Solutions_Manual_for_Organic_Chemistry_6th_Ed 21

Solutions_Manual_for_Organic_Chemistry_6th_Ed 21 - 1 -1 8...

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Unformatted text preview: 1 -1 8 continued (d) - .. � 0· Na+ :O - H + H-S-H .. stronger acid stronger base (e) H � /' I • •CH3 - �� H + C H3-O: stronger base H stronger acid + " _ equilibrium favors PRODUCTS __ --- .. H - O - H + Na+ : S-H conjugate acid conjugate base weaker base weaker acid CH3 - � - H H conjugate base weaker base + + .. --- CH 3 - O-H •• conjugate acid weaker acid • :•0: � I ..... CH3-C = •O• . conjugate base •• weaker base - equilibrium favors PRODUCTS stronger acid stronger base conjugate acid weaker acid equilibrium favors PRODUCTS i :0: II • C H 3 -C - •.0: . (g) :0: II .� CH 3 - C - O - H .. weaker acid ~ + :0: - . . II : O-S-CH 3 • • II :0 : - ... ..... ..f--l� equilibrium favors REACTANTS :0 : II 0 ==S-CH3 •• I :0 : weaker base -_ ... .. ..f--... I • 0•==S-CH3 • • II :0 : :0:- } t :0 : II H - O-S-CH3 • • II :0 : conjugate acid stronger acid 1-1 9 Solutions for (a) and (b) are presented in the Solved Problem in the text. Here, the newly formed bonds are shown in bold. H I • /" H-B-H (c) H - B - H + CH 3-O• - CH 3 " 1'1+ CH 3-O - CH 3 acid H �base f--l� _ :0: : 0: II I C H 3 -C -�: ...... ... CH 3-C = � conjugate base stronger base (d) coI�:O - H I: CH 3-C - H + acid •• •• base .. :0: I CH 3-r - H :O - H 11 •• ...
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This note was uploaded on 02/27/2010 for the course CHEM 140 taught by Professor Wade during the Spring '10 term at Whitman.

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