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प्रश्न
What is the action of following on ethyl bromide?
alcoholic sodium hydroxide
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उत्तर
When ethyl bromide is boiled with an alcoholic solution of sodium hydroxide \[\ce{(NaOH)}\], an alkene is formed due to the elimination of hydrogen atoms from carbon and halogen from α-carbon, which is called dehydrohalogenation.
\[\ce{\underset{(Ethyl bromide)}{\overset{\beta\phantom{.............}\alpha\phantom{............}}{CH3 - CH2 - Br}} + \underset{(alc.)}{NaOH} ->[\Delta] \underset{(Ethene)}{CH2 = CH2} + NaBr + H2O}\]
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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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\[\begin{array}{cc}\ce{CH3\phantom{................}}\\
|\phantom{...................}\\
\ce{CH3 - C - CH2 - Cl ->[Na/dry ether] A}\\
|\phantom{...................}\\
\ce{CH3\phantom{................}}
\end{array}\]
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|\\
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|\\
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\[\begin{array}{cc}
\ce{CH3}\phantom{.................}\\
|\phantom{...................}\\
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|\phantom{..................}\\
\ce{CH3}\phantom{................}
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\[\begin{array}{cc}
\ce{CH3}\phantom{................}\\
|\phantom{...................}\\
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|\phantom{...................}\\
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\[\begin{array}{cc}
\ce{CH3}\phantom{...............}\\
|\phantom{..................}\\
\ce{CH3 - C - CH2 - Cl ->[Na/dry ether]A}\\
|\phantom{...................}\\
\ce{CH3}\phantom{.................}
\end{array}\]
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\[\begin{array}{cc}
\ce{CH3}\phantom{.................}\\
|\phantom{...................}\\
\ce{CH3 - C - CH2 - Cl ->[Na/dry ether] A}\\
|\phantom{...................}\\
\ce{CH3}\phantom{.................}
\end{array}\]
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