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Kolbe Electrolysis for Alkenes

Explain the m...

Question

Explain the mechanism of the following reaction:
$C H_{3} - C H_{2} - O H H^{+} - - \to 443 K C H_{2} = C H_{2} + H_{2}^{-} O$

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Solution

The –OH group in the alcohol donates two electrons to $H^{+}$ from the acid reagent, forming an alkyloxonium ion. This ion acts as a very good leaving group which leaves to form a carbocation. The deprotonated acid (the nucleophile) then attacks the hydrogen adjacent to the carbocation and form a double bond.Primary alcohols undergo bimolecular elimination $(E^{2}$ mechanism) while secondary and tertiary alcohols undergo unimolecular elimination ( $E^{1}$ mechanism). Oxygen donates two electrons to a proton from sulfuric acid $H_{2} S O_{4}$ , forming an alkyloxonium ion. Then the nucleophile $H S O_{4}^{-}$ back-side attacks one adjacent hydrogen and the alkyloxonium ion leaves in a concerted process, making a double bond.

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Q. Explain the mechanism of the following reactions:

C H_{3} - C H_{2} - O H H^{+} - -- \to 443 K C H_{2} = C H_{2} + H_{2} O

Q. Explain the mechanism of the following reaction.

2 C H_{3} - C H_{2} - O H H^{+} - -- \to 413 K C H_{3} - C H_{2} - O - C H_{2} - C H_{3} + H_{2} O

Write the mechanism of the reaction:

C H_{2} = C H_{2} + H X \to C H_{3} - C H_{2} - X

The following reaction proceeds through which mechanism:

C H_{2} = C H_{2} + H B r \to C H_{3} - C H_{2} - B r

Q. Which of the following is position isomer of

C H_{3} - C H_{2} - O | | C - C H_{2} - C H_{3}

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Kolbe Electrolysis for Alkenes

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