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A chiral $C_5H_{10}O$ ether reacts with hot $HI$ to give a $C_5H_{10}I_2$ product.Treatment of this with hot $KOH$ in ethanol produces $1$,$3$-pentadiene.What is the structure of the original ether?

$\begin{array}{1 1} a \\ b \\ c\\ d\end{array}$

1 Answer

Above ethers reacts with hot HI to give a $C_2H_{10}I_2$ product but only (c) will give 1,3-pentadiene reacting with hot KOH.
Mechanisum :Step I:Protonation of ether molecule
Step II:Iodide is a good nucleophile.It attacks the least substituted carbon of the oxonium ion.
Hence (c) is the correct answer.
answered Feb 25, 2014 by sreemathi.v
 
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