Question

An aldol condensation reaction was performed where benzaldehyde and cyclopentanone were used. The product is 2,5 dibenzylidenecyclopentanone or just 2-benzylidenecyclopentanone. The spectra correspond with the aldehyde being benzaldehyde but the ketone used could either be acetone, cyclopentanone or cyclohexane. I need to find the NMR spectra that correspond to cyclopentanone being the ketone in the aldol condensation. Which NMR spectra correspond to the correct product?-7.76 7.73 7.61 -7.41 -7.10 7.07 -7.76 -7.73 –7.61 -7.41 -7.10 -7.07 7.75 7.70 7.65 7.60 7.55 7.50 7.45 7.40 7.35 7.30 7.25 7- 188.95 - 143.34 --134.79 130.51 128.96 128.40 -125.41 wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww ww-7.81 7.48 27.46 7.41 7.34 -2.93 -1.79 -7.81 - 7.48 7.46 -- 7.41 -7.34 -1.79 1.84 1.82 1.80 1.78 1.76 1.74 17.9 7.8 7.7 7.6 7200 190 - 190.36 180 170 160 150 140 136.92 136.19 135.98 -130.36 128.57 128.37 130 120 MhWaonewwwwwwwwwwwwwwwwwwwwwwwwwwwwww7.60 7.59 -7.44 7.38 -3.11 -- -7.60 -7.59 -7.44 -7.38 7.65 7.60 7.55 7.50 7.45 7.40 7.35 F009 13.934 1.90 4.10-1 8.0 7.8 7.6200 – 196.37 190 180 170 160 ΝύφημισματοκομικητικήΝαμίμουμινΑσημακοποιωνδρομηνίαμαλλιανείμουμένη μετα 150 140 137.30 J-135.80

I think the second set is with cyclohexanone and not cyclopentanone but I could be wrong.

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Answer #1

(1) First set : 0 H NMR = 4 7010, 7.07 7.41 7061 vallue shows their is only aromatic region region ③ IBCNMR 7073, 7.76 s presset contain acetone so this first as a Ketone acetone (2) second set: - O 0 HNMR Data! - 1079 s pentat 107g integral - This m128.37 128.52130.36,135.98,136019, 136.92 s This all value shows presence of olefins & aromatic region igo 365 shows Carbonyl237 Third set i- OH NMR Datas a 3olls singlet 4.10 integral This may due to equivdlent group such as Chith attached to electr* so y observing probable is all data following structure: - a - 30115 C 13C NMR so the third set Cyclopentanone singlet oloi

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