JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 36

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

Explanation:
1. Butene has four carbons with the double bond between C-2 and C-3
2. A methyl group on C-2 gives \( \ce{(CH3)2C=CH-CH3} \)

Teacher's Note:
1. Find the principal functional group first, then the longest chain containing it, then number for the lowest locants.
2. Write substituent prefixes in alphabetical order.

Answer: (a) 1,4-Butanedioic acid

Explanation:
1. Succinic acid is \( \ce{HOOC-CH2-CH2-COOH} \)
2. A four-carbon chain with COOH at both ends is butane-1,4-dioic acid

Teacher's Note:
1. Eliminate options with an incorrect parent chain length first.
2. Then compare locants and alphabetical order among the remaining options.

Answer: (c) 3-Amino-5-heptenoic acid

Explanation:
1. C-1 is the COOH carbon and the chain has seven carbons
2. The NH2 is on C-3 and the double bond starts at C-5, so 3-aminohept-5-enoic acid

Teacher's Note:
1. Draw the carbon skeleton out in full before naming; it prevents mistakes with branches.
2. Check that the principal group has the lowest possible locant.

Answer: (d) 3-Chloro-1-propene

Explanation:
1. The double bond is given the lowest locant, so numbering starts at the \( \ce{CH2=} \) end
2. Cl is then on C-3: 3-chloroprop-1-ene (allyl chloride is a common name)

Teacher's Note:
1. Eliminate options with an incorrect parent chain length first.
2. Then compare locants and alphabetical order among the remaining options.

Answer: (b) 3-Buten-1-ol

Explanation:
1. OH takes the lowest locant, so C-1 is the \( \ce{CH2OH} \) carbon
2. The double bond then starts at C-3: but-3-en-1-ol

Teacher's Note:
1. Draw the carbon skeleton out in full before naming; it prevents mistakes with branches.
2. Check that the principal group has the lowest possible locant.

Answer: (b) \( \ce{CH3-CH2-C#C-CH2-COOH} \)

Explanation:
1. C-1 is the COOH carbon; the triple bond lies between C-3 and C-4
2. This gives \( \ce{CH3CH2-C#C-CH2-COOH} \); (a) is hex-2-ynoic acid, (c) hex-4-ynoic acid and (d) is an alkene

Teacher's Note:
1. Eliminate options with an incorrect parent chain length first.
2. Then compare locants and alphabetical order among the remaining options.

Answer: (a) 3,3,3-Trichloropropanal

Explanation:
1. The aldehyde carbon is C-1 of a three-carbon chain
2. The three Cl atoms are on C-3

Teacher's Note:
1. Draw the carbon skeleton out in full before naming; it prevents mistakes with branches.
2. Check that the principal group has the lowest possible locant.

Answer: (a) 1-Chloro-3-methylbutane

Explanation:
1. The carbon bearing Cl is C-1 in a four-carbon chain
2. The methyl group is on C-3

Teacher's Note:
1. Eliminate options with an incorrect parent chain length first.
2. Then compare locants and alphabetical order among the remaining options.

Answer: (a) 3-Ethylhex-2-ene

Explanation:
1. The longest chain containing the double bond has six carbons
2. Numbering from the \( \ce{CH3-CH=} \) end gives C=C at C-2 and an ethyl on C-3

Teacher's Note:
1. Draw the carbon skeleton out in full before naming; it prevents mistakes with branches.
2. Check that the principal group has the lowest possible locant.

Answer: (d) 2-Chlorohex-4-en-2-ol

Explanation:
1. OH is the principal group and gets the lowest locant, so numbering starts at the \( \ce{CH3} \) next to C(OH)
2. OH and Cl are on C-2 and the double bond starts at C-4

Teacher's Note:
1. Find the principal functional group first, then the longest chain containing it, then number for the lowest locants.
2. Write substituent prefixes in alphabetical order.

Answer: (a) 2,4-Dimethylpentan-2-ol

Explanation:
1. The five-carbon chain is numbered so that OH has the lowest locant (2)
2. Methyl groups are on C-2 and C-4

Teacher's Note:
1. Draw the carbon skeleton out in full before naming; it prevents mistakes with branches.
2. Check that the principal group has the lowest possible locant.

Answer: (b) Pent-3-en-1-yne

Explanation:
1. The locant set (1, 3) is lower than (2, 4), so the triple bond gets C-1
2. Hence pent-3-en-1-yne

Teacher's Note:
1. Find the principal functional group first, then the longest chain containing it, then number for the lowest locants.
2. Write substituent prefixes in alphabetical order.

Answer: (c) 3-Methylbut-2-enoic acid

Explanation:
1. C-1 is the COOH carbon and the double bond is between C-2 and C-3
2. The extra methyl is on C-3

Teacher's Note:
1. Draw the carbon skeleton out in full before naming; it prevents mistakes with branches.
2. Check that the principal group has the lowest possible locant.

Answer: (b) 4-Hydroxypent-3-enoic acid

Explanation:
1. C-1 is the COOH carbon; the double bond is between C-3 and C-4
2. OH is on C-4

Teacher's Note:
1. Find the principal functional group first, then the longest chain containing it, then number for the lowest locants.
2. Write substituent prefixes in alphabetical order.

Answer: (a) 2-Butenoic acid

Explanation:
1. The four-carbon acid has C=C between C-2 and C-3
2. The name is but-2-enoic acid

Teacher's Note:
1. Draw the carbon skeleton out in full before naming; it prevents mistakes with branches.
2. Check that the principal group has the lowest possible locant.

Answer: (b) 2,2-Dimethylpentane

Explanation:
1. The longest chain has five carbons
2. Both methyls are on C-2 when numbered from the branched end

Teacher's Note:
1. Find the principal functional group first, then the longest chain containing it, then number for the lowest locants.
2. Write substituent prefixes in alphabetical order.

Answer: (a) I

Explanation:
1. Electronegativity of carbon increases with s-character
2. The terminal carbon is sp hybridised (50% s), more than sp3 (25% s) for the others

Teacher's Note:
1. Find the principal functional group first, then the longest chain containing it, then number for the lowest locants.
2. Write substituent prefixes in alphabetical order.

Answer: (d) 4,4-Dimethylpent-1-ene

Explanation:
1. The double bond is given the lowest locant (C-1)
2. Two methyl groups are then on C-4

Teacher's Note:
1. Draw the carbon skeleton out in full before naming; it prevents mistakes with branches.
2. Check that the principal group has the lowest possible locant.

Answer: (c) 1

Explanation:
1. A tertiary carbon is attached to three other carbons
2. Only the \( \ce{CH} \) of the isopropyl group qualifies; \( \ce{C(CH3)3} \) is quaternary

Teacher's Note:
1. Find the principal functional group first, then the longest chain containing it, then number for the lowest locants.
2. Write substituent prefixes in alphabetical order.

Answer: (d) 2-Methylpentane

Explanation:
1. An isopropyl group is \( \ce{(CH3)2CH-} \) attached to the rest of the molecule
2. 2-Methylpentane, \( \ce{(CH3)2CH-CH2CH2CH3} \), has one; the others have no such terminal group

Teacher's Note:
1. Draw the carbon skeleton out in full before naming; it prevents mistakes with branches.
2. Check that the principal group has the lowest possible locant.

Answer: (c) 2-Methylpentane-2,4-diol

Explanation:
1. The five-carbon chain has OH on C-2 and C-4 (locants 2,4); numbering from the end that gives the methyl locant 2
2. The methyl is on C-2, so 2-methylpentane-2,4-diol

Teacher's Note:
1. Find the principal functional group first, then the longest chain containing it, then number for the lowest locants.
2. Write substituent prefixes in alphabetical order.

Answer: (c) 3-Methylbut-2-en-1-ol

Explanation:
1. OH takes C-1; the double bond is between C-2 and C-3
2. The methyl group is on C-3

Teacher's Note:
1. Draw the carbon skeleton out in full before naming; it prevents mistakes with branches.
2. Check that the principal group has the lowest possible locant.

Answer: (b) Nonane

Explanation:
1. \( \ce{C9H20} \) fits \( \ce{CnH2n+2} \) with n = 9
2. Nine carbons are named non- so the alkane is nonane

Teacher's Note:
1. Draw the carbon skeleton out in full before naming; it prevents mistakes with branches.
2. Check that the principal group has the lowest possible locant.