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Chapter 12: Q16E (page 354)

How would you prepare 1-phenyl-2-butanone, C6H5CH2COCH2CH3, from benzyl bromide, C6H5CH₂Br? More than one step is required?

Short Answer

Expert verified

Thepreparation of 1-phenyl-2-butanone, C6H5CH2COCH2CH3, from benzyl bromide, C6H5CH₂Br can be explained.

Step by step solution

01

Formation of bonds

The metal Mg reacts with the benzyl bromide for the formation of benzyl magnesium bromide. There is a cleavage of C-Br bond and formation of C-Mg and Mg-Br bonds.

02

1-phenyl-2-butanone formation

The reaction of propanol with benzyl magnesium bromide results in the acidification for the formation of 1-phenylbutan-2-ol.

The double bond between C and O of propanol can be converted to the single bond of C-O and O-H for the formation of O-H and C-C bonds.

The oxidation of secondary alcohol to ketone.

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Most popular questions from this chapter

What functional groups might the following molecules contain?

(a) A compound with a strong absorption at 1710 cm-1

(b) A compound with a strong absorption at 1540 cm-1

(c) A compound with strong absorptions at 1720 cm-1 and 2500 to 3100cm-1

The infrared spectrum of the compound with the mass spectrum shown below has a medium-intensity peak at about 1650cm-1. There is also a

C-H out-of-plane bending peak near 880cm-1 . Propose a structure consistent with the data.

Question: Nitriles,R-C≡N, undergo a hydrolysis reaction when heated withaqueous acid. What is the structure of the compound produced by hydrolysis of propanenitrile, CH3CH2C≡N, if it has IR absorptions from 2500-3100cm-1and at 1710cm-1, and has M+=74?

Camphor, a saturated monoketone from the Asian camphor tree, is used among other things as a moth repellent and as a constituent of embalmingfluid. If camphor has M+=152.1201 by high-resolution mass spectrometry,what is its molecular formula? How many rings does camphor have?

Question: The nitrogen ruleof mass spectrometry says that a compound containing an odd number of nitrogen’s has an odd-numbered molecular ion.

Conversely, a compound containing an even number of nitrogen’s has

an even-numbered M+peak. Explain.
See all solutions

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