NaOH + EtOH will eliminate the Cl atom forming a double bond. At least, that's what I think. 3 could also be a viable answer since the OH can also attack the said double bond (this is probably not correct) and cause an OH to be added forming the compound in 3.
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This is an example of the subtle difference between different bases that are strong in water but may not be equally strong in other, less polar solvents. Alcoholic potassium hydroxide certainly does eliminate a proton from the substrate leading to product (4), but sodium hydroxide in ethanol is just a little weaker and uses nucleophilic substitution to form (1) instead.
answered Sep 22, 2023 at 18:55
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1$\begingroup$ Supporting data? Perhaps 4 less ring strain and conjugated double bond. This is a better to see the products and determine the mechanism than quess the mechanism and guess the products. $\endgroup$– jimchmstCommented Sep 23, 2023 at 19:51