Navigation
Explores the formation of unsaturated compounds via E1/E2 mechanisms and the reactivity of double bonds.
Why does a single molecule of 2-bromobutane choose to become a double-bonded alkene instead of a simple alcohol? The answer lies in a molecular 'tug-of-war' where temperature and base strength decide the winner.
In an elimination reaction, a molecule loses atoms (often a halide and a hydrogen) to form a double bond. This process is categorized into E1 (unimolecular) and E2 (bimolecular) mechanisms. In E2 reactions, a strong base pulls a proton from a -carbon while the leaving group departs simultaneously. But which hydrogen leaves? Zaitsev's Rule states that the major product is the most stable alkene—typically the one where the double bond is connected to the most alkyl groups. Think of it as the 'poor get poorer' rule: the carbon with fewer hydrogens is more likely to lose another one to create a highly substituted, stable double bond.
Predict the major product when 2-bromobutane reacts with ethoxide ().
1. Identify the -carbon (bearing the ) and the adjacent -carbons. 2. is a terminal ; is a group. 3. Removing H from gives 1-butene: . 4. Removing H from gives 2-butene: . 5. 2-butene is disubstituted and more stable. Result: 2-butene is the major product.
Quick Check
According to Zaitsev's rule, which is more stable: a trisubstituted alkene or a monosubstituted alkene?
Answer
A trisubstituted alkene is more stable because it has more alkyl groups stabilizing the double bond through hyperconjugation.
While elimination creates double bonds, electrophilic addition breaks them. When a hydrogen halide like adds to an unsymmetrical alkene, we use Markovnikov's Rule. This rule dictates that the hydrogen atom attaches to the carbon that already has the most hydrogen atoms (the 'rich get richer'). This happens because the reaction proceeds through the most stable carbocation intermediate. A tertiary carbocation () is far more stable than a primary one () due to inductive effects and hyperconjugation. By placing the on the less substituted carbon, the positive charge lands on the more substituted carbon, lowering the activation energy.
Reaction:
1. The electrophile attacks the double bond. 2. Path A: adds to , forming a primary carbocation ( is actually secondary, wait). 3. Let's re-evaluate: has 2 hydrogens, has 1 hydrogen. 4. Hydrogen adds to (the 'richer' carbon). 5. This creates a secondary carbocation on : . 6. attacks the carbocation. Result: 2-bromopropane.
Quick Check
In the addition of to 2-methylpropene, which carbon will the Chlorine atom bond to?
Answer
The Chlorine will bond to the central (tertiary) carbon because the Hydrogen will bond to the terminal group.
Alkyl halides often face a choice: undergo Substitution () or Elimination (). Three main factors tip the scales toward elimination: 1. Temperature: High heat favors elimination because it increases the entropy () of the system (producing three molecules from two). 2. Base Strength/Size: Strong, bulky bases like potassium tert-butoxide () are too 'fat' to reach the carbon for substitution, so they snatch a peripheral proton instead, forcing elimination. 3. Substrate Sterics: Tertiary halides are highly crowded, making substitution difficult and elimination the preferred path.
Predict the outcome of 2-iodopropane reacting with at vs .
1. At (lower temp), substitution () is significant, yielding ethyl isopropyl ether. 2. At (higher temp), the entropy term becomes more negative. 3. Elimination () becomes dominant. Result: Propene is the major product at high temperatures.
Which reagent would best favor an E2 reaction over an reaction for a primary alkyl halide?
What is the major product of the reaction between 1-methylcyclohexene and ?
Increasing the temperature of a reaction generally increases the yield of the substitution product over the elimination product.
Review Tomorrow
In 24 hours, try to draw the mechanism for the E2 reaction of 2-bromopentane and explain why 2-pentene is favored over 1-pentene.
Practice Activity
Find 5 unsymmetrical alkenes and practice adding to them using Markovnikov's rule, then 'reverse' the process by performing an elimination on the product.