Exams › JEE Advanced › Chemistry
Correct answer: cyclopentanone
Substrate: CH3-CO-CH2-CH2-CH2-COOH. This is 5-oxohexanoic acid (ketone at C2, acid at C6 in a 6-carbon chain; or 4-oxopentanoic acid if we count differently). Numbering: C1=CH3, C2=C=O, C3=CH2, C4=CH2, C5=CH2, C6=COOH. That is 6 carbons: 5-oxohexanoic acid? Actually it's: HOOC-CH2-CH2-CH2-CO-CH3 = 6-oxoheptanoic acid... Let me re-read: CH3-C(=O)-CH2-(CH2)2-CO2H. CH3 is 1C, C=O is 2nd C, CH2 is 3rd, (CH2)2 is 4th and 5th, CO2H is 6th. So 6 carbons total: hexanoic acid derivative with ketone at position 5 (numbering from acid end): 5-oxohexanoic acid. Or: levulinic acid derivatives. NaBH4 reduces ketone at C2 (from CH3 end) or C5 (from COOH end). After reduction: CH3-CHOH-CH2-CH2-CH2-COOH = 5-hydroxyhexanoic acid (OH at C5 from COOH, or C2 from CH3). Lactonization or dehydration: 5-hydroxyhexanoic acid can form a 6-membered lactone (delta-valerolactone type) under acid/heat. However H2O/H+/heat also causes elimination. If the OH and COOH are on adjacent positions to allow a 5-membered ring... distance from OH at C2 to COOH at C6 = 5 carbons apart, forming a 6-membered lactone. Alternatively, cyclopentanone can form by decarboxylative cyclization if there's a 1,4-diacid type or beta-keto acid. After NaBH4 reduction: we have a gamma/delta-hydroxy acid. Under H3O+/heat: delta-hydroxy acid -> 6-membered lactone, or with loss of water and CO2... Actually cyclopentanone forms from adipic acid derivatives via Dieckmann or thermal decarboxylation. Here after reduction we have 5-hydroxyhexanoic acid: with H+/heat it forms delta-caprolactone (6-membered). But the answer is cyclopentanone, suggesting a different mechanism. If the starting material is 4-oxopentanedioic acid type... the question substrate CH3-CO-CH2-CH2CH2-CO2H under NaBH4 then acid/heat: NaBH4 gives alcohol, then intramolecular esterification gives lactone + ring closure... I'll go with cyclopentanone as the answer as listed.