Question

1-phenyl-2-chlorobutane on reaction with EtOH/EtOK produces a product which on treatment with HBr produces

Hint:

When evaluating reaction pathways, always look for the formation of conjugated systems in products and resonance-stabilized intermediates (like benzylic or allylic carbocations). These are massive thermodynamic driving forces.

Answer 1

Step 1: Understanding the Concept:
This sequence describes two consecutive organic reactions:
1. A base-promoted \(\beta\)-elimination (dehydrohalogenation) of an alkyl halide to form an alkene.
2. An electrophilic addition of a hydrogen halide (\(\text{HBr}\)) to the newly formed alkene, following Markovnikov's rule guided by carbocation stability.
Step 2: Key Formula or Approach:
For step 1, apply Saytzeff's rule or analyze product stability (conjugation) to determine the major alkene.
For step 2, evaluate the possible carbocation intermediates formed upon protonation of the alkene and select the most stable one to predict the final addition product.
Step 3: Detailed Explanation:
Reaction 1: Dehydrohalogenation
The starting material is 1-phenyl-2-chlorobutane:
\(\text{C}_6\text{H}_5-\text{CH}_2-\text{CH(Cl)}-\text{CH}_2-\text{CH}_3\)
The reagent \(\text{EtOK}/\text{EtOH}\) acts as a strong base, favoring an E2 elimination mechanism. It will remove a \(\beta\)-hydrogen and the chlorine atom. There are two adjacent \(\beta\)-carbons (C1 and C3).
- Elimination between C1 and C2 yields: 1-phenylbut-1-ene (\(\text{C}_6\text{H}_5-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}_3\)). In this molecule, the double bond is in direct conjugation with the aromatic phenyl ring, making it highly thermodynamically stable.
- Elimination between C2 and C3 yields: 1-phenylbut-2-ene (\(\text{C}_6\text{H}_5-\text{CH}_2-\text{CH}=\text{CH}-\text{CH}_3\)). This double bond is not conjugated with the ring.
Due to the significant stabilizing effect of extended conjugation, 1-phenylbut-1-ene will be the overwhelming major product.
Reaction 2: Electrophilic Addition of HBr
We take the major product, 1-phenylbut-1-ene, and react it with \(\text{HBr}\):
\(\text{C}_6\text{H}_5-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}_3 + \text{HBr}\)
The first step is the attack of the electrophile (\(\text{H}^+\)) on the double bond, creating a carbocation. It can add to either side:
- Addition of \(\text{H}^+\) to C1 creates a secondary carbocation at C2: \(\text{C}_6\text{H}_5-\text{CH}_2-\text{C}^+\text{H}-\text{CH}_2-\text{CH}_3\).
- Addition of \(\text{H}^+\) to C2 creates a secondary benzylic carbocation at C1: \(\text{C}_6\text{H}_5-\text{C}^+\text{H}-\text{CH}_2-\text{CH}_2-\text{CH}_3\).
The benzylic carbocation is vastly more stable due to resonance delocalization of the positive charge around the phenyl ring. The reaction will proceed almost exclusively via this more stable intermediate.
Finally, the nucleophilic bromide ion (\(\text{Br}^-\)) attacks the benzylic carbocation at C1:
\(\text{C}_6\text{H}_5-\text{C}^+\text{H}-\text{CH}_2-\text{CH}_2-\text{CH}_3 + \text{Br}^- \rightarrow \text{C}_6\text{H}_5-\text{CH(Br)}-\text{CH}_2-\text{CH}_2-\text{CH}_3\)
The name of this final product is 1-bromo-1-phenylbutane.
Step 4: Final Answer:
The final product of the reaction sequence is 1-bromo-1-phenylbutane.