Question 10 of 26beginner🔧 ApplyShort Answer4 marks

Draw the structures of the following compounds. (i) 3-Methylbutanal (ii) p-Nitropropiophenone (iii) p-Methylbenzaldehyde (iv) 4-Methylpent-3-en-2-one (v) 4-Chloropentan-2-one (vi) 3-Bromo-4-phenylpentanoic acid (vii) p,p'-Dihydroxybenzophenone (viii) Hex-2-en-4-ynoic acid

(i)
Answer

CH₃-CH(CH₃)-CH₂-CHO (3-Methylbutanal: Butanal with methyl group at C-3)

(ii)
Answer

O₂N-C₆H₄-CO-CH₂-CH₃ (p-Nitropropiophenone: Propiophenone with NO₂ at para position)

(iii)
Answer

CH₃-C₆H₄-CHO (p-Methylbenzaldehyde: Benzaldehyde with CH₃ at para position)

(iv)
Answer

CH₃-CO-C(CH₃)=CH-CH₃ (4-Methylpent-3-en-2-one: Pent-3-en-2-one with methyl at C-4)

(v)
Answer

CH₃-CO-CH₂-CH(Cl)-CH₃ (4-Chloropentan-2-one: Pentan-2-one with Cl at C-4)

(vi)
Answer

CH₃-CH(Ph)-CH(Br)-CH₂-COOH (3-Bromo-4-phenylpentanoic acid)

(vii)
Answer

HO-C₆H₄-CO-C₆H₄-OH (p,p'-Dihydroxybenzophenone: Benzophenone with OH at para positions on both rings)

(viii)
Answer

HOOC-CH=CH-C≡C-CH₃ (Hex-2-en-4-ynoic acid: 6-carbon chain with double bond at C-2 and triple bond at C-4)

Explanation

This question tests IUPAC nomenclature understanding for aldehydes, ketones, and carboxylic acids. Students must identify the parent chain, functional groups, and substituents with their correct positions. For aromatic compounds (ii, iii, vii), para substitution means substituents are opposite each other on the benzene ring. For compounds with multiple functional groups (viii), both double and triple bonds must be positioned correctly. The numbering always starts from the functional group (aldehyde/carboxylic acid) or gives the lowest number to the carbonyl carbon (ketone).

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