CHEM 330. Problem set Indicate the polarity of the starred carbon atoms in the following molecules:

Transcription

1 CEM 330 Problem set 1 1. Indicate the polarity of the starred carbon atoms in the following molecules: (+) Br ( ) (+) ( ) 2. Identify appropriate synthons and suitable reagents for the formation of the starred bonds in the following molecules C N C N NaCN (-CN K) Cl or MgBr Et C 3 Et C 3 MgBr CEt C 3 CEt C 3 I Li CEt MeS MeS MeSNa 3. Estimate Δ for the following reactions (obtain BDE's from the Internet): C C σ 90 kcal/mol = π σ C C C= π C σ 60 C σ 100 C C σ 90 C σ 100 = π 100 C= π 100 σ 100 C= π 100 total Δ 50 kcal/mol σ C σ 95 kcal/mol C σ 60 C σ 100 σ 100 C σ 95 total Δ 35 kcal/mol

2 CEM 330 p. 2 PRBLEM SET 1 σ 104 kcal/mol C σ 95 kcal/mol 60 C σ 95 σ 104 C σ 95 total Δ 26 kcal/mol F F σ 136 kcal/mol F C σ 95 kcal/mol C F σ 108 kcal/mol 60 C σ 95 F σ 136 C F σ 108 total Δ 7 kcal/mol σ 50 kcal/mol C σ 60 C σ 100 σ 50 C σ 100 total Δ 90 kcal/mol C σ 95 kcal/mol σ 104 kcal/mol C σ C σ 95 σ 104 total Δ + 26 kcal/mol note: this estimate ignores a small, stabilizing resonance energy C σ 95 kcal/mol 2 2 x 2 2 σ 2 x 104 kcal/mol C σ C σ C σ 95 σ 104 C σ 95 σ 104 benzene res. E 36 total Δ + 16 kcal/mol SMe SMe S σ 87 kcal/mol C S σ 74 kcal/mol C σ 95 kcal/mol 60 C σ 95 S σ 87 C σ 74 total Δ 23 kcal/mol

3 CEM 330 p. 3 PRBLEM SET 1 4. Write an accurate mechanism for the following reaction: these conds. promote reversal of the Claisen rx. Et cat. EtNa Et, reflux then mild 3 + Et + Et Et Et Et Et + Et Et + Et Et 5. Provide an explanation for the following experimental observations and write accurate reaction mechanisms for each transformation: (a) treatment of compound 1 with excess NaEt causes isomerization to 2. non-enolizable β-ketoester: unstable in the presence of EtNa will undergo retro-dieckmann Et 1. Excess CEt NaEt, heat CEt 2 CEt Et CEt 1 2. Mild workup Et CEt Et CEt Et Et EtC Et (b) lactone 4 is obtained as one of the products of treatment of compound 3 with a catalytic amount of NaEt CEt cat. EtNa CEt Possibility 1: the mechanism of part a above could lead to: 3 Et CEt which then may advance to the final product in this way: 4 Et pka 17 CEt pka 25

4 CEM 330 p. 4 PRBLEM SET 1 Et CEt Et or CEt Et CEt CEt CEt CEt CEt 4 Et 6. Consider the following transformation Na 2 Et + EtNa Et + 2 Et A B (a) write a detailed step-by-step electronic mechanism for the reaction. Et Et Et Et Et Et Et Et Et Et Et (b) estimate the equilibrium concentration of the enolate of A in a solution that contains a 1 M instant concentration of both A and EtNa. If pka Et 25 and pka Et 17, then for the process: Et + Et Et + Et pkeq 8 and Keq 10 8 = [ Et ] [ enolate ] [ Et ] [ ester ] because the equilibrium is strongly shifted to the left, we conclude that [ Et ] and [ ester ] will

5 CEM 330 p. 5 PRBLEM SET 1 change very little from the original value of 1M when the system attains equilibrium. Therefore, the expression for Keq may be rewritten as: Keq = [ Et ] [ enolate ] [ Et ] [ ester ] 10 8 [ Et ] [ enolate ] 1 1 [ Et ] [ enolate ] Moreover, the stoichiometry of the reaction requires that 1 molecule of Et be formed for each molecule of enolate obtained during equilibration. Consequently, [ Et ] and [ enolate ] are numerically equal. So: Keq [ Et ] [ enolate ] [ enolate ] therefore [ enolate ] 10 4 M numerically equal (c) Assuming that the rate-limiting step for the conversion of A to B is the addition of a molecule of the enolate of A to an intact molecule of A, write a kinetic equation that describes the rate of the reaction as a function of a rate constant, k add d [ product ] rate = = k add [ enolate ] [ ester ] dt (d) Imagine replacing EtNa with t-buk in the above reaction, all other reaction parameters (concentration, solvent, temperature, etc.) being equal. Predict whether the reaction will proceed at a faster or a slower rate, and estimate the extent of such a rate increase or decrease. If we use t-buk (pka of t-bu 19) instead of EtNa (pka Et 17) the instant concentration of enolate will be greater: + Et Et + tbu pka 25 pka 19 pkeq 6 and Keq 10 6 : still unfavorable and shifted to the left the same logic developed above leads to the conclusion that: Keq [ enolate ] therefore [ enolate ] 10 3 M because the instant concentration of enolate is approximately 10 times greater than before, the reaction is predicted to occur at a rate 10 times faster (see equation developed in part c)