiomorphic groups.
2. Only the outer two are true stereocentres.