Final answer:
The compound C5H6O is likely to be 3-methyl-1-pentyne, characterized by IR absorption peaks at 3300 cm−1 (alkyne C≃H stretch) and 2102 cm−1 (C≃C triple bond). The NMR spectrum suggests a methyl group, and hydrogens adjacent to both a double bond and a triple bond.
Step-by-step explanation:
To identify the compound from its IR and proton NMR spectra, let's examine the given peaks. The IR spectrum shows sharp absorption at 3300 cm−1, which is indicative of a terminal alkyne group C≃H. This is further confirmed by the characteristic alkyne absorption at 2102 cm−1. The peak at 1634 cm−1 suggests a carbon-carbon double bond.
Turning to the NMR spectrum, the signal at δ 3.10 (1H, d, J = 2 Hz) may correspond to the hydrogen on a carbon adjacent to a triple bond due to the small coupling constant. The singlet at δ 3.79 (3H, s) is likely a methyl group, suggesting an O-CH3 group because it is not split by any adjacent hydrogens.
The doublet of doublets at δ 4.52 (1H, J = 6 Hz and 2 Hz) alongside the δ 6.38 (1H, d, J = 6 Hz) suggest these hydrogens are coupled to each other and one is adjacent to a double bond while the other is adjacent to a triple bond.
From these spectroscopic features, the compound is likely to be 3-methyl-1-pentyne. Its structure includes a terminal alkyne and a double bond, consistent with the observed IR absorptions, and the NMR peaks correspond to the unique hydrogens in this molecule.