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Chapter 16: PROBLEM 16.66 (page 639)

Question: The treatment of CH32C=CHCH2Brwith H2Oforms B (molecular formula C5H11O) as one of the products. Determine the structure of B from its 1H NMR and IR spectra.

Short Answer

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Answer

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01

Alkene

The alkene has a pi accepting tendency at which the nucleophile can attack and form a bond.

02

Spectroscopy of OH

The IR spectroscopy of OH ranges from 3200 to 3600 cm-1. The absorption signal of OH in the-H NMR is broad and appears as a singlet.

03

Explanation

The reaction of CH32C=CHCH2Brwith H2O forms an alcohol as a product:

Formation of product

The-H NMR spectroscopy of the product has five signals. The two geminal protons couple together and form a doublet of each.

The two equivalent methyls give separate singlet signals by coupling them together. The proton of OH gives singlet. The proton on the alkene gives doublet by coupling with vicinal protons.

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Most popular questions from this chapter

Question: One step in the synthesis of dodecahedrane (Section 4.11) involved reaction of the tetraene C with dimethylacetylene dicarboxylate (D) to afford two compounds having molecular formula C16H16O4 . This reaction has been called a domino Diels–Alder reaction. Identify the two products formed.

Question: Explain why the cyclopentadienide anion A gives only one signal in its 13C NMR spectrum.

Question: Draw additional resonance structures for each ion.

a.

b.

c.

d.

Question: The following reactions have been used to synthesize dieldrin and aldrin (named for Diels and Alder), two pesticides having a similar story to DDT (Section 7.4). Identify the lettered compounds in this reaction scheme.

Question: Addition of HBr to alleneCH2=C=CH2forms 2-bromoprop-1-ene rather than 3-bromoprop-1-ene, even though 3-bromoprop-1-ene is formed from an allylic carbocation. Considering the arrangement of orbitals in the allene reactant, explain this result.
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