Why does a car bonnet (hood) rise when you connect the clutch with a brake on? Is the rotational force to overcome the brakes moved to the opposite effect of moving the car chassis, until the brake is released?
 A: I assume your car is front wheel drive. 
The phenomenon is simply Newton's third law in disguise. The car exerts a torque on its forward axle and the wheels exert the same magnitude, opposite sense torque on the car. Normally, the torque is not so big, because as soon as it is exerted on the wheels by the car, the wheels push backward on the road and the car moves forward. However, with the brake on the hinder wheels, the car can't move forward nearly so freely, and the torque "winds" the car against the thrustback from its springs, particularly the hinder axle ones. You see this as the car's tilting so that its tail dips towards the ground. Draw a free-body diagram for the car, and one for the forward axle in the two cases where the brake (1) is and (2) is not applied to the hinder wheels and I think the behaviour in both cases should be clear.

Warning: People who laugh at my drawing skills will be mercilessly slashed into teeny weeny wincy tiny pieces!
Above I have drawn the FBD for the car and front axle when there is no brake on the hinder wheel. Newton II applied to horizontal components yields:
$$m\,a = F_D$$
where $m$ is the car's mass and $a$ its acceleration on engaging the clutch so that:
$$\tau = m\,a\,r$$
where $r$ is the radius of the front wheel. (Here I neglect inertia of the wheels). $m$ is maybe $1000{\rm kg}$ and $a$ maybe $0.2 g$. With $r = 0.2m$, the torque is then of the order of $400{\rm N\,m}$. Now look at the FBD with the brake applied to the hinder wheels:

There is now the fundamental difference that the hinder wheel cannot spin freely owing to the brake's "binding" it rigidly to the car's body. The car is now in a static situation as long as the hinder wheel does not skid. Now we still have $\tau = F_D\,r$, but $F_D$ is balanced by the friction from the hinder wheel. This can be rather more than $m\,a$. Therefore, the torque fleetingly rises to a much higher value on engaging the clutch: either the engine will stall, or the car will begin to rotate until the moment of the weight and the hinder wheel reaction force balances the torque $\tau$.
