Preparation of 2-butanone

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Nor Amirah Farhana Nawawi, Mia Organic Chemistry Lab Report 2 Jessica Sammons TITLE: Preparation of 2-butanone INTRODUCTION: The goal of this experiment was to prepare 2-butanone from 2-butanol. Chromic acid was used in this experiment to in order to prepare 2-butanol. Cr (VI) is rather 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. PROCEDURE: The experiment followed the instructions in the lab manual, except for the empty heating mantle, which our TA advised us to use with sand instead of heating it empty. RESULTS AND CALCULATIONS: Weight of…show more content…
There 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 longer there. The product is therefore almost pure without unreacted reactant. NMR analysis In a 2-butanone, there are three 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 left because it has nothing to do with the compound, and so now we are left with three peaks. The first peak, (we 'll start from the right) with chemical shift reading around has a ratio of three protons. It is also a triplet. 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 other 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. These criteria matches C-1 because it has no vicinal carbon (its vicinal

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