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Unformatted text preview: 1 3-23 Ha = [ 2.4 (singlet, IH) Hb = [ 3¡4 (doublet, 2 H¡ He = ¡ ¢ .8 (multipLet, 1 H¡ Hd = [ 0£9 (doUbLet, 6h¡ ¤ 3 ¥24 d CH3 H2 I H-O¢C £C-H I Q c b (a) The formUla C4h802 has one ele ment oF un satuRation . The iH singlet at [ ¢ 2. ¢ inDicates caRBoxylIC acid. The IH multiplet and the 6H doublet scReam isoPRoPyL GRoUp. ° Ch3• Ii Ch i CH¤C¥OH (b) The foRmuLa C9HloO has five elements of uNsatuRation. The 5 H patte¦ Between ¡ 7¡2§7 .4 indicates monosubstituted Benzene. The peak at [ 9.85 is unmistakably an aldehyde, t¨ying to Be a t©pLet ªecaUse «t is Weak¬y coupled o an adj acen c¦2. T he tW o tRiPlets at ¡ 2.7®3.0 aRe adj acen C§2 gRo¯ps . h2 ° V ¨ , ¨ J H I H2 h (c) The FoRmULa CS©802 has wo elements oF UnsatURatioN. A 3 § singLet at [ 2¡3 °s p±oBaBLy a CHª next to caRbonyl. The 2H quaRte² and the 3H t©pLe² are ceRtaIN to Be ethyL; with the Ch2 a ¡ 2.7, this alsO aPPeaRs to be next to caRBonyl....
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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.
- Spring '10