Solutions_Manual_for_Organic_Chemistry_6th_Ed 14

Solutions_Manual_for_Organic_Chemistry_6th_Ed 14 - 1 -8...

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1-8 continued H H h (b) (c) (d) (e) \ + C=C±N²O / 1 1 h H :0: . \ + C³ C´N´O: ± ² / ±1 1 h H :0: \ ± + CµC¶N´O: / ² 1 h h :0: major maJor minor T·EsE two foRms ·ave equivalent energy And are maJor becAuse t¸ey ·ave full octets, more bonds¹ And Less c·arge separation t·An t·e minor contrIbutor. ± h±C³OºH » ³ hºC±O´h 1 + ³ ¼ h mA½or M¾or (octets¿ more boÀds) + ± h± C=N²N: ´ µ h±C±N N: 1 ± h h maJor (negative c·Arge minor on eLectronegative atom) H´CµC N: ± ³ h± CÁC²N: 1 ² H H minor mAJor (negative c·arge on electronegative atom) ± ²
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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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