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EEJ MAIN Chemistry QUESTION #7127
Question 1
When $1$-phenyl-2-chloropropane is treated with alcoholic $KOH$, the major elimination product formed is:
  • $1$-phenylprop-$1$-ene✔️
  • $1$-phenylprop-$2$-ene
  • $3$-phenylprop-$1$-ene
  • $2$-phenylprop-$1$-ene
Correct Answer Explanation
Alcoholic $KOH$ induces dehydrohalogenation ($E2$ mechanism). According to Zaitsev's rule, the more substituted and conjugated alkene is preferred. $1$-phenylprop-$1$-ene is the major product because the resulting double bond is in conjugation with the benzene ring, providing extra stability.