Sample 1Medium
The acid-catalysed dehydration of ethanol at 443 K is shown in three steps. Step 1: ethanol takes up H⁺ to form protonated ethanol, CH3CH2–OH2⁺. Step 2: a water molecule leaves to give the ethyl carbocation, CH3CH2⁺. Step 3: the carbocation loses H⁺ to give ethene, CH2=CH2. Which step is the slow, rate-determining step?
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The answer is B.
Protonation (step 1) and deprotonation (step 3) are fast. Breaking the C–O bond to form the carbocation (step 2) is the slowest step and so determines the rate.
- A.
The scheme shows separate steps with a carbocation intermediate, so the reaction is not concerted.
- C.
Protonation of the alcohol by acid is a fast step.
- D.
Loss of a proton from the carbocation to form the double bond is fast.
NCERT: Class 12 Chemistry, Chapter 7