CH3-C(=O)-CH=CH2 gives two different products when treated with NaCN,HCN;
1)product -> H3C-C(OH)(CN)-CH=CH2 at temperature 5-10 degree C
2)product -> H3C-C(=O)-CH2-CH2-CN at temperature 80 degree C.
-Explain.
1
Expert's answer
2014-05-06T07:35:36-0400
The first reaction product occurs under kinetically controlled conditions. Its formation is reversible, so this product can be isolated only at low temperatures. The rate of the second product formation is lower, yet it is more thermodynamically stable. Therefore, under the thermodynamically controlled conditions this product will be accumulated in the reaction mixture.
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