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”)

http://www.chem.ucla.edu/harding/index.html

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



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


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



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



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

(%)



Guided Tour of Functional GroupsTerminal Alkene

CC

H

CH2CH2CH2CH3

H

H

100

Tran

smitt

ance

(%)

04000 3000 2000 1000 Stretching frequency (cm-1)1500


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


Guided Tour of Functional GroupsKetone


0

100

Tran

smitt

ance

(%)

CH3

O

sp3 C-H1709 cm-1

Very

str

ong


Infrared Spectroscopy Part 2Lecture Supplement page 122

100

Tran

smitt

ance

(%)



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


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

(%)


~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

(%)


Guided Tour of Functional GroupsCarboxylic Acid

CO CH2CH2CH2CH2CH3

O

H

very broad

1711 cm-1

100

Tran

smitt

ance

(%)


Guided Tour of Functional GroupsBenzene Ring

H3C

HH

H H

H

Sometimes two peaks

100

Tran

smitt

ance

(%)


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 (%

)


1700 cm-1

One ring or one pi bond

Step 1: Calculate DBE

Five Zone IR Spectrum AnalysisExample #1

100Tr

ansm

ittan

ce (%

)


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


100Tr

ansm

ittan

ce (%

)


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


100Tr

ansm

ittan

ce (%

)


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


100Tr

ansm

ittan

ce (%

)


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


100Tr

ansm

ittan

ce (%

)


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 (%

)


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


100Tr

ansm

ittan

ce (%

)


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


100Tr

ansm

ittan

ce (%

)


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


100Tr

ansm

ittan

ce (%

)


Zone 3 (2300-2000 cm-1) C8H7N DBE = 6Alkyne CC:Nitrile CN:

PossiblePossible}


100Tr

ansm

ittan

ce (%

)


Zone 4 (1850-1650 cm-1) C8H7N DBE = 6C=O: Absent - no peak; no oxygen in formula


100Tr

ansm

ittan

ce (%

)


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

Exam 1 Review: Tuesday, 5/1 YH3069 from 7-9pm (Ray) Winter 2012 Problems: 4–7, 15, 22. Winter 2011...

Documents

Transcript of Exam 1 Review: Tuesday, 5/1 YH3069 from 7-9pm (Ray) Winter 2012 Problems: 4–7, 15, 22. Winter 2011...