If you throw a bunch a uranium ore in one blob, nothing happens.
If you chemically purify the ore so that the only element present is uranium, still nothing happens.
The runaway chain reaction needed for a uranium-powered bomb involves U-235, an isotope having three fewer neutrons than the most common natural isotope U-238. According to Wikipedia,
The required material must be 85% or more of ${}^{235}\mathrm{U}$ and is known as weapons grade uranium...
Now U-235 and U-238 have essentially indistinguishable chemistry.1 How do you purify the material to be mostly U-235? The traditional technique is to make the gaseous compound uranium hexafluoride, $\mathrm{UF}_6$, and use some mass-based physical process to separate out the ${}^{235}\mathrm{UF}_6$ from the ${}^{238}\mathrm{UF}_6$.2 This is slow, tedious work. In fact, the Oak Ridge facility built to do this during WWII was by some measures the largest building in the world at the time. It extracted U-235 one tiny bit at a time, and even after months of operation, there was just enough material for a single bomb, with nothing left to spare to even test the device.
For plutonium weapons, there is a similar complication. Bombs are made from Pu-239. Pu-240 is too unstable, and if there is too much of it in your bomb, it leads to a premature reaction (by a small fraction of a second), scattering most of the fuel rather than detonating it. The problem is even more difficult given that plutonium isn't found naturally in large quantities - it is produced entirely as a byproduct in uranium-based reactors.
In the end, it takes a large industrial infrastructure to manufacture either type of fission bomb since you need large amounts of an isotopically pure substance.
1 Even the chemical difference between normal hydrogen, consisting of 1 proton, and deuterium, which is a proton and a neutron, is small. Changing the mass of the nucleus, even by a factor of 2, does very little. Now imagine changing the mass by a factor of only about 1%.
2 Note that there is only one natural isotope of fluorine, F-19, so the only difference in masses for the molecules comes from the uranium.
The WKB approximation states that in one dimension, the tunneling probability $P$ can be approximated as
$\ln P=-\frac{2\sqrt{2m}}{\hbar}\int_a^b \sqrt{V-E} dx$ ,
where the limits of integration $a$ and $b$ are the classical turning points, $m$ is the reduced mass, the electrical potential $V$ is a function of $x$, and $E$ is the total energy. Setting $V=kq_1q_2/x$, we have for the integral
$I=\int_a^b \sqrt{V-E} dx$
$=\frac{kq_1k_2}{\sqrt{E}}\int_{A}^1\sqrt{u^{-1}-1} du$ ,
where $A=a/b$. The indefinite integral equals $-u\sqrt{u^{-1}-1}+\tan^{-1}\sqrt{u^{-1}-1}$, and for $A\ll 1$ the definite integral is then $\pi/2$. The result is
$\ln P=-\frac{\pi kq_1q_2}{\hbar}\sqrt{\frac{2m}{E}} $ .
This result was obtained in Gamow 1938, and $G=-(1/2)\ln P$ is referred to as the Gamow factor or Gamow-Sommerfeld factor.
The fact that the integral $\int_A^1\ldots$ can be approximated as $\int_0^1\ldots$ tells us that the right-hand tail of the barrier dominates, i.e., it is hard for the nuclei to travel through the very long stretch of $\sim 1$ nm over which the motion is only mildly classically forbidden, but if they can do that, it's relatively easy for them to penetrate the highly classically forbidden region at $x\sim1$ fm. Surprisingly, the result can be written in a form that depends only on $m$ and $E$, but not on $a$, i.e., we don't even have to know the range of the strong nuclear force in order to calculate the result.
The generic WKB expression depends on $E$ through an expression of the form $V-E$, which might have led us to believe that with a 1 MeV barrier, it would make little difference whether $E$ was 1 eV or 1 keV, and fusion would be just as likely in trees and houses as in the sun. But because the tunneling probability is dominated by the tail of the barrier, not its peak, the final result ends up depending on $1/\sqrt{E}$.
Because $P$ increases extremely rapidly as a function of $E$, fusion is dominated by nuclei whose energies lie in the tails of the Maxwellian distribution. There is a narrow range of energies, known as the Gamow window, in which the product of $P$ and the Maxwell distribution is large enough to contibute significantly to the rate of fusion.
Gamow and Teller, Phys. Rev. 53 (1938) 608
Best Answer
The bottleneck in Solar fusion is getting two hydrogen nuclei, i.e. two protons, to fuse together.
Protons collide all the time in the Sun's core, but there is no bound state of two protons because there aren't any neutrons to hold them together. Protons can only fuse if one of them undergoes beta plus decay to become a neutron at the moment of the collision. The neutron and the remaining proton fuse to form a deuterium nucleus, and this can react with another proton to form $^{3}\text{He}$. The beta plus decay is mediated by the weak force so it's relatively slow process anyway, and the probability of the beta plus decay happening at just the right time is extremely low, which is why proton fusion is relatively slow in the Sun. It takes gazillions of proton-proton collisions to form a single deuterium nucleus.
Nuclear fusion weapons bombs fuse fast because they use a mixture of deuterium and tritium. They don't attempt to fuse $^{1}\text{H}$ so they don't have the bottleneck that the Sun has to deal with.