# Why can unsymmetrical tert-butyl ethers be prepared by dehydration of the constituent alcohols?

Practice problem 11.11 what stops R-O-R from forming at the same rate as the desired product?

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).
• @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. Aug 30 '17 at 14:33