JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 74

Answer: (b) homomers (identical molecules)

Explanation:
1. In I the priorities are Br > Cl > CH3 > H: Br (bottom), Cl (right), CH3 (top) run anticlockwise; H is on a horizontal bond, so reverse the result: R.
2. In II the priorities are Br (left), Cl (top), CH3 (right): clockwise; H is on a vertical bond, so the result stays R.
3. Both are R, so they are the same molecule.

Teacher's Note:
1. Assign R or S to each and compare.

Answer: (d) n-Butane

Explanation:
1. n-Butane has two types of hydrogen (C1 and C4 equivalent, C2 and C3 equivalent): 2 monochloro derivatives.
2. 2,2-Dimethylpentane gives 5, while benzene and neopentane give only one each.

Teacher's Note:
1. Count the types of equivalent hydrogen atoms.

Answer: (c) 4

Explanation:
1. With 4 carbons, N-methylpropan-1-amine and N-methylpropan-2-amine (\( \ce{CH3NHCH2CH2CH3} \) and \( \ce{CH3NHCH(CH3)2} \)) differ in the alkyl chain on nitrogen.
2. With 3 carbons only one secondary amine (CH3NHC2H5) exists, so 4 is the minimum.

Teacher's Note:
1. The alkyl group must be able to have a branched and an unbranched form (propyl and isopropyl).

Answer: (d) Chain isomerism

Explanation:
1. Both are \( \ce{C6H12O2} \) carboxylic acids with the group on the same type of carbon.
2. They differ in the arrangement of the carbon skeleton, so they are chain isomers.

Teacher's Note:
1. Different carbon skeleton and same functional group means chain isomerism.

Answer: (a) \( 2^{n-1} \)

Explanation:
1. The total number of stereoisomers is \( 2^n \) for an unsymmetrical compound.
2. They form \( 2^n/2 = 2^{n-1} \) enantiomeric pairs, and each pair gives one racemic mixture.

Teacher's Note:
1. Number of racemic mixtures equals number of enantiomeric pairs.

Answer: (a) identical

Explanation:
1. First structure: at the SH carbon, SH > C3 > CH3 > H gives S; at the Cl carbon, Cl > C2 > C2H5 > H gives S.
2. Second structure: both centres also work out as S.
3. Same constitution and same configuration at each centre: identical.

Teacher's Note:
1. Determine the configuration of each centre and compare.

Answer: (c) 3

Explanation:
1. Anthracene has three sets of equivalent positions: 1,4,5,8; 2,3,6,7; and 9,10.
2. One chloro derivative is possible for each set, giving 3 isomers.

Teacher's Note:
1. Count the sets of equivalent hydrogen atoms.
2. Use the symmetry of the molecule.

Answer: (b) identical

Explanation:
1. Both have the constitution CH3-CH(OH)-CH(Br)-C2H5 (3-bromopentan-2-ol).
2. The first is (2S,3S): at the OH carbon OH > C3 > CH3 > H gives S, and at the Br carbon Br > C2 > C2H5 > H gives S.
3. The second also works out as (2S,3S), so the molecules are identical.

Teacher's Note:
1. Assign the configuration at each stereocentre and compare.

Answer: (a) 3

Explanation:
1. But-1-ene, but-2-ene and 2-methylpropene.
2. Cis and trans but-2-ene are stereoisomers, so they are not counted separately.

Teacher's Note:
1. Ignore cis-trans isomers when asked for structural isomers.

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

Explanation:
1. Constitutional isomers differ in connectivity (R).
2. Stereoisomers differ only in spatial arrangement (Q).
3. Enantiomers are object and non-superimposable mirror image (S).
4. Diastereomers are stereoisomers that are not mirror images (P).

Teacher's Note:
1. Connectivity differs for constitutional isomers only.

Answer: (c) Z

Explanation:
1. Top carbon: NH2 (N) outranks CH3 (C). Bottom carbon: Cl outranks NO2 (Cl has higher atomic number than N).
2. NH2 and Cl are on the same side, so the configuration is Z.

Teacher's Note:
1. Higher atomic number gets the higher priority.

Answer: (b) Gauche

Explanation:
1. In the gauche conformer the OH and NO2 groups are close enough to form an intramolecular hydrogen bond.
2. This extra stabilisation makes gauche more stable than anti, while eclipsed forms remain strained.

Teacher's Note:
1. Intramolecular hydrogen bonding can stabilise a gauche form.

Answer: (a) \( \ce{-OH > -COOH > -CHO > -CH2OH} \)

Explanation:
1. The first atom decides: O (in OH) outranks C.
2. Among carbon groups, compare the substituents on C: COOH (O,O,O) > CHO (O,O,H) > CH2OH (O,H,H).

Teacher's Note:
1. Compare the first atom, then the set of atoms attached to it.

Answer: (c) (i), (ii) and (iv)

Explanation:
1. Pent-3-en-1-yne (HC#C-CH=CH-CH3), acetaldoxime (CH3CH=NOH) and 1-chloropropene (ClCH=CHCH3) have different groups on each doubly bonded atom.
2. Pent-1-ene and 2-methylbut-2-ene have two identical groups on one carbon, and but-2-yne is linear.

Teacher's Note:
1. Check each doubly bonded atom for two different groups.

Answer: (c) 2,3-Dimethylpentane and 2,4-dimethylpentane are position isomers

Explanation:
1. Butanone is \( \ce{C4H8O} \) and propanal is \( \ce{C3H6O} \); their molecular formulae differ, so they are not isomers.
2. Propanamide and 3-aminopropanal have the same formula but different functional groups, so they are functional isomers only.
3. 2,3- and 2,4-Dimethylpentane have the same chain and differ in the position of the methyl group: position isomers.

Teacher's Note:
1. Isomers must have the same molecular formula first.

Answer: (d) Neopentane

Explanation:
1. \( 14n+2=72 \) gives n = 5, so the alkane is \( \ce{C5H12} \).
2. The branched isomer with all 12 hydrogens equivalent is neopentane, \( \ce{C(CH3)4} \), which gives one monochloro derivative.
3. Isopentane gives four monochloro derivatives.

Teacher's Note:
1. Find the alkane from the molar mass, then choose the one with equivalent hydrogens.

Answer: (b) 5

Explanation:
1. Anisole (methoxybenzene).
2. Benzyl alcohol.
3. o-, m- and p-cresol (3 isomers).
4. Total = 1 + 1 + 3 = 5.

Teacher's Note:
1. Consider ether, alcohol and phenol types.

Answer: (c) 9

Explanation:
1. X = 2 (butanal and 2-methylpropanal).
2. Y = 3 (n-pentane, isopentane, neopentane).
3. Z = 4 (but-1-ene, cis-but-2-ene, trans-but-2-ene and 2-methylpropene).
4. Sum = 2 + 3 + 4 = 9.

Teacher's Note:
1. Count the geometrical isomers in (c).

Answer: (c) 2

Explanation:
1. In a geminal dihalide both chlorine atoms are on the same carbon.
2. They are 1,1-dichloropropane and 2,2-dichloropropane: 2 isomers.

Teacher's Note:
1. Geminal means both halogens on the same carbon.

Answer: (d) \( \ce{(CH3)2C=NOH} \)

Explanation:
1. \( \ce{(CH3)2C=NOH} \) has two identical methyl groups on carbon, so the OH can only lie on one equivalent side.
2. The other three have different groups (or ring constraints) on both doubly bonded atoms.

Teacher's Note:
1. Each doubly bonded atom must carry two different groups (a lone pair counts as a group on N).