Step 1: Recall the three steps of acid-catalysed dehydration of alcohol.
Step 1: Protonation of the OH group by the acid (fast equilibrium). Step 2: Loss of water from the protonated alcohol to form a carbocation (slow, rate-determining). Step 3: Loss of a proton from the carbocation to give the alkene (fast).
Step 2: Identify the slowest step.
The slowest (rate-determining) step is Step 2, in which the protonated alcohol loses water to form a carbocation intermediate. This step requires breaking the C-O bond and has the highest activation energy.
Step 3: Name the intermediate formed in the slowest step.
When the protonated alcohol ($R-OH_2^+$) loses water, a carbocation ($R^+$) is formed. A carbocation has a positively charged carbon with only six electrons around it.
Step 4: Eliminate wrong options.
Protonated alcohol forms in Step 1 (fast, not the slow step). Carbanion (negatively charged carbon) is not formed in this acid-catalysed ionic mechanism. Free radical requires homolytic bond cleavage (UV or heat) and is not part of this mechanism.
Step 5: Confirm the carbocation answer.
The carbocation formed in Step 2 then quickly loses a proton to give the alkene in Step 3. The rate of dehydration depends on carbocation stability: 3 degrees > 2 degrees > 1 degree, which is why tertiary alcohols dehydrate fastest.
Step 6: State the final answer.
The intermediate formed during the slowest step in dehydration of alcohol is a carbocation. \[ \boxed{\text{Carbocation}} \]