Tutorial 4 - 1 H NMR BLOW UP 13 C NMR (ignore peak at 77...

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Chemistry 2OA3 Nov. 9, 2010 Tutorial 4 1. (a) This is a tricky question, but makes use of all the reactions you have learned with alkenes (hydration, hydroboration, elimination, addition, etc.). The big problem is to find a series of reactions that will allow one alcohol isomer to be converted into another. You will need to use alkene intermediates to get there. For each chiral molecule, assign R or S to any chiral centres. NOTE: there may be more than one answer. For each reaction, you only need to draw one isomer. However, indicate if the reaction will produce more than one product. H 2 SO 4 /H 2 O Dilute H 2 SO 4 /H 2 O Dilute H 2 SO 4 , conc hot conc
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(b) Now do it all over again in the reverse order. 2. What is the structure of the compound of formula C 7 H 7 OBr, that has the following spectra?
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Unformatted text preview: 1 H NMR BLOW UP 13 C NMR (ignore peak at 77 ppm: due to CDCl 3 ) 2H 2H 3H , THF 3) Aromatic compound A (formula C 8 H 8 ) is treated with bromine to give a compound B (formula C 8 H 8 Br 2 ). After treatment with excess NaNH 2 and heat, compound C was prepared, which contains and alkyne group (molecular formula C 8 H 6 ). Treatment with MeLi then MeI gave D (C 9 H 8 ). Reduction with H 2 in the presence of Lindlaars catalyst gave product E. The spectra for an isomer of E (molecular formula C 9 H 10 ), compound F are shown below. i) show the sequence of products from Compound A to E ii) explain how the spectra confirms the structure of F. What would you expect the 1 H NMR of E to look like. 13 C NMR 1 H NMR 5H 1H 2H 2H IR...
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This note was uploaded on 05/16/2011 for the course CHEM BIO 20a3 taught by Professor Stover during the Fall '10 term at McMaster University.

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Tutorial 4 - 1 H NMR BLOW UP 13 C NMR (ignore peak at 77...

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