lab 10 questions

lab 10 questions - methyl groups' bond is stretched in...

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1. The reactions in part one yield non-soluble alkyl halides in acetone. If water is present, the product of NaBr/Cl becomes soluble in the solvent and therefore the reaction reaches equilibrium. While in acetone the product precipitates and the reaction continues to shift to the right. 2. By adding heat to the reaction the equilibrium would be shifted to the right. Therefore, a greater yield of product would be observed. 3. Both compounds are secondary alkyl halides, and therefore undergo an Sn2 reaction in acetone. 1-bromopropane because in its transition state, the two separate methyl groups become planar. This is still the case in the case of bromocyclopropane, however the two
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Unformatted text preview: methyl groups' bond is stretched in order to make the transition state planar. This stretching of the bond requires energy and therefore slows the reactions. 4. see attached diagram on page 4 5. Sn2: (CH3)3-COH + K (metal) --> (CH3)3-OK (potassium tert-butoxide) (CH3)3-COK + C2H5-Br --> (CH3)3-C-O-C2H5 Sn1: (CH3)3-C-Br + 2 EtOH --> (CH3)3-C-OC2H5 + Br- + H2O-C2H5+ (protonated ethanol) (CH3)3-C+ is formed. Ethanol then acts as a nucleophile… (CH3)3-C-O+(H)-C2H5 then a proton transfer yields: C2H5-OH2(+). .....
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This note was uploaded on 09/18/2011 for the course CHEM 3AL taught by Professor Li during the Spring '08 term at Berkeley.

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