Now, lithium 6 deuteride is another matter!
Step 1: with a slow neutron, like stray neutron from cosmic rays or from spontaneous fission of a nearby U nucleus
n+6Li=α+t+4,78 MeV, of which about 2,73 MeV are to triton
This is overwhelmingly the only fate of a slow neutron. There are possible reactions for chain loss:
n+6Li=7Li+γ
n+d=t+γ
β decay
but all of these are very low probability, because 6Li is a very good neutron absorber
Step 2: with the said fast triton
t+d=α+n+17,5 MeV. If triton was relatively slow, the breakdown of energy is 14 MeV for neutron and 3,5 MeV for α, but fast tritons allow a but wider spread of neutron energy.
Now THAT step does include a significant probability of chain loss. The triton may collide with electrons and lithium 6 nuclei, lose energy to ionization or undergo elastic or braking radiation with deuterons without fusing. So the said 2,73 MeV triton, or faster triton if a faster neutron was involved in step 1, may thermalize without fusion, and in cold deuteride this loses the chain.
No matter how low the probability of chain loss, chain reaction cannot happen unless there is some way for chain to branch.
Branching step seems to be
Step 3: with a fast neutron most likely
d+n+2,2 MeV=p+2n
The same might happen to α which also has enough energy. But I suspect that it would have less cross-sections even at same energy, and α starts at quarter the energy of a fission neutron.
So... if the lithium 6 deuteride is cold then a thermalized triton means chain loss. If the deuteride is hot then a thermalized triton eventually still finds a deuteron and propagates chain.
What is the neutron multiplication factor in cold lithium deuteride?