Examination of the Fischer esterification mechanism continues

Examination of the Fischer esterification mechanism continues. Overall reaction (ungraded): Part 1 involved MeOH addition to form the key tetrahedral intermediate. Part 2 involves loss of H2O from the tetrahedral intermediate to form the ester. Intermediate species (from Part 1): Add two curved arrows for the next step. For all remaining steps in the mechanism: Draw each species (organic and inorganic) resulting from the previous step. Each step will include all of, and only, the atoms given in the first step (i.e., do not use extra Make sure you include all species produced by the previous step. Be sure to: Draw Cl-. Include all nonbonding electrons on O's and Cl to give complete octets. Include the charge, lone pair, and two H's on the bound water O (i.e., protonated alcohol). Use two curved arrows to show the collapse of the tetrahedral intermediate: an OH lone pair moves to form a C=O double bond while a molecule of water leaves, thereby maintaining an octet around that C. (Depending on how you drew the mechanism, the drawing area below might not be needed. Do not draw the final products.)