Please answer all the parts Compound A is an aldehyde. The 'H-NMR spectrum below is that of compound A with integrations above each peak. It hasbeen expanded to show the peaks more clearly. As a result, the peak at about 9.5 ppm is not shown.Question 13'H NMR32Determine and draw the structure of compound B.Create OscerSketch Answer 13Determine and draw the structure of compound E.Question 14Create OscerSketch Answer 14Determine and draw the structure of compound F.Question 15Create OscerSketch Answer 15Save Answers10 Submissions LeftSubmit Answers

According to our guidelines, I can only answer first question because you have posted multiple questions. If you want any specific question to be solved then please specify the question number or post only that question. Note: Also in question 14 and question 15 1H NMR of the compounds are not given. To determine the structure from 1H NMR , the number of hydrogens can be estimated from the NMR spectra. As integration of the peaks give the number of protons responsible for the intensity of the peak. Multiplicity of the peak gives the number of non equivalent protons present in the neighborhood of the proton. Multiplicity = n + 1 Where n = number of non equivalent protons.The predicted structure for compound A is given below in an image. The IUPAC name of the structure A is 2-methylbutanal .