Identify the following compound from its IR and proton NMR spectra.

C6H10O:

NMR: 3.31 (3H, s); 2.41 (1H, s); 1.43 (6H, s)
IR: 2110, 3300 cm^-1 (sharp)

To identify the compound from its IR and proton NMR spectra, we can analyze the peaks and patterns in both spectra.

First, let's examine the proton NMR spectrum:

- The peak at 3.31 ppm corresponds to a singlet (s) signal from three hydrogen atoms (3H) in the molecule.
- The peak at 2.41 ppm is another singlet signal (s) from one hydrogen atom (1H).
- The peak at 1.43 ppm is a singlet signal (s) from six hydrogen atoms (6H).

Now, let's analyze the IR spectrum:

- There are two major peaks at 2110 cm^-1 and 3300 cm^-1. The sharp peak at 2110 cm^-1 suggests the presence of a carbon-carbon triple bond (C≡C), which is common in alkynes. The broad peak at 3300 cm^-1 suggests the presence of an alcohol or amine functional group (-OH or -NH).

Based on the proton NMR spectrum, we can determine the number of hydrogen atoms in the molecule. From the given information, the total number of hydrogen atoms is 3 + 1 + 6 = 10.

From the IR spectrum, we can deduce the presence of a carbon-carbon triple bond (C≡C) and a functional group with either an alcohol (-OH) or an amine (-NH).

Putting this information together, the compound C6H10O that fits the given spectra is 2-hexyne (2-hexyn-1-ol), which has a triple bond (C≡C) and an alcohol functional group (-OH).

To identify the compound from its IR and proton NMR spectra, we need to analyze each spectrum and look for characteristic peaks and signals.

Starting with the proton NMR spectrum:

1. The signal at 3.31 ppm indicates a singlet (s) and represents three hydrogen atoms. This suggests the presence of a methyl group (-CH3).

2. The signal at 2.41 ppm is another singlet (s) and corresponds to one hydrogen atom. This signal suggests the presence of a methylene group (-CH2-).

3. The signal at 1.43 ppm is a singlet (s) and represents six hydrogen atoms. This indicates the presence of a different methyl group (-CH3).

Overall, we have a total of 10 hydrogen atoms.

Now, let's consider the IR spectrum:

1. The sharp absorption at 3300 cm^-1 is indicative of an -OH group (hydroxyl group). This suggests the presence of an alcohol functional group.

2. The absorption at 2110 cm^-1 indicates the presence of a triple bond (C≡C). This suggests the presence of an alkyne functional group.

Based on the information obtained from both spectra, we can conclude that the compound consists of two methyl groups, one methylene group, an alcohol group, and a triple bond. By considering the molecular formula C6H10O, the compound that fits this description is 2-methyl-3-pentyne, which has the structure CH3-C≡C-CH2-CH(CH3)2.

diethyl ether