Practice MCQs for JEE Chemistry Organic Chemistry Some Basic Principles and Techniques
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Multiple Choice Questions
(a) 2,3-Dimethylhexan-5-ol
(b) 3-Ethyl-4-methylpentan-2-ol
(c) 3,4-Dimethylhexan-2-ol
(d) 2-Ethyl-3-methylpentan-4-ol
Show Answer & Explanation
Answer: (c) 3,4-Dimethylhexan-2-ol
Explanation:
1. The principal group is OH, so choose the longest chain containing the carbon bearing OH.
2. The longest chain has 6 carbons (the ethyl group is part of the chain).
3. Number from the OH end: OH on C2, methyl groups on C3 and C4.
Teacher's Note:
1. Include the longest chain, not the drawn main line.
2. Give the lowest locant to the principal group.
(a) 3-(Hydroxymethyl)pent-4-enenitrile
(b) 3-Hydroxymethylpent-2-enenitrile
(c) 4-Hydroxy-3-vinylbutanenitrile
(d) 3-Ethenyl-4-hydroxybutanenitrile
Show Answer & Explanation
Answer: (a) 3-(Hydroxymethyl)pent-4-enenitrile
Explanation:
1. Nitrile outranks alcohol and the C of CN is C1.
2. The chain C1 to C5 contains the double bond between C4 and C5 and is the longest.
3. The CH2OH group on C3 is a hydroxymethyl substituent: 3-(hydroxymethyl)pent-4-enenitrile.
Teacher's Note:
1. The carbon of the nitrile group is counted in the chain.
2. Choose the longest chain containing the principal group.
(a) 2-Chloro-4-(chloromethyl)-3-propylheptane
(b) 1,6-Dichloro-2-methyl-5-propylheptane
(c) 5-(1-Chloroethyl)-1-chloro-2-methylnonane
(d) 1-Chloro-5-(1-chloroethyl)-2-methyloctane
Show Answer & Explanation
Answer: (d) 1-Chloro-5-(1-chloroethyl)-2-methyloctane
Explanation:
1. Longest chain has 8 carbons: ClCH2-CH(CH3)-CH2-CH2-CH(CHClCH3)-CH2CH2CH3.
2. Where two chains of equal length exist, choose the one with more substituents.
3. Locants: 1-chloro, 2-methyl, 5-(1-chloroethyl); alphabetical order puts chloro before chloroethyl before methyl.
Teacher's Note:
1. Equal-length chains: take the more substituted one.
2. Complex substituents are named in brackets and alphabetised by their first letter.
(a) 3,3,4,4-Tetramethylpentane
(b) 2,2,3,3-Tetramethylpentane
(c) 2,2,3-Trimethyl-3-ethylbutane
(d) 2,3,3-Trimethyl-2-ethylbutane
Show Answer & Explanation
Answer: (b) 2,2,3,3-Tetramethylpentane
Explanation:
1. The longest chain has 5 carbons.
2. Number from the end that gives the lowest locants: 2,2,3,3.
Teacher's Note:
1. Lowest locant set decides the numbering.
(a) Ethyl 3-cyanobutanoate
(b) Ethyl 3-isocyanobutanoate
(c) Ethyl 2-isocyanobutanoate
(d) 3-Isocyanobutyl ethanoate
Show Answer & Explanation
Answer: (b) Ethyl 3-isocyanobutanoate
Explanation:
1. The ester is the principal group, named as ethyl butanoate (the acid part has 4 carbons).
2. The NC group is isocyano, located on C3.
Teacher's Note:
1. -NC is isocyano, -CN is cyano.
(a) 2,3-Dicyanopentane
(b) 3-Cyano-2-ethylbutanenitrile
(c) 2-Methyl-3-ethylbutanedinitrile
(d) 2-Ethyl-3-methylbutanedinitrile
Show Answer & Explanation
Answer: (d) 2-Ethyl-3-methylbutanedinitrile
Explanation:
1. Both CN carbons are in the main chain, which has 4 carbons (butanedinitrile).
2. Ethyl on C2 and methyl on C3, cited alphabetically.
Teacher's Note:
1. Include both nitrile carbons in the chain.
(a) 1,1-Dibromo-3,3,3-trichloropropane
(b) 3,3-Dibromo-1,1,1-trichloropropane
(c) 1,1,1-Tribromo-3,3-dichloropropane
(d) 1,3-Dibromo-1,3,3-trichloropropane
Show Answer & Explanation
Answer: (a) 1,1-Dibromo-3,3,3-trichloropropane
Explanation:
1. The chain is propane and all substituents are halogens.
2. Both numberings give the locant set {1,1,1,3,3}, so the tie is broken by alphabetical order: bromo is cited first and gets the lower locants.
3. Name: 1,1-dibromo-3,3,3-trichloropropane.
Teacher's Note:
1. When locant sets tie, the substituent cited first alphabetically gets the lowest number.
(a) 3-Ethynylhex-4-enal
(b) 4-Ethenylhex-2-ynal
(c) 3-Ethenylhex-4-ynal
(d) 3-(Prop-1-ynyl)pent-4-enal
Show Answer & Explanation
Answer: (c) 3-Ethenylhex-4-ynal
Explanation:
1. The aldehyde carbon is C1.
2. The longest chain through the triple bond has 6 carbons: C1 CHO, C2, C3, C4 to C6 with a triple bond at C4.
3. The vinyl group on C3 is the ethenyl substituent.
Teacher's Note:
1. Principal group first, then longest chain.
(a) 3-Ethenylbut-1-enal
(b) 2-Vinylbut-3-en-1-one
(c) 2-Ethenylbut-3-enal
(d) 3-Formylpenta-1,4-diene
Show Answer & Explanation
Answer: (c) 2-Ethenylbut-3-enal
Explanation:
1. CHO is C1; the chain is C1 to C4 with the double bond at C3.
2. The second vinyl group sits on C2.
Teacher's Note:
1. The aldehyde carbon always takes locant 1.
(a) 1-Chloro-3-methylbut-3-en-2-ol
(b) 4-Chloro-2-methylbut-1-en-3-ol
(c) 1-Chloro-2-methylbut-3-en-3-ol
(d) 2-Methyl-1-chlorobut-1-en-3-ol
Show Answer & Explanation
Answer: (a) 1-Chloro-3-methylbut-3-en-2-ol
Explanation:
1. OH is the principal group and gets the lowest locant: C2.
2. Chain: ClCH2 (C1), CH(OH) (C2), C(CH3)= (C3), =CH2 (C4).
3. Name: 1-chloro-3-methylbut-3-en-2-ol.
Teacher's Note:
1. Lowest locant goes to the OH group before the double bond.
(a) 2-Methylhex-4-yne
(b) 4-Methylpent-2-yne
(c) 2-Methylhex-2-yne
(d) 5-Methylhex-2-yne
Show Answer & Explanation
Answer: (d) 5-Methylhex-2-yne
Explanation:
1. Longest chain has 6 carbons containing the triple bond.
2. Number from the end nearer the triple bond: C2-C3 triple bond, methyl on C5.
Teacher's Note:
1. Number to give the multiple bond the lowest locant.
(a) 5-Bromo-7-chlorooct-2-ene-4,6-diol
(b) 4-Bromo-2-chlorooct-6-ene-3,5-diol
(c) 4-Bromo-2-chlorooct-6-en-3,5-diol
(d) 2-Bromo-4-chlorooct-7-ene-3,5-diol
Show Answer & Explanation
Answer: (b) 4-Bromo-2-chlorooct-6-ene-3,5-diol
Explanation:
1. The main chain has 8 carbons with two OH groups; the OH groups get the lowest locants: 3,5 (not 4,6).
2. The double bond is then at C6, Br on C4 and Cl on C2.
3. Name: 4-bromo-2-chlorooct-6-ene-3,5-diol.
Teacher's Note:
1. The principal group gets the lowest locants before the double bond.
(a) 3-Hydroxybutanoic acid
(b) 2-Hydroxybutanoic acid
(c) 1-Hydroxybutanoic acid
(d) 2-Hydroxypropanoic acid
Show Answer & Explanation
Answer: (b) 2-Hydroxybutanoic acid
Explanation:
1. The COOH carbon is C1 in a 4-carbon chain.
2. OH is on C2.
Teacher's Note:
1. COOH carbon is C1.
(a) 2-Formyl-3-methyl-4-oxopentanoic acid
(b) 3-Formyl-2-methyl-5-oxopentanoic acid
(c) 3,4-Diformylhexanoic acid
(d) 3-Formyl-4-methyl-5-oxopentanoic acid
Show Answer & Explanation
Answer: (d) 3-Formyl-4-methyl-5-oxopentanoic acid
Explanation:
1. Carboxylic acid is the principal group: C1 is COOH.
2. Chain: C1 COOH, C2 CH2, C3 CH(CHO), C4 CH(CH3), C5 CHO (written as oxo).
3. Substituents: 3-formyl, 4-methyl, 5-oxo.
Teacher's Note:
1. Aldehyde groups are prefixes (formyl or oxo) when an acid is present.
(a) Cyclopentanecarboxamide
(b) Cyclopentylamide
(c) Cyclopentanamide
(d) Cyclopentanecarbonamine
Show Answer & Explanation
Answer: (a) Cyclopentanecarboxamide
Explanation:
1. The amide group is attached to a ring carbon, so the suffix is carboxamide.
2. Ring parent is cyclopentane: cyclopentanecarboxamide.
Teacher's Note:
1. Ring + CONH2 = cycloalkanecarboxamide.
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Multiple Choice Questions (MCQs) for JEE Chemistry Organic Chemistry Some Basic Principles and Techniques
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