Practice problem 11.11 what stops R-O-R from forming at the same rate as the desired product?
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In order to understand why this is OK, you have to understand the mechanism by which this occurs. It is essentially an $\mathrm{S_N1}$ reaction that is most favoured by the conditions (acidic + polar protic solvent).
So, for the formation of an ether $\ce{^tBu-O-R}$ where $\ce{R}$ is a primary alkyl group:
- competing formation of $\ce{R-O-R}$ is slow because primary carbocations are very unstable, and
- competing formation of $\ce{^tBu-O-^tBu}$ is slow because $\ce{^tBuOH}$ is a poor nucleophile (tertiary alcohols are very bulky).
answered Aug 28, 2017 at 15:05
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$\begingroup$ What stops an SN2 mechanism from operating to form ROR? $\endgroup$– xasthorCommented Aug 28, 2017 at 15:23
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$\begingroup$ @xasthor Water is a bad leaving group for an SN2 reaction – the tert-butyl cation is stable enough to promote loss of water (in an SN1 sense), then ROH can attack (tBuOH is a terrible nucleophile).. there really is only 1 favoured outcome in this reaction.. $\endgroup$ Commented Aug 28, 2017 at 18:12
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1$\begingroup$ @NotEvans, H2O+ is not a bad leaving group. Besides, if it was a bad leaving group, it would not only affect the SN2 rate but also affect the SN1 rate $\endgroup$– xasthorCommented Aug 29, 2017 at 9:50
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$\begingroup$ SN2 displacement of a primary alcohol (as in an SN2) without prior activation is not a favourable/well known reaction. Look in any organic chemistry textbook (I suggest Clayden...)- it's well explained and it may help to clarify the basics for you :) $\endgroup$ Commented Aug 29, 2017 at 10:28
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1$\begingroup$ @xasthor I think what you are saying is that after protonation, it is possible for $\ce{ROH2+}$ to react in an SN2. Yes, it is possible - but then there is no good nucleophile (alcohols are really quite poor). Compared to the SN1 reaction which has everything going for it (very polar solvent + tertiary cation), I don't think it's likely that this SN2 can be competitive in terms of rate. Maybe, at most, you get some side product. If you compare to generic ROH + R'OH, which simply produces a hopeless mixture... this is much better. $\endgroup$ Commented Aug 30, 2017 at 14:33