Structure Elucidation of a C5H10O Liquid
A colorless liquid of molecular formula C5H10O (M+ = 86) is submitted for structure determination. Combustion of 0.215 g gives 0.540 g CO2 and 0.221 g H2O. The IR spectrum has a strong sharp band at 1716 cm−1 but no broad absorption near 3300 cm−1 and no pair of bands near 2720 and 2820 cm−1. The 1H NMR spectrum shows 0.92 ppm (triplet, 3H), 1.62 ppm (sextet, 2H), 2.13 ppm (singlet, 3H), and 2.42 ppm (triplet, 2H). The 13C NMR spectrum has five signals, including one at 208 ppm. A classmate proposes 3-pentanone because it also has a ketone carbonyl; another proposes pentanal. The student must use the coupling patterns and signal count to choose among the isomers and then predict the outcome of sodium borohydride reduction and a Tollens test.
1. Determine the structure of the unknown and explain how the 13C signal count rules out 3-pentanone.
Read the worked answer
Combustion gives C:H ≈ 5:10 and M+ 86 gives C5H10O. Degrees of unsaturation = (2×5 + 2 − 10)/2 = 1, consistent with one carbonyl. IR 1716 cm−1 is a ketone carbonyl; no 3300 cm−1 OH and no 2720/2820 cm−1 aldehyde C–H. 1H NMR: 0.92 ppm triplet 3H and 1.62 ppm sextet 2H indicate a terminal CH3CH2– unit; 2.42 ppm triplet 2H is a CH2 next to carbonyl; 2.13 ppm singlet 3H is a methyl ketone CH3CO–. That gives CH3COCH2CH2CH3, 2-pentanone. 13C has five signals: carbonyl, CH3CO, CH2 alpha, CH2 beta, CH3. 3-Pentanone, CH3CH2COCH2CH3, has a plane of symmetry and only three 13C signals, and its 1H NMR would show a 6H triplet and 4H quartet, not a 3H singlet. Pentanal would show an aldehyde proton near 9–10 ppm and two IR C–H bands near 2720 and 2820 cm−1.
2. Predict the products of NaBH4 reduction of the unknown and explain whether the product is optically active.
Read the worked answer
NaBH4 reduces the ketone to a secondary alcohol: 2-pentanone → 2-pentanol. The carbonyl carbon becomes a stereocenter (C2 attached to OH, H, CH3, and propyl). NaBH4 hydride attack occurs from either face of the planar carbonyl with similar probability, so the product is a racemic mixture of (R)- and (S)-2-pentanol. A racemate is not optically active overall because the two enantiomers rotate plane-polarized light equally and oppositely. If a chiral hydride reagent or enzyme were used, one face could be favored and the product could be optically active.
3. What would the Tollens test show for the unknown, and how would the result differ for pentanal? Explain the chemistry.
Read the worked answer
The unknown is a ketone, so Tollens reagent (Ag(NH3)2+ in aqueous ammonia) gives no silver mirror; ketones are not oxidized by the mild Tollens reagent. Pentanal is an aldehyde and would give a positive test: the aldehyde is oxidized to pentanoate while Ag+ is reduced to metallic silver, forming a mirror. The reaction requires the carbonyl carbon to retain a hydrogen that can be removed during oxidation; in 2-pentanone the carbonyl carbon has two carbon substituents and no aldehydic H, so oxidation is not favorable. This distinguishes the ketone from the aldehyde.