Exam 1 Review: Tuesday, 5/1 YH3069 from 7-9pm (Ray) Winter 2012 Problems: 4–7, 15, 22. Winter 2011...
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Transcript of Exam 1 Review: Tuesday, 5/1 YH3069 from 7-9pm (Ray) Winter 2012 Problems: 4–7, 15, 22. Winter 2011...
Exam 1 Review: Tuesday, 5/1 YH3069 from 7-9pm (Ray)
• Winter 2012Problems: 4–7, 15, 22.
• Winter 2011Problems: 1–11, 23, 28, 29.
• Spring 2011, Lecture 1Problems: 1–5, 12, 20.
• Spring 2011, Lecture 2Problems: 1, 2, 9–13, 15, 16, 20–22.
• Fall 2011Problems: 8, 20–24.
Suggested problems to attempt before attending the review:
Infrared Spectroscopy Part 1Lecture Supplement page 112
Midterm 1• 1 hour exam (in class on Friday, May 4)• Will cover:
– Intro & Review up through Carbohydrates (Mass Spectrometry and IR will not be on exam)
• Last name A-K A-P in CS50• Last name L-Z Q-Z in Franz 1260• Tools
– Pen and/or pencil– Eraser– Model kit– No calculators or cell phone
How should I study?• Review past “Exam 1”s on Hardinger’s website
http://www.chem.ucla.edu/harding/index.html(on left frame, click “Ch14C” then in middle frame click “Current and Past Exam and Keys”)
Exam 1 Review: Tuesday, 5/1 YH3069 from 7-9pm (Ray)
• Winter 2012Problems: 4–7, 15, 22.
• Winter 2011Problems: 1–11, 23, 28, 29.
• Spring 2011, Lecture 1Problems: 1–5, 12, 20.
• Spring 2011, Lecture 2Problems: 1, 2, 9–13, 15, 16, 20–22.
• Fall 2011Problems: 8, 20–24.
Suggested problems to attempt before attending the review:
Infrared Spectroscopy (IR)Molecular Vibrations
Fundamental principleAbsorption of photons causes changes in molecular vibrations
Molecular vibrations•Bonded atoms move around in space•Very fast: One vibration cycle = ~10-15 seconds
Bending (H-O-H)•Motion not along bond axis
Stretching (H-Cl)•Atoms move along bond axis
Movie files: “HCl stretch.mov” (left) and “water bend.mov” (right)
Molecular Vibrations
Vibration energy
Ground statelower energy
add energy
Excited statehigher energy
• vibration energy causes average bond length
water_groundstate.mov water_excitedstate.mov
Molecular VibrationsVibration energy
Excited vibrational state
E = h
For bond vibrations:E = dependent on atoms and bond order = ~5 kcal mol-1 = lower energy than red light photons = infrared photons
= stretching frequencyUnit = wavenumber = cm-1
Ground vibrational state
Vibr
atio
nal s
tate
ene
rgy
• Vibrational energy is quantized (only certain energy values are possible)
The Infrared Spectrum
Many photons absorbed
Spectrum = plot of photon energy versus photon quantity
Num
ber o
f pho
tons
abs
orbe
d
Stretching frequencyProportional to photon energy
Typical infrared spectrum:
Fewphotons
absorbed
Molecular Structure from IR Spectrum
•Structure controls number of photons absorbed (IR spectrum y-axis)
•Structure controls stretching frequency (IR spectrum x-axis)
How does spectrum give information about molecular structure?
Structure versus Photon Quantity
Chance of photon absorption controlled by change in dipole moment ()
+ X Y -
Useful approximation:Consider only one bond
From quantum mechanics:
Intensity of IR peak Vector sum of bond dipoles
Bond Dipoles Control Absorption Intensity
Bond dipole ~ (magnitude of electronegativity difference) x (bond length)
• EN causes bond dipole
• bond length causes bond dipole
+ X Y -
bond dipole causes absorption
In practical terms:
•Highly polar bond strong peak
•Symmetrical (nonpolar) or nearly symmetrical bond peak weak or absent
}
Absorption Intensity versus Bond Dipoles
C=O peak strongH3C CH3
O
CH3H3C
H3C CH3
C=C peak absent or weak
HH
H3C CH3
C=C peak present
Examples
2000 1500Wavenumbers
2000 1500Wavenumbers
2000 1500Wavenumbers
Structure versus Stretching Frequency
•Stretching frequency of two masses on a spring
Stretching frequency =1
2cf
mA + mB
mAmB
1/2
bond order
stretching frequencyincreasing
spring stiffness C-CC=CCC
atom masses
Functional groups determine IR stretching frequencies
atoms bond
Hooke’s Law (1660)
100
Tran
smitt
ance
(%)
04000 3000 1500 Stretching frequency (cm-1)10002000
Characteristic Stretching FrequenciesThe Five Zones
IR spectrum divided into five zones (groups) of important absorptions
1 2 3 4 5 Fingerprint region
Characteristic Stretching FrequenciesThe Five Zones
Bond Stretching Frequency Intensity and Shape
Zone 1: 3700-3200 cm-1
Alcohol O-H 3650-3200 cm-1 usually strong and broad
Alkyne C-H 3340-3250 cm-1 usually strong and sharp
Amine or amide N-H 3500-3200 cm-1 medium; often broad
Zone 2: 3200-2700 cm-1
Aryl* or vinyl** sp2 C-H 3100-3000 cm-1 variable
Alkyl sp3 C-H 2960-2850 cm-1 variable
Aldehyde C-H ~2900, ~2700 cm-1 medium; two peaks
Carboxylic acid O-H 3000-2500 cm-1 usually strong; very broad
* attached to benzene ring **attached to alkene
Characteristic Stretching FrequenciesThe Five Zones
Bond Stretching Frequency Intensity and Shape
Zone 3: 2300-2000 cm-1
Alkyne CC 2260-2000 cm-1 variable and sharp
Nitrile CN 2260-2220 cm-1 variable and sharp
Zone 4: 1850-1650 cm-1
Ketone C=O 1750-1705 cm-1 strong
Ester C=O 1750-1735 cm-1 strong
Aldehyde C=O 1740-1720 cm-1 strong
Carboxylic acid C=O 1725-1700 cm-1 strong
Amide C=O 1690-1650 cm-1 strong
C=O frequencies 20-40 cm-1 lower when conjugated to a pi bond
Characteristic Stretching FrequenciesThe Five Zones
Bond Stretching Frequency Intensity and Shape
Zone 5: 1680-1450 cm-1
Alkene C=C 1680-1620 cm-1 variable
Benzene C=C~1600 cm-1 and
~1500-1450 cm-1
variable;1600 cm-1 often two peaks
Fingerprint region (below 1450 cm-1): Not useful for Chem 14C
What do I need to know from this table?
•Functional groups in each zone Learn by working lots of problems
•Do not memorize stretching frequencies; table given on exam
•Complete table: Lecture Supplement and Thinkbook, inside front cover
Guided Tour of Functional GroupsTerminal Alkyne
C CH CH2CH2CH2CH3
100
Tran
smitt
ance
(%)
04000 3000 1500 Stretching frequency (cm-1)10002000
1 2 3 4 5 Fingerprint region
Guided Tour of Functional GroupsTerminal Alkene
CC
H
CH2CH2CH2CH3
H
H
100
Tran
smitt
ance
(%)
04000 3000 2000 1000 Stretching frequency (cm-1)1500
1 2 3 4 5 Fingerprint region
Guided Tour of Functional GroupsAlcohol
4000 3000 2000 1500 1000 Stretching frequency (cm-1)
0
100
Tran
smitt
ance
(%)
H O CH2CH2CH2CH2CH2CH3
broad
C-O
1 2 3 4 5 Fingerprint region
Guided Tour of Functional GroupsKetone
4000 3000 2000 1500 1000 Stretching frequency (cm-1)
0
100
Tran
smitt
ance
(%)
CH3
O
sp3 C-H1709 cm-1
Very
str
ong
1 2 3 4 5 Fingerprint region
Infrared Spectroscopy Part 2Lecture Supplement page 122
100
Tran
smitt
ance
(%)
04000 3000 1500 Stretching frequency (cm-1)10002000
1 2 3 4 5 Fingerprint region
Infrared Spectroscopy Part 1 Summary
•Infrared photons cause excitation of molecular vibrations•Photon absorption probability controlled by bond polarity•Energy of photons absorbed depends on:
} Functional groups
•IR spectrum divided into five zones•Each zone analyzed for absence or presence of functional groups•Stretching frequency, peak shape both important
Bond orderMasses of atoms bonded
Alcohol O-H usually gives broad peakC=O stretch gives strong peak
Guided Tour of Functional GroupsKetone
4000 3000 2000 1500 1000 Stretching frequency (cm-1)
0
100
Tran
smitt
ance
(%)
CH3
O
sp3 C-H1709 cm-1
Very
str
ong
Guided Tour of Functional GroupsAldehyde
CH CH2CH2CH2CH2CH3
O
~2900 cm-1
usually obscured
very
str
ong
1718 cm-1
100
Tran
smitt
ance
(%)
04000 3000 2000 1500 1000 Stretching frequency (cm-1)
~2700 cm-1
Guided Tour of Functional Groups
In general: Conjugation with C-C pi bond lowers C=O stretching frequency by 20-40 cm-1
CH3
O
CH3
O
CH3
O1709 cm-1
Effect of pi bond conjugation?
1667 cm-1 1686 cm-1
Guided Tour of Functional GroupsEster
CCH3O CH2CH2CH2CH2CH3
O
1743 cm-1
100
Tran
smitt
ance
(%)
04000 3000 2000 1500 1000 Stretching frequency (cm-1)
Guided Tour of Functional GroupsCarboxylic Acid
CO CH2CH2CH2CH2CH3
O
H
very broad
1711 cm-1
100
Tran
smitt
ance
(%)
04000 3000 2000 1500 1000 Stretching frequency (cm-1)
Guided Tour of Functional GroupsBenzene Ring
H3C
HH
H H
H
Sometimes two peaks
100
Tran
smitt
ance
(%)
04000 3000 2000 1500 1000 Stretching frequency (cm-1)
Five Zone IR Spectrum AnalysisExample #1: C6H12O2
DBE = C - (H/2) + (N/2) + 1
= 6 - (12/2) + (0/2) + 1
= 1
100Tr
ansm
ittan
ce (%
)
04000 3000 2000 1500 1000 Stretching frequency (cm-1)
1700 cm-1
One ring or one pi bond
Step 1: Calculate DBE
Five Zone IR Spectrum AnalysisExample #1
100Tr
ansm
ittan
ce (%
)
04000 3000 2000 1500 1000 Stretching frequency (cm-1)
1700 cm-1
Step 2: Analyze IR Spectrum
PresentAbsent - no N in formula
Zone 1 (3700-3200 cm-1) C6H12O2 DBE = 1
Alcohol O-H:
Terminal alkyne C-H:Amine or amide N-H:
Absent - not enough DBE; no peak ~2200 cm-1
Five Zone IR Spectrum AnalysisExample #1
100Tr
ansm
ittan
ce (%
)
04000 3000 2000 1500 1000 Stretching frequency (cm-1)
1700 cm-1
Zone 2 (3200-2700 cm-1) C6H12O2 DBE = 1Aryl/vinyl sp2 C-H:
Alkyl sp3 C-H:Aldehyde C-H:
Carboxylic acid O-H:
Absent - not enough DBE
Absent - no 2700 cm-1
Absent - not broad enough
Present
Five Zone IR Spectrum AnalysisExample #1
100Tr
ansm
ittan
ce (%
)
04000 3000 2000 1500 1000 Stretching frequency (cm-1)
1700 cm-1
Zone 3 (2300-2000 cm-1) C6H12O2 DBE = 1Alkyne CC:
Nitrile CN:
Absent - no peaks; not enough DBE
Absent - no peaks; not enough DBE
Five Zone IR Spectrum AnalysisExample #1
100Tr
ansm
ittan
ce (%
)
04000 3000 2000 1500 1000 Stretching frequency (cm-1)
Zone 4 (1850-1650 cm-1) C6H12O2 DBE = 1
1700 cm-1
C=O:Possibilities: Ketone Ester - not enough oxygens Aldehyde - no 2700 cm-1 peak Carboxylic acid - zone 2 not broad Amide - no nitrogen
Present @ 1700 cm-1
Verify with13C-NMR
Five Zone IR Spectrum AnalysisExample #1
100Tr
ansm
ittan
ce (%
)
04000 3000 2000 1500 1000 Stretching frequency (cm-1)
1700 cm-1
Zone 5 (1680-1450 cm-1) C6H12O2 DBE = 1Benzene ring:Alkene C=C:
Absent - no peak ~1600 cm-1; not enough DBEAbsent - no peak ~1600 cm-1; not enough DBE (C=C plus C=O)
Actual structure:OHO
Five Zone IR Spectrum AnalysisExample #2: C8H7N
100Tr
ansm
ittan
ce (%
)
04000 3000 2000 1500 1000 Stretching frequency (cm-1)
Step 1: Calculate DBEDBE = C - (H/2) + (N/2) + 1 = 8 - (7/2) + (1/2) + 1 = 6 Six rings and/or pi bonds
Possible benzene ring
Five Zone IR Spectrum AnalysisExample #2
100Tr
ansm
ittan
ce (%
)
04000 3000 2000 1500 1000 Stretching frequency (cm-1)
Step 2: Analyze IR Spectrum
Zone 1 (3700-3200 cm-1) C8H7N DBE = 6Absent - no oxygen in formulaAbsent - peaks too smallAbsent - peaks too small
“No amine/amide” = false conclusion Example: (CH3)3N has no N-H
Five Zone IR Spectrum AnalysisExample #2
100Tr
ansm
ittan
ce (%
)
04000 3000 2000 1500 1000 Stretching frequency (cm-1)
Zone 2 (3200-2700 cm-1) C8H7N DBE = 6Aryl/vinyl sp2 C-H:
Alkyl sp3 C-H:Aldehyde C-H:
Carboxylic acid O-H:
Present - peaks > 3000 cm-1
Present - peaks < 3000 cm-1
Absent - no 2700 cm-1; no C=O in zone 4Absent - not broad enough; no C=O in zone 4
Five Zone IR Spectrum AnalysisExample #2
100Tr
ansm
ittan
ce (%
)
04000 3000 2000 1500 1000 Stretching frequency (cm-1)
Zone 3 (2300-2000 cm-1) C8H7N DBE = 6Alkyne CC:Nitrile CN:
PossiblePossible}
Five Zone IR Spectrum AnalysisExample #2
100Tr
ansm
ittan
ce (%
)
04000 3000 2000 1500 1000 Stretching frequency (cm-1)
Zone 4 (1850-1650 cm-1) C8H7N DBE = 6C=O: Absent - no peak; no oxygen in formula
Five Zone IR Spectrum AnalysisExample #2
100Tr
ansm
ittan
ce (%
)
04000 3000 2000 1500 1000 Stretching frequency (cm-1)
Zone 5 (1680-1450 cm-1) C8H7N DBE = 6Benzene ring:Alkene C=C:
Present - peaks ~1600 cm-1 and ~1500 cm-1
Absent - not enough DBE for alkene plus benzene plus triple bond
Actual structure: CH2CC N