JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 73

Answer: (b) 14

Explanation:
1. Alcohols: \( \ce{C5H11OH} \) has 8 isomers (8 pentyl groups).
2. Ethers: methyl butyl 4, ethyl propyl 2, total 6.
3. Total = 8 + 6 = 14.

Teacher's Note:
1. Count alcohols and ethers separately.

Answer: (d) Dipole moment II > I; boiling point II > I; melting point I > II; stability I > II

Explanation:
1. The cis isomer is polar, with a higher dipole moment and boiling point.
2. The trans isomer is symmetrical, packs better in the crystal (higher melting point) and is more stable because of lower steric strain.

Teacher's Note:
1. Trans: higher melting point, stability; cis: higher dipole moment, boiling point.

Answer: (a) Metamerism

Explanation:
1. Both are \( \ce{C13H12O3S} \) with the same functional group, a sulfonate ester.
2. They differ in the groups attached on the two sides of the polyvalent group -SO2-O-, so they are metamers.

Teacher's Note:
1. Metamers differ in the groups attached to a polyvalent functional group.

Answer: (a) 4,4-Dimethylcyclohexa-2,5-dien-1-one

Explanation:
1. Keto-enol tautomerism needs an alpha hydrogen on an sp3 carbon.
2. Both alpha carbons of 4,4-dimethylcyclohexa-2,5-dienone are sp2 carbons with no such hydrogen.
3. The other three have alpha hydrogens (or are already the enol).

Teacher's Note:
1. Look for sp3 alpha C-H.

Answer: (a) II and III are identical

Explanation:
1. Taking OH > CHO > CH3 > H: I is R, II is S and III is R.
2. So I and II are enantiomers (their equimolar mixture is inactive and they have identical boiling points), while II and III are also enantiomers, not identical molecules.

Teacher's Note:
1. Assign R/S to each projection.

Answer: (b) (i), (iii) and (iv)

Explanation:
1. \( \ce{CCl4} \) is tetrahedral, trans-1,2-dichloroethene is centrosymmetric and p-dichlorobenzene is symmetrical: bond dipoles cancel in all three.
2. trans-Pent-2-ene has two different alkyl groups (CH3 and C2H5), so a small net dipole remains.

Teacher's Note:
1. Zero dipole needs complete cancellation of bond dipoles.

Answer: (d) All of these: a racemic mixture is resolvable; a meso isomer is optically inactive due to internal compensation; a racemic mixture is optically inactive due to external compensation

Explanation:
1. A racemic mixture can be resolved into its enantiomers.
2. A meso isomer is inactive because one half of the molecule rotates light oppositely to the other (internal compensation).
3. A racemic mixture is inactive because the rotations of separate enantiomer molecules cancel (external compensation).

Teacher's Note:
1. Internal: within one molecule. External: between different molecules.

Answer: (d) The first one (top CH3; H, Cl; H, CH3; bottom Cl)

Explanation:
1. Assign R/S to each centre of each structure.
2. First: (S,S), chiral and not meso.
3. Second: (R,S), meso. Third: (R,S), meso. Fourth: (S,R), meso.

Teacher's Note:
1. A meso form has opposite configurations at the two similar centres.

Answer: (d) 4-Methylcyclohex-2-en-1-one

Explanation:
1. 4-Methylcyclohex-2-en-1-one has an alpha hydrogen (C6) for keto-enol tautomerism and a chiral C4.
2. Butan-2-ol is only chiral, while butanone and pentane-2,4-dione are achiral.

Teacher's Note:
1. Need an alpha H and a chiral centre.

Answer: (c) R

Explanation:
1. Priorities: Br > CH=CH2 > CH2CH3 > H.
2. Br (right), CH=CH2 (bottom), CH2CH3 (left) run clockwise and H is on a vertical bond (away from the viewer): R.

Teacher's Note:
1. A vertical lowest-priority group needs no reversal.

Answer: (b) cis-1,3-Dimethylcyclohexane

Explanation:
1. cis-1,3-Dimethylcyclohexane has two chiral centres and a plane of symmetry: meso.
2. The trans-1,2 and trans-1,3 isomers are chiral, and 1,1-dimethylcyclohexane has no stereocentre.

Teacher's Note:
1. Meso: chiral centres plus a plane of symmetry.

Answer: (a) Z

Explanation:
1. Left carbon: Cl outranks F. Right carbon: D outranks H (isotope with higher mass).
2. Cl and D are on the same side: Z.

Teacher's Note:
1. Isotopes are ranked by mass number.

Answer: (d) Fully eclipsed (two methyl groups eclipsed)

Explanation:
1. The two bulky methyl groups eclipse each other, giving maximum torsional and steric strain.
2. It is about 19 kJ per mol above the anti form.

Teacher's Note:
1. Fully eclipsed has CH3 and CH3 at a dihedral angle of 0 degrees.

Answer: (c) R, S

Explanation:
1. Compound I: priorities NH2 > COOH > CH3 > H. NH2 (right) to COOH (top) to CH3 (bottom) runs anticlockwise, but H is on a horizontal bond (towards the viewer), so reverse: R.
2. Compound II: CH2SH (S,H,H) outranks COOH (O,O,O) because S has the higher atomic number, so the order is NH2 > CH2SH > COOH > H. The path is clockwise, reversed by the horizontal H: S.

Teacher's Note:
1. On a Fischer projection, a horizontal lowest-priority group reverses the apparent direction.
2. Compare the first point of difference: S beats O.

Answer: (b) cis-1,2-dichloroethene > trans-1,2-dichloroethene, and but-1-yne > but-2-yne

Explanation:
1. In the cis isomer the two C-Cl dipoles add up, while in the trans isomer they cancel (zero).
2. But-2-yne is symmetrical with zero dipole moment, while but-1-yne has a net dipole.

Teacher's Note:
1. Cancellation of bond dipoles decides the order.

Answer: (c) A-S, B-R, C-Q, D-P

Explanation:
1. A meso compound has chiral centres but is achiral overall (S).
2. Enantiomers are non-superimposable mirror images (R).
3. Diastereomers are stereoisomers that are not mirror images (Q).
4. A racemic mixture is an equimolar mixture of enantiomers (P).

Teacher's Note:
1. Match each term with its definition.

Answer: (a) \( \ce{CH3-C#C-CH3} \)

Explanation:
1. But-2-yne is symmetrical: the two C-CH3 bond dipoles point in opposite directions and cancel.
2. The other three molecules are unsymmetrical and have a net dipole moment.

Teacher's Note:
1. Symmetrical linear molecules have zero dipole moment.

Answer: (d) 2

Explanation:
1. The ring carbon of the first double bond carries two identical ring arms, so that double bond shows no geometrical isomerism.
2. The second double bond, \( \ce{-CH=CH-} \), has different groups on each carbon, so it can be cis or trans.
3. Total = 2.

Teacher's Note:
1. A double bond shows isomerism only if each carbon has two different groups.

Answer: (a) 3

Explanation:
1. Only unbranched alkenes give n-hexane.
2. They are hex-1-ene, hex-2-ene and hex-3-ene, so 3.

Teacher's Note:
1. Hydrogenation does not change the carbon skeleton.

Answer: (b) E

Explanation:
1. On the first carbon Cl has higher priority than CH3; on the second carbon I has higher priority than Br.
2. Cl and I lie on opposite sides, so the configuration is E.

Teacher's Note:
1. Give priority by atomic number, then compare sides.

Answer: (b) \( \ce{C6H5-CH=C(CH3)Cl} \)

Explanation:
1. Cis-trans labels need two identical (or analogous) groups, one on each double-bond carbon.
2. In the trisubstituted alkene \( \ce{C6H5CH=C(CH3)Cl} \) there are three different groups besides H, so only E/Z can be used.

Teacher's Note:
1. Use E/Z for tri- and tetra-substituted alkenes.

Answer: (b) 7

Explanation:
1. N-methyl, N-methyl with a butyl group: 4 isomers (four butyl groups).
2. N-methyl, N-ethyl with a propyl group: 2 isomers.
3. Triethylamine: 1 isomer.
4. Total = 4 + 2 + 1 = 7.

Teacher's Note:
1. Divide the six carbons among three alkyl groups on N.

Answer: (d) 8

Explanation:
1. Primary amines (butyl group, 4 isomers): \( \ce{C4H9NH2} \).
2. Secondary amines: methylpropyl (2 isomers, propyl and isopropyl) and diethyl (1) = 3.
3. Tertiary amine: ethyldimethylamine = 1.
4. Total = 4 + 3 + 1 = 8.

Teacher's Note:
1. The butyl group has 4 isomers, the propyl group 2.

Answer: (d) E

Explanation:
1. Top carbon: 14CH3 outranks CH3 (same atoms, higher mass).
2. Bottom carbon: OCH3 outranks 14CH3 (O above C).
3. The higher priority groups are on opposite sides: E.

Teacher's Note:
1. Compare atomic number first, mass number only if atoms are identical.

Answer: (b) Functional group isomerism

Explanation:
1. Both have the molecular formula \( \ce{C5H8O2} \).
2. One is a lactone (ester) and the other is a carboxylic acid, so they are functional isomers.

Teacher's Note:
1. Esters and carboxylic acids with the same formula are functional isomers.