Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 20: 68E (page 678)
The 1H and 13C NMR spectra below belong to a compouns with formula C6H10O2. Propose a structure for this compound.
Short Answer
The structure for the compound which satisfies the spectral data has been found.
Step by step solution
Finding the peak for absorption of vinylic proton
The molecular formula for the compound is C6H10O2. On ignoring hydrogen for the parent compound, the formula will be converted into C6H14. There is 4 hydrogen in the molecule and has two unsaturation units. The carboxyl carbonyl is one and the other is a double bond due to the peak for the vinylic proton absorption.
Finalizing the compound as an isolated methyl group
The three proton singlet in spectra at 1.8 未 represents an isolated methyl group. The singlet broad signal at 12.5 未 represents the carboxylic acid compound. The 3 proton triplet at 1.05 未 represents attachment of methyl group to methylene group and observed 2 proton multiplet at 2.25 未. The observation of multiplet indicates the attachment of the methyl group.
The peak at 174.19 未 represents carboxyl carbon. The signal at 22.24 未 and 11.79 未 represents methylene and methyl group for double-bonded carbon. Two carbons in the double bond represent 126.62 未 and the other 146.71 未 represents the carbonyl group. The isolated methyl group indicates the signal at 12.89 未.
Structure for the compound
The structure for the compound can be drawn as,
Over 30 million students worldwide already upgrade their learning with 91影视!
