How to Convert 2-Bromobutane to Ethane (Step-Down)
In Organic Chemistry, when we need to decrease the number of carbon atoms in a parent chain, we perform a "Step-Down" conversion. However, a brilliant step-down reaction is one where no carbon atoms are actually wasted, but rather split into useful, smaller molecules.
Today, we will convert a C4 molecule (2-Bromobutane) into two C2 molecules (Ethane) using a sequence of highly tested named reactions for JEE and NEET.
Video Tutorial: The C4 to C2 Cleavage
Watch Abhishek Sengar sir from CHEMCA explain this elegant three-step conversion pathway.
Step-by-Step Reaction Mechanism
-
Dehydrohalogenation (Saytzeff's Rule): We treat 2-Bromobutane with strong base (Alcoholic KOH). According to Saytzeff's rule, the more substituted (and therefore more stable) alkene is formed as the major product.
CH3-CH(Br)-CH2-CH3 → (Alc. KOH, Δ) CH3-CH=CH-CH3 (But-2-ene) -
Reductive Ozonolysis: We cleave the double bond in But-2-ene right down the middle using Ozone followed by Zinc dust and water. Since the alkene is perfectly symmetrical, it yields two identical molecules of Ethanal (Acetaldehyde).
CH3-CH=CH-CH3 → 1) O3 2) Zn/H2O 2 CH3-CHO (Ethanal) -
Clemmensen Reduction: Finally, we reduce the aldehyde group directly down to an alkane. Zinc amalgam with concentrated HCl strips the oxygen and adds hydrogens, turning Ethanal into Ethane.
2 CH3-CHO → (Zn(Hg) + conc. HCl) 2 CH3-CH3 (Ethane)Note: Sir mentions you could also use the Wolff-Kishner reduction (NH2NH2/KOH) or HI/Red Phosphorus here to achieve the exact same result!
Fig: Breaking the double bond and adding Oxygen atoms to each side.
Practice Questions for JEE & NEET
Examiners love tweaking the reagents slightly to test if you're paying attention. Can you spot the traps?
Question 1: In Step 1, what would be the major product if we reacted 2-Bromobutane with a bulky base like Potassium tert-butoxide (t-BuOK) instead of Alcoholic KOH?
Answer: But-1-ene (Hofmann Product)
Due to severe steric hindrance, a bulky base like t-BuOK cannot easily abstract the internal hydrogen to form the more stable Saytzeff product. Instead, it abstracts the more accessible terminal hydrogen, leading to the less substituted Hofmann product (But-1-ene) as the major product.
Question 2: In Step 2, what would be the final product if we performed Oxidative Ozonolysis using (O3 followed by H2O2) instead of Reductive Ozonolysis?
Answer: Ethanoic Acid (Acetic Acid)
Zinc acts as a reducing agent that stops the oxidation at the Aldehyde stage (Ethanal). If we use H2O2 (an oxidizing agent) without Zinc, the Aldehyde will be further oxidized into a Carboxylic Acid.
The product would be 2 molecules of CH3COOH.
Excellent job highlighting Saytzeff's rule here. If the elimination had accidentally formed the terminal But-1-ene instead of the symmetrical But-2-ene, the ozonolysis step would have yielded a messy mixture of methanal and propanal instead of pure ethanal. Great attention to regiochemistry!"
ReplyDelete