Keto-Enol Tautomerism: Key Points

The reason for the equilibrium lying to the left is due to bond energies. The keto form has a C–H, C–C, and C=O bond whereas the enol has a C=C, C–O an O–H bond. The sum of the first three is about 359 kcal/mol (1500 kJ/mol) and the second three is 347 kcal/mol (1452 kJ/mol). The keto form is therefore more thermodynamically stable by 12 kcal/mol (48 kJ/mol).

Although the keto form is most stable for aldehydes and ketones in most situations, there are several factors that will shift the equilibrium toward the enol form.  The same factors that stabilize alkenes or alcohols will also stabilize the enol form. There are two strong factors and three subtle factors.

1. Aromaticity.  Phenols can theoretically exist in their keto forms, but the enol form is greatly favored due to aromatic stabilization.

2. Hydrogen Bonding. Nearby hydrogen bond acceptors stabilize the enol form. When a Lewis basic group is nearby, the enol form is stabilized by internal hydrogen bonding.

Here are three more subtle effects in keto-enol tautomerism:

3. Solvent. Solvent can also play an important role in the relative stability of the enol form. For example, in benzene, the enol form of 2,4-pentanedione predominates in a 94:6 ratio over the keto form, whereas the numbers almost reverse completely in water. What’s going on? In a polar protic solvent like water, the lone pairs will be involved in hydrogen bonding with the solvent, making them less available to hydrogen bond with the enol form.

 4. Conjugation . π systems are a little like Cheerios in milk: given the choice, they want to connect together than hang out in isolation.  So in the molecule depicted, the more favorable tautomer will be the one on the left, where the double bond is a connected by conjugation to the phenyl.

 5. Substitution. In the absence of steric factors, increasing substitution at carbon will stabilize the enol form. Enols are alkenes too – so any factors that stabilize alkenes, will stabilize enols as well. All else being equal, double bonds increase in thermodynamic stability as substitution is increased. So in the above example, the enol on the left should be the more stable one. As you might suspect, “all things being equal” sounds like a big caveat. It is – all else is rarely equal. But that’s a topic for another day – or, more likely, another course.

 

Sources: “March’s Advanced Organic Chemistry”, “Solvents and solvent effects in Organic Chemistry”, by Christian Riechart. EDIT: Commenter Natalia helpfully points out that Carey & Sundberg A is a great resource for this topic (section 7.3 in my 4th edition) and she is right.

 One really interesting observation is that when you dissolve acetone in D2O, you slowly get incorporation of deuterium at the alpha carbon. This is the enol tautomer at work – it reacts with a proton/deuteron source at the alpha carbon and regenerates the ketone.

Under normal conditions, this is a pretty slow reaction. But when you add a bit of acid, suddenly the rate of deuterium incorporation drastically increases. What’s going on here?

Let’s call the rate for conversion of ketone into enol K1 and the rate for conversion of enol into ketone K2.  The cool thing is, acid speeds up BOTH reactions for keto-enol tautomerism – both the forward and the reverse reaction. Here’s how it works.

1)  For K1 (keto to enol), acid protonates the carbonyl, making the carbonyl carbon more electrophilic. The more electrophilic the carbonyl  the stronger an acid it becomes at the alpha carbon (remember, trifluoroacetone is more acidic than acetone). So deprotonation at the alpha carbon becomes much easier in the presence of acid, which results in the enol form.

2) For K2, acid also speeds up the conversion of the enol to the keto form. The enol form is nucleophilic at the alpha carbon, and reacts with the strongest electrophile around. An acid – say, D2SO4  – is a much, much stronger electrophile than D2O, so this is going to increase the rate of the enol to ketone conversion.

ring chain tautomerism

tautomerism mcq

ELEMENTS OF SYMMETRY -EXAMPLES

ELEMENTS  OF SYMMETRY

mcq on fundamental particles

phenolsethers summery

Electrophilic substitution reactons of haloarenes

Electrophilic substitution reactons of haloarenes

The halogen atom is ortho and para 

directing and therefore, substitution takes 

place at ortho and para positions giving 

electrophilic substitution reactions at the benzene ring.

In the resonance structures of chlorobenzene the ortho and para positions get negatively charged, i.e. the electron density is relatively more at ortho and para positions. The incoming electrophile is more likely to attack these positions. But, because of steric hindrance at the ortho position, the para-product usually predominates over the ortho product. Also, the halogen atom has -I effect because it is a electron withdrawing group. As a result, it tends to deactivate the benzene ring. Therefore, the electrophilic substitution reactions of chlorobenzene occur slowly and under drastic conditions compared to benzene. Thus,

• Haloarenes undergo electrophilic substitution reactions slowly as compared to benzene.

• Halogen group is ortho and para directing (para-product usually predominates over the ortho product).

Halogenation

Haloarenes react with halogens in the presence of ferric salt as catalyst to give ortho para isomers.

Nitration

Haloarenes with concentrated HNO₃ in the presence of concentrated H₂SO₄.

Sulphonation

Sulphonation occurs when haloarene is treated with concentrated H₂SO₄.

Alkylation and acylation

The alkylation and acylation reaction, known as 'Friedel-Craft reaction', is carried by treating haloarene with alkyl chloride or acyl chloride in the presence of a catalyst like anhydrous aluminium chloride. For example,

Haloform reaction

Haloform reaction

NETT       REACTION :

Haloform Reaction is a type of organic reaction where haloform is produced by halogenation of methyl ketone in the presence of a base. It is a kind of nuclophilic substitution reaction. This reaction is used in quanitatve analysis to indicate the presence of methyl ketone.

If the hypohalite used in this reaction is sodium hypoiodite, the product is iodoform (CHI3). Such a haloform reaction is also known as iodoform test. As iodoform is a yellow crystalline solid with a characteristic odor and melting point it can be handled much more easily handled in the the laboratory.

                              


                          

• When methyl ketones are treated with the halogen in basic solution, polyhalogenaton followed by cleavage of the methyl group occurs.

• The products are the carboxylate and trihalomethane, otherwise known as haloform.

• The reaction proceeds via successively faster halogenations at the α-position until the 3 H have been replaced.

• The halogenations get faster since the halogen stablises the enolate negative charge and makes it easier to form.

• Then a nucleophilic acyl substitution by hydroxide displaces the anion CX₃ as a leaving group that rapidly protonates.

• This reaction is often performed using iodine and as a chemical test for identifying methyl ketones. Iodoform is yellow and precipitates under the reaction conditions.

MECHANISM OF THE HALOFORM REACTION OF METHYL KETONES
Step 1: First, an acid-base reaction. Hydroxide functions as a base and removes the acidic α-hydrogen giving the enolate.
Step 2: The nucleophilic enolate reacts with the iodine giving the halogenated ketone and an iodide ion.
Step 3: Steps 1 and 2 repeat twice more yielding the trihalogenated ketone.
Step 4: The hydroxide now reacts as a nucleophile at the electrophilic carbonyl carbon, with the C=O becoming a C-O single bond and the oxygen is now anionic.
Step 5: Reform the favourable C=O and displace a leaving group, the trihalomethyl system which is stabilised by the 3 halogens. This gives the carboxylic acid.
Step 6: An acid-base reaction. The trihalomethyl anion is protonated by the carboxylic acid, giving the carboxylate and the haloform (trihalomethane).

nucleophillic substitution examples

substitution reactions summary with examples

      
                   

For each reaction below, determine whether the primary reaction is S_N1, S_N2, E1, or E2, and then draw the product.

Note: Me = methyl (CH₃)

problem set 1

SOLUTION

a) NaCN is charged! (Na⁺ and CN^-), so it's S_N2 or E2. CN is not a strong base, so it's S_N2.

b) KOtBu (potassium tert-butoxide) is charged, so it's S_N2 or E2. ^-OtBu is a strong base, so if anything is more bulky than 1º it will go E2. ^-OtBu is 3º, so it will definitely go E2 (KOtBu is a classic E2 reagent).

c) NaOMe is charged so E2 or S_N2. NaOMe is a strong base, so if anything >1º it will go E2. ^-OMe is 1º (actually, not even 1º), but the alkyl halide is 2º, so it will go E2.

d) NaOMe is charged so E2 or S_N2. NaOMe is a strong base, so if anything >1º it will go E2. But in this case there is no bulk whatsoever- nothing is >1º! NaOMe is 1 and the alkyl halide is also 1º, so it will go S_N2.

e) Methanol (MeOH) is neutral so probably E1 or S_N1. Methanol is a weak base and there's no bulk, so S_N1. In general water and alcohol do a mixture of S_N1 and E1 with alkyl halides (mostly S_N1).

f) H₂SO₄ is acidic so probably E1 or S_N1. Can't be S_N1 though because there is no nucleophile in H₂SO₄. (HSO₄^- is a very weak nucleophile). An alcohol with H₂SO₄ or H₃PO₄ is a dehydration reaction- E1.

g) H₂SO₄ is acidic so probably E1 or S_N1. In this case we have a nucleophile- Cl^-, so it will go S_N1.

h) Amines are neutral but they don't so S_N1/E1- they tend to go S_N2/E2, because they are basic (an amine solution has a basic pH). This amine is really bulky so it will go E2.

problem set 2

Rank the following compounds in order of decreasing reactivity with water (solvolysis). (1 = most reactive)

This is an S_N1 reaction. We know this because none of the reagents have charges (H₂O is neutral; if it were HO^-, it would probably be S_N2 or E2). So a carbocation will be formed in this reaction, and the compound whose carbocation is the most stable will react the fastest. So the 3º bromide will react fastest with water.

Problem # 3

Rank the following compounds in order of decreasing reactivity with NaI in acetone. (1 = most reactive)

This is an S_N2 reaction. We know this because NaI is an S_N2 reagent- charges give it away; when we see charges (Na⁺ is positive and I^- is negative) it's probably an S_N2 reaction. So we want reactants that are less substituted: methyl is more reactive than 1º, which is more reactive than 2º, etc. So methyl bromide reacts the fastest with NaI.

Problem #4

Rank the following electrophiles in order of decreasing reactivity with NaN₃ in DMF. (1 = most reactive)

Good leaving groups are stable. Larger ions tend to be more stable than smaller atoms (due to a smaller charge:size ratio. See problem 288), so when going down the periodic table, stability increases. I^- is more stable than Br^-, which is more stable than Cl^-, etc. So I^- is the best leaving group, and 2-iodobutane will react the fastest with a nucleophile.

Problem # 5

Indicate the reagents necessary to carry out each transformation.

For a), the wedge remains a wedge, so we have to do two S_N2 reactions (wedge to dash to wedge again). So we use PBr₃ to turn the OH into a Br. (Bromination of an alcohol with PBr₃ is an S_N2 reaction and so inverts stereochemistry).

For b), the wedge becomes a dash, so we can only do one SN2 reaction. So instead of using PBr₃ to make the OH a better leaving group, we use RSO₂Cl, which doesn't break the carbon-oxygen bond and so doesn't invert the stereochemistry.

NUCLEOPHILLIC SUBSTITUTIONS WITH EXAMPLES

SUBSTITUTION by making a new bond AT THE SAME TIME as breaking the old bond

• substitution requires a bond to be broken AND a new bond to be formed

• the LOWEST energy way of doing this (unless precluded by steric or other effects, see later) is to MAKE THE NEW BOND (getting some energy "back") at the same time as BREAKING THE OLD BOND, this is SN2

Example

This is fundamentally just a Lewis acid/base reaction of the kind we saw when we were learning about Lewis acid/base reactions, the Lewis base has the high energy chemically reactive electrons, which are used to make a new bond to the Lewis acid, and a stronger bond is formed (C-O in the example above) and a weaker bond is broken (C-Br above)

• HO– is the Lewis Base and Nucleophile

• the halide is the Lewis acid/electrophile

• the Br– anion is the Leaving Group

The halide AND the nucleophile (2 molecules) are involved in the rate determining step and so the reaction rate depends upon the concentration of them both, the reactions is kinetically SECOND (2nd) order

Examples of SN2 Reactions; Give the major organic product in the following reactions

• we understand these SN2 reactions a simple Lewis acid/base processes

• identify the Lewis base/NUCLEOPHILE as the reactant with the high energy electrons

• the Lewis acid/nucleophile must react with the Lewis acid/ELECTROPHILE

Factors Controlling SN2 Reactivity: Leaving Group Ability

Good leaving groups are:

• stable/less reactive as an anion

• which means generally weak bases (have weak X-H bonds)

• polarize the C–LG bond (and are polarizable to make strong partial bonds in transition state)

Recall

Similarly

• INCREASING leaving group ability going down the periodic

• anion stability increases I– > Br– > Cl–

• the bond strength increases C-I < C-Br < C-Cl, i.e. the iodide anion is stable because when it reacts it makes WEAK BONDS

2.2  Factors Controlling SN2 Reactivity: Solvent Effects

Polar APROTIC  (NON-hydrogen-bonding) Solvents: There are Several, you need to know these

• Polar protic solvents have WEAKER intermolecular ion-dipole forces with dissolved ions

• The ion-dipole interactions certainly solvent and stabilize ions, but the H-bonding effect particularly on anions is MISSING

• There is a usually a SMALLER energy difference between reactants and the transition state when SN2 reactions are performed in polar APROTIC solvents, SN2 reactions in polar APROTIC solvents are usually faster than in polar protic solvents 

• you NEED TO KNOW the polar APROTIC solvents, this is not easy because they are different structures, learn them by working with them

Factors Controlling SN2 Reactivity: Nucleophilicity

Summary: Nucleophilicity order is sometimes difficult to remember, but favored by.....

1. anion over neutral

2. less electronegative over more electronegative

3. small atomic size over large atomic size in aprotic solvents

4. large atomic size over small atomic size in protic solvents

• These factors are all quite predictable except GOING DOWN THE PERIODIC TABLE (again!)

Factors Controlling SN2 Reactivity:  Steric Effects

First Order Nucleophilic Substitution (SN1) Reaction

• here the solvent "helps" to break the C–Br bond, the reaction is a solvolysis reaction (lysis - bond breaking)

We need a new substitution MECHANISM to account for this: The SN1 Mechanism

• the SN1 reaction requires a polar protic solvent to stabilize the ionic (cation and halide) intermediates

• usually requires heat (energy) to break the C–X bond unimolecularly

• ONLY the halide (not the nucleophile) involved in the R.D.S., thus SN1 (1 means only 1 reactant in the R.D.S.)

• requires a stable intermediate cation, NO SN1 for methyl or primary halides

3.1  Stereochemistry of SN1 Reactions: Racemization (?)

Example

Distinguishing SN1 and SN2 Reactions

• NOTE: the factors above favor the reactions by making them go faster, e.g. SN2 is FASTER at a primary carbon, SN1 is faster at a tertiary carbon, SN1 is faster in polar protic solvents etc.

• However, weak nucleophiles do not favor SN1 because they make Sn1 reactions faster, they don't, but they do make competing SN2 reactions SLOWER

• SN2 reactions are not precluded by polar protic solvents, they are just faster in aprotic solvents

Examples:  assign the mechanism of the following reactions to SN1 or SN2

Example Problems: Give the major organic product of reactions

• polar aprotic solvent, strong nucleophile, SN2, Br- better leaving group

• 1 equivalent means exactly the same number of nucleophiles as organic reactants, which in this context means that there is only enough nucleophile to substitute one of the halide leaving groups

• polar protic solvent and heat, no strong nucleophile and allylic halide, must be SN1. 

• polar aprotic solvent, strong nucleophile, SN2, allylic position more reactive

• 1 EQUIVALENT will ONLY REACT at the carbon where SN2 will be fastest