The major product obtained in the following reaction is :
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- \(\mathrm{C_6H_5CH=CHC_6H_5}\)
- \((+)\text{-}\mathrm{C_6H_5CH(O\text{-}t\text{-}Bu)CH_2C_6H_5}\)
- \((-)\text{-}\mathrm{C_6H_5CH(O\text{-}t\text{-}Bu)CH_2C_6H_5}\)
- \((\pm)\text{-}\mathrm{C_6H_5CH(O\text{-}t\text{-}Bu)CH_2C_6H_5}\)
The Correct Option is A
Solution and Explanation
Potassium tert-butoxide is a strong but very bulky base, so it is a poor nucleophile; with heat it brings about E2 dehydrohalogenation rather than SN1/SN2 substitution (ether formation).
The \(\beta\)-hydrogen on \(\mathrm{CH_2C_6H_5}\) is benzylic and relatively acidic, and the alkene formed is conjugated with both phenyl rings (stilbene), so elimination is strongly favoured.
Product: \(\mathrm{C_6H_5CH=CHC_6H_5}\) (trans-stilbene, achiral).
The ethers in options (2), (3), (4) are substitution products and are not formed.
Answer: (1) \(\mathrm{C_6H_5CH=CHC_6H_5}\)
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