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Or£{21111C ChCmiStry Carbonyls and Alcohols Carbonyl Reactivity
Example 5.12
Whichof the following is the MOST electrophiliccarbonylcompound?
A. Amide B. Ester
C. Aldehyde D. Acid anhydride
Solution
The reactivity of a carbonyl is dictated by the leaving group on the carbon of the carbonyl. As the acidity of the conjugate acid of the leaving group increases, so does the reactivity of that particular carbonyl. In the choices above, the conjugate acids of the leaving groups are an amine, an alcohol, H2, and a carboxylic acid respectively. The most acidic is the carboxylic acid, so the anhydride is the most reactive carbonyl.
Deprotonation of oc-Protons
The hydrogen on the alpha carbon (the carbon adjacent to the carbonyl carbon) is acidic (its pKa is approximately 19), so it can be removed using a strong base.
Enolates are formed when a hydrogen on the alpha carbon is deprotonated. The enolate can regain a proton at either the carbon or the oxygen. If it is protonated at the oxygen, an enol is formed. There is an equilibrium between the ketone and
enol. The conversion from a ketone into an enol is known as tautomerization, because a ketone and its enol are tautomers, structural isomers that vary in the position of a 7t-bond and a hydrogen. The tautomerization of acetone is shown in Figure 5-19.
Ketone Carbanion Enolate Enol
CH,
! base
Figure 5-19
The carbanion that forms is a good nucleophile. When an alkyl halide is added to the solution, the carbanion can attack the alkyl halide in a nucleophilic substitution reaction to form a new carbon-carbon bond. This results in a longer ketone. The halide can be any halide, but the reaction works best with an alkyl iodide compound. Alkyl bromides and chlorides yield more O-alkylation side products than alkyl iodides. The generic reaction is shown in Figure 5-20.
Ketone Carbanion Longer Ketone
Figure 5-20
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OrgaillC ChemiStry CarbonylsandAlcohols Carbonyl Reactivity
When the carbonyl is asymmetric, the possibility of two different enolates arises (each formed by deprotonating a different alpha hydrogen). If one of the two alpha carbons is more substituted, then it is both more sterically hindered and more stable. This creates a situation where the reaction can be dictated by temperature and base size. At low temperature with a bulky base, the less hindered (less substituted) alpha carbon gets deprotonated to form the so called kinetic enolate (lower energy transition state). This is referred to as kinetic control.
At a higher temperature with a small base, the more hindered (more substituted) alpha carbon gets deprotonated to form the so called thermodynamic enolate (leading to themore stable product). This is referred to as thermodynamic control.
We shall apply this concept to aldol condensation reactions of asymmetric
ketones later on in this section.
Example 5.13
What is the final product after acetone is treated first with NaH, followed by iodoethane, and subsequently followed by workup?
A. 2-Methyl-2-butanol B. 3-Methyl-2-butanone
C. 2-Pentanone D. 2-Butanone Solution
This reaction is similar to the generic reaction in Figure 5-20. The carbon chain length is increased by two carbons (from three to five) when the electrophile is ethyl iodide. This eliminates choice D. The product is a ketone, so choice A is eliminated. The final product is 2-pentanone, as shown below, which makes
choice C the best answer.
:o: :o: :o:
II __ II II
CH3 B£ CH3 H2C ^CH3
H1>
H^^I base ^ ^ H3CH2C
n2^ L 2-Pentanone
/
H3C
Oxidation and Reduction
Oxidation and reduction are recurring in organic chemistry, so working from a logic-based foundation is key. If the oxophilic carbon (carbon containing abond to oxygen) hashydrogens, it canbe oxidized. Primary alcohols are oxidized into aldehydes, which can be further oxidized into carboxylic acids. Secondary alcohols are oxidized into ketones. Tertiary alcohols cannot be oxidized (the alcohol carbon has no hydrogen to lose). Reduction is defined as the opposite of oxidation, so the reverse ofeach reaction just mentioned represents reduction.
To make the processes more clear, we shall define oxidation and reduction in
terms of bonds to oxygen and bonds to hydrogen. More than just oxygen containing compounds do redox chemistry. When two cysteine residues form a crosslink, they undergo dehydrogenation (loss of hydrogen), an oxidative process. When the rc-bond in a fatty acid is hydrogenated to form an aliphatic chain, it has undergone a reductive process.
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OrgaillC ChCmistry Carbonyls and Alcohols Carbonyl Reactivity
Oxidation is defined as an increase in oxidation state, which is caused by either losing a bond to a lesselectronegative atom (in mostcases hydrogen) or gaininga bond to a more electronegative atom (in most cases oxygen). Reduction is definedas a decreasein oxidation state, which is caused by either gaining a bond to a less electronegative atom (in most cases hydrogen) or losing a bond to a more electronegative atom (in most cases oxygen).
Oxidation Reduction
Gain of bonds to O Loss of bonds to O Loss of bonds to H Gain of bonds to H Increase in oxidation state Decrease in oxidation state
Determining oxidation states using the method learned in general chemistry is a matter of assigning oxygen a -2 (because it is more electronegative than the atoms to which it bonded, and it makes two bonds) and a +1 to hydrogen (because it is less electronegative than the atoms to which it bonded, and it makes one bond).
The oxidation state of any remaining atoms is found by difference. In organic chemistry, oxidation states for specific atoms can easily be found by considering electron sharing in each bond. If the bond is between two atoms of unequal electronegativity, then the more electronegative atom is assigned a -1 and the less electronegative atom is assigned a +1. The oxidation state of an atom is found by summing the numbers from all of the bonds and any formal charge it may have. Figure 5-21 shows this method as it applies to the oxidation state of carbon 2 in 2-propanol and acetone.
[O] ^
lost a bond to H
gained a bond to O Figure 5-21
Each of the four bonds to carbon is analyzed for its relative electronegativity compared to the atoms to which it is bonded. Bonds to hydrogen give a negative to carbon and a positive to hydrogen, because carbon is more electronegative than hydrogen. Bonds to oxygen give a positive to carbon and a negative to oxygen, because carbon is less electronegative than oxygen. Both carbons in a carbon-carbon bond get zero, because there is no difference in electronegativity.
In a secondary alcohol,the oxophiliccarbon has an oxidationstate of 0 while in a ketone, the oxophilic carbon has an oxidation state of +2. This means that the carbon was oxidized by two electrons, which is predictable, because it has lost a bond to hydrogen (oxidizing it by one electron) and has gained a bond to oxygen (oxidizing it by another electron). Figure 5-22 shows that oxygen and hydrogen do not change oxidation state when going from 2-pentanol to acetone.
0C0?0CH3
O A -2
H3C"0~0C0^CH2 @
Figure 5-22
Oxygen has a -2 oxidation state, as is expected. Hydrogen has a +1 oxidation state as expected. Oxidation states should be made this simple.
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OrgailiC Chemistry Carbonyls and Alcohols Carbonyl Reactivity
Oxidizing and Reducing Agents
Oxidizing agents are rich in oxygen (given that they deliver oxygen to the reactant) and reducing agents are rich in hydrogen (given that they deliver hydrogen to the reactant). This perspective is applicable in both organic chemistry and biochemistry. It should be noted that NAD+ acts like Cr03 with pyridine, NADH acts like NaBH4, and FADH2 acts like H2/Pd. Ethanol can be oxidized into ethanal using either Cr03 in the presence of HC1 and pyridine (known as PCC) or by NAD+. In the reduction half-reaction, NAD+ is reduced to NADH, because for every oxidation half-reaction, there is a reduction half- reaction. The coupledhalf-reactions are shown in Figure 5-23.
Oxidized by 2 e"
Reduced by 2 e
NAD+ NADH + H+
Figure 5-23
Oxidizing agents cause oxidation (and in doing so, they get reduced). Oxidizing agents are rich in oxygen, poor in hydrogen, and have an atom in a high oxidation state with high electron affinity. Reducing agents cause reduction (and in doing so, they get oxidized). Reducing agents are poor in oxygen, rich in hydrogen, and have anatom ina low oxidation state with low ionization energy.
Some common oxidizing and reducing agentsare listed below.
Oxidizing Agents fRich in O: Poor in H) Reducing Agents TPoor in O:Rich in H)
KMn04 Cr03 O3 UAIH4 NaBH4 H2NNH2
RCO3H ROOR Cl2 H2/Pd HCl/Zn HOCH2CH2SH
NAD+ NADP+ FAD NADH NADPH FADH2
Typical oxidation and reduction reactions involve the conversion between
alcohols and carbonyl compounds. Figure 5-24 is a summary of common oxidation-reduction reactions in organic chemistry (1° alcohol -ằ aldehyde -ằ
acid and2° alcohol ->ketone - 3* alcohols undergo no oxidation reactions).
Primary alcohols have twohydrogens to lose, so they are oxidized twice (first into an aldehyde and then into a carboxylic acid.) The carboxylic acid is reduced back into a primary alcohol using LiAlH4.
H
O
R J-OH oxidizing agent || oxidizing agent
R- H
H
alcohol reducing; agent aldehyde
O
R- OH
carboxylic acid reducing agent
Secondary alcohols haveonehydrogen to lose, so they are oxidized once.
O
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H
ằ+OH oxidizing agent
R- •R' R'
2° alcohol reducing agent ketone Figure 5-24
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Organic Chemistry carbonyls and Aicohois
Oxidation
From an organic chemistry perspective, oxidation is the process of gaining oxygen (and/or losing hydrogen) at the carbon which has an electronegative atom attached. The first rule is to count the number ofcarbon-oxygen bonds to the carbon of interest in the reactant. As the number ofcarbon-oxygen bonds increases, the oxidation level (andoxidation state) ofthecompound increases. In terms ofoxidation levels, every carbon-oxygen bond counts as one. Ifyou lose a carbon-hydrogen bond and replace it with a carbon-oxygen bond, the oxidation level has increased by one while the oxidation state of the carbon has increased by two. The oxidation of a primary alcohol into an aldehyde and then into a carboxylic acid is shown in Figure 5-25.
HO. H
H3C/oCNH
Oxidation State:
0 +1 -1 -1 = -1
Cr03/HCl Pyridine
O
H3C/0 ^H
Oxidation State:
0 + 1+1-1 =+1
K2Cr?07
—-—^-^-ằ
H2S04(aq)
O +111+1
HaC^"0 ^OH
Oxidation State:
0 + 1 + 1 + 1=+3
Figure 5-25
Youshould be able to recognize reagents, the reaction type, and correctly predict the product for redox reactions. As a point of interest, Cr04 with HCl in pyridine is known as pyridinium chlorochromate, or PCC for short. Most organic compounds lose carbon-hydrogen bonds and gain carbon-oxygen bonds when oxidized, so the reagent (oxidizing agent) must be rich in oxygen. The two
most common oxidizing agents in organic chemistry are CrO^" and Mn04".
Some common oxidation reactions of alcohols are shown in Figure 5-26.
R
\
H
Primary alcohol
OH
"PCC"
CrQ3Cr
Pyridinium N+-H
H*7 S.
H
Primary alcohol
DH K2Cr2°7,
OH
H
^>C_ OH
R R
Secondary alcohol Tertiary alcohol
H2S04(aq)
KMn04 KOH(aq)
[O]
Figure 5-26
O
II
R H
Aldehyde
O
II
R OH
Carboxylic acid
O
II
R R
Ketone No Reaction
(No Hs to lose)
Carbonyl Reactivity
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Organic Chemistry carbonyisandAicohois Carbonyl Reactivity These reactions should be known well enough that you can recognize them in a group of many reactions. For other oxidation reactions, you should simply recognize the oxidizing agent and know that the product is more oxidized than the reactant. Figure 5-27 shows some oxidation reactions of compounds other than alcohols. The common theme is that one of the reactants is rich in oxygen.
R H
HH R
Disubstituted alkene
(H3C)2S
HIO,
R H
\ ^OH
H R
Vicinal diol
H3COH/H20
o - C6H5C03H
Ketone
R
>-O
H
Aldehyde
R
H
Aldehyde
Lactone (Ester)or
Figure 5-27
03 has excessive oxygen;
alkene reactant is being
oxidized
HI04 has excessive oxygen; diol reactant is being oxidized
C6H5C03H has excessive oxygen; ketone reactant is being oxidized
The three oxidation reactions shown in Figure 5-27 are ozonolysis of an alkene, oxidative cleavage of a vicinal diol using periodic acid, and the Baeyer-Villiger oxidation of a ketone respectively. These reactions are rare enough that they don't need to be memorized, but you should recognize that each is an oxidation.
More important than memorizing organic chemistry reactions is the ability to predict results and explain observations based on the data given about these reactions in the passages associatedwith them. For instance, excessive oxygen in the reactantshould providea clue that the organicmolecule is being oxidized.
Example 5.14
What is the majororganic product for the following reaction?
OH OH
KMnQ4^
OH"(aq)
A- O O B' O OH c O O D* OH O
Solution
The secondary alcohol is oxidized into a ketone, which eliminates choice D. The primary alcohol is oxidized all the way to a carboxylicacid, which makes choice
C the best answer.
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OrgaillC ChemiStry Carbonyls and Alcohols Carbonyl Reactivity
Example 5.15
Which of the following compounds is NOT an oxidizing agent?
A. Peroxybenzoic acid (C6H5CO3H) B. Ozone (O3)
C. Diisobutylaluminum hydride [((H3C)2CHCH2)2A1H]
D. Chromic acid (H2Cr04)
Solution
Organic oxidizing agents generally contain oxygen atoms, like choices A, B, and D. Diisobutylaluminum hydride (DIBAH) contains no oxygens, and in fact contains hydrogens. DIBAH is a selective reducing agent that is weaker than lithium aluminum hydride (LLAIH4). The best answer is choice C.
Example 5.16
Oxidation of an aldehyde into a carboxylic acid requires the addition of:
A. an oxidizing agent.
B. a reducing agent.
C. a strong base.
D. a strong acid.
Solution
Oxidation requires the addition of an oxidizing agent to carry out the oxidation.
Oxidation can be carried out in either acidic or basic conditions, so choices C and D are eliminated. The best answer for this question is choice A.
Example 5.17
When a primary alcohol reacts with potassium permanganate, the primary alcohol is acting as which of the following in the reaction?
A. An oxidizing agent B. A reducing agent C. A strong base D. A strong acid
Solution
Because the alcohol is being oxidized, it is losing electrons. This means that it is causing the reduction of Mn04" into Mn02- Reducing agents get oxidized, so the primary alcohol is the reducing agent. The best answer is choice B.
Example 5.18
Oxidation of a secondary alcohol leads to:
A. an aldehyde.
B. a carboxylic acid.
C. a ketone.
D. a tertiary alcohol.
Solution
As shown in Figure 5-26, a secondary alcohol is oxidized into a ketone and no further, because it has only one hydrogen to lose to oxidation. The best answer for this question is choice C.
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Organic Chemistry Carbonyls and Alcohols Carbonyl Reactivity
Example 5.19
Which of the following compounds CANNOT be oxidized by chromic acid?
A. 2-Methyl-3-pentanol B. Buteraldehyde
C. 3-Pentanol D. Phenol Solution
Primary and secondary alcohols can be oxidized while tertiary alcohols and phenols cannot be oxidized. Choices A and C are secondary alcohols, so they can be oxidized into ketones. Aldehydes are oxidized into carboxyKc acids, so choice B is eliminated. This leaves choice D, phenol, as the best answer. The carbon bonded to oxygen in phenol has no hydrogen atoms, so it cannot be oxidized.
Example 5.20
To oxidize a primary alcohol into an aldehyde without further oxidation into a carboxylic acid, what reagent should you add in conjunction with chromic acid?
A. Pyridine and hydrochloric acid B. Toluene and hydrochloric acid C. Pyridine and sulfuric acid
D. Toluene and sulfuric acid Solution
A primary alcohol is oxidized into an aldehyde and no further when pyridinium
chlorochromate is added to the solution. This is known as PCC or Collin's
reagent. The absence of water from solution prevents the aldehyde from further oxidation intoa carboxylic acid. Thebestanswerfor this question is choice A.
Reduction
Reduction is the process of losing oxygen (and/or gaining hydrogen) at the carbon which has an electronegative atom attached. As with oxidation, the first step is to count the number of carbon-oxygen bonds in the reactant to the carbon of interest. As the number of carbon-oxygen bonds decreases, the oxidation level decreases (as does the oxidation state). If you lose a carbon-oxygen bond and replace it witha carbon-hydrogen bond, the oxidation level decreases by oneand the oxidation state of carbon decreases by two. From a ketone or aldehyde to an alcohol, the oxidation statedecreases by two and the leveldecreases by one. The reduction of an esterintoa primary alcohol is shownin Figure 5-28.
O HO H
+1"+1 1. LiAlH4(thf) +FM
y0^ * xttt ^7 *• x-0^ + HOCH2CH3
H^^OCH^H, ZNH*°™ H3C^^H
Oxidation State: 0 +1 + 1-+1 = +3 Oxidation State: 0 + 1-1-1=-1
Figure 5-28
The two most common reducing agents in carbonyl chemistry are LLAIH4 and NaBH4. Aluminumis less electronegative than boron, so LiAlH4 is more apt to donate an H" to the carbonyl carbon, and thus is more reactive than NaBH4- As such, NaBH4 reduces only ketones and aldehydes while LiAlH4 reduces all carbonylcompounds. Figures 5-29a and 5-29b show these carbonyl reductions.
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Organic Chemistry carbonyls and Alcohols
R' H
Aldehyde
O
R" R' Ketone
O O
II
Carboxylic acid
O
II
R OR'
Ester O
II
Aldehyde
O
II
R R'
Ketone
1. LiAlH4(thf) 2. NH4Cl(aq)
1. LiAlH4(thf) 2. NH4Cl(aq)
1. NaBH4(et2Q)^
2. NH4Cl(aq)
1. NaBH4(et2Q) 2. NH4Cl(aq)
HO, H
\ ?
R/C^H
Primary alcohol
HO.
.C. + HOR'
R H
Primary alcohol
H
HO. H
+ H,0
R H
Primary alcohol
HO, H
R R'
Secondary alcohol Figure 5-29a
HCl/Zn(Hg)^ H
or ^
1. HS(CH2)2SH
H
2. Raney Ni H2NNH2^
KOH(aq) *
2 equivalents HSCH2CH2OH
R' H
Alkane
H H
^ C \ + N2 + H20
W R'
Alkane
r
SH HS*
Reduced Linkage
H H
*~ R*/ Y'/H
S-CH2CH2OH S—CH2CH2OH Disulfide linkage
R H
H2 FADH2
• o r
H Pd
H CH2COzH
Unsaturated fatty acid
H CH2C02H
Saturated fatty acid Figure 5-29b
The aldehyde-to-alkane reactions are Clemmensen reduction and Raney nickel reduction, respectively. They reduce either an aldehyde or ketone to an alkane.
Carbonyl Reactivity
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Organic Chemistry Carbonyls and Alcohols Carbonyl Reactivity
Example 5.21
What is the major organic product after pentanal is treated with sodium borohydride (NaBHi)?
A. A primary alcohol B. A secondary alcohol
C. A hemiacetal
D. A carboxylic acid
Solution
Aldehydes can be oxidized into a carboxylic acid or reduced into a primary alcohol. Sodium borohydride is a reducing agent, so an alcohol is formed. This eliminates choices C and D. The carbonyl in an aldehyde is on the C-l carbon, so the hydroxyl group forms on C-l carbon, resulting in the formation of a primary
alcohol. Choice A is the best answer.
Example 5.22
Acetone, when reacted with lithium aluminum hydride and quenched with weakly acidicwater, yields which of the following products?
A. Propane B. 1-Propanol C. 2-Propanol D. Propene
Solution
Acetone is a ketone, thus when it is reduced, it forms a secondary alcohol. The only secondary alcohol listed is choice C, so choice C is the best answer.
Example 5.23
To carry out the following synthetic transformation, what reagent should be
added?
Q Q O
CH2OH
tY- - &
A. LiAlH4 B. NaBH4 C. H2N=NH2 D. FADH2
Solution
In this reaction, an aldehyde has been reduced into a primary alcohol on a compound where an ester functional group is unaffected. FADH2 carries out hydrogenation of a 7c-bond, so choice D is eliminated. Hydrazine, H2NNH2 reduces aldehydes and ketones to alkanes, so choice C is eliminated. Both LiAlH4 and NaBH4 reduce aldehydes to primary alcohols, but LiAlH4 would
also reduce the ester. This eliminates choice A and makes choice B the best answer. NaBH4 is selective for aldehydes and ketones.
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OrganiC ChemiStry Carbonyls and Alcohols Carbonyl Reactivity
Example 5.24
What is the oxidation state of aluminum in LiAlH4?
A. +3 B. +1 C. -3 D. -5 Solution
The term hydride implies that hydrogen has a negativecharge associated with it.
Lithium is +1, so the Al must have an oxidation state of +3 to have the charge on the compound be zero. The best answer is choice A.
Example 5.25
What is the major organic product for the following reaction?
O O
1. NaBH4(thf)
A A 2.NH4Cl(aq)- •
r i
A- OH OH B' O OH C O O D- OH O
yX^ sX^J -X^KoH /k/JkoH
Solution
Sodium borohydride, NaBHj, reduces aldehydes and ketones only. Both the ketone and the aldehyde functional groups are reduced to alcohols (secondary and primary respectively). This yields a 1,3-diol, so the best answer is choice A.
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