JEE Chemistry Organic Chemistry Some Basic Principles And Techniques MCQs Set 12

Answer: (d) 2-phenylethyl 2-phenylethanoate

Explanation:
1. The acid part is C₆H₅CH₂COOH, whose chain has two carbons, so it is 2-phenylethanoate
2. The alcohol part is –CH₂CH₂C₆H₅, which is 2-phenylethyl
3. Alkyl group first, then the acid name: 2-phenylethyl 2-phenylethanoate. Answer: (d)

Teacher's Note:
1. In an ester name the alkyl group of the alcohol is written first.
2. The acid part uses the suffix -oate with its locants.

Answer: (d) 1-bromo-1-chloro-2,2,2-trifluoro-1-iodoethane

Explanation:
1. The parent is ethane with six halogen substituents
2. Either numbering gives the locant set {1,1,1,2,2,2}, so the tie is broken by alphabetical order: the substituent cited first (bromo) gets the lowest locant
3. Alphabetical order: bromo, chloro, fluoro, iodo
4. Name: 1-bromo-1-chloro-2,2,2-trifluoro-1-iodoethane. Answer: (d)

Teacher's Note:
1. First use the lowest set of locants; if it is a tie, give the lower number to the substituent that comes first alphabetically.
2. Prefixes like tri are ignored for alphabetising.

Answer: (c) Eight

Explanation:
1. Pentane gives 3 alkyl groups (1-pentyl, 2-pentyl, 3-pentyl)
2. 2-Methylbutane gives 4 (2-methylbutyl, 3-methylbutan-2-yl, 2-methylbutan-2-yl, 3-methylbutyl)
3. 2,2-Dimethylpropane gives 1 (neopentyl)
4. Total = 3 + 4 + 1 = 8. Answer: (c)

Teacher's Note:
1. Count the distinct positions of the free valency on each pentane skeleton.
2. Skeletons are pentane, isopentane and neopentane.

Answer: (c) 4-methoxy-2-nitrobenzenecarbaldehyde

Explanation:
1. The aldehyde is the principal group, so the ring carbon bearing –CHO is C1
2. NO₂ is at C2 and OCH₃ is at C4 (para)
3. Substituents in alphabetical order: methoxy, nitro
4. Name: 4-methoxy-2-nitrobenzenecarbaldehyde (4-methoxy-2-nitrobenzaldehyde). Answer: (c)

Teacher's Note:
1. The principal functional group decides the parent name and numbering.
2. Number so that the substituents get the lowest locants.

Answer: (b) (b)

Explanation:
1. For a ring attached to –COCl the name is cycloalkane + carbonyl chloride: cyclohexanecarbonyl chloride
2. The name cyclohexaneoyl chloride is wrong because the ring is not part of a chain ending in –oyl
3. The other three names follow the rule of ring name plus suffix, with the ring carbon bearing the principal group numbered 1. Answer: (b)

Teacher's Note:
1. When a functional group is attached to a ring, the suffix is added after the word carbo (carboxamide, carbonitrile, carbonyl chloride).
2. Esters with a ring acid are named alkyl cycloalkanecarboxylate.

Answer: (c) 5-bromo-3-chloro-1-ethyl-2-methylcyclohex-1-ene

Explanation:
1. Ring atoms in order: C(CH₃)=C(C₂H₅)–CH₂–CHBr–CH₂–CHCl– back to C(CH₃)
2. The double bond takes locants 1 and 2
3. Option 1: C1 = C–CH₃ gives locants {1,2,4,6}; option 2: C1 = C–C₂H₅, C2 = C–CH₃, then CHCl is C3, CH₂ is C4, CHBr is C5, which gives {1,2,3,5}
4. The lower set is {1,2,3,5}. Alphabetical order: bromo, chloro, ethyl, methyl. Answer: (c)

Teacher's Note:
1. Number the double bond as 1 and 2 first, then pick the direction giving the lowest locant set for the substituents.
2. Name substituents alphabetically.

Answer: (c) benzene-1,2-dicarbaldehyde

Explanation:
1. Two –CHO groups attached directly to a ring are named with the suffix -carbaldehyde
2. Locants 1 and 2 for adjacent carbons
3. Name: benzene-1,2-dicarbaldehyde (common name phthalaldehyde). Answer: (c)

Teacher's Note:
1. Aldehyde groups attached to a ring use carbaldehyde.
2. A second -CHO group is not named as a formyl substituent when the suffix can express both.

Answer: (a) 4

Explanation:
1. Chlorotoluenes: ortho, meta and para = 3
2. Benzyl chloride, C₆H₅CH₂Cl = 1
3. Total = 4. Answer: (a)

Teacher's Note:
1. Place Cl on the ring (three positions) or on the CH3 group.
2. Benzenoid means containing a benzene ring.

Answer: (a) Butanedioic acid

Explanation:
1. Butanedioic acid, HOOC–CH₂–CH₂–COOH, has free rotation about all C–C bonds and no double bond, so no cis-trans isomerism
2. 1,2-Dimethylcyclopropane has restricted rotation in the ring
3. 1,2-Dichloroethene and but-2-enedioic acid (maleic and fumaric acids) have a C=C bond with two different groups on each carbon. Answer: (a)

Teacher's Note:
1. Cis-trans isomerism needs restricted rotation and two different groups on each carbon.
2. Rings can provide the restriction just as double bonds do.

Answer: (c) Metamerism

Explanation:
1. Diethyl ether is C₂H₅–O–C₂H₅ and methyl propyl ether is CH₃–O–C₃H₇
2. Both are ethers with the formula C₄H₁₀O but different alkyl groups on the two sides of the oxygen
3. This is metamerism. Answer: (c)

Teacher's Note:
1. Metamerism occurs in compounds with a polyvalent functional group like -O-, -S- or -NH-.
2. The functional group is the same in both compounds.

Answer: (c) 4

Explanation:
1. Pentan-1-ol
2. 2-Methylbutan-1-ol
3. 3-Methylbutan-1-ol
4. 2,2-Dimethylpropan-1-ol
5. That makes 4 primary alcohols. Answer: (c)

Teacher's Note:
1. A primary alcohol has the -OH on a carbon attached to only one other carbon.
2. Go through the pentane, isopentane and neopentane skeletons.

Answer: (d) none of these

Explanation:
1. Phenol is C₆H₅OH and benzyl alcohol is C₆H₅CH₂OH, with different molecular formulas, so they are not isomers of any kind
2. They differ in the nature of the –OH carbon (aromatic sp² versus aliphatic sp³), so they belong to different classes and are not true homologues
3. So none of the given descriptions fits. Answer: (d)

Teacher's Note:
1. Isomers must have the same molecular formula.
2. Homologues belong to the same class and differ by a CH2 unit.

Answer: (c) 8

Explanation:
1. There are three double bonds, each of which can show geometrical isomerism since each carbon has two different groups
2. The molecule is unsymmetrical, so the number of isomers is \( 2^3 = 8 \). Answer: (c)

Teacher's Note:
1. For an unsymmetrical polyene, isomers = 2 raised to the number of stereogenic double bonds.
2. Symmetrical ends would reduce the count.

Answer: (c) 4

Explanation:
1. The ring carbons bearing OH and NH₂ are both stereocentres
2. The two substituents are different, so the molecule has no plane of symmetry
3. Number of stereoisomers = \( 2^2 = 4 \) (two cis and two trans, each as an enantiomeric pair). Answer: (c)

Teacher's Note:
1. Count stereocentres, then check for symmetry or meso forms.
2. With different groups at the 1 and 3 positions no meso form exists.

Answer: (a) (i) 4, (ii) 16, (iii) 10

Explanation:
1. (i) Two different chiral centres: \( 2^2 = 4 \)
2. (ii) Four different chiral centres: \( 2^4 = 16 \)
3. (iii) Four chiral centres but the molecule is symmetrical, with n = 4 even: \( 2^{n-1} + 2^{n/2-1} = 8 + 2 = 10 \) (includes meso forms). Answer: (a)

Teacher's Note:
1. For a symmetrical molecule with an even number of stereocentres, use 2^(n-1) + 2^(n/2-1).
2. Meso forms reduce the count below 2^n.