Transcript Document
Aldehydes and Ketones
Part II
Carbonyl vs. Cyano Group
O C
carbonyl
C N
cyano These groups appear to be totally different.
Carbonyl vs. Cyano Group O
N C
C
They are similar: Carbon bonded to a heteroatom. O makes two bonds, N makes three bonds. So, we simply have an extra
bond in the cyano group.
Discussion Questions for Online Students
Answer the following questions for both the carbonyl and cyano groups.
Where would H + bond?
Where would a nucleophile attack?
How would H 2 O add?
How many H 2 O molecules can add?
Online Students Answer Questions Offline, Send Solutions to Instructor as an Attachment to an e-mail Not accepted after Feb 8, 2002
Hydrolysis of a Nitrile
•
An alcohol contains a hydroxyl (-OH) group.
•
A nitrile contains a cyano (-CN) group.
•
Nitrile is the family name; a nitrile contains a cyano group. You recognize that a nitr ile contains nitr ogen.
Hydrolysis of a Nitrile
• • • •
Hydrolysis means clevage (olysis) by water (hydro).
How does water cleave a nitrile?
Water adds to
has two
bonds. A cyano group bonds, so two water molecules add to one cyano group, making a gem diol.
What do gem-diols do?
How does Water Add to a Nitrile?
• • • •
Water can be considered H + and OH .
The OH adds to the C of the cyano.
The H + A
adds to the N of the cyano.
bond remains, so the process is repeated: OH to C and H + to N.
•
The result: a gem-diol (and amino group).
Addition of Water to a Nitrile
R C N nitrile + H acid-catalyzed hydrolysis OH H + OH R C N H Step 1 (OH adds to C and H adds to N) H OH H + Step 2 (OH adds to C and H adds to N) The gem diol goes to an amide in limited water and to an acid in excess water.
OH OH R C OH OH
+
H 2 O H N H 2 R C N OH H 2 geminal triol geminal diol H 2 O O R C OH carboxylic acid H 2 O Step 3 (gem diol losses water) O OH R C N H 2 OH geminal diol R C N H 2 amide The amide can be converted to an acid by hydrolysis.
Compare
•
The addition of water to a carbonyl group of an aldehyde or ketone
•
• With
The addition of water to the cyano group of a nitrile
4 bonds O
C
+
H
OH O C H 4 bonds OH Bonding occurs between + and - poles.
4 bonds N
C
+
H
OH N C H 4 bonds OH
Go Back to the Previous Slide and Ask Yourself the Following Question
•
What are the similarities and differences in the addition of water to a carbonyl and cyano group?
Key to Both Reactions
• •
Identifying the positive and negative poles of the reactants Getting the first step correct by bonding the positive end (H + ) of the reagent water to the negative end (O) of the substrate and the negative end (OH ) of the reagent to the positive end (C) of the substrate, while keeping four bonds to carbon.
Comparison
•
Similarities
• Mode of addition is the same.
• H adds to heteroatom • OH adds to carbon • A bond is cleaved •
Differences
• Carbonyl has one bond, cyano group has two • Water (H, OH) add twice
For Cyano: A Second H 2 O Adds 4 bonds
N H C
OH
+
H
OH 4 bonds N C H OH 2 OH
gem-diol
When you ask, "how does the substrate react?" The answer is: just like any addition to a polar double bond.
A Geminal Diol Loses Water
• • •
Two groups bonded to the same carbon atom are geminal (twins).
A geminal or gem-diol is two ols (OH groups) bonded to the same C atom.
Gem-diols are generally less stable than the corresponding carbonyl group, so gem-diols lose water to form or reform a carbonyl group.
N H 2 C OH OH
gem-diol
-
H 2 O N H 2 C
=
O O General Principle: Gem-diols lose H 2 O.
C N H 2
acid amide
Oxidation-Reduction Reactions
•
General Principles:
•
Oxidation is the gain of oxygen or loss of hydrogen.
Reduction is the gain of hydrogen or loss of oxygen.
Oxidation is an Increase in Oxidation Number
The
oxidation number
of any covalently bonded atom is found by assigning valence electrons to that atom and subtracting that number of valence electrons from the atom’s group number in the periodic table.
• The group number gives that atom’s normal number of valence electrons.
• The oxidation number is the difference between the normal number of valence electrons and the number of assigned electrons.
Procedure for Finding an Oxidation Number of a Covalent Atom
• • • •
Show all valence electrons with a Lewis Structure .
Assign bonding valence electrons to the more electronegative atom of the pair sharing those electrons.
Assign nonbonding valence electrons atom that has (owns) them .
to the Subtract the sum of assigned bonding and nonbonding electrons from the atom’s group number in the periodic table.
Oxidation Numbers for Water
Lewis Structure
H O H
Assign bonding electrons: O is more electronegative than H, so in any bond between O and H, O is assigned or given all of the bonding electrons. For water, O shares four bonding electrons with H. O gets all = 4. H = 0.
Assign nonbonding electrons. O has 4, H = O.
Sum of assigned electrons for O = 4 + 4 = 8 O's group number = 6.
Oxidation Number of oxygen = 6 - 8 = -2 Oxidation Number of each H = 1 - 0 = +1
Oxidation Numbers for Methane
Lewis Structure
H H C H H
Assign bonding electrons: C is more electronegative than H, so in any bond between C and H, C is assigned or given all of the bonding electrons. For methane, C shares eight bonding electrons with H. C gets all = 8. H = 0.
Assign nonbonding electrons. C has 0, H = O.
Sum of assigned electrons for C = 8 + 0 = 8 C's group number = 4.
Oxidation Number of carbon = 4 - 8 = -4 Oxidation Number of each H = 1 - 0 = +1
Oxidation Numbers for Carbon Dioxide Lewis Structure
O C O
Assign bonding electrons: O is more electronegative than C, so in any bond between C and O, O is assigned or given all of the bonding electrons. For CO 2 , C shares eight bonding electrons with O. O gets all = 8. C = 0.
Assign nonbonding electrons. C has 0, each O = 4.
Sum of assigned electrons for C = 0, O = 8 C's group number = 4. O's group number = 6.
Oxidation Number of carbon = 4 - 0 = +4 Oxidation Number of each O = 6 - 8 = -2
Alternative Method
• • •
H will be +1 in hydrocarbons and derivatives, and O will be –2.
Since the sum of oxidation numbers is the overall charge of the chemical species, the oxidation state of a single carbon can be found mathematically.
However, for two or more carbon atoms this method gives an average oxidation number.
Oxidation Numbers of Underlined Atoms CH 3 CH 2 CH 3 = H H 3 C C CH 3 H assigned = 6, Ox.No. = -2 O O CH 3 CCH 3 = H 3 C C CH 3 assigned = 2, Ox. No. = +2
Oxidation Numbers of Underlined Atoms CH 3 OH sum = 0 = 4 - 2 + x x = -2 CH 2 O sum = 0 = 2 - 2 + x x = 0 x = oxidation number of carbon.
Find the Oxidation Number of the Underlined Atom in Each of the following Structures.
CH
4
CH
3
OH CH
2
O HCO
2
H CO
2
Solutions
CH 4
-4
CH 3 OH
-2
CH 2 O HCO 2 H
0 +2
CO 2
+4
Preparation of Aldehydes and Ketones
• •
Oxidation of I o alcohols to aldehydes Oxidation of II o alcohols to ketones
R Preparation of Aldehydes from I o Alcohols
-1
OH mild oxidation CH 2 C H I o alcohol H strong oxidation
+1
O R CH 2 aldehyde C H R
+3
O CH 2 acid C OH
Preparation of Aldehydes from I o Alcohols
• • • • •
Requires a mild oxidizing agent.
A strong oxidizing agent makes an acid.
The oxidation number of C changes from –1 to + 1.
The alcohol loses two H’s.
Loss of H’s is oxidation.
Mild Oxidizing Agents
•
PCC = pyridinum chlorochromate (pyridine poisons chromium +6 somewhat, reducing its oxidizing power)
•
Swern (a man’s name) = DMSO (dimethylsulfoxide), oxalyl chloride, and triethylamine.
Complete the following Equations
OH PCC OH Swern
Preparation of Aldehydes from I o Alcohols R
-1
O H
PCC
CH 2 C H I o alcohol H
or Swern
mild oxidation
-2
H
+1
R CH 2 O C H aldehyde Oxidation is an increase in oxidation number!
(-1 to +1 is an increase)
Oxidation Methodology
• •
Carbon atoms that contain oxygen and hydrogen are candidates for further oxidation.
•
Remove an H bonded to the C that contains O and replace the H with OH.
When you get a gem-diol, dehydrate (lose water and make a carbonyl group.
Oxidizable Carbon Atoms
C
OH C
H
H I o alcohol
C
OH C
C
H II o alcohol O C H aldehyde Carbon atoms in the red square are oxidizable.
Mild Oxidation of I
o
Alcohol
OH OH
C
C
H
mild (O)
C
H I o alcohol C OH
H
gem-diol - H 2 O
C
O C
H
aldehyde (O) = oxidation (net loss of 2 Hs)
Strong Oxidizing Agents
• • • • •
KMnO 4 and K 2 Cr 2 O 7 are strong oxidizing agents.
They contain Mn +7 and Cr +6 , respectively.
The transition metal ions are reduced, generally to Mn +4 and Cr +3 . KMnO Mn +4 4 is purple and K is brown and Cr +3 2 Cr 2 O 7 is orange.
is green.
Strong Oxidation of I
o
Alcohol
OH OH
C
C
H
strong (O)
C
H I o alcohol C OH OH gem-diol - H 2 O
C
O C acid OH (O) = oxidation (net loss of 2 Hs, gain 1 O)
Strong Oxidation of Aldehyde
O C H aldehyde strong (O) O C acid OH (O) = Oxidation (net gain of 1 O)
Mild or Strong Oxidation of II o Alcohol
C
OH C
C
(O) H II o alcohol
C
OH C
C
OH gem-diol
C
(O) = Oxidation (net loss of 2 Hs) O C C ketone
Summary of Oxidation Reactions
• • • •
I o Alcohols can be oxidized to aldehydes with mild oxidizing agents and to acids with strong oxidizing agents.
Aldehydes can be oxidized to acids with strong oxidizing agents.
II o Alcohols can be oxidized to ketones with mild or strong oxidizing agents.
Methodology is the same for all.
Ozone (O
3
) as an Oxidizing Agent
• • • • •
Ozone, an allotrope of oxygen, is a powerful oxidizing agent.
It cleaves both the sigma and pi bonds of a double bond.
Each half of the clevage gains an oxygen with a double bond (i.e., a carbonyl group).
Aldehydes formed by this reaction will be oxidized to acids unless a reducing agent is added before workup of the reaction.
Reducing agents are zinc (dust) or dimethylsulfide (Me 2 S).
Ozonolysis
• • •
Ozonolysis literally means clevage with ozone. Step 1 of an ozonolysis is a reaction of a double bond with O 3 .
If Zn/HCl or (CH 3 ) 2 S is used in Step 2, then any aldehydes produced will not be oxidized to acids.
If Zn/HCl or (CH 3 ) 2 S is not used in Step 2, then any aldehydes produced will be oxidized to acids.
Oxidative vs. Reductive Workup
•
When Zn/HCl or Me 2 S is not added, the reaction is an oxidative workup (aldehydes will be oxidized, thus oxidative).
•
When Zn/HCl or Me 2 S is added, the reaction is a reductive workup (aldehydes will not be oxidized, because these reducing agents neutralize any excess ozone, thus reductive).
Ozonolysis of (Z)-2-Butene with Oxidative Workup CH 3 C H C CH 3 H (1) O 3 (2) H 2 O CH H 3 C C CH 3 H CH 3 HO C
O O
acids C CH 3 OH CH 3 H C
O O
aldehydes C CH 3 H
Summary Ozonolysis
• • • •
Ozone cleaves double bonds C=C to give two carbonyl compounds.
If one or both products are aldehydes, they may be isolated with a reductive workup, or they may be further oxidized to acids with an oxidative workup.
If one or both products are ketones, they are not further oxidized. Thus, ozonolysis is a way to convert alkenes into aldehydes and ketones.
Ozonolysis of (Z)-2-Butene with Reductive Workup CH 3 C H C CH 3 H (1) O 3 (2) Zn/HCl or (CH 3 ) 2 S CH 3 C H C CH 3 H CH 3 H C
O O
aldehydes C CH 3 H