Question
Use the following spectra to determine the compound. Label the IR. The H-NMR has 5 peaks and the C-NMR has 4 peaks. Molecular ion peak (weight) is 100. No M+2 present.

100 LCO SD- 緋ouram,# 1180 4S00 3000 e001 1000 spo
media%2Fbbb%2Fbbb4072d-4d03-47bb-b46e-b9
media%2F527%2F527bff3a-792b-4446-bb7c-db
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Answer #1

Interpretation of spectrum

IR

2850-2950 cm-1 for aliphatic C-H stretch

1620 cm-1 for C=O stretch

2770 and 2850 cm-1 for aldehydic stretching

1H NMR

H1 : 0.9 ppm (triplet) for CH3 next to CH2

H2 : 1.6 ppm (doublet of pentet) for CH2 protons between CH3 and CH

H3 : 2.1 ppm (pentet) for CH between two CH2's

H4 : 9.6 ppm (singlet) for aldehydic proton

13C

11.44 ppm for CH3 carbon

21.45 for CH2 carbon

54.95 ppm for CH carbon

205.64 ppm for C=O carbon

Final structure of compound,

H, H, H2H, H1 H1 /

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