I often ask students: why is the residual proton resonance of benzene-d6 so much broader than that of chloroform-d? It is surprising how few can provide an answer. The figure below shows the signals from both solvents on the same scale. The answer is below.
The proton signal for the 99.6% D benzene-d6 is due to C6D5H. The lone proton on the benzene ring is coupled to the deuterons on the ring. If you look closely at the resonance, you can see partially resolved coupling as bumps on the side of the signal and a distinctly non-Lorentzian shape. If the coupling of the proton to the meta and para deuterons can be neglected, we would expect the resonance to be a 1:2:3:2:1 quintet due to coupling to the ortho deuterons.
The signal from 99.8% D chloroform-d on the other hand is due to CHCl3 (ordinary chloroform). There is no H-D coupling and as a result the resonance is sharper.