Which would have the lowest frequency the C=O bond in heptanal, in hept-2-enal, or in benzaldehyde?
Frequency in the context of bond vibrations is typically associated with the energy of the vibration. The bond vibration frequency of a carbonyl group (C=O) can be affected by various factors such as resonance, induction, conjugation, and steric strain. For the C=O bond vibration frequency to be lower, the bond must have a lower force constant, meaning the bond is less stiff or there is more electron delocalization/bond lengthening that can lower the vibrational energy.
Examining the structures you've mentioned:
1. Heptanal is an aliphatic aldehyde with a straight-chain structure and no conjugation. Its C=O bond should display a typical aldehyde carbonyl stretch without much influence from adjacent bonds.
2. Hept-2-enal has a C=C double bond at the second carbon, and this double bond can have a conjugative interaction with the C=O group. This conjugation will slightly decrease the double-bond character of the C=O bond and lower the frequency compared to a non-conjugated C=O bond due to the electron-donating effect of the alkene.
3. Benzaldehyde has a C=O group conjugated with an aromatic ring (phenyl group). The resonance in the aromatic system provides greater delocalization of electrons and a higher degree of conjugation for the C=O bond compared to heptanal or hept-2-enal. The resonance with the benzene ring will lower the C=O vibration frequency more significantly than in hept-2-enal.
As a result, we can infer that the carbonyl stretching frequency would likely be lowest in benzaldehyde due to the resonance stabilization provided by the aromatic ring. Heptanal would have the highest frequency due to lack of conjugation, and hept-2-enal would have a frequency that falls between the other two, although closer to heptanal given that the conjugative effect is less pronounced than with an aromatic ring.