Chem 351 Fall 2001 Final Mechanism

Part 7: MECHANISMS

I This reaction converts an alcohol into an alkyl chloride, usually via an SN2 type mechanism. Remember -OH is a poor leaving group. Here the O becomes attached to the S and eventually leaves as part of sulfur dioxide, SO₂when the chloride ion attacks as a nucleophile.

Mechanism for the reaction of an alcohol with thionyl chloride giving an alkyl chloride

II Let's look at the reaction, the starting material is a secondary alcohol, we treat it with sulfuric acid / heat to get an alkene.... an elimination. Since it is an alcohol, we are probably looking at an E1 reaction.... the carbocation intermediate is confirmed by the rearrangement that is apparent due to the change in the carbocation skeleton. First make the OH a better leaving group by protonation, then lose the water giving the carbocation. A 1,2-alkyl shift gives the new skeleton. The red numbers should help you keep track of this change, note the location of the +ve charge in the rearrangement product. Finally, water is shown reacting as base, removing a proton to give the alkene.

E1 with carbocation rearrangement

III

This is based on the substitution experiment and the theories behind SN1 and SN2 reactions. First challenge is to draw the structures correctly (see right). Both systems are alkyl halides with reasonable leaving groups (halide ions).
In SN2, where the Nu attacks and displaces the leaving group and the same time. The key factor is steric hindrance. Since both these systems are primary, then the approach of the Nu is unhindered and reaction is fast.
In SN1, the rate determining step is the loss of the leaving group creating the carbocation. The key factor is the stability of the carbocation. Although both are primary (which is normally unfavourable), here both are resonance stabilised.

IV
Here we have an cyclic alkyl bromide reacting with a strong base to give an alkene.... an elimination. Alkyl halides react with strong bases via the E2 mechanism. The key factor in an E2 is that the usual scenario has the C-H and C-LG bonds antiperiplanar (180^o) with respect to each other. This implies that the LG and the H are both in axial positions. Note that since the t-butyl substituent is large and has such a strong preference to be equatorial, that the rings can't ring flip.

In the first case, the trans isomer, the bonds that are antiperiplanar to the C-Br bond are the two C-C bonds shown in green. Since the ring flip is prevented the reaction occurs much more slowly, probably via an E1 reaction.

In the second case, the cis isomer, there are two C-H bonds that are antiperiplanar to the C-Br bond (shown in green) so the E2 is facilitated and occurs rapidly.