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Chem. 1B – 12/6 Lecture Announcements I • Exam #3 - Results – Average = 77.8% – Best average so far for Chem 1B – Only a few questions had poor performance (reducing agent and bonus) – Performance on work out problem was not great – Overall class average now ~ 70% Score Range # Students 90-104 28 80s 43 70s 26 60s 18 50s 7 <50 5 Solutions posted Announcements II • Lab – Lab Final Wednesday and Thursday in Lab – On Material Since Lab Midterm • Mastering – Ch. 20 assignment (Organic Chemistry) due 12/10 Announcements III • Will post a practice quiz on organic chemistry • Final Exam – Thurs., 12/15 12:45 to 2:45 – 40 multiple choice questions – no work out problem – About 4 questions each for Ch. 14, 15, 17, and 24 and about 8 questions each for Ch. 16, 18, and 20 – Less calculation intensive than exams 1 and 2 Announcements IV • Thursday’s Lecture – will have teaching evaluations at end • Today’s Lecture – Organic Chemistry (Ch. 20) • Alkynes (triple bonds) • Reactions • Aromatics • Functional Groups Chapter 20 Organic Chemistry • Alkynes – Contain at least 1 carbon-carbon triple bond – Naming (replace –ane ending with –yne with number referring to end of triple bond closest to the #1 carbon) – Triple bond uses sp hybridization and leads to a linear structure – Example: • • CH3CH=CHCH3 is 2-butyne (linear) Carbon skeleton structure Chapter 20 Organic Chemistry • Alkynes – cont. – Alkynes are considerably more energetic than alkenes – Used less by organic chemists (harder to synthesize, fewer uses) – Used by Dr. Spence in ene-diyne compounds (generates cyclic radical) Chapter 20 Organic Chemistry • Some Basic Hydrocarbon Reactions – Combustion (all types, but alkynes generate more energy than alkenes and alkanes generate the least energy) HxCy + O2(g) → CO2(g) + H2O(g) (unbalanced, but be able to balance) – Halogenation of Alkanes • • Example CH3CH3(g) + Cl2(g) → CH3CH2Cl(g) + HCl(g) Products are typically more stable than reactants (C-X bonds are pretty stable) Chapter 20 Organic Chemistry • Some Basic Hydrocarbon Reactions – Halogenation of Alkanes – cont. • Occurs by “free radical” mechanism: Cl2(g) + heat or light → 2Cl•(g) (where “•” shows free radical) Cl•(g) + CH3CH3(g) → HCl(g) + CH3CH2•(g) CH3CH2•(g) + Cl2(g) → CH3CH2Cl(g) + Cl•(g) • Free radical reactions are hard to control, so will also produce related compounds (e.g. CH2ClCH2Cl(g)) • Syngas reactions also are free radical reactions Chapter 20 Organic Chemistry • Some Basic Hydrocarbon Reactions – Alkene Reactions (addition to double bond) • Hydrogenation (also works with alkynes) – – – H H H H H Example: CH2=CH2(g) + H2(g) → CH3CH3(g) Requires H2 at high pressure and catalyst Practical example: making margarine from seed oil H ----- H C C H CH H C H H Chapter 20 Organic Chemistry • Some Basic Hydrocarbon Reactions – Alkene Reactions (addition to double bond) • Halogenation – – – – Example: CH3CH=CH2(g) + Cl2(g) → CH3CHClCH2Cl(g) Can also use HX (hydrohalide gas) as reactant In this case both H and X are added to alkene carbons CH3CH=CH2(g) + HCl(g) → CH3CHClCH3(g) + CH3CH2CH2Cl (g) both products possible, but only one observed Chapter 20 Organic Chemistry • Some Basic Hydrocarbon Reactions – Alkene Reactions – Halogenation – cont. • • • Why is only CH3CHClCH3(g) observed? Markovnikov’s Rule (H added to side with most Hs) What is expected product of HCl(g) + ? Chapter 20 Organic Chemistry • Questions 1. Give the name for the following compounds: a) CH3CH=C(CH3)2 b) CH2=C(CH3)CH=CH2 c) (CH3)2CHC=CH 2. Predict the product of HBr(g) + 3. What type of product is produced by hydrogenation of alkenes? a) alkanes b) alkynes c) dienes d) halocarbons Chapter 20 Organic Chemistry • Aromatic compounds – Benzene, the simplest aromatic compound – Formula = C6H6 – Structure (see below) H H H H = H • H However, all C – C bonds are the same length (due to all sp2 hybridization which perfectly matches 120° bond angle for hexagon) Chapter 20 Organic Chemistry • Aromatic compounds – cont. – While adding double bonds makes compounds less thermodynamically stable, benzene and other aromatic compounds (compounds containing benzene ring) are relatively stable both thermodynamically and kinetically – Some due to “resonance stabilization” – Due to stability, reactions are different than alkene reactions Chapter 20 Organic Chemistry • Aromatic compounds – cont. – Substituted aromatics (benzene ring plus substituent) – Examples of monosubstituted aromatics Cl OH methylbenzene = toluene chlorobenzene hydroxybenzene = phenol 2-butylbenzene or 2-phenylbutane Chapter 20 Organic Chemistry • Aromatic compounds – cont. – Disubstituted aromatics – Number around ring starting with earlier (alphabetical) constituent Cl 3 2 4 OH 5 1 name = 1-hydroxy-2methylbenzene 1, 2 disubstitution is also known as “ortho” 6 hydroxy before methyl so right C = #1 name = ? 1,3-disubstitution = “meta” and 1-4-disubstitution = “para” Chapter 20 Organic Chemistry • Aromatic compounds – cont. – Substituent Naming (other than alkyl) Substituent Name -Br Bromo- -Cl Chloro- -OH Hydroxy- -NH2 Amino- Chapter 20 Organic Chemistry • Aromatic compounds – cont. – Reactions: substitution reactions – Examples: Cl FeCl3 + Cl2 + HCl (catalyst) Cl replaces H AlCl3 + CH3Cl + HCl (catalyst) CH3 replaces H Chapter 20 Organic Chemistry • 1. a) b) c) 2. More Questions Predict the product of the following reactions: (CH3)CH=CH2 + Br2 → trans (CH3)CH=CH(CH3) + H2(g) → (CH3)2C=CH(CH3) + HBr → Give the names of the following compounds: NH2 Br Chapter 20 Organic Chemistry • Functional Groups • What you need to know: • • • Class name Identification Polar groups • Alcohols ROH (R = rest of molecule) • • OH group is polar Example: CH3OH (methanol) • Ethers ROR • • Example (CH3CH2OCH2CH3 = diethyl ether) Not as polar as alcohols Chapter 20 Organic Chemistry • Functional Groups – cont. • Ketones (R-C-R) O • • example: acetone somewhat polar • Carboxylic Acids (R-C-OH) O • Polar and acidic • Amines (R3N) – note here one or two Rs can be Hs • Polar and basic