How to Convert Ethane to 2-Methylpropan-1-ol
Organic Chemistry conversions can feel like solving a puzzle. The trick to mastering them for JEE and NEET is simple: Always count your carbon atoms first!
In this classic conversion, our reactant is Ethane (2 carbon atoms) and our target product is 2-Methylpropan-1-ol (4 carbon atoms). Because the number of carbon atoms has increased, we know right away this will involve a Step-Up reaction.
Video Tutorial: The Synthesis Pathway
Watch Abhishek Sengar sir from CHEMCA break down this multi-step pathway, involving named reactions like the Wurtz reaction and Hydroboration-Oxidation.
Step-by-Step Reaction Mechanism
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Chlorination (Activation): Ethane is an inert alkane. We first activate it by converting it to an alkyl halide using free radical halogenation in the presence of sunlight.
CH3-CH3 → (Cl2 / hν) CH3-CH2-Cl (Chloroethane) -
Wurtz Reaction (Step-Up): We double the carbon chain by treating Chloroethane with Sodium in Dry Ether.
2 CH3-CH2-Cl + 2Na → (Dry Ether) CH3-CH2-CH2-CH3 (n-Butane) -
Isomerization (Branching): The target molecule has a branch. We isomerize the straight-chain n-butane using Anhydrous Aluminium Chloride.
CH3-CH2-CH2-CH3 → (Anhy. AlCl3 / HCl) CH3-CH(CH3)-CH3 (Isobutane) -
Selective Bromination: Bromine selectively attacks the 3° (tertiary) hydrogen, which is the most stable free radical intermediate.
CH3-CH(CH3)-CH3 → (Br2 / hν) CH3-C(Br)(CH3)-CH3 (2-Bromo-2-methylpropane) -
Dehydrohalogenation: We eliminate the Bromine to form a double bond using a strong base.
CH3-C(Br)(CH3)-CH3 → (Alc. KOH, Δ) CH2=C(CH3)-CH3 (2-Methylpropene) -
Hydroboration-Oxidation: Finally, to place the -OH group at the primary (terminal) carbon, we add water via Anti-Markovnikov's rule.
CH2=C(CH3)-CH3 → 1) B2H6/THF 2) H2O2 / OH− HO-CH2-CH(CH3)-CH3
Fig: Structure of the final product: 2-Methylpropan-1-ol
Practice Questions for JEE & NEET
Master these core reagents to guarantee full marks in organic conversion maps!
Question 1: Which reagent would you use to convert 2-Methylpropene into 2-Methylpropan-2-ol (Markovnikov addition of water)?
Answer: Acid-Catalyzed Hydration
To place the -OH group on the more substituted carbon (the inner tertiary carbon), you would use dilute acid. The reagent is written as: Dil. H2SO4 or H2O / H+.
Question 2: What happens if you treat Chloroethane with Aqueous KOH instead of Alcoholic KOH?
Answer: Substitution occurs instead of Elimination.
Alcoholic KOH acts as a strong base and causes dehydrohalogenation (Elimination), creating a double bond (Alkene).
Aqueous KOH acts as a nucleophile and causes Nucleophilic Substitution (SN2), replacing the Chlorine with an -OH group, resulting in Ethanol (an Alcohol).
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This made a complex topic look so easy!
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