Comparing Solvent Effects on SN1 and SN2 Reactions

Question

The major product of the synthesis, 3-methoxy-3-methylheptane, is produced by an SN1 mechanism. Complete the reaction mechanism by adding curved arrows showing electron flow to demonstrate the formation of the major product (3-methoxy-3-methylheptane).

Answer 1

What's the reaction? Is it CH₃CH₂C(CH₃)(Cl)CH₂CH₂CH₃ + CH₃OH and maybe heat? That's called a solvolysis reaction.

 

In the first step you'd draw a curved arrow from the bond between C-Cl, ending (arrow head) on the Cl. Then the Cl comes off as Cl^- , leaving you with a carbocation, with a + charge on that C atom.

 

Now show a curved arrow from one of the CH₃OH oxygen lone pairs to the space between the O and the + C atoms. What you get then is a completed bond between the C and O, but now the + charge is on the O.

 

Now show another CH₃OH molecule near that + charge, with an arrow from one of its O lone pairs with the arrow head between that O and the H attached to the carbocation. At the same time, draw an arrow from the bond between H and the + O, ending on the + O.

 

In the final stage you have the product, and the + charge is now on the O of that second CH₃O⁺H₂ molecule.
 
 
S_N1 mechanisms produce racemization of configuration, so if you had the R-enantiomer (or S) to start with, you'd get an almost equal mixture of the R and S-enantiomers at the end. Although the intermediate carbocation is flat, and the CH₃OH can attack from either side, there is more steric hindrance from the Cl^- leaving group on one side, so attack from the opposite side to the Cl^- is slightly preferred. That's why you would get possibly slightly more of the inverted product. Often the answers to these questions simply require "racemization" as the answer, however.