Question

I need help with the lab’s question below and H-NMR ASAP. Please explain and show all the work. I have posted lab guidelines as well for the reference.

.nl H20令 9:19 AM X Chemistry 3342 Laboratory Manual Background (Week Two) Equal molar amounts of acid (Y) and alcohol (Y) starting material will be used in the second acid-catalyzed esterification reaction we will complete. When the reaction takes place one molecule or ester and one molecule of water will form, ie., the amount of ester formed is equal to the amount of water formed in the reaction. If (X) molecules of acid (and alcohol) consumed, then the amount of ester (and water) s (X), and the amount of acid and alcohol left over is (Y)- (X): +H,O Start 0 Fquilibrium Y-X Y-X Thus, if we are able to calculate a mole ratio of ester to alcohol we can calculate the equilibrium constant for the eaction Our experimental procedure (i.e., extractions) l remove the carboxylic acid and water from the reaction leaving a mixture of ester and alcohol. The mole ratio of ester to alcohol that can be relatively quantified using 'H-NMR. Procedure (Week Two) Place 3.0 mL of propionic acid in a clean/dry 50 mL pear-shaped flask containing a stir bar. Add two drops of concentrated H2SO4 to the 50 mL pear-shaped flask. Finally, add 3.7 mL of isobutyl alcohol to the pear shaped flask. Equip the flask with a( IN - Open With Print

Answer 1

Answer:

Reaction scheme [mixture of]

NMR Assignment:

Proton (d)ppm                 Integration calc.               Identity of proton(s)                                      Comments         

Ha:         3.4                          1.00/2 = 0.50                      CH2 of ester part, doublet

Hb:         1.5                          0.503/1 = 0.50                    CH of ester part multiplet

Hc:          0.5                          3.76/                                      CH3 of ester part of molecule                     Refer Hf calculation

                                                                                                And alcohol parts

                                                                                                Alcohol correspond 0.147*6= 0.882

                                                                                                Ester + Total – alcohol

                                                                                                3.76-0.882=2.78

                                                                                                2.78/6= 0.46 (close to 0.50)

Hd:         1.9                          1.00/2 =0.50                       CH2 of acid part quartet

He:         0.75                        1.540/3= 0.51                     CH3 of acid part, triplet

Hf:          2.8                          0.294/2 =0.147                  CH2 of unreacted alcohol

Hg:         1.3                          0.150/1=0.15                      CH of unreacted alcohol

Equilibrium:

([Product]/[Unreacted])²             =             ([Ester] /[Alcohol])²        ([0.50]/[0.147])²

=1.913* 10-3