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I am looking at this problem:

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I thought the KOH in alcohol would support a nucleophilic attack of hydroxide on the haloalkane, so the product I would get is B (elimination of HCl). However, this seems to be incorrect.

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    $\begingroup$ KOH in alcohol is a reasonably strong base. By far the most acidic proton in the 4-chlorobutan-1-ol starting material is the -OH so the first species generated is the alkoxide of the chlorobutanol $\endgroup$
    – Waylander
    Commented Mar 13 at 10:05
  • $\begingroup$ Yes, the formation of the [5] ring is energetically favoured, Cl- is a good leaving group. This will be a fast and irreversible reaction $\endgroup$
    – Waylander
    Commented Mar 13 at 13:56
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    $\begingroup$ @DUDE_WITH_J would prefer if you write questions, as much as possible, in clear language and with minimal grammatical errors. We don't demand perfection; but please maintain so that questions remain salvageable. There are various examples of good questions being asked in a bad way. $\endgroup$
    – Harikrishnan M
    Commented Mar 14 at 2:49
  • $\begingroup$ Few helpful links to help with formatting: Chem / MathJaX formatting, Upright Vs Italics. Welcome to ChemSE! $\endgroup$
    – Harikrishnan M
    Commented Mar 14 at 2:51

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