Answers to Problem Set 5

4.7.

a.  Rate = (k₁k₂/k_-1){[S][R₃N]/[R₃NH⁺]}

Apply SSA to HCCBr, and equilibrium condition to the vinyl anion.

b.  You need to define rate. 

In terms of the individual products:

d[ROH]/dt = {k₁k₂[S][H₂O]}/{k_-1[^-OPNB] + k₂[H₂O] + k₃}

d[Alkene]/dt = {k₁k₃[S]}/{k_-1[^-OPNB] + k₂[H₂O] + k₃}

Note that if k_-1[^-OPNB] is presumed to be small, these add together and simplify to:

-d[ROPNB]/dt = k1[S].

(S is substrate in this example.)

c.  If the sigma complex is a steady-state intermediate, then k₁ is slow.

d[P]/dt = {k₁k₂[S][Br₂]}/{k_-1[Br^-] + k₂}

Simplification requires estimating the relative magnitude of k_-1[Br^-] and k₂; the rapid equilibrium of the final step indicates that [Br^-] is always small and therefore k₂ dominates.  Thus, the simplified rate expression is:

d[P]/dt = k₁[S][Br₂]

d.  The restriction to make no assumption about relative magnitude of rate constants requires that you analyze both the possibility that k₁ is slow and that k₂ is slow.

k₁ slow:  d[P]/dt = {k₁k₂[S][Cr]}/{k_-1 + k₂[Cr]} (by application of the SSA to the enol)  If k_-1 is small, this reduces to k₁[S].

k₂ slow: d[P]/dt = {(k₁k₂/k_-1)[S][Cr]}  Note that in the above expression, assuming k₂ is small (our condition here), we get the same expression.

(Again, S is the keto substrate.)

4.11. The minimal interpretation of each piece of data is:

1. Proton transfer is not the source of the second hydrogen in the benzyl alcohol CH₂ group; it is likely coming from a second molecule of benzaldehyde.
2. Either both products exchange rapidly with water (benzoic acid does, but benzyl alcohol does not), or a common intermediate does. Reversible hydration of benzaldehyde accomplishes this transformation prior to reaction.
3. The transition state is composed of two molecules of benzaldehyde and one of hydroxide. However, the structure of the transition state must be derived from other data.
4. There is a significant increase in electron density at the carbonyl carbon during the rate-limiting step. This would be accomplished by a nucleophilic addition of some sort.
5. The significant inverse solvent KIE implies proton transfer from solvent must occur prior to the transition state.

A model for the transition state consistent with this data is:

5.6. The data provided can be minimally interpreted as follows:

1. The composition of the transition state is protonated reactant, or transfer of the proton to the reactant.
2. This is a kinetic isotope effect, which I have not discussed in class. For A, this is a primary KIE, indicating proton transfer is the rate-limiting step. For B, it is a solvent isotope effect (D⁺ is a stronger acid than H⁺); the observation indicates the rate-limiting step is loss of water. You should be able to construct a mechanism without this data.
3. Cleavage of the C-O bond cannot occur before the RDS for B.
4. This rules out reaction of a common intermediate as the RDS.

The alcohol must go through the cyclopropylcarbinyl species in order to avoid a primary carbocation. An alternative is to eliminate to the Z diene, then isomerize as above, since the Z diene isomerizes faster than the alcohol.

5.8. This system is controlled exclusively by the timing of nucleophilic attack relative to formation of tight ion pairs, solvent-separated ion pairs, and free ions. Species 2 arises from neighboring group participation by the double bond; attack of the nucleophile on the cyclopropane (at the original C-OTs site) gives 1 with retention. Attack of the tight ion pair (prior to interaction with the double bond) gives inversion. From the rate data, we can surmise that the RDS is always generation of the TIP.

6.13.  Addition of Br⁺ can occur from either face; anti addition of bromide leads to the two expected trans products.  If Br⁺ addition occurs anti to the acetate, intramolecular neighboring group participation leads to formation of two unexpected cis dibromides via a bicyclo-[4.3.0] system. 

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Last updated: 10/09/03