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I got this question from the PYQs of a college entrance exam. According to the answer key, 4-chloro-3-methylpent-1-ene reacts with NaOH in water to give 4-hydroxy-3-methylpent-1-ene via the SN2 mechanism.

I think this reaction should undergo the E2 mechanism to give a conjugated diene.(3-methylpenta-1,3-diene).

My thought process

I took 3 factors into consideration:

Substrate : Secondary alkyl halide, so both SN2 and E2 are possible.

Nucleophile/Base: OH- is a strong non-bulky base and a strong nucleophile; therefore, both SN2 and E2 are possible.

Solvent: Polar protic solvent generally favours E2 over SN2. Source

Polar protic solvents tend to favor elimination (E2) over substitution (SN2).

Polar aprotic solvents tend to favor substitution (SN2) relative to elimination (E2).

Considering all three factors, I concluded that the E2 mechanism should provide the majority yield.

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  • $\begingroup$ Might be hard to say in this case, since steric effects might be significant. The best way to check would be to run the actual reaction. All you have here are general preferences where you don't know the magnitude of the effect, so at the end of the day, there is no hard prediction. $\endgroup$
    – Zhe
    Commented Apr 3 at 15:27
  • $\begingroup$ @Zhe, could you elaborate a bit on the steric effect? I don't get where the steric factor comes into play, as both the leaving group and the base are non-bulky. $\endgroup$
    – AADETYA JAISWAL
    Commented Apr 3 at 17:56
  • $\begingroup$ That doesn't mean the substitution around the site of reactivity isn't bulky itself... $\endgroup$
    – Zhe
    Commented Apr 4 at 13:19

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