Number of lines in the ${ }^{19} \mathrm{F}$ NMR spectrum of $\mathrm{F}_{2} \mathrm{C}(\mathrm{Br})-\mathrm{C}(\mathrm{Br}) \mathrm{Cl}_{2}$, at $-120^{\circ} \mathrm{C}$ assuming it a mixture of static conformations given below, are 5
(a)one
(b)two
(c)four
(d)five
Answer
Answer: D ✓ checked by 4AB · confidence medium
Explanation
Static rotamers: in the conformer with Br anti to Br the two F are mirror-equivalent (one line); in the two gauche conformers (enantiomers, identical spectra) the F atoms are diastereotopic and give an AB quartet (four lines) — five lines in all (d).