Deduce the structure of an unknown compound using the data below: Molecular Formula: C6H12O IR: 1705 cm-1 1H NMR: no absorptions
greater than δ 3 ppm 13C NMR: δ 24.4, δ 26.4, δ 44.2, and δ 212.6 ppm. Resonances at δ 44.2 and 212.6 have very low intensity.
1 answer:
Answer:
The compound elucidated from the spectral data is <u>4-methyl penta-2-none</u>
Explanation:
- <u>1700 cm-1 from IR data</u> suspects aldehyde/ketone or carboxylic acid. However,since the peak is not a stretched vibration, it implies an aldehyde or ketone present.
- <u>3 ppm H NMR</u> confirms O-CH3 bond
- <u>24.4 13C NMR</u> confirms CH3-R bond
- 26.4 13C NMR confirms H3C-(R)C(H)-R
- 44.2 13C NMR Confirms C=
- 2126 13C NMR confirms aldehyde C=O bond
<u>The deduced structure is 4-methyl penta-2-one (see attached) </u>given multiple CH3 atoms.
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Answer:
the final volume of the gas is = 1311.5 mL
Explanation:
Given that:
a sample gas has an initial volume of 61.5 mL
The workdone = 130.1 J
Pressure = 783 torr
The objective is to determine the final volume of the gas.
Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.
Converting the external pressure to atm ; we have
External Pressure :
The workdone W = V
The change in volume ΔV=
ΔV =
ΔV =
ΔV = 1.25 L
ΔV = 1250 mL
Recall that the initial volume = 61.5 mL
The change in volume V is
multiply through by (-), we have:
= 1250 mL + 61.5 mL
= 1311.5 mL
∴ the final volume of the gas is = 1311.5 mL