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 Mastering 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) Example: CH2=CH2(g) + H2(g) → CH3CH3(g) Requires H2 at high pressure and catalyst Practical example: making margarine from seed oil C H H ----- H C 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 Give the name for the following compounds: CH3CH=C(CH3)2 b) CH2=C(CH3)CH=CH2 c) (CH3)2CHC=CH 2. Predict the product of HBr(g) + 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) = However, all C – C bonds are the same length (due to all sp2 hybridization which perfectly matches 120° bond angle for hexagon) H H H H H H

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 chlorobenzene methylbenzene = toluene 2-butylbenzene or 2-phenylbutane hydroxybenzene = phenol

Chapter 20 Organic Chemistry Aromatic compounds – cont. Disubstituted aromatics Number around ring starting with earlier (alphabetical) constituent Cl 3 2 name = 1-hydroxy-2-methylbenzene 4 OH 1 1, 2 disubstitution is also known as “ortho” 5 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 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 → 2. 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) Polar and acidic Amines (R3N) – note here one or two Rs can be Hs Polar and basic