Question

How can${}^{\mathbf{1}}\mathbf{H}$NMR be used to prove that the addition of HBr to propene follows the rule that says that the electrophile adds to the sp² carbon bonded to the most hydrogens?

Short answer

Formation of 2-bromopropane

Step-by-step solution

Step-by-Step Solution Step 1: The addition  of HBr  to propene

The addition of HBr to propene is an electrophilic substitution reaction. HBr provides an electrophile${\mathbf{H}}^{\mathbf{+}}$. This electrophile attacks the double bond to form a primary and secondary carbocation. As secondary carbocation is most stable compared to the primary carbocation, therefore, the major product formed is 2-bromopropane.

The signals  in NMR

2-bromopropane gives one doublet and one septet signal in ${}^1\mathrm{H}$NMR spectra. But if the electrophile adds to the ${\mathrm{sp}}^2$carbon bonded to the least number of hydrogens it gives three signals in the ${}^1\mathrm{H}$NMR spectrum.

Formation of 2-bromopropane