I've seen A=5 and 8 described as bottlenecks in nucleosynthesis, because there are no stable nuclei with these mass numbers. One of my students suggested that you could, e.g., fuse 7Li+2H to produce 9Be, thereby bypassing the A=8 bottleneck. I've read that A=8 is actually surmounted by doing 4He+4He->8Be, then 8Be+4He (during the extremely short time that the 8Be holds together). Am I correct in thinking that 7Li+2H doesn't happen because by the time a star has built up an appreciable amount of 7Li, it has already used up all its 2H? What about the A=5 bottleneck?