Question

Question: The treatment of ${\left({\mathrm{CH}}_3\right)}_2\mathrm{C}={\mathrm{CHCH}}_2\mathrm{Br}$with ${\mathrm{H}}_2\mathrm{O}$forms B (molecular formula ${\mathrm{C}}_5{\mathrm{H}}_{11}\mathrm{O}$) as one of the products. Determine the structure of B from its ¹H NMR and IR spectra.

Short answer

Answer

Step-by-step solution

Alkene

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

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.

Explanation

The reaction of ${\left({\mathrm{CH}}_3\right)}_2\mathrm{C}={\mathrm{CHCH}}_2\mathrm{Br}$with 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.