chem 233: organic laboratory i prelab lecture university...

University of Illinois at ChicagoUICCHEM 233: Organic

Laboratory I Prelab Lecture

Lab One: Characterization of Organic Solids and Liquids by Elemental Analysis, Melting Point,

Boiling Point and Infrared Spectroscopy

UICUniversity of Illinois at Chicago

© 2009, Dr. Chad L. LandrieCHEM 233: Organic Chemistry Laboratory 1

SlidePrelab Lecture: Lab 1

Goals and Objectives

2

Identify Unknown Organic Compounds by:

• Infrared Spectroscopy: Determine functional groups; obtain IR of both a liquid film and solid with ATR accessory

• IHD: Determine number of double bonds, triple bonds and/or rings in the molecule

• Elemental Analysis: Used to determine empirical formula

• Melting Point and Boiling Point: Physical properties used to confirm hypothetical structures by comparison to known




Infrared Spectroscopy

3

Primary Function = Identify Functional Groups(See “IR Primer” from last week for review.)

Example

O-H

C=O C=C

C(sp2)-H




Index of Hydrogen Deficiency (IHD)

4

Primary Function: Indicates the number of double/triple bonds and/or rings.

• Also Known As = Degrees of Unsaturation• Saturation = Maximum Number of Hydrogens

• Saturation for Hydrocarbons = CnH2n+2

ExamplesIHD = 1 (one double bond)

IHD = 1 (one ring)

IHD = 4 (three double bonds & one ring)




Calculating IHD from Molecular Formula

5

1. Hydrocarbons (CnHx) and Oxygenates (CnHxOy): Ignore O(2n+2) - X

2

2. Compounds with N (CnHxNy): Subtract # N from # H(2n+2) - (X-Y)

2

IHD =

IHD =

3. Halogens (CnHxXy): Add # Halogens to # H(2n+2) - (X+Y)

2IHD =




Complex Example

6

Combine Previous Rules Together When Necessary:

H2N

OCH3

Br

F

MF = C7H7BrFNO

(2*7+2) - (7-1+2)2

IHD = = 4




Review: Determining Molecular Formula from Elemental Analysis and Molecular Formula

7

Elemental Analysis + Molecular Weight Molecular Formula

C: 62.04%H: 10.41%O: 27.55%

C: 62.04g/12g/mol = 5.17 molH: 10.41g/1g/mol = 10.41 molO: 27.55g/16g/mol = 1.72 mol

100% 100g

C: 5.17/1.72 = 3.0H: 10.41/1.72 = 6.0O: 1.72 /1.72 = 1.0

1. Convert percent into moles; assume % is grams. 2. Normalize; divide by smallest value

C: 3.0 x 1.0 = 3.0H: 6.0 x 1.0 = 6.0O: 1.0 x 1.0 = 1.0

3. Multiply by Integers if necessary

Emperical Formula:C3H6O

4. Divide given MW by empirical MW

Emperical MW:58 g/mol

(Given MW = 174 g/mol)

174/58 = 3 3 x C3H6O = C9H18O3

Molecular Formula




Measuring Boiling Point of Liquids

8

Follow Miniscale Procedure on Page 41 of your textbook.

Process:•Boiling point = when equilibrium vapor pressure

of the liquid equal the atmospheric pressure (Patm = Pº).

•Assemble apparatus as shown in figure to the left. Perform in hood.

• Place thermometer just above level of the liquid.•Add one boiling stone to liquid to ensure

smooth boiling and prevent “bumping” (uncontrolled boiling).

•Heat liquid with heating mantle until a vigorous boil is observed.

•Record temperature of vapors. Cool slightly, then repeat two more times.

←refluxing vapor

←boiling liquid




Melting Point

9

Compare measured melting points with those found in a chemical catalog, but be careful:

• More than one compound may have the same mp.

• Impurities or mixtures lead to depressed mps (except in cases of eutectic mixtures).

• Heating the apparatus too rapidly will result in depressed readings. Perform two mps to save time--a quick estimate and a careful measurement (1ºC/sec) close to estimate.




Measuring Melting Point

10

Teaching Assistant will Demonstrate this Technique

Caution: Every student turn apparatus OFF when finished. Overheating will damage the equipment.

University of Illinois at ChicagoUIC

CHEM 233: Organic Laboratory I

Lab 2: Separation of Dyes & Spinach Pigments by Column and Thin Layer Chromatography

OHOH

OH

OH

trans

N

N N

NO

OMeO

Mg2+

R

Chlorophyll a




Some Limited Definitions

2

Chromatography: the process of partitioning a solute between two immiscible phases--one stationary, one mobile.

Chromatography: the process of separating mixtures of two or more compounds based on their relative affinities to two

immiscible phases--one stationary, one mobile.

Chromatography: a process in which a chemical mixture carried by a liquid or gas is separated into components as a result of differential distribution of the solutes as they flow around or over a stationary liquid or solid phase. (Webster)




Partition Coefficient (Kc)

3

Volume of blue box represents the concentration of solute

dissolved in the mobile phase.

Volume of red box represents the concentration of solute dissolved in/adsorbed on the

stationary phase.

Solutes are in equilibrium with mobile phase and

stationary Phase.

The equilibrium constant for chromatography is called the

partition coefficient (Kc).

[solute in mobile phase]

[solute in stationary phase]Kc =

[A]M

[A]S

Kc =

Larger Kc = solute has greater affinity for stationary phase and/or smaller affinity for mobile phase

Smaller Kc = solute has smaller affinity for stationary phase and/or greater affinity for mobile phase




Schematic Representation of Chromatographic Separation of Two Solutes with Different Kc

4

Kc = 2

X

YKc = 0.5

Distance Traveled Increasing

X

Y

X

Y

Mobile Phase (moving left to right)

Stationary Phase Conclusions:

1. Solutes with larger Kc values (i.e. X) move slower because they “spend more time” in the stationary phase, which is not moving, compared to molecules with smaller Kc.

2. Longer distances traveled leads to better separation (distance between X and Y) of solutes.

3. Solutes with small differences in Kc will be harder to separate.

4. Solutes with large differences in Kc will be easier to separate.

5. Changing the composition of the stationary phase or the mobile phase will change Kc for both solutes, X and Y, although not necessarily equally. (This conclusion is not represented in the scheme to the left.)

Tim

e =

nT

ime

= 2

nT

ime

= 3

n




Types of Chromatography

5

Type Mobile Phase Stationary Phase Solid Support

Column Liquid Packed Solid (i.e. SiO2, Al2O3)

None

Thin Layer (TLC) Liquid Bound Solid(i.e. SiO2, Al2O3)

Typically, thin aluminum or glass sheet

Packed ColumnGas-Liquid (GLC)

gas/vapor Liquid(i.e. carbowax)

Typically, Diatomaceous Earth (Celite = Diatoms)

Capillary ColumnGas-Liquid (GLC)

gas/vapor LiquidNone:

liquid is ADsorbed onto inside wall of column

High-Performance Liquid (HPLC)

Liquid Packed Solid None

Types Used in CHEM 233 = SiO2 & Al2O3 are polar stationary phases.




Eluting Power of Some Liquid* Mobile Phases

6* Note: Not all mobile phases are liquids. Gases, such as He, can also be used as mobile phases.

Eluent: The sample and/or mobile phase contained within or exiting a chromatographic device.

Eluting Power: The ability of a solvent to move a sample through a stationary phase of a chromatographic device.

Incr

easi

ng e

lutin

g po

wer

with

po

lar

stat

iona

ry p

hase

s

Incr

easi

ng e

lutin

g po

wer

with

no

np

ola

r st

atio

nary

pha

ses

watermethanolethanol

1-propanolacetone

ethyl acetatediethyl etherchloroform

dichloromethanetoluenehexane

petroleum ether

Two or more liquid solvents may be mixed together in

varying ratios to obtain mobile phases with intermediate

eluting powers. Mixtures of ethyl acetate and hexanes are

common mobile phases in organic chemistry for column and thin-layer chromatography.




Properties of Two Solid* Stationary Phases Used in Column and Thin-Layer Chromatography

7

Silica Gel• Polymer of SiO2 x H2O

• Porous solid• Slightly acidic

• Polar stationary phase• Can be made nonpolar by replacing H with long alkyl

chains (for use in reverse phase chromatography)• Commonly used as a desiccant and food preservative

Alumina• Al2O3

• Porous solid• Comes in three forms: acidic, basic & neutral

• Polar stationary phase• More polar than silica gel

• Commonly used as an abrasive or polishing agent• Major component of rubies and sapphire (corundum)




Polarity is the Major Factor Affecting Kc in Column Chromatography and TLC

8

Polarity: Separation of charge; as in a molecular dipole.

O H

Polar Molecule(large molecular dipole)

H2C H

Nonpolar Molecule(small molecular dipole)

More polar compounds have greater affinity for polar stationary phases than less polar compounds.

“Polar attracts polar.”=

Nonpolar compounds have greater affinity for nonpolar stationary

phases than more polar compounds.“Nonpolar attracts nonpolar.”=




Relative Polarities of Functional Groups

9

Elutropic Series (pg 178)

RCO2H > ROH > RNH2 > RR’C=O > RCO2R’ > ROR’ > C=C > R-X

carboxylic acid alcohol amine ketone ester ether alkene halide

Decreasing Polarity

Decreasing Retention Time on a Column

Increasing Retention Factor (Rf) on a TLC Plate

Decreasing Affinity for Polar Stationary Phase (Decreasing Kc)




10

What conclusion can you draw about the relative eluting power of liquid mobile phases (for polar stationary phases)

and the elutropic series of functional groups?




Column Chromatography

11

sand

Silica Gel(SiO2)Polar

StationaryPhase

plug

Less PolarMore Polar

Mixture

Increasing Time

More Polar

Less PolarMore Polar

Elute WithSolvent

Less Polar =Smaller Kc =

Shorter Retention Time

More Polar =Smaller Kc =

Longer Retention Time

Incr

easin

g Di

stan

ce

Trav

eled

by

Solu

te




Thin Layer Chromatography (TLC)

12

Principles/Process:

• Silica gel (stationary phase) is bound to a planar support, often glass or aluminum--in our case a thin

sheet of aluminum.

• Two solutes A and C are are spotted onto the TLC plate with a small capillary or mp tube. Smaller spots =

better separation.

• A and C are spotted together in the middle. (For function of co-spot B, see manual pg. 37)

• Mobile phase rises up the TLC plate by capillary action.

• Solutes are in equilibrium with mobile phase and stationary phase; they travel upward with mobile phase

• A and C travel different distances, depending on their relative attraction to the stationary phase (Kc).

mobile phase

t = 0 min

A B Corigin

developing chamber

mobile phase

t = 5 min

A B C

solventfront

Technique Notes:

1. Origin must be above level of mobile phase.

2. Developing chamber must be covered to prevent eluent from evaporating off TLC plate.

3. After elution and before visualization, TLC plates must be dried thoroughly to remove mobile phase.




Spinach Pigments

13

N

N N

NO

OMeO

Mg2+

R

CH2

O

O

Chlorophyll a

R =

N

N N

NO

OMeO

Mg2+

R

Chlorophyll b

HO

N

NH HN

NO

OMeO

R

Pheophytin a

N

NH HN

NO

OMeO

R

Pheophytin b

HO

ß-Carotene

Lycopene

OH

OH

Carotenes Xanthophylls

HOLutein

HO Zeaxanthin

Carotenoids

Based on the description in the course manual, determine the color of each pigment in your groups.

What is the relative order of polarity for the six pigments?




TLC Separation of Spinach Pigments

14

dX

ds

Your best TLC should separate spinach pigments into six distinct spots corresponding to the pigments below:

•pheophytin a (one spot, may not be seen)•pheophytin b (one spot)•chlorophyll a (one spot)•chlorophyll b (one spot)•carotenes (group of spots close together?)•xanthophylls (group of spots close together?)

Circle the six spots on your TLC and determine the identity of each based on color and relative polarities (Rf-values). Read the background info in course manual for guidance.

Retention Factor (Rf)

Rf =dx

ds



Lab 3: Separation of Ethyl Acetate and Butyl Acetate by Simple Distillation. Analysis of fractions by Gas-Liquid

Chromatography.




Simple Distillation: Applications & Goals

2

Applications:

1. Separate volatile liquid from non-volatile solute (i.e.

H2O and NaCl)

2. Separate two volatile liquids when their boiling

point difference is > 50 ºC.

Today’s Goals:

1. Separate a 1:1 mixture of ethyl acetate and butyl

acetate by simple distillation.

2. Use gas chromatography to determine whether

separation was successful.

H3C O

O

CH3

Ethyl AcetateH3C O

O

Butyl AcetateCH3




Equilibrium Vapor Pressure is Directly Proportional to Temperature

3

Conclusions:

• b.p. = when equilibrium vapor pressure of a liquid (Pº) equals the atmospheric pressure (Patm).

• b.p. = rapid/spontaneous vaporation throughout the liquid (bubbles).

• Compounds with higher Pº have lower bps and vice versa.

• More volatile liquids = lower bps and vice versa.

volatile (chemistry) adj. = readily vaporizes at relatively low temperatures; high relative equilibrium vapor pressure (Pº) at relatively low temperatures.

Temp (ºC)

Pres

s. (T

orr) Boiling Point

(Patm = Pº)760

100

Pº (equilibrium vapor pressure of water)

78

Pº (equilibrium vapor pressure of ethanol)




Simple Distillation Apparatus*

4

{thermometer adapter

distillate

water in(from faucet)

water out(to drain)

stillpot(round bottom flask)

stillhead

vacuum adapter(used here as a driptip)

West Condenser

Notes:

1. Assemble apparatus vertically--not leaning.

2. Thermometer should be slightly below entrance to condenser to ensure the end is immersed in vapors.

3. Use Keck Clips to secure horizontal joints in order to prevent separation.

4. All joints must be secured tightly to prevent vapors from escaping apparatus.

5. Do not use grease; can contaminate samples.

6. Use a gentle flow of water through condenser.

7. Always disassemble apparatus when finished to prevent “joint freezing.”

* Locations of Bunsen clamps are not shown.




Boiling Points of Mixtures

5

bp (mixture): when total pressure above a liquid (Ptot) equals the atmospheric pressure (Patm).

Vaporization

bp: Ptot = Patm

Daltons Law: total pressure above a liquid (Ptot) = sum of partial pressures (Px) of each component in the mixture.

Ptot = PA + PB + PC . . .

Raoult’s Law: partial pressure (Px) is the mole fraction (NX) multiplied by the equilibrium vapor pressure (Pº)

Px = Nx*Pº

mole fraction (NX): # moles (n) of one component in the mixture divided by the total n of all components.

Nx = nx / (nx + nY)

As T increase, the Pred and Pblue increases until Ptot equals the

atmospheric pressure--at which point the mixture boils.




Which compound would you expect to have a higher equilibrium vapor pressure? Why?

Ethyl Acetate vs. Butyl Acetate

6

Based on your previous answer, which would you expect to have a higher boiling point?

Which compound will be the first to exit the simple distillation apparatus? Why?

H3C O

O

CH3

Ethyl AcetateH3C O

O

Butyl AcetateCH3




Collect Three Fractions

7

Construct a graph similar* to the one below in your course manual according to the instructions.

Fraction 1(relatively constant

temperature)

Fraction 2(dramatic change in

temperature)Fraction 3

(relatively constant temperature)

Tem

pera

ture

(ºC)

Volume of Distillate (mL) 30

*If your graph does not look like this, your thermometer may be misplaced or you may be distilling faster than 1-2 drops/second.




Composition of Each Fraction Determined by Gas-Liquid Chromatography (GLC)

8

Type Mobile Phase Stationary Phase Solid Support

Column Liquid Packed Solid (i.e. SiO2, Al2O3)

None

Thin Layer (TLC) Liquid Bound Solid(i.e. SiO2, Al2O3)

Typically, thin aluminum or glass sheet

Packed ColumnGas-Liquid (GLC)

gas/vapor(e.g., Helium)

Liquid(i.e. carbowax)

Typically, Diatomaceous Earth (Celite = Diatoms)

Capillary ColumnGas-Liquid (GLC)

gas/vapor LiquidNone:

liquid is ADsorbed onto inside wall of column

High-Performance Liquid (HPLC)

Liquid Packed Solid None

Type Used Today




Column & Thin-Layer Chromatography:• Kc depends primarily on polarity of solute, mobile phase

and stationary phase.

Gas-Liquid Chromatography:•Kc depends on polarity of solute and stationary phase.•Mobile phase is always He or H2. (No effect on Kc)• Kc also depends on equilibrium vapor pressure (Pº).• More volatile solutes = lower bp = higher Pº =

larger [A]M = smaller Kc = shorter retention time

Gas-Liquid Chromatography (GLC) Principles

9

Kc = 2

X

YKc = 0.5

Distance Traveled Increasing

X

Y

X

Y

Mobile Phase (moving left to right)

Stationary Phase

Tim

e =

nT

ime

= 2

nT

ime

= 3

n

Similar Principles as Lab 2

[A]M

[A]S

Kc =

Comparison




Analysis of a Sample GLC

10

Area = 12 Area = 6

Time (sec)10 20 30 40 50 60 70 80 90 100

CH3

toluenebp = 110 ºCɛ = 2.4

ethyl benzenebp = 136 ºCɛ = 2.5

GLC Graph

Conclusions:

1. Dielectric constants (ɛ) are close, therefore the polarity of toluene and ethylbenzene must be very similar. Polarity, then, does not affect Kc in this example.

2. Greater MW of ethylbenzene = greater Van Der Waals forces (VWF) = lower Pº = higher bp (less volatile).

3. Ethylbenzene has a significantly higher bp than toluene, therefore we can conclude ethylbenzene has a lower Pº.

4. Since: lower Pº = lower [A]M = larger Kc, then ethylbenzene moves through column slower (longer retention time.)

5. Another way of saying 4: toluene has a higher Pº which means it “spends” more time in the mobile phase compared to ethylbenzene; therefore, toluene will move through the column faster (shorter retention time).

CH3




Areas are Measured by a Thermal Conductivity Detector (TCD) in Our GCs

11

thermal conductivity = the ability of a substance to conduct/transfer heat

1. gas inlet (from column)2. gas outlet3. seal4. filament wire5. electrical connections

• Electric current is passed through the filament wire (4).

• A solute (sample) passes through the cell with the mobile phase (1→2).

• Because of the solute’s thermal conductivity, the temperature of the filament (4) decreases.

• A change in filament temperature causes a change in resistance of the filament.

• This resistance change is measured electrically. The electric signal is measured by a computer.

• Increasing concentration of solute/sample = larger change in resistance = larger detector response (area under curve).

Thermal Conductivity Cell




Determining Composition from GLC Areas

12

Area = 12 Area = 6

Time (sec)10 20 30 40 50 60 70 80 90 100

We will determine areas with the aid of the LoggerPro software; however, other “manual” methods are discussed in your textbook.

Ideal % Composition Calculation:

mol % toluene =12

toluene

ethylbenzene

12 + 6* 100% = 67%

However, the thermal conductivity of each substance is slightly different, so a correction factor must be applied when using a thermal conductivity detector (TCD), which we are.Table of Correction Factors: textbook, pg 201.

% Composition Calculation(w/ correction factors):

mol % toluene =12(0.86)

12(0.86) + 6(0.77)* 100% = 69%




GLC Basic Equipment

13

Injection Ports

PackedColumns(coiled)




Packed Column

14

Stain

less S

teel T

ubing

Mobile Phase(Helium Gas)

Stationary Phase(A liquid Adsorbed onto a

packed solid support. Several types and polarities. See text

and manual for examples.)

Solid Support(Typically, diatomaceous earth.

Tradenames = Celite® or Chromosorb®. Celite is also commonly used as a filter aid.)

The eluent (mobile phase and solute) travel through the column in between the liquid-coated solid

support particles.



Lab 4: Separation of Ethyl Acetate and Butyl Acetate by Fractional Distillation. Analysis of fractions by Gas-Liquid

Chromatography.




Simple Distillation: Applications & Goals

2

Applications:

1. Separate mixtures of volatile liquids not separable by simple dist. (∆bp < 50 ºC)

2. Separate hydrocarbons in crude oil with fractionating

tower (refining).

Today’s Goals:

1. Separate a 1:1 mixture of ethyl acetate and butyl

acetate by simple distillation.

2. Compare GLC results from fractional to simple.

H3C O

O

CH3

Ethyl AcetateH3C O

O

Butyl AcetateCH3




Fractional Distillation Apparatus*

3

{thermometer adapter

distillate

water in(from faucet)

water out(to drain)

stillpot(round bottom flask)

stillhead

vacuum adapter(used here as a driptip)

West Condenser

Notes:

1. See notes for Lab 3 setup.

2. Pipette bulbs = seal air inside outer jacket of Hempel column = insulated = uniform temperature gradient.

3. Raschig rings = increase surface area = increase number of theoretical plates = better separation.

4. Copper wire plug may be necessary for some Hempel columns to prevent Raschig rings from falling in stillpot.

5. Use 1-2 boiling stones in stillpot.

6. Stillpot should not be larger than 100 mL.* Locations of Bunsen clamps are not shown.

Hempel Column

pipettebulb

Raschigrings




Temperature Gradient Required for Fractionation

4

Lower Temp(T = bp of component

currently distilling)

Higher Temp(T = bp mixture

in stillpot)

Tem

pera

ture

Gra

dien

t

Process:

1. Temperature gradient is usually established automatically so long as column is insulated.

2. Many vaporizations and condensations take place inside Hempel column.

4. The vapor phase becomes more and more concentrated in the more volatile component with each vaporization/condensation cycle since the more volatile component has higher Pº and since the Nx increases with each cycle.(P = Pº * Nx; Raoult’s Law)

5. Vaporization/condensation cycle = theoretical plate. More theoretical plates = better separation.




Temperature-Composition Diagram

5

Temperature

b.p. of Bb.p. of A

% Composition

50 77 90 9833102% A% B 2103350779098

vapor line: calculatedusing Raoult's Law

liquid line: boiling pointof the liquid mixture

Component Amore volatile

higher Eq. Vapor Pressure (P°)lower boiling point

Component Bless volatile

lower Eq. Vapor Pressure (P°)higher boiling point

Condensationvapor to liquid

composition does not changesurface of column or packing

Vaporizationliquid to vapor

> P° = more vaporizationcomposition changes

Raoutl's Law: P=Na*P°

Each “step” =one theoretical plate




Theoretical Plates

6

Factors Affecting Theoretical Plates (N)

1. temperature gradient2. surface area (↑SA=↑N)3. column length (↑L=↑N)4. rate of gas flow (GC only)

HETP = height equivalent to a theoretical plate

• measure of the efficiency of a fractionating column or GC column

• lower HETP = less height needed for 1 N = more efficient column

HETP =column height or length (L)

theoretical plates (N)




Collect Three Fractions

7

Construct a graph similar* to the one below in your course manual according to the instructions.

Fraction 1(relatively constant

temperature)

Fraction 2(dramatic change in

temperature)Fraction 3

(relatively constant temperature)

Tem

pera

ture

(ºC)


*If your graph does not look like this, your thermometer may be misplaced or you may be distilling faster than 1-2 drops/second.




Comparison of V (distillate) vs. T Graphs

8

Tem

pera

ture

(ºC)


Tem

pera

ture

(ºC)


Which graph below represents the most efficient separation of a binary mixture? Why?Which most likely represents simple distillation

and which fractional distillation? Why?



Lab 5: Steam Distillation of Monoterpenes Carvone and Limonene from Caraway Seeds. Analysis of Products by

Infrared Spectroscopy and TLC.

O

H

(R)-Carvone

O

H

(S)-Carvone




Midterm Exam Reminder

2

Midterm Exam = _

• Sample exam online: www.chadlandrie.com (shared files page).

• Review past homework questions & assigned reading.

• Review background info provided in course manual.

• Review prelab lecture notes.

• Exam will be given during your lab time in the same room.

• Topics cover both theory and techniques of lab experiments.

http://www.chadlandrie.com





Steam Distillation: Applications & Goals

3

Applications:

1. Last separation/purification technique in part

one of CHEM 233 course.

2. Industrial = isolation of volatile essential oils from

plant material.

3. General = isolation of oils/liquids with high bps.

Today’s Goals:

1. Isolate carvone from caraway seeds by steam dist.

Determine enantiomer.

2. Obtain IR of isolated oil; compare to known IRs of

carvone.

3. Obtain TLC of isolated oil with a co-spot of authentic.




Steam Distillation: Advantages & Requirements

4

Advantages:

1. Distill liquids with high boiling points at

temperatures < 100ºC.

2. Avoid high temp simple/fractional distillation, which could cause decomposition.

3. Alternative = vacuum dist.

Requirements:

1. Non-reactive with H2O.

2. Immiscible with H2O.

3. Stable (does not decompose) at 100 ºC.

4. Pº ≥ 5 Torr at 100 ºC.




Steam Distillation Apparatus

5

stillheadClaisenadapter

separatoryfunnel

(used here asa dropping funnel)

water &caraway seeds

replacementwater(hot!!)

Notes:

1. No need for boiling stones; caraway seeds provide sufficient sites for nucleation.

2. Steam is generated in situ by boiling water in the stillpot.

3. Replacement water (hot) is added at the same rate distillate is exiting the condenser.

4. Separatory funnel is actually being used during the distillation as a dropping funnel.

5. Collect approximately 75 mL of distillate. Distillate will be cloudy since carvone and water are immiscible.

6. Volatile components in the caraway seeds are co-distilling with the water vapor.




Initial Considerations

6

1. Which compound above has the highest Pº? Lowest?

2. In a mixture of water and carvone, which would you expect to distill out first? Why?

3. Do you expect carvone to be soluble in water? Why or why not?

O

H

O

H

(R)-Carvone (S)-Carvone

HOH

water




Principles: Mathematical

7

immiscible = not soluble in all proportions

For immiscible component (x) in a heterogeneous mixture with H2O:

Raoult’s Law: Px = Nx * Pxº

Px = Pxº

Since x is not soluble in water, it does not depend on its mole fraction in the mixture. This relationship applies to each

component in mixture, including water.

Dalton’s law still applies:

PT = Pxº + PH2Oº

PT = Px + PH2O

or

Conclusion:

bp (mixture): PT = Patm

The total vapor pressure (PT) is always higher than the most volatile component--

always H2O in steam distillation.

The bp of the mixture is always lower than the lowest boiling component--again,

always H2O (100ºC) in steam distillation.




Vapor Pressure-Temperature Diagram for Water, Carvone and Limonene

0

100

200

300

400

500

600

700

800

900

0 50 100 150 200 250Temperature

(0C)

Pre

ssu

re (

To

rr)

Principles: Graphical

8

760atmospheric pressure

PT

PH2O

Plim

Plim

boiling point (< 100 ºC): PT = (PH2O + Plim + Pcar) = 760 Torr

O




Procedure Notes on Extraction

9

H2O layer(d = 1.0 g/mL)

CH2Cl2 layer(d = 1.33 g/mL)

• Your TA will demonstrate proper use of a separatory

funnel for extractions.

• CH2Cl2 and water are immiscible.

• Less dense liquid (H2O in this case) is top layer.

• Like dissolves like; therefore, carvone (organic)

is soluble in CH2Cl2 (organic).




Analysis of Isolated Oil

10

Thin-Layer Chromatography • Compare Rf value of isolated oil with

authentic sample of carvone.• Find a solvent system (mixture of EtOAc &

hexanes) where Rf < 0.6.• Use a co-spot.

• Are the Rf values of the isolated and authentic samples of carvone the same or

different? What does the result imply?• Is the oil mainly one component or a

mixture of many components?

Infrared Spectroscopy

isolated oil(dilute in small

amount of CH2Cl2)

co-spot authentic sampleof carvone

• Obtain IR of isolated oil. You may need volatiles cover since carvone is, well, volatile.

• Compare your IR spectra with those of carvone on next slides.

• Identify singnals for C=C, C=O in your spectra.

Smell!!Does your oil smell like spearmint [(R)-carvone] or does it smell like caraway [(S)-carvone]? Consult with your friends; maybe the receptors in your nose can’t tell the difference!! IR and TLC cannot tell you which enantiomer of carvone you’ve isolated--only your nose. Optical rotation can differentiate between enantiomers, but we do not have a polarimeter.




IR of Both Enantiomers of Carvone

11

(S)-(+)-Carvone

(R)-(-)-Carvone

(S)-carvone is the enantiomer of (R)-carvone

Enantiomers have the same physical properities including IR vibrational

frequencies, mp, bp, Rf, etc.

IR spectra for both enantiomers of carvone--

and for any two enantiomeric

compounds--are identical.



Lab 6: Base Extraction of Benzoic Acid from Acetanilide followed by Recrystallization and mp Determination.

CH2Cl2

HN

O

O

O

Na

OH

H2O

OH

O

+ OH O

O

+ H2O

(pKa = 4.20) (pKa = 15.7)

Keq

OH

O

crude solid:• not pure/clean• depressed mp

• possibly discolored

HN CH3

Opure solid:

• white crystals• sharp mp

• narrow mp range

H2O

amine(conjugate base)

ammonium cation(conjugate acid)

+ +R NH2 H OH2protonation

R NH3




Todays Goals

2

1. Separate ~1.0 g of a 1:1 w/w mixture of benzoic acid and acetanilide by base-extraction.

2. Purify each isolated solid by recrystallization from boiling water.

3. Measure melting points and compare to reported values.

4. Determine % yield obtained for each solid based on mass of starting mixture.

O

OH

benzoic acid(m.p. = 121-123 ºC)

HN CH3

O

acetanilide(m.p. = 111-115 ºC)




General Scheme for:Base Extraction of 1:1 Benzoic Acid/Acetanilide

3

OH

O

HN

O

CH2Cl2CH2Cl2

HN

O

O

O

Na

OHAdd 3 MNaOH (aq)

H2O

separatelayers

HN

O

CH2Cl2

H2O

O

O

Na

1. Neutralize/ Protonate with 3 M HCl 2. Vacuum filter precipitated benzoic acid

evaporateCH2Cl2

OH

O



HN



CH3

O

1. recrystallize from boiling water 2. vacuum filter

OH

O

pure solid:• white crystals

• sharp mp• narrow mp range

HN CH3

Opure solid:

• white crystals• sharp mp

• narrow mp range

1. recrystallize from boiling water 2. vacuum filter

CH2Cl2 is denser than water = bottom layer




Three Readily Ionizable Functional Groups That are Separable by Acid or Base Extractions

4

R

O

O HR

O

OOH+ + H2O

carboxylic acid(conjugate acid)

carboxylate anion(conjugate base)

OOH H2O

phenol(conjugate acid)

phenoxide anion(conjugate base)

O+ +

H2O

amine(conjugate base)

ammonium cation(conjugate acid)

+ +R NH2

H

H OH2

deprotonation

deprotonation

protonationR NH3

R R

*Hydroxide and hydronium were chosen as general examples; other specific acids and bases may be used. The counterion on the left of the equilibria will depend upon the specific acid or base on the left.




Solubility--”Like Dissolves Like”

5

R

O

O

ORR NH3

R NH2O HR R

O

O H

water soluble• ionic and very polar

• small organic ion

water insoluble; organic solvent soluble• nonionic, but polar functional groups

• i.e. organic solvent = CH2Cl2

Water Polarity

• VSEPR = water is bent; bond angles of 104.5º

• strong molecular dipole• inorganic and very polar (Ɛ=80)

• well suited to dissolve ionic compounds, including small organic anions and cations

HO

H




Choosing an Effective Extracting Acid or Base

6

For an Effective Acid or Base Extraction, Keq must be > 1.0

Base Extraction

pKeq = pKa (acid left) - pKa (acid right)

Keq = 10-[pKa (acid left) - pKa (acid right)]

Acid Extraction

HBHAB A–Keq

organic base

extractingacid (left)

extractingconjugate

base

organicconjugateacid (right)

BHHA B– A–Keq

organic acid (left)

extractingbase

extractingconjugateacid (right)

organicconjugate

base

• remember: p = -log10

• For derivation of this equation, see text page 154.

Example: Example:

pKeq = 4.20 - 15.7

Keq = 10-(-11.5) = 3.16 x 1011

Since Keq is large (>>1), aqueous hydroxide is a good extracting solvent for benzoic acid.

pKeq = -8.0 - 5.20

Keq = 10-(-13.2) = 1.58 x 1013

Since Keq is large (>>1), aqueous HCl is a good extracting solvent for 4-methoxyaniline.

OH

O

+ OH O

O

+ H2O

(pKa = 4.20) (pKa = 15.7)

KeqN

H3CO

HH

HCl+

(pKa = -8.0)

N+ Cl

(pKa = 5.20)H3CO

HH

HKeq




Several Smaller Extractions are More Effective than One Large Extraction with Same Volume

7

FA = Vo

KVx + Vo

FA = fraction of solute remaining in original solventK = partition coefficient = ([Ax]/[Ao]) → larger K = more efficient extracting solventVo = volume of original solventVx = volume of extracting solvent (per extraction)n = number of extractions

( )n

FA = 20

2*5 + 20( )3

FA = 20

2*15 + 20( )1

FA = 0.30 FA = 0.40

Example one: 3 extractions x 5 mL each (15 mL total) from 20 mL of Vo. K = 2.

Example two: 1 extractions x 15 mL each (15 mL total) from 20 mL of Vo. K = 2.

**30% of solute remains in the original solvent.** **40% of solute remains in the original solvent.**




General Requirements for Extracting Solvent

8

1. Does not react irreversibly with the solute.

2. Immiscible with original solvent.(i.e. H2O and CH2Cl2)

3. Selectively removes desired component.(i.e. large K for component to be extracted and small K for the rest.)

4. Easly separated from the solute.(i.e. cyrstallizatio/precipitation then filtration,distillation)




Recrystallization

9

Desired Product(yellow/large circles)

+Impurities

(small/blue circles)

Dissolve product and impurities in a minimum amount of solvent at high

temperature.

Cool solution. Product crystallizes out of

solution. Impurities remain dissolved.

Vacuum filter solution through a Buchner funnel. Mother liquor (filtrate)

contains impurities.

filtrate

Requirements for Recrystallization Solvent

1. Not reactive with desired product(s).

2. Desired product is completely soluble at elevated temperature, but only slightly soluble–or not soluble at

all–at room temperature.

3. Undesired components (impurities) are highly soluble at all temperatures (small temperature coefficient).

4. Easily removed from crystalline product by filtration and evaporation.



Lab 7: Preparation of Alkyl Chlorides by Nucleophilic Aliphatic Substitution.

carbocationintermediate

ClCl

H2OH

alkyl chloridealkeneSN1

E1

HCl




Goals & Data to Collect

2

product name(IUPAC and common)

yield NA NA

substitution(3º, 2º or 1º)

b.p. (determined by reading head temperature during distillation)

AgNO3 Precip. Test (1-3)(1=slowest; 3=fastest)

NaI Precip. Test (1-3)(1=slowest; 3=fastest)

H3CH3C

CH3

OH

2-methyl-2-propanol(tert-butanol)

HClH2O

H3CH3C

CH3

ClH3C

CH3

Cl H3C

CH3Cl

Work individually for this lab. You will prepare tert-butyl chloride. The other alkyl chlorides will be provided for you.

Provided Provided




Procedure Notes

3

ice-bath

round-bottom flask

vacuum adapter(“drip-tip”)

1. Follow procedure in textbook, page 465. Work individually.

2. After extraction and drying the organic layer, isolate product by simple distillation not short-path.

3. Products are volatile. Use an ice-bath to prevent product from evaporating. Also, collect product in round bottom flask attached to drip-tip to also prevent evaporation.

4. Preweigh the collection flask for easy measurement of mass at the end. Note: you will need to look up the densities of the reactants in order to calculate % yield.

5. Perform chemical tests in test-tubes.

6. **Caution: products are highly flammable!**




Procedure Warning

4

The organic layer is washed with a saturated solution of sodium bicarbonate (NaHCO3) to neutralize excess HCl.

FYI: This equilibrium (albeit a very large Keq) also explains why H2O is

slightly acidic in the presence of CO2.

HO O

O

NaH Cl

HO OH

O+ NaCl

carbonic acid

decomposition

CO2 (gas) + H2O

sodium bicarbonate

• CO2 gas will rapidly build up pressure in a closed separatory funnel. Vent often! TA will demonstrate proper venting technique.

• Q: Why would aqueous NaOH not be an appropriate base to neutralize excess HCl?




Nucleophilic Substitution of Alcohols

5

H OOHHCl

Cl +H

1º alcohol 1º alkyl chloride

H OOHHCl

Cl+

H

3º alcohol 3º alkyl chloride

SN2SubstitutionNucleophilic2nd Order

SN1SubstitutionNucleophilic1st Order




Mechanisms and Rates

6

• rate = k[oxonium ion][chloride]

• rate is 2nd order/bimolecular (two reactants in RDS)

• SN2 = no carbocation intermediate

OHH Cl

protonationO H

H

oxonium ion

+ Cl

RDSH OCl +

H

SN2

• rate = k[oxonium ion]

• rate is 1st order/unimolecular (one reactant in RDS)

• SN2 = carbocation intermediate(stability: 3º>2º>>1º)

SN1

OHH Cl

protonation O H

H

RDS

heterolysis

CH3

CH3H3C


ClCl

oxonium ion




E1 Competes with SN1

7

• E1 is reversible: Markovnikoff addition of HCl to alkene regenerates carbocation intermediate.

• Since addition of chloride to carbocation is irreversible, E1 does not interfere with SN1 in our experimental conditions.


ClCl

H2OH

alkyl chloridealkeneSN1

E1

HCl




Silver Nitrate Test for Alkyl Chlorides

8

ClAg+ NO3

-

Lewis base

Ag+ = Lewis acid H3C CH3

CH3


O N O

OAgCl(s)

O N O

O

alkyl nitrate

+ +

(precipitatesin EtOH)

RDS: SN1

• Positive test = observe AgCl precipitate in EtOH.• Ag+ is a strong enough Lewis acid (e- acceptor) to remove Cl-. • As a result, carbocation intermediates are formed. Mechanism =

“SN1-like.”• Therefore, reactivity of alkyl chlorides = 3º>2º>>1º.




Sodium Iodide Test for Alkyl Chlorides

9

ClNa+ I–

RDS: SN2I + NaCl(s)

(precipitatesin acetone)

FYI: Also known as the Finkelstein reaction.

• Positive test = observe NaCl precipitate in acetone.• Na+ is not a strong enough Lewis acid (e- acceptor) to remove Cl-. • As a result, carbocation intermediates are not formed. Mechanism =

SN2.• Therefore, reactivity of alkyl chlorides = 1º>2º>>3º.




Summary: Classification Tests for Alkyl Chlorides

10

Test SolventObserved

PrecipitateMechanism

Reactivity of R–X

Silver Nitrate(AgNO3)

ethanol AgCl SN1 3º>2º>>1º

Sodium Iodide(NaI)

acetone NaCl SN2 1º>2º>>3º




Preparation is Key to Success

11

• Labs 8-10 are technically challenging and can only be completed within the class time if you are adequately prepared.

• Preparation includes a thorough understanding of the background information and procedural steps before you enter the lab.

• Write a clear procedure that you intend to follow in your prelab notebook entry so that you can get to work right away on those days.

• Working within these time constraints will be good practice for the practical exam, which is also challenging.



CH3

O

H3C

H3C OH

H3C

H3C OH

H H2SO4, H2O, HgO

reflux, 30 min

Hg2+

2-methyl-3-butyn-2-ol (1)cyclic mercuroniumion intermediate (2)

3-hydroxy-3-methyl-2-butanone (3)

34

4H

H3C

H3C OH

3

Lab 8: Hydration of a Terminal Alkyne. Preparation of 3-Hydroxy-3-methyl-2-butanone.

CH3

N

H3C

H3C OH

semicarbazone 8

HN

O

NH2

m.p. = 162-163 ºC

H3C

semicarbazone 9

H

NNH

O

NH2

m.p. = 222-223 ºC

CH3




Safety First

2

• Mercuric oxide and semicarbazide are highly toxic.

• Wear gloves, protective clothing and goggles at all times.

• Place ALL waste in waste bottles–nothing down the sink.

HgO H2N NH

ONH2




Procedure Notes

3

Apparatus for Heating at Reflux

Reflux Condenser(use Hempel column)

H2O in

H2O out

Notes:1. Since there are no stirplates, hand-swirl the

r.b. flask until ready for reflux. Caution: flask may be hot since the reaction is exothermic.

2. Collect product by steam distillation directly from the reaction flask (next slide).

3. Extract product from steam-distillate with CH2Cl2 according to textbook procedure. After drying, transfer combined organic layers to a PREWEIGHED r.b. flask.

4. Remove CH2Cl2 by simple distillation only (no vacuum). Do NOT distill final product as the textbook instructs. ALL CH2Cl2 must be removed for successful derivatization.

5. Weigh r.b. flask+product (calculate a “crude” yield) before proceeding with the semicarbazide derivitization.

Reflux: the state of a boiling liquid whose vapors are continually returned by condensation; typically used to hold reactions at the boiling point of the solvent for extended periods of time




Steam Distillation Apparatus

4

stillheadClaisenadapter

separatoryfunnel

(used here asa dropping funnel)

water &reaction mixture

replacementwater

(boiling hot!!)

collect 30-40 mLof distillate

West condenser

Notes:

1. No need for boiling stones; mercuric oxide provides sufficient sites for nucleation.

2. Steam is generated in situ by boiling water in the stillpot.

3. Replacement water is added at the same rate distillate is exiting the condenser.

4. Separatory funnel is actually being used during the distillation as a dropping funnel.

5. Collect approximately 50 mL of distillate. Distillate will be cloudy since product and water are immiscible.

6. Volatile product is co-distilling with the water vapor.




Hydration of a Terminal Alkyne is Regioselective

5

C

OH3C

H3C OH

H3C

H3C OH

H H2SO4, H2O, HgO

reflux, 30 min

Hg2+

2-methyl-3-butyn-2-ol (1)cyclic mercuroniumion intermediate (2)

34

4 HH3C

H3C OH

3

Hg1+

4 HH3C

H3C OH

3

2º vinyliccarbocation

O HH

4 HH3C

H3C OH

3

O Hg1+

Hg1+

4 HH3C

H3C OH

3


H

H+Hg

H H

1+

H+

CH3

OH3C

H3C OH3-hydroxy-3-methyl-2-butanone (3)

Hg2+

+ Hg2+

regioselective: preferential reaction at one site of a functional group over other sites that could undergo the same reaction

• Carbon-3 is the most electrophilic since a 1º vinylic carbocation is not a contributing resonance structure. It would be too high in energy.

• Water (nucleophile) adds preferentially (regioselective) to the most substituted–and consequently the most electrophilic–carbon atom in the alkyne (C-3).




• Hg2+ forms a stable (low energy) mercuronium intermediate that can react further with water.

• Without Hg2+, the alkyne would have to be protonated to give a free 2º vinylic carbocation. Unstabilized vinylic carbocations are high energy species and thus unlikely intermediates.

Why is Hg2+ Catalyst Needed?

6

H3C

H3C OH

H

H2SO4, H2O

Hg2+

2-methyl-3-butyn-2-ol (1)

cyclic mercuroniumion intermediate (2)

HH3C

H3C OH

Hg1+

HH3C

H3C OH


Hg2+

HgO

H2SO4, H2O

no HgOno reactionX

H+

H3C

H3C OH

H

H2º vinylic

carbocation

unstable intermediate

stable intermediate




• Semicarbazones are typically crystalline solids.• The m.p.s of many semicarbazones are reported in the literature.• Therefore, m.p. analysis of semicarbazone derivatives can confirm the

product of hydration.

Semicarbazone Derivitization Confirms Product

7

H3C

H3C OH

H

2-methyl-3-butyn-2-ol (1)

34

CH3

OH3C

H3C OH

3-hydroxy-3-methyl-2-butanone (3)

H3C

H3C OH

3-hydroxy-3-methyl-butanal (7)

H

O

CH3

NH3C

H3C OH

semicarbazone 8

H3C

semicarbazone 9

H

N

HN

O

NH2

NH

O

NH2

m.p. = 162-163 ºC

m.p. = 222-223 ºC

semicarbazidederivitization

H2SO4, H2O, HgO

reflux, 30 min

MarkovnikovAddition

anti-MarkovnikovAddition

semicarbazidederivitization

CH3




Mehcanism: Semicarbazide Condensation

8

NH

O

NH2H2N N

H

O

NH2N

H

HN

O

NH2H2N

H

• Only the beta-nitrogen atom (second atom adjacent to the carbonyl carbon) of semicarbazide undergoes addition to the ketone or aldehyde. It’s the most nucleophilic.

• The electron pairs on the alpha nitrogen atoms (first atom adjacent to the carbonyl carbon) are delocalized through resonance. Therefore, they are less available for bonding to other atoms, which makes that nitrogen atom less nucleophilic.

R

O

R NH

O

NH2H2N+

most nucleophilicnitrogen atom

semicarbazide

CH3

NH3C

H3C OH

HN

O

NH2

+ H2O




Recrystallization

9

Desired Product(yellow/large circles)

+Impurities

(small/blue circles)

Dissolve product and impurities in a minimum amount of solvent at high

temperature.

Cool solution. Product crystallizes out of

solution. Impurities remain dissolved.

Vacuum filter solution through a Buchner funnel. Mother liquor (filtrate)

contains impurities.

Requirements for Recrystallization Solvent

1. Not reactive with desired product(s).

2. Desired product is completely soluble at elevated temperature, but only slightly soluble–or not soluble at all–at room temperature.

3. Undesired components (impurities) are highly soluble at all temperatures (small temperature coefficient).

4. Easily removed from crystalline product by filtration and evaporation. filtrate




IR Spectrum of 2-Methyl-3-butyn-2-ol

10

CH3C

H3C OH

CH

5001000150020002500300035004000Wavenumbers

40

50

60

70

80

Csp3–HCsp–HO–H

• C≡C band not seen• dipole moment very small

• What starting material bands to you expect to disappear in the reaction product?

• What bands do you expect to appear in the reaction product?

3364

32

95

2984



Lab 9: Preparation of Ethyl Oct-3-ynoate (13) through a Cu-Catalyzed Diazo Coupling Reaction to a Terminal Alkyne.

C C H C OEt

ONN

H

+ C OEt

O

HH

CC

a

b

baa

a b

bdiazocoupling

(insertion)+ N2

11 12 13



Me C C H

O

OEtNN

H+

CuI, CH3CN, rt

Diazocoupling:This Week

O

OEt

Ni(OAc)2, NaBH4Ethylenediamine

EtOH, H2

Hydrogenation:Next Week

C OEt

OCMe

34

HH

CCMe3

4

O

OEtCCMe3

4

11 12 13

13 14: ethyl (Z)-oct-3-enoate

• Primary goal = synthesize 14 through hydrogenation of 13, which is prepared by diazocoupling reaction

• 14 is a surrogate/substitute for biodiesel (shorter chain = more volatile)• Your material 14 will be tested by Prof. Bergstrom’s group at UIC to

determine whether the alkene in the side chain is responsible for increased NOx emissions observed in biodiesels.




Safety First

3

• Ethyl diazoacetate is highly toxic.• Ethyl diazoaceate is explosive when heated, contacts acid.• Add this reagent in the hood, behind glass.• Wear gloves, protective clothing and goggles at all times.

• Place ALL waste in waste bottles–nothing down the sink.

O

OEtNN

H

O

OEtN2

H

ethyl diazoaceate (EDA)




Tasks to Complete Today

4

1. As a group, fill in the reaction table in your data page using the values given on the following slide. Calculate missing quantities such as mmol and equivalents. Your TA will check this table.

2. Designate someone to record data in the group data page. This will be turned into your TA at the end of the day with your group’s IR spectra. Your TA will use this page to evaluate the results section of your lab report.

3. Prepare 13 according to the procedure in the course manual.

4. Remove solvent, weigh product and obtain IR spectra. Save remaining product for next week’s reaction. Seal flask tightly.




Diazocoupling Reaction Table

5

compound M.W. (g/mol) d (g/mL) or M (mmol/mL)

Rxn weightor volume

(include unit)

You calculate!

mmol equivalents(molar ratio)

CuI(catalyst) 190.45 – 0.306 g

CH3CN(acetonitrile) 41.05 d = 0.786 20 mL

1-hexyne (11)only 97%! 82.14 d = 0.715 2.0 mL

ethyl diazoacetate (12)only 85%! 114.10 d = 1.085 2.0 mL

1.0(limiting rgt)

Me C C H

O

OEtNN

H+

CuI, CH3CN, rt O

OEtCCMe3

4

1-hexyne (11) ethyl diazoacetate (12) ethyl 3-octynoate (13)



Lab 10: Stereoselective Reduction of an Alkyne

aa

g

g

f f, ed

d

b

be

Z

H3C

H H

H HH H

O

O CH3

H HH H

Z

catalyst surface

H H

M0

catalyst surface MII

H H


H H

oxidativeaddition

d

σ∗


H H

coordinationinsertion

reductiveelimination

H Hsyn addition

A

BD

C




Safety First

2

1. Hydrogen (H2) is highly flammable. Do not use H2 balloons near heat or open flames. Avoid static electricity.Avoid H2/air mixtures.

2. NaBH4 is toxic and flammable. Minimize exposure to air.

3. Ni(OAc)2 is a carcinogen. Wear protective clothing, goggles and gloves at all times.

4. Post-reaction catalyst may be flammable when “dry.” Do not filter to complete “dryness.”




NMR Analysis: Did The Diazo Coupling Work?(This Analysis was Performed the Previous Semester)

3

a

aa

a

g

g

g

g

f

f

f, e

f, ed

d

d

b

b

b

b

d

e

e

H3C

H H

H H

H H

HX

X

Y

O

O CH3

H H

H

NN

Y

Z

H3C

H H

H HH H

O

O CH3

H HH H

Z

residual benzenesolvent (C6H6)




5001000150020002500300035004000Wavenumbers

40

60

80

100

IR Analysis: Did Your Diazo Coupling Work?

4

5001000150020002500300035004000Wavenumbers

20

40

60

80

100

5001000150020002500300035004000Wavenumbers

40

60

80

100

HCC

C OCH2CH3

ON

H

N

OCH2CH3

OCC

N=N=N C=O

Csp–H

C=O




Stereoselective Ni-Catalyzed Hydrogenation

5


EtOH, H2C OEt

OCMe

34

HHO

OEtCCMe3

4

13: ethyl 3-octynoate 14: ethyl cis-3-octenoate

OHCCMe3

4

16: 3-hexyn-1-ol


EtOH, H2C OHCMe34

HH

17: ethyl cis-3-hexen-1-ol

AlternativeHydrogenation

1. If your diazo coupling was successful. . .

2. If you did not obtain 13, fear not . . .




Theoretical Yield of 13

6

Last week you started with 2.0 mL of 12, which is the limiting reagent. So, . . .

EDA bottle indicated only 85%

EDA density

EDA Mol. Wt.

reaction stoich.ratio

13Mol. Wt.

Starting vol. of 12

13Theoretical

Yield

2.0 mL 12 × 0.85 mL 121 mL 12

×1.085 g 121 mL 12

×1 mol 12

114.10 g 12×

1 mol 131 mol 12

×168.23 g 13

1 mol 13= 2.72 g 13

Me C C H

O

OEtNN

H+

CuI, CH3CN, rt O

OEtCCMe3

4

1-hexyne (11) ethyl diazoacetate (12) ethyl 3-octynoate (13)

C6H10Mol. Wt.: 82.14360

density: 0.715

C4H6N2O2Mol. Wt.: 114.10264

density: 1.085

C10H16O2Mol. Wt.: 168.23284




Reaction Table

7

If your measured mass of 13 is less than 2.72 g, enter that value in the table below. Do not use a starting mass higher than 2.72 g, which is the theoretical yield of 13.

compound M.W. (g/mol) d (g/mL)or M (mol/L)

Rxn weightor volume

(include unit)mmol equivalents

(molar ratio)

Ni(OAc)2•4H2O 248.86 – 1.0

ethanol(EtOH)

46.07 0.789 g/mL 50 mL – –

NaBH4 in ethanol (EtOH)

37.83 1.0 M (mol/L) 1.0of NaBH4

NaOH (aq) 40.00 2.0 M (mol/L) 0.10

ethylenediamine 60.10 0.90 g/mL 4.0

ethyl 3-octynoate (13)

168.23 – 2.72 g 16.2 1.0

ethanol(EtOH)

46.07 0.789 g/mL 15 mL – –


EtOH, H2C OEt

OCMe

34

HHO

OEtCCMe3

4

13: ethyl 3-octynoate 14: ethyl cis-3-octenoate




NMR Interpretation

8

1. Course website: www.chadlandrie.com → CHEM 233 → NMR Spectra → Lab 10 Analysis

2. Additional prelab lecture notes that were not presented today. (Download lab 10 lecture notes from website.)

You must download your NMR spectra from the course website. For help interpreting the 1H-NMR spectra of your product, use the following resources:






Distinguishing Between Cis and Trans Alkenes Using 1H-NMR Coupling Constants (J-values)

9

cis: Jb,c = 4-12 Hz

O

O CH3

HbHcH3C

Hd Hd Ha Ha

trans: Jb,c = 14-19 Hz

Ja,b

Jb,c Jb,c

Jc,d

Ja,b

Jb,c

Jc,d

Jb,c

O

O CH3Ha Ha

H3CHc

HbHd Hd

5.73 ppm

HcHb

~ 5.45 ppm*If Hb and Hc are not well resolved and/or overlap with each

other, it may be difficult or impossible to determine Jb,c.




Determining Coupling Constants (J-values)

10

5.74

23 p

pm5.

7223

ppm

5.70

23 p

pm

5.70

90 p

pm5.

6890

ppm

5.66

90 p

pm

Ja,b

Jb,c

J-value (Hz) = ∆ δH (ppm) * spectrometer frequency (MHz)

Ja,b = (5.7423 - 5.7223) * 300 MHzJa,b = 6.0 Hz

Examples

Jb,c = (5.7423 - 5.7090) * 300 MHzJa,b = 10.0 Hz




NMR Analysis: Did The Hydrogenation Work?(Previous Semester’s Results)

11

a

ag

f

f, e

d

d

b

bec

H3C

H H

H HH H

O

O CH3

H HH H

c g

O

O CH3

HHH3C

H H H H

HH

HH

H Hg

c

a

b

d

e

f

YZ

bc d

f, e

a

g

Y Z




Cis or Trans?

12

Z YO

O CH3

HHH3C

H H H H

HH

HH

H H

cd

ZY

JYZJYc

JYd JZYJZd

JZc

** 500 MHz Spectrometer **

JYZ = (5.7412-5.7197)*500 = 10.8 Hz

JYc = (5.7588-5.7445)*500 = 7.2 Hz

JYd = (5.7588-5.7557)*500 = 1.6 Hz

JZY = (5.4803-5.4587)*500 = 10.8 Hz

JZd = (5.4984-5.4837)*500 = 7.4 Hz

JZc = (5.4984-5.4950)*500 = 1.7 Hzcis = 4-12 Hz

chem 233: organic laboratory i prelab lecture university...

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