what difference would you expect to see when comparing the IR spectra of 1-butanol with 1-bromobutane and 2-butanol with 2-bromobutane?
The correct answer and explanation is:
Correct Answer:
When comparing the IR spectra of 1-butanol with 1-bromobutane and 2-butanol with 2-bromobutane, the key difference is the presence or absence of the broad O–H stretch around 3200–3600 cm⁻¹. The alcohols (1-butanol and 2-butanol) will show this broad O–H stretching absorption, while the alkyl halides (1-bromobutane and 2-bromobutane) will lack it. Additionally, alkyl halides show a characteristic C–Br stretching vibration around 500–600 cm⁻¹, which will be absent in the alcohols.
Explanation:
Infrared (IR) spectroscopy is a technique that measures how molecules absorb infrared light, revealing specific functional groups based on characteristic vibrations. In this case, we are comparing alcohols and their corresponding alkyl bromide derivatives.
1-butanol and 2-butanol are both alcohols and thus contain hydroxyl (O–H) functional groups. The IR spectra of both will display a broad and strong absorption band between 3200 and 3600 cm⁻¹. This broad peak results from hydrogen bonding in the O–H group and is a clear indicator of the alcohol functional group.
On the other hand, 1-bromobutane and 2-bromobutane are alkyl halides and do not have O–H bonds. Therefore, their IR spectra will not show the broad O–H stretch. Instead, they will feature a C–Br bond stretch, usually observed in the range of 500 to 600 cm⁻¹. This absorption is less intense and occurs at a lower frequency, due to the heavier bromine atom.
Both alcohols and alkyl halides share common alkane C–H stretches between 2850 and 2960 cm⁻¹ and C–C stretches, which makes the O–H and C–Br bands the distinguishing features.
In conclusion, the main difference between the IR spectra of butanols and bromobutanes lies in the presence of the O–H stretch in alcohols and the C–Br stretch in alkyl halides. These differences allow for clear identification of functional groups in each compound.