Which isomer of $C_4{H}_9{\mathrm{Br}}_1$ would show only one peak in its $1_H$ NMR spectrum?

• A. sec-butyl bromide (${\mathrm{CH}}_3\mathrm{CH}{2}_2\mathrm{CHBr}\mathrm{CH}{3}_3$)
• B. tert-butyl bromide (${\mathrm{(CH}}_3$)$3_{}$$\mathrm{CBr}$)
• C. isobutyl bromide ($\mathrm{(CH}{3}_3$)$\mathrm{CH}{2}_2\mathrm{CH}{2}_2\mathrm{Br}$)
• D. n-butyl bromide ($\mathrm{CH}{3}_3\mathrm{CH}{2}_2\mathrm{CH}{2}_2\mathrm{CH}{2}_2\mathrm{Br}$)

Answer: (B)

Verified step by step guidance

1. Identify the molecular formula and the possible isomers of C\_4H\_9Br given: sec-butyl bromide, tert-butyl bromide, isobutyl bromide, and n-butyl bromide.
2. Recall that the number of peaks in a \( ^1H \) NMR spectrum corresponds to the number of chemically non-equivalent sets of protons (hydrogens) in the molecule.
3. Analyze the symmetry of each isomer: a molecule with high symmetry will have fewer unique proton environments, potentially leading to fewer NMR peaks.
4. Focus on tert-butyl bromide, which has a central carbon bonded to three identical methyl groups and one bromine atom, creating a highly symmetrical environment where all methyl protons are equivalent.
5. Conclude that the isomer with only one type of proton environment (and thus only one peak in the \( ^1H \) NMR) is the one with the highest symmetry, which is tert-butyl bromide.