a student performed an experiment, using a cocktail peanut, before it was burned the peanut half weighed .353 g. After burning t
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I have attached an image of the IR spectrum required to answer this question.
Looking at the IR, we can look for any clear major stretches that stand out. Immediately, looking at the spectrum, we see an intense stretch at around 1700 cm⁻¹. A stretch at this frequency is due to the C=O stretch of a carbonyl. Therefore, we know our answer must contain a carbonyl, so it could still be a ketone, aldehyde, carboxylic, ester, acid chloride or amide. However, if we look in the 3000 range of the spectrum, we see some unique pair of peaks at 2900 and 2700. These two peaks are characteristic of the sp² C-H stretch of the aldehyde.
Therefore, we can already conclude that this spectrum is due to an aldehyde based on the carbonyl stretch and the accompanying sp² C-H stretch.
Looking at the IR, we can look for any clear major stretches that stand out. Immediately, looking at the spectrum, we see an intense stretch at around 1700 cm⁻¹. A stretch at this frequency is due to the C=O stretch of a carbonyl. Therefore, we know our answer must contain a carbonyl, so it could still be a ketone, aldehyde, carboxylic, ester, acid chloride or amide. However, if we look in the 3000 range of the spectrum, we see some unique pair of peaks at 2900 and 2700. These two peaks are characteristic of the sp² C-H stretch of the aldehyde.
Therefore, we can already conclude that this spectrum is due to an aldehyde based on the carbonyl stretch and the accompanying sp² C-H stretch.