The most general Lorentz transformation that is connected to the identity is given by the conjugation by $\exp(-A)$ where
$$ A = \frac 12 \omega_{\mu\nu} \gamma^\mu \gamma^\nu $$
and $\omega_{\mu\nu}$ is an antisymmetric tensor containing $D(D-1)/2$ parameters. The group of all such transformations is isomorphic to $Spin(D-1,1)$. If $\omega$ only contains one component $0\mu$, then it is a boost, and the nonzero numerical value of $\omega$ is the rapidity - the "hyperbolic angle" $\eta$ such that $v/c=\tanh\eta$.
If only one doubly spatial component of $\omega$ is nonzero, then this component $\omega_{\mu\nu} = -\omega_{\nu\mu}$ is obviously the angle itself. Note that the spatial-spatial terms in $A$ are anti-Hermitean, producing unitary transformations; the mixed temporal-spatial terms in $A$ are Hermitean and they don't product unitary transformations on the 4-component space of spinors (but they become unitary if they're promoted to transformations of the full Hilbert space of quantum field theory).
In 4 dimensions, a general antisymmetric matrix $4\times 4$ contains 6 independent parameters and has eigenvalues $\pm i a, \pm i b$, so in 3+1 dimensions, one can always represent a general Lorentz transformation as a rotation around an axis in the 4-dimensional space followed by a boost in the complementary transverse 2-plane. This is the counterpart of the statement that any $SU(2)$ rotations in 3 dimensions is a rotation around a particular axis by an angle.
If you allowed $A$ to contain something else than $\gamma^{\mu\nu}$ matrices which generate the Lorentz group, you could get other groups. Only for a properly chosen subset of allowed values of $\omega$, you would get a closed group from the resulting exponentials (under multiplication). In particular, if you allowed $A$ to be an arbitrary complex combination of any products of gamma matrices, well, then you would allow $A$ to be any complex $4\times 4$ matrix, and its exponentials would produce the full group $GL(4,C)$ - surprising, Carl? ;-) It's not a terribly useful groups in physics because actions are usually not invariant under this "full group", are they?
Also, there are not too many groups in between $Spin(3,1)$ and $GL(4,C)$ - I guess that there's no proper group of $GL(4,C)$ that has a proper $Spin(3,1)$ subgroup. Obviously, there are many subgroups of $Spin(3,1)$ - such as $Spin(3)$, $Spin(1,1)\times Spin(2)$, and others.