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CHEM 1040
Lori Jones

ORGANIC SYNTHESIS: O CH C3 Æ 3 Æ H3C NH CH 3 To produce an amide, we need an amine plus either an ester or an acid halide. To produce the amine from ethane: 1) CH 3H + 3l /h2 → CH CH Cl3+ H2l (substitution reaction or halogenation) (ethane) (chloroethane) 2) CH CH Cl + NH – → CH CH NH + Cl – (substitution reaction) 3 2 2 3 2 2 (chloroethane) (ethanamine/ethylamine) To produce an ester from ethane: 1) CH 3H + 3l /h2 → CH CH Cl3+ H2l (substitution reaction or halogenation) (ethane) (chloroethane) 2) CH CH Cl + OH → CH CH OH + Cl – (nucleophilic substitution reaction) 3 2 3 2 (chloroethane) (ethanol) 2– + 3) CH 3CH 2H + Cr O 2 7 /H → CH CO3 Ooxidation) (ethanol) (ethanoic acid) + 4) CH C3OH + CH CH OH3+ H2→ CH C(O)OCH CH +3H O (subst2tuti3n re2ction or (ethanoic acid) (ethyl ethanoate) condensation reaction) To produce an acid halide: 1) CH 3H + 3l /h2 → CH CH Cl3+ H2l (substitution reaction or halogenation) (ethane) (chloroethane) – – 2) CH 3CH C2 + OH → CH CH OH 3 Cl2 (nucleophilic substitution reaction) (chloroethane) (ethanol) 2– + 3) CH C3 OH2+ Cr O 2 7 /H → CH CO3 (Oxidation) (ethanol) (ethanoic acid) 4) CH C3OH + SOCl → CH 2(O)Cl +3HCl + SO 2 (substitution reaction) (ethanoic acid) (acetyl chloride) To produce the amide: O CH 3C(O)OCH CH 2 CH3CH NH 3 2 2 H3C NH CH 3 + HOCH CH2 3 (substitution reaction) O CH C(O)Cl + CH CH NH → H3C NH CH 3 + HCl (substitution reaction) 3 3 2 2 1 Review Questions from Last Class: 1. CH CH3CH )CHCH3 2 + ?? → CH CH(CH 3CH(Br)CH3 3 2. What is the most likely major product when 2-methylpentane is treated with Br /hv? 2 3. What is the product of the oxidation of 2-propanol? 4. What is the product of the reaction of the following with H O/H SO ? 2 2 4 CH 2 i) 1-methylcyclopentene ii) 5. What is the product of 1-bromo-3-methylbutane + NaOH? 6. i) ii) benzaldehyde + ethanol → 7. Which of the following are chiral? 2– 8. Draw the Lewis structure for C 2 and give the formal charges for each carbon atom. 9. Draw the Lewis structure for nitrite ion, give the formal charges and predict the shape. 10. How many sigma and pi bonds are present in H CSCN?3 11. A solution of bleach, sodium hypochlorite, labeled “0.030M NaClO”, had a pH of 10.00. Calculate K aor HClO. 12. Write NIE’s for the reactions that occur when perchloric acid and sodium hydroxide are added to a C 2H 5H C3/C H 2H 5uffe2. 13. Calculate the pH after the addition of 0.0250 moles of HCl to 1.00 L of a buffer solution of 0.0425 M acetic acid and 0.0596 M sodium acetate. Assume no change in volume. 14. For the titration curve below, determine the pK fob the base. A) 2.0 B) 3.0 C) 5.3 D) 7.0 E) 8.7 15. a) What is the pH at the stoichiometric point of 0.130 M HCOOH (aqtitrated with 0.130 M KOH (aq) b) Which indicator should be used for the titration above? A) methyl red (pK In= 5.0) C) bromothymol blue (pK = 7.In B) alizarin yellow R (pK = In.1) D) cresol red (pK =In.9) 2 Answers for the Review Questions: 1. H3C CH 3 CH3 CH3 CH 3 + HBr → H3C H3C + H3C CH2 → Br Reactant - alkene Iotermediate (3-methyl-1-butene) (2 carbocation) Product – alkyl halide (2-bromo-3-methylbutane) + This is an addition r eaction. The electrophile, H , opens up the pi bond of the alkene by bonding with two electrons. For the major product, the H goes to the side with the most H’s (to produce a methyl group on the end). We now have a s econdary carbocation intermediate due to the formal charge left on the carbon atom. The nucleophile, Br , is then attracted to the 2 carbocation to complete the structure. 2. H3C CH 3 H3C CH3 H3C • + Br2/hv → CH3 CH CH 3 3 + HBr + Br ⋅ → Reactant - alkane Intermediate CH3 (2-methylpentane) o Br (3 free radical) Major Product – alkyl halide (2-bromo-2-methylpentane) This is a substitution reaction. Light (or heat) breaks the bromine molecule into two free radicals of bromine. One radical removes a H atom from the al kane. The preference is a tertiary hydrogen over a secondary or primary. This produces HBr and a tertia ry free radical. The sec ond bromine radical then collides with the tertiary free radical to produce th e major product. Other produ cts are formed in this reaction, but because the tertiary free radical is the most stable form, this will then produce the major product 2-bromo-2-methylpentane. (Other monobr ominated species would include: 1-bromo-2- methylpentane (intermediate = 1 o free radical), 3-bromo-2-methylpentane (intermediate = 2 free radical), 2- o o bromo-4-methylpentane (intermediate = 2 free radical), and 1- bromo-4-methylpentane (intermediate = 1 free radical), – a total of 5 possible monobrominated species) 3. Reaction (oxidation of 2-propanol): O OH 2– + + Cr2O 7 /H → H C CH 3 3 H 3 CH 3 Product - ketone Reactant – 2 oalcohol (propanone or acetone) (2-propanol) 3 4. i) H3C OH H3C + H2O/H 2O →4 Product – cycloalkanol Reactant – cycloalkene (1-methylcyclopentanol) (1-methylcyclopentene) ii) H2C H3C OH + H O/H SO → 2 2 4 Product – cycloalkanol Reactant – cycloalkane + double bond (1-methylcyclopentanol) (methylenecyclopentane) 5. Br CH 3 HO CH 3 CH 3 + NaOH → CH 3 + NaBr Reactant – alkyl halide Product – alcohol (1-bromo-3-methylbutane) (3-methyl-1-butanol) 6. i) Step #1: OH Na O CH 3 + 2Na(s) → H C CH 2 3 3
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