1. Chem. 1B – 12/6 Lecture

2. 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

3. 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

4. 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

5. 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

6. 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

7. 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)

8. 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)

9. 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

10. 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

11. 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

12. 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) + ?

13. 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

14. 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

15. 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

16. 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

17. 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”

18. Chapter 20 Organic Chemistry
Aromatic compounds – cont.
Substituent Naming (other than alkyl)
Substituent
Name
-Br
Bromo-
-Cl
Chloro-
-OH
Hydroxy-
-NH2
Amino-

19. 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

20. 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

21. 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

22. 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