The $^{31}\mathrm{P}(^{1}\mathrm{H})$ NMR spectrum of 2,2,6,6-$\mathrm{N_4P_4Cl_4(NMe_2)_4}$ is expected to show
(a)two triplets
(b)two doublets
(c)one doublet and one triplet
(d)one quartet and one doublet
Answer
Answer: A ✓ checked by 4AB · confidence high
Explanation
2,2,6,6-substitution gives two PCl₂ and two P(NMe₂)₂ phosphorus atoms alternating around the ring; each P couples to two equivalent neighbours of the other kind, so each type appears as a triplet: two triplets.