Compound A has molecular formula C8H8O. An IR spectrum of | StudySoup

Textbook Solutions for Organic Chemistry

Chapter 19 Problem 19.74

Question

Compound A has molecular formula C8H8O. An IR spectrum of compound A exhibits a signal at 1680 cm-1 . The 1 H NMR spectrum of compound A exhibits a group of signals between 7.5 and 8 ppm (with a combined integration of 5) and one upfield signal with an integration of 3. Compound A is converted into compound B under the following conditions: Compound A Compound B Zn(Hg), HCl, heat When compound B is treated with Br2 and AlBr3, two different monobromination products are obtained. Identify both of these products and predict which one will be the major product. Explain your reasoning.

Solution

Step 1 of 4

First, note that the molecular formula indicates a hydrogen deficiency index of 5 . Given that we only have eight total carbons in the molecule, it's almost a certainty that we have a benzene ring present. Recall that benzene accounts for four elements of unsaturation ( 3 double bonds plus the ring). That leaves one double bond somewhere in the structure of the compound A.

The IR signal at \(1680 \mathrm{~cm}^{-1}\), combined with the presence of oxygen in the molecular formula, indicates a ketone that lies in conjugation with a benzene ring. Normally we would expect a ketone to exhibit its IR signal at a higher wavenumber, but conjugation with the benzene ring reduces the strength of the carbonyl bond and shifts the IR signal to its present location.

 

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Title Organic Chemistry 2 
Author David R. Klein
ISBN 9781118454312

Compound A has molecular formula C8H8O. An IR spectrum of

Chapter 19 textbook questions

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