(3)(1 pt) If acid/base reactions are as fast as everyone says they are, why didyou have to heat under reflux in the deprotonation step?(4)(2 pts) Label the 'H-NMR provided below to assign the different signals tothe matching protons. (The numbers above the signals are theintegrations). Use letters to label the different signals/protons. The H onthe alcohol is not present in the spectra.3.0باليō (ppm)2.0-OH1.00.0

Question 3: Acid-base reactions, especially those involving deprotonation, are indeed generally very…

Answered: (3) (1 pt) If acid/base reactions are…

Organic Chemistry

9th Edition

ISBN:9781305080485

Author:John E. McMurry

Publisher:John E. McMurry

Chapter20: Carboxylic Acids And Nitriles

Section20.SE: Something Extra

Problem 67AP

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On the NMR provided for the product, what protons on the product structure correspond to signals A-C? You do not need to assign which specific H's are in each signal. On the NMR for the product, what protons on the product structure correspond to signal D? On the NMR for the product, what protons on the product structure correspond to signal E?
Please give clear handwritten ans
Hc Ha oHe Hb -Hd i) rank the labelled protons in order of increasing pka. ") Explain your answer in Ci).
Which one of the following compounds (from Figure #7) produced the 'H NMR spectrum (from Figure #8) shown below? Figure #7 inte compound A compound B compound C compound D Please click here if image does not display. Figure #8 3 203938 PPM 10 9 8 7 5 4 3 2 1 8 (ppm) Please click here if image does not display. O compound A O compound D O compound C O compound B
For the 1 H NMR spectrum of acetyl salicylic acid shown below, assign each signal to the proton to which it corresponds (the COOH proton does not appear on this spectrum).
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Deduce the structure of an unknown compound using the data. C5H₁O: 10 NMR: 69.8 (1 H, s), 8 1.1 (9 H, s) Deduce the structure of the unknown compound. Select /||||||| G C Draw Rings More Erase Q2 Q
draw the structure of the nmr (ethyl acetate) then circle the protons on the structures and correlate the 1H's to the resonances (peaks) on the 1H NMR spectra.
When comparing 1H NMR spectra of benzene and methylbenzene (toluene), how to determine which spectra belongs to which compound? O Signals on 1H NMR spectra of toluene will be near 11- 12 ppm., while for benzene these signals will be near 6.5-8.0 ppm: O Both spectra will be the same O IH NMR spectra of toluene will be more intense compare to benzene one O 1H NMR spectra of benzene will have just one signal, while the one of toluene will have more than one
Bb.84.
look at this reaction (first image) If Product A is formed in this reaction, we expect to see a signal of the hydrogen atom of the alcohol functional group in the 1H NMR spectrum... (second image) Explain how this HSQC spectrum establishes that the signal of the hydrogen atom with the chemical shift of 2.22 ppm is the hydrogen atom of the alcohol functional group!
Methyl ethyl ether (shown below) has an -OCH 2 - group, an -OCH 3 - group, and a methyl group. The ' H NMR spectrum displays a quartet at 3.5 ppm, a signal at 3.0 ppm, and a signal at 1.2 ppm. Using the data in the table, what is the correct assignment of the spectrum? H H. H. Proton Environment Chemical Shift (8,ppm) Methylene, -C H 2 - 0.3 - 1.2 Methyl, -C H 3 1.1 - 1.9 Alkoxy, -OC H 2/3 3.3 - 4.0 -OC H 3 = 3.5 ppm (quartet), -OC H 2 = 3.0 ppm (singlet), -C H 3 = 1.2 ppm (triplet) -OC H 3 = 1.2 ppm (singlet), -OCH 2 3.5 ppm (quartet), -C H 3 = 3.0 ppm (triplet) -OC H 3 = 3.0 ppm (singlet), -OCH 2= 3.5 ppm (quartet), -C H 3 = 1.2 ppm (triplet) -OC H 3 = 3.5 ppm (quartet), -OC H 2 1.2 ppm (triplet), -C H 3 = 3.0 ppm (quartet) -OC H 3 = 1.2 ppm (singlet), -OCH 2= 3.0 ppm (doublet), -C H 3 = 3.5 ppm (quartet) -OC H 3 = 3.0 ppm (triplet), -OC H 2 = 3.5 ppm (quartet), -C H 3 = 1.2 ppm (triplet)

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