I suspect that on the first bounce a lot of energy went into making the tictac rotate, then on the second bounce the tictac stopped rotating and the energy went back into linear motion.
I don't have any tictacs to hand, but if you'd like to weigh one of yours we could work out how fast it would have to rotate to absorb 30cm worth of bounce potential energy.
Response to comment: actually I didn't need the mass, I just need the length. Since I forgot to ask for the length of a tictac let's assume it's 1cm.
If you look at http://www.web-formulas.com/Physics_Formulas/Moment_of_Inertia.aspx the moment of inertia of a beam of length L is:
$$I = \frac{mL^2}{12}$$
Strictly speaking the tictac is better modelled as a cylinder, but treating it as a beam is a reasonable approximation and makes the calculation easier. Anyhow, let's suppose you drop the tictac from a height $h$ so it's kinetic energy when it hits the ground is $mgh$. The kinetic energy of a rotating body is $0.5Iw^2$, so if all the kinetic energy is converted into rotation we get:
$$ \frac{1}{2}Iw^2 = mgh $$
so
$$ w = \sqrt{\frac{2mgh}{I}} $$
Substitute $I = mL^2/12$ and we get:
$$ w = \sqrt{\frac{24gh}{L^2}} $$
Now feed in $d$ = 1 metre and $L$ = 0.01 metres and we get $w$ is about 1,500 radians per second, which is a frequency of rotation of about 250Hz.
OK, I think it's pretty unlikely that the tictac is rotating 250 times a second, so I think your explanation is far more likely i.e. that you're converting vertical motion into horizontal motion. In my defence, you didn't say that the tictac was bouncing along horizontally. If you've ever watch a rugby ball or American football ball bouncing along the ground you'll know that the height of the bounces is very irregular.