NMR

1

• NMR
• Chemical Shift
• Protons in a strong magnetic field will absorb
  Radio Frequency radiation at a specific frequency
  (Resonance).
• The frequency of absorption (in Hz) relative to
  the frequency of the NMR instrument in MHz) is
  reported in units of PPM and is a function of the
  electromagnetic force required to induce the
  proton to resonate.
• The resonance signal, reported in PPM, is plotted
  on the X-axis of the NMR chart. It location is
  referred to as the Chemical Shift.
• Chemical Equivalence Protons having the same
  chemical and magnetic environment will produce
  one NMR signal.
• The valence electrons in the vicinity of a proton
  provide Diamagnetic Shielding of the proton,
  i.e., the electron density about the proton.
• Electronegative Groups decrease electron density
  by withdrawing electrons from the area around the
  proton thus less electromagnetic energy is
  required to induce resonance. The NMR signal thus
  moves down field increasing the Chemical Shift.
• Electron Donating Groups donate electrons to the
  area around the proton, increasing electron
  density. Thus, the NMR signal moves upfield
  decreasing the Chemical Shift.
• Spin Spin Splitting (the n1) rule
• The number of peaks produced by equivalent
  Protons on a carbon atom is determined by the
  number of protons on the adjacent carbon
  atoms.Example Two protons on a carbon atom
  attached to two other carbon atoms with a total
  of 5 protons will act as a group to produce n1
  (516) peaks, i.e., a sextet.
• Peak Areas Area under absorption peak is
  proportional to the number of protons present.
  Compute relative peak area by measuring the
  stairstep heights. See Pavia on page 915 or web
  site notes.
• Look for Methyl groups around 1.0 PPM, Methylene
  Groups (CH2) around 2.0 PPM Aromatic Protons
  around 7.0 PPM. Aldehyde proton around 9.5 PPM
  Carboxylic Acid Proton around 10.5 PPM.
• The presence of nearby Electronegative elements
  (Oxygen, Halogens, Nitrogen groups) will decrease
  the electron density causing the signal to move
  downfield, i,e,. increase the Chemical Shift.

2

• NMR (Cont)
• Aromatic Rings and other unsaturated molecules
• Some proton types have chemical shifts not
  explained by the Electronegativity of the
  attached groups.
• Aryl (benzene ring), Alkene (C), Alkyne (C ),
  and Aldehyde (OCH) related protons have
  resonance effects not expected from
  electron-withdrawing effects of electronegative
  elements.
• The movement of ? electrons in an unsaturated
  molecule also generates a magnetic field that can
  influence the applied magnetic field.
• The relative shielding and deshielding of protons
  in groups with ? electrons is dependent on the
  orientation of the molecule with respect to the
  applied magnetic field
• Activating groups will add electron density and
  provide additional shielding of the protons.
• Deactivating Groups will decrease electron
  density resulting in decreased shielding.
• In a Benzene ring , the ? electrons are induced
  to circulate around the ring by the applied
  magnetic field, creating a ring current, which in
  turn produces a magnetic field further
  influencing the shielding of the ring protons.
• The presence of ring current causes the applied
  magnetic field to become non-uniform (diamagnetic
  anisotropy) in the vicinity of the benzene ring.
• The effect of the anisotropic field is to further
  deshield the benzene protons increasing the
  chemical shift. i.e., absorption signal moves to
  the left (downfield)on the chart.
• Thus, protons attached to the benzene ring are
  influenced by three (3) magnetic fields
• Strong Applied Magnetic Field
• Normal Shielding by the Valence Electrons
• Anisotropic Effect from Ring Current

3

• NMR (Cont)
• Protons on atoms other than Carbon atoms (acids,
  alcohols, amines, amides, etc) also produce
  Absorptions, usually singlets, with highly
  varying Chemical Shifts.
• Carbon 13
• Spin-Spin Splitting is different than in H1 NMR
• The splitting peaks produced by a given carbon
  atom are based on the number of protons attached
  to that carbon atom.
• Coupled Decoupled (protons removed) spectra.
• Most Carbon atoms, except truly chemically
  equivalent carbons, produce unique absorptions
  thus, C-13 is mostly used to identify the number
  of unique carbon atoms in the molecule.
• NMR Analysis Approach
• Check for signal at Chemical Shift 9-10 ppm
  Aldehyde present
• Check for signal at Chemical Shift 10-11 ppm
  Carboxylic Acid
• Check for presence of Methyl groups at Chemical
  Shift in vicinity of 1.0 ppm.
• The signal for a Methyl group may be moved
  downfield under the influence of an
  electronegative group or atom.
• Look for peak area integration values near the
  signal. The value represents the relative number
  of protons on the carbon atom. If the value is a
  multiple of 3, there is more than one
  equivalent carbon atom involved, i.e., 2 or 3
  equivalent Methyl groups (Iso or Tertiary
  groups).
• If the Methyl group signal is a multiplet, the
  number of peaks in the multiplet is equal to the
  total number of protons on the carbon atoms
  attached to the Methyl Carbon plus 1 (n1 rule).
• There may be multiple Methyl groups present,
  some of which could be equivalent producing one
  signal (singlet or multiplet) and other
  non-unique Methyl groups each producing its own
  signal (singlet or multiplet).

4

• NMR (Cont)
• NMR Analysis Approach (Cont)
• Check for the presence of Methylene groups.
• The same process used to determine the number of
  Methyl groups is the same for Methylene
  groups, except here the Area Integration values
  would be multiples of 2.
• The presence of Benzene rings is indicated by
  signals with Chemical Shifts of about 7.
• The signals for Benzene rings are often complex
  because of the anisotropic effect of the ring
  current and substitution patterns on the ring.
• The Area integration value often gives some idea
  of how many of the six (6) ring protons have been
  substituted.
• There may be other singlets not associated with
  the Methyl or Methylene groups that are
  associated with protons attached to atoms other
  than carbon OH, NH, NH2, etc.
• If available
• Evaluate the IR for functional groups present.
• Evaluate the Mass Spectra for the Molecular
  Weight and the presence of Chlorine (2 peaks, 31
  abundance ratio), Bromine (2 peaks, 11 abundance
  ratio), or Nitrogen where the Molecular Ion Peak
  value is Odd.
• Assemble the various fragments into possible
  compound configurations.

5

• NMR (Cont)
• NMR Analysis Approach (Cont)
• NMR Analysis Example
  2-Phenylbutane

Note Each of the fragments has one or more
unresolved sigma bonds. Try matching
an unresolved bond in one fragment with an
unresolved bond in another fragment.
Some fragments may be accounted for as
partial fragments of other
fragments. If any unresolved bonds
persist, check the molecular weight to see a
symmetrical molecule might be involved.
6

• Infrared Spectrophotometry (IR)
• Principal Absorptions
• Carbonyl Group - 1715 cm-1 (Aldehydes, Ketones,
  Acids, Esters, Amides, Anhydrides)
• C-O - 1000 - 1300 cm-1 (Alcohols, Ethers,
  Esters, Acids)
• Saturated - C-H (Alkanes) - Right side of 3000
  cm-1
• Unsaturated - C-H (Alkenes) - Left side of
  3000 cm-1
• Unsaturated - Aromatic versus Alkenes
• Aromatic - Overtone area
  1667 to 2000 cm-1 Out of Plane (OOP)
  690 to 900 cm-1 4 sharp
  absorptions in pairs 1450 to 1600 cm-1
• Alkynes - CC (2150 cm-1) C--H (Terminal)
  (3300 cm-1)
• Nitriles - CN (2250 cm-1)
• Nitro (NO2) - 1300 1390 cm-1 1500 to 1600
  cm-1
• Amines - 3300 to 3500 cm-1 Primary Amine (Two
  Bands) Secondary Amine (One Band, often weak)
  Tertiary Amine (No Absorptions)
• Analysis Approach
• Carbonyl Group - Confirm presence or absence of
  Carbonyl (CO) group, a strong absorption at
  about 1715 cm-1, actual range could be from 1600
  1820 cm-1.
• Saturation / Unsaturation
• Saturation - Confirm presence or absence of
  saturated alkane(-C-H) structures, i.e.,
  multiple absorptions between 2900 3000 cm-1.
• Unsaturation - Confirm presence or absence of
  unsaturated(C-H) structures, i.e., multiple
  absorptions between 3100 3000 cm-1.

7

• Determine whether unsaturated structures belong
  to alkene or aromatic structures.
• Aromaticity is confirmed as follows
• 1- 4 weak absorptions in aromatic overtone region
  from 1700 2000 cm-1
• 4 sharp absorptions that occur in pairs from 1450
  1600 cm-1.
• Out of Plane (OOP) bending occurs at 690 900
  cm-1.
• Determine substitution patterns on benzene ring
  using table on page 897 of the Pavia text.
• Alkene structures are confirmed as follows
• Out of plane (OOP) bending from 650 1000 cm-1
• CC stretch from 1600 -1675 cm-1
• Determine substitution patterns on the alkene
  structures from table on page 895 of the Pavia
  text.
• Hydroxyl (OH) group
• Alcohols / Phenols Stretch (sharp peak) from
  3600 - 3650 cm-1
• Acids Stretch, usually very broad (from
  hydrogen bonds) at2500 - 3300 cm-1
• C-O group
• Alcohols Stretch in range of 1000 1300 cm-1
• Ethers Stretch in range of 1000 1300 cm-1
• Esters Stretch (sometimes 2) in range of
  1000 1300 cm-1
• Amines
• N-H Stretch occurs in the range 3300 3500 cm-1.

8

• Nitro Compounds
• NO stretch is usually two strong bands at 1300
  -1390 cm-1 1500 1600 cm-1
• Nitriles
• CN stretch (sharp) absorption near 2250 cm-1
• Alkynes
• C-H (terminal alkyne) stretch (sharp) is usually
  near 3300 cm-1
• C-CC-C stretch (sharp) is usually near 2150
  cm-1

9

• Mass Spectrometry
• Mass Spectrum with Molecular Ion Peak(s).Note
  Sometimes the Mass Spectrum is not given, just
  the value(s) for the Molecular Ion
  peak (M).
• Molecular Ion Peak (M)
• Peak(s) farthest to the right on the Mass
  Spectrum
• Represents the Molecular Weight
• Molecular Ion peak values that are Odd indicate
  the presence of an Odd number of Nitrogen atoms
  in the compound
• Two Molecular Ion peaks with a relative abundance
  ratio of 31 indicate the presence of a single
  Chlorine
• Two Molecular Ion peaks with a relative abundance
  ratio of 11 indicate the presence of a single
  Bromine.
• Ultraviolet Visual Spectrometry
• UV-Vis Molar Absorptivity (Molar Extinction
  Coefficient ? log ?
• Conjugate systems (alternating double bonds -
  ?,? - Unsaturated ketones, Dienes,
  Polyenes)show values of ? log ? in the
  range ? 10,000 100,000 (Log ?
  4 5)
• Aromatic Conjugated Systems show values of ? and
  log ? in the range ? 1000 10,000
  (Log ? 3 4)
• Carbonyl (CO) compounds show values of ? and log
  ? in the range ? 10 100
  (Log ? 1.5 2.5)
• Nitro (-1ON O) compounds show values of ? in
  the range ? lt10
  ( Log ? lt 1.0)

10

• Partial Elemental Analysis
• A partial elemental analysis of the compound
  usually provides Carbon
  Hydrogen.
• The student must complete this analysis and give
  the Molecular Formula of the compound.
• Example From the Mass Spectrum the Molecular
  Weight is 58.080
  C - 62.0 H 10.4
• 0.620 58.080 36.03 / 12.01
  ?3 ? 3 Carbons atoms
• 0.104 58.080 6.05 / 1.001
  ?6 ? 6 Hydrogen atoms
• Determine mass of remaining elements in the
  compound
• 58.080 (36.03 6.05) 15.95
  16 ? 1 Oxygen
• ? Molecular Formula C3H6O
• Note The molecular ion peak(s), molar
  absorptivity coefficient (from UV- VIS), and
  the principal functional groups from the IR
  provide the information necessary to
  identify any additional elements present in
  the compound.

11

• The Laboratory Process
• Know the Stoichiometric Reactions and the
  Reaction Mechanisms.
• All synthesis experiments require the
  determination of reactant masses, moles,
  stoichiometric molar ratios, limiting reagent,
  and theoretical yield.
• Note These items usually placed at the beginning
  of a lab report.
• Products are sometimes purified by
  Recrystallization
• Note Know the details of this process and the
  properties of a good solvent.
• Unknown Liquids are usually first purified by
  Simple Distillation
• Describe the data collection process in Simple
  Distillation to explain how the pure component
  is separated from the impurities.
• Explain why measured boiling points usually dont
  match the literature boiling points.
• Intermediate product mixtures are often washed
  or extracted
• Explain the Extraction process
• What equipment is used?
• What reagents were used?
• What is venting
• Did product change phases?
• The Refractive Index is determined for most
  organic liquids, either unknowns or synthesized
  products.
• How is the measured value corrected for
  temperature?
• ND20 NDRT ?t ( 0.00045)
• ?t Room Temp (RT) - 20

12

• The Chem 315 Experiments
• Melting Point Refractive Index
• Melting Point
• Define Melting Point
• What is melting point range?
• What happens to the melting point and melting
  point range of a mixture as the concentration of
  one of the compounds increases relative to the
  concentration of the other compounds in the
  mixture?
• What is the relationship between the amount of
  contaminant in a mixture and the Eutectic Point
• Refractive Index
• Define Refractive Index
• What is the impact of temperature on Refractive
  Index?
• How is the measured value corrected for
  temperature?
• Recrystallization
• What is the purpose of Recrystallization?
• Distinguish between Recrystallization and
  Precipitation.
• What constitutes a Good solvent know at least
  5 properties.
• Describe the recrystallization process.

13

• The Chem 315 Experiments (Cont)
• Gas Chromatography Acetate Mixture
• What are the principles of separation in Gas
  Chromatography?
• Liquid Phase
• Carrier Gas
• Moving Gas Phase
• Stationary Liquid Phase
• Time a compound stays in the Vapor Phase is a
  function of the Vapor Pressure of the compound.
• The more volatile (Low Boiling Point / Higher
  Vapor Pressure) compounds arrive at the end of
  the column first and pass into the detector.
• What are the factors affecting the separation of
  compounds?
• Boiling Points (vapor pressure) of compounds.
• Flow Rate of Carrier Gas.
• Choice of Liquid Phase Molecular weights,
  functional groups, and polarities of component
  molecules are factors in selecting liquid phase.
• Length of Column Similar compounds require
  longer columns than dissimilar compounds.
  Isomeric mixtures often require quite long
  columns.
• What is retention Time?
• The period following injection that is required
  for a compound to pass through the column to the
  point where the detector current is maximum, i.e.
  maximum pen deflection or maximum peak height.
• For a given set of constant conditions (carrier
  gas, flow rate of carrier gas, column
  temperature, column length, liquid phase,
  injection port temperature), the retention time
  of any given compound is always constant.

14

• The Chem 315 Experiments (Cont)
• Gas Chromatography Acetate Mixture (Cont)
• What is the relationship between the area under a
  Gas Chromatography peak and the Mole content of
  the mixture?
• The area under a gas chromatograph peak is
  proportional to the amount (moles) of the
  compounds eluted.
• The molar percentage composition of a mixture can
  be approximated by comparing the relative areas
  of the peaks in the chromatogram.
• This method assumes that the detector is equally
  sensitive to all compounds and its response is
  linear.
• Triangulation Method of Determining Area Under
  Peak.
• Multiply the height of peak (in mm) above the
  baseline by the width of the peak at half the
  height.
• The Baseline is a straight line connecting
  side arms of the peak. Best if peaks
  are symmetrical.
• Add all of the areas to get the total area.
• Divide each area by total area to get mole
  fraction.

15

• The Chem 315 Experiments (Cont)
• Simple Fractional Distillation
• Pure Substance
• Temperature remains constant during distillation
  process so long as both vapor and liquid are
  present
• Liquid Mixture
• Temperature increases throughout process because
  composition of vapor changes continuously.
• Composition of vapor in equilibrium with the
  heated solution is different from the composition
  of the solution.
• What is Simple Distillation?
• Single vaporization- condensation cycle of a
  mixture that produces a distillate that is always
  impure at any temperature range between the range
  of boiling points of the components.
• What is Fractional Distillation?
• Accomplishes the same thing as multiple simple
  sequential vaporization-condensation cycles, by
  inserting a Fractionating Column between the
  Distillation Flask and the Distilling Head.
• Gas Chromatography of Distillates
• Determine the Mole of the original sample
  mixture and the fractions obtained from the
  Simple Fractional Distillations along using Gas
  Chromatography and Refractive Index.
• Compare the Mole results obtained from the
  three methods with an evaluation of the best
  method to physically separate the mixture
  components and determine the Mole composition of
  the mixture.

16

• The Chem 315 Experiments (Cont)
• Synthesis of t-Butyl Chloride
• Reaction of t-Butyl Alcohol (or t-Pentyl Alcohol)
  with conc. HCL to form t-Butyl Chloride (or
  t-Pentyl Chloride).
• Stoichiometric Equation and Reaction Mechanism
• Three-step Sn1 Nucleophilic Substitution
  Reaction.
• This is a First Order Rate Reaction where the
  Rate of Formation of t-Butyl Chloride (t-Pentyl
  Chloride) is dependent only on the concentration
  of the Alcohol it is independent of the amount
  of acid (HCL) used.
• The yield (mass or moles) of the washed and
  driedt-Butyl (t-Pentyl) Chloride product is
  compared to the theoretical amount of product
  expected, which is computed from a Limiting
  Reagent calculation using the Stoichiometric
  Molar Ratio.
• The Limiting Reagent is that reactant whose
  mass (on a molar equivalent basis) is totally
  consumed in the reaction leaving an excess of the
  other reactant.
• The Limiting Reagent, thus, determines the
  maximum amount (in moles) of product that can be
  expected.

17

• The Chem 315 Experiments (Cont)
• Qualitative Organic Analyses
• Solubility (H2O, H2SO4)
• Beilstein (Flame) for Alkyl Aryl Halides
• Silver Nitrate/Ethanol (Sn1 reaction) for Alkyl
  Aryl Halides
• Compounds equipped with good leaving groups (H2O,
  CL, Br, I)
• Compounds that can form stable carbocations
• Benzyl gt Allyl gt Tertiary (3o) gt
  Secondary (2o) gt
• Primary (1o) gt Methyl gt Vinyl gt Aryl
  (Aromatic).
• Allyl, Benzyl, Tertiary Halides give positive
  test (White PPT).
• Primary Secondary Alkyl Halides test positive
  when heated (100oC).
• Aromatic and many Vinyl Substituted Halides do
  not give positive tests.
• Sodium Iodide/Acetone (Sn2 reaction) for Alkyl
  Aryl Halides
• Relative Halide reactivity for an Sn2 reaction is
  the opposite of an Sn1 reaction, that is
• Vinyl gt Methyl gt Primary (1o) gt Secondary (2o) gt
  Tertiary (3o) gt Allyl gt Benzyl gt Aryl
  (Aromatic)
• Primary Alkyl Chlorides and Bromides would react
  with the Sodium Iodide in a reaction to
  substitute the Chloride Bromide ions with the
  Iodide ions. An immediate precipitate is formed.
• Secondary Alkyl Halides will give a precipitate
  when heated to 50oC and then cooled.
• Tertiary Alkyl Halides will also give a
  precipitate when heated to 50oC and then cooled.
• Unreactive Aryl (Benzyl and Aromatic) Chlorides
  Bromides would not produce a reaction even after
  heating, thus no precipitate.

18

• The Chem 315 Experiments (Cont)
• Qualitative Organic Analyses (Cont)
• Bromine/Methylene Chloride for Unsaturated CC
  CC
• Addition reaction of Bromine (Br2), a red liquid,
  to a compound containing a double or triple bond
  produces a colorless Dibromide.
• The double (or triple bond) must be sufficiently
  electron-rich to initiate the reaction.
  Therefore, minimal electron withdrawing groups
  (Deactivators), such as Carboxyl Groups attached
  to molecule, would hinder the reaction.
• Unsubstituted Aromatic compounds do not react
  with the Bromine reagent.
• Even if the ring has substituted activating
  groups (donate electrons to the ring) the
  reaction would be a substitution and not an
  addition.
• KMnO4 (Baeyer Test) for Unsaturated CC CC
• Potassium Permanganate (KMNO4) (Purple Color) is
  an oxidizing agent.
• Following the oxidation of an unsaturated
  compound, the Permanganate ion is reduced to
  Manganese Dioxide (MnO4), a brown precipitate.
• Note Other easily oxidized compounds
  Aldehydes, some Alcohols, Phenols, and Aromatic
  Amines should be accounted for in your
  analysis.
• Ignition for Aromatic CC bonds
• Positive test is a sooty yellow flame.
• Note The Sooty flame usually comes off
  fairly quickly. Look for it moving
  quickly away and upward from the
  yellow/blue flame area.
• Positive test is indicative of a high degree of
  Unsaturation and is probably Aromatic

19

• The Chem 315 Experiments (Cont)
• Qualitative Organic Analyses (Cont)
• Acetyl Chloride for Alcohols
• Acid Chlorides react with Alcohols to form
  esters.
• Acetyl Chloride forms Acetate esters.
• This test does not work well with solid alcohols.
• Positive test is evolution of Heat and Hydrogen
  Chloride (HCl) gas.
• Phenols also react with Acetyl Chloride and
  should be eliminated prior to testing for
  Alcohols.
• Amines also react with Acetyl Chloride to produce
  heat and also should be eliminated prior to
  testing.
• Lucas Test for Alcohols
• Primary Alcohols dissolve in reagent giving clear
  solution.
• Secondary Alcohols produce cloudiness after about
  3-5 minutes. May need to heat slightly.
• Tertiary, Benzylic, and Allylic alcohols produce
  immediate cloudiness eventually, an immiscible
  Alkyl Halide separates into a separate layer.
• Chromic Acid for Alcohols
• Distinguish Primary Secondary Alcohols from
  Tertiary Alcohols.
• Chromic Acid (Cr6) oxidizes Primary and
  Secondary Alcohols to Carboxylic Acids and
  Ketones, respectively.
• Chromium (6) - orange is reduced to Chromium
  (3) green.
• Tertiary Alcohols do not react with Chromic Acid.

20

• The Chem 318 Experiments
• Bromination of Toluene
• Electrophilic Aromatic Substitution
• The Methyl group is electron donating (ring
  activating) favors O/P subst.
• Meta vs Ortho/Para Substitution Which prevails
  in this experiment and why?
• Positively charged Bromine Cations (the
  Electrophile) how are the Bromine cations
  produced, i.e., the reaction between Bromine and
  Iron (Fe)
• Nitration of Methyl Benzoate
• Carbo-Methoxy group is electron withdrawing
  therefore it favors Meta substitution of the
  Nitronium ions - Why?
• Reaction producing the Nitronium Ions (H2SO4
  HNO3.
• Water Impact of too much water in reaction.
• Temperature Keep low or high Why?
• Grignard
• Bromobenzene Magnesium ? Phenyl Magnesium
  Bromide
• Grignard CO2 (Electrophile) ? Benzoic Acid
• Grignard Ketone ? Tertiary Alcohol
• Grignard Aldehyde ? Secondary Alcohol
• Grignard Formaldehyde ? Primary Alcohol
• Function of Ether (Stabilize the Grignard)
• Impact of too much heat (Production of Biphenyl)

21

• The Chem 318 Experiments (Cont)
• Qualitative Analysis
• Classification Tests for Aldehydes Ketones
• Solubility in Water - Insoluble gt 5 Carbons
• Solubility in H2SO4 - Most Aldehydes Ketones
• 2,4-Dimethylphenylhydrazine - Aldehydes Ketones
• Yellow ppt Unconjugated Red ppt
  Highly Conjugated
• Chromic Acid - Reduction of Cr 6 (orange)
  to Cr3 (green ppt) for Aldehydes Note
  Primary Secondary Alcohols, which must be
  eliminated first, also test positive.
• Tollins Reagent - Silver Mirror for
  Aldehydes Note Secondary Alcohols with
  Alpha Carbon adjacent to Carbonyl also test
  positive.
• Iodoform Test - Yellow ppt for Methyl
  Ketones Note Acetaldehyde Sec Alcohols
  with Alpha Carbon adjacent to Carbonyl also
  test positive.
• Synthesis of Iso-Pentyl Acetate (Banana Oil)
• Acid Catalyzed Esterification
• Excess Acetic acid required to maximize yield,
  i.e., move equilibrium to right.
• Purpose of washing with Sodium Bicarbonate
  Remove Acid
• Purpose of washing with Satd Sodium Chloride
  Remove Water

22

• The Chem 318 Experiments (Cont)
• Dibenzalacetone
• Claisen-Schmidt Reaction involving formation of
  an Enolate ion.
• Two moles Benzaldehyde react with 1 mole acetone
  to form 1 mole of Dibenzalacetone.
• The Molar ratio of Benzaldehyde to Acetone is
  21, which must be taken into account when
  determining the Limiting Reagent.
• Product is washed with distilled water to remove
  all traces of Sodium Hydroxide.
• Acetanilide
• Amines vs Amides Amines are weakly basic and
  Amides are less basic than Amines.
• Amides stronger acidity is related to the strong
  electron withdrawing effect of the Carbonyl
  Oxygen.
• Amines are easily oxidized, but can be protected
  by first converting them to an Amide.
• The Synthesis of Acetanilide, an Amide, from
  Aniline, an Amine, is by Nucleophilic Acyl
  Substitution, an Addition/Elimination reaction.
• Acetic Anhydride acts as the positively charged
  (Carbonyl Carbon) Electrophile.