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(a) Synthesis from an Appropriate Alkene:

To synthesize 2-butanol from an alkene, use but-2-ene (CH?–CH=CH–CH?) as the starting alkene.

Reaction:

Alkene Used: But-2-ene: CH3–CH=CH–CH3

Reagent: Dilute H2SO4 (or water in the presence of an acid catalyst)

Reaction Type: Acid-catalyzed hydration (Markovnikov’s rule applies)

Reaction Mechanism:

  1. Protonation of the double bond to form a more stable carbocation.

  2. Water adds to the carbocation.

  3. Deprotonation gives the alcohol.

Final Answer: To synthesize the given alcohol (butan-2-ol), take but-2-ene and react it with dilute sulfuric acid (H?SO?) in water. The Markovnikov addition of water across the double bond gives 2-butanol.

 

(b)  This reaction is the acid-catalyzed dehydration of ethanol (CH?CH?OH) to form ethene (CH?=CH?) and water, occurring at 443 K in the presence of an acid (H+).

Mechanism (Stepwise):

Step 1: Protonation of Ethanol

The lone pair on the oxygen atom of ethanol accepts a proton (H+) from the acid catalyst.

CH3CH2OH + H+ → CH3CH3OH+2

This forms a protonated alcohol (oxonium ion).

Step 2: Loss of Water Molecule

The protonated ethanol loses a molecule of water, forming a carbocation.

CH3CH2OH+2→CH3CH2++H2O

A primary carbocation (CH3CH2+) is formed.

Note: In this case, rearrangement is not possible since it’s only a 2-carbon chain.

Step 3: Elimination of a Proton (Deprotonation)

A base (like water or another alcohol molecule) abstracts a β-hydrogen from the carbocation, forming a double bond (alkene).

CH3CH2+ → CH2=CH2 + H+

Final Answer:

Mechanism for dehydration of ethanol:

  1. Protonation of –OH group → CH3CH2OH2+

  2. Loss of H?O → formation of CH3CH2+

  3. Deprotonation → formation of ethene (CH2=CH2)

Posted by

Saumya Singh

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