As a matter of fact, chemical reactions can reduce mass just like nuclear reactions. I find it hard to accept, but it's true.
When the molecules in the dynamite explode, bonds between atoms are broken and reformed in different configurations. The result of this is that the net electrical potential energy in the resulting molecules is less than the electrical potential energy of the original stick. Now here's the cool part – that means it has less mass. Like, literally, less mass. Like, if you let the heat, light, and sound dissipate, and ultracarefully collect and weigh all the reaction products (impossible in practice, of course), it would weigh a tiny bit less than the original stick.
I find it helpful to think of a simpler example. Take two hydrogen atoms, and an oxygen atom. Allow them to run into each other. Their electron orbitals merge and hybridize. As their electrons settle into their new, shared, lower-energy state, they release photons. These photons carry away energy, and therefore mass, from the atoms. The resulting $\mathrm{H_2O}$ molecule literally weighs less than the two hydrogen and one oxygen beforehand.
Weird!
The mass of a free neutron is 939.566 MeV/c$^2$ (almost 1 GeV/c$^2$, so that's probably where your instructor got the "1" value), and the mass of a free proton is 938.272 MeV/c$^2$. A free neutron will decay into a free proton, free electron ($\beta^-$), and an anti-neutrino, $\bar{\nu}$. The mass of the electron is 0.511 MeV/c$^2$, and of the anti-neutrino, practically zero.
In the center-of-mass (CoM) reference frame (the rest-frame of the neutron), the total energy to start with is the mass energy of the neutron: 939.566 MeV.
After the decay, the mass energy of the products is 938.783 Mev, so there remains 0.783 MeV of energy to be shared as kinetic energy between the proton and the anti-neutrino. The net momentum in the CoM frame must be zero, but with three particles involved, the energy is not split uniquely among the three. The $\beta^-$ and $\bar{\nu}$ carry most of the kinetic energy, but again, not uniquely split.
The non-unique energy and momentum of the $\beta^-$ is what led physicists to consider the existence of the third particle in the decay.
Best Answer
The topic of "Energy Conservation" really depends on the particular "theory", paradigm, that you're considering — and it can vary quite a lot.
A good hammer to use to hit this nail is Noether's Theorem: see, e.g., how it's applied in Classical Mechanics.
The same principle can be applied to all other theories in Physics, from Thermodynamics and Statistical Mechanics all the way up to General Relativity and Quantum Field Theory (and Gauge Theories).
Thus, the lesson to learn is that Energy is only conserved if there's translational time symmetry in the problem.
Which brings us to General Relativity: in several interesting cases in GR, it's simply impossible to properly define a "time" direction! Technically speaking, this would imply a certain global property (called "global hyperbolicity") which not all 4-dimensional spacetimes have. So, in general, Energy is not conserved in GR.
As for quantum effects, Energy is conserved in Quantum Field Theory (which is a superset of Quantum Mechanics, so to speak): although it's true that there can be fluctuations, these are bounded by the "uncertainty principle", and do not affect the application of Noether's Theorem in QFT.
So, the bottom line is that, even though energy is not conserved always, we can always understand what this non-conservation mean via Noether's Theorem. ;-)