Transcript Chem. 31 * 9/15 Lecture

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