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5-chloro-1, 3 - dimethylcyclohexene is the substrate. I understand that hydroboration oxidation occurs in 2 steps the first being addition of Boron as an electrophile and the second being oxidation by peroxide ion. Since the solution now contains Hydroxide ion now this will attack the carbon bonded to chlorine by SN2 mechanism. But in my book , in the final product chlorine is not replaced with OH. I don't understand why.

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    $\begingroup$ You assume that the peroxide/base reaction and an SN2 reaction have similar rates. Moreover, E2 elimination is also a real possibility rather than displacement of chlorine. $\endgroup$ – user55119 Mar 9 '18 at 22:02
  • $\begingroup$ What solvent is the reaction run in? $\endgroup$ – Waylander Mar 9 '18 at 22:11
  • $\begingroup$ @Waylander I don't know I just started learning organic chemistry and this question appears with the substrate and an arrow on which B2H6 and OH-, H2O2 are written. Maybe the question is ill posed. $\endgroup$ – vishal mishra Mar 10 '18 at 8:37
  • $\begingroup$ @user55119 2° carbon and SN2/E2 conditions elimination product will be major right? Why this was down voted I will edit the question to make it clear. $\endgroup$ – vishal mishra Mar 10 '18 at 8:39
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    $\begingroup$ My point, as Waylander has noted, is that the basic peroxide oxidation is faster (often needs cooling) than an SN2 reaction with hydroxide. The SN2 reaction on a halocyclohexane will compete with an E2 reaction. It's all about conditions (relative rates). $\endgroup$ – user55119 Mar 10 '18 at 16:12
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OH- is not that great a nucleophile for a secondary chloride.

If we assume an ethereal solvent, most likely THF, much of the OH- will be in aqueous phase and not available to react with the product in the organic phase. If the reaction is kept cool (as it should be) E2 is unlikely to be a major pathway hence the product is 5-chloro-1,3-dimethylcyclohexan-2-ol

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  • $\begingroup$ Why this reaction should be kept cool? $\endgroup$ – vishal mishra Mar 10 '18 at 11:33
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    $\begingroup$ The oxidative work up is exothermic. If it gets too hot pathways like E2 or OH- attack can become relevant. $\endgroup$ – Waylander Mar 10 '18 at 11:44

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