The goal of this experiment was to prepargon 2-butanone from 2-butanol. Chromic venomous was used in this experiment to in regularise to prepare 2-butanol. Cr (VI) is kind of orange, but Cr (III) is dark green ? therefore by oxidizing the alcohol (2-butanol), an orange Cr (IV) is reduced to green. NMR and IR tests were taken to determine the result, and the crystallized derivative of this product was obtained.
mathematical operation: The experiment followed the instructions in the lab manual, except for the drop heating mantle, which our TA advised us to use with sandpaper instead of heating it empty.
QuestionsQ1)IR analysisWhen analyzing the IR spectra, these are few that were observed. at that place is a peak around 1700 cm-1, indicating the carbonyl group, C = O bond (1600 ? 1800 cm-1). This proved that the product had turned to carbonyl on stead of the alcohol, O ? H bond that it used to. The O ? H bond, which would be very easy to recognize in an IR because of its broad stretch around 3200 ? 3500 cm-1, is no semipermanent there. The product is therefore almost pure without unreacted reactant.
NMR analysisIn a 2-butanone, there are terce protons at C-1, none at C-2 (the C=O bond is here), two at C-3, and three at C-4. All the protons / hydrogen give different signals, or have different chemical shifts, because there is no symmetry in this molecule. Referring to the attachment of the H NMR result, there are four significant peaks ? ignore the first on the leave because it has nothing to do with the compound, and so now we are remaining with three peaks. The first peak, (we?ll start from the sound) with chemical shift rendition around ~∂ 1.0, has a ratio of three protons. It is also a ternion. Being to the upmost upfield means this carbon must be the furthest from the carbonyl group. This entire characteristic matches C-4. The triplet shown indicates there are two different protons in the vicinal carbon, which is C-3 (no, there?s no C-5.) Remember: triplet = 3 = n + 1. Therefore n = 2.
The second peak from the right shows a singlet with three protons, with chemical shift ~∂ 2.0. These criteria matches C-1 because it has no vicinal carbon (its vicinal carbon is the carbonyl, C=O) as C-2 already has four bonds; so, singlet it is. Finally, the third peak, which is the one with a foursome, has two protons, and being the most down - ~∂ 2.3. This must be the exclusively carbon left, C-3. And yes, C-3 has two hydrogens, the peak breaks into a quartet to mirror the three protons on the vicinal carbon, which is C-4 that we had just discussed above. The new(prenominal) vicinal carbon is C-2, and it gives no signal (of course).
It is the most downfield because the proton in C-2 is most deshielded because of the electron-withdrawing carbonyl, and is also attached to another carbon. The other protons on C-1 is less deshielded as it only attached to the carbonyl group, and C-4 is the least deshielded as it attached to the alkane without attaching to an electron-withdrawing group.
DerivativeThe derivative was do because 2-butanone is present in liquid form under get on temperature. So, 2,4-dinitrophenylhydrazone was made to determine that product earlier was rightfully 2-butanone. The veridical melting point of 2,4-dinitrophenylhydrazone is 117˚C, and the experimental melting point was 90˚C - 96˚C, which is very close to the actual. This shows that the reaction really makes 2-butanone.
Q2)The oxidation with chromate is not general for all alcohols, as only secondary and primary alcohol would be oxidized this musical mode ? not a tertiary alcohol.
Q3)If 1-butanol were oxidized, the product that would be produced is butanoic acid; a carboxylic acid.
BibliographyBook title: Organic Chemistry seventh editionWriter: Francis A. CareyPublisher: Mc Graw Hillwebsite: wikipedia
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