What is the product after treating ethene successively with chlorine and water, sulfuric acid, and potassium hydroxide? Question:  


  Ethene reacts with $\ce{Cl2}$ in $\ce{H2O}$ to form A. A on reaction with $\ce{H2SO4}$ followed by heating gives B. B on reaction with alcoholic $\ce{KOH}$ gives C. Find C.


My answer is:  

A: $\ce{CH2(OH)CH2Cl}$
B: $\ce{CH2=CHCl}$
C: ethyne.  

Explanation:

Ethene on reaction with chlorine water undergoes addition reaction to give A. A on reaction with $\ce{H2SO4}$ followed by heating undergoes dehydration to give B. B on reaction with alcoholic $\ce{KOH}$, undergoes elimination to give ethyne.

But the answer given is $\ce{CH2=CH-O-CH=CH2}$.

Where I am going wrong and what are the products?
The first reaction is a hydrochlorination reaction:



The second reaction is an acid catalyzed ether synthesis. The the solution is full of other 2-chloroethanol, which is capable of acting as a nucleophile in an $\mathrm{S_N2}$ reaction to eject $\ce{H2O}$. This is reaction will have a much lower energy barrier than the dehydration you proposed, which requires the extraction of an aliphatic hydrogen. The ether synthesis mechanism can be seen here:



The final step is a series of two $\mathrm{E2}$ reactions (shown condensed here):

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