Question 1

Why does the carbonyl stretch appear at different frequencies in aldehydes versus ketones?

Accepted Answer

Aldehydes show a strong carbonyl stretch around 1725 cm⁻¹, while ketones absorb at lower frequencies. This difference reflects the distinct electronic environments of the C=O bond in each functional group. The C=O bond vibration requires different energy levels depending on the substituents attached to the carbonyl carbon.

Question 2

What IR peaks distinguish an aldehyde from other carbonyl compounds?

Accepted Answer

Aldehydes display two characteristic weak bands at 2700–2800 cm⁻¹ from the aldehydic C–H bond stretch, in addition to the carbonyl absorption. This combination of the C=O stretching and the aldehydic C–H peaks is diagnostic for identifying aldehyde functional groups in IR spectroscopy.

Question 3

How does conjugation affect the IR absorption frequency of a carbonyl group?

Accepted Answer

Conjugating the carbonyl group with a double bond or aromatic ring delocalizes π electrons, inducing partial single bond character. This reduces the bond's stretching energy, causing absorption at lower frequencies than nonconjugated carbonyls. The effect is observable as a shift to lower wavenumbers in the IR spectrum.

Question 4

Why do smaller cyclic ketones show higher carbonyl stretching frequencies?

Accepted Answer

Decreasing ring size increases ring strain in cyclic ketones, which raises the carbonyl absorption frequency. Cyclopropanone, the smallest and most strained cyclic ketone, exhibits the highest stretching frequency. Larger rings experience lower strain and therefore absorb at lower frequencies than smaller rings.

Question 5

What are the two main electronic transitions observed in aldehydes and ketones using UV-Vis spectroscopy?

Accepted Answer

Aldehydes and ketones exhibit n-π* and π-π* transitions. The π-π* transition is stronger but occurs below 200 nm, outside the observable range of typical UV spectrometers. The n-π* transition is weaker because the nonbonding orbital of oxygen and the π* antibonding orbital are perpendicular, making this a forbidden transition.

Question 6

How does conjugation shift UV-Vis absorption wavelengths in aldehydes and ketones?

Accepted Answer

Conjugation causes the π-π* absorption band to shift to higher wavelengths within the observable range of UV spectrometers. Each conjugated double bond adds approximately 30 nm to the absorption wavelength. This allows detection of conjugated carbonyl compounds that would otherwise absorb below 200 nm.

Question 7

Why is the n-π* transition considered a forbidden transition in carbonyl compounds?

Accepted Answer

The n-π* transition is forbidden because the nonbonding orbital on oxygen and the anti-bonding π* orbital on the C=O bond are perpendicular to each other. This perpendicular arrangement prevents orbital overlap, making the transition much less probable and resulting in weak absorption intensity.

醛和酮的红外和紫外-可见光谱

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