Write a detailed mechanism for the following reaction. Draw step one of the mechanism which forms the initial enolate. Draw only the enolate form of the reaction, NOT the carbanion. Include lone pairs and formal charges in the mechanism. Draw out explicitly only the ONE hydrogen that is removed in this mechanistic step. DO NOT USE ABBREVIATIONS SUCH AS OEt for OCH2CH3.

the answer is c i took the test

answer carbon and three oxygen atoms;

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o