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A student obtained a H NMR spectrum of her dibenzalacetone product (see page 6) and deduced that she 7. had made the trans,t

На H. (706-7.12 ppm) Signal split by H J 16 Hz. .(7.72-7.78 ppm) Signal split by H to a doublet J 16 Hz with 7.30 7.00 7.70 7
A student obtained a 'H NMR spectrum of her dibenzalacetone product (see page 6) and deduced that she 7. had made the trans,trans-stereoisomer. Is her conclusion correct? Explain. (1.5 pts)
На H. (706-7.12 ppm) Signal split by H J 16 Hz. .(7.72-7.78 ppm) Signal split by H to a doublet J 16 Hz with 7.30 7.00 7.70 7.50 7 10 7 20 7.00 8.00 7 80 7.60
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Answer #1

yes, the conclusion is correct .because the coupling constant lies between value J=16Hz-18Hz for trans isomers. both the hydrogens Ha and Hb coupling with coupling constant J=16Hz .

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A student obtained a 'H NMR spectrum of her dibenzalacetone product (see page 6) and deduced that she 7. had made the trans,trans-stereoisomer. Is her conclusion correct? Explain. (1.5 pts)...
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