In the below reaction (reaction of 1-bromo-2-phenoxyethane with KOH at hugh temperature):
$$\ce{Ph-O-CH2-CH2-Br ->[\ce{KOH}][200\ ^\circ C]P}$$
According to me KOH will help in displacing bromide ion and making a carbocation. This carbocation would be attacked by the double bonds of the benzene ring. So the product should be benzotetrahydrofuran:
But the product is given as phenoxyethene (phenyl vinyl ether):
I could not understand how is this possible.
First, let's take a look at your proposed carbocation intermediate:
$$\ce{Ph-O-CH2-CH2+}$$
This is a primary carbocation, which makes it very unstable. It's unlikely to form, so the reaction can't involve going through this stage. Primary halides are much more likely to undergo $\ce{S_N2}$ or $\ce{E2}$ reactions. Whether it's the former or the latter will depend on the nucleophilic and basic charachter of the other reagent. Bases (specially if sterically hindered) tend to favor $\ce{E2}$, while nucleophiles of low basicity tend to favor $\ce{S_N2}$. However, many reagents may and will give a mix of both products.
With all this information, since $\ce{KOH}$ is a very strong base and the reaction happens at high temperature, the result will be the $\ce{E2}$ product, in your second picture.
