Solutions_Manual_for_Organic_Chemistry_6th_Ed 200

Solutions_Manual_for_Organic_Chemistry_6th_Ed 200 - 9 -8...

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Unformatted text preview: 9 -8 contin ued (b) ( c) ( d) 9-9 i 9H H20 H3C-C-C::C-H NaNH2 H-C:::C-H Ph yCH3 Ph OH NaNH2 CH3-C:::C-H NaNH2 .- CH3-C:::C-CHCH2CH2CH3 H-C:::C-H CH3I 2) HCCH2CH2CH3 H20 OH NaNH2 CH3-C:::C-H NaNH2 H-C:::C-H 2) CH3I 2) CH3CCH2CH3 CH3-C:::C -CCH2CH3 CH3 H20 H (Br H H H � -ooH :� H -C=C-C-C�C� H -C �C-CH2CH2CH3 :OH C=C I) r Br H�oo B{) \CH2CH2CH3 1 t: H HI ) H HI H -C C-CH2CH3 HO-H H-C=C=C-CH2CH3 H-C=C-C-CH2CH3 U� H 1t :OH � H-OH H H-C=C=C-CH2CH3 H -C-C= C-CH2CH3 H-C-C=C-CH2CH3 H H H 1) 2) .. ----l.-� I o 1) 2) � 1) I 3) 1) II o 1) � � I 3) II o I . 1 1 1 1 .. \ ,1 _ � 1 =(= 00 00 - 00 --- � _ 1 _ o 1 Ii / _ 00 � 00 } - -0 0 1 00 � 1 } � U I I - 9 -1 0 To determine the equilibrium constant in the reaction: !J. G = RT In tenninal alkyne � internal alkyne !J.G - 1 7.0 kJ/mole Keg = (- 4.0 kcallmole) RT e = ( !1G) Keg e = ( -(-17,000) (8.314)(473) ) = e4.32 = 75 R = 8 .3 1 4 JIK-mole [internal] [terminal] 193 75 98.7% i nternal 1 .3 % tenninal ...
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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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