Drawing Mechanisms For Organic Chemistry

When you write a mechanism, you do not have to include the reaction (energy) diagram, just the steps showing all the intermediates. Here are the conventions for writing a particular mechanism:

Show all intermediates that you know about as separate sequential drawings (part E gives tips for figuring out what might come next).
Link all intermediates by straight arrows, double if you know the step is reversible and single if you know it is not. Each set of arrows followed by a new structure is a step.
Show one change in bonding for each step (e.g. for E1: ionization, removal of proton), unless you know that more than one bond is changed in a given step (e.g. E2).
If there are steps that you have little evidence about because they are after the rate determining step, use analogies to other known reactions to fill in the blanks (e.g. loss of a proton after an acid-catalyzed reaction)
If necessary, add an intermediate to the set you know about, again using analogies to other known reactions, to ensure that only one bond-making / bond-breaking occurs for each step.
If there are no known intermediates, sketch the transition state and label it as such (see F).

Here is an annotated example using the dehydration of an alcohol:

Equilibrium 1: reaction is acid-catalyzed; spectroscopy shows the conjugate acid of the alcohol, intermediate 1, is formed very fast - proton transfers are almost never rate-determining steps for other reactions.
Equilibrium 2: the rate determining step (acid and alcohol concentrations affect the rate). Evidence for a carbocation, intermediate 2? With all alcohols, some substitution is observed, more if the acid is something like HBr, whose conjugate base is nucleophilic; with some alcohols, rearrangement occurs. Both of these observations are consistent with carbocation formation (and not with concerted, carbanion or radical reactions)
Equilibrium 3: This reaction cannot be readily observed under these reaction conditions since it is after the rate-determining step. However, we observe separately that alkenes dissolve in concentrated sulfuric acid, and thus must undergo an acid-base reaction themselves (protonation) to form soluble ions, which must be carbocations.