Compound A, with molecular formula $\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}$, shows two signals in its $^{13} \mathrm{C}$ NMR spectrum. Compound $\mathrm{B}$, an isomer of compound A, shows four signals, and in the proton-coupled mode, the signal farthest downfield is a doublet. Identify compounds $A$ and $B$.