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प्रश्न
Explain the dehydrohalogenation reaction of 2-chlorobutane.
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उत्तर
When 2-chlorobutane is boiled with an alcoholic solution of potassium hydroxide, it undergoes the elimination of the hydrogen atoms from β-carbon and chlorine atoms from α-carbon resulting in the formation of But-2-ene and But-1-ene.
Reaction:
But according to Saytzeff's rule, in dehydrohalogenation reaction, the preferred product is that alkene which has a greater number of alkyl groups attached to doubly bonded carbon atoms.
Hence, But-2-ene is obtained as the major product in the dehydrohalogenation reaction of 2-chlorobutane.
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संबंधित प्रश्न
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\[\begin{array}{cc}\ce{CH3 - CH = CH3 ->[Red P/Br2] A ->[Ag2O/H2O]B}\\|\phantom{.........................}\\
\ce{OH\phantom{.......................}}\end{array}\]
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\[\ce{\underset{\text{(I)}}{CH3 - CH2 - CH2 - Cl}}\]
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\[\begin{array}{cc}
\ce{CH3}\\
|\\
\ce{CH3-C-Cl}\phantom{...}\\
|\\
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\[\begin{array}{cc}
\ce{CH3}\phantom{.......}\\
|\phantom{.........}\\
\ce{CH3 - c - CH2 - Cl ->[Na/dry ether]}\\
|\phantom{.........}\\
\ce{CH3}\phantom{.......}
\end{array}\]
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Complete the following reaction giving major product.
\[\begin{array}{cc}
\ce{CH3}\phantom{................}\\
|\phantom{...................}\\
\ce{CH3 - C - CH2 - Cl ->[Na/dry ether] A}\\
|\phantom{...................}\\
\ce{CH3}\phantom{................}\\
\end{array}\]
