Draw the product formed by the reaction of potassium t~butoxide with ([S.2S)-[~bromo ~methyl-I-phenylbutane (shown). Clearly show the stereochemistry 0f the producl (CHshCo K’ Hjcun H,CHzc (CH;)COH
The Correct Answer and Explanation is:
The reaction of potassium tert-butoxide with (2S,3S)-3-bromo-2-methyl-1-phenylbutane undergoes an E2 elimination mechanism. Let’s go step-by-step to determine the structure and stereochemistry of the alkene product formed.
Step 1: Analyze the Substrate
The substrate is (2S,3S)-3-bromo-2-methyl-1-phenylbutane, which has the following structure:
- A four-carbon chain with a bromine on carbon 3 and a methyl group on carbon 2
- A phenyl group attached to carbon 1
- Chiral centers at carbon 2 and carbon 3, both in the S configuration
Step 2: Understand the Base
Potassium tert-butoxide ((CH₃)₃CO⁻) is a strong, bulky base, which promotes E2 elimination by abstracting a β-hydrogen anti-periplanar to the leaving group (Br).
Step 3: Predict the Mechanism
In an E2 reaction:
- The base removes a β-hydrogen anti to the leaving group (Br)
- The electrons from the β-hydrogen form a π-bond between carbon 2 and 3
- The leaving group (Br⁻) departs from carbon 3
Since the base is bulky, it prefers to abstract the least hindered β-hydrogen — usually leading to the less substituted (Hofmann) alkene. However, if stereoelectronic factors favor the more substituted (Zaitsev) alkene, that will form.
In this molecule:
- β-hydrogens are on carbon 2
- The hydrogen anti to the Br must be positioned opposite in a staggered conformation
- The resulting double bond forms between C-2 and C-3
Step 4: Determine the Product
The double bond forms between C-2 and C-3, and since carbon 2 has a methyl group and carbon 3 is now substituted with the chain, the product is:
(E)-2-methyl-1-phenyl-1-butene
The stereochemistry of the product is E (trans) because:
Final Answer: Structure
(E)-2-methyl-1-phenyl-1-butene
mathematicaCopyEdit Ph
\
C=C
/ \
CH3 CH2CH3
Conclusion
The strong bulky base potassium tert-butoxide promotes an E2 reaction by removing a β-hydrogen anti to the Br group. The result is formation of a double bond with E stereochemistry, giving (E)-2-methyl-1-phenyl-1-butene as the major product.The reaction of potassium tert-butoxide with (2S,3S)-3-bromo-2-methyl-1-phenylbutane undergoes an E2 elimination mechanism. Let’s go step-by-step to determine the structure and stereochemistry of the alkene product formed.
