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CHEM 1040
Lori Jones

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Answers for the Final Review Questions: 1. H C CH3 CH 3 3 CH CH3 + HBr → 3 H3C H3C H3C CH2 + → Br Reactant - alkene Intermediate (3-methyl-1-butene) (2 carbocation) Product– alkyl halide (2-bromo-3-methylbutane) This is an addition reaction. The electrophile, H, opens up the pi bond of the alkene by bonding with two electrons and the H goes to the side with the most H’s (to produce a methyl group on the end). We now have a secondary carbocation intermediate due to the formal charge left on the carbon – o atom. The nucleophile, Br , is then attracted to the 2 carbocation to complete the structure. 2. H C CH H C CH H C 3 3 3 3 3 + Br 2hv fi • CH 3 CH CH3 3 + HBrfi Reactant - alkane Intermediate CH 3 (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 bromine radicals. One radical removes an H atom from the alkane. The preference is a tertiary hydrogen over a secondary or primary. This produces HBr and a tertiary free radical as an intermediate. The second bromine radical then reacts with the tertiary free radical to produce the major product. Other products 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 monobrominated species would include: 1- bromo-2-methylpentane (intermediate = 1 free radical), 3-bromo-2-methylpentane (intermediate = 2 o o free radical), 2-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– + + Cr 2 7 /H → H3C CH 3 H 3 CH 3 Product - ketone o Reactant – 2 alcohol (propanone or acetone) (2-propanol) 1 4. i) H C 3 OH H3C + H2O/H 2O 4 Product – cycloalkanol Reactant – cyclalkene (1-methylcyclopentanol) (1-methylcyclopentene) ii) H2C H3C OH + H 2/H 2O →4 Product – cycloalkanol Reactant – cyclalkene (1-methylcyclopentanol) (methylenecyclopentane) 5. benzaldehyde + ethanol → hemiacetal O H H HO O + CH 3H O2 → CH 3 Reactant – aldehyde Product – hemiacetal (benzaldehyde) (1-methylcyclopentanol) 6. Which of the following are chiral? 1,2-dichlorbutane 2-aminopropanoic acid H H N H3C OH Cl Cl CH 3 * CH H * O * 3 * H H H H H HO H H H O H Achiral Chiral Chiral Chiral 4-methyl-1-cyclohexanol is achiral (no carbon has four different groupsdue to the symmetry within this molecule – subsitutents are directly across from one another). 2-methyl-1-cyclopentanolcontains two asymmetric carbons. One carbon is bonded to –OH, –H, –CH 2 (left-hand side of ring) and –CH(3H )(right-hand side of ring). The second carbon is bonded to –CH , –H, –CH – (right-hand side of ring) and –CH(OH)(left-hand side of ring). 3 2 For the last two molecules, the four different groups have been circled to show the asymmetric carbons. 2 7. 1) Synthesis of acetone from propane to produce a ketone from an alkane, we must first convert to the alkane into a 2º alkyl halide and then substitute the halogen with an alcohol group. Finally we can then oxidize the 2º alcohol. a) CH 3H CH2+ C3 /hν → 2H CH(Cl)CH +3HCl 3 (substitution rxn or halogenation) (propane) (2-chloropropane – major product) b) CH C3(Cl)CH + Na3H → CH CH(OH)CH + 3aCl 3 (substitution reaction) (2-chloropropane) (2-propanol) c) CH CH(OH)CH + Cr O 2–/H → CH C(O)CH (oxidation) 3 3 2 7 3 3 (2-propanol) (propanone or acetone) (2 H’s removed to make room for π bond) (2) Synthesis of N-ethylethanamide from ethane: O CH 3H à 3 à H3C NH CH3 To produce an amide, we need an amine plus either an ester or an acid halide. To produce the amine from ethane: 1) CH CH + Cl /hν → CH CH Cl + HCl (substitution rxn or halogenation) 3 3 2 3 2 (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 CH + Cl /hν → CH CH Cl + HCl (substitution reaction or halogenation) 3 3 2 3 2 (ethane) (chloroethane) 2) CH CH Cl + OH → CH CH OH + Cl – (nucleophilic substitution reaction) 3 2 3 2 (chloroethane) (ethanol) 2– + 3) CH C3 OH2+ Cr O 2 7 /H → CH COO3 (oxidation) (ethanol) (ethanoic acid) + 4) CH C3OH + CH CH OH3+ H 2 CH C(O)OCH CH + 3 O 2 3 2 (substitution reaction or (ethanoic acid) (ethyl ethanoate) condensation reaction) To produce an acid halide: 1) CH CH + Cl /hν → CH CH Cl + HCl (substitution reaction or halogenation) 3 3 2 3 2 (ethane) (chloroethane) – – 2) CH 3H C2 + OH → CH CH OH +3Cl 2 (nucleophilic substitution reaction) (chloroethane) (ethanol) 2– + 3) CH C3 OH2+ Cr O 2 7 /H → CH COO3 (oxidation) (ethanol) (ethanoic acid) 3 4) CH C3OH + SOCl → CH C(2)Cl+ HCl3+ SO 2 (substitution reaction) (ethanoic acid) (acetyl c
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