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 given molecules are SO2 and BrF5.
Explanation:
Consider the molecule SO2:
The central atom is S.
The number of domains on S in this molecule is three.
Domain geometry is trigonal planar.
But there is a lone pair on the central atom.
So, according to VSEPR theory,
the molecular geometry becomes bent or V-shape.
Hybridization on the central atom is
.
Consider the molecule BrF5:
The central atom is Br.
The number of domains on the central atom is six.
Domain geometry is octahedral.
But the central atom has a lone pair of electrons.
So, the molecular geometry becomes square pyramidal.
The hybridization of the central atom is .
The shapes of SO2 and BrF5 are shown below:
