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Chemistry
Why Does Compound C Exhibit a Higher Rate of Solvolysis Than Compound D?
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[QUOTE="James Pelezo, post: 5488335, member: 554167"] Sorry, I missed the solvent in the original post;i.e., 50%EtOH/HOH solution. Now, I'm not sure why you are defining the solvent as a substrate. Typically in chemistry, the substrate is the object (compound) of interest in the reaction process. These would be the unsaturated halides shown in the original post. The solvent (again, I missed in the original post) is the 50:50 EtOH/HOH mixture. The presence of the ethanol is there only to increase the solubility of the unsaturated halides as they react with the water. I tend to agree with your conclusion that the Phenyl substituted structure would function to destabilize the Carbon - Chloride bond more than the 1-Chloro-3-methyl-2-butene which shows no resonance. This is an interesting reaction and because the halogen is attached to a 1[SUP]o[/SUP]-carbon, I'd assume it would proceed by an Sn2 process. I've run t-Butyl Chloride vs t-Butyl Bromide in a 50:50 Isopropyl Alcohol/Water mix and the t-BuBr rate > t-BuCl rate (k(t-BuBr) ~ 1.5 x 10[SUP]-[/SUP][SUP]4[/SUP] M⋅sec[SUP]-[/SUP]1 vs. k(t-BuCl) ~ 1.25 x 10[SUP]-[/SUP][SUP]5[/SUP]). However, the halogens are attached to 3[SUP]o[/SUP]-carbons and does proceed by an Sn1 process. The tertiary halide reactions in alcohol/water solvent is a relatively well known reaction and easy to run, but an Sn2 would be more difficult to control (in my humble opinion). I'd like to see the results of an actual trial, but I still agree with the phenyl substituted unsaturated halide as being the faster b/c of resonance. Good question. [/QUOTE]
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Why Does Compound C Exhibit a Higher Rate of Solvolysis Than Compound D?
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