Free Energy, Entropy, Thermodynamic Equilibrium

Entropy, Free Energy and Thermodynamic Equilibrium

Chemical reactions are performed by mixing the reactants and regulating external conditions such as temperature and pressure. Two basic questions though arise:

1. Is it possible for the reaction to occur at the selected conditions?
2. If the reaction proceeds, what determines the ratio of products and reactants at equilibrium? 

Both questions are answered by chemical thermodynamics:

Thermodynamics can tell us whether a proposed reaction is spontaneous (possible) under particular conditions even before the actual experiment.
Thermodynamics can also predict the ratio of products and reactants at equilibrium provided that the reaction is spontaneous.

Note: Thermodynamics cannot answer though how fast a reaction will proceed.

 

After many years of observation scientists concluded that the characteristic common to all spontaneous processes (processes that occur without outside intervention) is an increase in the property called entropy (S).

 

How entropy is defined?

A precise, quantitative definition of entropy was proposed by the Austrian physicist Ludwig Boltzmann in the late 19th century. According to this definition entropy is related to probability:

If a system has several states available to it, the one that can be achieved in the greatest number of ways (has the largest number of microstates) is the one most likely to occur. The state with the greatest probability has the highest entropy.

S = k_B . lnΩ

Where,

k_B is Boltzmann’s constant (R/N_A)

Ω is the number of microstates corresponding to a given state (including both position and energy)

Note: The above definition of entropy is not useful in a practical sense for the typical types of samples used by chemists because those samples contain so many components (for example 1 mole of gas contains 6.022 x 10²³ individual particles).

 

How entropy is associated with chemical processes?

Entropy changes, ΔS – not S – are associated with changes of state (from solid to liquid, liquid to gas…). Since a change of state – for example from solid to liquid – at a substance’s melting point is a reversible process, we can calculate the change in entropy for this process by using the equation:

ΔS = q_rev / T = ΔΗ / Τ      (at constant temperature T and pressure P)

Where:

ΔS change in entropy that occurs during the change of state

q_rev  = ΔΗ / Τ  energy required for the reversible process to occur (for example energy required to melt 1 mole of solid at the melting point, ΔΗ is the enthalpy change of fusion)

T is the temperature where the change of state occurs (melting point, boiling point)

Phase Changes - Energy Changes - Heating Curves

 

Many important properties of liquids and solids relate to the ease with which they change from one state to another. Water for example, when heated it evaporates that is changes from liquid to the gas state. In general, each state of matter (solid, liquid, gas)  can change into either of the the other two states. Figure I.1 shows these transformations which are called phase changes or changes of state.

What happens when a solid is heated? Typically, it melts to form a liquid. If the heating continues, the liquid at some point boils and forms the vapor phase (gas). This process can be represented by a heating curve: a plot of temperature versus time for a process where energy is added at a constant rate. The heating curve of water is shown in Fig. I.2.

There are five separate zones on the graph (heating curve) of Fig. I.2:

 

Zone 1 (Ice):

As energy flows into the ice, the random vibrations of the water molecules increase as the temperature rises from -20 °C to 0 °C. Eventually, the molecules become so energetic that they break loose from their solid lattice positions and the change from solid to liquid occurs. This is indicated by a plateau at 0 °C on the heating curve. At this temperature, called the melting point, all the added energy is used to break the ice structure by breaking the hydrogen bonds, thus increasing the potential energy of the water molecules. The enthalpy change that occurs at the melting point when a solid melts is called the heat of fusion or enthalpy of fusion ΔΗ_fus. The temperature remains constant until all the solid has changed to liquid.

The general equation for calculating heat energy required to change the temperature of a solid is:

Q = m * c_s * ΔΤ      (1)

Where: Q heat energy (Joules)

c_s specific heat of the solid (Joules/g°C)

ΔΤ temperature change (°C)

 

Notes:

Specific heat of a solid cs is the amount of heat energy that changes the temperature of 1.0 g of a solid by 1.0 °C.

Each substance has its own specific heat. The specific heat of ice is for example 2.1 Joules/g°C.

 

Zone 2 (Ice & Water):

In zone 2, the temperature remains constant at 0 °C. At this temperature, called the melting point, all the added energy is used to disrupt the ice structure by breaking the hydrogen bonds and potential energy is increasing. The attractive forces that hold particles in fixed positions in the solid must be overcome to form the liquid. The heat absorbed in this case is called the heat of fusion or enthalpy of fusion and is symbolized ΔΗ_fusion.

Each substance has its own heat of fusion. The heat of fusion of ice is 340 Joules/g. Exactly the same amount of heat is given up when 1.0 g of water is changed to ice. This heat is called the heat of crystallization.

The general equation for calculating heat energy to change a solid to a liquid is:

Q = m * ΔΗ_fusion     (2)

Where: Q heat energy (Joules, J)

m mass of solid (g)

ΔH heat or enthalpy of fusion (J/g)

 

Zone 3 (Water):

The temperature is again changing as soon as all the solid (ice in this case) has changed to liquid. Then it begins to increase again starting from 0 °C up to 100 °C. The particles of a liquid are in constant motion and they are not held together as tightly as the particles of a solid. To change the temperature of a liquid heat energy must be added according to equation (1) and where m is the mass of 1.0 g of water in this case, where cs is the specific heat of water (c_s)water = 4.2 J/g°C and ΔΤ is the temperature change.

 

Zone 4 (Water & Steam):

At 100 °C the liquid water reaches its boiling point, and the temperature again remains constant as the added energy is used to vaporize the liquid. The heat absorbed is called heat of vaporiza-tion(ΔΗ_vapor). This heat is increasing the potential energy of the molecules of the liquid. Each substance has its own heat of vaporization. The heat of vaporization for water is 2270 J/g. Exactly the same amount of heat is given up when 1.0 g of water vapor is changed to liquid water. This heat is called the heat of condensation.

The general equation for calculating heat energy to change a liquid to a gas is:

Q = m * ΔΗ_vapor     (3)

Where: Q heat energy (Joules, J)

m mass of solid (g)

ΔHvapor heat or enthalpy of vaporization (J/g)

 

Notes:

Each substance has its own heat of vaporization. The heat of vaporization for water is 2270 J/g.

 

Zone 5 (Steam):

When all the liquid is changed to vapor the temperature again begins to rise. Note that phase changes are physical changes. No chemical bonds have been broken but intermolecular forces have been overcome. On the average, gaseous molecules are many times further apart from each other than molecules of solids and liquids.

To change the temperature of a gas, heat energy must be added. The amount of heat energy that changes the temperature of 1.0 g of a gas by 1.0 °C is called its specific heat (c_s)gas. Each substance has its own specific heat. The specific heat of steam is 2.02 J/g°C.

To change the temperature of a gas heat energy must be added according to equation (1) and where m is the mass of 1.0 g of steam in this case, where cs is the specific heat of steam (c_s)steam = 2.02 J/g°C and ΔΤ is the temperature change.

 

Note:

All substances have the same basic heating curve graphs (five zones). The differences are going to be the transition temperatures, and the values for specific heats c_s and ΔΗ’s.

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Relevant Posts

Free energy, entropy and thermodynamic equilibrium

Gas Laws - Ideal Gas Law

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References

1. P. Atkins, J. de Paula, “Physical Chemistry”, 9th Edition, W. H. Freeman (2009)
2. I. N. Levine, “Physical Chemistry”, 6th Edition, McGraw-Hill (2008)
3. S. S. Zumdahl, “Chemical Principles”, 6th Edition, Houghton Mifflin Company (2009)
4. A. W. Adamson, A. P. Gast, “Physical Chemistry of Surfaces”, John Wiley & Sons (1997

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Key Terms

phase changes, changes of state, heating curve, heat of fusion, enthalpy of fusion, ,ΔΗ, , specific heat

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The chemistry of enolate ions – Reactions with alkyl halides

The chemistry of enolate ions – Alkylations - Enolate ion reactions with alkyl halides

Enolates are reactive nucleophiles. Although the major enolate Lewis contributor shows concentration of electron density on the electronegative oxygen (Fig. I.1) when it reacts as a nucleophile, it behaves like the electron density is concentrated on the α-carbon next to carbonyl group.

However, from the resonance forms shown in Fig. I.1, it is clear that enolates are capable of reacting as both carbon and oxygen nucleophiles (resonance structures II and I respectively). Enolates react with alkyl halides, aldehydes/ketones and esters and these reactions are shown below:

 

Reactions of enolates with alkyl halides

Enolates are reactive nucleophiles and react with alkyl halides by the S_N2 and E2 mechanism. The electrophiles (alkyl halides in this case) need to be S_N2 reactive and this means that react in the order (Fig. I.2):

 

      

Primary, benzylic and allyl alkyl halides are among the best alkylating agents. More branched halides tend to prefer to undergo unwanted E2 elimination reactions. Tertiary alkyl halides are practically unreactive for enolate alkylation.

Since enolates react as nucleophiles with alkyl halides the following rules must be obeyed for the S_N2 reaction to occur:

Electrophiles –alkyl halides in this case - cannot be tertiary
The leaving group X in the alkyl halide RX must be a good leaving group (-Br, -I, OTs, OMs or OTf)
The nucleophile – enolate anion – must be a good nucleophile
The solvent must be an aprotic polar solvent

Note: If the electrophile is tertiary then an E2 reaction occur (enolates are moderate bases)

Let us examine the reaction of cyclohexane-1,3-dione with ethyl iodide. The reaction product is an alkylated diketone at the 2 position (2-ethylcyclohexane-1,3-dione) (Fig. I.3).

 

 

    

The basic steps of the above reaction mechanism are as follows (Fig. I.3):

The CH3O- anion (base) attacks the most acidic H (the –H atom between the two carbonyls). Electrons of the C-H bond make a double bond (π bond) and electrons of the C=O bond move to oxygen
The nucleophilic C=C π bond attacks carbon of the C-I bond in CH3CH2I and the very good leaving group –I leaves.  Alternatively, the resonance structure of the corresponding enolate can be drawn and the electron-pair on the α-carbon to the carbonyl group attacks the C-atom having the –I leaving group. In the same step, electrons on the oxygen atom move back to remake a π bond. The result is a new C-C bond (alkylation) in-between the 2 carbonyl atoms.

 

Note: An important consideration that affects all alkylations – like the ones presented above - is the choice of base:

A strong base can be selected to deprotonate the starting material completely. There is complete conversion of the starting material to the anion before addition of the electrophile (i.e. an alkyl halide, halogen…), which is added in a subsequent step. The choice of a strong base is practically more demanding but the electrophile and base never meet each other, so their compatibility is not a concern.
A weaker base may be used in the presence of the electrophile. The weaker base will not deprotonate the starting material completely and therefore only a small amount of anion will be formed but that small amount will react with the electrophile. This choice is easier practically (mixing the starting material, base and electrophile) but works only if the base and electrophile are compatible.

 

In general, the reaction conditions and the base must be chosen with care since an enolate tends to react with another enolate ion by attacking the C=O carbon forming dimers (aldol condensation) or polymers. The condensation problem described above (enolate reacting with unenolized carbonyl under basic conditions) does not exist if there is no unenolized carbonyl compound present. This can be achieved by choosing the following:

an enolate anion that will be sufficiently stable to survive until the alkylation is complete
a sufficiently strong base (pka at least 3 or 4 units higher than pka of the carbonyl compound) that will ensure that all of the starting carbonyl will be converted into the corresponding enolate.

 

Is there such a sufficiently strong base for making enolates?

One of the best bases for making enolates is LDA made from diisopropylamine and butyl lithium. The reaction is shown in Fig. I.4. LDA came into general use in the 1970’s but today more modern species  are used such as lithium isopropylcyclohexylamine (LICA) and lithiumtetramethylpiperidide (LTMP) which are even more hindered and less nucleophilic as a result. LDA will deprotonate virtually all ketones and esters that have an acidic proton to form the corresponding lithium enolates rapidly, completely, and irreversibly even at low temperatures (-78 C) required for some of these reactive species to survive. Therefore, any possibility for condensation reactions (enolate reacting with unenolized carbonyl) is minimized.

What is the mechanism of the alkylation of lithium enolates when react with alkyl halides?

The reaction of these lithium enolates with alkyl halides is one of the most important C-C bond-forming reactions in chemistry. Carbon-carbon bond formation is important since this is the basis for the construction of the molecular framework of organic molecules by synthesis.

The mechanism of the reaction is S_N2 in which the carbon nucleophile displaces a halide (from the electrophile, alkyl halide RX) with inversion of configuration at the alkylating group. The mechanism is the same as the one described in Fig. I.3 except that the base is LDA instead of CH3O-.

Note: The electrophile (RX) can also be any unhindered electrophile with a good leaving group X (such as X= OTs, OMs or OTf).

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Relevant Posts

Carbocations: Factors affecting their stability

Carbocations and factors affecting instability

Carbocations: Stability. formation and reactions

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References

1. A.J. Kresge, Pure Appl. Chem., 63, 213 (1991)
2. B. Capon, The Chemistry of Enols, Wiley, NY, 307–322 (1990)
3. S.E. Biali et al., J. Am. Chem. Soc. 107, 1007 (1985).

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Key Terms

carbonyl compounds, keto-enol reaction, keto form, enol form, α-hydrogen, enolization, enolates, tautomerism

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Food Chemistry: Antioxidants and Oxidation Reactions in Foods

Antioxidants and Oxidation Reactions in Foods

An antioxidant is a substance that delays the onset or slows the rate of oxidation. It is also used to extent the self-life of a food. Antioxidants play an important role in food chemistry.

Antioxidants can be divided into two categories:

Naturally occurring (vitamin E, vitamin C, β-carotene) (Fig. I1)
Synthetic (2-BHA, 3-BHA, BHT) (Fig. I1)

The unsaturated bonds present in all fats and oils represent active centers that react with oxygen. This reaction leads to the formation of primary, secondary, and tertiary oxidation products that may make the fat or fat-containing foods unsuitable for consumption. This deterioration in flavor of fats and fatty foods is described as rancidity. Oxidation, in nearly all cases, leads to flavors that are not desirable, so we normally strive to reduce or avoid oxidation during storage and processing of food products. Most food components, are vulnerable to oxidation, and oxidation may change their flavor, color, and nutritive value.

 

What are the main factors that affect the rate of oxidation?

amount of oxygen present
degree of unsaturation of the compounds
presence of antioxidants
presence of catalysts for the oxidation (i.e. copper, heme-containing molecules)
light exposure
temperature of storage
nature of packaging material.

 

 

What is the mechanism of the oxidation reaction?

The oxidation reaction (autoxidation) can be divided into the following three steps:

Initiation: Hydrogen is abstracted from an olefinic compound to yield a free radical. The removal of hydrogen occurs at the carbon atom next to the double bond in the presence of a catalyst (metal) or light.

RH → R* + H*

Propagation: Once a free radical has been formed it will combine with oxygen to form a peroxy-free radical, which can in turn abstract hydrogen from another unsaturated molecule to yield a peroxide and a new free radical (propagation reaction).

R* + O₂ → RO₂*

RO₂* + RH → ROOH + R*

The above reaction may be repeated up to several thousand times and has the nature of a chain reaction.

Termination: Termination occurs if the free radicals react with themselves to yield nonactive products.

R* + R* → R-R

R* + RO₂* → RO₂R

nRO₂* → (RO₂)n

 

What are the oxidation products produced when food products are oxidized?

Oxidation is a chain reaction with free radicals as reactive intermediates. Products generated can be divided as:

Primary oxidation products (hydroperoxides)
Secondary oxidation products (carbonyls)
Tertiary oxidation products (fatty acids)

The peroxides (ROOH) formed in the propagation step of the free radical reaction are the primary oxidation products. These oxidation products are generally unstable and decompose into the secondary oxidation products, which include a variety of compounds, including carbonyls, which are the most important.

The peroxides do not affect the flavor of foods. Their oxidation products though –aldehydes and ketones amongst other – known as secondary oxidation products are mainly responsible for flavor deterioration. As the aldehydes are themselves oxidized, fatty acids are formed. These free fatty acids may be considered tertiary oxidation products.

 

How antioxidants prevent food oxidation? Which are the main naturally occurring antioxidants? Which are the main synthetic antioxidants?

Antioxidants are added to foods such as oils, fats and butter as they react with oxygen-containing free radicals and prevent oxidation. Amongst the most important naturally occurring antioxidants are:

Vitamin E (Fig. I.1), a fat soluble vitamin, is a very effective natural antioxidant. It is found in foods such as nuts, wheat germ, seeds, whole grain and in vegetable oils
Vitamin C (Fig. I.2), is present in all animal and plant cells, mostly in free form, and it is probably bound to protein as well. Vitamin C is particularly abundant in rose hips, black and red currants, strawberries, parsley, oranges, lemons (in peels more than in pulp), grapefruit, a variety of cabbages and potatoes
β-Carotene (provitamin A), occurs in large amounts in apricots, cherries, cantaloups, carrots and peaches.
The element selenium, is present in fish, shellfish, meat, eggs, grain and chicken.

 

Since for economic reasons, it is not always possible to use natural antioxidants several synthetic antioxidants have been prepared. Most of them are phenols with a hydroxyl group attached to the benzene ring.

Amongst them the most important synthetic antioxidants are:

BHA (Fig. I.2) Commercial BHA is a mixture of two isomers, 2- and 3-tert-butyl-4-hydroxyanisole
tert-Butylhydroquinone (TBHQ) (Fig. I.2) is a particularly powerful antioxidant used, for example, for stabilization of soya oil.
Propyl gallate, (PG) (Fig. I.2) is very active in fats and oils since it is enriched at the surface of fat and come in contact with air

 

In order to be used as an antioxidant, a synthetic compound has to meet the following requirements:

it should not be toxic
it has to be highly active at low concentrations (0.01–0.02%)
it has to concentrate on the surface of the fat or oil phase.

Utilization of antioxidants is often regulated by governments through controls on the use of food additives. In North America incorporation of antioxidants is permitted at a maximum level of 0.01% for any one antioxidant, and a maximum of 0.02% for any combination. The regulations related to permitted levels often vary from country to country.

 

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Relevant Posts

Food Chemistry: Chemical reactions in cooking

Caramelization in Cooking - Caramelization Reactions

Food Preservatives: Sulfites and SO2

 

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References

1. E.N. Frankel, “Lipid Oxidation”, The Oily Press: Dundee, U.K. 1998
2. H-D. Belitz et al., “Food Chemistry”, 4th Edition, Springer Verlag, 2009
3. P. Barham et al., Chem. Rev., 110, 2313 (2010)
4. L.H. Skibsted, “Lipid Oxidation Pathways”, AOCS Press: Urbana, Illinois: 2008

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Key Terms

antioxidants, rancidity, oxidation of foods, vitamin E, vitamin C, ,BHA, , PG, TBHQ

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Keto-Enol Tautomerism – Enolization – Reactions and Mechanism

Acid or Base catalyzed Enolization – Reactions and Mechanism

Carbonyl compounds (aldehydes, ketones, carboxylic esters, carboxylic amides) react as electrophiles at the sp² hybridized carbon atoms and as nucleophiles if they contain an H-atom in the α-position relative to their C=O or C=N bonds. This is because this H is acidic and it can be removed by a base leaving behind an electron pair for nucleophilic attacks.    

For most compounds in organic chemistry all the molecules have the same structure – even if this structure cannot satisfactory represented by a Lewis formula – but for many compounds there is a mixture of two or more structurally distinct compounds that are in rapid equilibrium. This phenomenon is called tautomerism.

Tautomerism is the phenomenon that occurs in any reaction that simply involves the intramolecular transfer of a proton. An equilibrium is established between the two tautomers (structurally distinct compounds) and there is a rapid shift back and forth between the distinct compounds.

A very common form of tautomerism is that between a carbonyl compound containing an α-hydrogen and its enol form (Fig. I.1).

Fig. I.1: A keto-enol reaction

 

An enol is exactly what the name implies: an ene-ol. It has a C=C double bond (diene) and an OH group (alcohol) joined directly to it.

Notice that in the above reaction as in any keto-enol reaction there is no change in pH since a proton is lost from carbon and gained on oxygen. The reaction is known as enolization as it is the conversion of a carbonyl compound into its enol.

Notice also that in the above reaction the product is almost the same as the starting material since the only change is the transfer of one proton and the shift of the double bond.

In simple cases (R₂ = H, alkyl, OR, etc.) the equilibrium of the keto-enol reaction lies well to the left (keto structure) (Table I.1). The reason can be seen by examining the bond energies in Table I.2.

 

Compound	Enol Content, %
Acetone	6 * 10-7
PhCOCH3	1.1 * 10-6
CH3CHO	6 * 10-5
Cyclohexanone	4 * 10-5
Ph2CHCHO	9.1
PhCOCH2COCH3	89.2

Table I.1: The enol content of some carbonyl compounds

 

If keto-enol reactions are common for aldehydes and ketones why don’t simple aldehydes and ketones exist as enols?

IR and NMR Spectra of carbonyl compounds show no signs of enols. The equilibrium lies well over towards the keto form (the equilibrium constant k for cyclohexanone is about 10-5).

	Bond (Energy, kJ/mol)			Sum ( kJ/mol)
keto form	C-H (413)	C-C (350)	C=O (740)	1503
enol form	C=C (620)	C-O (367)	O-H (462)	1449

Table I.2: Bond energies in the keto and in the enol form. The keto form is thermodynamically more stable than the enol form by approximately 50 kJ/mol

The approximate sum of the bond energies in the keto form is 1503 kJ/mol while in the enol form 1449. Therefore, the keto form is thermodynamically more stable than the enol form by approximately 50 kJ/mol.

In most cases, enol forms cannot be isolated since they are less stable and are formed in minute quantities. However, there are some exceptions and in certain cases a larger amount of the enol form is present and it can be even the predominant species:

Molecules in which the enolic double bond is in conjugation with another double bond (cases are shown in Table I.1 like Ph₂CHCHO and PhCOCH₂COCH₃)
Molecules that contain two or more bulky aryl groups (Fig. I.2). Compound I in Fig. I.2 (a substituted enol) is the major species in equilibrium (~95%) while the keto form is the minor species (~5%). In cases like this steric hindrance destabilizes the keto form (the two substituted aryl groups are 109° apart) while in the enol form 120° apart.

 

Fig. I.2: A keto-enol reaction. The enol form (I) is the major species in this case since the keto form is destabilized by steric hindrance (the substituted aryl groups are closer in the keto form – the C-C angle is 109° and this is unfavorable due to steric hindrance)

 

Is there experimental evidence that keto-enol reactions are common for aldehydes and ketones?

If the NMR spectrum of a simple carbonyl compound in D₂O is obtained – such as pinacolone’s (CH₃)₃CCOCH₃ – the signal for protons next to the carbonyl group very slowly disappears. The isolated compound’s mass spectrum (after the above mentioned reaction with D₂O is over) shows that those hydrogen atoms have been replaced by deuterium atoms. There is a peak at (M+1)⁺ or (M+2)⁺ or (M+3)⁺ instead of M⁺. The reaction is shown in Fig. I.3: 

Fig. I.3: Evidence for a keto-enol reaction when pinacolone (CH₃)₃CCOCH₃ reacts with D₂O. When the enol form of the pinacolone reverts to the keto form it picks up a deuteron instead of a proton because the solution consists almost entirely of D₂O.

 

What mechanism can be proposed for the above reaction?

Enolization is a slow process in neutral solution, even in D₂O, and is catalyzed by acid or base in order to happen.

In the acid-catalyzed reaction the molecule is first protonated on oxygen and then loses the C-H proton in a second step (Fig. I.4). When the enol form reverts to the keto – since this is an equilibrium process – it picks up a deuteron instead of a proton since the solution is D₂O. 

 

Fig. I.4: The acid-catalyzed keto-enol reaction mechanism. If D2O is the solvent then the α-hydrogens to carbonyl group are replaced by deuterium.

 

In the base-catalyzed reaction the C-H proton is removed first by the base (for example hydroxide ion OH^-, OD^- in our case) and the proton (or D⁺ in our case) added to the oxygen atom in a second step (Fig. I.5).

Fig. I.5: The base-catalyzed keto-enol reaction mechanism. If D2O is the solvent then the α-hydrogens to carbonyl group are replaced by deuterium.

 

Notice that the enolization reactions in Fig. I.4 and Fig. I.5 are catalytic. In the acid-catalyzed mechanism the D⁺ (or H⁺ if water is the solvent) is regenerated at the end (catalyst). In the base-catalyzed mechanism OD^- (or OH^- if water is the solvent) is regenerated at the end (catalyst).

The enolate ion generated from the enol (Fig. I.6) in the base-catalyzed mechanism is nucleophilic due to:

Oxygen’s small atomic radius
Formal negative charge

An enolate ion is an ion with a negative charge on oxygen with adjacent C-C double bond.

 

Fig. I.6: Enolate ion resonance contributors. Although the major contributor is resonace structure I when it reacts as a nucleophile structure II is more reactive.

 

Enolates are reactive nucleophiles. Although the major enolate Lewis contributor shows concentration of electron density on the electronegative oxygen when it reacts as a nucleophile, it behaves like the electron density is concentrated on the α-carbon next to carbonyl group.

Enolates react with alkyl halides, aldehydes/ketones and esters and these reactions are shown in the post entitled “The chemistry of enolate ions – Enolate ion reactions”.

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Relevant Posts

Carbocations: Factors affecting their stability

Carbocations and factors affecting instability

Carbocations: Stability. formation and reactions

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References

1. A.J. Kresge, Pure Appl. Chem., 63, 213 (1991)
2. B. Capon, The Chemistry of Enols, Wiley, NY, 307–322 (1990)
3. S.E. Biali et al., J. Am. Chem. Soc. 107, 1007 (1985).

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Key Terms

carbonyl compounds, keto-enol reaction, keto form, enol form, α-hydrogen, enolization, enolates, tautomerism

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