Identify the following compound from its IR and proton NMR spectra. C5H6O: IR: 3300 (sharp), 2102, 1634 cm–1 NMR: δ 3.10 (1H, d, J = 2 Hz); δ 3.79 (3H, s); δ 4.52 (1H, doublet of doublets, J = 6 Hz and 2 Hz); δ 6.38 (1H, d, J = 6 Hz)

To identify the compound based on its IR and proton NMR spectra, we need to analyze the given data and interpret the information provided.

First, let's analyze the IR spectrum. IR spectroscopy provides information about the functional groups present in a compound. The significant peaks in the IR spectrum are:

- 3300 cm–1 (sharp): This peak indicates the presence of a hydroxyl group (-OH). Hydroxyl groups are characteristic of alcohols and carboxylic acids.
- 2102 cm–1: This peak is present in the carbonyl region, which suggests the presence of a carbonyl group. The location of this peak is closer to the region of alkynes (C≡C) rather than aldehydes/ketones (C=O).
- 1634 cm–1: This peak indicates the presence of a carbonyl group, typically associated with aldehydes or ketones.

Based on the information from the IR spectrum, we can conclude that the compound contains an aldehyde or ketone functional group and may also have a hydroxyl group.

Now, let's analyze the proton NMR spectrum. Proton NMR spectroscopy provides information about the chemical environment of the hydrogen atoms in the compound. The significant peaks in the proton NMR spectrum are:

- δ 3.10 (1H, d, J = 2 Hz): This peak represents a proton that is split into a doublet (d) due to coupling with neighboring protons. The presence of coupling suggests the presence of adjacent protons or a CH group.
- δ 3.79 (3H, s): This peak represents a singlet (s) and indicates the presence of three chemically equivalent protons (3H) in a distinct environment. This signal is commonly associated with a methyl group (-CH3).
- δ 4.52 (1H, doublet of doublets, J = 6 Hz and 2 Hz): This peak indicates a proton that is split by two different coupling constants. The doublet of doublets pattern indicates the presence of adjacent protons with different coupling constants. This signal is commonly seen with a proton (H) coupling with protons on a neighboring carbon (CH2).
- δ 6.38 (1H, d, J = 6 Hz): This peak represents a proton that is split into a doublet (d) due to coupling with a proton in a distinct environment. This pattern indicates that this proton is next to a proton in a different chemical environment.

Based on the information from the proton NMR spectrum, we can conclude that the compound contains an aldehyde or ketone group, a methyl group (-CH3), and a CH2 group.

Combining the information from both the IR and proton NMR spectra, the compound with the molecular formula C5H6O is likely to be 2-methyl-3-pentyn-2-ol.