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‘There could be ART in Organic Synthesis’ declared the inimitable monarch of organic synthesis, Professor R.B. Woodward. His school unveiled several elegant approaches covering a variety of complex structures and broke new grounds to define the art of organic synthesis. ‘If organic synthesis is a branch of science, what is the LOGIC of organic synthesis?’ marveled several others. The development of the concept of logical approaches towards synthesis has been evolving over the past several decades. A few stalwarts focused their attention on this theme and attempted to evolve a pattern to define this logic. There is no doubt that all of us who dabble with synthesis contribute our small bit in the magnificent direction. A few names stand out in our minds for their outstanding contributions. Notable contributions came from the schools of J.A. Marshal, E.J. Wenkert, G. Stock, S Hanessian, E.E. van Tamalen, S. Masamune, R.B. Woodward, E.J. Corey and several others. More focused on this theme were the contributions from the school of E.J. Corey.
The period 1960 – 1990 witnessed the evolution of this thought and the concept bloomed into a full-fledged topic that now merits a separate space in college curriculum. Earlier developments focused on the idea of ANTITHETIC APPROACHES and perfected the art of DISCONNECTION via RETROSYNTHESIS. This led to logical approaches for the construction of SYNTHETIC TREES that summarized various possible approaches for the proposed Target structure. All disconnections may not lead to good routes for synthesis. Once the synthetic tree was constructed, the individual branches were analyzed critically. The reactions involved were looked into, to study their feasibility in the laboratory, their mechanistic pathways were analyzed to understand the conformational and stereochemical implications on the outcome of each step involved and the time / cost factors of the proposed routes were also estimated. The possible areas of pitfall were identified and the literature was critically scanned to make sure that the steps contemplated were already known or feasible on the basis of known chemistry. In some cases, model compounds were first constructed to study the feasibility of the particular reaction, before embarking on the synthesis of the complex molecular architecture. Thus a long process of logical planning is now put in place before the start of the actual synthetic project. In spite of all these careful and lengthy preparations, an experienced chemist is still weary of the Damocles Sword of synthesis viz., the likely failure of a critical step in the proposed route(s), resulting in total failure of the entire project. All achievements are 10% inspiration and 90% perspiration. For these brave molecular engineers, sometimes also called chemists, these long-drawn programs and possible perils of failures are still worth, for the perspiration is enough reward.
A sound knowledge of mechanistic organic chemistry, detailed information on the art and science of functional group transformations, bond formation and cleavage reactions, mastery over separation and purification techniques and a sound knowledge of spectroscopic analysis are all essential basics for the synthesis of molecules. A synthetic chemist should also be aware of developments in synthetic strategies generated over the years for different groups of compounds, which include Rules and guidelines governing synthesis. Since organic chemistry has a strong impact on the development of other sister disciplines like pharmacy, biochemistry and material science, an ability to understand one or more of these areas and interact with them using their terminologies is also an added virtue for a synthetic chemist. With achievements from synthesis of strained molecules (once considered difficult (if not impossible) to synthesize, to the synthesis of complex, highly functionalized and unstable molecules, an organic chemist could now confidently say that he could synthesize any molecule that is theoretically feasible. This is the current status of the power of organic synthesis. Based on the task assigned to the chemist, he would select a Target molecule for investigation and devise suitable routes for synthesis.
Disconnection of bonds
Having chosen the TARGET molecule for synthesis, the next exercise is to draw out synthetic plans that would summarize all reasonable routes for its synthesis. During the past few decades, chemists have been working on a process called RETROSYNTHESIS. Retrosynthesis could be described as a logical Disconnection at strategic bonds in such a way that the process would progressively lead to easily available starting material(s) through several synthetic plans. Each plan thus evolved, describes a ‘ROUTE’ based on a retrosynthesis. Each disconnection leads to a simplified structure. The logic of such disconnections forms the basis for the retroanalysis of a given target molecule. Natural products have provided chemists with a large variety of structures, having complex functionalities and stereochemistry. This area has provided several challenging targets for development of these concepts. The underlining principle in devising logical approaches for synthetic routes is very much akin to the following simple problem. Let us have a look of the following big block, which is made by assembling several small blocks (Fig 1.4.2.1). You could easily see that the large block could be broken down in different ways and then reassembled to give the same original block.
Fig 1.4.2.1
Now let us try and extend the same approach for the synthesis of a simple molecule. Let us look into three possible ‘disconnections’ for a cyclohexane ring as shown in Fig 1.4.2.2.
Fig 1.4.2.2
In the above analysis we have attempted to develop three ways of disconnecting the six membered ring. Have we thus created three pathways for the synthesis of cyclohexane ring? Do such disconnections make chemical sense? The background of an organic chemist should enable him to read the process as a chemical reaction in the reverse (or ‘retro-‘) direction. The dots in the above structures could represent a carbonium ion, a carbanion, a free radical or a more complex reaction (such as a pericyclic reaction or a rearrangement). Applying such chemical thinking could open up several plausible reactions. Let us look into path b, which resulted from cleavage of one sigma bond. An anionic cyclisation route alone exposes several candidates as suitable intermediates for the formation of this linkage. The above analysis describes only three paths out of the large number of alternate cleavage routes that are available. An extended analysis shown below indicates more such possibilities (Fig 1.4.2.3). Each such intermediate could be subjected to further disconnection process and the process continued until we reach a reasonably small, easily available starting materials. Thus, a complete ‘SYNTHETIC TREE’ could be constructed that would summarize all possible routes for the given target molecule.
Fig 1.4.2.3
Efficiency of a route
A route is said to be efficient when the ‘overall yield’ of the total process is the best amongst all routes investigated. This would depend not only on the number of steps involved in the synthesis, but also on the type of strategy followed. The strategy could involve a ‘linear syntheses’ involving only consequential steps or a ‘convergent syntheses’ involving fewer consequential steps. Fig 1.4.3.1 shown below depicts a few patterns that could be recognized in such synthetic trees. When each disconnection process leads to only one feasible intermediate and the process proceeds in this fashion
Fig 1.4.3.1
all the way to one set of starting materials (SM), the process is called a Linear Synthesis. On the other hand, when an intermediate could be disconnected in two or more ways leading to different intermediates, branching occurs in the plan. The processes could be continued all the way to SMs. In such routes different branches of the synthetic pathways converge towards an intermediate. Such schemes are called Convergent Syntheses.
The flow charts shown below (Fig 1.4.3.2) depicts a hypothetical 5-step synthesis by the above two strategies. Assuming a very good yield (90%) at each step (this is rarely seen in real projects), a linier synthesis gives 59% overall yield, whereas a convergent synthesis gives 73% overall yield for the same number of steps..
Contributors
• Prof. R Balaji Rao (Department of Chemistry, Banaras Hindu University, Varanasi) as part of Information and Communication Technology | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/11%3A_Alkynes/11.12%3A_Synthesis.txt |
Solutions to exercises
You are undoubtedly already familiar with the general idea of oxidation and reduction: you learned in general chemistry that when a compound or atom is oxidized it loses electrons, and when it is reduced it gains electrons. You also know that oxidation and reduction reactions occur in pairs: if one species is oxidized, another must be reduced at the same time - thus the term 'redox reaction'.
Most of the redox reactions you have seen previously in general chemistry probably involved the flow of electrons from one metal to another, such as the reaction between copper ion in solution and metallic zinc:
Cu+2(aq) + Zn(s) → Cu(s) + Zn+2(aq)
In organic chemistry, redox reactions look a little different. Electrons in an organic redox reaction often are transferred in the form of a hydride ion - a proton and two electrons. Because they occur in conjunction with the transfer of a proton, these are commonly referred to as hydrogenation and dehydrogenation reactions: a hydride plus a proton adds up to a hydrogen (H2) molecule. Be careful - do not confuse the terms hydrogenation and dehydrogenation with hydration and dehydration - the latter refer to the gain and loss of a water molecule (and are not redox reactions), while the former refer to the gain and loss of a hydrogen molecule.
When a carbon atom in an organic compound loses a bond to hydrogen and gains a new bond to a heteroatom (or to another carbon), we say the compound has been dehydrogenated, or oxidized. A very common biochemical example is the oxidation of an alcohol to a ketone or aldehyde:
When a carbon atom loses a bond to hydrogen and gains a bond to a heteroatom (or to another carbon atom), it is considered to be an oxidative process because hydrogen, of all the elements, is the least electronegative. Thus, in the process of dehydrogenation the carbon atom undergoes an overall loss of electron density - and loss of electrons is oxidation.
Conversely, when a carbon atom in an organic compound gains a bond to hydrogen and loses a bond to a heteroatom (or to another carbon atom), we say that the compound has been hydrogenated, or reduced. The hydrogenation of a ketone to an alcohol, for example, is overall the reverse of the alcohol dehydrogenation shown above. Illustrated below is another common possibility, the hydrogenation (reduction) of an alkene to an alkane.
Hydrogenation results in higher electron density on a carbon atom(s), and thus we consider process to be one of reduction of the organic molecule.
Notice that neither hydrogenation nor dehydrogenation involves the gain or loss of an oxygen atom. Reactions which do involve gain or loss of one or more oxygen atoms are usually referred to as 'oxygenase' and 'reductase' reactions, and are the subject of section 16.10 and section 17.3.
For the most part, when talking about redox reactions in organic chemistry we are dealing with a small set of very recognizable functional group transformations. It is therefore very worthwhile to become familiar with the idea of 'oxidation states' as applied to organic functional groups. By comparing the relative number of bonds to hydrogen atoms, we can order the familiar functional groups according to oxidation state. We'll take a series of single carbon compounds as an example. Methane, with four carbon-hydrogen bonds, is highly reduced. Next in the series is methanol (one less carbon-hydrogen bond, one more carbon-oxygen bond), followed by formaldehyde, formate, and finally carbon dioxide at the highly oxidized end of the group.
This pattern holds true for the relevant functional groups on organic molecules with two or more carbon atoms:
Alkanes are highly reduced, while alcohols - as well as alkenes, ethers, amines, sulfides, and phosphate esters - are one step up on the oxidation scale, followed by aldehydes/ketones/imines and epoxides, and finally by carboxylic acid derivatives (carbon dioxide, at the top of the oxidation list, is specific to the single carbon series).
Notice that in the series of two-carbon compounds above, ethanol and ethene are considered to be in the same oxidation state. You know already that alcohols and alkenes are interconverted by way of addition or elimination of water (section 14.1). When an alcohol is dehydrated to form an alkene, one of the two carbons loses a C-H bond and gains a C-C bond, and thus is oxidized. However, the other carbon loses a C-O bond and gains a C-C bond, and thus is considered to be reduced. Overall, therefore, there is no change to the oxidation state of the molecule.
You should learn to recognize when a reaction involves a change in oxidation state in an organic reactant . Looking at the following transformation, for example, you should be able to quickly recognize that it is an oxidation: an alcohol functional group is converted to a ketone, which is one step up on the oxidation ladder.
Likewise, this next reaction involves the transformation of a carboxylic acid derivative (a thioester) first to an aldehyde, then to an alcohol: this is a double reduction, as the substrate loses two bonds to heteroatoms and gains two bonds to hydrogens.
An acyl transfer reaction (for example the conversion of an acyl phosphate to an amide) is not considered to be a redox reaction - the oxidation state of the organic molecule is does not change as substrate is converted to product, because a bond to one heteroatom (oxygen) has simply been traded for a bond to another heteroatom (nitrogen).
It is important to be able to recognize when an organic molecule is being oxidized or reduced, because this information tells you to look for the participation of a corresponding redox agent that is being reduced or oxidized- remember, oxidation and reduction always occur in tandem! We will soon learn in detail about the most important biochemical and laboratory redox agents.
Template:ExampleStart
Exercise 16.1: is an aldol condensation a redox reaction? Explain.
Exercise 16.2: Is the reaction catalyzed by squalene epoxidase a redox reaction? How about squalene cyclase? Explain.
Template:ExampleEnd
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/12%3A_Oxidation_and_Reduction/12.01%3A_Introduction.txt |
Liquid Ammonia Solutions
A remarkable feature of the alkali metals is their ability to dissolve reversibly in liquid ammonia. Just as in their reactions with water, reacting alkali metals with liquid ammonia eventually produces hydrogen gas and the metal salt of the conjugate base of the solvent—in this case, the amide ion (NH2) rather than hydroxide:
$\ce{M(s) + NH_3(l) \rightarrow 1/2 H_2(g) + M^+(am) + NH_2^{-} (am)} \label{21.20}$
where the (am) designation refers to an ammonia solution, analogous to (aq) used to indicate aqueous solutions. Without a catalyst, the reaction in Equation \ref{21.20} tends to be rather slow. In many cases, the alkali metal amide salt ($\ce{MNH2}$) is not very soluble in liquid ammonia and precipitates, but when dissolved, very concentrated solutions of the alkali metal are produced. One mole of Cs metal, for example, will dissolve in as little as 53 mL (40 g) of liquid ammonia. The pure metal is easily recovered when the ammonia evaporates.
Solutions of alkali metals in liquid ammonia are intensely colored and good conductors of electricity due to the presence of solvated electrons (e, $\ce{NH3}$), which are not attached to single atoms. A solvated electron is loosely associated with a cavity in the ammonia solvent that is stabilized by hydrogen bonds. Alkali metal–liquid ammonia solutions of about 3 M or less are deep blue (Figure $2$) and conduct electricity about 10 times better than an aqueous $\ce{NaCl}$ solution because of the high mobility of the solvated electrons. As the concentration of the metal increases above 3 M, the color changes to metallic bronze or gold, and the conductivity increases to a value comparable with that of the pure liquid metals.
The most common sources of the hydride nucleophile are lithium aluminum hydride ($\ce{LiAlH4}$) and sodium borohydride ($\ce{NaBH4}$). Note! The hydride anion is not present during this reaction; rather, these reagents serve as a source of hydride due to the presence of a polar metal-hydrogen bond. Because aluminum is less electronegative than boron, the Al-H bond in $\ce{LiAlH4}$ is more polar, thereby, making $\ce{LiAlH4}$ a stronger reducing agent.
12.03: Reduction of Alkenes
Addition of hydrogen to a carbon-carbon double bond is called hydrogenation. The overall effect of such an addition is the reductive removal of the double bond functional group. Regioselectivity is not an issue, since the same group (a hydrogen atom) is bonded to each of the double bond carbons. The simplest source of two hydrogen atoms is molecular hydrogen (H2), but mixing alkenes with hydrogen does not result in any discernible reaction. Although the overall hydrogenation reaction is exothermic, a high activation energy prevents it from taking place under normal conditions. This restriction may be circumvented by the use of a catalyst, as shown in the following diagram.
An example of an alkene addition reaction is a process called hydrogenation.In a hydrogenation reaction, two hydrogen atoms are added across the double bond of an alkene, resulting in a saturated alkane. Hydrogenation of a double bond is a thermodynamically favorable reaction because it forms a more stable (lower energy) product. In other words, the energy of the product is lower than the energy of the reactant; thus it is exothermic (heat is released). The heat released is called the heat of hydrogenation, which is an indicator of a molecule’s stability.
Catalysts are substances that changes the rate (velocity) of a chemical reaction without being consumed or appearing as part of the product. Catalysts act by lowering the activation energy of reactions, but they do not change the relative potential energy of the reactants and products. Finely divided metals, such as platinum, palladium and nickel, are among the most widely used hydrogenation catalysts. Catalytic hydrogenation takes place in at least two stages, as depicted in the diagram. First, the alkene must be adsorbed on the surface of the catalyst along with some of the hydrogen. Next, two hydrogens shift from the metal surface to the carbons of the double bond, and the resulting saturated hydrocarbon, which is more weakly adsorbed, leaves the catalyst surface. The exact nature and timing of the last events is not well understood.
As shown in the energy diagram, the hydrogenation of alkenes is exothermic, and heat is released corresponding to the ΔE (colored green) in the diagram. This heat of reaction can be used to evaluate the thermodynamic stability of alkenes having different numbers of alkyl substituents on the double bond. For example, the following table lists the heats of hydrogenation for three C5H10 alkenes which give the same alkane product (2-methylbutane). Since a large heat of reaction indicates a high energy reactant, these heats are inversely proportional to the stabilities of the alkene isomers. To a rough approximation, we see that each alkyl substituent on a double bond stabilizes this functional group by a bit more than 1 kcal/mole.
Alkene Isomer (CH3)2CHCH=CH2
3-methyl-1-butene
CH2=C(CH3)CH2CH3
2-methyl-1-butene
(CH3)2C=CHCH3
2-methyl-2-butene
Heat of Reaction
( ΔHº )
–30.3 kcal/mole –28.5 kcal/mole –26.9 kcal/mole
From the mechanism shown here we would expect the addition of hydrogen to occur with syn-stereoselectivity. This is often true, but the hydrogenation catalysts may also cause isomerization of the double bond prior to hydrogen addition, in which case stereoselectivity may be uncertain. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/12%3A_Oxidation_and_Reduction/12.02%3A_Reducing_Agents.txt |
Introduction
In the late 1970’s the lipid hypothesis came in to existences. The lipid hypothesis states that eating saturated fats leads to elevated LDL(Low Density Lipoprotein) which is perceived to be bad cholesterol. This will result in coronary heart disease which is hardening and narrowing of arteries resulting in heart attack. Fats were eventually classified in to 2 categories, “healthy fats” and “unhealthy fats”. Unhealthy fats where perceived to be saturated fats, healthy fats where perceived to be unsaturated fats.
A meta-analysis of 72 studies with over 103,052 people have found no validity in the lipid hypothesis. The conclusion of the Meta-Analysis was,“In contrast to current recommendations, this systematic review found no evidence that saturated fat increases the risk of coronary disease, or that polyunsaturated fats have a cardio protective effect.”[1]
Dietary fats play a critical role in human health. They help keep cells healthy, help with brain development, help with the use of fat soluble vitamins, and they help cushion organs protecting them against blunt trauma. Fats come in multiple forms, Saturated, Unsaturated and trans fats just to name a few.
Hydrogenation Reaction
Unsaturated fatty acids may be converted to saturated fatty acids by the relatively simple hydrogenation reaction. Recall that the addition of hydrogen to an alkene (unsaturated) results in an alkane (saturated). A simple hydrogenation reaction is:
H2C=CH2 + H2 → CH3CH3
alkene plus hydrogen yields an alkane
Saturated Fats
Saturated fats are solid at room temperature due to their molecular shape. The term saturated is in reference to an Sp3 carbon chain that has its remaining Sp3 orbitals bonded with hydrogen atoms. Thus the term “saturated”. It’s “saturated” with hydrogen.
Saturated fats have a chain like structure which allows them to stack very well forming a solid at room temperature. Unsaturated fats are not linear due to double bonded carbons which results in a different molecular shape because the Sp2 carbons are trigonal planar, not tetrahedral(Sp3 carbons) as the carbons are in saturated fats. This change in structure will cause the fat molecules to not stack very well resulting in fats that are liquid at room temperature. Butter is mostly saturated fat, that’s why it’s solid at room temperature. Olive Oil is liquid at room temperature, thus it’s an unsaturated fat. An unsaturated fat can be made in to a saturated fat by a Hydrogenation process.
These are similar molecules, differing in their melting points. If the compound is a solid at room temperature, you usually call it a fat. If it is a liquid, it is often described as an oil.
Their melting points are largely determined by the presence of carbon-carbon double bonds in the molecule. The higher the number of carbon-carbon double bonds, the lower the melting point.
If there aren't any carbon-carbon double bonds, the substance is said to be saturated. A typical saturated fat might have the structure:
Molecules of this sort are usually solid at room temperature.
If there is only one carbon-carbon double bond in each of the hydrocarbon chains, it is called a mono-unsaturated fat (or mono-unsaturated oil, because it is likely to be a liquid at room temperature.)
A typical mono-unsaturated oil might be:
If there are two or more carbon-carbon double bonds in each chain, then it is said to be polyunsaturated.
For example:
For simplicity, in all these diagrams, all three hydrocarbon chains in each molecule are the same. That doesn't have to be the case - you can have a mixture of types of chain in the same molecule.
Trans Fat
A major health concern during the hydrogenation process is the production of trans fats. Trans fats are the result of a side reaction with the catalyst of the hydrogenation process. This is the result of an unsaturated fat which is normally found as a cis isomer converts to a trans isomer of the unsaturated fat. Isomers are molecules that have the same molecular formula but are bonded together differently. Focusing on the Sp2 double bonded carbons, a cis isomer has the hydrogens on the same side. Due to the added energy from the hydrogenation process, the activation energy is reached to convert the cis isomers of the unsaturated fat to a trans isomer of the unsaturated fat. The effect is putting one of the hydrogens on the opposite side of one of the carbons. This results in a trans configuration of the double bonded carbons. The human body doesn’t recognized trans fats.
Although the trans fatty acids are chemically "monounsaturated" or "polyunsaturated" they are considered so different from the cis monounsaturated or polyunsaturated fatty acids that they can not be legally designated as unsaturated for purposes of labeling. Most of the trans fatty acids (although chemically still unsaturated) produced by the partial hydrogenation process are now classified in the same category as saturated fats.
The major negative is that trans fat tends to raise "bad" LDL- cholesterol and lower "good" HDL-cholesterol, although not as much as saturated fat. Trans fat are found in margarine, baked goods such as doughnuts and Danish pastry, deep-fried foods like fried chicken and French-fried potatoes, snack chips, imitation cheese, and confectionery fats.
Margarine manufacture
Some margarine is made by hydrogenating carbon-carbon double bonds in animal or vegetable fats and oils. You can recognise the presence of this in foods because the ingredients list will include words showing that it contains "hydrogenated vegetable oils" or "hydrogenated fats".
The impression is sometimes given that all margarine is made by hydrogenation - that's simply not true.
Vegetable oils often contain high proportions of polyunsaturated and mono-unsaturated fats (oils), and as a result are liquids at room temperature. That makes them messy to spread on your bread or toast, and inconvenient for some baking purposes.
You can "harden" (raise the melting point of) the oil by hydrogenating it in the presence of a nickel catalyst. Conditions (like the precise temperature, or the length of time the hydrogen is passed through the oil) are carefully controlled so that some, but not necessarily all, of the carbon-carbon double bonds are hydrogenated.
This produces a "partially hydrogenated oil" or "partially hydrogenated fat".
You need to hydrogenate enough of the bonds to give the final texture you want. However, there are possible health benefits in eating mono-unsaturated or polyunsaturated fats or oils rather than saturated ones - so you wouldn't want to remove all the carbon-carbon double bonds.
The downside of hydrogenation as a means of hardening fats and oils
There are some probable health risks from eating hydrogenated fats or oils. Consumers are becoming more aware of this, and manufacturers are increasingly finding alternative ways of converting oils into spreadable solids.
One of the problems arises from the hydrogenation process.
The double bonds in unsaturated fats and oils tend to have the groups around them arranged in the "cis" form.
The relatively high temperatures used in the hydrogenation process tend to flip some of the carbon-carbon double bonds into the "trans" form. If these particular bonds aren't hydrogenated during the process, they will still be present in the final margarine in molecules of trans fats.
The consumption of trans fats has been shown to increase cholesterol levels (particularly of the more harmful LDL form) - leading to an increased risk of heart disease.
Any process which tends to increase the amount of trans fat in the diet is best avoided. Read food labels, and avoid any food which contains (or is cooked in) hydrogenated oil or hydrogenated fat.
Problems
1. After the hydrogenation of an unsaturated fatty acid, would it exist at room temperature as a liquid or solid?
2. Write the hydrogenation reaction for linoleic acid to hydrogenate all of the double bonds. What is the new name for this fatty acid? Hint: count carbons.
3. Compare a "hard" type margarine vs. a "soft" margarine. Which has the most double bonds?
4. Which is the most saturated?
Contributors
• Charles Ophardt, Professor Emeritus, Elmhurst College; Virtual Chembook
• Antonio Rodriguez
• Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/12%3A_Oxidation_and_Reduction/12.04%3A_Application-_Hydrogenation_of_Oils.txt |
Reactions between alkynes and catalysts are a common source of alkene formation. Because alkynes differ from alkenes on account of their two procurable π bonds, alkynes are more susceptible to additions. Aside from turning them into alkenes, these catalysts affect the arrangement of substituents on the newly formed alkene molecule. Depending on which catalyst is used, the catalysts cause anti- or syn-addition of hydrogens. Alkynes can readily undergo additions because of their availability of two π bonds.
Hydrogenation of an Alkyne
Alkynes can be fully hydrogenated into alkanes with the help of a platinum catalyst. However, the use of two other catalysts can be used to hydrogenate alkynes to alkanes. These catalysts are: Palladium dispersed on carbon (Pd/C) and finely dispersed nickel (Raney-Ni).
Hydrogenation of an Alkyne to a Cis-Alkene
Because hydrogenation is an interruptible process involving a series of steps, hydrogenation can be stopped, using modified catalysts (e.g., Lindlar’s Catalyst) at the transitional alkene stage. Lindar’s catalyst has three components: Palladium-Calcium Carbonate, lead acetate and quinoline. The quinoline serves to prevent complete hydrogenation of the alkyne to an alkane. Lindlar’s Catalyst transforms an alkyne to a cis-alkene.
Hydrogenation of an Alkyne to a Trans-Alkene
Alkynes can be reduced to trans-alkenes with the use of sodium dissolved in an ammonia solvent. An Na radical donates an electron to one of the P bonds in a carbon-carbon triple bond. This forms an anion, which can be protonated by a hydrogen in an ammonia solvent. This prompts another Na radical to donate an electron to the second P orbital. Soon after this anion is also protonated by a hydrogen from the ammonia solvent, resulting in a trans-alkene.
Contributors
• Ravjot Takhar (UCD)
12.06: The Reduction of Polar CX Bonds
Reduction of Halides
Alkyl halides can be reduced to alkanes through reaction with hydrife reagents, most commonly LiAlH4
Hydride addition to epoxides
The hydride attacks the less substituted side of the epoxide. The H and OH are anti to each other.
12.07: Oxidizing Agents
The laboratory oxidation of an alcohol to form an aldehyde or ketone is mechanistically different from the biochemical oxidations with NAD(P)+ that we saw earlier in this chapter. The general picture of laboratory oxidations is illustrated below. Essentially what happens is that the hydroxide hydrogen of the alcohol is replaced by a leaving group (X in the figure below).
Then, a base can abstract the proton bound to the alcohol carbon, which results in elimination of the X leaving group and formation of a new carbon-oxygen double bond. As you can see by looking closely at this general mechanism, tertiary alcohols cannot be oxidized in this way – there is no hydrogen to abstract in the final step!
A common method for oxidizing secondary alcohols to ketones uses chromic acid (H2CrO4) as the oxidizing agent. Chromic acid, also known as Jones reagent, is prepared by adding chromium trioxide (CrO3) to aqueous sulfuric acid.
A mechanism for the chromic acid oxidation of a ketone is shown below.
Note that the chromium reagent has lost two bonds to oxygen in this reaction, and thus has been reduced (it must have been reduced - it is the oxidizing agent!).
Ketones are not oxidized by chromic acid, so the reaction stops at the ketone stage. In contrast, primary alcohols are oxidized by chromic acid first to aldehydes, then straight on to carboxylic acids.
It is actually the hydride form of the aldehyde that is oxidized (recall from section 11.3 that aldehydes in aqueous solution exist in rapid equilibrium with their hydrate forms).
One of the hydroxyl groups of the hydrate attacks chromic acid, and the reaction proceeds essentially as shown for the oxidation of a secondary alcohol.
Under some conditions, chromic acid will even oxidize a carbon in the benzylic position to a carboxylic acid (notice that a carbon-carbon bond is broken in this transformation).
A number of other common oxidizing agents are discussed below.
The pyridinium chlorochromate (PCC) and Swern oxidation reactions are useful for oxidizing primary alcohols to aldehydes. Further oxidation of the aldehyde to the carboxylic acid stage does not occur with these reagents, because the reactions are carried out in anhydrous (water-free) organic solvents such as dichloromethane, and therefore the hydrate form of the aldehyde is not able to form.
The Swern oxidation uses dimethylsulfoxide and oxalyl chloride, followed by addition of a base such as triethylamine. The actual oxidizing species in this reaction is the dimethylchlorosulfonium ion, which forms from dimethylsulfoxide and oxalyl chloride.
You will be asked to propose a mechanism for these reactions in the end of chapter problems.
Pyridinium chlorochromate is generated by combining chromium trioxide, hydrochloric acid, and pyridine.
The PCC and Swern oxidation conditions can both also be used to oxidize secondary alcohols to ketones.
Silver ion, Ag(I), is often used to oxidize aldehydes to carboxylic acid. Two common reaction conditions are:
The set of reagents in the latter reaction conditions are commonly known as ‘Tollens’ reagent’.
Alkenes are oxidized to cis-1,2-diols by osmium tetroxide (OsO4). The stereospecificity is due to the formation of a cyclic osmate ester intermediate. Osmium tetroxide is used in catalytic amounts, and is regenerated by N-methylmorpholine-N-oxide.
cis-1,2-diol compounds can be oxidized to dialdehydes (or diketones, depending on the substitution of the starting diol) using periodic acid:
Alkenes can also be oxidized by treatment with ozone, O3. In ozonolysis, the carbon-carbon double bond is cleaved, and the alkene carbons are converted to aldehydes:
Dimethyl sulfide or zinc is added in the work-up stage of the reaction in order to reduce hydrogen peroxide, which is formed in the reaction, to water.
Alternatively, hydrogen peroxide and aqueous base can be added in the workup to obtain carboxylic acids:
Potassium permanganate (KMnO4) is another very powerful oxidizing agent that will oxidize primary alcohols and aldehydes to carboxylic acids. KMnO4 is also useful for oxidative cleavage of alkenes to ketones and carboxylic acids:
Finally, alkenes can be oxidized to epoxides using a 'peroxyacid' such as m-chloroperoxybenzoic acid. Notice the presence of a third oxygen in the peroxyacid functional group.
The mechanism is similar to that of the biological epoxidation catalyzed by squalene epoxidase (section 16.10A), with the π electrons in the alkene double bond attacking the 'outer' oxygen of the peroxyacid and cleaving the reactive O-O peroxide bond.
Uncatalyzed epoxidation of an asymmetric alkene generally results in two diastereomeric epoxide products, with the epoxide adding either from above or below the plane of the alkene.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/12%3A_Oxidation_and_Reduction/12.05%3A_Reduction_of_Alkynes.txt |
Oxacyclopropane rings, also called epoxide rings, are useful reagents that may be opened by further reaction to form anti vicinal diols. One way to synthesize oxacyclopropane rings is through the reaction of an alkene with peroxycarboxylic acid.
Oxacyclopropane Synthesis by Peroxycarboxylic Acid
Oxacyclopropane synthesis by peroxycarboxylic acid requires an alkene and a peroxycarboxylic acid as well as an appropriate solvent. The peroxycarboxylic acid has the unique property of having an electropositive oxygen atom on the COOH group. The reaction is initiated by the electrophilic oxygen atom reacting with the nucleophilic carbon-carbon double bond. The mechanism involves a concerted reaction with a four-part, circular transition state. The result is that the originally electropositive oxygen atom ends up in the oxacyclopropane ring and the COOH group becomes COH.
Mechanism
Peroxycarboxylic acids are generally unstable. An exception is meta-chloroperoxybenzoic acid, shown in the mechanism above. Often abbreviated MCPBA, it is a stable crystalline solid. Consequently, MCPBA is popular for laboratory use. However, MCPBA can be explosive under some conditions.
Peroxycarboxylic acids are sometimes replaced in industrial applications by monoperphthalic acid, or the monoperoxyphthalate ion bound to magnesium, which gives magnesium monoperoxyphthalate (MMPP). In either case, a nonaqueous solvent such as chloroform, ether, acetone, or dioxane is used. This is because in an aqueous medium with any acid or base catalyst present, the epoxide ring is hydrolyzed to form a vicinal diol, a molecule with two OH groups on neighboring carbons. (For more explanation of how this reaction leads to vicinal diols, see below.) However, in a nonaqueous solvent, the hydrolysis is prevented and the epoxide ring can be isolated as the product. Reaction yields from this reaction are usually about 75%. The reaction rate is affected by the nature of the alkene, with more nucleophilic double bonds resulting in faster reactions.
Example
Since the transfer of oxygen is to the same side of the double bond, the resulting oxacyclopropane ring will have the same stereochemistry as the starting alkene. A good way to think of this is that the alkene is rotated so that some constituents are coming forward and some are behind. Then, the oxygen is inserted on top. (See the product of the above reaction.) One way the epoxide ring can be opened is by an acid catalyzed oxidation-hydrolysis. Oxidation-hydrolysis gives a vicinal diol, a molecule with OH groups on neighboring carbons. For this reaction, the dihydroxylation is anti since, due to steric hindrance, the ring is attacked from the side opposite the existing oxygen atom. Thus, if the starting alkene is trans, the resulting vicinal diol will have one S and one R stereocenter. But, if the starting alkene is cis, the resulting vicinal diol will have a racemic mixture of S, S and R, R enantiomers.
The Synthesis of Disparlure using epoxidation
The gypsy moth (Porthytria dispar) is a serious pest of the forests. In 1976 B.A. Bierl et.al., (Science, 170,88 (1970)) isolated the sex pheromone from extracts of 78,000 tips of the last two abdominal segments of female moths. The structure was assigned as 1.5.1. Later, the precursor molecule – the cis-olefin was also
Fig 1.5.1
isolates from the same source. A laboratory bioassay from synthetic materials showed that just 2 pg of 1.5.1 was enough to elicit bioactivity. Since the availability of the molecule from natural sources was very minute even for structure elucidation problems and study its anticipated role as pest control molecule, there was intense interest in an efficient synthesis of this molecule. Some disconnections for this simple molecule are depicted in Fig 1.5.2.
Fig 1.5.2
Epoxides could be made from corresponding olefins. In this case, the olefin should be Z-olefin. When synthesis of such olefins are not stereospecific, direct epoxidation using peroxides would yield a mixture of α- and β-epoxide from both isomeric olefins. To avoid such mixtures at the last stage, one should introduce selectivity at an early stage of the synthesis.
The first attempt was directed towards synthesis of the appropriate olefin and epoxidation (B.A. Bierl et.al., (Science, 170, 88 (1970)). The stereoselectivity was unsatisfactory (Fig 1.5.3). This necessitated extensive purification.
References
1. Royals, E. 1954. Advanced Organic Chemistry. New York: Prentice Hall. 948 p.
2. Streitwieser, A. and C. Heathcock. 1981. Introduction to Organic Chemistry. 2nd ed. New York: Macmillan Publishing Co. 1258 p.
3. Vollhardt, K. and N. Schore. 2007. Organic Chemistry: Structure and Function. 5th ed. New York: W.H. Freeman and Company. 1254 p.
4. Wheland, G. 1949. Advanced Organic Chemistry. 3rd ed. New York: John Wiley & Sons. 871 p.
Problems
1. Predict the product of the reaction of cis-2-hexene with MCPBA (meta-chloroperoxybenzoic acid)
a) in acetone solvent.
b) in an aqueous medium with acid or base catalyst present.
2. Predict the product of the reaction of trans-2-pentene with magnesium monoperoxyphthalate (MMPP) in a chloroform solvent.
3. Predict the product of the reaction of trans-3-hexene with MCPBA in ether solvent.
4. Predict the reaction of propene with MCPBA.
a) in acetone solvent
b) after aqueous work-up.
5. Predict the reaction of cis-2-butene in chloroform solvent.
Answers
1. a) Cis-2-methyl-3-propyloxacyclopropane
b) Racemic (2R,3R)-2,3-hexanediol and (2S,3S)-2,3-hexanediol
2. Trans-3-ethyl-2-methyloxacyclopropane.
3. Trans-3,4-diethyloxacyclopropane.
4. a) 1-ethyl-oxacyclopropane
b) Racemic (2S)-1,2-propandiol and (2R)-1,2-propanediol
5. Cis-2,3-dimethyloxacyclopropane
Contributors
• Kristen Perano | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/12%3A_Oxidation_and_Reduction/12.08%3A_Epoxidation.txt |
Anti Dihydroxylation
Epoxides may be cleaved by aqueous acid to give glycols that are often diastereomeric with those prepared by the syn-hydroxylation reaction described above. Proton transfer from the acid catalyst generates the conjugate acid of the epoxide, which is attacked by nucleophiles such as water in the same way that the cyclic bromonium ion described above undergoes reaction. The result is anti-hydroxylation of the double bond, in contrast to the syn-stereoselectivity of the earlier method. In the following equation this procedure is illustrated for a cis-disubstituted epoxide, which, of course, could be prepared from the corresponding cis-alkene. This hydration of an epoxide does not change the oxidation state of any atoms or groups.
Syn Dihydroxylation
Osmium tetroxide oxidizes alkenes to give glycols through syn addition. A glycol, also known as a vicinal diol, is a compound with two -OH groups on adjacent carbons.
Introduction
The reaction with $OsO_4$ is a concerted process that has a cyclic intermediate and no rearrangements. Vicinal syn dihydroxylation complements the epoxide-hydrolysis sequence which constitutes an anti dihydroxylation of an alkene. When an alkene reacts with osmium tetroxide, stereocenters can form in the glycol product. Cis alkenes give meso products and trans alkenes give racemic mixtures.
$OsO_4$ is formed slowly when osmium powder reacts with gasoues $O_2$ at ambient temperature. Reaction of bulk solid requires heating to 400 °C:
$Os_{(s)} + 2O_{2\;(g)} \rightarrow OS_4$
Since Osmium tetroxide is expensive and highly toxic, the reaction with alkenes has been modified. Catalytic amounts of OsO4 and stoichiometric amounts of an oxidizing agent such as hydrogen peroxide are now used to eliminate some hazards. Also, an older reagent that was used instead of OsO4 was potassium permanganate, $KMnO_4$. Although syn diols will result from the reaction of KMnO4 and an alkene, potassium permanganate is less useful since it gives poor yields of the product because of overoxidation.
Mechanism
• Electrophilic attack on the alkene
• Pi bond of the alkene acts as the nucleophile and reacts with osmium (VIII) tetroxide (OsO4)
• 2 electrons from the double bond flows toward the osmium metal
• In the process, 3 electron pairs move simultaneously
• Cyclic ester with Os (VI) is produced
• Reduction
• H2S reduces the cyclic ester
• NaHSO4 with H2O may be used
• Forms the syn-1,2-diol (glycol)
Example: Dihydroxylation of 1-ethyl-1-cycloheptene
Hydroxylation of alkenes
Dihydroxylated products (glycols) are obtained by reaction with aqueous potassium permanganate (pH > 8) or osmium tetroxide in pyridine solution. Both reactions appear to proceed by the same mechanism (shown below); the metallocyclic intermediate may be isolated in the osmium reaction. In basic solution the purple permanganate anion is reduced to the green manganate ion, providing a nice color test for the double bond functional group. From the mechanism shown here we would expect syn-stereoselectivity in the bonding to oxygen, and regioselectivity is not an issue.
When viewed in context with the previously discussed addition reactions, the hydroxylation reaction might seem implausible. Permanganate and osmium tetroxide have similar configurations, in which the metal atom occupies the center of a tetrahedral grouping of negatively charged oxygen atoms. How, then, would such a species interact with the nucleophilic pi-electrons of a double bond? A possible explanation is that an empty d-orbital of the electrophilic metal atom extends well beyond the surrounding oxygen atoms and initiates electron transfer from the double bond to the metal, in much the same fashion noted above for platinum. Back-bonding of the nucleophilic oxygens to the antibonding π*-orbital completes this interaction. The result is formation of a metallocyclic intermediate, as shown above.
Chemical Highlight
Antitumor drugs have been formed by using dihydroxylation. This method has been applied to the enantioselective synthesis of ovalicin, which is a class of fungal-derived products called antiangiogenesis agents. These antitumor products can cut off the blood supply to solid tumors. A derivative of ovalicin, TNP-470, is chemically stable, nontoxic, and noninflammatory. TNP-470 has been used in research to determine its effectiveness in treating cancer of the breast, brain, cervix, liver, and prostate.
Problems
Questions:
1. Give the major product.
2. What is the product in the dihydroxylation of (Z)-3-hexene?
3. What is the product in the dihydroxylation of (E)-3-hexene?
4. Draw the intermediate of this reaction.
5. Fill in the missing reactants, reagents, and product.
Solutions
1. A syn-1,2-ethanediol is formed. There is no stereocenter in this particular reaction. The OH groups are on the same side.
2. Meso-3,4-hexanediol is formed. There are 2 stereocenters in this reaction.
3. A racemic mixture of 3,4-hexanediol is formed. There are 2 stereocenters in both products.
4. A cyclic osmic ester is formed.
5. The Diels-Alder cycloaddition reaction is needed in the first box to form the cyclohexene. The second box needs a reagent to reduce the intermediate cyclic ester (not shown). The third box has the product: 1,2-cyclohexanediol.
12.10: Oxidative Cleavage of Alkenes
Ozonolysis is a method of oxidatively cleaving alkenes or alkynes using ozone (\(O_3\)), a reactive allotrope of oxygen. The process allows for carbon-carbon double or triple bonds to be replaced by double bonds with oxygen. This reaction is often used to identify the structure of unknown alkenes. by breaking them down into smaller, more easily identifiable pieces. Ozonolysis also occurs naturally and would break down repeated units used in rubber and other polymers. On an industrial scale, azelaic acid and pelargonic acids are produced from ozonolysis.
Introduction
The gaseous ozone is first passed through the desired alkene solution in either methanol or dichloromethane. The first intermediate product is an ozonide molecule which is then further reduced to carbonyl products. This results in the breaking of the Carbon-Carbon double bond and is replaced by a Carbon-Oxygen double bond instead.
Reaction Mechanism
Step 1:
The first step in the mechanism of ozonolysis is the initial electrophilic addition of ozone to the Carbon-Carbon double bond, which then form the molozonide intermediate. Due to the unstable molozonide molecule, it continues further with the reaction and breaks apart to form a carbonyl and a carbonyl oxide molecule.
Step 2:
The carbonyl and the carbonyl oxide rearranges itself and reforms to create the stable ozonide intermediate. A reductive workup could then be performed to convert convert the ozonide molecule into the desired carbonyl products.
12.11: Oxidative Cleavage of Alkynes
Alkynes can also undergo oxidative cleavage. Internal alkynes form carboxylic acids (RCOOH) and terminal alkynes form carboxylic acids and CO2. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/12%3A_Oxidation_and_Reduction/12.09%3A_Dihydroxylation.txt |
This page looks at the oxidation of alcohols using acidified sodium or potassium dichromate(VI) solution. This reaction is used to make aldehydes, ketones and carboxylic acids, and as a way of distinguishing between primary, secondary and tertiary alcohols.
Oxidizing the different types of alcohols
The oxidizing agent used in these reactions is normally a solution of sodium or potassium dichromate(VI) acidified with dilute sulfuric acid. If oxidation occurs, the orange solution containing the dichromate(VI) ions is reduced to a green solution containing chromium(III) ions. The electron-half-equation for this reaction is
$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
Oxidizing agents
K2Cr2O7 potassium dichromate
CrO3 Chromium Trioxide
Both of these are used along with
H2SO4, H2O
1o alcohol → Carboxylic acid
2o alcohol → Ketone
3o alcohol → No reaction
Primary alcohols
Primary alcohols can be oxidized to either aldehydes or carboxylic acids depending on the reaction conditions. In the case of the formation of carboxylic acids, the alcohol is first oxidized to an aldehyde which is then oxidized further to the acid.
Full oxidation to carboxylic acids
You need to use an excess of the oxidizing agent and make sure that the aldehyde formed as the half-way product stays in the mixture. The alcohol is heated under reflux with an excess of the oxidizing agent. When the reaction is complete, the carboxylic acid is distilled off. The full equation for the oxidation of ethanol to ethanoic acid is:
$3CH_3CH_2OH + 2Cr_2O_7^{2-} + 16H+ \rightarrow 3CH_3COOH + 4Cr^{3+} + 11H_2O$
The more usual simplified version looks like this:
$CH_3CH_2OH + 2[O] \rightarrow CH_3COOH + H_2O$
Alternatively, you could write separate equations for the two stages of the reaction - the formation of ethanal and then its subsequent oxidation.
$CH_3CH_2OH + [O] \rightarrow CH_3CHO + H_2O$
$CH_3CHO + [O] \rightarrow CH_3COOH$
This is what is happening in the second stage:
Secondary alcohols
Secondary alcohols are oxidized to ketones - and that's it. For example, if you heat the secondary alcohol propan-2-ol with sodium or potassium dichromate(VI) solution acidified with dilute sulfuric acid, you get propanone formed. Playing around with the reaction conditions makes no difference whatsoever to the product. Using the simple version of the equation and showing the relationship between the structures:
If you look back at the second stage of the primary alcohol reaction, you will see that an oxygen "slotted in" between the carbon and the hydrogen in the aldehyde group to produce the carboxylic acid. In this case, there is no such hydrogen - and the reaction has nowhere further to go.
Tertiary alcohols
Tertiary alcohols are not oxidized by acidified sodium or potassium dichromate(VI) solution - there is no reaction whatsoever. If you look at what is happening with primary and secondary alcohols, you will see that the oxidizing agent is removing the hydrogen from the -OH group, and a hydrogen from the carbon atom attached to the -OH. Tertiary alcohols don't have a hydrogen atom attached to that carbon.
You need to be able to remove those two particular hydrogen atoms in order to set up the carbon-oxygen double bond.
Using these reactions as a test for the different types of alcohol
First you have to be sure that you have actually got an alcohol by testing for the -OH group. You would need to show that it was a neutral liquid, free of water and that it reacted with solid phosphorus(V) chloride to produce a burst of acidic steamy hydrogen chloride fumes. You would then add a few drops of the alcohol to a test tube containing potassium dichromate(VI) solution acidified with dilute sulfuric acid. The tube would be warmed in a hot water bath.
Picking out the tertiary alcohol
In the case of a primary or secondary alcohol, the orange solution turns green. With a tertiary alcohol there is no color change. After heating:
Distinguishing between the primary and secondary alcohols
You need to produce enough of the aldehyde (from oxidation of a primary alcohol) or ketone (from a secondary alcohol) to be able to test them. There are various things which aldehydes do which ketones don't. These include the reactions with Tollens' reagent, Fehling's solution and Benedict's solution, and are covered on a separate page.
These tests can be a bit of a bother to carry out and the results are not always as clear-cut as the books say. A much simpler but fairly reliable test is to use Schiff's reagent. Schiff's reagent is a fuchsin dye decolorized by passing sulfur dioxide through it. In the presence of even small amounts of an aldehyde, it turns bright magenta.
It must, however, be used absolutely cold, because ketones react with it very slowly to give the same color. If you heat it, obviously the change is faster - and potentially confusing. While you are warming the reaction mixture in the hot water bath, you can pass any vapours produced through some Schiff's reagent.
• If the Schiff's reagent quickly becomes magenta, then you are producing an aldehyde from a primary alcohol.
• If there is no color change in the Schiff's reagent, or only a trace of pink color within a minute or so, then you are not producing an aldehyde, and so haven't got a primary alcohol.
Because of the color change to the acidified potassium dichromate(VI) solution, you must therefore have a secondary alcohol. You should check the result as soon as the potassium dichromate(VI) solution turns green - if you leave it too long, the Schiff's reagent might start to change color in the secondary alcohol case as well.
Formation of Aldehydes using PCC
Pyridinium chlorochromate (PCC) is a milder version of chromic acid.
PCC oxidizes alcohols one rung up the oxidation ladder, from primary alcohols to aldehydes and from secondary alcohols to ketones. Unlike chromic acid, PCC will not oxidize aldehydes to carboxylic acids. Similar to or the same as: $CrO_3$ and pyridine (the Collins reagent) will also oxidize primary alcohols to aldehydes. Here are two examples of PCC in action.
• If you add one equivalent of PCC to either of these alcohols, you obtain the oxidized version. The byproducts (featured in grey) are Cr(IV) as well as pyridinium hydrochloride.
• One has to be careful with the amount of water present in the reaction. If water were present, it can ad to the aldehyde to make the hydrate, which could be further oxidized by a second equivalent of PCC were it present. This is not a concern with ketones, since there is no H directly bonded to C.
How does it work? Oxidation reactions of this sort are actually a kind of elimination reaction. We’re going from a carbon-oxygen single bond to a carbon-oxygen double bond. The elimination reaction can occur because we’re putting a good leaving group on the oxygen, namely the chromium, which will be displaced when the neighboring C-H bond is broken with a base.
The first step is attack of oxygen on the chromium to form the Cr-O bond. Secondly, a proton on the (now positive) OH is transferred to one of the oxygens of the chromium, possibly through the intermediacy of the pyridinium salt. A chloride ion is then displaced, in a reaction reminiscent of a 1,2 elimination reaction, to form what is known as a chromate ester.
The C-O double bond is formed when a base removes the proton on the carbon adjacent to the oxygen. [aside: I've drawn the base as Cl(-) although there are certainly other species which could also act as bases here (such as an alcohol). It is also possible for pyridine to be used as the base here, although only very low concentrations of the deprotonated form will be present under these acidic conditions.] The electrons from the C-H bond move to form the C-O bond, and in the process break the O-Cr bond, and Cr(VI) becomes Cr(IV) in the process (drawn here as O=Cr(OH)2 ).
Real life notes: If you end up using PCC in the lab, don’t forget to add molecular sieves or Celite or some other solid to the bottom of the flask, because otherwise you get a nasty brown tar that is a real major pain to clean up. The toxicity and mess associated with chromium has spurred the development of other alternatives like TPAP, IBX, DMP, and a host of other neat reagents you generally don’t learn about until grad school.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
James Ashenhurst (MasterOrganicChemistry.com)
12.14: Application- The Oxidation of Ethanol
Ethanol is oxidized in the liver to acetaldehyde:
The acetaldehyde is in turn oxidized to acetic acid (HC2H3O2), a normal constituent of cells, which is then oxidized to carbon dioxide and water. Even so, ethanol is potentially toxic to humans. The rapid ingestion of 1 pt (about 500 mL) of pure ethanol would kill most people, and acute ethanol poisoning kills several hundred people each year—often those engaged in some sort of drinking contest. Ethanol freely crosses into the brain, where it depresses the respiratory control center, resulting in failure of the respiratory muscles in the lungs and hence suffocation. Ethanol is believed to act on nerve cell membranes, causing a diminution in speech, thought, cognition, and judgment.
Rubbing alcohol is usually a 70% aqueous solution of isopropyl alcohol. It has a high vapor pressure, and its rapid evaporation from the skin produces a cooling effect. It is toxic when ingested but, compared to methanol, is less readily absorbed through the skin.
12.15: Sharpless Epoxidation
Epoxides are very useful intermediates in organic synthesis. Because most naturally occurring molecules (including those with medicinal properties) are chiral, control of stereochemistry is one of the most important challenges facing a synthetic chemist attempting to synthesize a naturally occurring molecule in the laboratory. In what was arguably one of the most important discoveries in synthetic organic chemistry in recent decades, Barry Sharpless of Stanford University reported in 1980 that he and his colleagues had developed a method to stereoselectively epoxidize asymmetric alkenes which contained an alcohol in the allylic position. The ‘Sharpless asymmetric oxidation’ is achieved with the use of a chiral catalyst composed of (+) or (-) diethyltartrate and an organotitanium compound (J. Am. Chem. Soc. 1980, 102, 5974). Depending on which stereoisomer of diethyltartrate is used, the peroxyacid oxygen tends to add to either the top or bottom plane of the alkene.
This technique allows for the specific introduction of two new stereocenters at an alkene position, which as you can imagine makes it an extremely useful synthetic tool.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/12%3A_Oxidation_and_Reduction/12.12%3A_Oxidation_of_Alcohols.txt |
In chemistry, a radical (more precisely, a free radical) is an atom, molecule, or ion that has unpaired valence electrons or an open electron shell, and therefore may be seen as having one or more "dangling" covalent bonds.
With some exceptions, these "dangling" bonds make free radicals highly chemically reactive towards other substances, or even towards themselves: their molecules will often spontaneously dimerize or polymerize if they come in contact with each other. Most radicals are reasonably stable only at very low concentrations in inert media or in a vacuum.
Free radicals may be created in a number of ways, including synthesis with very dilute or rarefied reagents, reactions at very low temperatures, or breakup of larger molecules. The latter can be affected by any process that puts enough energy into the parent molecule, such as ionizing radiation, heat, electrical discharges, electrolysis, and chemical reactions. Indeed, radicals are intermediate stages in many chemical reactions.
History
The first organic free radical identified was triphenylmethyl radical. This species was discovered by Moses Gomberg in 1900 at the University of Michigan USA. Historically, the term radical in radical theory was also used for bound parts of the molecule, especially when they remain unchanged in reactions. These are now called functional groups. For example, methyl alcohol was described as consisting of a methyl "radical" and a hydroxyl "radical". Neither are radicals in the modern chemical sense, as they are permanently bound to each other, and have no unpaired, reactive electrons; however, they can be observed as radicals in mass spectrometry when broken apart by irradiation with energetic electrons.
Depiction in chemical reactions
In this chapter, we will learn about some reactions in which the key steps involve the movement of single electrons. You may recall from way back in section 6.1A that single electron movement is depicted by a single-barbed'fish-hook' arrow (as opposed to the familiar double-barbed arrows that we have been using throughout the book to show two-electron movement).
Single-electron mechanisms involve the formation and subsequent reaction of free radical species, highly unstable intermediates that contain an unpaired electron. We will learn in this chapter how free radicals are often formed from homolytic cleavage, an event where the two electrons in a breaking covalent bond move in opposite directions.
(In contrast, essentially all of the reactions we have studied up to now involve bond-breaking events in which both electrons move in the same direction: this is called heterolytic cleavage).
We will also learn that many single-electron mechanisms take the form of a radical chain reaction, in which one radical causes the formation of a second radical, which in turn causes the formation of a third radical, and so on.
The high reactivity of free radical species and their ability to initiate chain reactions is often beneficial - we will learn in this chapter about radical polymerization reactions that form useful materials such as plexiglass and polyproylene fabric. We will also learn about radical reactions that are harmful, such as the degradation of atmospheric ozone by freon, and the oxidative damage done to lipids and DNA in our bodies by free radicals species. Finally, we will see how some enzymes use bound metals to catalyze high e
The geometry and relative stability of carbon radicals
As organic chemists, we are particularly interested in radical intermediates in which the unpaired electron resides on a carbon atom. Experimental evidence indicates that the three bonds in a carbon radical have trigonal planar geometry, and therefore the carbon is considered to be sp2-hybridized with the unpaired electron occupying the perpendicular, unhybridized 2pzorbital. Contrast this picture with carbocation and carbanion intermediates, which are both also trigonal planar but whose 2pz orbitals contain zero or two electrons, respectively.
The trend in the stability of carbon radicals parallels that of carbocations (section 8.4B): tertiary radicals, for example, are more stable than secondary radicals, followed by primary and methyl radicals. This should make intuitive sense, because radicals, like carbocations, can be considered to be electron deficient, and thus are stabilized by the electron-donating effects of nearby alkyl groups. Benzylic and allylic radicals are more stable than alkyl radicals due to resonance effects - an unpaired electron can be delocalized over a system of conjugated pi bonds. An allylic radical, for example, can be pictured as a system of three parallel 2pz orbitals sharing three electrons.
Trends in radical stability
Allyic & Benzlic > 3o > 2o > 1o > Methyl
Template:ExampleStart
Exercise 17.1: Draw the structure of a benzylic radical compound, and then draw a resonance form showing how the radical is stabilized.
Template:ExampleEnd
With enough resonance stabilization, radicals can be made that are quite unreactive. One example of an inert organic radical structure is shown below.
In this molecule, the already extensive resonance stabilization is further enhanced by the ability of the chlorine atoms to shield the radical center from external reagents. The radical is, in some sense, inside a protective 'cage'.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/13%3A_Radical_Reactions/13.01%3A_Introduction.txt |
The three phases of radical chain reactions
Because of their high reactivity, free radicals have the potential to be both extremely powerful chemical tools and extremely harmful contaminants. Much of the power of free radical species stems from the natural tendency of radical processes to occur in a chain reaction fashion. Radical chain reactions have three distinct phases: initiation, propagation, and termination.
The initiation phase describes the step that initially creates a radical species. In most cases, this is a homolytic cleavage event, and takes place very rarely due to the high energy barriers involved. Often the influence of heat, UV radiation, or a metal-containing catalyst is necessary to overcome the energy barrier.
Molecular chlorine and bromine will both undergo homolytic cleavage to form radicals when subjected to heat or light. Other functional groups which also tend to form radicals when exposed to heat or light are chlorofluorocarbons, peroxides, and the halogenated amide N-bromosuccinimide (NBS).
The propagation phase describes the 'chain' part of chain reactions. Once a reactive free radical is generated, it can react with stable molecules to form new free radicals. These new free radicals go on to generate yet more free radicals, and so on. Propagation steps often involve hydrogen abstraction or addition of the radical to double bonds.
Chain termination occurs when two free radical species react with each other to form a stable, non-radical adduct. Although this is a very thermodynamically downhill event, it is also very rare due to the low concentration of radical species and the small likelihood of two radicals colliding with one another. In other words, the Gibbs free energy barrier is very high for this reaction, mostly due to entropic rather than enthalpic considerations. The active sites of enzymes, of course, can evolve to overcome this entropic barrier by positioning two radical intermediates adjacent to one another.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
13.03: Halogenation of Alkanes
Methane and chlorine
If a mixture of methane and chlorine is exposed to a flame, it explodes - producing carbon and hydrogen chloride. This is not a very useful reaction! The reaction we are going to explore is a more gentle one between methane and chlorine in the presence of ultraviolet light - typically sunlight. This is a good example of a photochemical reaction - a reaction brought about by light.
CH4 + Cl2 + energy → CH3Cl + HC
CH4+Cl2CH3Cl+HCl
The organic product is chloromethane. One of the hydrogen atoms in the methane has been replaced by a chlorine atom, so this is a substitution reaction. However, the reaction doesn't stop there, and all the hydrogens in the methane can in turn be replaced by chlorine atoms. Multiple substitution is dealt with on a separate page, and you will find a link to that at the bottom of this page.
Substitution reactions happen in which hydrogen atoms in the methane are replaced one at a time by chlorine atoms. You end up with a mixture of chloromethane, dichloromethane, trichloromethane and tetrachloromethane.
The original mixture of a colorless and a green gas would produce steamy fumes of hydrogen chloride and a mist of organic liquids. All of the organic products are liquid at room temperature with the exception of the chloromethane which is a gas.
If you were using bromine, you could either mix methane with bromine vapor, or bubble the methane through liquid bromine - in either case, exposed to UV light. The original mixture of gases would, of course, be red-brown rather than green.
You wouldn't choose to use these reactions as a means of preparing these organic compounds in the lab because the mixture of products would be too tedious to separate. The mechanisms for the reactions are explained on separate pages.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/13%3A_Radical_Reactions/13.02%3A_General_Features_of_Radical_Reactions.txt |
lkanes (the most basic of all organic compounds) undergo very few reactions. One of these reactions is halogenation, or the substitution of a single hydrogen on the alkane for a single halogen to form a haloalkane. This reaction is very important in organic chemistry because it opens a gateway to further chemical reactions.
Introduction
While the reactions possible with alkanes are few, there are many reactions that involve haloalkanes. In order to better understand the mechanism (a detailed look at the step by step process through which a reaction occurs), we will closely examine the chlorination of methane. When methane (CH4) and chlorine (Cl2) are mixed together in the absence of light at room temperature nothing happens. However, if the conditions are changed, so that either the reaction is taking place at high temperatures (denoted by Δ) or there is ultra violet irradiation, a product is formed, chloromethane (CH3Cl).
Energetics
Why does this reaction occur? Is the reaction favorable? A way to answer these questions is to look at the change in enthalpy ($\Delta{H}$) that occurs when the reaction takes place.
ΔH = (Energy put into reaction) – (Energy given off from reaction)
If more energy is put into a reaction than is given off, the ΔH is positive, the reaction is endothermic and not energetically favorable. If more energy is given off in the reaction than was put in, the ΔH is negative, the reaction is said to be exothermic and is considered favorable. The figure below illustrates the difference between endothermic and exothermic reactions.
ΔH can also be calculated using bond dissociation energies (ΔH°):
$\Delta{H} = \sum \Delta{H^°} \text{ of bonds broken} - \sum \Delta{H^°} \text{ of bonds formed}$
Let’s look at our specific example of the chlorination of methane to determine if it is endothermic or exothermic:
Since, the ΔH for the chlorination of methane is negative, the reaction is exothermic. Energetically this reaction is favorable. In order to better understand this reaction we need to look at the mechanism ( a detailed step by step look at the reaction showing how it occurs) by which the reaction occurs.
Radical Chain Mechanism
The reaction proceeds through the radical chain mechanism. The radical chain mechanism is characterized by three steps: initiation, propagation and termination. Initiation requires an input of energy but after that the reaction is self-sustaining. The first propagation step uses up one of the products from initiation, and the second propagation step makes another one, thus the cycle can continue until indefinitely.
Step 1: Initiation
Initiation breaks the bond between the chlorine molecule (Cl2). For this step to occur energy must be put in, this step is not energetically favorable. After this step, the reaction can occur continuously (as long as reactants provide) without input of more energy. It is important to note that this part of the mechanism cannot occur without some external energy input, through light or heat.
Step 2: Propagation
The next two steps in the mechanism are called propagation steps. In the first propagation step, a chlorine radical combines with a hydrogen on the methane. This gives hydrochloric acid (HCl, the inorganic product of this reaction) and the methyl radical. In the second propagation step more of the chlorine starting material (Cl2) is used, one of the chlorine atoms becomes a radical and the other combines with the methyl radical.
The first propagation step is endothermic, meaning it takes in heat (requires 2 kcal/mol) and is not energetically favorable. In contrast the second propagation step is exothermic, releasing 27 kcal/mol. Since the second propagation step is so exothermic, it occurs very quickly. The second propagation step uses up a product from the first propagation step (the methyl radical) and following Le Chatelier's principle, when the product of the first step is removed the equilibrium is shifted towards it's products. This principle is what governs the unfavorable first propagation step's occurance.
Step 3: Termination
In the termination steps, all the remaining radicals combine (in all possible manners) to form more product (CH3Cl), more reactant (Cl2) and even combinations of the two methyl radicals to form a side product of ethane (CH3CH3).
Problems with the Chlorination of Methane
The chlorination of methane does not necessarily stop after one chlorination. It may actually be very hard to get a monosubstituted chloromethane. Instead di-, tri- and even tetra-chloromethanes are formed. One way to avoid this problem is to use a much higher concentration of methane in comparison to chloride. This reduces the chance of a chlorine radical running into a chloromethane and starting the mechanism over again to form a dichloromethane. Through this method of controlling product ratios one is able to have a relative amount of control over the product.
References
1. Matyjaszewski, Krzysztof, Wojciech Jakubowski, Ke Min, Wei Tang, Jinyu Huang, Wade A. Braunecker, and Nicolay V. Tsarevsky. "Diminishing Catalyst Concentration in Atom Transfer Radical Polymerization with Reducing Agents." Proceedings of the National Academy of Sciences of the United States of America 103 (2006): 15309-5314.
2. Morgan, G. T. "A State Experiment in Chemical Research." Science 72 (1930): 379-90.
3. Phillips, Francis C. "# Researches upon the Chemical Properties of Gases." Researches upon the Chemical Properties of Gases 17 (1893): 149-236.
Problems
Answers to these questions are in an attached slide
1. Write out the complete mechanism for the chlorination of methane.
2. Explain, in your own words, how the first propagation step can occur without input of energy if it is energetically unfavorable.
3. Compounds other than chlorine and methane go through halogenation with the radical chain mechanism. Write out a generalized equation for the halogenation of RH with X2including all the different steps of the mechanism.
4. Which step of the radical chain mechanism requires outside energy? What can be used as this energy?
5. Having learned how to calculate the change in enthalpy for the chlorination of methane apply your knowledge and using the table provided below calculate the change in enthalpy for the bromination of ethane.
Compound Bond Dissociation Energy (kcal/mol)
CH3CH2-H 101
CH3CH2-Br 70
H-Br 87
Br2 46
Contributors
• Kristen Kelley and Britt Farquharson | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/13%3A_Radical_Reactions/13.04%3A_The_Mechanism_of_Halogenation.txt |
When alkanes larger than ethane are halogenated, isomeric products are formed. Thus chlorination of propane gives both 1-chloropropane and 2-chloropropane as mono-chlorinated products. Four constitutionally isomeric dichlorinated products are possible, and five constitutional isomers exist for the trichlorinated propanes. Can you write structural formulas for the four dichlorinated isomers?
$CH_3CH_2CH_3 + 2Cl_2 \rightarrow \text{Four} \; C_3H_6Cl_2 \; \text{isomers} + 2 HCl$
The halogenation of propane discloses an interesting feature of these reactions. All the hydrogens in a complex alkane do not exhibit equal reactivity. For example, propane has eight hydrogens, six of them being structurally equivalent primary, and the other two being secondary. If all these hydrogen atoms were equally reactive, halogenation should give a 3:1 ratio of 1-halopropane to 2-halopropane mono-halogenated products, reflecting the primary/secondary numbers. This is not what we observe. Light-induced gas phase chlorination at 25 ºC gives 45% 1-chloropropane and 55% 2-chloropropane.
CH3-CH2-CH3 + Cl2 → 45% CH3-CH2-CH2Cl + 55% CH3-CHCl-CH3
The results of bromination ( light-induced at 25 ºC ) are even more suprising, with 2-bromopropane accounting for 97% of the mono-bromo product.
CH3-CH2-CH3 + Br2 → 3% CH3-CH2-CH2Br + 97% CH3-CHBr-CH3
These results suggest strongly that 2º-hydrogens are inherently more reactive than 1º-hydrogens, by a factor of about 3:1. Further experiments showed that 3º-hydrogens are even more reactive toward halogen atoms. Thus, light-induced chlorination of 2-methylpropane gave predominantly (65%) 2-chloro-2-methylpropane, the substitution product of the sole 3º-hydrogen, despite the presence of nine 1º-hydrogens in the molecule.
(CH3)3CH + Cl2 → 65% (CH3)3CCl + 35% (CH3)2CHCH2Cl
If you are uncertain about the terms primary (1º), secondary (2º) & tertiary (3º) Click Here.
It should be clear from a review of the two steps that make up the free radical chain reaction for halogenation that the first step (hydrogen abstraction) is the product determining step. Once a carbon radical is formed, subsequent bonding to a halogen atom (in the second step) can only occur at the radical site. Consequently, an understanding of the preference for substitution at 2º and 3º-carbon atoms must come from an analysis of this first step.
First Step: R3CH + X· → R3C· + H-X
Second Step: R3C· + X2 → R3CX + X·
Since the H-X product is common to all possible reactions, differences in reactivity can only be attributed to differences in C-H bond dissociation energies. In our previous discussion of bond energy we assumed average values for all bonds of a given kind, but now we see that this is not strictly true. In the case of carbon-hydrogen bonds, there are significant differences, and the specific dissociation energies (energy required to break a bond homolytically) for various kinds of C-H bonds have been measured. These values are given in the following table.
R (in R–H) methyl ethyl i-propyl
t-butyl
phenyl benzyl allyl vinyl
Bond Dissociation Energy
(kcal/mole)
103 98 95 93 110 85 88 112
The difference in C-H bond dissociation energy reported for primary (1º), secondary (2º) and tertiary (3º) sites agrees with the halogenation observations reported above, in that we would expect weaker bonds to be broken more easily than are strong bonds. By this reasoning we would expect benzylic and allylic sites to be exceptionally reactive in free radical halogenation, as experiments have shown. The methyl group of toluene, C6H5CH3, is readily chlorinated or brominated in the presence of free radical initiators (usually peroxides), and ethylbenzene is similarly chlorinated at the benzylic location exclusively. The hydrogens bonded to the aromatic ring (referred to as phenyl hydrogens above) have relatively high bond dissociation energies and are not substituted.
C6H5CH2CH3 + Cl2 → C6H5CHClCH3 + HCl | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/13%3A_Radical_Reactions/13.05%3A_Chlorination_of_Other_Alkanes.txt |
A Free Radical Substitution Reaction
This page gives you the facts and a simple, uncluttered mechanism for the free radical substitution reaction between methane and bromine. This reaction between methane and bromine happens in the presence of ultraviolet light - typically sunlight. This is a good example of a photochemical reaction - a reaction brought about by light.
$CH_4 + Br_2 \rightarrow CH_3Br + HBr$
The organic product is bromomethane. One of the hydrogen atoms in the methane has been replaced by a bromine atom, so this is a substitution reaction. However, the reaction doesn't stop there, and all the hydrogens in the methane can in turn be replaced by bromine atoms.
The mechanism
The mechanism involves a chain reaction. During a chain reaction, for every reactive species you start off with, a new one is generated at the end - and this keeps the process going.The over-all process is known as free radical substitution, or as a free radical chain reaction.
• Chain initiation: The chain is initiated (started) by UV light breaking a bromine molecule into free radicals.
Br22Br
• Chain propagation reactions: These are the reactions which keep the chain going.
CH4 + BrCH3 + HBr
CH3 + Br2CH3Br + Br
• Chain termination reactions: These are reactions which remove free radicals from the system without replacing them by new ones.
2BrBr2
CH3 + BrCH3Br
CH3 + CH3CH3CH3
Selectivity
When alkanes larger than ethane are halogenated, isomeric products are formed. Thus chlorination of propane gives both 1-chloropropane and 2-chloropropane as mono-chlorinated products. Four constitutionally isomeric dichlorinated products are possible, and five constitutional isomers exist for the trichlorinated propanes. Can you write structural formulas for the four dichlorinated isomers?
$CH_3CH_2CH_3 + 2Cl_2 \rightarrow \text{Four} \; C_3H_6Cl_2 \; \text{isomers} + 2 HCl$
The halogenation of propane discloses an interesting feature of these reactions. All the hydrogens in a complex alkane do not exhibit equal reactivity. For example, propane has eight hydrogens, six of them being structurally equivalent primary, and the other two being secondary. If all these hydrogen atoms were equally reactive, halogenation should give a 3:1 ratio of 1-halopropane to 2-halopropane mono-halogenated products, reflecting the primary/secondary numbers. This is not what we observe. Light-induced gas phase chlorination at 25 ºC gives 45% 1-chloropropane and 55% 2-chloropropane.
CH3-CH2-CH3 + Cl2 → 45% CH3-CH2-CH2Cl + 55% CH3-CHCl-CH3
The results of bromination ( light-induced at 25 ºC ) are even more suprising, with 2-bromopropane accounting for 97% of the mono-bromo product.
CH3-CH2-CH3 + Br2 → 3% CH3-CH2-CH2Br + 97% CH3-CHBr-CH3
These results suggest strongly that 2º-hydrogens are inherently more reactive than 1º-hydrogens, by a factor of about 3:1. Further experiments showed that 3º-hydrogens are even more reactive toward halogen atoms. Thus, light-induced chlorination of 2-methylpropane gave predominantly (65%) 2-chloro-2-methylpropane, the substitution product of the sole 3º-hydrogen, despite the presence of nine 1º-hydrogens in the molecule.
(CH3)3CH + Cl2 → 65% (CH3)3CCl + 35% (CH3)2CHCH2Cl
Contributors
Jim Clark (Chemguide.co.uk)
13.09: Application- The Ozone Layer and CFCs
The high reactivity of free radicals and the multiplicative nature of radical chain reactions can be useful in the synthesis of materials such as polyethylene plastic - but these same factors can also result in dangerous consequences. You are probably aware of the danger posed to the earth's protective stratospheric ozone layer by the use of chlorofluorocarbons (CFCs) as refrigerants and propellants in aerosol spray cans. Freon-11, or CFCl3, is a typical CFC that was widely used until fairly recently. It can take months or years for a CFC molecule to drift up into the stratosphere from the surface of the earth, and of course the concentration of CFCs at this altitude is very low. Ozone, on the other hand, is continually being formed in the stratosphere. Why all the concern, then, about destruction of the ozone layer - how could such a small amount of CFCs possibly do significant damage? The problem lies in the fact that the process by which ozone is destroyed is a chain reaction, so that a single CFC molecule can initiate the destruction of many ozone molecules before a chain termination event occurs.
Although there are several different processes by which the ozone destruction process might occur, the most important is believed to be the chain reaction shown below.
To address the problem of ozone destruction, scientists are developing new organohalogen refrigerant compounds that are less stable than the older CFCs like Freon-11, in the hope that the new compounds will break down in the lower atmosphere before they reach an altitude where they can harm the ozone layer. Most of the new compounds contain carbon-hydrogen bonds, which are subject to homolytic cleavage initiated by hydroxide radicals present in the lower atmosphere.
This degradation occurs before the refrigerant molecules have a chance to drift up to the stratosphere where the ozone plays its important protective role. The degradation products are quite unstable and quickly degrade further, by a variety of mechanisms, into relatively harmless by-products. The hydroxide radical is sometimes referred to as an atmospheric 'detergent' due to its ability to degrade refrigerants and other volatile organic pollutants which have escaped into the atmosphere.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/13%3A_Radical_Reactions/13.06%3A_Chlorination_versus_Bromination.txt |
When halogens are in the presence of unsaturated molecules such as alkenes, the expected reaction is addition to the double bond carbons resulting in a vicinal dihalide (halogens on adjacent carbons). However, when the halogen concentration is low enough, alkenes containing allylic hydrogens undergo substitution at the allylic position rather than addition at the double bond. The product is an allylic halide (halogen on carbon next to double bond carbons), which is acquired through a radical chain mechanism.
Why Substitution of Allylic Hydrogens?
As the table below shows, the dissociation energy for the allylic C-H bond is lower than the dissociation energies for the C-H bonds at the vinylic and alkylic positions. This is because the radical formed when the allylic hydrogen is removed is resonance-stabilized. Hence, given that the halogen concentration is low, substitution at the allylic position is favored over competing reactions. However, when the halogen concentration is high, addition at the double bond is favored because a polar reaction outcompetes the radical chain reaction.
Radical Allylic Bromination (Wohl-Ziegler Reaction)
Preparation of Bromine (low concentration)
NBS (N-bromosuccinimide) is the most commonly used reagent to produce low concentrations of bromine. When suspended in tetrachloride (CCl4), NBS reacts with trace amounts of HBr to produce a low enough concentration of bromine to facilitate the allylic bromination reaction.
Allylic Bromination Mechanism
Step 1: Initiation
Once the pre-initiation step involving NBS produces small quantities of Br2, the bromine molecules are homolytically cleaved by light to produce bromine radicals.
Step 2: Propagation
One bromine radical produced by homolytic cleavage in the initiation step removes an allylic hydrogen of the alkene molecule. A radical intermediate is generated, which is stabilized by resonance. The stability provided by delocalization of the radical in the alkene intermediate is the reason that substitution at the allylic position is favored over competing reactions such as addition at the double bond.
The intermediate radical then reacts with a Br2 molecule to generate the allylic bromide product and regenerate the bromine radical, which continues the radical chain mechanism. If the alkene reactant is asymmetric, two distinct product isomers are formed.
Step 3: Termination
The radical chain mechanism of allylic bromination can be terminated by any of the possible steps shown below.
Radical Allylic Chlorination
Like bromination, chlorination at the allylic position of an alkene is achieved when low concentrations of Cl2 are present. The reaction is run at high temperatures to achieve the desired results.
Industrial Uses
Allylic chlorination has important practical applications in industry. Since chlorine is inexpensive, allylic chlorinations of alkenes have been used in the industrial production of valuable products. For example, 3-chloropropene, which is necessary for the synthesis of products such as epoxy resin, is acquired through radical allylic chlorination (shown below).
Problems (Answers are attached as a file)
1. Cyclooctene undergoes radical allylic bromination. Write out the complete mechanism including reactants, intermediates and products.
2. Predict the two products of the allylic chlorination reaction of 1-heptene.
3. What conditions are required for allylic halogenation to occur? Why does this reaction outcompete other possible reactions such as addition when these conditions are met?
4. Predict the product of the allylic bromination reaction of 2-benzylheptane. (Hint: How are benzylic hydrogens similar to allylic hydrogens?)
5. The reactant 5-isopropyl-1-hexene generates the products 3-bromo-5-isopropyl-1-hexene and 1-bromo-5-isopropyl-2-hexene. What reagents were used in this reaction?
13.11: Application- Oxidation of Unsaturated Lipids
Fats and oils that are in contact with moist air at room temperature eventually undergo oxidation and hydrolysis reactions that cause them to turn rancid, acquiring a characteristic disagreeable odor. One cause of the odor is the release of volatile fatty acids by hydrolysis of the ester bonds. Butter, for example, releases foul-smelling butyric, caprylic, and capric acids. Microorganisms present in the air furnish lipases that catalyze this process. Hydrolytic rancidity can easily be prevented by covering the fat or oil and keeping it in a refrigerator.
Another cause of volatile, odorous compounds is the oxidation of the unsaturated fatty acid components, particularly the readily oxidized structural unit
in polyunsaturated fatty acids, such as linoleic and linolenic acids. One particularly offensive product, formed by the oxidative cleavage of both double bonds in this unit, is a compound called malonaldehyde.
Rancidity is a major concern of the food industry, which is why food chemists are always seeking new and better antioxidants, substances added in very small amounts (0.001%–0.01%) to prevent oxidation and thus suppress rancidity. Antioxidants are compounds whose affinity for oxygen is greater than that of the lipids in the food; thus they function by preferentially depleting the supply of oxygen absorbed into the product. Because vitamin E has antioxidant properties, it helps reduce damage to lipids in the body, particularly to unsaturated fatty acids found in cell membrane lipids.
• Anonymous | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/13%3A_Radical_Reactions/13.10%3A_Radical_Halogenation_at_an_Allylic_Carbon.txt |
Protons and other electrophiles are not the only reactive species that initiate addition reactions to carbon-carbon double bonds. Curiously, this first became evident as a result of conflicting reports concerning the regioselectivity of HBr additions. As noted earlier, the acid-induced addition of HBr to 1-butene gave predominantly 2-bromobutane, the Markovnikov Rule product. However, in some early experiments in which peroxide contaminated reactants were used, 1-bromobutane was the chief product. Further study showed that an alternative radical chain-reaction, initiated by peroxides, was responsible for the anti-Markovnikov product. This is shown by the following equations.
The weak O–O bond of a peroxide initiator is broken homolytically by thermal or hight energy. The resulting alkoxy radical then abstracts a hydrogen atom from HBr in a strongly exothermic reaction. Once a bromine atom is formed it adds to the π-bond of the alkene in the first step of a chain reaction. This addition is regioselective, giving the more stable carbon radical as an intermediate. The second step is carbon radical abstraction of another hydrogen from HBr, generating the anti-Markovnikov alkyl bromide and a new bromine atom. Each of the steps in this chain reaction is exothermic, so once started the process continues until radicals are lost to termination events.
This free radical chain addition competes very favorably with the slower ionic addition of HBr described earlier, especially in non-polar solvents. It is important to note, however, that HBr is unique in this respect. The radical addition process is unfavorable for HCl and HI because one of the chain steps becomes endothermic (the second for HCl & the first for HI).
Other radical addition reactions to alkenes have been observed, one example being the peroxide induced addition of carbon tetrachloride shown in the following equation
RCH=CH2 + CCl4 (peroxide initiator) > RCHClCH2CCl3
The best known and most important use of free radical addition to alkenes is probably polymerization. Since the addition of carbon radicals to double bonds is energetically favorable, concentrated solutions of alkenes are prone to radical-initiated polymerization, as illustrated for propene by the following equation. The blue colored R-group represents an initiating radical species or a growing polymer chain; the propene monomers are colored maroon. The addition always occurs so that the more stable radical intermediate is formed.
RCH2(CH3)CH· + CH3CH=CH2 >RCH2(CH3)CH-CH2(CH3)CH· + CH3CH=CH2 >RCH2(CH3)CHCH2(CH3)CH-CH2(CH3)CH· > etc.
Anti-Markovnikov rule describes the regiochemistry where the substituent is bonded to a less substituted carbon, rather than the more substitued carbon. This process is quite unusual, as carboncations which are commonly formed during alkene, or alkyne reactions tend to favor the more substitued carbon. This is because substituted carbocation allow more hyperconjugation and indution to happen, making the carbocation more stable.
Introduction
This process was first explained by Morris Selig Karasch in his paper: 'The Addition of Hydrogen Bromide to Allyl Bromide' in 1933.1 Examples of Anti-Markovnikov includes Hydroboration-Oxidation and Radical Addition of HBr. A free radical is any chemical substance with unpaired electron. The more substituents the carbon is connected to, the more substituted is that carbon. For example: Tertiary carbon (most substituted), Secondary carbon (medium substituted), primary carbon (least substituted)
Anti-Markovnikov Radical Addition of Haloalkane can ONLY happen to HBr and there MUST be presence of Hydrogen Peroxide (H2O2). Hydrogen Peroxide is essential for this process, as it is the chemical which starts off the chain reaction in the initiation step. HI and HCl cannot be used in radical reactions, because in their radical reaction one of the radical reaction steps: Initiation is Endothermic, as recalled from Chem 118A, this means the reaction is unfavorable. To demonstrate the anti-Markovnikov regiochemistry, I will use 2-Methylprop-1-ene as an example below:
Initiation Steps
Hydrogen Peroxide is an unstable molecule, if we heat it, or shine it with sunlight, two free radicals of OH will be formed. These OH radicals will go on and attack HBr, which will take the Hydrogen and create a Bromine radical. Hydrogen radical do not form as they tend to be extremely unstable with only one electron, thus bromine radical which is more stable will be readily formed.
Propagation Steps
The Bromine Radical will go on and attack the LESS SUBSTITUTED carbon of the alkene. This is because after the bromine radical attacked the alkene a carbon radical will be formed. A carbon radical is more stable when it is at a more substituted carbon due to induction and hyperconjugation. Thus, the radical will be formed at the more substituted carbon, while the bromine is bonded to the less substituted carbon. After a carbon radical is formed, it will go on and attack the hydrogen of a HBr, which a bromine radical will be formed again.
Termination Steps
There are also Termination Steps, but we do not concern about the termination steps as they are just the radicals combining to create waste products. For example two bromine radical combined to give bromine. This radical addition of bromine to alkene by radical addition reaction will go on until all the alkene turns into bromoalkane, and this process will take some time to finish.
Problems
Please give the product(s) of the reactions below:
1. CH3-C(CH3)=CH-CH3 + HBr + H2O2 ==> ?
2. CH3-C(CH3)=CH-CH3 + HI + H2O2 ==> ?
3. CH3-C(CH3)=CH-CH3 + HCl + H2O2 ==> ?
4. CH3-CH=CH-CH3 + HBr + H2O2 ==> ?
5. CH3-C(CH3)=CH-CH3 + HBr ==> ?
Answers
1. CH3-CH(CH3)-CHBr-CH3 (Anti-Markovnikov)
2. CH3-C(CH3)I-CH2-CH3 (Markovnikov)
3. CH3-C(CH3)Cl-CH2-CH3 (Markovnikov)
4. CH3-CHBr-CH-CH3 or CH3-CH-CHBr-CH3 (Both molecules are the same)
5. CH3-C(CH3)Br-CH2-CH3 (Markovnikov) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/13%3A_Radical_Reactions/13.13%3A_Radical_Addition_Reactions_to_Double_Bonds.txt |
All the monomers from which addition polymers are made are alkenes or functionally substituted alkenes. The most common and thermodynamically favored chemical transformations of alkenes are addition reactions. Many of these addition reactions are known to proceed in a stepwise fashion by way of reactive intermediates, and this is the mechanism followed by most polymerizations. A general diagram illustrating this assembly of linear macromolecules, which supports the name chain growth polymers, is presented here. Since a pi-bond in the monomer is converted to a sigma-bond in the polymer, the polymerization reaction is usually exothermic by 8 to 20 kcal/mol. Indeed, cases of explosively uncontrolled polymerizations have been reported.
It is useful to distinguish four polymerization procedures fitting this general description.
• Radical Polymerization The initiator is a radical, and the propagating site of reactivity (*) is a carbon radical.
• Cationic Polymerization The initiator is an acid, and the propagating site of reactivity (*) is a carbocation.
• Anionic Polymerization The initiator is a nucleophile, and the propagating site of reactivity (*) is a carbanion.
• Coordination Catalytic Polymerization The initiator is a transition metal complex, and the propagating site of reactivity (*) is a terminal catalytic complex.
Radical Chain-Growth Polymerization
Virtually all of the monomers described above are subject to radical polymerization. Since this can be initiated by traces of oxygen or other minor impurities, pure samples of these compounds are often "stabilized" by small amounts of radical inhibitors to avoid unwanted reaction. When radical polymerization is desired, it must be started by using a radical initiator, such as a peroxide or certain azo compounds. The formulas of some common initiators, and equations showing the formation of radical species from these initiators are presented below.
By using small amounts of initiators, a wide variety of monomers can be polymerized. One example of this radical polymerization is the conversion of styrene to polystyrene, shown in the following diagram. The first two equations illustrate the initiation process, and the last two equations are examples of chain propagation. Each monomer unit adds to the growing chain in a manner that generates the most stable radical. Since carbon radicals are stabilized by substituents of many kinds, the preference for head-to-tail regioselectivity in most addition polymerizations is understandable. Because radicals are tolerant of many functional groups and solvents (including water), radical polymerizations are widely used in the chemical industry.
In principle, once started a radical polymerization might be expected to continue unchecked, producing a few extremely long chain polymers. In practice, larger numbers of moderately sized chains are formed, indicating that chain-terminating reactions must be taking place. The most common termination processes are Radical Combination and Disproportionation. These reactions are illustrated by the following equations. The growing polymer chains are colored blue and red, and the hydrogen atom transferred in disproportionation is colored green. Note that in both types of termination two reactive radical sites are removed by simultaneous conversion to stable product(s). Since the concentration of radical species in a polymerization reaction is small relative to other reactants (e.g. monomers, solvents and terminated chains), the rate at which these radical-radical termination reactions occurs is very small, and most growing chains achieve moderate length before termination.
The relative importance of these terminations varies with the nature of the monomer undergoing polymerization. For acrylonitrile and styrene combination is the major process. However, methyl methacrylate and vinyl acetate are terminated chiefly by disproportionation.
Another reaction that diverts radical chain-growth polymerizations from producing linear macromolecules is called chain transfer. As the name implies, this reaction moves a carbon radical from one location to another by an intermolecular or intramolecular hydrogen atom transfer (colored green). These possibilities are demonstrated by the following equations
Chain transfer reactions are especially prevalent in the high pressure radical polymerization of ethylene, which is the method used to make LDPE (low density polyethylene). The 1º-radical at the end of a growing chain is converted to a more stable 2º-radical by hydrogen atom transfer. Further polymerization at the new radical site generates a side chain radical, and this may in turn lead to creation of other side chains by chain transfer reactions. As a result, the morphology of LDPE is an amorphous network of highly branched macromolecules. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/13%3A_Radical_Reactions/13.14%3A_Polymers_and_Polymerization.txt |
Allylic Carbocation
Conjugation occurs when p orbital on three or more adjacent atoms can overlap Conjugation tends to stabilize molecules
Allylic carbocations are a common conjugated system.
The positive charge of a carbocation is contained in a P orbital of a sp2 hybridized carbon. This allows for overlap with double bonds. The positive charge is more stable because it is spread over 2 carbons.
Molecular Orbitals of an Allylic Carbocation
The stability of the carbocation of propene is due to a conjugated π electron system. A "double bond" doesn't really exist. Instead, it is a group of 3 adjacent, overlapping, non-hybridized p orbitals we call a conjugated π electron system. You can clearly see the interactions between all three of the p orbitals from the three carbons resulting in a really stable cation. It all comes down to where the location of the electron-deficient carbon is.
Molecular orbital descriptions can explain allylic stability in yet another way using 2-propenyl. Fig.6
Fig.6 Shows the 3 possible Molecular orbitals of 2-propenyl
If we just take the π molecular orbital and not any of the s, we get three of them. π1 is bonding with no nodes, π2 is nonbonding (In other words, the same energy as a regular p-orbital) with a node, and π3 is antibonding with 2 nodes (none of the orbitals are interacting). The first two electrons will go into the π1 molecular orbital, regardless of whether it is a cation, radical, or anion. If it is a radical or anion, the next electron goes into the π2 molecular orbital. The last anion electron goes into the nonbonding orbital also. So no matter what kind of carbon center exists, no electron will ever go into the antibonding orbital.
The Bonding orbitals are the lowest energy orbitals and are favorable, which is why they are filled first. Even though the nonbonding orbitals can be filled, the overall energy of the system is still lower and more stable due to the filled bonding molecular orbitals.
This figure also shows that π2 is the only molecular orbital where the electrion differs, and it is also where a single node passes through the middle. Because of this, the charges of the molecule are mainly on the two terminal carbons and not the middle carbon.
This molecular orbital description can also illustrate the stability of allylic carbon centers in figure 7.
Fig.7 diagram showing how the electrons fill based on the Aufbau principle.
The π bonding orbital is lower in energy than the nonbonding p orbital. Since every carbon center shown has two electrons in the lower energy, bonding π orbitals, the energy of each system is lowered overall (and thus more stable), regardless of cation, radical, or anion.
1,3-Dienes
Conjugated double bonds are separated by a single bond. 1,3-dienes are an excellent example of a conjugated system. Each carbon in 1,3 dienes are sp2 hybridized and therefore have one p orbital. The four p orbitals in 1,3-butadiene overlap to form a conjugated system.
Conjugated vs. Nonconjugated Dienes
Conjugated dienes are two double bonds separated by a single bond
Nonconjugated (Isolated) Dienes are two double bonds are separated by more than one single bond.
When using electrostatic potential maps, it is observed that the pi electron density overlap is closer together and delocalized in conjugated dienes, while in non conjugated dienes the pi electron density is located differently across the molecule. Since having more electron density delocalized makes the molecule more stable conjugated dienes are more stable than non conjugated
For example in 1,3-butadiene the carbons with the single bond are sp2 hybridized unlike in nonconjugated dienes where the carbons with single bonds are sp3 hybridized. This difference in hybridization shows that the conjugated dienes have more 's' character and draw in more of the pi electrons, thus making the single bond stronger and shorter than an ordinary alkane C-C bond (1.54Å).
Stability of Conjugated Dienes
Conjugated dienes are more stable than non conjugated dienes (both isolated and cumulated) due to factors such as delocalization of charge through resonance and hybridization energy. This can also explain why allylic radicals are much more stable than secondary or even tertiary carbocations. This is all due to the positioning of the pi orbitals and ability for overlap to occur to strengthen the single bond between the two double bonds.
The resonance structure shown below gives a good understanding of how the charge is delocalized across the four carbons in this conjugated diene. This delocalization of charges stablizes the conjugated diene:
Along with resonance, hybridization energy effect the stability of the compound. For example in 1,3-butadiene the carbons with the single bond are sp2 hybridized unlike in nonconjugated dienes where the carbons with single bonds are sp3 hybridized. This difference in hybridization shows that the conjugated dienes have more 's' character and draw in more of the pi electrons, thus making the single bond stronger and shorter than an ordinary alkane C-C bond (1.54Å).
Molecular Orbitals of 1,3 Dienes
A molecular orbital model for 1,3-butadiene is shown below. Note that the lobes of the four p-orbital components in each pi-orbital are colored differently and carry a plus or minus sign. This distinction refers to different phases, defined by the mathematical wave equations for such orbitals. Regions in which adjacent orbital lobes undergo a phase change are called nodes. Orbital electron density is zero in such regions. Thus a single p-orbital has a node at the nucleus, and all the pi-orbitals shown here have a nodal plane that is defined by the atoms of the diene. This is the only nodal surface in the lowest energy pi-orbital, π1. Higher energy pi-orbitals have an increasing number of nodes.
' | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/14%3A_Conjugation_Resonance_and_Dienes/14.01%3A_Conjugation.txt |
Conjugation occurs when p orbital on three or more adjacent atoms can overlap Conjugation tends to stabilize molecules. Allylic carbocations are a common conjugated system.
The positive charge of a carbocation is contained in a P orbital of a sp2 hybrizied carbon. This allows for overlap with double bonds. The positive charge is more stable because it is spread over 2 carbons.
The true structure of the conjugated allyl carbocation is a hybrid of of the two resonance structure so the positive charge is delocalized over the two terminal carbons. This delocalization stablizes the allyl carbocation making it more stable than a normal primary carbocation.
14.03: Common Examples of Resonance
1) Three atoms in a A=B-C where C is and atom with a p orbital.
There are two major resonance possible. The two structure differ in the lcoation of the double bond. The anion, cation, or radical is stabilized by declocaliztion.
Examples
Carboxylate Anion
Allylic carbocation
Allyic radical
2) Conjugated double bonds
The benzene ring has two resonance structures which can be drawn by moving elections in a cyclic manner.
Conjugated double bonds contain multiple resonance structures.
3) Cations adjacent of an atom with lone pair electons.
Because heteroatoms such as oxygen and nitrogen are more electronegative than carbon, you might expect that they would by definition be electron withdrawing groups that destabilize carbocations. In fact, the opposite is often true: if the oxygen or nitrogen atom is in the correct position, the overall effect is carbocation stabilization. This is due to the fact that although these heteroatoms are electron withdrawing groups by induction, they are electron donating groups by resonance, and it is this resonance effect which is more powerful. (We previously encountered this same idea when considering the relative acidity and basicity of phenols and aromatic amines in section 7.4). Consider the two pairs of carbocation species below:
In the more stable carbocations, the heteroatom acts as an electron donating group by resonance: in effect, the lone pair on the heteroatom is available to delocalize the positive charge. In the less stable carbocations the positively-charged carbon is more than one bond away from the heteroatom, and thus no resonance effects are possible. In fact, in these carbocation species the heteroatoms actually destabilize the positive charge, because they are electron withdrawing by induction
4) Double bonds with one atom more electronegative that the other
Multiple resonance structures are possible which causes a charge separation in the molecule.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
14.04: The Resonance Hybrid
Resonance contributors for the carboxylate group
The convention of drawing two or more resonance contributors to approximate a single structure may seem a bit clumsy to you at this point, but as you gain experience you will see that the practice is actually very useful when discussing the manner in which many functional groups react. Let’s next consider the carboxylate ion (the conjugate base of a carboxylic acid). As our example, we will use formate, the simplest possible carboxylate-containing molecule. The conjugate acid of formate is formic acid, which causes the painful sting you felt if you have ever been bitten by an ant.
Usually, you will see carboxylate groups drawn with one carbon-oxygen double bond and one carbon-oxygen single bond, with a negative formal charge located on the single-bonded oxygen. In actuality, however, the two carbon-oxygen bonds are the same length, and although there is indeed an overall negative formal charge on the group, it is shared equally between the two oxygens. Therefore, the carboxylate can be more accurately depicted by a pair of resonance contributors. Alternatively, a single structure can be used, with a dashed line depicting the resonance-delocalized π bond and the negative charge located in between the two oxygens.
Let’s see if we can correlate these drawing conventions to a valence bond theory picture of the bonding in a carboxylate group. We know that the carbon must be sp2-hybridized, (the bond angles are close to 120˚, and the molecule is planar), and we will treat both oxygens as being sp2-hybridized as well. Both carbon-oxygen sbonds, then, are formed from the overlap of carbon sp2 orbitals and oxygen sp2 orbitals.
In addition, the carbon and both oxygens each have an unhybridized 2pz orbital situated perpendicular to the plane of the sigma bonds. These three 2pz orbitals are parallel to each other, and can overlap in a side-by-side fashion to form a delocalized pi bond.
Resonance contributor A shows oxygen #1 sharing a pair of electrons with carbon in a pi bond, and oxygen #2 holding a lone pair of electrons in its 2pz orbital. Resonance contributor B, on the other hand, shows oxygen #2 participating in the pi bond with carbon, and oxygen #1 holding a lone pair in its 2pz orbital. Overall, the situation is one of three parallel, overlapping 2pz orbitals sharing four delocalized pi electrons. Because there is one more electron than there are 2pz orbitals, the system has an overall charge of –1. This is the kind of 3D picture that resonance contributors are used to approximate, and once you get some practice you should be able to quickly visualize overlapping 2pz orbitals and delocalized pi electrons whenever you see resonance structures being used. In this text, carboxylate groups will usually be drawn showing only one resonance contributor for the sake of simplicity, but you should always keep in mind that the two C-O bonds are equal, and that the negative charge is delocalized to both oxygens.
Major vs minor resonance contributors
Different resonance contributors do not always make the same contribution to the overall structure of the hybrid. If one resonance structure is more stable (lower in energy) than another, then the first will come closer to depicting the ‘real’ (hybrid) structure than the second. In the case of carboxylates, contributors A and B are equivalent to each other in terms of their relative stability and therefore their relative contribution to the hybrid structure. However, there is a third resonance contributor ‘C’ that we have not considered yet, in which the carbon bears a positive formal charge and both oxygens are single-bonded and bear negative charges.
Structure C is relatively less stable, and therefore makes a less important contribution to the overall structure of the hybrid relative to A and B.
How do we know that structure C is the less stable, and thus the ‘minor’ contributor? There are four basic rules which you need to learn in order to evaluate the relative stability of different resonance contributors. We will number them 5-8 so that they may be added to in the 'rules for resonance' list from section 2.2C.
5) The carbon in contributor C does not have an octet – in general, resonance contributors in which a carbon does not fulfill the octet rule are relatively less important.
6) In structure C, a separation of charge has been introduced that is not present in A or B. In general, resonance contributors in which there is a greater separation of charge are relatively less important.
7) In structure C, there are no double bonds. In general, a resonance structure with a lower number of multiple bonds is relatively less important.
There is one more important rule that does not apply to this particular example, but we will list it here in the interest of completeness (we will soon see an actual example).
8) The resonance contributor in which a negative formal charge is located on a more electronegative atom (such as oxygen or nitrogen) is more stable than one in which the negative charge is located on a less electronegative atom (such as carbon).
When discussing other examples of resonance contributors, we will often see cases where one form is less stable – but these minor resonance contributors can still be very relevant in explaining properties of structure or reactivity, and should not be disregarded.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/14%3A_Conjugation_Resonance_and_Dienes/14.02%3A_Resonance_and_Allylic_Carbocations.txt |
Let’s see if we can correlate these drawing conventions to a valence bond theory picture of the bonding in a carboxylate group. We know that the carbon must be sp2-hybridized, (the bond angles are close to 120˚, and the molecule is planar), and we will treat both oxygens as being sp2-hybridized as well. Both carbon-oxygen sbonds, then, are formed from the overlap of carbon sp2 orbitals and oxygen sp2 orbitals.
In addition, the carbon and both oxygens each have an unhybridized 2pz orbital situated perpendicular to the plane of the sigma bonds. These three 2pz orbitals are parallel to each other, and can overlap in a side-by-side fashion to form a delocalized pi bond.
Resonance contributor A shows oxygen #1 sharing a pair of electrons with carbon in a pi bond, and oxygen #2 holding a lone pair of electrons in its 2pz orbital. Resonance contributor B, on the other hand, shows oxygen #2 participating in the pi bond with carbon, and oxygen #1 holding a lone pair in its 2pz orbital. Overall, the situation is one of three parallel, overlapping 2pz orbitals sharing four delocalized pi electrons. Because there is one more electron than there are 2pz orbitals, the system has an overall charge of –1. This is the kind of 3D picture that resonance contributors are used to approximate, and once you get some practice you should be able to quickly visualize overlapping 2pz orbitals and delocalized pi electrons whenever you see resonance structures being used. In this text, carboxylate groups will usually be drawn showing only one resonance contributor for the sake of simplicity, but you should always keep in mind that the two C-O bonds are equal, and that the negative charge is delocalized to both oxygens.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
14.06: Conjugated Dienes
When considering compounds having two or more double bonds in a molecule, it is useful to identify three distinct ways in which these functions may be oriented with respect to each other. First, the double bonds may be separated by one or more sp3-hybridized carbon atoms, as in 1,5-hexadiene. In this circumstance each double bond behaves independently of the other, and we refer to them as isolated. A second relationship has the double bonds connected to each other by a single bond, as in 1,3-hexadiene, and we refer to this arrangement as conjugated. Finally, two double bonds might share a carbon atom, as in 1,2-hexadiene. The central carbon atom in such a system is sp-hybridized, and we call such double bonds cumulated.
Dienes can adopt two possible conformations through rotation about the single bond joining the two double bonds: the s-cis and the s-trans conformations.
The energy barrier to isomerization is normally low, and the s-trans conformer is often more stable than the s-cis conformer.
Naming Dienes
First identify the longest chain containing both carbons with double bonds in the compound. Then give the lowest possible number for the location of the carbons with double bonds and any other functional groups present (remember when naming alkenes that some groups take priority such as alcohols). Do not forget stereochemistry or any other orientation of the double bond such as (E/Z,cis or trans).
Examples:
Different conformations of Conjugated Dienes
There are two different conformations of conjugated dienes which are s-cis and s-trans conformations. s-cis is when the double bonds are cis in reference to the single bond and s-trans is when the two double bonds are trans in reference to the single bond. The cis conformation is less stable due to the steric interation of hydrogens on carbon. One important use of the cis conformation of a conjugated diene is that it is used diels-alder cycloaddition reactions. Even though the trans conformation is more stable the cis conformation is used because of the molecule's ability to interconvert and rotate about the single bond.
14.07: Interesting Dienes and Polyenes
Conjugated dienes (alkenes with two double bonds and a single bond in between) can be polymerized to form important compounds like rubber. This takes place, in different forms, both in nature and in the laboratory. Interactions between double bonds on multiple chains leads to cross-linkage which creates elasticity within the compound.
Polymerization of 1,3-Butadiene
For rubber compounds to be synthesized, 1,3-butadiene must be polymerized. Below is a simple illustration of how this compound is formed into a chain. The 1,4 polymerization is much more useful to polymerization reactions.
Above, the green structures represent the base units of the polymers that are synthesized and the red represents the bonds between these units which form these polymers. Whether the 1,3 product or the 1,4 product is formed depends on whether the reaction is thermally or kinetically controlled.
Synthetic Rubber
The most important synthetic rubber is Neoprene which is produced by the polymerization of 2-chloro-1,3-butadiene.
In this illustration, the dashed lines represent repetition of the same base units, so both the products and reactants are polymers. The reaction proceeds with a mechanism similar to the Friedel-Crafts mechanism. Cross-linkage between the chlorine atom of one chain and the double bond of another contributes to the overall elasticity of neoprene. This cross-linkage occurs as the chains lie next to each other at random angles, and the attractions between double bonds prevent them from sliding back and forth.
Colored molecules
The counjugated double bonds in beta-carotene produce the orange color in carrots. The conjugated double bons in lycopene produce the red color in tomatoes.
ß carotene
lycopene
Problem
Draw out the mechanism for the natural synthesis of rubber from 3-methyl-3-butenyl pyrophosphate and 2-methyl-1,3-butadiene. Show the movement of electrons with arrows.
14.08: The CarbonCarbon Bond Length in 13-Butadiene
The conjugated diene has 2 double bonds with one single C-C bond between them. This structure offers stability because the two pi bonds can transfer electrons through the two carbons that are sp2 hybridized with a single bond which results in electron delocalization. Extended P orbital sharing makes this diene more stable than the isolated dienes. The more stable molecule also has lower energy and a shorter bond length.
Contributors
• Natasha Singh
14.09: Stability of Conjugated Dienes
Conjugated dienes are more stable than non conjugated dienes (both isolated and cumulated) due to factors such as delocalization of charge through resonance and hybridization energy. This can also explain why allylic radicals are much more stable than secondary or even tertiary carbocations. This is all due to the positioning of the pi orbitals and ability for overlap to occur to strengthen the single bond between the two double bonds.
The resonance structure shown below gives a good understanding of how the charge is delocalized across the four carbons in this conjugated diene. This delocalization of charges stablizes the conjugated diene:
Along with resonance, hybridization energy effect the stability of the compound. For example in 1,3-butadiene the carbons with the single bond are sp2 hybridized unlike in nonconjugated dienes where the carbons with single bonds are sp3 hybridized. This difference in hybridization shows that the conjugated dienes have more 's' character and draw in more of the pi electrons, thus making the single bond stronger and shorter than an ordinary alkane C-C bond (1.54Å).
Another useful resource to consider are the heats of hydrogenation of different arrangements of double bonds. Since the higher the heat of hydrogenation the less stable the compound, it is shown below that conjugated dienes (~54 kcal) have a lower heat of hydrogenation than their isolated (~60 kcal) and cumulated diene (~70 kcal) counterparts.
Here is an energy diagram comparing different types of bonds with their heats of hydrogenation to show relative stability of each molecule:
The stabilization of dienes by conjugation is less dramatic than the aromatic stabilization of benzene. Nevertheless, similar resonance and molecular orbital descriptions of conjugation may be written.
• Shravan Rao | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/14%3A_Conjugation_Resonance_and_Dienes/14.05%3A_Electron_Delocalization_Hybridization_and_Geometry.txt |
Addition reactions of isolated dienes proceed more or less as expected from the behavior of simple alkenes. Thus, if one molar equivalent of 1,5-hexadiene is treated with one equivalent of bromine a mixture of 5,6-dibromo-1-hexene, 1,2,5,6-tetrabromohexane and unreacted diene is obtained, with the dibromo compound being the major product (about 50%)
fig
Similar reactions of conjugated dienes, on the other hand, often give unexpected products. The addition of bromine to 1,3-butadiene is an example. As shown below, a roughly 50:50 mixture of 3,4-dibromo-1-butene (the expected product) and 1,4-dibromo-2-butene (chiefly the E-isomer) is obtained. The latter compound is remarkable in that the remaining double bond is found in a location where there was no double bond in the reactant. This interesting relocation requires an explanation.
CH2=CH-CH=CH2 + Br2 BrCH2CHBr-CH=CH2 + BrCH2CH=CHCH2Br
3,4-dibromo-1-butene 1,4-dibromo-2-butene
The expected addition product from reactions of this kind is the result of 1,2-addition, i.e. bonding to the adjacent carbons of a double bond. The unexpected product comes from 1,4-addition, i.e. bonding at the terminal carbon atoms of a conjugated diene with a shift of the remaining double bond to the 2,3-location. (These numbers refer to the four carbons of the conjugated diene and are not IUPAC nomenclature numbers.) Product compositions are often temperature dependent: at 40 oC, 85% of the product mixture in the addition reaction above is the 1,4 product, whereas at 0 oC, only about 30% is the 1,4 product.
Bonding of an electrophilic atom or group to one of the end carbon atoms of a conjugated diene ( carbon #1 in the figure below) generates an allyl cation intermediate. Such cations are stabilized by charge delocalization, and it is this delocalization that accounts for the 1,4-addition product produced in such addition reactions. As shown in the diagram, the positive charge is distributed over carbons #2 and #4 so it is at these sites that the nucleophilic component bonds. Note that resonance stabilization of the allyl cation is greater than comparable stabilization of 1,3-butadiene, because charge is delocalized in the former, but created and separated in the latter.
An explanation for the temperature influence is shown in the following energy diagram for the addition of HBr to 1,3-butadiene. The initial step in which a proton bonds to carbon #1 is the rate determining step, as indicated by the large activation energy (light gray arrow). The second faster step is the product determining step, and there are two reaction paths (colored blue for 1,2-addition and magenta for 1,4-addition). The 1,2-addition has a smaller activation energy than 1,4-addition - it occurs faster than 1,4 addition, because the bromide nucleophile is closer to carbon #2 then to carbon #4. However, the 1,4-product is more stable than the 1,2-product. At low temperatures, the products are formed irreversibly and reflect the relative rates of the two competing reactions. This is termed kinetic control. At higher temperatures, equilibrium is established between the products, and the thermodynamically favored 1,4-product dominates.
When a conjugated diene is attacked by an electrophile, the resulting products are a mixture of 1,2 and 1,4 isomers. Kinetics and Thermodynamics control a reaction when there are two products under different reaction conditions. The Kinetic product (Product A) will be formed fast, and the Thermodynamic product (Product B) will be formed more slowly. Usually the first product formed is the more stable favored product, but in this case, the slower product formed is the more stable product; Product B.
Introduction
Like nonconjugated dienes, conjugated dienes are subject to attack by electrophiles. In fact, conjugated electrophiles experience relatively greater kinetic reactivity when reacted with electrophiles than nonconjugated dienes do. Upon electrophilic addition, the conjugated diene forms a mixture of two products—the kinetic product and the thermodynamic product—whose ratio is determined by the conditions of reaction. A reaction yielding more thermodynamic product is under thermodynamic control, and likewise, a reaction that yields more kinetic product is under kinetic control.
Conclusion
The reactivity of conjugated dienes (hydrocarbons that contain two double bonds) varies depending on the location of double bonds and temperature of the reaction.These reactions can produce both thermodynamic and kinetic products. Isolated double bonds provide dienes with less stability thermodynamically than conjugated dienes. However, they are more reactive kinetically in the presence of electrophiles and other reagents. This is a result of Markovnikov addition to one of the double bonds. A carbocation is formed after a double bond is opened. This carbocation has two resonance structures and addition can occur at either of the positive carbons.
Practice Problems
1. Write out the products of 1,2 addition and 1,4- addition of a) HBr and Br. b) DBr to 1,3-cyclo-hexadiene. What is unusual about the products of 1,2- and 1,4- addition of HX to unsubstituted cyclic 1,3-dienes?
2. Is the 1,2-addition product formed more rapidly at higher temperatures, even though it is the 1,4-addition product that predominates under these conditions?
3. Why is the 1,4-addition product the thermodynamically more stable product?
4. Out of the following radical cations which one is not a reasonable resonance structure?
5. Addition of 1 equivalent of Bromine to 2,4-hexadiene at 0 degrees C gives 4,5-dibromo-2-hexene plus an isomer. Which of the following is that isomer:
1. 5,5-dibromo-2-hexene
2. 2,5-dibromo-3-hexene
3. 2,2-dibromo-3-hexene
4. 2,3-dibromo-4-hexene
6. Which of the following will be the kinetically favored product from the depicted reaction?
7. Addition of HBr to 2,3-dimethyl-1,3-cyclohexadiene may occur in the absence or presence of peroxides. In each case two isomeric C8H13Br products are obtained. Which of the following is a common product from both reactions?
8. and 9.
8. The kinetically controlled product in the above reaction is:
1. 3-Chloro-1-Butene
2. 1-Chloro-2-Butene
9. For the reaction in question 8, which one is the result of 1,4-addition?
1. 3-Chloro-1-Butene
2. 1-Chloro-2-Butene
Answers to Problems
1. A) Same product for both modes of addition.
B) Both cis and trans isomers will form.
Addition of the HX to unsubstituted cycloalka-1,3-dienes in either 1,2- or 1,4- manner gives the same product becasuse of symmetry.
2. Yes. the Kinetic Product will still form faster but in this case there will be enough energy to form the thermodynamic product because the thermodynamic product is still more stable.
3. The 1,4- product is more thermodynamically stable because there are two alkyl groups on each side of the double bond. This form offers stability to the overall structure.
4. All of these isomers are viable.
5. D
6. C
7. D
8. A
9. B | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/14%3A_Conjugation_Resonance_and_Dienes/14.10%3A_Electrophilic_Addition-_12-_Versus_14-Addition.txt |
Kinetic and Thermodynamic Products
Kinetic products form the fastest. They usually occur at or below 0°C. This is also known as the 1,2-adduct because the substituents are added to the first and second carbons. Kinetic products contain a terminal double bond and the reaction is irreversible. Thermodynamic products form at higher temperatures, generally greater than 40 °C. These are known are the 1,4-adducts because they add to the first and fourth carbons. Thermodynamic products contain an internal double bond and the reaction is reversible. Also, when reactions are carried out, thermodynamic products are more stable than kinetic products because they are more substituted.
Figure 1: Conditions at 0 °C:
Final Products:
• 1,2 Addition (Kinetic controlled product)
• 1,4 Addition (Thermodynamic controlled product)
The reason for the two products is the difference in their activation energies. The reaction will go through to completion on the easiest path and in this case that is the Kinetic path. This path has a much lower Activation Energy which means that less is required to have the product formed. This product- Product A, is not the most stable form however. Over time, the amount of Product B- the more stable one, will increase and the Product A will decrease.
Figure 2: Potential Energy profiles
Understanding the Reaction Coordinate Diagram
1. Major product at low temperatures- 1, 2 Bromobutene: The main product at low temperatures is Kinetically controlled. This means that there is not enough energy to overcome the Activation Energy of the Thermodynamic product even though it is the more stable product.
2. 98% product: At a low temperature the amount of energy in the reaction is not enough to get a large amount of the product over into the Thermodynamic Isomer. One of the two factors that influences the outcome of the reaction is the rate of product formation which is the Kinetic control. The Proximity Effect makes it so that the when the H is taken from the Br by the double bond, the Br is left close to the C2 and, in order to take away the positive charge, the C2 bonds with it quickly forming 1,2-Bromobutene.
3. 2% product:The Thermodynamic Isomer is not favored at low temperatures because the Rate of Reaction is based on the Activation Energy. The Activation Energy of the Thermodynamic Isomer is higher than the Kinetic and therefore will happen a lot slower.
4. The Intermediate is low in energy due to charge delocalization: The charge is delocalized across the three Carbons so the positive charge is not only on one Carbon. The Br will attack at the 2 and 4 positions.
5. Kinetic Higher Energy product: The product formed Kinetically is less stable than the Thermodynamic product. The double bond in the Kinetic Product is only connected to one single bond which means that it has left of an Allylic effect on the overall structure.
6. Thermodynamic Lower Energy Product: The Thermodynamic product has two Alkyl groups bonded to the double bond. This has a stabilizing effect on the molecule.
7. Major Product at high temperatures- 1,4 bromobutene: This is the Thermodynamic Product that does not occur rapidly due to the Resonance Structure required to form the final product. If the reaction is left to go through equilibrium this product will predominate. It is not based on the physical conditions of the reaction and that is why it is the Thermodynamic product.
Table 1: Conjugated Dienese: Kinetics vs. Thermodynamic Conditions
Kinetic and Thermodynamic Product Ratios
To ensure the greatest possible yield of thermodynamic products, the reaction should be carried out at a temperature of 40°C or greater. This is known as thermodynamic control. At higher temperatures and longer reaction times, thermodynamic products are favored. On the contrary, at lower temperatures, one would tend to see a greater yield of kinetic products. These products are generally formed at or around 0°C. Carrying out reactions around these temperatures is known as kinetic control and kinetic products form before thermodynamic products.
Since the thermodynamic product contains an internal double bond, it is more stable than the kinetic product, and this is due to hyperconjugation with neighboring atoms. Additionally, a higher activation energy results in the thermodynamic product forming slower than the kinetic product. Therefore, a thermodynamically controlled reaction gives a more stable product and kinetically controlled reaction gives a less stable product.
Figure: The various temperature conditions that may instigate a change or speed up the reaction
The conjugated diene has 2 double bonds with one single C-C bond between them. This structure offers stability because the two pi bonds can transfer electrons through the two carbons that are sp2 hybridized with a single bond which results in electron delocalization. Extended P orbital sharing makes this diene more stable than the isolated dienes. The more stable molecule also has lower energy and a shorter bond length.
Figure 4:
The p electrons reach out to the electrophile and form a bond that in turn forms a Carbocation. The Markovnikov Addition states that the most stable carbocation is most likely to be formed with the charge going on the more substituted carbon. The difference between a conjugated diene and an alkene is that there is still a double bond left after the reaction has completed.
• The Kinetic product is formed more quickly because it has a more stable carbocation to begin with. Due to resonance forms, the most stable isomer is the one with the double bond in the center of the molecule. As the reaction completes under normal conditions, the more likely product is the one formed with the most stable carbocation.
• The Kinetic isomer product. While the carbocation is more stable, the final product is not. The double bond ends up on a primary carbon. It would have more stabilizing effect if it were in the middle which is why it tries to get there.
Contributors
• Natasha Singh
14.12: The DielsAlder Reaction
The unique character of conjugated dienes manifests itself dramatically in the Diels-Alder Cycloaddition Reaction. A cycloaddition reaction is the concerted bonding together of two independent pi-electron systems to form a new ring of atoms. When this occurs, two pi-bonds are converted to two sigma-bonds, the simplest example being the hypothetical combination of two ethene molecules to give cyclobutane. This does not occur under normal conditions, but the cycloaddition of 1,3-butadiene to cyanoethene (acrylonitrile) does, and this is an example of the Diels-Alder reaction. The following diagram illustrates two cycloadditions, and introduces several terms that are useful in discussing reactions of this kind.
In the hypothetical ethylene dimerization on the left, each reactant molecule has a pi-bond (colored orange) occupied by two electrons. The cycloaddition converts these pi-bonds into new sigma-bonds (colored green), and this transformation is then designated a [2+2] cycloaddition, to enumerate the reactant pi-electrons that change their bonding location.
The Diels-Alder reaction is an important and widely used method for making six-membered rings, as shown on the right. The reactants used in such reactions are a conjugated diene, simply referred to as the diene, and a double or triple bond coreactant called the dienophile, because it combines with (has an affinity for) the diene. The Diels-Alder cycloaddition is classified as a [4+2] process because the diene has four pi-electrons that shift position in the reaction and the dienophile has two.
The Diels-Alder reaction is a single step process, so the diene component must adopt an s-cisconformation in order for the end carbon atoms (#1 & #4) to bond simultaneously to the dienophile. For many acyclic dienes the s-trans conformer is more stable than the s-cis conformer (due to steric crowding of the end groups), but the two are generally in rapid equilibrium, permitting the use of all but the most hindered dienes as reactants in Diels-Alder reactions. In its usual form, the diene component is electron rich, and the best dienophiles are electron poor due to electron withdrawing substituents such as CN, C=O & NO2. The initial bonding interaction reflects this electron imbalance, with the two new sigma-bonds being formed simultaneously, but not necessarily at equal rates.
Mechanism
We end this chapter with a discussion of a type of reaction that is different from anything we have seen before. In the Diels-Alder cycloaddition reaction, a conjugated diene reacts with an alkene to form a ring structure.
In a Diels-Alder reaction, the alkene reacting partner is referred to as the dienophile. Essentially, this process involves overlap of the 2p orbitals on carbons 1 and 4 of the diene with 2p orbitals on the two sp2-hybridized carbons of the dienophile. Both of these new overlaps end up forming new sigma bonds, and a new pi bond is formed between carbon 2 and 3 of the diene.
One of the most important things to understand about this process is that it is concerted – all of the electron rearrangement takes place at once, with no carbocation intermediates.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/14%3A_Conjugation_Resonance_and_Dienes/14.11%3A_Kinetic_Versus_Thermodynamic_Products.txt |
The Diels-Alder reaction is enormously useful for synthetic organic chemists, not only because ring-forming reactions are useful in general but also because in many cases two new stereocenters are formed, and the reaction is inherently stereospecific. A cis dienophile will generate a ring with cis substitution, while a trans dienophile will generate a ring with trans substitution:
In order for a Diels-Alder reaction to occur, the diene molecule must adopt what is called the s-cis conformation:
The s-cis conformation is higher in energy than the s-trans conformation, due to steric hindrance. For some dienes, extreme steric hindrance causes the s-cis conformation to be highly strained, and for this reason such dienes do not readily undergo Diels-Alder reactions.
Cyclic dienes, on the other hand, are ‘locked’ in the s-cis conformation, and are especially reactive. The result of a Diels-Alder reaction involving a cyclic diene is a bicyclic structure:
Here, we see another element of stereopecificity: Diels-Alder reactions with cyclic dienes favor the formation of bicyclic structures in which substituents are in the endo position.
The endo position on a bicyclic structure refers to the position that is inside the concave shape of the larger (six-membered) ring. As you might predict, the exo position refers to the outside position.
The rate at which a Diels-Alder reaction takes place depends on electronic as well as steric factors. A particularly rapid Diels-Alder reaction takes place between cyclopentadiene and maleic anhydride.
We already know that cyclopentadiene is a good diene because of its inherent s-cis conformation. Maleic anhydride is also a very good dienophile, because the electron-withdrawing effect of the carbonyl groups causes the two alkene carbons to be electron-poor, and thus a good target for attack by the pi electrons in the diene.
In general, Diels-Alder reactions proceed fastest with electron-donating groups on the diene (eg. alkyl groups) and electron-withdrawing groups on the dienophile.
Alkynes can also serve as dienophiles in Diels-Alder reactions:
Below are just three examples of Diels-Alder reactions that have been reported in recent years:
The Diels-Alder reaction is just one example of a pericyclic reaction: this is a general term that refers to concerted rearrangements that proceed though cyclic transition states. Two well-studied intramolecular pericyclic reactions are known as the Cope rearrangement . . .
. . .and the Claisen rearrangement (when an oxygen is involved):
Notice that the both of these reactions require compounds in which two double bonds are separated by three single bonds.
Pericyclic reactions are rare in biological chemistry, but here is one example: the Claisen rearrangement catalyzed by chorismate mutase in the aromatic amino acid biosynthetic pathway.
The study of pericyclic reactions is an area of physical organic chemistry that blossomed in the mid-1960s, due mainly to the work of R.B. Woodward, Roald Hoffman, and Kenichi Fukui. The Woodward-Hoffman rules for pericyclic reactions (and a simplified version introduced by Fukui) use molecular orbital theory to explain why some pericyclic processes take place and others do not. A full discussion is beyond the scope of this text, but if you go on to study organic chemistry at the advanced undergraduate or graduate level you are sure to be introduced to this fascinating area of inquiry.
Stereochemistry of the Diels-Alder reaction
We noted earlier that addition reactions of alkenes often exhibited stereoselectivity, in that the reagent elements in some cases added syn and in other cases anti to the the plane of the double bond. Both reactants in the Diels-Alder reaction may demonstrate stereoisomerism, and when they do it is found that the relative configurations of the reactants are preserved in the product (the adduct). The following drawing illustrates this fact for the reaction of 1,3-butadiene with (E)-dicyanoethene. The trans relationship of the cyano groups in the dienophile is preserved in the six-membered ring of the adduct. Likewise, if the terminal carbons of the diene bear substituents, their relative configuration will be retained in the adduct. Using the earlier terminology, we could say that bonding to both the diene and the dienophile is syn. An alternative description, however, refers to the planar nature of both reactants and terms the bonding in each case to be suprafacial (i.e. to or from the same face of each plane). This stereospecificity also confirms the synchronous nature of the 1,4-bonding that takes place.
The essential characteristics of the Diels-Alder cycloaddition reaction may be summarized as follows:
(i) The reaction always creates a new six-membered ring. When intramolecular, another ring may also be formed.
(ii) The diene component must be able to assume a s-cis conformation.
(iii) Electron withdrawing groups on the dienophile facilitate reaction.
(iv) Electron donating groups on the diene facilitate reaction.
(v) Steric hindrance at the bonding sites may inhibit or prevent reaction.
(vi) The reaction is stereospecific with respect to substituent configuration in both the dienophile and the diene.
These features are illustrated by the following eight examples, one of which does not give a Diels-Alder cycloaddition.
There is no reaction in example D because this diene cannot adopt an s-cis orientation. In examples B, C, F, G & H at least one of the reactants is cyclic so that the product has more than one ring, but the newly formed ring is always six-membered. In example B the the same cyclic compound acts as both the diene colored blue) and the dienophile (colored red). The adduct has three rings, two of which are the five-membered rings present in the reactant, and the third is the new six-membered ring (shaded light yellow). Example C has an alkyne as a dienophile (colored red), so the adduct retains a double bond at that location. This double bond could still serve as a dienophile, but in the present case the diene is sufficiently hindered to retard a second cycloaddition. The quinone dienophile in reaction F has two dienophilic double bonds. However, the double bond with two methyl substituents is less reactive than the unsubstituted dienophile due in part to the electron donating properties of the methyl groups and in part to steric hindrance. The stereospecificity of the Diels-Alder reaction is demonstrated by examples A, E & H. In A & H the stereogenic centers lie on the dienophile, whereas in E these centers are on the diene. In all cases the configuration of the reactant is preserved in the adduct.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/14%3A_Conjugation_Resonance_and_Dienes/14.13%3A_Specific_Rules_Governing_the_DielsAlder_Reaction.txt |
Electromagnetic radiation such as visible light is commonly treated as a wave phenomenon, characterized by a wavelength or frequency. Wavelength is defined on the left below, as the distance between adjacent peaks (or troughs), and may be designated in meters, centimeters or nanometers (10-9 meters). Frequency is the number of wave cycles that travel past a fixed point per unit of time, and is usually given in cycles per second, or hertz (Hz). Visible wavelengths cover a range from approximately 400 to 800 nm. The longest visible wavelength is red and the shortest is violet. Other common colors of the spectrum, in order of decreasing wavelength, may be remembered by the mnemonic: ROY G BIV. The wavelengths of what we perceive as particular colors in the visible portion of the spectrum are displayed and listed below. In horizontal diagrams, such as the one on the bottom left, wavelength will increase on moving from left to right.
• Violet: 400 - 420 nm
• Indigo: 420 - 440 nm
• Blue: 440 - 490 nm
• Green: 490 - 570 nm
• Yellow: 570 - 585 nm
• Orange: 585 - 620 nm
• Red: 620 - 780 nm
When white light passes through or is reflected by a colored substance, a characteristic portion of the mixed wavelengths is absorbed. The remaining light will then assume the complementary color to the wavelength(s) absorbed. This relationship is demonstrated by the color wheel shown below. Here, complementary colors are diametrically opposite each other. Thus, absorption of 420-430 nm light renders a substance yellow, and absorption of 500-520 nm light makes it red. Green is unique in that it can be created by absoption close to 400 nm as well as absorption near 800 nm.
Early humans valued colored pigments, and used them for decorative purposes. Many of these were inorganic minerals, but several important organic dyes were also known. These included the crimson pigment, kermesic acid, the blue dye, indigo, and the yellow saffron pigment, crocetin. A rare dibromo-indigo derivative, punicin, was used to color the robes of the royal and wealthy. The deep orange hydrocarbon carotene is widely distributed in plants, but is not sufficiently stable to be used as permanent pigment, other than for food coloring. A common feature of all these colored compounds, displayed below, is a system of extensively conjugated $\pi$-electrons.
The Electromagnetic Spectrum
The visible spectrum constitutes but a small part of the total radiation spectrum. Most of the radiation that surrounds us cannot be seen, but can be detected by dedicated sensing instruments. This electromagnetic spectrum ranges from very short wavelengths (including gamma and x-rays) to very long wavelengths (including microwaves and broadcast radio waves). The following chart displays many of the important regions of this spectrum, and demonstrates the inverse relationship between wavelength and frequency (shown in the top equation below the chart).
The energy associated with a given segment of the spectrum is proportional to its frequency. The bottom equation describes this relationship, which provides the energy carried by a photon of a given wavelength of radiation.
To obtain specific frequency, wavelength and energy values use this calculator.
UV-Visible Absorption Spectra
To understand why some compounds are colored and others are not, and to determine the relationship of conjugation to color, we must make accurate measurements of light absorption at different wavelengths in and near the visible part of the spectrum. Commercial optical spectrometers enable such experiments to be conducted with ease, and usually survey both the near ultraviolet and visible portions of the spectrum. For a description of a UV-Visible spectrometer Click Here.
The visible region of the spectrum comprises photon energies of 36 to 72 kcal/mole, and the near ultraviolet region, out to 200 nm, extends this energy range to 143 kcal/mole. Ultraviolet radiation having wavelengths less than 200 nm is difficult to handle, and is seldom used as a routine tool for structural analysis.
The energies noted above are sufficient to promote or excite a molecular electron to a higher energy orbital. Consequently, absorption spectroscopy carried out in this region is sometimes called "electronic spectroscopy". A diagram showing the various kinds of electronic excitation that may occur in organic molecules is shown on the left. Of the six transitions outlined, only the two lowest energy ones (left-most, colored blue) are achieved by the energies available in the 200 to 800 nm spectrum. As a rule, energetically favored electron promotion will be from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO), and the resulting species is called an excited state. For a review of molecular orbitals click here.
When sample molecules are exposed to light having an energy that matches a possible electronic transition within the molecule, some of the light energy will be absorbed as the electron is promoted to a higher energy orbital. An optical spectrometer records the wavelengths at which absorption occurs, together with the degree of absorption at each wavelength. The resulting spectrum is presented as a graph of absorbance (A) versus wavelength, as in the isoprene spectrum shown below. Since isoprene is colorless, it does not absorb in the visible part of the spectrum and this region is not displayed on the graph. Absorbance usually ranges from 0 (no absorption) to 2 (99% absorption), and is precisely defined in context with spectrometer operation.
Electronic transitions
Let’s take as our first example the simple case of molecular hydrogen, H2. As you may recall from section 2.1A, the molecular orbital picture for the hydrogen molecule consists of one bonding σ MO, and a higher energy antibonding σ* MO. When the molecule is in the ground state, both electrons are paired in the lower-energy bonding orbital – this is the Highest Occupied Molecular Orbital (HOMO). The antibonding σ* orbital, in turn, is the Lowest Unoccupied Molecular Orbital (LUMO).
If the molecule is exposed to light of a wavelength with energy equal to ΔE, the HOMO-LUMO energy gap, this wavelength will be absorbed and the energy used to bump one of the electrons from the HOMO to the LUMO – in other words, from the σ to the σ* orbital. This is referred to as a σ - σ* transition. ΔE for this electronic transition is 258 kcal/mol, corresponding to light with a wavelength of 111 nm.
When a double-bonded molecule such as ethene (common name ethylene) absorbs light, it undergoes a π - π* transition. Because π- π* energy gaps are narrower than σ - σ* gaps, ethene absorbs light at 165 nm - a longer wavelength than molecular hydrogen.
The electronic transitions of both molecular hydrogen and ethene are too energetic to be accurately recorded by standard UV spectrophotometers, which generally have a range of 220 – 700 nm. Where UV-vis spectroscopy becomes useful to most organic and biological chemists is in the study of molecules with conjugated pi systems. In these groups, the energy gap for π -π* transitions is smaller than for isolated double bonds, and thus the wavelength absorbed is longer. Molecules or parts of molecules that absorb light strongly in the UV-vis region are called chromophores.
Let’s revisit the MO picture for 1,3-butadiene, the simplest conjugated system (see section 2.1B). Recall that we can draw a diagram showing the four pi MO’s that result from combining the four 2pz atomic orbitals. The lower two orbitals are bonding, while the upper two are antibonding.
Comparing this MO picture to that of ethene, our isolated pi-bond example, we see that the HOMO-LUMO energy gap is indeed smaller for the conjugated system. 1,3-butadiene absorbs UV light with a wavelength of 217 nm.
As conjugated pi systems become larger, the energy gap for a π - π* transition becomes increasingly narrow, and the wavelength of light absorbed correspondingly becomes longer. The absorbance due to the π - π* transition in 1,3,5-hexatriene, for example, occurs at 258 nm, corresponding to a ΔE of 111 kcal/mol.
In molecules with extended pi systems, the HOMO-LUMO energy gap becomes so small that absorption occurs in the visible rather then the UV region of the electromagnetic spectrum. Beta-carotene, with its system of 11 conjugated double bonds, absorbs light with wavelengths in the blue region of the visible spectrum while allowing other visible wavelengths – mainly those in the red-yellow region - to be transmitted. This is why carrots are orange.
The conjugated pi system in 4-methyl-3-penten-2-one gives rise to a strong UV absorbance at 236 nm due to a π - π* transition. However, this molecule also absorbs at 314 nm. This second absorbance is due to the transition of a non-bonding (lone pair) electron on the oxygen up to a π* antibonding MO:
This is referred to as an n - π* transition. The nonbonding (n) MO’s are higher in energy than the highest bonding p orbitals, so the energy gap for an n - π* transition is smaller that that of a π - π* transition – and thus the n - π* peak is at a longer wavelength. In general, n - π* transitions are weaker (less light absorbed) than those due to π - π* transitions.
Template:ExampleStart
Exercise 4.3: How large is the π - π* transition in 4-methyl-3-penten-2-one?
Exercise 4.4: Which of the following molecules would you expect absorb at a longer wavelength in the UV region of the electromagnetic spectrum? Explain your answer.
Solution
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
Template:ExampleEnd | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/14%3A_Conjugation_Resonance_and_Dienes/14.15%3A_Conjugated_Dienes_and_Ultraviolet_Light.txt |
Because of the low hydrogen to carbon ratio in aromatic compounds (note that the H:C ratio in an alkane is >2), chemists expected their structural formulas would contain a large number of double or triple bonds. Since double bonds are easily cleaved by oxidative reagents such as potassium permanganate or ozone, and rapidly add bromine and chlorine, these reactions were applied to these aromatic compounds. Surprisingly, products that appeared to retain many of the double bonds were obtained, and these compounds exhibited a high degree of chemical stability compared with known alkenes and cycloalkenes (aliphatic compounds). On treatment with hot permanganate solution, cinnamaldehyde gave a stable, crystalline C7H6O2 compound, now called benzoic acid. The H:C ratio in benzoic acid is <1, again suggesting the presence of several double bonds. Benzoic acid was eventually converted to the stable hydrocarbon benzene, C6H6, which also proved unreactive to common double bond transformations, as shown below. For comparison, reactions of cyclohexene, a typical alkene, with these reagents are also shown (green box). As experimental evidence for a wide assortment of compounds was acquired, those incorporating this exceptionally stable six-carbon core came to be called "aromatic".
If benzene is forced to react by increasing the temperature and/or by addition of a catalyst, It undergoes substitution reactions rather than the addition reactions that are typical of alkenes. This further confirms the previous indication that the six-carbon benzene core is unusually stable to chemical modification. The conceptual contradiction presented by a high degree of unsaturation (low H:C ratio) and high chemical stability for benzene and related compounds remained an unsolved puzzle for many years. Eventually, the presently accepted structure of a regular-hexagonal, planar ring of carbons was adopted, and the exceptional thermodynamic and chemical stability of this system was attributed to resonance stabilization of a conjugated cyclic triene.
Benzene:
Here, two structurally and energetically equivalent electronic structures for a stable compound are written, but no single structure provides an accurate or even an adequate representation of the true molecule. The six-membered ring in benzene is a perfect hexagon (all carbon-carbon bonds have an identical length of 1.40 Å). The cyclohexatriene contributors would be expected to show alternating bond lengths, the double bonds being shorter (1.34 Å) than the single bonds (1.54 Å). An alternative representation for benzene (circle within a hexagon) emphasizes the pi-electron delocalization in this molecule, and has the advantage of being a single diagram. In cases such as these, the electron delocalization described by resonance enhances the stability of the molecules, and compounds composed of such molecules often show exceptional stability and related properties.
Evidence for the enhanced thermodynamic stability of benzene was obtained from measurements of the heat released when double bonds in a six-carbon ring are hydrogenated (hydrogen is added catalytically) to give cyclohexane as a common product. In the following diagram cyclohexane represents a low-energy reference point. Addition of hydrogen to cyclohexene produces cyclohexane and releases heat amounting to 28.6 kcal per mole. If we take this value to represent the energy cost of introducing one double bond into a six-carbon ring, we would expect a cyclohexadiene to release 57.2 kcal per mole on complete hydrogenation, and 1,3,5-cyclohexatriene to release 85.8 kcal per mole. These heats of hydrogenation would reflect the relative thermodynamic stability of the compounds. In practice, 1,3-cyclohexadiene is slightly more stable than expected, by about 2 kcal, presumably due to conjugation of the double bonds. Benzene, however, is an extraordinary 36 kcal/mole more stable than expected. This sort of stability enhancement is now accepted as a characteristic of all aromatic compounds.
A molecular orbital description of benzene provides a more satisfying and more general treatment of "aromaticity". We know that benzene has a planar hexagonal structure in which all the carbon atoms are sp2 hybridized, and all the carbon-carbon bonds are equal in length. As shown below, the remaining cyclic array of six p-orbitals ( one on each carbon) overlap to generate six molecular orbitals, three bonding and three antibonding. The plus and minus signs shown in the diagram do not represent electrostatic charge, but refer to phase signs in the equations that describe these orbitals (in the diagram the phases are also color coded). When the phases correspond, the orbitals overlap to generate a common region of like phase, with those orbitals having the greatest overlap (e.g. π1) being lowest in energy. The remaining carbon valence electrons then occupy these molecular orbitals in pairs, resulting in a fully occupied (6 electrons) set of bonding molecular orbitals. It is this completely filled set of bonding orbitals, or closed shell, that gives the benzene ring its thermodynamic and chemical stability, just as a filled valence shell octet confers stability on the inert gases.
The Molecular Orbitals of Benzene
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
15.02: The Structure of Benzene
Among the many distinctive features of benzene, its aromaticity is the major contributor to why it is so unreactive. This section will try to clarify the theory of aromaticity and why aromaticity gives unique qualities that make these conjugated alkenes inert to compounds such as Br2 and even hydrochloric acid. It will also go into detail about the unusually large resonance energy due to the six conjugated carbons of benzene.
The delocalization of the p-orbital carbons on the sp2 hybridized carbons is what gives the aromatic qualities of benzene.
This diagram shows one of the molecular orbitals containing two of the delocalized electrons, which may be found anywhere within the two "doughnuts". The other molecular orbitals are almost never drawn.
• Benzene, C6H6, is a planar molecule containing a ring of six carbon atoms, each with a hydrogen atom attached.
The six carbon atoms form a perfectly regular hexagon. All of the carbon-carbon bonds have exactly the same lengths - somewhere between single and double bonds.
• There are delocalized electrons above and below the plane of the ring.
• The presence of the delocalized electrons makes benzene particularly stable.
• Benzene resists addition reactions because those reactions would involve breaking the delocalization and losing that stability.
• Benzene is represented by this symbol, where the circle represents the delocalized electrons, and each corner of the hexagon has a carbon atom with a hydrogen attached.
Basic Structure of Benzene
Because of the aromaticity of benzene, the resulting molecule is planar in shape with each C-C bond being 1.39 Å in length and each bond angle being 120°. You might ask yourselves how it's possible to have all of the bonds to be the same length if the ring is conjugated with both single (1.47 Å) and double (1.34 Å), but it is important to note that there are no distinct single or double bonds within the benzene. Rather, the delocalization of the ring makes each count as one and a half bonds between the carbons which makes sense because experimentally we find that the actual bond length is somewhere in between a single and double bond. Finally, there are a total of six p-orbital electrons that form the stabilizing electron clouds above and below the aromatic ring.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/15%3A_Benzene_and_Aromatic_Compounds/15.01%3A_Background.txt |
Unlike aliphatic organics, nomenclature of benzene-derived compounds can be confusing because a single aromatic compound can have multiple possible names (such as common and systematic names) be associated with its structure. In these sections, we will analyze some of the ways these compounds can be named.
Simple Benzene Naming
Some common substituents, like NO2, Br, and Cl, can be named this way when it is attached to a phenyl group. Long chain carbons attached can also be named this way. The general format for this kind of naming is:
(positions of substituents (if >1)- + # (di, tri, ...) + substituent)n + benzene.
For example, chlorine (Cl) attached to a phenyl group would be named chlorobenzene (chloro + benzene). Since there is only one substituent on the benzene ring, we do not have to indicate its position on the benzene ring (as it can freely rotate around and you would end up getting the same compound.)
Figure 8. Example of simple benzene naming with chlorine and NO2 as substituents.
Figure 9. More complicated simple benzene naming examples - Note that standard nomenclature priority rules are applied here, causing the numbering of carbons to switch. See Nomenclature of Organic Compounds for a review on naming and priority rules.
Ortho-, Meta-, Para- (OMP) Nomenclature for Disubstituted Benzenes
Instead of using numbers to indicate substituents on a benzene ring, ortho- (o-), meta- (m-), or para (p-) can be used in place of positional markers when there are two substituents on the benzene ring (disubstituted benzenes). They are defined as the following:
• ortho- (o-): 1,2- (next to each other in a benzene ring)
• meta- (m): 1,3- (separated by one carbon in a benzene ring)
• para- (p): 1,4- (across from each other in a benzene ring)
Using the same example above in figure 9a (1,3-dichlorobenzene), we can use the ortho-, meta-, para- nomenclature to transform the chemical name into m-dichlorobenzene, as shown in the figure below.
Figure 10. Transformation of 1,3-dichlorobenzene into m-dichlorobenzene.
Here are some other examples of ortho-, meta-, para- nomenclature used in context:
However, the substituents used in ortho-, meta-, para- nomenclature do not have to be the same. For example, we can use chlorine and a nitro group as substituents in the benzene ring.
In conclusion, these can be pieced together into a summary diagram, as shown below:
Base Name Nomenclature
In addition to simple benzene naming and OMP nomenclature, benzene derived compounds are also sometimes used as bases. The concept of a base is similar to the nomenclature of aliphatic and cyclic compounds, where the parent for the organic compound is used as a base (a name for its chemical name. For example, the following compounds have the base names hexane and cyclohexane, respectively. See Nomenclature of Organic Compounds for a review on naming organic compounds.
Benzene, similar to these compounds shown above, also has base names from its derived compounds. Phenol (C6H5OH), as introduced previously in this article, for example, serves as a base when other substituents are attached to it. This is best illustrated in the diagram below.
Figure 14. An example showing phenol as a base in its chemical name. Note how benzene no longer serves as a base when an OH group is added to the benzene ring.
Alternatively, we can use the numbering system to indicate this compound. When the numbering system is used, the carbon where the substituent is attached on the base will be given the first priority and named as carbon #1 (C1). The normal priority rules then apply in the nomenclature process (give the rest of the substituents the lowest numbering as you could).
Figure 15. The naming process for 2-chlorophenol (o-chlorophenol). Note that 2-chlorophenol = o-chlorophenol.
Below is a list of commonly seen benzene-derived compounds. Some of these mono-substituted compounds (labeled in red and green), such as phenol or toluene, can be used in place of benzene for the chemical's base name.
Figure 16. Common benzene derived compounds with various substituents.
Common vs. Systematic (IUPAC) Nomenclature
According to the indexing preferences of the Chemical Abstracts, phenol, benzaldehyde, and benzoic acid (labeled in red in Figure 16) are some of the common names that are retained in the IUPAC (systematic) nomenclature. Other names such as toluene, styrene, naphthalene, or phenanthrene can also be seen in the IUPAC system in the same way. While the use of other common names are usually acceptable in IUPAC, their use are discouraged in the nomenclature of compounds.
Nomenclature for compounds which has such discouraged names will be named by the simple benzene naming system. An example of this would include toluene derivatives like TNT. (Note that toluene by itself is retained by the IUPAC nomenclature, but its derivatives, which contains additional substituents on the benzene ring, might be excluded from the convention). For this reason, the common chemical name 2,4,6-trinitrotoluene, or TNT, as shown in figure 17, would not be advisable under the IUPAC (systematic) nomenclature.
To correctly name TNT under the IUPAC system, the simple benzene naming system should be used:
Figure 18. Systematic (IUPAC) name of 2,4,6-trinitrotoluene (common name), or TNT.
Note that the methyl group is individually named due to the exclusion of toluene from the IUPAC nomenclature.
Figure 19. The common name 2,4-dibromophenol, is shared by the IUPAC systematic nomenclature.
Only substituents phenol, benzoic acid, and benzaldehyde share this commonality.
Since the IUPAC nomenclature primarily rely on the simple benzene naming system for the nomenclature of different benzene derived compounds, the OMP (ortho-, meta-, para-) system is not accepted in the IUPAC nomenclature. For this reason, the OMP system will yield common names that can be converted to systematic names by using the same method as above. For example, o-Xylene from the OMP system can be named 1,2-dimethylbenzene by using simple benzene naming (IUPAC standard).
The Phenyl and Benzyl Groups
The Phenyl Group
As mentioned previously, the phenyl group (Ph-R, C6H5-R) can be formed by removing a hydrogen from benzene and attaching a substituent to where the hydrogen was removed. To this phenomenon, we can name compounds formed this way by applying this rule: (phenyl + substituent). For example, a chlorine attached in this manner would be named phenyl chloride, and a bromine attached in this manner would be named phenyl bromide. (See below diagram)
Figure 20. Naming of Phenyl Chloride and Phenyl Bromide
While compounds like these are usually named by simple benzene type naming (chlorobenzene and bromobenzene), the phenyl group naming is usually applied to benzene rings where a substituent with six or more carbons is attached, such as in the diagram below.
Figure 21. Diagram of 2-phenyloctane.
Although the diagram above might be a little daunting to understand at first, it is not as difficult as it seems after careful analysis of the structure is made. By looking for the longest chain in the compound, it should be clear that the longest chain is eight (8) carbons long (octane, as shown in green) and that a benzene ring is attached to the second position of this longest chain (labeled in red). As this rule suggests that the benzene ring will act as a function group (a substituent) whenever a substituent of more than six (6) carbons is attached to it, the name "benzene" is changed to phenyl and is used the same way as any other substituents, such as methyl, ethyl, or bromo. Putting it all together, the name can be derived as: 2-phenyloctane (phenyl is attached at the second position of the longest carbon chain, octane).
The Benzyl Group
The benzyl group (abbv. Bn), similar to the phenyl group, is formed by manipulating the benzene ring. In the case of the benzyl group, it is formed by taking the phenyl group and adding a CH2 group to where the hydrogen was removed. Its molecular fragment can be written as C6H5CH2-R, PhCH2-R, or Bn-R. Nomenclature of benzyl group based compounds are very similar to the phenyl group compounds. For example, a chlorine attached to a benzyl group would simply be called benzyl chloride, whereas an OH group attached to a benzyl group would simply be called benzyl alcohol.
Figure 22. Benzyl Group Nomenclature
Additionally, other substituents can attach on the benzene ring in the presence of the benzyl group. An example of this can be seen in the figure below:
Figure 23. Nomenclature of 2,4-difluorobenzyl chloride. Similar to the base name nomenclatures system, the carbon in which th base substitutent is attached on the benzene ring is given the first priority and the rest of the substituents are given the lowest number order possible.
Similar to the base name nomenclature system, the carbon in which the base substituent is attached on the benzene ring is given the first priority and the rest of the substituents are given the lowest number order possible. Under this consideration, the above compound can be named: 2,4-difluorobenzyl chloride.
Commonly Named Benzene Compounds Nomenclature Summary Flowchart
Summary Flowchart (Figure 24). Summary of nomenclature rules used in commonly benzene derived compounds.
As benzene derived compounds can be extremely complex, only compounds covered in this article and other commonly named compounds can be named using this flowchart.
Determination of Common and Systematic Names using Flowchart
To demonstrate how this flowchart can be used to name TNT in its common and systematic (IUPAC) name, a replica of the flowchart with the appropriate flow paths are shown below:
References
1. Nicolaou, K. C., & Montagnon, T. (2008). Molecules That Changed the World. KGaA, Weinheim: Wiley-VCH. p. 54
2. Pitman, V. (2004). Aromatherapy. Great Britain, UK: Nelson Thornes. p.135-136
3. Burton, G. (2000). Chemical Ideas. Bicester, Oxon: Heinemann. p.290-292
4. Vollhardt, K. P.C. & Shore, N. (2007). Organic Chemistry (5th Ed.). New York: W. H. Freeman. p. 667-669
5. Schnaubelt, K. (1999). Medical Aromatherapy. Berkeley, CA: Frog Books. p. 211-213
6. Patrick, G. L. (2004). Organic Chemistry. New York, NY: Taylor & Francis. p. 135-136
7. Talbott, S. M. (2002). A Guide to Understanding Dietary Supplements. Binghamton, NY: Haworth Press. p. 616-619
8. Lifton, R. J. (2000). The Nazi doctors. New York, NY: Basic Books. p. 255-261
9. Myers, R. L., & Myers, R. L. (2007). The 100 most important chemical compounds. Westport, CT: Greenwood Publishing Group. p. 281-282
Practice Problems
Q1) (True/False) The compound above contains a benzene ring and thus is aromatic.
Q2) Benzene unusual stability is caused by how many conjugated pi bonds in its cyclic ring? ____
Q3) Menthol, a topical analgesic used in many ointments for the relief of pain, releases a peppermint aroma upon exposure to the air. Based on this conclusion, can you imply that a benzene ring is present in its chemical structure? Why or why not?
Q4)
Q5) At normal conditions, benzene has ___ resonance structures.
Q6) Which of the following name(s) is/are correct for the following compound?
a) nitrohydride benzene
b) phenylamine
c) phenylamide
d) aniline
e) nitrogenhydrogen benzene
f) All of the above is correct
Q7) Convert 1,4-dimethylbenzene into its common name.
Q8) TNT's common name is: ______________________________
Q9) Name the following compound using OMP nomenclature:
Q10) Draw the structure of 2,4-dinitrotoluene.
Q11) Name the following compound:
Q12) Which of the following is the correct name for the following compound?
a) 3,4-difluorobenzyl bromide
b) 1,2-difluorobenzyl bromide
c) 4,5-difluorobenzyl bromide
d) 1,2-difluoroethyl bromide
e) 5,6-difluoroethyl bromide
f) 4,5-difluoroethyl bromide
Q13) (True/False) Benzyl chloride can be abbreviated Bz-Cl.
Q14) Benzoic Acid has what R group attached to its phenyl functional group?
Q15) (True/False) A single aromatic compound can have multiple names indicating its structure.
Q16) List the corresponding positions for the OMP system (o-, m-, p-).
Q17) A scientist has conducted an experiment on an unknown compound. He was able to determine that the unknown compound contains a cyclic ring in its structure as well as an alcohol (-OH) group attached to the ring. What is the unknown compound?
a) Cyclohexanol
b) Cyclicheptanol
c) Phenol
d) Methanol
e) Bleach
f) Cannot determine from the above information
Q18) Which of the following statements is false for the compound, phenol?
a) Phenol is a benzene derived compound.
b) Phenol can be made by attaching an -OH group to a phenyl group.
c) Phenol is highly toxic to the body even in small doses.
d) Phenol can be used as a catalyst in the hydrogenation of benzene into cyclohexane.
e) Phenol is used as an antiseptic in minute doses.
f) Phenol is amongst one of the three common names retained in the IUPAC nomenclature.
Answer Key to Practice Questions
Q1) False, this compound does not contain a benzene ring in its structure.
Q2) 3
Q3) No, a substance that is fragrant does not imply a benzene ring is in its structure. See camphor example (figure 1)
Q4) No reaction, benzene requires a special catalyst to be hydrogenated due to its unusual stability given by its three conjugated pi bonds.
Q5) 2
Q6) b, d
Q7) p-Xylene
Q8) 2,4,6-trinitrotoluene
Q9) p-chloronitrobenzene
Q10)
Q11) 4-phenylheptane
Q12) a
Q13) False, the correct abbreviation for the benzyl group is Bn, not Bz. The correct abbreviation for Benzyl chloride is Bn-Cl.
Q14) COOH
Q15) True. TNT, for example, has the common name 2,4,6-trinitrotoluene and its systematic name is 2-methyl-1,3,5-trinitrobenzene.
Q16) Ortho - 1,2 ; Meta - 1,3 ; Para - 1,4
Q17) The correct answer is f). We cannot determine what structure this is since the question does not tell us what kind of cyclic ring the -OH group is attached on. Just as cyclohexane can be cyclic, benzene and cycloheptane can also be cyclic.
Q18) d
• David Lam | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/15%3A_Benzene_and_Aromatic_Compounds/15.03%3A_Nomenclature_of_Benzene_Derivatives.txt |
The chemical shifts of aromatic protons
Some protons resonate much further downfield than can be accounted for simply by the deshielding effect of nearby electronegative atoms. Vinylic protons (those directly bonded to an alkene carbon) and aromatic (benzylic) protons are dramatic examples.
We'll consider the aromatic proton first. Recall that in benzene and many other aromatic structures, a sextet of pelectrons is delocalized around the ring. When the molecule is exposed to B0, these pelectrons begin to circulate in a ring current, generating their own induced magnetic field that opposes B0. In this case, however, the induced field of the pelectrons does not shield the benzylic protons from B0 as you might expect– rather, it causes the protons to experience a stronger magnetic field in the direction of B0 – in other words, it adds to B0 rather than subtracting from it.
To understand how this happens, we need to understand the concept of diamagnetic anisotropy (anisotropy means `non-uniformity`). So far, we have been picturing magnetic fields as being oriented in a uniform direction. This is only true over a small area. If we step back and take a wider view, however, we see that the lines of force in a magnetic field are actually anisotropic. They start in the 'north' direction, then loop around like a snake biting its own tail.
If we are at point A in the figure above, we feel a magnetic field pointing in a northerly direction. If we are at point B, however, we feel a field pointing to the south.
In the induced field generated by the aromatic ring current, the benzylic protons are at the equivalent of ‘point B’ – this means that the induced current in this region of space is oriented in the same direction as B0.
In total, the benzylic protons are subjected to three magnetic fields: the applied field (B0) and the induced field from the pelectrons pointing in one direction, and the induced field of the non-aromatic electrons pointing in the opposite (shielding) direction. The end result is that benzylic protons, due to the anisotropy of the induced field generated by the ring current, appear to be highly deshielded. Their chemical shift is far downfield, in the 6.5-8 ppm region.
Characteristic NMR Absorption of Benzene Derivatives
Hydrogens directly attached to an arene ring show up about 7-9 PPM in the NMR. This is called the aromatic region. Hydrogen environments directly bonded to an arene ring show up about 2.5 PPM.
Charateristic IR Absorption of Benzene Derivatives
Arenes produce an characteristic C=C absorption about 1680-1600 1/cm
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
15.06: Benzenes Unusual Stability
If benzene is forced to react by increasing the temperature and/or by addition of a catalyst, It undergoes substitution reactions rather than the addition reactions that are typical of alkenes. This further confirms the previous indication that the six-carbon benzene core is unusually stable to chemical modification. The conceptual contradiction presented by a high degree of unsaturation (low H:C ratio) and high chemical stability for benzene and related compounds remained an unsolved puzzle for many years. Eventually, the presently accepted structure of a regular-hexagonal, planar ring of carbons was adopted, and the exceptional thermodynamic and chemical stability of this system was attributed to resonance stabilization of a conjugated cyclic triene.
Benzene:
Here, two structurally and energetically equivalent electronic structures for a stable compound are written, but no single structure provides an accurate or even an adequate representation of the true molecule. The six-membered ring in benzene is a perfect hexagon (all carbon-carbon bonds have an identical length of 1.40 Å). The cyclohexatriene contributors would be expected to show alternating bond lengths, the double bonds being shorter (1.34 Å) than the single bonds (1.54 Å). An alternative representation for benzene (circle within a hexagon) emphasizes the pi-electron delocalization in this molecule, and has the advantage of being a single diagram. In cases such as these, the electron delocalization described by resonance enhances the stability of the molecules, and compounds composed of such molecules often show exceptional stability and related properties.
Evidence for the enhanced thermodynamic stability of benzene was obtained from measurements of the heat released when double bonds in a six-carbon ring are hydrogenated (hydrogen is added catalytically) to give cyclohexane as a common product. In the following diagram cyclohexane represents a low-energy reference point. Addition of hydrogen to cyclohexene produces cyclohexane and releases heat amounting to 28.6 kcal per mole. If we take this value to represent the energy cost of introducing one double bond into a six-carbon ring, we would expect a cyclohexadiene to release 57.2 kcal per mole on complete hydrogenation, and 1,3,5-cyclohexatriene to release 85.8 kcal per mole. These heats of hydrogenation would reflect the relative thermodynamic stability of the compounds. In practice, 1,3-cyclohexadiene is slightly more stable than expected, by about 2 kcal, presumably due to conjugation of the double bonds. Benzene, however, is an extraordinary 36 kcal/mole more stable than expected. This sort of stability enhancement is now accepted as a characteristic of all aromatic compounds.
A molecular orbital description of benzene provides a more satisfying and more general treatment of "aromaticity". We know that benzene has a planar hexagonal structure in which all the carbon atoms are sp2 hybridized, and all the carbon-carbon bonds are equal in length. As shown below, the remaining cyclic array of six p-orbitals ( one on each carbon) overlap to generate six molecular orbitals, three bonding and three antibonding. The plus and minus signs shown in the diagram do not represent electrostatic charge, but refer to phase signs in the equations that describe these orbitals (in the diagram the phases are also color coded). When the phases correspond, the orbitals overlap to generate a common region of like phase, with those orbitals having the greatest overlap (e.g. π1) being lowest in energy. The remaining carbon valence electrons then occupy these molecular orbitals in pairs, resulting in a fully occupied (6 electrons) set of bonding molecular orbitals. It is this completely filled set of bonding orbitals, or closed shell, that gives the benzene ring its thermodynamic and chemical stability, just as a filled valence shell octet confers stability on the inert gases. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/15%3A_Benzene_and_Aromatic_Compounds/15.04%3A_Spectroscopic_Properties.txt |
In 1931, German chemist and physicist Erich Hückel proposed a theory to help determine if a planar ring molecule would have aromatic properties. His rule states that if a cyclic, planar molecule has 4n+2 π electrons, it is considered aromatic. This rule would come to be known as Hückel's Rule.
Four Criteria for Aromaticity
When deciding if a compound is aromatic, go through the following checklist. If the compound does not meet all the following criteria, it is likely not aromatic.
1. The molecule is cyclic (a ring of atoms)
2. The molecule is planar (all atoms in the molecule lie in the same plane)
3. The molecule is fully conjugated (p orbitals at every atom in the ring)
4. The molecule has $4n+2\, π$ electrons ($n=0$ or any positive integer)
According to Hückel's Molecular Orbital Theory, a compound is particularly stable if all of its bonding molecular orbitals are filled with paired electrons. This is true of aromatic compounds, meaning they are quite stable. With aromatic compounds, 2 electrons fill the lowest energy molecular orbital, and 4 electrons fill each subsequent energy level (the number of subsequent energy levels is denoted by n), leaving all bonding orbitals filled and no anti-bonding orbitals occupied. This gives a total of 4n+2 π electrons. You can see how this works with the molecular orbital diagram for the aromatic compound, benzene, below. Benzene has 6 π electrons. Its first 2 π electrons fill the lowest energy orbital, and it has 4 π electrons remaining. These 4 fill in the orbitals of the succeeding energy level. Notice how all of its bonding orbitals are filled, but none of the anti-bonding orbitals have any electrons.
To apply the 4n+2 rule, first count the number of π electrons in the molecule. Then, set this number equal to 4n+2 and solve for n. If is 0 or any positive integer (1, 2, 3,...), the rule has been met. For example, benzene has six π electrons:
\begin{align*} 4n + 2 &= 6 \[4pt] 4n &= 4 \[4pt] n &= 1 \end{align*}
For benzene, we find that $n=1$, which is a positive integer, so the rule is met.
Perhaps the toughest part of Hückel's Rule is figuring out which electrons in the compound are actually π electrons. Once this is figured out, the rule is quite straightforward. π electrons lie in p orbitals. Sp2 hybridized atoms have 1 p orbital each. So if every molecule in the cyclic compound is sp2 hybridized, this means the molecule is fully conjugated (has 1 p orbital at each atom), and the electrons in these p orbitals are the π electrons. A simple way to know if an atom is sp2 hybridized is to see if it has 3 attached atoms and no lone pairs of electrons. This video provides a very nice tutorial on how to determine an atom's hybridization. In a cyclic hydrocarbon compound with alternating single and double bonds, each carbon is attached to 1 hydrogen and 2 other carbons. Therefore, each carbon is sp2 hybridized and has a p orbital. Let's look at our previous example, benzene:
Each double bond (π bond) always contributes 2 π electrons. Benzene has 3 double bonds, so it has 6 π electrons.
Aromatic Ions
Hückel's Rule also applies to ions. As long as a compound has 4n+2 π electrons, it does not matter if the molecule is neutral or has a charge. For example, cyclopentadienyl anion is an aromatic ion. How do we know that it is fully conjugated? That is, how do we know that each atom in this molecule has 1 p orbital? Let's look at the following figure. Carbons 2-5 are sp2 hybridized because they have 3 attached atoms and have no lone electron pairs. What about carbon 1? Another simple rule to determine if an atom is sp2 hybridized is if an atom has 1 or more lone pairs and is attached to an sp2 hybridized atom, then that atom is sp2 hybridized also. This video explains the rule very clearly. Therefore, carbon 1 has a p orbital. Cyclopentadienyl anion has 6 π electrons and fulfills the 4n+2 rule.
Heterocyclic Aromatic Compounds
So far, you have encountered many carbon homocyclic rings, but compounds with elements other than carbon in the ring can also be aromatic, as long as they fulfill the criteria for aromaticity. These molecules are called heterocyclic compounds because they contain 1 or more different atoms other than carbon in the ring. A common example is furan, which contains an oxygen atom. We know that all carbons in furan are sp2 hybridized. But is the oxygen atom sp2 hybridized? The oxygen has at least 1 lone electron pair and is attached to an sp2 hybridized atom, so it is sp2 hybridized as well. Notice how oxygen has 2 lone pairs of electrons. How many of those electrons are π electrons? An sp2 hybridized atom only has 1 p orbital, which can only hold 2 electrons, so we know that 1 electron pair is in the p orbital, while the other pair is in an sp2 orbital. So, only 1 of oxygen's 2 lone electron pairs are π electrons. Furan has 6 π electrons and fulfills the 4n+2 rule.
Problems
Using the criteria for aromaticity, determine if the following molecules are aromatic:
Answers
1. Aromatic - only 1 of S's lone pairs counts as π electrons, so there are 6 π electrons, n=1
2. Not aromatic - not fully conjugated, top C is sp3 hybridized
3. Not aromatic - top C is sp2 hybridized, but there are 4 π electrons, n=1/2
4. Aromatic - N is using its 1 p orbital for the electrons in the double bond, so its lone pair of electrons are not π electrons, there are 6 π electrons, n=1
5. Aromatic - there are 6 π electrons, n=1
6. Not aromatic - all atoms are sp2 hybridized, but only 1 of S's lone pairs counts as π electrons, so there 8 π electrons, n=1.5
7. Not aromatic - there are 4 π electrons, n=1/2
8. Aromatic - only 1 of N's lone pairs counts as π electrons, so there are 6 π electrons, n=1
9. Not aromatic - not fully conjugated, top C is sp3 hybridized
10. Aromatic - O is using its 1 p orbital for the elections in the double bond, so its lone pair of electrons are not π electrons, there are 6 π electrons, n=1
• Ramie Hosein | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/15%3A_Benzene_and_Aromatic_Compounds/15.07%3A_The_Criteria_for_Aromaticity_-_Huckels_Rule.txt |
Aromatic Compounds with more than one ring
Benzene rings may be joined together (fused) to give larger polycyclic aromatic compounds. A few examples are drawn below, together with the approved numbering scheme for substituted derivatives. The peripheral carbon atoms (numbered in all but the last three examples) are all bonded to hydrogen atoms. Unlike benzene, all the C-C bond lengths in these fused ring aromatics are not the same, and there is some localization of the pi-electrons.
The six benzene rings in coronene are fused in a planar ring; whereas the six rings in hexahelicene are not joined in a larger ring, but assume a helical turn, due to the crowding together of the terminal ring atoms. This helical configuration renders the hexahelicene molecule chiral, and it has been resolved into stable enantiomers.
Aromatic Heterocycles
Many unsaturated cyclic compounds have exceptional properties that we now consider characteristic of "aromatic" systems. The following cases are illustrative:
Compound
Structural
Formula
Reaction
with Br2
Thermodynamic
Stabilization
1,3-Cyclopentadiene
Addition ( 0 ºC )
Slight
1,3,5-Cycloheptatriene
Addition ( 0 ºC )
Slight
1,3,5,7-Cyclooctatetraene
Addition ( 0 ºC )
Slight
Benzene
Substitution
Large
Pyridine
Substitution
Large
Furan
Substitution ( 0 ºC )
Moderate
Pyrrole
Substitution
Moderate
Benzene is the archetypical aromatic compound. It is planar, bond angles=120º, all carbon atoms in the ring are sp2 hybridized, and the pi-orbitals are occupied by 6 electrons. The aromatic heterocycle pyridine is similar to benzene, and is often used as a weak base for scavenging protons. Furan and pyrrole have heterocyclic five-membered rings, in which the heteroatom has at least one pair of non-bonding valence shell electrons. By hybridizing this heteroatom to a sp2 state, a p-orbital occupied by a pair of electrons and oriented parallel to the carbon p-orbitals is created. The resulting planar ring meets the first requirement for aromaticity, and the π-system is occupied by 6 electrons, 4 from the two double bonds and 2 from the heteroatom, thus satisfying the Hückel Rule.
Four illustrative examples of aromatic compounds are shown above. The sp2 hybridized ring atoms are connected by brown bonds, the π-electron pairs and bonds that constitute the aromatic ring are colored blue. Electron pairs that are not part of the aromatic π-electron system are black. The first example is azulene, a blue-colored 10 π-electron aromatic hydrocarbon isomeric with naphthalene. The second and third compounds are heterocycles having aromatic properties. Pyridine has a benzene-like six-membered ring incorporating one nitrogen atom. The non-bonding electron pair on the nitrogen is not part of the aromatic π-electron sextet, and may bond to a proton or other electrophile without disrupting the aromatic system. In the case of thiophene, a sulfur analog of furan, one of the sulfur electron pairs (colored blue) participates in the aromatic ring π-electron conjugation. The last compound is imidazole, a heterocycle having two nitrogen atoms. Note that only one of the nitrogen non-bonding electron pairs is used for the aromatic π-electron sextet. The other electron pair (colored black) behaves similarly to the electron pair in pyridine.
Charged Aromatic Compounds
Carbanions and carbocations may also show aromatic stabilization. Some examples are:
The three-membered ring cation has 2 π-electrons and is surprisingly stable, considering its ring strain. Cyclopentadiene is as acidic as ethanol, reflecting the stability of its 6 π-electron conjugate base. Salts of cycloheptatrienyl cation (tropylium ion) are stable in water solution, again reflecting the stability of this 6 π-electron cation. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/15%3A_Benzene_and_Aromatic_Compounds/15.08%3A_Examples_of_Aromatic_Compounds.txt |
Aromaticity
Molecular orbital theory is especially helpful in explaining the unique properties of a class of compounds called aromatics. Benzene, a common organic solvent, is the simplest example of an aromatic compound.
Although it is most often drawn with three double bonds and three single bonds, it is known that all of the carbon-carbon bonds in benzene are exactly the same length - 1.38 Å. This is shorter than a typical carbon-carbon single bond (about 1.54 Å), and slightly longer than a typical carbon-carbon double bond (about 1.34 Å).
In addition, the π bonds in benzene are significantly less reactive than isolated or conjugated π bonds in most alkenes. To illustrate this unique stability, we will make use of the idea of ‘heat of hydrogenation’. The carbon-carbon double bond in an alkene can be converted to a single bond through a process called ‘catalytic hydrogenation’ –essentially adding a molecule of H2 to the double bond.
We will learn more about how this process occurs, both in the laboratory and in living cells, later in the text (section 16.5). For now, what is important to understand is that the hydrogenation process is exothermic: the alkane is lower in energy than the alkene, so hydrogenating the double bond results in the release of energy in the form of heat. Converting one mole of cyclohexene to cyclohexane, for example, releases 28.6 kilocalories. If the benzene molecule is considered to be a six-membered ring with three isolated double bonds, the heat of hydrogenation should theoretically be three times this value, or 85.8 kcal/mol. The actual heat of hydrogenation of benzene, however, is only 49.8 kcal/mol, or 36 kcal/mol less than what we would expect if using the isolated double bond model. Something about the structure of benzene makes these π bonds especially stable. This ‘something’ has a name: it is called ‘aromaticity’.
What exactly is this ‘aromatic’ property that makes the pbonds in benzene so much less reactive than those in alkenes? In a large part, the answer to this question lies in the fact that benzene is a cyclic molecule in which all of the ring atoms are sp2-hybridized. This allows the π electrons to be delocalized in molecular orbitals that extend all the way around the ring, above and below the plane of the ring. For this to happen, of course, the ring must be planar – otherwise the 2pz orbitals couldn’t overlap properly. Benzene is indeed known to be a flat molecule.
Do all cyclic molecules with alternating single and double bonds have this same aromatic stability? Quite simply, the answer is ‘no’. The eight-membered cyclooctatetraene ring shown below is not flat, and its π bonds are much more reactive than those of benzene.
Clearly it takes something more to be aromatic, and this can best be explained with molecular orbital theory. Let’s look at an energy diagram for the molecular orbitals containing the π electrons in benzene.
Quantum mechanical calculations conclude that the six molecular orbitals in benzene, formed from six atomic 2pz orbitals, occupy four separate energy levels. Ψ1 and Ψ6* have unique energy levels, while the Ψ2- Ψ3 and Ψ4*-Ψ5* pairs are degenerate (more than one orbital at the same energy level). When we use the aufbau principle to fill up these orbitals with the six π electrons in benzene, we see that the bonding orbitals are completely filled, and the antibonding orbitals are empty. This gives us a good clue to the source of the special stability of benzene: a full set of bonding MO’s is similar in many ways to the ‘full shell’ of electrons possessed by the very stable noble gases like helium, neon, and argon.
Now, let’s do the same thing for cyclooctatetraene, which we have already learned is not aromatic.
The result of molecular orbital calculations tells us that the lowest and highest energy MOs (Ψ1 and Ψ8*) have unique energy levels, while the other six come in degenerate pairs. Notice that y4and y5are at the same energy level as the isolated 2pz atomic orbitals: these are therefore neither bonding nor antibonding, rather they are referred to as nonbonding MOs. Filling up the MOs with the eight π electrons in the molecule, we find that the last two electrons are unpaired and fall into the two degenerate nonbonding orbitals. Because we don't have a perfect filled shell of bonding MOs, our molecule is not aromatic. As a consequence, each of the double bonds in cyclooctatetraene acts more like an isolated double bond.
Here, then, are the conditions that must be satisfied for a molecule to be considered aromatic:
1. It must have a cyclic structure.
2. The ring must be planar.
3. Each atom in the ring must be sp2-hybridized, so that π electrons can be delocalized around the ring.
4. The number of π electrons in the ring must be such that, in the ground state of the molecule, all bonding MOs are completely filled, and all nonbonding and antibonding MOs are completely empty.
It turns out that, in order to satisfy condition #4, the ring must contain a specific number of π electrons. The set of possible numbers is quite easy to remember - the rule is simply 4n+2, where n is any positive integer (this is known as the Hückel rule, named after Erich Hückel, a German scientist who studied aromatic compounds in the 1930’s). Thus, if n = 0, the first Hückel number is (4 x 0) + 2, or 2. If n = 1, the Hückel number is (4 x 1) + 2, or 6 (the Hückel number for benzene). The series continues with 10, 14, 18, 22, and so on. Cyclooctatetraene has eight π electrons, which is not a Hückel number. Because 6 is such a common Hückel number, chemists often use the term 'aromatic sextet'.
Benzene is best visualized as a planar ring made up of carbon-carbon sbonds, with two ‘donut-like’ rings of fully delocalized π electron density above and below the plane of the ring (the fact that there is a ring of π electron density on both sides of the molecule stems from the fact that the overlapping p orbitals have two lobes, and the electron density is located in both). This general picture is valid not just for benzene but for all other aromatic structures as well.
Let’s look at some different aromatic compounds other than benzene. Pyridine and pyrimidine both fulfill all of the criteria for aromaticity.
In both of these molecules, the nitrogen atoms are sp2 hybridized, with the lone pair occupying an sp2 orbital and therefore not counted among the aromatic sextet. The Hückel number for both pyridine and pyrimidine is six.
Rings do not necessarily need to be 6-membered in order to have six π electrons. Pyrrole and imidizole, for example, are both aromatic 5-membered rings with six π electrons.
The nitrogen atoms in both of these molecules are sp2-hybridized (as they must be for the rings to be aromatic). In pyrrole, the lone pair can be thought of as occupying a 2pz orbital, and thus both of these electrons contribute to the aromatic π system. In imidazole, one lone pair occupies a 2pz orbital and is part of the aromatic sextet, while the second occupies one of the sp2 orbitals and is not part of the sextet.
Molecules with more then one ring can also fulfill the Hückel criteria, and often have many of the same properties as monocyclic aromatic compounds, including a planar structure. Indole (a functional group in the amino acid tryptophan) and purine (a functional group in guanine and adenine DNA/RNA bases) both have a total of ten π electrons delocalized around two rings.
The nitrogen in indole and the N9 nitrogen in purine both contribute a pair of electrons to the π system. The N1, N3, and N7 nitrogens of purine, in contrast, hold their lone pair in sp2 orbitals, outside of the aromatic system.
Example
Exercise 2.2: Are the following molecules likely to be aromatic? Explain, using Huckel’s criteria. and orbital drawings. Hint: Ions can also be aromatic!
Solution
Up to now we have been talking about molecules in which the entire structure makes up an aromatic system. However, in organic chemistry we will more often encounter examples of molecules which have both aromatic and nonaromatic parts. Toluene, a common organic solvent (which is much safer to use than benzene) is simply a benzene ring with a methyl substituent. Benzyaldehyde is benzene with an aldehyde substituent, and phenol is benzene with a hydroxyl substituent.
In these ‘substituted benzene’ compounds, the entire molecule is not aromatic, just the benzene ring part.
When a benzene ring is part of a larger molecule, it is called a ‘phenyl’ group. The amino acid phenylalanine, for example, contains a phenyl group. The amino acids tyrosine, tryptophan, and histidine contain phenol, indole, and imidazole groups, respectively.
Pyridoxine, commonly known as vitamin B6, is a substituted pyridine.
The DNA and RNA bases are based on pyrimidine (cytosine, thymine, and uracil) and purine (adenine and guanine).
The flat, aromatic structure of these bases plays a critical role in the overall structure and function of DNA and RNA.
15.11: BuckminsterfullereneIs It Aromatic
If we extend the structure of corannulene by adding similar cycles of five benzene rings, the curvature of the resulting molecule should increase, and eventually close into a sphere of carbon atoms. The archetypical compound of this kind (C60) has been named buckminsterfullerene because of its resemblance to the geodesic structures created by Buckminster Fuller. It is a member of a family of similar carbon structures that are called fullerenes. These materials represent a third class of carbon allotropes. Alternating views of the C60 fullerene structure are shown on the right, together with a soccer ball-like representation of the 12 five and 20 six-membered rings composing its surface. Precise measurement by Atomic Force Microscopy (AFM) has shown that the C-C bond lengths of the six-membered rings are not all equal, and depend on whether the ring is fused to a five or six-membered beighbor. By clicking on this graphic, a model of C60 will be displayed.
Although C60 is composed of fused benzene rings its chemical reactivity resembles that of the cycloalkenes more than benzene. Indeed, exposure to light and oxygen slowly degrade fullerenes to cage opened products. Most of the reactions thus far reported for C60 involve addition to, rather than substitution of, the core structure. These reactions include hydrogenation, bromination and hydroxylation. Strain introduced by the curvature of the surface may be responsible for the enhanced reactivity of C60.
. Larger fullerenes, such as C70, C76, C82 & C84have ellipsoidal or distorted spherical structures, and fullerene-like assemblies up to C240 have been detected. A fascinating aspect of these structures is that the space within the carbon cage may hold atoms, ions or small molecules. Such species are called endohedral fullerenes. The cavity of C60 is relatively small, but encapsulated helium, lithium and atomic nitrogen compounds have been observed. Larger fullerenes are found to encapsulate lanthanide metal atoms.
Interest in the fullerenes has led to the discovery of a related group of carbon structures referred to as nanotubes. As shown in the following illustration, nanotubes may be viewed as rolled up segments of graphite. The chief structural components are six-membered rings, but changes in tube diameter, branching into side tubes and the capping of tube ends is accomplished by fusion with five and seven-membered rings. Many interesting applications of these unusual structures have been proposed. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/15%3A_Benzene_and_Aromatic_Compounds/15.09%3A_What_Is_the_Basis_of_Huckels_Rule.txt |
Over reaction of Aniline and Phenol
The strongest activating and ortho/para-directing substituents are the amino (-NH2) and hydroxyl (-OH) groups. Direct nitration of phenol (hydroxybenzene) by dilute nitric acid gives modest yields of nitrated phenols and considerable oxidative decomposition to tarry materials; aniline (aminobenzene) is largely destroyed. Bromination of both phenol and aniline is difficult to control, with di- and tri-bromo products forming readily. Because of their high nucleophilic reactivity, aniline and phenol undergo substitution reactions with iodine, a halogen that is normally unreactive with benzene derivatives. The mixed halogen iodine chloride (ICl) provides a more electrophilic iodine moiety, and is effective in iodinating aromatic rings having less powerful activating substituents.
C6H5–NH2 + I2 + NaHCO3 p-I–C6H4–NH2 + NaI + CO2 + H2O
By acetylating the heteroatom substituent on phenol and aniline, its activating influence can be substantially attenuated. For example, acetylation of aniline gives acetanilide (first step in the following equation), which undergoes nitration at low temperature, yielding the para-nitro product in high yield. The modifying acetyl group can then be removed by acid-catalyzed hydrolysis (last step), to yield para-nitroaniline. Although the activating influence of the amino group has been reduced by this procedure, the acetyl derivative remains an ortho/para-directing and activating substituent.
C6H5–NH2 + (CH3CO)2O
pyridine (a base)
C6H5–NHCOCH3
HNO3 , 5 ºC
p-O2N–C6H4–NHCOCH3
H3O(+) & heat
p-O2N–C6H4–NH
The following diagram illustrates how the acetyl group acts to attenuate the overall electron donating character of oxygen and nitrogen. The non-bonding valence electron pairs that are responsible for the high reactivity of these compounds (blue arrows) are diverted to the adjacent carbonyl group (green arrows). However, the overall influence of the modified substituent is still activating and ortho/para-directing.
Some limitations of Friedel-Crafts Alkylation
There are possibilities of carbocation rearrangements when you are trying to add a carbon chain greater than two carbons. The rearrangements occur due to hydride shifts and methyl shifts. For example, the product of a Friedel-Crafts Alkylation will show an iso rearrangement when adding a three carbon chain as a substituent. One way to resolve these problems is through Friedel-Crafts Acylation.
Also, the reaction will only work if the ring you are adding a substituent to is not deactivated. Friedel-Crafts fails when used with compounds such as nitrobenzene and other strong deactivating systems.
Friedel-Crafts reactions cannot be preformed then the aromatic ring contains a NH2, NHR, or NR2 substituent. The lone pair electrons on the amines react with the Lewis acid AlCl3. This places a positive charge next to the benzene ring, which is so strongly activating that the Friedel-Crafts reaction cannot occur.
Lastly, Friedel-Crafts alkylation can undergo polyalkylation. The reaction adds an electron donating alkyl group, which activates the benzene ring to further alkylation.
This problem does not occure during Friedel-Crafts Acylation because an acyl group is deactivating. The prevents further actylations.
16.02: Disubstituted Benzenes
Orientational Interaction of Substituents
When a benzene ring has two substituent groups, each exerts an influence on subsequent substitution reactions. The activation or deactivation of the ring can be predicted more or less by the sum of the individual effects of these substituents. The site at which a new substituent is introduced depends on the orientation of the existing groups and their individual directing effects. We can identify two general behavior categories, as shown in the following table. Thus, the groups may be oriented in such a manner that their directing influences act in concert, reinforcing the outcome; or are opposed (antagonistic) to each other. Note that the orientations in each category change depending on whether the groups have similar or opposite individual directing effects.
Antagonistic or Non-Cooperative
Reinforcing or Cooperative
D = Electron Donating Group (ortho/para-directing)
W = Electron Withdrawing Group (meta-directing)
Reinforcing or Cooperative Substitutions
The products from substitution reactions of compounds having a reinforcing orientation of substituents are easier to predict than those having antagonistic substituents. For example, the six equations shown below are all examples of reinforcing or cooperative directing effects operating in the expected manner. Symmetry, as in the first two cases, makes it easy to predict the site at which substitution is likely to occur. Note that if two different sites are favored, substitution will usually occur at the one that is least hindered by ortho groups.
The first three examples have two similar directing groups in a meta-relationship to each other. In examples 4 through 6, oppositely directing groups have an ortho or para-relationship. The major products of electrophilic substitution, as shown, are the sum of the individual group effects. The strongly activating hydroxyl (–OH) and amino (–NH2) substituents favor dihalogenation in examples 5 and six.
Antagonistic or Non-Cooperative Substitutions
Substitution reactions of compounds having an antagonistic orientation of substituents require a more careful analysis. If the substituents are identical, as in example 1 below, the symmetry of the molecule will again simplify the decision. When one substituent has a pair of non-bonding electrons available for adjacent charge stabilization, it will normally exert the product determining influence, examples 2, 4 & 5, even though it may be overall deactivating (case 2). Case 3 reflects a combination of steric hindrance and the superior innate stabilizing ability of methyl groups relative to other alkyl substituents. Example 6 is interesting in that it demonstrates the conversion of an activating ortho/para-directing group into a deactivating meta-directing "onium" cation [–NH(CH3)2(+) ] in a strong acid environment. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/16%3A_Electrophilic_Aromatic_Substitution/16.01%3A_Limitations_on_Electrophilic_Substitution_Reactions_with_Substituted_Benzenes.txt |
When we synthesize benzene derivatives with two or more substituents the effect of directing groups must be taken into account. Often the order of reactions can change the products produced.
From benzene synthesize:
Two reactions an acylation and a bromination. In this case the two substituents are meta to each other. This means that the effects of a meta directing groups must be utilized. Of the two reactions the acylation puts a meta director on the benzene ring. The means the acylation need to come first.
If the reaction is reversed an ortho/para directing bromine is added first. The end products are different.
16.04: Halogenation of Alkyl Benzenes
The benzylic C-H bonds weaker than most sp3 hybridized C-H. This is because the radical formed from homolysis is resonance stabilized.
Resonance stabilization of the benzylic radical
Because of the weak C-H bonds, benzylic hydrogens can form benzylic halides under radical conditions.
NBS as a Bromine Source
NBS (N-bromosuccinimide) is the most commonly used reagent to produce low concentrations of bromine. When suspended in tetrachloride (CCl4), NBS reacts with trace amounts of HBr to produce a low enough concentration of bromine to facilitate the allylic bromination reaction.
Allylic Bromination Mechanism
Step 1: Initiation
Once the pre-initiation step involving NBS produces small quantities of Br2, the bromine molecules are homolytically cleaved by light to produce bromine radicals.
Step 2 and 3: Propagation
Step 4: Termination
16.05: Oxidation and Reduction of Substituted Benzenes
Oxidation of Alkyl Side-Chains
The benzylic hydrogens of alkyl substituents on a benzene ring are activated toward free radical attack, as noted earlier. Furthermore, SN1, SN2 and E1 reactions ofbenzylic halides, show enhanced reactivity, due to the adjacent aromatic ring. The possibility that these observations reflect a general benzylic activation is supported by the susceptibility of alkyl side-chains to oxidative degradation, as shown in the following examples (the oxidized side chain is colored). Such oxidations are normally effected by hot acidic pemanganate solutions, but for large scale industrial operations catalyzed air-oxidations are preferred. Interestingly, if the benzylic position is completely substituted this oxidative degradation does not occur (second equation, the substituted benzylic carbon is colored blue).
C6H5CH2CH2CH2CH3 + KMnO4 + H3O(+) & heat
C6H5CO2H + CO2
p-(CH3)3C–C6H4CH3 + KMnO4 + H3O(+) & heat
p-(CH3)3C–C6H4CO2H
These equations are not balanced. The permanganate oxidant is reduced, usually to Mn(IV) or Mn(II). Two other examples of this reaction are given below, and illustrate its usefulness in preparing substituted benzoic acids.
Reduction of Nitro Groups and Aryl Ketones
Electrophilic nitration and Friedel-Crafts acylation reactions introduce deactivating, meta-directing substituents on an aromatic ring. The attached atoms are in a high oxidation state, and their reduction converts these electron withdrawing functions into electron donating amino and alkyl groups. Reduction is easily achieved either by catalytic hydrogenation (H2 + catalyst), or with reducing metals in acid. Examples of these reductions are shown here, equation 6 demonstrating the simultaneous reduction of both functions. Note that the butylbenzene product in equation 4 cannot be generated by direct Friedel-Crafts alkylation due to carbocation rearrangement. The zinc used in ketone reductions, such as 5, is usually activated by alloying with mercury (a process known as amalgamation).
Several alternative methods for reducing nitro groups to amines are known. These include zinc or tin in dilute mineral acid, and sodium sulfide in ammonium hydroxide solution. The procedures described above are sufficient for most cases.
16.06: Multistep Synthesis
From benzene make m-bromoaniline:
In this reaction three reactions are required.
1) A nitration
2) A conversion from the nitro group to an amine
3) A bromination
Because the end product is meta a meta directing group must be utilized. Of the nitro, bromine, and amine group, only the nitro group is meta direction. This means that the first step need to be the nitration and not the bromination. Also, the conversion of the nitro group to an amine must occur last because the amine group is ortho/para direction.
From benzene make p-nitropropylbenzene :
In this reaction three reactions are required.
1) A Friedel Crafts acylation
2) A conversion from the acyl group to an alkane
3) A nitration
Because the propyl group has more than two carbons, it must be added in two steps. A Friedel Crafts acylation followed by a Clemmensen Reduction. Remeber that Friedel Crafts reactions are hindered if the benzene ring is strongly deactivated. This means that the acyl group must go on first. Because the end product is para a para directing group must be utilized. Of the nitro, acyl, and alkane group, only the alkane group is meta direction. This means that the acyl group must be converted to an alkane prior to the nitration step.
'
16.07: Electrophilic Aromatic Substitution
Bensene contains six pi electrons which are delocalized in six p orbitals above and below the plane of the benzene ring.
The six pi electrons obey Huckel's rule so benzene is especially stable. This means that the aromatic ring want to be retained during reactions. Because of this benzene does not undergo addition like other unsaturated hydrocarbons.
Benzene can undergo electrophilic aromatic substitution because aromaticity is maintained. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/16%3A_Electrophilic_Aromatic_Substitution/16.03%3A_Synthesis_of_Benzene_Derivatives.txt |
The General Mechanism
Step 1 (Slow)
The e- in the pi bond attacks the electrophile
One carbon gets a positive charge the other forms a C-E bond
This forms the arenium ion.
The arenium ion is conjugated but not aromatic.
Step 2 (Fast)
The LPE on a base attacks the hydrogen.
This causes the e- in the C-H bond to form a C-C double bond and aromaticity is reformed
A Detailed discussion of the Mechanism for Electrophilic Substitution Reactions of Benzene
A two-step mechanism has been proposed for these electrophilic substitution reactions. In the first, slow or rate-determining, step the electrophile forms a sigma-bond to the benzene ring, generating a positively charged benzenonium intermediate. In the second, fast step, a proton is removed from this intermediate, yielding a substituted benzene ring. The following four-part illustration shows this mechanism for the bromination reaction. Also, an animated diagram may be viewed.
Preliminary step: Formation of the strongly electrophilic bromine cation
Step 1: The electrophile forms a sigma-bond to the benzene ring, generating a positively charged benzenonium intermediate
Step 2: A proton is removed from this intermediate, yielding a substituted benzene ring
This mechanism for electrophilic aromatic substitution should be considered in context with other mechanisms involving carbocation intermediates. These include SN1 and E1 reactions of alkyl halides, and Brønsted acid addition reactions of alkenes.
To summarize, when carbocation intermediates are formed one can expect them to react further by one or more of the following modes:
1. The cation may bond to a nucleophile to give a substitution or addition product.
2. The cation may transfer a proton to a base, giving a double bond product.
3. The cation may rearrange to a more stable carbocation, and then react by mode #1 or #2.
SN1 and E1 reactions are respective examples of the first two modes of reaction. The second step of alkene addition reactions proceeds by the first mode, and any of these three reactions may exhibit molecular rearrangement if an initial unstable carbocation is formed. The carbocation intermediate in electrophilic aromatic substitution (the benzenonium ion) is stabilized by charge delocalization (resonance) so it is not subject to rearrangement. In principle it could react by either mode 1 or 2, but the energetic advantage of reforming an aromatic ring leads to exclusive reaction by mode 2 (ie. proton loss).
Other Examples of Electophilic Aromatic Substitution
Many other substitution reactions of benzene have been observed, the five most useful are listed below (chlorination and bromination are the most common halogenation reactions). Since the reagents and conditions employed in these reactions are electrophilic, these reactions are commonly referred to as Electrophilic Aromatic Substitution. The catalysts and co-reagents serve to generate the strong electrophilic species needed to effect the initial step of the substitution. The specific electrophile believed to function in each type of reaction is listed in the right hand column.
Reaction Type Typical Equation Electrophile E(+)
Halogenation: C6H6 + Cl2 & heat
FeCl3 catalyst
——> C6H5Cl + HCl
Chlorobenzene
Cl(+) or Br(+)
Nitration: C6H6 + HNO3 & heat
H2SO4 catalyst
——> C6H5NO2 + H2O
Nitrobenzene
NO2(+)
Sulfonation: C6H6 + H2SO4 + SO3
& heat
——> C6H5SO3H + H2O
Benzenesulfonic acid
SO3H(+)
Alkylation:
Friedel-Crafts
C6H6 + R-Cl & heat
AlCl3 catalyst
——> C6H5-R + HCl
An Arene
R(+)
Acylation:
Friedel-Crafts
C6H6 + RCOCl & heat
AlCl3 catalyst
——> C6H5COR + HCl
An Aryl Ketone
RCO(+)
Contributors
• Prof. Steven Farmer (Sonoma State University)
• William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/16%3A_Electrophilic_Aromatic_Substitution/16.08%3A_The_General_Mechanism.txt |
Halogenation is an example of electrophillic aromatic substitution. In electrophilic aromatic substitutions, a benzene is attacked by an electrophile which results in substition of hydrogens. However, halogens are not electrophillic enough to break the aromaticity of benzenes, which require a catalyst to activate.
Activation of Halogen
(where X= Br or Cl, we will discuss further in detail later why other members of the halogen family Flourine and Iodine are not used in halogenation of benzenes)
Hence, Halogen needs the help and aid of Lewis Acidic Catalysts to activate it to become a very strong electrophile. Examples of these activated halogens are Ferric Hallides (FeX3) Aluminum Halides (AlX3) where X= Br or Cl. In the following examples, the halogen we will look at is Bromine.
In the example of bromine, in order to make bromine electrophillic enough to react with benzene, we use the aid of an aluminum halide such as aluminum bromide.
With aluminum bromide as a Lewis acid, we can mix Br2 with AlBr3 to give us Br+. The presence of Br+ is a much better electrophile than Br2 alone. Bromination is acheived with the help of AlBr3 (Lewis acid catalysts) as it polarizes the Br-Br bond. The polarization causes polarization causes the bromine atoms within the Br-Br bond to become more electrophillic. The presence of Br+ compared to Br2 alone is a much better electrophille that can then react with benzene.
As the bromine has now become more electrophillic after activation of a catalyst, an electrophillic attack by the benzene occurs at the terminal bromine of Br-Br-AlBr3. This allows the other bromine atom to leave with the AlBr3 as a good leaving group, AlBr4-.
After the electrophilic attack of bromide to the benzene, the hydrogen on the same carbon as bromine substitutes the carbocation in which resulted from the attack. Hence it being an electrophilic aromatic SUBSTITUTION. Since the by-product aluminum tetrabromide is a strong nucleophile, it pulls of a proton from the Hydrogen on the same carbon as bromine.
In the end, AlBr3was not consumed by the reaction and is regenerated. It serves as our catalyst in the halogenation of benzenes.
Dissociation Energies of Halogens and its Effect on Halogenation of Benzenes
The electrophillic bromination of benzenes is an exothermic reaction. Considering the exothermic rates of aromatic halogenation decreasing down the periodic table in the Halogen family, Flourination is the most exothermic and Iodination would be the least. Being so exothermic, a reaction of flourine with benzene is explosive! For iodine, electrophillic iodination is generally endothermic, hence a reaction is often not possible. Similar to bromide, chlorination would require the aid of an activating presence such as Alumnium Chloride or Ferric Chloride. The mechanism of this reaction is the same as with Bromination of benzene.
Problems
1. What reagents would you need to get the given product?
What reagents would you need to gete given product
2. What product would result from the given reagents?
3. What is the major product given the reagents below?
4. Draw the formatin of Cl+ from AlCl3 and Cl2
5. Draw the mechanism of the reaction between Cl+ and a benzene.
Solutions
1. Cl2 and AlCl3 or Cl2 and FeCl3
2. No Reaction
3.
4.
5.
Contributors
• Catherine Nguyen | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/16%3A_Electrophilic_Aromatic_Substitution/16.09%3A_Halogenation.txt |
Nitration and sulfonation of benzene are two examples of electrophilic aromatic substitution. The nitronium ion (NO2+) and sulfur trioxide (SO3) are the electrophiles and individually react with benzene to give nitrobenzene and benzenesulfonic acid respectively.
Nitration of Benzene
The source of the nitronium ion is through the protonation of nitric acid by sulfuric acid, which causes the loss of a water molecule and formation of a nitronium ion.
Sulfuric Acid Activation of Nitric Acid
The first step in the nitration of benzene is to activate HNO3with sulfuric acid to produce a stronger electrophile, the nitronium ion.
Because the nitronium ion is a good electrophile, it is attacked by benzene to produce Nitrobenzene.
Mechanism
(Resonance forms of the intermediate can be seen in the generalized electrophilic aromatic substitution)
Sulfonation of Benzene
Sulfonation is a reversible reaction that produces benzenesulfonic acid by adding sulfur trioxide and fuming sulfuric acid. The reaction is reversed by adding hot aqueous acid to benzenesulfonic acid to produce benzene.
Mechanism
To produce benzenesulfonic acid from benzene, fuming sulfuric acid and sulfur trioxide are added. Fuming sulfuric acid, also refered to as oleum, is a concentrated solution of dissolved sulfur trioxide in sulfuric acid. The sulfur in sulfur trioxide is electrophilic because the oxygens pull electrons away from it because oxygen is very electronegative. The benzene attacks the sulfur (and subsequent proton transfers occur) to produce benzenesulfonic acid.
Reverse Sulfonation
Sulfonation of benzene is a reversible reaction. Sulfur trioxide readily reacts with water to produce sulfuric acid and heat. Therefore, by adding heat to benzenesulfonic acid in diluted aqueous sulfuric acid the reaction is reversed.
Further Applications of Nitration and Sulfonation
Nitration is used to add nitrogen to a benzene ring, which can be used further in substitution reactions. The nitro group acts as a ring deactivator. Having nitrogen present in a ring is very useful because it can be used as a directing group as well as a masked amino group. The products of aromatic nitrations are very important intermediates in industrial chemistry.
Because sulfonation is a reversible reaction, it can also be used in further substitution reactions in the form of a directing blocking group because it can be easily removed. The sulfonic group blocks the carbon from being attacked by other substituents and after the reaction is completed it can be removed by reverse sulfonation. Benzenesulfonic acids are also used in the synthesis of detergents, dyes, and sulfa drugs. Bezenesulfonyl Chloride is a precursor to sulfonamides, which are used in chemotherapy.
Outside Links
Aromatic Sulfonation
Aromatic Nitration
Problems
1. What is/are the required reagent(s)for the following reaction:
2. What is the product of the following reaction:
3. Why is it important that the nitration of benzene by nitric acid occurs in sulfuric acid?
4. Write a detailed mechanism for the sulfonation of benzene, including all resonance forms.
5. Draw an energy diagram for the nitration of benzene. Draw the intermediates, starting materials, and products. Label the transition states. (For questions 1 and 2 see Electrophilic Aromatic Substitution for hints)
For other problems involving Electrophilic Aromatic Substitution and similar reactions see:
• Electrophilic Aromatic Substitution
• Activating and Deactivating Benzene Rings
• Electrophilic Attack on Disubstituted Benzenes
Solutions
1. SO3 and H2SO4 (fuming)
2.
3. Sulfuric acid is needed in order for a good electrophile to form. Sulfuric acid protonates nitric acid to form the nitronium ion (water molecule is lost). The nitronium ion is a very good electrophile and is open to attack by benzene. Without sulfuric acid the reaction would not occur.
4.
5. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/16%3A_Electrophilic_Aromatic_Substitution/16.10%3A_Nitration_and_Sulfonation.txt |
Friedel-Crafts Alkylation
Friedel-Crafts Alkylation was first discovered by French scientist Charles Friedel and his partner, American scientist James Crafts, in 1877. This reaction allowed for the formation of alkyl benzenes from alkyl halides, but was plagued with unwanted supplemental activity that reduced its efficiency.
The mechanism takes place as follows:
Step one:
The first step creates a cabocation that acts as the electrophile in the reaction. This step activates the haloalkane. Secondary and teriary halides only form the free cabocation in the step.
Step two
The second step has an electrophilic attack on the benzene that results in multiple resonance forms. The halogen reactions with the intermediate and picks up the hydrogen to eliminate the positive charge.
Finish
The final step shown above is the results of the end of step and shows the final products.
The reactivity of haloalkanes increases as you move up the periodic table and increase polarity. This means that an RF haloalkane is most reactive followed by RCl then RBr and finally RI. This means that the Lewis acids used as catalysts in Friedel-Crafts Alkylation reactions tend have similar halogen combinations such as BF3, SbCl5, AlCl3, SbCl5, and AlBr3, all of which are commonly used in these reactions.
Some limitations of Friedel-Crafts Alkylation
There are possibilities of carbocation rearrangements when you are trying to add a carbon chain greater than two carbons. The rearrangements occur due to hydride shifts and methyl shifts. For example, the product of a Friedel-Crafts Alkylation will show an iso rearrangement when adding a three carbon chain as a substituent. Also, the reaction will only work if the ring you are adding a substituent to is not deactivated. For a look at substituents that activating or deactivating Benzene Rings.
The three key limitations of Friedel-Crafts alkylation are:
1. Carbocation Rearrangement - Only certain alkylbenzenes can be made due to the tendency of cations to rearrange.
2. Compound Limitations - Friedel-Crafts fails when used with compounds such as nitrobenzene and other strong deactivating systems.
3. Polyalkylation - Products of Friedel-Crafts are even more reactive than starting material. Alkyl groups produced in Friedel-Crafts Alkylation are electron-donating substituents meaning that the products are more susceptible to electrophilic attack than what we began with. For synthetic purposes, this is a big dissapointment.
To remedy these limitations, a new and improved reaction was devised: The Friedel-Crafts Acylation. (also known as Friedel-Crafts Alkanoylation).
Friedel-Crafts Acylation
The goal of the reaction is the following:
The very first step involves the formation of the acylium ion which will later react with benzene:
The second step involves the attack of the acylium ion on benzene as a new electrophile to form one complex:
The third step involves the departure of the proton in order for aromaticity to return to benzene:
During the third step, AlCl4 returns to remove a proton from the benzene ring, which enables the ring to return to aromaticity. In doing so, the original AlCl3 is regenerated for use again, along with HCl. Most importantly, we have the first part of the final product of the reaction, which is a ketone. Thie first part of the product is the complex with aluminum chloride as shown:
The final step involves the addition of water to liberate the final product as the acylbenzene:
Because the acylium ion (as was shown in step one) is stabilized by resonance, no rearrangement occurs (Limitation 1). Also, because of of the deactivation of the product, it is no longer susceptible to electrophilic attack and hence, is no longer susceptible to electrophilic attack and hence, no longer goes into further reactions (Limitation 3). However, as not all is perfect, Limitation 2 still prevails where Friedel-Crafts Acylation fails with strong deactivating rings.
16.12: Substituted Benzenes
When substituted benzene compounds undergo electrophilic substitution reactions of the kind discussed above, two related features must be considered:
I. The first is the relative reactivity of the compound compared with benzene itself. Experiments have shown that substituents on a benzene ring can influence reactivity in a profound manner. For example, a hydroxy or methoxy substituent increases the rate of electrophilic substitution about ten thousand fold, as illustrated by the case of anisole in the virtual demonstration (above). In contrast, a nitro substituent decreases the ring's reactivity by roughly a million. This activation or deactivation of the benzene ring toward electrophilic substitution may be correlated with the electron donating or electron withdrawing influence of the substituents, as measured by molecular dipole moments. In the following diagram we see that electron donating substituents (blue dipoles) activate the benzene ring toward electrophilic attack, and electron withdrawing substituents (red dipoles) deactivate the ring (make it less reactive to electrophilic attack).
The influence a substituent exerts on the reactivity of a benzene ring may be explained by the interaction of two effects:
The first is the inductive effect of the substituent. Most elements other than metals and carbon have a significantly greater electronegativity than hydrogen. Consequently, substituents in which nitrogen, oxygen and halogen atoms form sigma-bonds to the aromatic ring exert an inductive electron withdrawal, which deactivates the ring (left-hand diagram below).
The second effect is the result of conjugation of a substituent function with the aromatic ring. This conjugative interaction facilitates electron pair donation or withdrawal, to or from the benzene ring, in a manner different from the inductive shift. If the atom bonded to the ring has one or more non-bonding valence shell electron pairs, as do nitrogen, oxygen and the halogens, electrons may flow into the aromatic ring by p-π conjugation (resonance), as in the middle diagram. Finally, polar double and triple bonds conjugated with the benzene ring may withdraw electrons, as in the right-hand diagram. Note that in the resonance examples all the contributors are not shown. In both cases the charge distribution in the benzene ring is greatest at sites ortho and para to the substituent.
In the case of the nitrogen and oxygen activating groups displayed in the top row of the previous diagram, electron donation by resonance dominates the inductive effect and these compounds show exceptional reactivity in electrophilic substitution reactions. Although halogen atoms have non-bonding valence electron pairs that participate in p-π conjugation, their strong inductive effect predominates, and compounds such as chlorobenzene are less reactive than benzene. The three examples on the left of the bottom row (in the same diagram) are examples of electron withdrawal by conjugation to polar double or triple bonds, and in these cases the inductive effect further enhances the deactivation of the benzene ring. Alkyl substituents such as methyl increase the nucleophilicity of aromatic rings in the same fashion as they act on double bonds. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/16%3A_Electrophilic_Aromatic_Substitution/16.11%3A_FriedelCrafts_Alkylation_and_FriedelCrafts_Acylation.txt |
II. The second factor that becomes important in reactions of substituted benzenes concerns the site at which electrophilic substitution occurs. Since a mono-substituted benzene ring has two equivalent ortho-sites, two equivalent meta-sites and a unique para-site, three possible constitutional isomers may be formed in such a substitution. If reaction occurs equally well at all available sites, the expected statistical mixture of isomeric products would be 40% ortho, 40% meta and 20% para. Again we find that the nature of the substituent influences this product ratio in a dramatic fashion. Bromination of methoxybenzene (anisole) is very fast and gives mainly the para-bromo isomer, accompanied by 10% of the ortho-isomer and only a trace of the meta-isomer. Bromination of nitrobenzene requires strong heating and produces the meta-bromo isomer as the chief product.
Some additional examples of product isomer distribution in other electrophilic substitutions are given in the table below. It is important to note here that the reaction conditions for these substitution reactions are not the same, and must be adjusted to fit the reactivity of the reactant C6H5-Y. The high reactivity of anisole, for example, requires that the first two reactions be conducted under very mild conditions (low temperature and little or no catalyst). The nitrobenzene reactant in the third example is very unreactive, so rather harsh reaction conditions must be used to accomplish that reaction.
Y in C6H5–Y
Reaction
% Ortho-Product
% Meta-Product
% Para-Product
–O–CH3 Nitration 30–40 0–2 60–70
–O–CH3 F-C Acylation 5–10 0–5 90–95
–NO2 Nitration 5–8 90–95 0–5
–CH3 Nitration 55–65 1–5 35–45
–CH3 Sulfonation 30–35 5–10 60–65
–CH3 F-C Acylation 10–15 2–8 85–90
–Br Nitration 35–45 0–4 55–65
–Br Chlorination 40–45 5–10 50–60
These observations, and many others like them, have led chemists to formulate an empirical classification of the various substituent groups commonly encountered in aromatic substitution reactions. Thus, substituents that activate the benzene ring toward electrophilic attack generally direct substitution to the ortho and para locations. With some exceptions, such as the halogens, deactivating substituents direct substitution to the meta location. The following table summarizes this classification.
Orientation and Reactivity Effects of Ring Substituents
Activating Substituents
ortho & para-Orientation
Deactivating Substituents
meta-Orientation
Deactivating Substituents
ortho & para-Orientation
–O(–)
–OH
–OR
–OC6H5
–OCOCH3
–NH2
–NR2
–NHCOCH3
–R
–C6H5
–NO2
–NR3(+)
–PR3(+)
–SR2(+)
–SO3H
–SO2R
–CO2H
–CO2R
–CONH2
–CHO
–COR
–CN
–F
–Cl
–Br
–I
–CH2Cl
–CH=CHNO2
The information summarized in the above table is very useful for rationalizing and predicting the course of aromatic substitution reactions, but in practice most chemists find it desirable to understand the underlying physical principles that contribute to this empirical classification. We have already analyzed the activating or deactivating properties of substituents in terms of inductive and resonance effects, and these same factors may be used to rationalize their influence on substitution orientation.
The first thing to recognize is that the proportions of ortho, meta and para substitution in a given case reflect the relative rates of substitution at each of these sites. If we use the nitration of benzene as a reference, we can assign the rate of reaction at one of the carbons to be 1.0. Since there are six equivalent carbons in benzene, the total rate would be 6.0. If we examine the nitration of toluene, tert-butylbenzene, chlorobenzene and ethyl benzoate in the same manner, we can assign relative rates to the ortho, meta and para sites in each of these compounds. These relative rates are shown (colored red) in the following illustration, and the total rate given below each structure reflects the 2 to 1 ratio of ortho and meta sites to the para position. The overall relative rates of reaction, referenced to benzene as 1.0, are calculated by dividing by six. Clearly, the alkyl substituents activate the benzene ring in the nitration reaction, and the chlorine and ester substituents deactivate the ring.
From rate data of this kind, it is a simple matter to calculate the proportions of the three substitution isomers. Toluene gives 58.5% ortho-nitrotoluene, 37% para-nitrotoluene and only 4.5% of the meta isomer. The increased bulk of the tert-butyl group hinders attack at the ortho-sites, the overall product mixture being 16% ortho, 8% meta and 75% para-nitro product. Although chlorobenzene is much less reactive than benzene, the rate of ortho and para-substitution greatly exceeds that of meta-substitution, giving a product mixture of 30% ortho and 70% para-nitrochlorobenzene. Finally, the benzoic ester gave predominantly the meta-nitro product (73%) accompanied by the ortho (22%) and para (5%) isomers, as shown by the relative rates. Equivalent rate and product studies for other substitution reactions lead to similar conclusions. For example, electrophilic chlorination of toluene occurs hundreds of times faster than chlorination of benzene, but the relative rates are such that the products are 60% ortho-chlorotoluene, 39% para and 1% meta-isomers, a ratio similar to that observed for nitration.
The manner in which specific substituents influence the orientation of electrophilic substitution of a benzene ring is shown in the following interactive diagram. As noted on the opening illustration, the product-determining step in the substitution mechanism is the first step, which is also the slow or rate determining step. It is not surprising, therefore, that there is a rough correlation between the rate-enhancing effect of a substituent and its site directing influence. The exact influence of a given substituent is best seen by looking at its interactions with the delocalized positive charge on the benzenonium intermediates generated by bonding to the electrophile at each of the three substitution sites. This can be done for seven representative substituents by using the selection buttons underneath the diagram.
In the case of alkyl substituents, charge stabilization is greatest when the alkyl group is bonded to one of the positively charged carbons of the benzenonium intermediate. This happens only for ortho and para electrophilic attack, so such substituents favor formation of those products. Interestingly, primary alkyl substituents, especially methyl, provide greater stabilization of an adjacent charge than do more substituted groups (note the greater reactivity of toluene compared with tert-butylbenzene).
Nitro (NO2), sulfonic acid (SO3H) and carbonyl (C=O) substituents have a full or partial positive charge on the atom bonded to the aromatic ring. Structures in which like-charges are close to each other are destabilized by charge repulsion, so these substituents inhibit ortho and para substitution more than meta substitution. Consequently, meta-products predominate when electrophilic substitution is forced to occur.
Halogen ( X ), OR and NR2 substituents all exert a destabilizing inductive effect on an adjacent positive charge, due to the high electronegativity of the substituent atoms. By itself, this would favor meta-substitution; however, these substituent atoms all have non-bonding valence electron pairs which serve to stabilize an adjacent positive charge by pi-bonding, with resulting delocalization of charge. Consequently, all these substituents direct substitution to ortho and para sites. The balance between inductive electron withdrawal and p-π conjugation is such that the nitrogen and oxygen substituents have an overall stabilizing influence on the benzenonium intermediate and increase the rate of substitution markedly; whereas halogen substituents have an overall destabilizing influence. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/16%3A_Electrophilic_Aromatic_Substitution/16.13%3A_Electrophilic_Aromatic_Substitution_of_Substituted_Benzenes.txt |
The manner in which specific substituents influence the orientation of electrophilic substitution of a benzene ring is shown in the following interactive diagram. As noted on the opening illustration, the product-determining step in the substitution mechanism is the first step, which is also the slow or rate determining step. It is not surprising, therefore, that there is a rough correlation between the rate-enhancing effect of a substituent and its site directing influence. The exact influence of a given substituent is best seen by looking at its interactions with the delocalized positive charge on the benzenonium intermediates generated by bonding to the electrophile at each of the three substitution sites. This can be done for seven representative substituents by using the selection buttons underneath the diagram.
In the case of alkyl substituents, charge stabilization is greatest when the alkyl group is bonded to one of the positively charged carbons of the benzenonium intermediate. This happens only for ortho and para electrophilic attack, so such substituents favor formation of those products. Interestingly, primary alkyl substituents, especially methyl, provide greater stabilization of an adjacent charge than do more substituted groups (note the greater reactivity of toluene compared with tert-butylbenzene).
Nitro (NO2), sulfonic acid (SO3H) and carbonyl (C=O) substituents have a full or partial positive charge on the atom bonded to the aromatic ring. Structures in which like-charges are close to each other are destabilized by charge repulsion, so these substituents inhibit ortho and para substitution more than meta substitution. Consequently, meta-products predominate when electrophilic substitution is forced to occur.
Halogen ( X ), OR and NR2 substituents all exert a destabilizing inductive effect on an adjacent positive charge, due to the high electronegativity of the substituent atoms. By itself, this would favor meta-substitution; however, these substituent atoms all have non-bonding valence electron pairs which serve to stabilize an adjacent positive charge by pi-bonding, with resulting delocalization of charge. Consequently, all these substituents direct substitution to ortho and para sites. The balance between inductive electron withdrawal and p-π conjugation is such that the nitrogen and oxygen substituents have an overall stabilizing influence on the benzenonium intermediate and increase the rate of substitution markedly; whereas halogen substituents have an overall destabilizing influence.
16.15: Orientation Effects in Substituted Benzenes
Substituted rings are divided into two groups based on the type of the substituent that the ring carries:
• Activated rings: the substituents on the ring are groups that donate electrons.
• Deactivated rings: the substituents on the ring are groups that withdraw electrons.
Introduction
Examples of activating groups in the relative order from the most activating group to the least activating:
-NH2, -NR2 > -OH, -OR> -NHCOR> -CH3 and other alkyl groups
with R as alkyl groups (CnH2n+1)
Examples of deactivating groups in the relative order from the most deactivating to the least deactivating:
-NO2, -CF3> -COR, -CN, -CO2R, -SO3H > Halogens
with R as alkyl groups (CnH2n+1)
The order of reactivity among Halogens from the more reactive (least deactivating substituent) to the least reactive (most deactivating substituent) halogen is:
F> Cl > Br > I
The order of reactivity of the benzene rings toward the electrophilic substitution when it is substituted with a halogen groups, follows the order of electronegativity. The ring that is substituted with the most electronegative halogen is the most reactive ring ( less deactivating substituent ) and the ring that is substituted with the least electronegatvie halogen is the least reactive ring ( more deactivating substituent ), when we compare rings with halogen substituents. Also the size of the halogen effects the reactivity of the benzene ring that the halogen is attached to. As the size of the halogen increase, the reactivity of the ring decreases.
The direction of the reaction
The activating group directs the reaction to the ortho or para position, which means the electrophile substitute the hydrogen that is on carbon 2 or carbon 4. The deactivating group directs the reaction to the meta position, which means the electrophile substitute the hydrogen that is on carbon 3 with the exception of the halogens that is a deactivating group but directs the ortho or para substitution.
Substituents determine the reaction direction by resonance or inductive effect
Resonance effect is the conjugation between the ring and the substituent, which means the delocalizing of the $\pi$ electrons between the ring and the substituent. Inductive effect is the withdraw of the sigma ( the single bond ) electrons away from the ring toward the substituent, due to the higher electronegativity of the substituent compared to the carbon of the ring.
Activating groups (ortho or para directors)
When the substituents like -OH have an unshared pair of electrons, the resonance effect is stronger than the inductive effect which make these substituents stronger activators, since this resonance effect direct the electron toward the ring. In cases where the subtituents is esters or amides, they are less activating because they form resonance structure that pull the electron density away from the ring.
By looking at the mechanism above, we can see how groups donating electron direct the ortho, para electrophilic substition. Since the electrons locatinn transfer between the ortho and para carbons, then the electrophile prefer attacking the carbon that has the free electron.
Inductive effect of alkyl groups activates the direction of the ortho or para substitution, which is when s electrons gets pushed toward the ring.
Deactivating group (meta directors)
The deactivating groups deactivate the ring by the inductive effect in the presence of an electronegative atom that withdraws the electrons away from the ring.
we can see from the mechanism above that when there is an electron withdraw from the ring, that leaves the carbons at the ortho, para positions with a positive charge which is unfavorable for the electrophile, so the electrophile attacks the carbon at the meta positions.
Halogens are an exception of the deactivating group that directs the ortho or para substitution. The halogens deactivate the ring by inductive effect not by the resonance even though they have an unpaired pair of electrons. The unpaired pair of electrons gets donated to the ring, but the inductive effect pulls away the s electrons from the ring by the electronegativity of the halogens.
Substituents determine the reactivity of rings
The reaction of a substituted ring with an activating group is faster than benzene. On the other hand, a substituted ring with a deactivated group is slower than benzene.
Activating groups speed up the reaction because of the resonance effect. The presence of the unpaired electrons that can be donated to the ring, stabilize the carbocation in the transition state. Thus; stabilizing the intermediate step, speeds up the reaction; and this is due to the decrease of the activating energy. On the other hand, the deactivating groups, withdraw the electrons away from the carbocation formed in the intermediate step, thus; the activation energy is increased which slows down the reaction.
The CH3Group is and ortho, para Director
Alkyl groups are Inductive activators
With o/p attack the form a tertiary arenium carbocation which speeds up the reaction
The O-CH3 Group is an ortho, para Director
Ortho and Para producst produces a resonance structure which stabilizes the arenium ion. This causes the ortho and para products for form faster than meta. Generally, the para product is preferred because of steric effects.
Acyl groups are meta Directors
Acyl groups are resonance deactivators. Ortho and para attack produces a resonance structure which places the arenium cation next to and additional cation. This destabilizes the arenium cation and slows down ortho and para reaction. By default the meta product forms faster because it lacks this destablizing resonance structure.
Problems
1. Predict the direction of the electrophile substition on these rings:
2. Which nitration product is going to form faster?
nitration of aniline or nitration of nitrobenzene?
3. Predict the product of the following two sulfonation reactions:
A.
B.
4. Classify these two groups as activating or deactivating groups:
A. alcohol
B. ester
5. By which effect does trichloride effect a monosubstituted ring?
Answers
1. The first substitution is going to be ortho and/or para substitution since we have a halogen subtituent. The second substition is going to be ortho and/or para substitution also since we have an alkyl substituent.
2. The nitration of aniline is going to be faster than the nitration of nitrobenzene, since the aniline is a ring with NH2 substituent and nitrobenzene is a ring with NO2 substiuent. As described above NH2is an activating group which speeds up the reaction and NO2 is deactivating group that slows down the reaction.
3. A. the product is
B. the product is
4. A. alcohol is an activating group.
B. ester is a deactivating group.
5. Trichloride deactivate a monosubstitued ring by inductive effect. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/16%3A_Electrophilic_Aromatic_Substitution/16.14%3A_Why_Substituents_Activate_or_Deactivate_a_Benzene_Ring.txt |
The resonance effect described here is undoubtedly the major contributor to the exceptional acidity of carboxylic acids. However, inductive effects also play a role. For example, alcohols have pKa's of 16 or greater but their acidity is increased by electron withdrawing substituents on the alkyl group. The following diagram illustrates this factor for several simple inorganic and organic compounds (row #1), and shows how inductive electron withdrawal may also increase the acidity of carboxylic acids (rows #2 & 3). The acidic hydrogen is colored red in all examples.
Water is less acidic than hydrogen peroxide because hydrogen is less electronegative than oxygen, and the covalent bond joining these atoms is polarized in the manner shown. Alcohols are slightly less acidic than water, due to the poor electronegativity of carbon, but chloral hydrate, Cl3CCH(OH)2, and 2,2,2,-trifluoroethanol are significantly more acidic than water, due to inductive electron withdrawal by the electronegative halogens (and the second oxygen in chloral hydrate). In the case of carboxylic acids, if the electrophilic character of the carbonyl carbon is decreased the acidity of the carboxylic acid will also decrease. Similarly, an increase in its electrophilicity will increase the acidity of the acid. Acetic acid is ten times weaker an acid than formic acid (first two entries in the second row), confirming the electron donating character of an alkyl group relative to hydrogen, as noted earlier in a discussion of carbocation stability. Electronegative substituents increase acidity by inductive electron withdrawal. As expected, the higher the electronegativity of the substituent the greater the increase in acidity (F > Cl > Br > I), and the closer the substituent is to the carboxyl group the greater is its effect (isomers in the 3rd row). Substituents also influence the acidity of benzoic acid derivatives, but resonance effects compete with inductive effects. The methoxy group is electron donating and the nitro group is electron withdrawing (last three entries in the table of pKa values).
For additional information about substituent effects on the acidity of carboxylic acids Click Here
17.02: Substituted Benzoic Acids
Electron-withdrawing groups
The conjugate base of benzoic acid is destabilized by electron-donating groups. This makes the acid less acidic
Electron-withdrawing groups deactivate the benzene ring to electrophilic attack and make benzoic acids more acidic.
Electron-donating groups
The conjugate base of benzoic acid is stabilized by electron-withdrawing groups. This makes the acid more acidic
Electron-withdrawing groups activate the benzene ring to electrophilic attack and make benzoic acids less acidic.
17.03: Extraction
An acid-base extraction is a type of liquid-liquid extraction. It typically involves different solubility levels in water and an organic solvent. The organic solvent may be any carbon-based liqiuid that does not dissolve very well in water; common ones are ether, ethyl acetate, or dichloromethane.
Acid-base extraction is typically used to separate organic compounds from each other based on their acid-base properties. The method rests on the assumption that most organic compounds are more soluble in organic solvents than they are in water. However, if the organic compound is rendered ionic, it becomes more soluble in water than in the organic solvent. These compounds can easily be made into ions either by adding a proton (an H+ ion), making the compound into a positive ion, or by removing a proton, making the compound into a negative ion.
Suppose you have a mixture of two compounds. There is a neutral one which doesn't react with any acids or bases. There is also a basic one, which reacts with acids by picking up a proton. In this case, a proton might be added via reaction with a strong mineral acid (represented by HX in the drawing). Suppose an aqueous solution of mineral acid, such as HCl, were shaken vigorously with an ethereal solution of an organic base and an organic neutral. The proton would be transferred to a basic compound, but not to a neutral one. The basic compound would become ionic, and more water-soluble.
Note that in the drawing, the ether is represented in yellow, whereas the water is shown in blue. The water is on the bottom in this case because water has a higher density than ether, so it will sink to the bottom (along with anything dissolved in it). Some organic solvents do have a higher density than water, so the aqueous solution would float to the top in those case.
As a result, the ethereal solution would contain only the neutral compound, not the basic one. The neutral compound could be isolated simply by evaporating the ether.
However, as a practical matter, the ether would have to be dried first. What's the difference between evaporating and drying? Have you ever been to the beach or taken a shower? Drying refers to the removal of water. This step is necessary because ether tends to dissolve a lot of water in it. Once the ether has been evaporated, there would be some neutral compound, but it would be mixed with water.
Water removal is most easily done by adding a drying agent, such as magnesium sulfate or sodium sulfate. The water sticks to these solids, which are then filtered off.
Now the neutral compound is alone in the ether. Evaporation of the ether gives the pure, neutral compound.
However, the basic compound is stuck in the water, and it isn't the same compound anymore. It's an ion, now. If we want the original compound in a pure form, we need to take that proton away. That can be done by adding a mineral base, such as sodium hydroxide.
The mineral base will remove the proton, leaving the original organic compound. The organic compound is uncharged and not as soluble in water anymore. It will go back into the ether layer.
Conversely, we might have a mixture of an acidic organic compound and a neutral compound to start out with. In that case, we would add a mineral base in the first place, to take a proton away from the acidic compound. The mineral base might be something like sodium hydroxide or sodium bicarbonate. In the drawing, it is just represented as Na+ B-.
The acidic compound becomes ionic and water-soluble when it loses a proton. That leaves the neutral compound alone.
To get that acidic compound back, we would add a mineral acid such as hydrochloric acid in order to restore the missing proton.
Just as in the other case, the ether layer containing a pure compound could be separated, dried and evaporated in order to provide the pure compound.
Acidity
But how do we know whether something is an organic acid or a base? Common structural features of organic acids and bases are displayed below.
Note that the terms, "strong acid" and "weak acid", are relative with organic compounds. Sometimes, the term, "strong acid", designates a compound that completely ionizes in solution, so that it automatically gives up a H+ ion and forms an ionic compound. Hydrochloric acid, HCl, in water is a good example. That isn't true here; none of these acids ionize very easily on their own, and they appear in solution just as they do above, with just a small minority of molecules forming H+ and an anion. In this case, the term just compares one group of acidic compounds (called carboxylic acids) to another group of acidic compounds (called phenols). Carboxylic acids are more likely to give up protons than are phenols, so carboxylic acids are referred to in this context as "strong" and phenols as "weak".
The carboxylic acid group contains a C=O (a carbonyl) with an additional OH group attached to the carbon. Examples are shown below.
When carboxylic acids are treated with mineral bases such as sodium hydroxide, the carboxylic OH group gives up a proton to the hydroxide, forming a water molecule. The electrons in the O-H bond stay behind, putting a negative charge on the resulting carboxylate anion. The salt that forms is much more water soluble.
This reaction is completely reversible. A mineral acid, such as HCl, could provide protons to the carboxylate anion. The carboxylate ion would use a pair of electrons to bind to a proton, and the compound would become a neutral (as in uncharged) carboxylic acid again.
Phenols also contain an OH group, but instead of being attached to a C=O group, the OH is attached to a benzene (a six-carbon ring with three double bonds). Examples are shown below.
Phenols react with bases in the same way as do carboxylic acids, just not so as easily.
Because phenols do not react as easily as carboxylic acids, there are situations in which a carboxylic acid would react with a base but a phenol would not. For example, carboxylic acids react even with weak bases such as sodium bicarbonate (baking soda).
Phenols, on the other hand, do no such thing.
If the OH is attached to a carbon in an organic compound, but it is not attached to either a C=O or a benzene ring, it is not acidic enough to be removed to an appreciable extent. That is true even if there is a carbonyl or a benzene somewhere else in the molecule. As a result, acid-base extraction is not possible in these cases.
Organic bases are compounds that contain nitrogen atoms. In order to be basic, the nitrogen atom must have a lone pair. The lone pair is needed in order to make a bond with the proton.
Once the lone pair has donated to the proton to form a bond with it, the nitrogen compound becomes positively charged. It then becomes more water-soluble.
If the nitrogen does not have a lone pair, it is unable to bond to a proton. However, some compounds that do have a lone pair on the nitrogen still can't donate their lone pair to make a bond to the hydrogen. Most often that's because of a very electronegative oxygen atom nearby. The attraction of the oxygen for the lone pair makes the lone pair less able to donate to another atom. There can also be other reasons, especially involving electron delocalization or aromaticity that makes the lone pair unavailable for bonding.
17.04: Sulfonic Acids
Sulfonic acids are similar to carboxylic acids and have the general structure of RSO3H. Sulfonic acids are very strong acids (pKa ~ -7). The most common sulfonic acid is p-toluenesulfonic acid.
The conjugate bases of sulfonic acids are called sulfonate anions and are resonance stabilized. Consequently, sulfonate anions make good leaving groups.
The sulfonate anion has three major resonance structures. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/17%3A_Carboxylic_Acids_and_the_Acidity_of_the_OH_Bond/17.01%3A_Inductive_Effects_in_Aliphatic_Carboxylic_Acids.txt |
Common amino acids
There are 20 common amino acids. They are composed of C, H, O, N and S atoms. They are structurally and chemically different, and also differ in size and volume. Some are branched structures, some are linear, some have ring structures. One of the 20 common amino acids is actually an imino acid. A typical grouping of their chemical nature is as follows:
• Nonpolar (hydrocarbons and one sulfur-containing amino acid). Dispersion forces and hydrophobic effects predominate in their interactions. They cannot H-bond with water and these side chains have a characteristic hydrophobic effect in water.
• Polar uncharged. Contain functional groups that can H-bond with water and other amino acids. Include C, H, O, N and S atoms.
• Acidic. Contain a carboxylic acid functional group with a negative charge at neutral pH. Can H-bond with water, can form ionic interactions, and can also serve as nucleophiles or participate in acid-base chemistry.
• Basic. Nitrogen containing bases (e.g. guanidino, imidazole or amino groups) with a net positive charge at neutral pH. Can serve as proton donors in chemical reactions, and form ionic interactions.
The amino acids have a name, as well as a three letter or single letter mnemonic code:
Strereochemistry in amino acids
With the exception of glycine, all the 19 other common amino acids have a uniquely different functional group on the central tetrahedral alpha carbon (i.e. Ca )
• The C a is termed "chiral" to indicate there are four different constituents and that the Ca is asymmetric
• Since the C a is asymmetric there exists two possible, non-superimposable, mirror images of the amino acids:
• How are these two uniquely different structures distinguished?
The D, L system
Glyceraldehyde contains a chiral carbon, and therefore, there are two enantiomers of this molecule.
• One is labeled the "L" form, and the other the "D" form
• This is the frame of reference used to describe amino acid enantiomers as being either the "L" or "D" form
Even though the two enantiomers would seem to be essentially equivalent to each other, all common amino acids are found in the "L" enantiomer in living systems
• When looking down the H-C
• a bond towards the Ca there is a mnemonic to identify the L-enantiomer of amino acids (note: in this view the three functional groups are pointing away from you, and not towards you; the H atom is omitted for clarity - but it would be in front of the C)
• Starting with the carbonyl functional group, and going clockwise around the C
• a of the L-enantiomer, the three functional groups spell out the word CORN.
• If you follow the above instructions, it will spell out CONR (a silly, meaningless word) for the D-enantiomer
The R,S system of naming chiral centers
• A relative ranking of the "priority" of various functional groups is given as:
SH > OH > NH2 > COOH > CHO > CH2OH > CH3 > H
• A chiral center has four different functional groups. Identify the functional group with the lowest priority
• View the chiral center down the bond from the chiral center to the lowest priority atom
• don't confuse this with the CORN mnemonic method of identifying the L-amino acid chirality by viewing from the H to the Ca )
• Assign priorities to the three other functional groups connected to the chiral center, using the above ranking
• If the priorities of these other groups goes in a clockwise rotation, the chirality is "R". If the priorities of these other groups goes counterclockwise, the chirality is "S". (Note that this assignment has nothing to do with optical activity, and is not using L-glyceraldehyde as a reference molecule)
Spectroscopic properties of amino acids
This refers to the ability of amino acids to absorb or emit electromagnetic energy at different wavelengths (i.e. energies)
• No amino acids absorb light in the visible spectrum (i.e. they are "colorless").
• If proteins have color (e.g. hemoglobin is red) it is because they contain a bound, non-protein atom, ion or molecule; iron in this case)
• All amino acids absorb in the infrared region (longer wavelengths, weaker energy than visible light)
• Some amino acids absorb in the ultraviolet spectrum (shorter wavelengths, higher energy than visible light)
• Absorption occurs as electrons rise to higher energy states
• Electrons in aromatic ring structures absorb in the u.v. spectrum. Such structures comprise the side chains of
• tryptophan, tyrosine and phenylalanine.
Amino acids as zwitterions
An amino acid has both a basic amine group and an acidic carboxylic acid group.
There is an internal transfer of a hydrogen ion from the -COOH group to the -NH2 group to leave an ion with both a negative charge and a positive charge. This is called a zwitterion.
This is the form that amino acids exist in even in the solid state. If you dissolve the amino acid in water, a simple solution also contains this ion. A zwitterion is a compound with no overall electrical charge, but which contains separate parts which are positively and negatively charged.
Adding an alkali to an amino acid solution
If you increase the pH of a solution of an amino acid by adding hydroxide ions, the hydrogen ion is removed from the -NH3+ group.
You could show that the amino acid now existed as a negative ion using electrophoresis. In its simplest form, electrophoresis can just consist of a piece of moistened filter paper on a microscope slide with a crocodile clip at each end attached to a battery. A drop of amino acid solution is placed in the center of the paper.
Although the amino acid solution is colourless, its position after a time can be found by spraying it with a solution of ninhydrin. If the paper is allowed to dry and then heated gently, the amino acid shows up as a coloured spot. The amino acid would be found to travel towards the anode (the positive electrode).
Adding an acid to an amino acid solution
If you decrease the pH by adding an acid to a solution of an amino acid, the -COO- part of the zwitterion picks up a hydrogen ion.
This time, during electrophoresis, the amino acid would move towards the cathode (the negative electrode).
Shifting the pH from one extreme to the other
Suppose you start with the ion we've just produced under acidic conditions and slowly add alkali to it. That ion contains two acidic hydrogens - the one in the -COOH group and the one in the -NH3+ group. The more acidic of these is the one in the -COOH group, and so that is removed first - and you get back to the zwitterion.
So when you have added just the right amount of alkali, the amino acid no longer has a net positive or negative charge. That means that it wouldn't move towards either the cathode or anode during electrophoresis. The pH at which this lack of movement during electrophoresis happens is known as the isoelectric point of the amino acid. This pH varies from amino acid to amino acid. If you go on adding hydroxide ions, you will get the reaction we've already seen, in which a hydrogen ion is removed from the -NH3+ group.
You can, of course, reverse the whole process by adding an acid to the ion we've just finished up with. That ion contains two basic groups - the -NH2 group and the -COO- group. The -NH2 group is the stronger base, and so picks up hydrogen ions first. That leads you back to the zwitterion again.
. . . and, of course, you can keep going by then adding a hydrogen ion to the -COO- group.
The Isoelectric Point
As defined above, the isoelectric point, pI, is the pH of an aqueous solution of an amino acid (or peptide) at which the molecules on average have no net charge. In other words, the positively charged groups are exactly balanced by the negatively charged groups. For simple amino acids such as alanine, the pI is an average of the pKa's of the carboxyl (2.34) and ammonium (9.69) groups. Thus, the pI for alanine is calculated to be: (2.34 + 9.69)/2 = 6.02, the experimentally determined value. If additional acidic or basic groups are present as side-chain functions, the pI is the average of the pKa's of the two most similar acids. To assist in determining similarity we define two classes of acids. The first consists of acids that are neutral in their protonated form (e.g. CO2H & SH). The second includes acids that are positively charged in their protonated state (e.g. -NH3+). In the case of aspartic acid, the similar acids are the alpha-carboxyl function (pKa = 2.1) and the side-chain carboxyl function (pKa = 3.9), so pI = (2.1 + 3.9)/2 = 3.0. For arginine, the similar acids are the guanidinium species on the side-chain (pKa = 12.5) and the alpha-ammonium function (pKa = 9.0), so the calculated pI = (12.5 + 9.0)/2 = 10.75.
Why isn't the isoelectric point of an amino acid at pH 7?
When an amino acid dissolves in water, the situation is a little bit more complicated than we tend to pretend at this level. The zwitterion interacts with water molecules - acting as both an acid and a base. As an acid:
The -NH3+ group is a weak acid and donates a hydrogen ion to a water molecule. Because it is only a weak acid, the position of equilibrium will lie to the left.
As a base:
The -COO- group is a weak base and takes a hydrogen ion from a water molecule. Again, the equilibrium lies to the left.
When you dissolve an amino acid in water, both of these reactions are happening. However, the positions of the two equilibria aren't identical - they vary depending on the influence of the "R" group. In practice, for the simple amino acids we have been talking about, the position of the first equilibrium lies a bit further to the right than the second one. That means that there will be rather more of the negative ion from the amino acid in the solution than the positive one.
In those circumstances, if you carried out electrophoresis on the unmodified solution, there would be a slight drift of amino acid towards the positive electrode (the anode). To stop that, you need to cut down the amount of the negative ion so that the concentrations of the two ions are identical. You can do that by adding a very small amount of acid to the solution, moving the position of the first equilibrium further to the left. Typically, the pH has to be lowered to about 6 to achieve this. For glycine, for example, the isoelectric point is pH 6.07; for alanine, 6.11; and for serine, 5.68.
Contributors
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/17%3A_Carboxylic_Acids_and_the_Acidity_of_the_OH_Bond/17.05%3A_Amino_Acids.txt |
Structure of the carboxyl acid group
Carboxylic acids are organic compounds which incorporate a carboxyl functional group, CO2H. The name carboxyl comes from the fact that a carbonyl and a hydroxyl group are attached to the same carbon.
The carbon and oxygen in the carbonyl are both sp2 hybridized which give a carbonyl group a basic trigonal shape. The hydroxyl oxygen is also sp2 hybridized which allows one of its lone pair electrons to conjugate with the pi system of the carbonyl group. This make the carboxyl group planar an can represented with the following resonance structure.
Carboxylic acids are named such because they can donate a hydrogen to produce a carboxylate ion. The factors which affect the acidity of carboxylic acids will be discussed later.
17.07: Nomenclature
The IUPAC system of nomenclature assigns a characteristic suffix to these classes. The –e ending is removed from the name of the parent chain and is replaced -anoic acid. Since a carboxylic acid group must always lie at the end of a carbon chain, it is always is given the #1 location position in numbering and it is not necessary to include it in the name.
Many carboxylic acids are called by the common names. These names were chosen by chemists to usually describe a source of where the compound is found. In common names of aldehydes, carbon atoms near the carboxyl group are often designated by Greek letters. The atom adjacent to the carbonyl function is alpha, the next removed is beta and so on.
Formula
Common Name
Source
IUPAC Name
Melting Point
Boiling Point
HCO2H formic acid ants (L. formica) methanoic acid 8.4 ºC 101 ºC
CH3CO2H acetic acid vinegar (L. acetum) ethanoic acid 16.6 ºC 118 ºC
CH3CH2CO2H propionic acid milk (Gk. protus prion) propanoic acid -20.8 ºC 141 ºC
CH3(CH2)2CO2H butyric acid butter (L. butyrum) butanoic acid -5.5 ºC 164 ºC
CH3(CH2)3CO2H valeric acid valerian root pentanoic acid -34.5 ºC 186 ºC
CH3(CH2)4CO2H caproic acid goats (L. caper) hexanoic acid -4.0 ºC 205 ºC
CH3(CH2)5CO2H enanthic acid vines (Gk. oenanthe) heptanoic acid -7.5 ºC 223 ºC
CH3(CH2)6CO2H caprylic acid goats (L. caper) octanoic acid 16.3 ºC 239 ºC
CH3(CH2)7CO2H pelargonic acid pelargonium (an herb) nonanoic acid 12.0 ºC 253 ºC
CH3(CH2)8CO2H capric acid goats (L. caper) decanoic acid 31.0 ºC 219 ºC
Example (Common Names Are in Red)
Naming carboxyl groups added to a ring
When a carboxyl group is added to a ring the suffix -carboxylic acid is added to the name of the cyclic compound. The ring carbon attached to the carboxyl group is given the #1 location number.
Naming carboxylates
Salts of carboxylic acids are named by writing the name of the cation followed by the name of the acid with the –ic acid ending replaced by an –ate ending. This is true for both the IUPAC and Common nomenclature systems.
Naming carboxylic acids which contain other functional groups
Carboxylic acids are given the highest nomenclature priority by the IUPAC system. This means that the carboxyl group is given the lowest possible location number and the appropriate nomenclature suffix is included. In the case of molecules containing carboxylic acid and alcohol functional groups the OH is named as a hydroxyl substituent. However, the l in hydroxyl is generally removed.
In the case of molecules containing a carboxylic acid and aldehydes and/or ketones functional groups the carbonyl is named as a "Oxo" substituent.
In the case of molecules containing a carboxylic acid an amine functional group the amine is named as an "amino" substituent.
When carboxylic acids are included with an alkene the following order is followed:
(Location number of the alkene)-(Prefix name for the longest carbon chain minus the -ane ending)-(an –enoic acid ending to indicate the presence of an alkene and carboxylic acid)
Remember that the carboxylic acid has priority so it should get the lowest possible location number. Also, remember that cis/tran or E/Z nomenclature for the alkene needs to be included if necessary.
Naming dicarboxylic acids
For dicarboxylic acids the location numbers for both carboxyl groups are omitted because both functional groups are expected to occupy the ends of the parent chain. The ending –dioic acid is added to the end of the parent chain. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/17%3A_Carboxylic_Acids_and_the_Acidity_of_the_OH_Bond/17.06%3A_Structure_and_Bonding.txt |
Physical Properties of Some Carboxylic Acids
The table at the beginning of this page gave the melting and boiling points for a homologous group of carboxylic acids having from one to ten carbon atoms. The boiling points increased with size in a regular manner, but the melting points did not. Unbranched acids made up of an even number of carbon atoms have melting points higher than the odd numbered homologs having one more or one less carbon. This reflects differences in intermolecular attractive forces in the crystalline state. In the table of fatty acids we see that the presence of a cis-double bond significantly lowers the melting point of a compound. Thus, palmitoleic acid melts over 60º lower than palmitic acid, and similar decreases occur for the C18 and C20 compounds. Again, changes in crystal packing and intermolecular forces are responsible.
The factors that influence the relative boiling points and water solubilities of various types of compounds were discussed earlier. In general, dipolar attractive forces between molecules act to increase the boiling point of a given compound, with hydrogen bonds being an extreme example. Hydrogen bonding is also a major factor in the water solubility of covalent compounds To refresh your understanding of these principles Click Here. The following table lists a few examples of these properties for some similar sized polar compounds (the non-polar hydrocarbon hexane is provided for comparison).
Physical Properties of Some Organic Compounds
Formula
IUPAC Name
Molecular Weight
Boiling Point
Water Solubility
CH3(CH2)2CO2H butanoic acid 88 164 ºC very soluble
CH3(CH2)4OH 1-pentanol 88 138 ºC slightly soluble
CH3(CH2)3CHO pentanal 86 103 ºC slightly soluble
CH3CO2C2H5 ethyl ethanoate 88 77 ºC moderately soluble
CH3CH2CO2CH3 methyl propanoate 88 80 ºC slightly soluble
CH3(CH2)2CONH2 butanamide 87 216 ºC soluble
CH3CON(CH3)2 N,N-dimethylethanamide 87 165 ºC very soluble
CH3(CH2)4NH2 1-aminobutane 87 103 ºC very soluble
CH3(CH2)3CN pentanenitrile 83 140 ºC slightly soluble
CH3(CH2)4CH3 hexane 86 69 ºC insoluble
The first five entries all have oxygen functional groups, and the relatively high boiling points of the first two is clearly due to hydrogen bonding. Carboxylic acids have exceptionally high boiling points, due in large part to dimeric associations involving two hydrogen bonds. A structural formula for the dimer of acetic acid is shown here. When the mouse pointer passes over the drawing, an electron cloud diagram will appear. The high boiling points of the amides and nitriles are due in large part to strong dipole attractions, supplemented in some cases by hydrogen bonding.
17.09: Spectroscopic Properties
IR
The carboxyl group is associated with two characteristic infrared stretching absorptions which change markedly with hydrogen bonding. The spectrum of a CCl4 solution of propionic acid (propanoic acid), shown below, is illustrative. Carboxylic acids exist predominantly as hydrogen bonded dimers in condensed phases. The O-H stretching absorption for such dimers is very strong and broad, extending from 2500 to 3300 cm-1. This absorption overlaps the sharper C-H stretching peaks, which may be seen extending beyond the O-H envelope at 2990, 2950 and 2870 cm-1. The smaller peaks protruding near 2655 and 2560 are characteristic of the dimer. In ether solvents a sharper hydrogen bonded monomer absorption near 3500 cm-1 is observed, due to competition of the ether oxygen as a hydrogen bond acceptor. The carbonyl stretching frequency of the dimer is found near 1710 cm-1, but is increased by 25 cm-1 or more in the monomeric state. Other characteristic stretching and bending absorptions are marked in the spectrum.
NMR
The combination of anisotropy and electronegativity causes the O-H hydrogen in a carboxylic acid to be very deshielded.
Hydrogen environments adjacent to a carboxylic acid are shifted to the region of 2.5-3.0 ppm.Deshielding occurs due to the fact that the sp2 hybridized carbon the the carboxylic acid is more electronegative than a sp3 hybridized carbon.
17.10: Interesting Carboxylic Acids
Carboxylic acids are widespread in nature, often combined with other functional groups. Simple alkyl carboxylic acids, composed of four to ten carbon atoms, are liquids or low melting solids having very unpleasant odors. The fatty acids are important components of the biomolecules known as lipids, especially fats and oils. As shown in the following table, these long-chain carboxylic acids are usually referred to by their common names, which in most cases reflect their sources. A mnemonic phrase for the C10 to C20 natural fatty acids capric, lauric, myristic, palmitic, stearic and arachidic is: "Curly, Larry & Moe Perform Silly Antics" (note that the names of the three stooges are in alphabetical order).
Interestingly, the molecules of most natural fatty acids have an even number of carbon atoms. Analogous compounds composed of odd numbers of carbon atoms are perfectly stable and have been made synthetically. Since nature makes these long-chain acids by linking together acetate units, it is not surprising that the carbon atoms composing the natural products are multiples of two. The double bonds in the unsaturated compounds listed on the right are all cis (or Z).
FATTY ACIDS
Saturated
Formula
Common Name
Melting Point
CH3(CH2)10CO2H lauric acid 45 ºC
CH3(CH2)12CO2H myristic acid 55 ºC
CH3(CH2)14CO2H palmitic acid 63 ºC
CH3(CH2)16CO2H stearic acid 69 ºC
CH3(CH2)18CO2H arachidic acid 76 ºC
Unsaturated
Formula
Common Name
Melting Point
CH3(CH2)5CH=CH(CH2)7CO2H palmitoleic acid 0 ºC
CH3(CH2)7CH=CH(CH2)7CO2H oleic acid 13 ºC
CH3(CH2)4CH=CHCH2CH=CH(CH2)7CO2H linoleic acid -5 ºC
CH3CH2CH=CHCH2CH=CHCH2CH=CH(CH2)7CO2H linolenic acid -11 ºC
CH3(CH2)4(CH=CHCH2)4(CH2)2CO2H arachidonic acid -49 ºC
The following formulas are examples of other naturally occurring carboxylic acids. The molecular structures range from simple to complex, often incorporate a variety of other functional groups, and many are chiral. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/17%3A_Carboxylic_Acids_and_the_Acidity_of_the_OH_Bond/17.08%3A_Physical_Properties.txt |
Prostaglandins were first discovered and isolated from human semen in the 1930s by Ulf von Euler of Sweden. Thinking they had come from the prostate gland, he named them prostaglandins. It has since been determined that they exist and are synthesized in virtually every cell of the body. Prostaglandins, are like hormones in that they act as chemical messengers, but do not move to other sites, but work right within the cells where they are synthesized.
Introduction
Prostaglandins are unsaturated carboxylic acids, consisting of of a 20 carbon skeleton that also contains a five member ring. They are biochemically synthesized from the fatty acid, arachidonic acid. See the graphic on the left. The unique shape of the arachidonic acid caused by a series of cis double bonds helps to put it into position to make the five member ring.
Structure of prostaglandin E2 (PGE2)
Prostaglandin Structure
Prostaglandins are unsaturated carboxylic acids, consisting of of a 20 carbon skeleton that also contains a five member ring and are based upon the fatty acid, arachidonic acid. There are a variety of structures one, two, or three double bonds. On the five member ring there may also be double bonds, a ketone, or alcohol groups. A typical structure is on the left graphic.
Functions of Prostaglandins
There are a variety of physiological effects including:
1. Activation of the inflammatory response, production of pain, and fever. When tissues are damaged, white blood cells flood to the site to try to minimize tissue destruction. Prostaglandins are produced as a result.
2. Blood clots form when a blood vessel is damaged. A type of prostaglandin called thromboxane stimulates constriction and clotting of platelets. Conversely, PGI2, is produced to have the opposite effect on the walls of blood vessels where clots should not be forming.
3. Certain prostaglandins are involved with the induction of labor and other reproductive processes. PGE2 causes uterine contractions and has been used to induce labor.
4. Prostaglandins are involved in several other organs such as the gastrointestinal tract (inhibit acid synthesis and increase secretion of protective mucus), increase blood flow in kidneys, and leukotriens promote constriction of bronchi associated with asthma.
Ball-and-stick model of the aspirin molecule, as found in the solid state. Single-crystal X-ray diffraction data from Kim, Y.; Machida, K.; Taga, T.; Osaki, K. (1985). "Structure Redetermination and Packing Analysis of Aspirin Crystal". Chem. Pharm. Bull. 33 (7): 2641-2647. ISSN 1347-5223.
Effects of Aspirin and other Pain Killers
When you see that prostaglandins induce inflammation, pain, and fever, what comes to mind but aspirin. Aspirin blocks an enzyme called cyclooxygenase, COX-1 and COX-2, which is involved with the ring closure and addition of oxygen to arachidonic acid converting to prostaglandins. The acetyl group on aspirin is hydrolzed and then bonded to the alcohol group of serine as an ester. This has the effect of blocking the channel in the enzyme and arachidonic can not enter the active site of the enzyme. By inhibiting or blocking this enzyme, the synthesis of prostaglandins is blocked, which in turn relives some of the effects of pain and fever. Aspirin is also thought to inhibit the prostaglandin synthesis involved with unwanted blood clotting in coronary heart disease. At the same time an injury while taking aspirin may cause more extensive bleeding.
Contributors
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook
17.12: Preparation of Carboxylic Acids
The carbon atom of a carboxyl group has a high oxidation state. It is not surprising, therefore, that many of the chemical reactions used for their preparation are oxidations. Such reactions have been discussed in previous sections of this text, and the following diagram summarizes most of these. To review the previous discussion of any of these reaction classes simply click on the number (1 to 4) or descriptive heading for the group.
Two other useful procedures for preparing carboxylic acids involve hydrolysis of nitriles and carboxylation of organometallic intermediates. As shown in the following diagram, both methods begin with an organic halogen compound and the carboxyl group eventually replaces the halogen. Both methods require two steps, but are complementary in that the nitrile intermediate in the first procedure is generated by a SN2 reaction, in which cyanide anion is a nucleophilic precursor of the carboxyl group. The hydrolysis may be either acid or base-catalyzed, but the latter give a carboxylate salt as the initial product.
In the second procedure the electrophilic halide is first transformed into a strongly nucleophilic metal derivative, and this adds to carbon dioxide (an electrophile). The initial product is a salt of the carboxylic acid, which must then be released by treatment with strong aqueous acid.
An existing carboxylic acid may be elongated by one methylene group, using a homologation procedure called the Arndt-Eistert reaction. To learn about this useful method Click Here.
Contributors
• William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/17%3A_Carboxylic_Acids_and_the_Acidity_of_the_OH_Bond/17.11%3A_Aspirin_Arachidonic_Acid_and_Prostaglandins.txt |
Salt Formation
Because of their enhanced acidity, carboxylic acids react with bases to form ionic salts, as shown in the following equations. In the case of alkali metal hydroxides and simple amines (or ammonia) the resulting salts have pronounced ionic character and are usually soluble in water. Heavy metals such as silver, mercury and lead form salts having more covalent character (3rd example), and the water solubility is reduced, especially for acids composed of four or more carbon atoms.
RCO2H + NaHCO3 RCO2(–) Na(+) + CO2 + H2O
RCO2H + (CH3)3N: RCO2(–) (CH3)3NH(+)
RCO2H + AgOH RCO2δ(-) Agδ(+) + H2O
Carboxylic acids and salts having alkyl chains longer than six carbons exhibit unusual behavior in water due to the presence of both hydrophilic (CO2) and hydrophobic (alkyl) regions in the same molecule. Such molecules are termed amphiphilic (Gk. amphi = both) or amphipathic. Depending on the nature of the hydrophilic portion these compounds may form monolayers on the water surface or sphere-like clusters, called micelles, in solution.
Substitution of the Hydroxyl Hydrogen
This reaction class could be termed electrophilic substitution at oxygen, and is defined as follows (E is an electrophile). Some examples of this substitution are provided in equations (1) through (4).
RCO2–H + E(+) RCO2E + H(+)
If E is a strong electrophile, as in the first equation, it will attack the nucleophilic oxygen of the carboxylic acid directly, giving a positively charged intermediate which then loses a proton. If E is a weak electrophile, such as an alkyl halide, it is necessary to convert the carboxylic acid to the more nucleophilic carboxylate anion to facilitate the substitution. This is the procedure used in reactions 2 and 3. Equation 4 illustrates the use of the reagent diazomethane (CH2N2) for the preparation of methyl esters. This toxic and explosive gas is always used as an ether solution (bright yellow in color). The reaction is easily followed by the evolution of nitrogen gas and the disappearance of the reagent's color. This reaction is believed to proceed by the rapid bonding of a strong electrophile to a carboxylate anion.
The nature of SN2 reactions, as in equations 2 & 3, has been described elsewhere. The mechanisms of reactions 1 & 4 will be displayed by clicking the "Toggle Mechanism" button below the diagram.
Alkynes may also serve as electrophiles in substitution reactions of this kind, as illustrated by the synthesis of vinyl acetate from acetylene. Intramolecular carboxyl group additions to alkenes generate cyclic esters known as lactones. Five-membered (gamma) and six-membered (delta) lactones are most commonly formed. Electrophilic species such as acids or halogens are necessary initiators of lactonizations. Even the weak electrophile iodine initiates iodolactonization of γ,δ- and δ,ε-unsaturated acids. Examples of these reactions will be displayed by clicking the "Other Examples" button.
Substitution of the Hydroxyl Group
Reactions in which the hydroxyl group of a carboxylic acid is replaced by another nucleophilic group are important for preparing functional derivatives of carboxylic acids. The alcohols provide a useful reference chemistry against which this class of transformations may be evaluated. In general, the hydroxyl group proved to be a poor leaving group, and virtually all alcohol reactions in which it was lost involved a prior conversion of –OH to a better leaving group. This has proven to be true for the carboxylic acids as well.
Four examples of these hydroxyl substitution reactions are presented by the following equations. In each example, the new bond to the carbonyl group is colored magenta and the nucleophilic atom that has replaced the hydroxyl oxygen is colored green. The hydroxyl moiety is often lost as water, but in reaction #1 the hydrogen is lost as HCl and the oxygen as SO2. This reaction parallels a similar transformation of alcohols to alkyl chlorides, although its mechanism is different. Other reagents that produce a similar conversion to acyl halides are PCl5 and SOBr2.
The amide and anhydride formations shown in equations #2 & 3 require strong heating, and milder procedures that accomplish these transformations will be described in the next chapter.
Reaction #4 is called esterification, since it is commonly used to convert carboxylic acids to their ester derivatives. Esters may be prepared in many different ways; indeed, equations #1 and #4 in the previous diagram illustrate the formation of tert-butyl and methyl esters respectively. The acid-catalyzed formation of ethyl acetate from acetic acid and ethanol shown here is reversible, with an equilibrium constant near 2. The reaction can be forced to completion by removing the water as it is formed. This type of esterification is often referred to as Fischer esterification. As expected, the reverse reaction, acid-catalyzed ester hydrolysis, can be carried out by adding excess water.
A thoughtful examination of this reaction (#4) leads one to question why it is classified as a hydroxyl substitution rather than a hydrogen substitution. The following equations, in which the hydroxyl oxygen atom of the carboxylic acid is colored red and that of the alcohol is colored blue, illustrate this distinction (note that the starting compounds are in the center).
H2O + CH3CO-OCH2CH3
H-substitution
CH3CO-OH + CH3CH2-OH
HO-substitution
CH3CO-OCH2CH3 + H2O
In order to classify this reaction correctly and establish a plausible mechanism, the oxygen atom of the alcohol was isotopically labeled as 18O (colored blue in our equation). Since this oxygen is found in the ester product and not the water, the hydroxyl group of the acid must have been replaced in the substitution. A mechanism for this general esterification reaction will be displayed on clicking the "Esterification Mechanism" button; also, once the mechanism diagram is displayed, a reaction coordinate for it can be seen by clicking the head of the green "energy diagram" arrow. Addition-elimination mechanisms of this kind proceed by way of tetrahedral intermediates (such as A and B in the mechanism diagram) and are common in acyl substitution reactions. Acid catalysis is necessary to increase the electrophilic character of the carboxyl carbon atom, so it will bond more rapidly to the nucleophilic oxygen of the alcohol. Base catalysis is not useful because base converts the acid to its carboxylate anion conjugate base, a species in which the electrophilic character of the carbon is reduced.
Since a tetrahedral intermediate occupies more space than a planar carbonyl group, we would expect the rate of this reaction to be retarded when bulky reactants are used. To test this prediction the esterification of acetic acid was compared with that of 2,2-dimethylpropanoic acid, (CH3)3CO2H. Here the relatively small methyl group of acetic acid is replaced by a larger tert-butyl group, and the bulkier acid reacted fifty times slower than acetic acid. Increasing the bulk of the alcohol reactant results in a similar rate reduction. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/17%3A_Carboxylic_Acids_and_the_Acidity_of_the_OH_Bond/17.13%3A_Reactions_of_Carboxylic_AcidsGeneral_Features.txt |
Comparing the strengths of weak acids
The strengths of weak acids are measured on the pKa scale. The smaller the number on this scale, the stronger the acid is.
Three of the compounds we shall be looking at, together with their pKa values are:
Remember - the smaller the number the stronger the acid. Comparing the other two to ethanoic acid, you will see that phenol is very much weaker with a pKa of 10.00, and ethanol is so weak with a pKa of about 16 that it hardly counts as acidic at all!
Acidity of Carboxylic Acids
The pKa 's of some typical carboxylic acids are listed in the following table. When we compare these values with those of comparable alcohols, such as ethanol (pKa = 16) and 2-methyl-2-propanol (pKa = 19), it is clear that carboxylic acids are stronger acids by over ten powers of ten! Furthermore, electronegative substituents near the carboxyl group act to increase the acidity.
Compound
pKa
Compound
pKa
HCO2H 3.75 CH3CH2CH2CO2H 4.82
CH3CO2H 4.74 ClCH2CH2CH2CO2H 4.53
FCH2CO2H 2.65 CH3CHClCH2CO2H 4.05
ClCH2CO2H 2.85 CH3CH2CHClCO2H 2.89
BrCH2CO2H 2.90 C6H5CO2H 4.20
ICH2CO2H 3.10 p-O2NC6H4CO2H 3.45
Cl3CCO2H 0.77 p-CH3OC6H4CO2H 4.45
Why should the presence of a carbonyl group adjacent to a hydroxyl group have such a profound effect on the acidity of the hydroxyl proton? To answer this question we must return to the nature of acid-base equilibria and the definition of pKa , illustrated by the general equations given below. These relationships were described in an previous section of this text.
We know that an equilibrium favors the thermodynamically more stable side, and that the magnitude of the equilibrium constant reflects the energy difference between the components of each side. In an acid base equilibrium the equilibrium always favors the weaker acid and base (these are the more stable components). Water is the standard base used for pKa measurements; consequently, anything that stabilizes the conjugate base (A:(–)) of an acid will necessarily make that acid (H–A) stronger and shift the equilibrium to the right. Both the carboxyl group and the carboxylate anion are stabilized by resonance, but the stabilization of the anion is much greater than that of the neutral function, as shown in the following diagram. In the carboxylate anion the two contributing structures have equal weight in the hybrid, and the C–O bonds are of equal length (between a double and a single bond). This stabilization leads to a markedly increased acidity, as illustrated by the energy diagram displayed by clicking the "Toggle Display" button.
mpounds like alcohols and phenol which contain an -OH group attached to a hydrocarbon are very weak acids. Alcohols are so weakly acidic that, for normal lab purposes, their acidity can be virtually ignored. However, phenol is sufficiently acidic for it to have recognizably acidic properties - even if it is still a very weak acid. A hydrogen ion can break away from the -OH group and transfer to a base.
The pKa of ethanol is about 17, while the pKa of acetic acid is about 5: this is a 1012-fold difference in the two acidity constants. In both compounds, the acidic proton is bonded to an oxygen atom. How can they be so different in terms of acidity?
We begin by considering the conjugate bases.
In both species, the negative charge on the conjugate base is held by an oxygen, so periodic trends cannot be invoked. For acetic acid, however, there is a key difference: a resonance contributor can be drawn in which the negative charge is localized on the second oxygen of the group. The two resonance forms for the conjugate base are equal in energy. What this means is that the negative charge on the acetate ion is not located on one oxygen or the other: rather it is shared between the two. Chemists use the term ‘delocalization of charge’ to describe this situation. In the ethoxide ion, by contrast, the negative charge is ‘locked’ on the single oxygen – it has nowhere else to go.
Recall the findamental idea that electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ than when they are confined to one atom. Here, a charge is being ‘spread out’ (in other words, delocalized) by resonance, rather than simply by the size of the atom involved.
The delocalization of charge by resonance has a very powerful effect on the reactivity of organic molecules, enough to account for the difference of over 12 pKa units between ethanol and acetic acid. The acetate ion is that much more stable than the ethoxide ion, all due to the effects of resonance delocalization.
The resonance effect also explains why a nitrogen atom is much more basic when it is in an amine, but not significantly basic when it is part of an amide group. Recall that in an amide, there is significant double-bond character to the carbon-nitrogen bond, due to a second resonance contributor in which the nitrogen lone pair is part of a p bond.
While the electron lone pair of an amine nitrogen is ‘stuck’ in one place, the lone pair on an amide nitrogen is delocalized by resonance. Notice that in this case, we are extending our central statement to say that electron density – in the form of a lone pair – is stabilized by resonance delocalization, even though there is not a negative charge involved. Here’s another way to think about it: the lone pair on an amide nitrogen is not available for bonding with a proton – these two electrons are too ‘comfortable’ being part of the delocalized pi-bonding system. The lone pair on an amine nitrogen, by contrast, is not part of a delocalized pi system, and is very ready to form a bond with any acidic proton that might be nearby.
Why is phenol acidic?
Compounds like alcohols and phenol which contain an -OH group attached to a hydrocarbon are very weak acids. Alcohols are so weakly acidic that, for normal lab purposes, their acidity can be virtually ignored. However, phenol is sufficiently acidic for it to have recognizably acidic properties - even if it is still a very weak acid. A hydrogen ion can break away from the -OH group and transfer to a base. For example, in solution in water:
Phenol is a very weak acid and the position of equilibrium lies well to the left. Phenol can lose a hydrogen ion because the phenoxide ion formed is stabilised to some extent. The negative charge on the oxygen atom is delocalised around the ring. The more stable the ion is, the more likely it is to form. One of the lone pairs on the oxygen atom overlaps with the delocalised electrons on the benzene ring.
This overlap leads to a delocalization which extends from the ring out over the oxygen atom. As a result, the negative charge is no longer entirely localized on the oxygen, but is spread out around the whole ion.
Spreading the charge around makes the ion more stable than it would be if all the charge remained on the oxygen. However, oxygen is the most electronegative element in the ion and the delocalized electrons will be drawn towards it. That means that there will still be a lot of charge around the oxygen which will tend to attract the hydrogen ion back again. That is why phenol is only a very weak acid.
Why is phenol a much stronger acid than cyclohexanol? To answer this question we must evaluate the manner in which an oxygen substituent interacts with the benzene ring. As noted in our earlier treatment of electrophilic aromatic substitution reactions, an oxygen substituent enhances the reactivity of the ring and favors electrophile attack at ortho and para sites. It was proposed that resonance delocalization of an oxygen non-bonded electron pair into the pi-electron system of the aromatic ring was responsible for this substituent effect. A similar set of resonance structures for the phenolate anion conjugate base appears below the phenol structures.
The resonance stabilization in these two cases is very different. An important principle of resonance is that charge separation diminishes the importance of canonical contributors to the resonance hybrid and reduces the overall stabilization. The contributing structures to the phenol hybrid all suffer charge separation, resulting in very modest stabilization of this compound. On the other hand, the phenolate anion is already charged, and the canonical contributors act to disperse the charge, resulting in a substantial stabilization of this species. The conjugate bases of simple alcohols are not stabilized by charge delocalization, so the acidity of these compounds is similar to that of water. An energy diagram showing the effect of resonance on cyclohexanol and phenol acidities is shown on the right. Since the resonance stabilization of the phenolate conjugate base is much greater than the stabilization of phenol itself, the acidity of phenol relative to cyclohexanol is increased. Supporting evidence that the phenolate negative charge is delocalized on the ortho and para carbons of the benzene ring comes from the influence of electron-withdrawing substituents at those sites.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/17%3A_Carboxylic_Acids_and_the_Acidity_of_the_OH_Bond/17.14%3A_Carboxylic_AcidsStrong_Organic_BrnstedLowry_Acids.txt |
A carbonyl group is a chemically organic functional group composed of a carbon atom double-bonded to an oxygen atom --> [C=O] The simplest carbonyl groups are aldehydes and ketones usually attached to another carbon compound. These structures can be found in many aromatic compounds contributing to smell and taste.
The Carbonyl Group
C=O is prone to additions and nucleophillic attack because or carbon's positive charge and oxygen's negative charge. The resonance of the carbon partial positive charge allows the negative charge on the nucleophile to attack the Carbonyl group and become a part of the structure and a positive charge (usually a proton hydrogen) attacks the oxygen. Just a reminder, the nucleophile is a good acid therefore "likes protons" so it will attack the side with a positive charge.
Before we consider in detail the reactivity of aldehydes and ketones, we need to look back and remind ourselves of what the bonding picture looks like in a carbonyl. Carbonyl carbons are sp2 hybridized, with the three sp2 orbitals forming soverlaps with orbitals on the oxygen and on the two carbon or hydrogen atoms. These three bonds adopt trigonal planar geometry. The remaining unhybridized 2p orbital on the central carbonyl carbon is perpendicular to this plane, and forms a ‘side-by-side’ pbond with a 2p orbital on the oxygen.
The carbon-oxygen double bond is polar: oxygen is more electronegative than carbon, so electron density is higher on the oxygen side of the bond and lower on the carbon side. Recall that bond polarity can be depicted with a dipole arrow, or by showing the oxygen as holding a partial negative charge and the carbonyl carbon a partial positive charge.
A third way to illustrate the carbon-oxygen dipole is to consider the two main resonance contributors of a carbonyl group: the major form, which is what you typically see drawn in Lewis structures, and a minor but very important contributor in which both electrons in the pbond are localized on the oxygen, giving it a full negative charge. The latter depiction shows the carbon with an empty 2p orbital and a full positive charge.
Some Carbonyl Compounds
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
18.02: General Reactions of Carbonyl Compounds
Nucleophilic Addition to Aldehydes and Ketones
The result of carbonyl bond polarization, however it is depicted, is straightforward to predict. The carbon, because it is electron-poor, is an electrophile: it is a great target for attack by an electron-rich nucleophilic group. Because the oxygen end of the carbonyl double bond bears a partial negative charge, anything that can help to stabilize this charge by accepting some of the electron density will increase the bond’s polarity and make the carbon more electrophilic. Very often a general acid group serves this purpose, donating a proton to the carbonyl oxygen.
The same effect can also be achieved if a Lewis acid, such as a magnesium ion, is located near the carbonyl oxygen.
Unlike the situation in a nucleophilic substitution reaction, when a nucleophile attacks an aldehyde or ketone carbon there is no leaving group – the incoming nucleophile simply ‘pushes’ the electrons in the pi bond up to the oxygen.
Alternatively, if you start with the minor resonance contributor, you can picture this as an attack by a nucleophile on a carbocation.
After the carbonyl is attacked by the nucleophile, the negatively charged oxygen has the capacity to act as a nucleophile. However, most commonly the oxygen acts instead as a base, abstracting a proton from a nearby acid group in the solvent or enzyme active site.
This very common type of reaction is called a nucleophilic addition. In many biologically relevant examples of nucleophilic addition to carbonyls, the nucleophile is an alcohol oxygen or an amine nitrogen, or occasionally a thiol sulfur. In one very important reaction type known as an aldol reaction (which we will learn about in section 13.3) the nucleophile attacking the carbonyl is a resonance-stabilized carbanion. In this chapter, we will concentrate on reactions where the nucleophile is an oxygen or nitrogen.
Nucleophilic Substitution of RCOZ (Z = Leaving Group)
Carbonyl compounds with leaving groups have reactions similar to aldehydes and ketones. The main difference is the presence of an electronegative substituent that can act as a leaving group during a nucleophile substitution reaction. Although there are many types of carboxylic acid derivatives known, this article focuses on four: acid halides, acid anhydrides, esters, and amides.
General mechanism
1) Nucleophilic attack on the carbonyl
2) Leaving group is removed
Although aldehydes and ketones also contain carbonyls, their chemistry is distinctly different because they do not contain suitable leaving groups. Once a tetrahedral intermediate is formed, aldehydes and ketones cannot reform their carbonyls. Because of this, aldehydes and ketones typically undergo nucleophilic additions and not substitutions.
The relative reactivity of carboxylic acid derivatives toward nucleophile substitutions is related to the electronegative leaving group’s ability to activate the carbonyl. The more electronegative leaving groups withdraw electron density from the carbonyl, thereby increasing its electrophilicity. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/18%3A_Introduction_to_Carbonyl_Chemistry_Organometallic_Reagents_Oxidation_and_Reduction/18.01%3A_Introduction.txt |
Solutions to exercises
You are undoubtedly already familiar with the general idea of oxidation and reduction: you learned in general chemistry that when a compound or atom is oxidized it loses electrons, and when it is reduced it gains electrons. You also know that oxidation and reduction reactions occur in pairs: if one species is oxidized, another must be reduced at the same time - thus the term 'redox reaction'.
Most of the redox reactions you have seen previously in general chemistry probably involved the flow of electrons from one metal to another, such as the reaction between copper ion in solution and metallic zinc:
Cu+2(aq) + Zn(s) → Cu(s) + Zn+2(aq)
In organic chemistry, redox reactions look a little different. Electrons in an organic redox reaction often are transferred in the form of a hydride ion - a proton and two electrons. Because they occur in conjunction with the transfer of a proton, these are commonly referred to as hydrogenation and dehydrogenation reactions: a hydride plus a proton adds up to a hydrogen (H2) molecule. Be careful - do not confuse the terms hydrogenation and dehydrogenation with hydration and dehydration - the latter refer to the gain and loss of a water molecule (and are not redox reactions), while the former refer to the gain and loss of a hydrogen molecule.
When a carbon atom in an organic compound loses a bond to hydrogen and gains a new bond to a heteroatom (or to another carbon), we say the compound has been dehydrogenated, or oxidized. A very common biochemical example is the oxidation of an alcohol to a ketone or aldehyde:
When a carbon atom loses a bond to hydrogen and gains a bond to a heteroatom (or to another carbon atom), it is considered to be an oxidative process because hydrogen, of all the elements, is the least electronegative. Thus, in the process of dehydrogenation the carbon atom undergoes an overall loss of electron density - and loss of electrons is oxidation.
Conversely, when a carbon atom in an organic compound gains a bond to hydrogen and loses a bond to a heteroatom (or to another carbon atom), we say that the compound has been hydrogenated, or reduced. The hydrogenation of a ketone to an alcohol, for example, is overall the reverse of the alcohol dehydrogenation shown above. Illustrated below is another common possibility, the hydrogenation (reduction) of an alkene to an alkane.
Hydrogenation results in higher electron density on a carbon atom(s), and thus we consider process to be one of reduction of the organic molecule.
Notice that neither hydrogenation nor dehydrogenation involves the gain or loss of an oxygen atom. Reactions which do involve gain or loss of one or more oxygen atoms are usually referred to as 'oxygenase' and 'reductase' reactions, and are the subject of section 16.10 and section 17.3.
For the most part, when talking about redox reactions in organic chemistry we are dealing with a small set of very recognizable functional group transformations. It is therefore very worthwhile to become familiar with the idea of 'oxidation states' as applied to organic functional groups. By comparing the relative number of bonds to hydrogen atoms, we can order the familiar functional groups according to oxidation state. We'll take a series of single carbon compounds as an example. Methane, with four carbon-hydrogen bonds, is highly reduced. Next in the series is methanol (one less carbon-hydrogen bond, one more carbon-oxygen bond), followed by formaldehyde, formate, and finally carbon dioxide at the highly oxidized end of the group.
This pattern holds true for the relevant functional groups on organic molecules with two or more carbon atoms:
Alkanes are highly reduced, while alcohols - as well as alkenes, ethers, amines, sulfides, and phosphate esters - are one step up on the oxidation scale, followed by aldehydes/ketones/imines and epoxides, and finally by carboxylic acid derivatives (carbon dioxide, at the top of the oxidation list, is specific to the single carbon series).
Notice that in the series of two-carbon compounds above, ethanol and ethene are considered to be in the same oxidation state. You know already that alcohols and alkenes are interconverted by way of addition or elimination of water (section 14.1). When an alcohol is dehydrated to form an alkene, one of the two carbons loses a C-H bond and gains a C-C bond, and thus is oxidized. However, the other carbon loses a C-O bond and gains a C-C bond, and thus is considered to be reduced. Overall, therefore, there is no change to the oxidation state of the molecule.
You should learn to recognize when a reaction involves a change in oxidation state in an organic reactant . Looking at the following transformation, for example, you should be able to quickly recognize that it is an oxidation: an alcohol functional group is converted to a ketone, which is one step up on the oxidation ladder.
Likewise, this next reaction involves the transformation of a carboxylic acid derivative (a thioester) first to an aldehyde, then to an alcohol: this is a double reduction, as the substrate loses two bonds to heteroatoms and gains two bonds to hydrogens.
An acyl transfer reaction (for example the conversion of an acyl phosphate to an amide) is not considered to be a redox reaction - the oxidation state of the organic molecule is does not change as substrate is converted to product, because a bond to one heteroatom (oxygen) has simply been traded for a bond to another heteroatom (nitrogen).
It is important to be able to recognize when an organic molecule is being oxidized or reduced, because this information tells you to look for the participation of a corresponding redox agent that is being reduced or oxidized- remember, oxidation and reduction always occur in tandem! We will soon learn in detail about the most important biochemical and laboratory redox agents.
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Exercise 16.1: is an aldol condensation a redox reaction? Explain.
Exercise 16.2: Is the reaction catalyzed by squalene epoxidase a redox reaction? How about squalene cyclase? Explain.
Template:ExampleEnd
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/18%3A_Introduction_to_Carbonyl_Chemistry_Organometallic_Reagents_Oxidation_and_Reduction/18.03%3A_A_Preview_of_Oxidation_and_Reduction.txt |
he most common sources of the hydride Nucleophile are lithium aluminum hydride (LiAlH4) and sodium borohydride (NaBH4). Note! The hydride anion is not present during this reaction; rather, these reagents serve as a source of hydride due to the presence of a polar metal-hydrogen bond. Because aluminum is less electronegative than boron, the Al-H bond in LiAlH4 is more polar, thereby, making LiAlH4 a stronger reducing agent.
Addition of a hydride anion (H:-) to an aldehyde or ketone gives an alkoxide anion, which on protonation yields the corresponding alcohol. Aldehydes produce 1º-alcohols and ketones produce 2º-alcohols.
In metal hydrides reductions the resulting alkoxide salts are insoluble and need to be hydrolyzed (with care) before the alcohol product can be isolated. In the sodium borohydride reduction the methanol solvent system achieves this hydrolysis automatically. In the lithium aluminum hydride reduction water is usually added in a second step. The lithium, sodium, boron and aluminum end up as soluble inorganic salts at the end of either reaction. Note! LiAlH4 and NaBH4 are both capable of reducing aldehydes and ketones to the corresponding alcohol.
Example 1
Mechanism
This mechanism is for a LiAlH4 reduction. The mechanism for a NaBH4 reduction is the same except methanol is the proton source used in the second step.
1) Nucleopilic attack by the hydride anion
2) The alkoxide is protonated
Properties of Hydride Sources
Two practical sources of hydride-like reactivity are the complex metal hydrides lithium aluminum hydride (LiAlH4) and sodium borohydride (NaBH4). These are both white (or near white) solids, which are prepared from lithium or sodium hydrides by reaction with aluminum or boron halides and esters. Lithium aluminum hydride is by far the most reactive of the two compounds, reacting violently with water, alcohols and other acidic groups with the evolution of hydrogen gas. The following table summarizes some important characteristics of these useful reagents.
It would be great to convert this table to text.
Problems
1) Please draw the products of the following reactions:
2) Please draw the structure of the molecule which must be reacted to produce the product.
3) Deuterium oxide (D2O) is a form of water where the hydrogens have been replaced by deuteriums. For the following LiAlH4 reduction the water typically used has been replaced by deuterium oxide. Please draw the product of the reaction and place the deuterium in the proper location. Hint! Look at the mechanism of the reaction.
1)
2)
3)
Contributors
• Prof. Steven Farmer (Sonoma State University)
• William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
18.05: The Stereochemistry of Carbonyl Reduction
Notice that in the course of the nucleophilic addition pictured above, the hybridization of the carbonyl carbon changes from sp2 to sp3, meaning that the bond geometry changes from trigonal planar to tetrahedral. It is also important to note that if the starting carbonyl is asymmetric (in other words, if the two R groups are not equivalent), then a new stereocenter has been created. The configuration of the new stereocenter depends upon which side of the carbonyl plane the nucleophile attacks from.
If the reaction is catalyzed by an enzyme, the stereochemistry of addition is tightly controlled, and leads to one specific stereoisomer - this is because the nucleophilic and electrophilic substrates are bound in a specific positions within the active site, so that attack must occur specifically from one side. If, however, the reaction occurs uncatalyzed in solution, then either side of the carbonyl is equally likely to be attacked, and the result will be a 50:50 racemic mixture.
This is the rule for most nonenzymatic reactions, but as with most rules, there are exceptions. If, for example, the geometry of the carbonyl-containing molecule is constrained in such a way that approach by the nucleophile is less hindered from one side, a 50:50 racemic mixture will not necessarily result. Consider camphor, the distinctive-smelling compound found in many cosmetics and skin creams.
Upon inspection it is clear that topside attack and bottom side attack by a nucleophile are nonequivalent in terms of steric hindrance. A relatively simple experiment shows that, when the incoming nucleophile is a hydride ion from the common synthetic reducing agent sodium borohydride (a reaction type we will study in a later chapter), the product of bottom side attack predominates by a ratio of about 6 to 1 (see section 16.4D for more details on this experiment). We can infer from this result that approach from the bottom (si) face of the carbonyl in camphor is less hindered.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/18%3A_Introduction_to_Carbonyl_Chemistry_Organometallic_Reagents_Oxidation_and_Reduction/18.04%3A_Reduction_of_Aldehydes_and_Ketones.txt |
Chemoselective Reductions
Enones present unique challenges as reducing agents can also attack the alkene giving a mixture of products. Methods to selectively reduce the ketone (Luche Reduction) and the alkene (Stryker Reduction) have been developed.
Acyclic Compounds
Reductions of aldehydes and ketones follow the same selectivity models as the addition of unstabilized carbon nucleophiles to these functional groups. Non-chelating reducing agents (NaBH4, LiAH4, etc.) show Felkin-Anh selectivity while reducing agents which can chelate ( Zn(BH4)2) show Cram Chelate selectivity.
Tetrahedron Lett., 1985, 26,5139-5142.
As seen in the example above, lithium will sometimes chelate.
Cyclohexanones
Hydrides can approach cyclohexanones from the axial or equatorial face of the ketone.
It has been obvserved that increasingly bulky hydride reagents prefer to attack from the equatorial face of the carbonyl. This is rationalized by the increased steric demand of a nucleophile approaching from the axial face of the carbonyl as it encounters the axial substituents (H in this case) at the 3 and 5 positions.
This argument would thus seem to always argue for attack from the equatorial face, but we must also take into consideration any developing torsional strain through the transition state. Attack from the axial face avoids developing eclipsing interactions between the C–O bond and the C–HE bonds at the 2 and 6 positions. Attack from the equatorial face forces the C–O bond to travel past the C–HE bonds to sit in the chair conformation. We can see this below as the dihedral angle in the starting cyclhexanone starts as a positive number and ends up as a negative number which shows that the C–O bond must have gone through an eclipsing conformation to reach its final position. Axial attack does not cause a change in sign of the dihedral angle, thus avoiding any eclipsing interactions in the transition state.
We can thus predict that small hydride reagents, such as LiAlH4, will prefer to attack from the axial face as the torsional strain in the transition state is the dominant interaction while large hydride reagents, such as H–BR4, will attack from the equatorial face as the steric interactions from the reagent's approach are now the dominant interaction.
Substrate Directed Reductions
J. Am. Chem. Soc., 1988, 110, 3560-3578.
Tetrahedron Lett., 1987, 28, 155-158.
Enantioselective Reductions
Chiral Boronates
If the selectivity from these reactions is the opposite of your desired product, you can use a Mitsunobu Reaction to invert the stereocenter.
Tar-B
Biological Reduction
Addition to a carbonyl by a semi-anionic hydride, such as NaBH4, results in conversion of the carbonyl compound to an alcohol. The hydride from the BH4- anion acts as a nucleophile, adding H- to the carbonyl carbon. A proton source can then protonate the oxygen of the resulting alkoxide ion, forming an alcohol.
Formally, that process is referred to as a reduction. Reduction generally means a reaction in which electrons are added to a compound; the compound that gains electrons is said to be reduced. Because hydride can be thought of as a proton plus two electrons, we can think of conversion of a ketone or an aldehyde to an alcohol as a two-electron reduction. An aldehyde plus two electrons and two protons becomes an alcohol.
Aldehydes, ketones and alcohols are very common features in biological molecules. Converting between these compounds is a frequent event in many biological pathways. However, semi-anionic compounds like sodium borohydride don't exist in the cell. Instead, a number of biological hydride donors play a similar role.
NADH is a common biological reducing agent. NADH is an acronym for nicotinamide adenine dinucleotide hydride. Insetad of an anionic donor that provides a hydride to a carbonyl, NADH is actually a neutral donor. It supplies a hydride to the carbonyl under very specific circumstances. In doing so, it forms a cation, NAD+.
Contributors
• Stephen Laws
• Michael Di Maso (UC Davis)
• Jared Shaw (UC Davis)
18.07: Reduction of Carboxylic Acids and Their Derivatives
Since relatively few methods exist for the reduction of carboxylic acid derivatives to aldehydes, it would be useful to modify the reactivity and solubility of LAH to permit partial reductions of this kind to be achieved. The most fruitful approach to this end has been to attach alkoxy or alkyl groups on the aluminum. This not only modifies the reactivity of the reagent as a hydride donor, but also increases its solubility in nonpolar solvents. Two such reagents will be mentioned here; the reactive hydride atom is colored blue.
Lithium tri-tert-butoxyaluminohydride (LtBAH), LiAl[OC(CH3)3]3H : Soluble in THF, diglyme & ether.
Diisobutylaluminum hydride (DIBAH), [(CH3)2CHCH2]2AlH : Soluble in toluene, THF & ether.
Each of these reagents carries one equivalent of hydride. The first (LtBAH) is a complex metal hydride, but the second is simply an alkyl derivative of aluminum hydride. In practice, both reagents are used in equimolar amounts, and usually at temperatures well below 0 ºC. The following examples illustrate how aldehydes may be prepared from carboxylic acid derivatives by careful application of these reagents. A temperature of -78 ºC is easily maintained by using dry-ice as a coolant.
Reduction of Acid Chlorides and Esters
Acid chlorides can be converted to aldehydes using lithium tri-tert-butoxyaluminum hydride (LiAlH(Ot-Bu)3). The hydride source (LiAlH(Ot-Bu)3) is a weaker reducing agent than lithium aluminum hydride. Because acid chlorides are highly activated they still react with the hydride source; however, the formed aldehyde will react slowly, which allows for its isolation.
General Reaction:
Example 1
Acid chlorides can be converted to aldehydes using lithium tri-tert-butoxyaluminum hydride (LiAlH(Ot-Bu)3). The hydride source (LiAlH(Ot-Bu)3) is a weaker reducing agent than lithium aluminum hydride. Because acid chlorides are highly activated they still react with the hydride source; however, the formed aldehyde will react slowly, which allows for its isolation.
General Reaction:
Example 1
Esters can be converted to aldehydes using diisobutylaluminum hydride (DIBAH). The reaction is usually carried out at -78 oC to prevent reaction with the aldehyde product.
Example 1:
Esters can be converted to 1o alcohols using LiAlH4, while sodium borohydride (\(NaBH_4\)) is not a strong enough reducing agent to perform this reaction.
Example 1:
Mechanism
1) Nucleophilic attack by the hydride
2) Leaving group removal
3) Nucleopilic attack by the hydride anion
4) The alkoxide is protonated
Reduction of Carboxylic Acids and Amides
Carboxylic acids can be converted to 1o alcohols using Lithium aluminum hydride (LiAlH4). Note that NaBH4 is not strong enough to convert carboxylic acids or esters to alcohols. An aldehyde is produced as an intermediate during this reaction, but it cannot be isolated because it is more reactive than the original carboxylic acid.
Example
Possible Mechanism
1) Deprotonation
2) Nucleopilic attack by the hydride anion
3) Leaving group removal
4) Nucleopilic attack by the hydride anion
5) The alkoxide is protonated
General Reaction
Example 1: Amide Reductions
Alkyl groups attached to the nitrogen do not affect the reaction.
Mechanism
1) Nucleophilic attach by the hydride
2) Leaving group removal
3) Nucleophilic attach by the hydride | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/18%3A_Introduction_to_Carbonyl_Chemistry_Organometallic_Reagents_Oxidation_and_Reduction/18.06%3A_Enantioselective_Carbonyl_Reductions.txt |
This page looks at ways of distinguishing between aldehydes and ketones using oxidizing agents such as acidified potassium dichromate(VI) solution, Tollens' reagent, Fehling's solution and Benedict's solution.
Why do aldehydes and ketones behave differently?
You will remember that the difference between an aldehyde and a ketone is the presence of a hydrogen atom attached to the carbon-oxygen double bond in the aldehyde. Ketones don't have that hydrogen.
The presence of that hydrogen atom makes aldehydes very easy to oxidize. Or, put another way, they are strong reducing agents. However, because ketones do not have that particular hydrogen atom, they are resistant to oxidation, and only very strong oxidizing agents like potassium manganate(VII) solution (potassium permanganate solution) oxidize ketones. However, they do it in a destructive way, breaking carbon-carbon bonds.
Provided you avoid using these powerful oxidizing agents, you can easily tell the difference between an aldehyde and a ketone. Aldehydes are easily oxidized by all sorts of different oxidizing agents and ketones are not.
What is formed when aldehydes are oxidized?
It depends on whether the reaction is done under acidic or alkaline conditions. Under acidic conditions, the aldehyde is oxidized to a carboxylic acid. Under alkaline conditions, this couldn't form because it would react with the alkali. A salt is formed instead.
Building equations for the oxidation reactions
If you need to work out the equations for these reactions, the only reliable way of building them is to use electron-half-equations. The half-equation for the oxidation of the aldehyde obviously varies depending on whether you are doing the reaction under acidic or alkaline conditions.
Under acidic conditions it is:
$RCHO + H_2O \rightarrow RCOOH + 2H^+ +2e^-$
. . . and under alkaline conditions:
$RCHO + 3OH^- \rightarrow RCOO^- + 2H_2O +2e^-$
These half-equations are then combined with the half-equations from whatever oxidizing agent you are using. Examples are given in detail below.
Specific examples
In each of the following examples, we are assuming that you know that you have either an aldehyde or a ketone. There are lots of other things which could also give positive results. Assuming that you know it has to be one or the other, in each case, a ketone does nothing. Only an aldehyde gives a positive result.
Using acidified potassium dichromate(VI) solution
A small amount of potassium dichromate(VI) solution is acidified with dilute sulphuric acid and a few drops of the aldehyde or ketone are added. If nothing happens in the cold, the mixture is warmed gently for a couple of minutes - for example, in a beaker of hot water.
ketone No change in the orange solution.
aldehyde Orange solution turns green.
The orange dichromate(VI) ions have been reduced to green chromium(III) ions by the aldehyde. In turn the aldehyde is oxidized to the corresponding carboxylic acid. The electron-half-equation for the reduction of dichromate(VI) ions is:
$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
Combining that with the half-equation for the oxidation of an aldehyde under acidic conditions:
$RCHO + H_2O \rightarrow RCOOH + 2H^+ +2e^-$
. . . gives the overall equation:
$2RCHO + Cr_2O_7^{2-} + 8H^+ \rightarrow 3RCOOH +2Cr^{3+}+ 4H_2O$
Using Tollens' reagent (the silver mirror test)
Tollens' reagent contains the diamminesilver(I) ion, [Ag(NH3)2]+. This is made from silver(I) nitrate solution. You add a drop of sodium hydroxide solution to give a precipitate of silver(I) oxide, and then add just enough dilute ammonia solution to redissolve the precipitate. To carry out the test, you add a few drops of the aldehyde or ketone to the freshly prepared reagent, and warm gently in a hot water bath for a few minutes.
ketone No change in the colourless solution.
aldehyde The colourless solution produces a grey precipitate of silver, or a silver mirror on the test tube.
Aldehydes reduce the diamminesilver(I) ion to metallic silver. Because the solution is alkaline, the aldehyde itself is oxidized to a salt of the corresponding carboxylic acid. The electron-half-equation for the reduction of of the diamminesilver(I) ions to silver is:
$Ag(NH_3)_2^+ + e^- \rightarrow Ag + 2NH_3$
Combining that with the half-equation for the oxidation of an aldehyde under alkaline conditions:
$RCHO + 3OH^- \rightarrow RCOO^- + 2H_2O +2e^-$
gives the overall equation:
$2Ag(NH_3)_2^+ + RCHO + 3OH^- \rightarrow 2Ag + RCOO^- + 4H_2O +2H_2O$
Using Fehling's solution or Benedict's solution
Fehling's solution and Benedict's solution are variants of essentially the same thing. Both contain complexed copper(II) ions in an alkaline solution.
• Fehling's solution contains copper(II) ions complexed with tartrate ions in sodium hydroxide solution. Complexing the copper(II) ions with tartrate ions prevents precipitation of copper(II) hydroxide.
• Benedict's solution contains copper(II) ions complexed with citrate ions in sodium carbonate solution. Again, complexing the copper(II) ions prevents the formation of a precipitate - this time of copper(II) carbonate.
Both solutions are used in the same way. A few drops of the aldehyde or ketone are added to the reagent, and the mixture is warmed gently in a hot water bath for a few minutes.
ketone No change in the blue solution.
aldehyde The blue solution produces a dark red precipitate of copper(I) oxide.
Aldehydes reduce the complexed copper(II) ion to copper(I) oxide. Because the solution is alkaline, the aldehyde itself is oxidized to a salt of the corresponding carboxylic acid. The equations for these reactions are always simplified to avoid having to write in the formulae for the tartrate or citrate ions in the copper complexes. The electron-half-equations for both Fehling's solution and Benedict's solution can be written as:
$2Cu^{2+}_{complexed} + 2OH^- + 2e^- \rightarrow Cu_2O + H_2O$
Combining that with the half-equation for the oxidation of an aldehyde under alkaline conditions:
$RCHO + 3OH^- \rightarrow RCOO^- + 2H_2O +2e^-$
to give the overall equation:
$RCHO + 2Cu^{2+}_{complexed} + 5OH^- \rightarrow RCOO^- + Cu_2O + 3H_2O$
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/18%3A_Introduction_to_Carbonyl_Chemistry_Organometallic_Reagents_Oxidation_and_Reduction/18.08%3A_Oxidation_of_Aldehydes.txt |
The alkali metals (Li, Na, K etc.) and the alkaline earth metals (Mg and Ca, together with Zn) are good reducing agents, the former being stronger than the latter. These same metals reduce the carbon-halogen bonds of alkyl halides. The halogen is converted to a halide anion, and the carbon bonds to the metal which has characteristics similar to a carbanion (R:-).
Formation of Organometallic Reagents
Many organometallic reagents are commercially available, however, it is often necessary to make then. The following equations illustrate these reactions for the commonly used metals lithium and magnesium (R may be hydrogen or alkyl groups in any combination).
• An Alkyl Lithium Reagent
$\ce{R3C-X} + \ce{2Li} \rightarrow \ce{R3C-Li} + \ce{LiX}$
• A Grignard Regent
$\ce{R3C-X} + \ce{Mg} \rightarrow \ce{R3C-MgX}$
Halide reactivity in these reactions increases in the order: Cl < Br < I and Fluorides are usually not used. The alkyl magnesium halides described in the second reaction are called Grignard Reagents after the French chemist, Victor Grignard, who discovered them and received the Nobel prize in 1912 for this work. The other metals mentioned above react in a similar manner, but Grignard and Alky Lithium Reagents most widely used. Although the formulas drawn here for the alkyl lithium and Grignard reagents reflect the stoichiometry of the reactions and are widely used in the chemical literature, they do not accurately depict the structural nature of these remarkable substances. Mixtures of polymeric and other associated and complexed species are in equilibrium under the conditions normally used for their preparation.
A suitable solvent must be used. For alkyl lithium formation pentane or hexane are usually used. Diethyl ether can also be used but the subsequent alkyl lithium reagent must be used immediately after preparation due to an interaction with the solvent. Ethyl ether or THF are essential for Grignard reagent formation. Lone pair electrons from two ether molecules form a complex with the magnesium in the Grignard reagent (As pictured below). This complex helps stabilize the organometallic and increases its ability to react.
These reactions are obviously substitution reactions, but they cannot be classified as nucleophilic substitutions, as were the earlier reactions of alkyl halides. Because the functional carbon atom has been reduced, the polarity of the resulting functional group is inverted (an originally electrophilic carbon becomes nucleophilic). This change, shown below, makes alkyl lithium and Grignard reagents excellent nucleophiles and useful reactants in synthesis.
Examples
Reaction of Organometallic Reagents with Various Carbonyls
Because organometallic reagents react as their corresponding carbanion, they are excellent nucleophiles. The basic reaction involves the nucleophilic attack of the carbanionic carbon in the organometallic reagent with the electrophilic carbon in the carbonyl to form alcohols.
Both Grignard and Organolithium Reagents will perform these reactions
Addition to formaldehyde gives 1o alcohols
Addition to aldehydes gives 2o alcohols
Addition to ketones gives 3o alcohols
Addition to carbon dioxide (CO2) forms a carboxylic acid
Examples
Going from Reactants to Products Simplified
Mechanism for the Addition to Carbonyls
The mechanism for a Grignard agent is shown. The mechanism for an organometallic reagent is the same.
1) Nucleophilic attack
2) Protonation
Organometallic Reagents as Bases
These reagents are very strong bases (pKa's of saturated hydrocarbons range from 42 to 50). Although not usually done with Grignard reagents, organolithium reagents can be used as strong bases. Both Grignard reagents and organolithium reagents react with water to form the corresponding hydrocarbon. This is why so much care is needed to insure dry glassware and solvents when working with organometallic reagents.
In fact, the reactivity of Grignard reagents and organolithium reagents can be exploited to create a new method for the conversion of halogens to the corresponding hydrocarbon (illustrated below). The halogen is converted to an organometallic reagent and then subsequently reacted with water to from an alkane.
Conjugate base anions of terminal alkynes (acetylide anions) are nucleophiles, and can do both nucleophilic substitution and nucleophilic addition reactions.
Formation of Acetylide Anions
Terminal alkynes are much more acidic than most other hydrocarbons. Removal of the proton leads to the formation of an acetylide anion, RC=C:-. The origin of the enhanced acidity can be attributed to the stability of the acetylide anion, which has the unpaired electrons in an sp hybridized orbital. The stability results from occupying an orbital with a high degree of s-orbital character.
There is a strong correlation between s-character in the orbital containing the non-bonding electrons in the anion and the acidity of hydrocarbons. The enhanced acidity with greater s-character occurs despite the fact that the homolytic C-H BDE is larger.
Compound Conjugate Base Hybridization "s Character" pKa C-H BDE (kJ/mol)
CH3CH3 CH3CH2- sp3 25% 50 410
CH2CH2 CH2CH- sp2 33% 44 473
HCCH HCC- sp 50% 25 523
Consequently, acetylide anions can be readily formed by deprotonation using a sufficiently strong base. Amide anion (NH2-), in the form of NaNH2 is commonly used for the formation of acetylide anions.
Nucleophilic Substitution Reactions of Acetylides
Acetylide anions are strong bases and strong nucleophiles. Therefore, they are able to displace halides and other leaving groups in substitution reactions. The product is a substituted alkyne.
Because the ion is a very strong base, the substitution reaction is most efficient with methyl or primary halides without substitution near the reaction center,
Secondary, tertiary or even bulky primary substrates will give elimination by the E2 mechanism.
Limitation of Organometallic Reagents
As discussed above, Grignard and organolithium reagents are powerful bases. Because of this they cannot be used as nucleophiles on compounds which contain acidic hydrogens. If they are used they will act as a base and deprotonate the acidic hydrogen rather than act as a nucleophile and attack the carbonyl. A partial list of functional groups which cannot be used are: alcohols, amides, 1o amines, 2o amines, carboxylic acids, and terminal alkynes.
Problems
1) Please write the product of the following reactions.
2) Please indicate the starting material required to produce the product.
3) Please give a detailed mechanism and the final product of this reaction
4) Please show two sets of reactants which could be used to synthesize the following molecule using a Grignard reaction.
Answers
1)
2)
3)
Nucleophilic attack
Protonation
4)
18.10: Reaction of Organometallic Reagents with Aldehydes and Ketones
Because organometallic reagents react as their corresponding carbanion, they are excellent nucleophiles. The basic reaction involves the nucleophilic attack of the carbanionic carbon in the organometallic reagent with the electrophilic carbon in the carbonyl to form alcohols.
Both Grignard and Organolithium Reagents will perform these reactions
Addition to formaldehyde gives 1o alcohols
Addition to aldehydes gives 2o alcohols
Addition to ketones gives 3o alcohols
Addition to carbon dioxide (CO2) forms a carboxylic acid
Examples
Going from Reactants to Products Simplified
Mechanism for the Addition to Carbonyls
The mechanism for a Grignard agent is shown. The mechanism for an organometallic reagent is the same.
1) Nucleophilic attack
2) Protonation
Nucleophilic Addition of Acetylides to Carbonyls
Acetylide anions will add to aldehydes and ketones to form alkoxides, which, upon protonation, give propargyl alcohols.
With aldehydes and non-symmetric ketones, in the absence of chiral catalyst, the product will be a racemic mixture of the two enantiomers.
The triple bond in the propargyl alcohol can be modified by using the reactivity of the alkyne. For example, Markovnikov and anti-Markovnikov hydration of the triple bond leads to formation of the hydroxy-substituted ketone and aldehyde, respectively, after enol-keto tautomerization. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/18%3A_Introduction_to_Carbonyl_Chemistry_Organometallic_Reagents_Oxidation_and_Reduction/18.09%3A_Organometallic_Reagents.txt |
Disconnection of bonds
Having chosen the TARGET molecule for synthesis, the next exercise is to draw out synthetic plans that would summarize all reasonable routes for its synthesis. During the past few decades, chemists have been working on a process called RETROSYNTHESIS. Retrosynthesis could be described as a logical Disconnection at strategic bonds in such a way that the process would progressively lead to easily available starting material(s) through several synthetic plans. Each plan thus evolved, describes a ‘ROUTE’ based on a retrosynthesis. Each disconnection leads to a simplified structure. The logic of such disconnections forms the basis for the retroanalysis of a given target molecule. Natural products have provided chemists with a large variety of structures, having complex functionalities and stereochemistry. This area has provided several challenging targets for development of these concepts. The underlining principle in devising logical approaches for synthetic routes is very much akin to the following simple problem. Let us have a look of the following big block, which is made by assembling several small blocks (Fig 1.4.2.1). You could easily see that the large block could be broken down in different ways and then reassembled to give the same original block.
Fig 1.4.2.1
Now let us try and extend the same approach for the synthesis of a simple molecule. Let us look into three possible ‘disconnections’ for a cyclohexane ring as shown in Fig 1.4.2.2.
Fig 1.4.2.2
In the above analysis we have attempted to develop three ways of disconnecting the six membered ring. Have we thus created three pathways for the synthesis of cyclohexane ring? Do such disconnections make chemical sense? The background of an organic chemist should enable him to read the process as a chemical reaction in the reverse (or ‘retro-‘) direction. The dots in the above structures could represent a carbonium ion, a carbanion, a free radical or a more complex reaction (such as a pericyclic reaction or a rearrangement). Applying such chemical thinking could open up several plausible reactions. Let us look into path b, which resulted from cleavage of one sigma bond. An anionic cyclisation route alone exposes several candidates as suitable intermediates for the formation of this linkage. The above analysis describes only three paths out of the large number of alternate cleavage routes that are available. An extended analysis shown below indicates more such possibilities (Fig 1.4.2.3). Each such intermediate could be subjected to further disconnection process and the process continued until we reach a reasonably small, easily available starting materials. Thus, a complete ‘SYNTHETIC TREE’ could be constructed that would summarize all possible routes for the given target molecule.
Fig 1.4.2.3
1.4.3 Efficiency of a route
A route is said to be efficient when the ‘overall yield’ of the total process is the best amongst all routes investigated. This would depend not only on the number of steps involved in the synthesis, but also on the type of strategy followed. The strategy could involve a ‘linear syntheses’ involving only consequential steps or a ‘convergent syntheses’ involving fewer consequential steps. Fig 1.4.3.1 shown below depicts a few patterns that could be recognized in such synthetic trees. When each disconnection process leads to only one feasible intermediate and the process proceeds in this fashion
Fig 1.4.3.1
all the way to one set of starting materials (SM), the process is called a Linear Synthesis. On the other hand, when an intermediate could be disconnected in two or more ways leading to different intermediates, branching occurs in the plan. The processes could be continued all the way to SMs. In such routes different branches of the synthetic pathways converge towards an intermediate. Such schemes are called Convergent Syntheses.
The flow charts shown below (Fig 1.4.3.2) depicts a hypothetical 5-step synthesis by the above two strategies. Assuming a very good yield (90%) at each step (this is rarely seen in real projects), a linier synthesis gives 59% overall yield, whereas a convergent synthesis gives 73% overall yield for the same number of steps..
Fig 1.4.3.2
1.4.4 Problem of substituents and stereoisomers
The situation becomes more complex when you consider the possibility of unwanted isomers generated at different steps of the synthesis. The overall yield drops down considerably for the synthesis of the right isomer. Reactions that yield single isomers (Diastereospecific reactions) in good yields are therefore preferred. Some reactions like the Diels Alder Reaction generate several stereopoints (points at which stereoisomers are generated) simultaneously in one step in a highly predictable manner. Such reactions are highly valued in planning synthetic strategies because several desirable structural features are introduced in one step. Where one pure enantiomer is the target, the situation is again complex. A pure compound in the final step could still have 50% unwanted enantiomer, thus leading to a drastic drop in the efficiency of the route. In such cases, it is desirable to separate the optical isomers as early in the route as possible, along the synthetic route. This is the main merit of the Chiron Approach, in which the right starting material is chosen from an easily available, cheap ‘chiral pool’. We would discuss this aspect after we have understood the logic of planning syntheses. Given these parameters, you could now decide on the most efficient route for any given target.
Molecules of interest are often more complex than the plain cyclohexane ring discussed above. They may have substituents and functional groups at specified points and even specific stereochemical points. Construction of a synthetic tree should ideally accommodate all these parameters to give efficient routes. Let us look into a slightly more complex example shown in Fig 1.4.4.1 . The ketone 1.4.4.1A is required as an intermediate in a synthesis. Unlike the plain cyclohexane discussed above, the substitution pattern and the keto- group in this molecule impose some restrictions on disconnection processes.
Fig 1.4.4.1
Cleavage a: This route implies attack of an anion of methylisopropylketone on a bromo-component. Cleavage b: This route implies simple regiospecific methylation of a larger ketone that bears all remaining structural elements. Cleavage c: This route implies three different possibilities. Route C-1 envisages an acylonium unit, which could come from an acid halide or an ester. Route C-2 implies an umpolung reaction at the acyl unit. Route C-3 suggests an oxidation of a secondary alcohol, which could be obtained through a Grignard-type reaction. Cleavage d: This implies a Micheal addition.
Each of these routes could be further developed backwards to complete the synthetic tree. These are just a few plausible routes to illustrate an important point that the details on the structure would restrict the possible cleavages to some strategic points. Notable contributions towards planning organic syntheses came from E.J. Corey’s school. These developments have been compiles by Corey in a book by the title LOgIC OF CHEMICAL SYNTHESIS. These and several related presentations on this topic should be taken as guidelines. They are devised after analyzing most of the known approaches published in the literature and identifying a pattern in the logic. They need not restrict the scope for new possibilities. Some of the important strategies are outlined below.
1.4.5 Preliminary scan
When a synthetic chemist looks at the given Target, he should first ponder on some preliminary steps to simplify the problem on hand. Is the molecule polymeric? See whether the whole molecule could be split into monomeric units, which could be coupled by a known reaction. This is easily seen in the case of peptides, nucleotides and organic polymers. This could also be true to other natural products. In molecules like C-Toxiferin 1 (1.4.5.1A) (Fig 1.4.5.1), the point of dimerisation is obvious. In several other cases, a deeper insight is required to identify the monomeric units, as is the case with Usnic acid (1.4.5.1B). In the case of the macrolide antibiotic Nonactin (1.4.5.1C), this strategy reduces the possibilities to the synthesis of a monomeric unit (1.4.5.1D). The overall structure has S4 symmetry and is achiral even though assembled from chiral precursors. Both (+)-nonactic acid and (−)-nonactic acid (1.4.5.1D) are needed to construct the macrocycle and they are joined head-to-tail in an alternating (+)-(−)-(+)-(−) pattern. (see J. Am. Chem. Soc., 131, 17155 (2009) and references cited therein).
Fig 1.4.5.1
Is a part of the structure already solved? Critical study of the literature may often reveal that the same molecule or a closely related one has been solved. R.B. Woodward synthesized (1.4.5.2C) as a key intermediate in an elegant synthesis of Reserpine (1.4.5.2A). The same intermediate compound (1.4.5.2C) became the key starting compound for Velluz et.al., in the synthesis of Deserpidine (1.4.5.2B) (Fig 1.4.5.2).
Fig 1.4.5.2
Such strategies reduce the time taken for the synthesis of new drug candidates. These strategies are often used in natural product chemistry and drug chemistry. Once the preliminary scan is complete, the target molecule could be disconnected at Strategic Bonds.
1.4.6 Strategic Bonds, Retrons and Transforms
STRATEGIC BONDS are the bonds that are cleaved to arrive at suitable Starting Materials (SM) or SYNTHONS. For the purpose of bond disconnection, Corey has suggested that the structure could be classified according to the sub-structures generated by known chemical reactions. He called the sub-structures RETRONS and the chemical transformations that generate these Retrons were called TRANSFORMS. A short list of Transforms and Retrons are given below (TABLE 1.4.6.1). Note that when Transforms generate Retrons, the product may have new STEREOPOINTS (stereochemical details) generated that may need critical appraisal.
Fig 1.4.6.1
The structure of the target could be such that the Retron and the corresponding Transforms could be easily visualized and directly applied. In some cases, the Transforms or the Retrons may not be obvious. In several syntheses, transformations do not simplify the molecule, but they facilitate the process of synthesis. For example, a keto- group could be generated through modification of a -CH-NO2 unit through a Nef reaction. This generates a new set of Retron / transforms pair. A few such transforms are listed below, along with the nomenclature suggested by Corey (Fig 1.4.6.2).
Fig 1.4.6.2
A Rearrangement Reaction could be a powerful method for generating suitable new sub-structures. In the following example, a suitable Pinacol Retron, needed for the rearrangement is obtained through an acyloin transform (Fig 1.4.6.3). Such rearrangement Retrons are often not obvious to inexperienced eyes.
Fig 1.4.6.3
Some transforms may be necessary to protect (acetals for ketones), modify (reduction of a ketone to alcohol to avoid an Aldol condensation during a Claisen condensation) or transpose a structural element such as a stereopoint (e.g. SN2 inversion, epimerization etc.,) or shifting a functional group. Such transforms do not simplify the given structural unit. At times, activation at specific points on the structure may be introduced to bring about a C-C bond formation and later the extra group may be removed. For example, consider the following retrosynthesis in which an extra ester group has been introduced to facilitate a Dieckmann Retron. In complex targets, combinations of such strategies could prove to be a very productive strategy in planning retrosynthesis. Witness the chemical modification strategy shown below for an efficient stereospecific synthesis of a trisubstituted olefin (Fig 1.4.6.4)
Fig 1.4.6.4
Fig 1.4.6.4 Examples for FGA / FGR strategies for complex targets
Amongst the molecular architectures, the bridged-rings pose a complex challenge in Structure-Based disconnection procedures. Corey has suggested guidelines for efficient disconnections of strategic bonds.
A bond cleavage for retrosynthesis should lead to simplified structures, preferably bearing five- or six-membered rings. The medium and large rings are difficult to synthesize stereospecifically. Amongst the common rings, a six-membered ring is easily approached and manipulated to large and small rings. Simultaneous cleavage of two bonds, suggesting cycloaddition – retrons are often more efficient. Some cleavages of strategic bonds are shown in Fig 1.4.6.5, suggesting good and poor cleavage strategies based on this approach. However, these guidelines are not restrictive.
Fig 1.4.6.5
Fig 1.4.6.5: Some cleavages at strategic bonds on bridged-ring systems.
Identifying Retron – Transform sets in a given target molecule is therefore a critical component in retrosynthesis. Such an approach could often generate several synthetic routes. The merit of this approach is that starting materials do not prejudice this logic. Retrosyntheses thus developed could throws open several routes that need further critical scrutiny on the basis of known facts.
Identification of Retrons / Transforms sets provided the prerequisite for computer assisted programs designed for generating retrosynthetic routes. A list of Retrons and the corresponding transforms were interlinked and the data was stored in the computer. All known reactions were thus analyzed for their Retron / Transform characteristics and documented. The appropriate literature citations were also documented and linked. Based on these inputs, computer programs were designed to generate retrosynthetic routes for any given structure. Several such programs are now available in the market to help chemists generate synthetic strategies. Given any structure, these programs generate several routes. Once the scientist identifies the specific routes of interest for further analysis, the program generates detailed synthetic steps, reagents required and the appropriate citations. In spite of such powerful artificial intelligence, the intelligence and intuitive genius of a chemist is still capable of generating a new strategy, not yet programmed. Again, human intelligence is still a critical input for the analysis the routes generated using a computer. Based on the experience of the chemists’ team, their projected aim of the project and facilities available, the routes are further screened.
Contributors
• Prof. R Balaji Rao (Department of Chemistry, Banaras Hindu University, Varanasi) as part of Information and Communication Technology | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/18%3A_Introduction_to_Carbonyl_Chemistry_Organometallic_Reagents_Oxidation_and_Reduction/18.11%3A_Retrosynthetic_Analysis_of_Grignard_Products.txt |
Compound 1.4.1.4A illustrates several important points in Protection / Deprotection protocol. Both the functional groups could react with a Grignard Reagent. Carboxylic acid group would first react with one mole of the Grignard Reagent to give a carboxylate anion salt. This anion does not react any further with the reagent. When two moles of Grignard Reagent are added to the reaction mixture, the second mole attacks the ketone to give a tertiary alcohol. On aqueous work-up, the acid group is regenerated. Thus, the first mole of the reagent provides a selective transient protection for the –COOH group. Once the acid group is esterified, such selectivity towards this reagent is lost. The reagent attacks at both sites. If reaction is desired only at the ester site, the keto- group should be selectively protected as an acetal. In the next step, the grignard reaction is carried out. Now the reagent has only one group available for reaction. On treatment with acid, the ketal protection in the intermediate compound is also hydrolyzed to regenerated the keto- group.
Fig 1.4.1.4
Protection of Aldehydes and Ketones
Since alcohols, aldehydes and ketones are the most frequently manipulated functional groups in organic synthesis, a great deal of work has appeared in their protection / deprotection strategies. In this discussion let us focus on the classes of protecting groups rather than an exhaustive treatment of all the protections.
Acetals
There are two general methods for the introduction of this protection. Transketalation is the method of choice when acetals (ketals) with methanol are desired. Acetone is the by-product, which has to be removed to shift the equilibrium to the right hand side. This is achieved by refluxing with a large excess of the acetonide reagent. Acetone formed is constantly distilled. In the case of cyclic diols, the water formed is continuously removed using a Dean-Stork condenser (Fig 1.4.1.6).
Fig 1.4.1.6
The rate of formation of ketals from ketones and 1,2-ethanediol (ethylene glycol), 1,3-propanediol and 2,2-dimethyl-1,3-propanediol are different. So is the deketalation reaction. This has enabled chemists to selectively work at one center. The following examples from steroid chemistry illustrate these points (Fig 1.4.1.7).
Fig 1.4.1.7
The demand for Green Chemistry processes has prompted search for new green procedures. Some examples from recent literature are given here (Fig 1.4.1.8).
Fig 1.4.1.8
Thioketals
Compared with their oxygen analogues, thioketals markedly differ in their chemistry. The formation as well as deprotection is promoted by suitable Lewis acids. The thioacetals are markedly stable under deketalation conditions, thus paving way for selective operations at two different centers. When conjugated ketones are involved, the ketal formation (as well as deprotection) proceeds with double bond migration. On the other hand, thioketals are formed and deketalated without double bond migration (Fig 1.4.1.9).
Fig 1.4.1.9
Silyl Ethers (R – OSiR3)
The oxygen – silicon sigma bond is stable to lithium and Grignard reagents, nucleophiles and hydride reagents but very unstable to water and mild aqueous acid and base conditions. A silyl ether of secondary alcohol is less reactive than that of a primary alcohol. The O – trimethylsilyl (O – SiMe3) was first protection of this class. (Fig 1.4.1.24).
Fig 1.4.1.24
Replacement of methyl group with other alkyl and aryl groups gives a large variety of silyl ether with varying degrees of stability towards hydrolysis (Fig 1.4.1.25).
Fig 1.4.1.25
The following examples illustrate the selectivity in formation and hydrolysis of this group (Fig 1.4.1.26).
Fig 1.4.1.26
Contributors
• Prof. R Balaji Rao (Department of Chemistry, Banaras Hindu University, Varanasi) as part of Information and Communication Technology | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/18%3A_Introduction_to_Carbonyl_Chemistry_Organometallic_Reagents_Oxidation_and_Reduction/18.12%3A_Protecting_Groups.txt |
Addition of Grignard reagents convert esters to 3o alcohols.
In effect the Grignard reagent adds twice.
Example 1:
Mechanism
1) Nucleophilic attack
2) Leaving group removal
3) Nucleophilic attack
4) Protonation
Example 1:
18.14: Reaction of Organometallic Reagents with Other Compounds
Organometallic reagents and carbon dioxide
Grignard reagents react with carbon dioxide in two stages. In the first, you get an addition of the Grignard reagent to the carbon dioxide. Dry carbon dioxide is bubbled through a solution of the Grignard reagent in ethoxyethane, made as described above. For example:
The product is then hydrolyzed (reacted with water) in the presence of a dilute acid. Typically, you would add dilute sulphuric acid or dilute hydrochloric acid to the solution formed by the reaction with the CO2. A carboxylic acid is produced with one more carbon than the original Grignard reagent. The usually quoted equation is (without the red bits):
Almost all sources quote the formation of a basic halide such as Mg(OH)Br as the other product of the reaction. That's actually misleading because these compounds react with dilute acids. What you end up with would be a mixture of ordinary hydrated magnesium ions, halide ions and sulfate or chloride ions - depending on which dilute acid you added.
18.15: -Unsaturated Carbonyl Compounds
Introduction
One of the largest and most diverse classes of reactions is composed of nucleophilic additions to a carbonyl group. Conjugation of a double bond to a carbonyl group transmits the electrophilic character of the carbonyl carbon to the beta-carbon of the double bond. These conjugated carbonyl are called enones or α, β unsaturated carbonyls. A resonance description of this transmission is shown below.
From this formula it should be clear that nucleophiles may attack either at the carbonyl carbon, as for any aldehyde, ketone or carboxylic acid derivative, or at the beta-carbon. These two modes of reaction are referred to as 1,2-addition and 1,4-addition respectively. A 1,4-addition is also called a conjugate addition.
Basic reaction of 1,2 addition
Here the nucleophile adds to the carbon which is in the one position. The hydrogen adds to the oxygen which is in the two position.
Basic reaction of 1,4 addition
In 1,4 addition the Nucleophile is added to the carbon β to the carbonyl while the hydrogen is added to the carbon α to the carbonyl.
Mechanism for 1,4 addition
1) Nucleophilic attack on the carbon β to the carbonyl
2) Proton Transfer
Here we can see why this addition is called 1,4. The nucleophile bonds to the carbon in the one position and the hydrogen adds to the oxygen in the four position.
3) Tautomerization
Going from reactant to products simplified
1,2 Vs. 1,4 addition
Whether 1,2 or 1,4-addition occurs depends on multiple variables but mostly it is determined by the nature of the nucleophile. During the addition of a nucleophile there is a competition between 1,2 and 1,4 addition products. If the nucleophile is a strong base, such as Grignard reagents, both the 1,2 and 1,4 reactions are irreversible and therefor are under kinetic control. Since 1,2-additions to the carbonyl group are fast, we would expect to find a predominance of 1,2-products from these reactions.
If the nucleophile is a weak base, such as alcohols or amines, then the 1,2 addition is usually reversible. This means the competition between 1,2 and 1,4 addition is under thermodynamic control. In this case 1,4-addition dominates because the stable carbonyl group is retained.
Gilman Reagents
Another important reaction exhibited by organometallic reagents is metal exchange. Organolithium reagents react with cuprous iodide to give a lithium dimethylcopper reagent, which is referred to as a Gilman reagent. Gilman reagents are a source of carbanion like nucleophiles similar to Grignard and Organo lithium reagents. However, the reactivity of organocuprate reagents is slightly different and this difference will be exploited in different situations. In the case of α, β unsaturated carbonyls organocuprate reagents allow for an 1,4 addition of an alkyl group. As we will see later Grignard and Organolithium reagents add alkyl groups 1,2 to α, β unsaturated carbonyls
Organocuprate reagents are made from the reaction of organolithium reagents and CuI
This acts as a source of R:-
Example
Nucleophiles which add 1,2 to α, β unsaturated carbonyls
Metal Hydrides
Grignard Reagents
Organolithium Reagents
18.16: SummaryThe Reactions of Organometallic Reagents
Nucleophilic Addition to Aldehydes and Ketones
The result of carbonyl bond polarization, however it is depicted, is straightforward to predict. The carbon, because it is electron-poor, is an electrophile: it is a great target for attack by an electron-rich nucleophilic group. Because the oxygen end of the carbonyl double bond bears a partial negative charge, anything that can help to stabilize this charge by accepting some of the electron density will increase the bond’s polarity and make the carbon more electrophilic. Very often a general acid group serves this purpose, donating a proton to the carbonyl oxygen.
The same effect can also be achieved if a Lewis acid, such as a magnesium ion, is located near the carbonyl oxygen.
Unlike the situation in a nucleophilic substitution reaction, when a nucleophile attacks an aldehyde or ketone carbon there is no leaving group – the incoming nucleophile simply ‘pushes’ the electrons in the pi bond up to the oxygen.
Alternatively, if you start with the minor resonance contributor, you can picture this as an attack by a nucleophile on a carbocation.
After the carbonyl is attacked by the nucleophile, the negatively charged oxygen has the capacity to act as a nucleophile. However, most commonly the oxygen acts instead as a base, abstracting a proton from a nearby acid group in the solvent or enzyme active site.
This very common type of reaction is called a nucleophilic addition. In many biologically relevant examples of nucleophilic addition to carbonyls, the nucleophile is an alcohol oxygen or an amine nitrogen, or occasionally a thiol sulfur. In one very important reaction type known as an aldol reaction (which we will learn about in section 13.3) the nucleophile attacking the carbonyl is a resonance-stabilized carbanion. In this chapter, we will concentrate on reactions where the nucleophile is an oxygen or nitrogen.
Nucleophilic Substitution of RCOZ (Z = Leaving Group)
Carbonyl compounds with leaving groups have reactions similar to aldehydes and ketones. The main difference is the presence of an electronegative substituent that can act as a leaving group during a nucleophile substitution reaction. Although there are many types of carboxylic acid derivatives known, this article focuses on four: acid halides, acid anhydrides, esters, and amides.
General reaction
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/18%3A_Introduction_to_Carbonyl_Chemistry_Organometallic_Reagents_Oxidation_and_Reduction/18.13%3A_Reaction_of_Organometallic_Reagents_with_Carboxylic_Acid_Derivatives.txt |
Disconnection of bonds
Having chosen the TARGET molecule for synthesis, the next exercise is to draw out synthetic plans that would summarize all reasonable routes for its synthesis. During the past few decades, chemists have been working on a process called RETROSYNTHESIS. Retrosynthesis could be described as a logical Disconnection at strategic bonds in such a way that the process would progressively lead to easily available starting material(s) through several synthetic plans. Each plan thus evolved, describes a ‘ROUTE’ based on a retrosynthesis. Each disconnection leads to a simplified structure. The logic of such disconnections forms the basis for the retroanalysis of a given target molecule. Natural products have provided chemists with a large variety of structures, having complex functionalities and stereochemistry. This area has provided several challenging targets for development of these concepts. The underlining principle in devising logical approaches for synthetic routes is very much akin to the following simple problem. Let us have a look of the following big block, which is made by assembling several small blocks (Fig 1.4.2.1). You could easily see that the large block could be broken down in different ways and then reassembled to give the same original block.
Fig 1.4.2.1
Now let us try and extend the same approach for the synthesis of a simple molecule. Let us look into three possible ‘disconnections’ for a cyclohexane ring as shown in Fig 1.4.2.2.
Fig 1.4.2.2
In the above analysis we have attempted to develop three ways of disconnecting the six membered ring. Have we thus created three pathways for the synthesis of cyclohexane ring? Do such disconnections make chemical sense? The background of an organic chemist should enable him to read the process as a chemical reaction in the reverse (or ‘retro-‘) direction. The dots in the above structures could represent a carbonium ion, a carbanion, a free radical or a more complex reaction (such as a pericyclic reaction or a rearrangement). Applying such chemical thinking could open up several plausible reactions. Let us look into path b, which resulted from cleavage of one sigma bond. An anionic cyclisation route alone exposes several candidates as suitable intermediates for the formation of this linkage. The above analysis describes only three paths out of the large number of alternate cleavage routes that are available. An extended analysis shown below indicates more such possibilities (Fig 1.4.2.3). Each such intermediate could be subjected to further disconnection process and the process continued until we reach a reasonably small, easily available starting materials. Thus, a complete ‘SYNTHETIC TREE’ could be constructed that would summarize all possible routes for the given target molecule.
Fig 1.4.2.3
1.4.3 Efficiency of a route
A route is said to be efficient when the ‘overall yield’ of the total process is the best amongst all routes investigated. This would depend not only on the number of steps involved in the synthesis, but also on the type of strategy followed. The strategy could involve a ‘linear syntheses’ involving only consequential steps or a ‘convergent syntheses’ involving fewer consequential steps. Fig 1.4.3.1 shown below depicts a few patterns that could be recognized in such synthetic trees. When each disconnection process leads to only one feasible intermediate and the process proceeds in this fashion
Fig 1.4.3.1
all the way to one set of starting materials (SM), the process is called a Linear Synthesis. On the other hand, when an intermediate could be disconnected in two or more ways leading to different intermediates, branching occurs in the plan. The processes could be continued all the way to SMs. In such routes different branches of the synthetic pathways converge towards an intermediate. Such schemes are called Convergent Syntheses.
The flow charts shown below (Fig 1.4.3.2) depicts a hypothetical 5-step synthesis by the above two strategies. Assuming a very good yield (90%) at each step (this is rarely seen in real projects), a linier synthesis gives 59% overall yield, whereas a convergent synthesis gives 73% overall yield for the same number of steps..
Fig 1.4.3.2
1.4.4 Problem of substituents and stereoisomers
The situation becomes more complex when you consider the possibility of unwanted isomers generated at different steps of the synthesis. The overall yield drops down considerably for the synthesis of the right isomer. Reactions that yield single isomers (Diastereospecific reactions) in good yields are therefore preferred. Some reactions like the Diels Alder Reaction generate several stereopoints (points at which stereoisomers are generated) simultaneously in one step in a highly predictable manner. Such reactions are highly valued in planning synthetic strategies because several desirable structural features are introduced in one step. Where one pure enantiomer is the target, the situation is again complex. A pure compound in the final step could still have 50% unwanted enantiomer, thus leading to a drastic drop in the efficiency of the route. In such cases, it is desirable to separate the optical isomers as early in the route as possible, along the synthetic route. This is the main merit of the Chiron Approach, in which the right starting material is chosen from an easily available, cheap ‘chiral pool’. We would discuss this aspect after we have understood the logic of planning syntheses. Given these parameters, you could now decide on the most efficient route for any given target.
Molecules of interest are often more complex than the plain cyclohexane ring discussed above. They may have substituents and functional groups at specified points and even specific stereochemical points. Construction of a synthetic tree should ideally accommodate all these parameters to give efficient routes. Let us look into a slightly more complex example shown in Fig 1.4.4.1 . The ketone 1.4.4.1A is required as an intermediate in a synthesis. Unlike the plain cyclohexane discussed above, the substitution pattern and the keto- group in this molecule impose some restrictions on disconnection processes.
Fig 1.4.4.1
Cleavage a: This route implies attack of an anion of methylisopropylketone on a bromo-component. Cleavage b: This route implies simple regiospecific methylation of a larger ketone that bears all remaining structural elements. Cleavage c: This route implies three different possibilities. Route C-1 envisages an acylonium unit, which could come from an acid halide or an ester. Route C-2 implies an umpolung reaction at the acyl unit. Route C-3 suggests an oxidation of a secondary alcohol, which could be obtained through a Grignard-type reaction. Cleavage d: This implies a Micheal addition.
Each of these routes could be further developed backwards to complete the synthetic tree. These are just a few plausible routes to illustrate an important point that the details on the structure would restrict the possible cleavages to some strategic points. Notable contributions towards planning organic syntheses came from E.J. Corey’s school. These developments have been compiles by Corey in a book by the title LOgIC OF CHEMICAL SYNTHESIS. These and several related presentations on this topic should be taken as guidelines. They are devised after analyzing most of the known approaches published in the literature and identifying a pattern in the logic. They need not restrict the scope for new possibilities. Some of the important strategies are outlined below.
Contributors
• Prof. R Balaji Rao (Department of Chemistry, Banaras Hindu University, Varanasi) as part of Information and Communication Technology | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/18%3A_Introduction_to_Carbonyl_Chemistry_Organometallic_Reagents_Oxidation_and_Reduction/18.17%3A_Synthesis.txt |
Organophosphorus ylides react with aldehydes or ketones to give substituted alkenes in a transformation called the Wittig reaction. This reaction is named for George Wittig who was awarded the Nobel prize for this work in 1979. A principal advantage of alkene synthesis by the Wittig reaction is that the location of the double bond is absolutely fixed, in contrast to the mixtures often produced by alcohol dehydration.
Preparation of Phosphorus Ylides
It has been noted that dipolar phosphorus compounds are stabilized by p-d bonding. This bonding stabilization extends to carbanions adjacent to phosphonium centers, and the zwitterionic conjugate bases derived from such cations are known as ylides. An ylide is defined as a compound with opposite charges on adjacent atoms both of which have complete octets. For the Wittig reaction discussed below an organophosphorus ylide, also called Wittig reagents, will be used. The ability of phosphorus to hold more than eight valence electrons allows for a resonance structure to be drawn forming a double bonded structure.
The stabilization of the carbanion provided by the phosphorus causes an increase in acidity (pKa ~35). Very strong bases, such as butyl lithium, are required for complete formation of ylides.
The ylides shown here are all strong bases. Like other strongly basic organic reagents, they are protonated by water and alcohols, and are sensitive to oxygen. Water decomposes phosphorous ylides to hydrocarbons and phosphine oxides, as shown.
Although many ylides are commercially available it is often necessary to create them synthetically. Ylides can be synthesized from an alkyl halide and a trialkyl phosphine. Typically triphenyl phosphine is used to synthesize ylides. Because a SN2 reaction is used in the ylide synthesis methyl and primary halides perform the best. Secondary halides can also be used but the yields are generally lower. This should be considered when planning out a synthesis which involves a synthesized Wittig reagent.
1) SN2 reaction
2) Deprotonation
The Wittig Reaction
The most important use of ylides in synthesis comes from their reactions with aldehydes and ketones, which are initiated in every case by a covalent bonding of the nucleophilic alpha-carbon to the electrophilic carbonyl carbon. Ylides react to give substituted alkenes in a transformation called the Wittig reaction. This reaction is named for George Wittig who was awarded the Nobel prize for this work in 1979. A principal advantage of alkene synthesis by the Wittig reaction is that the location of the double bond is absolutely fixed, in contrast to the mixtures often produced by alcohol dehydration.
Going from reactants to products simplified
Mechanism of the Wittig reaction
Following the initial carbon-carbon bond formation, two intermediates have been identified for the Wittig reaction, a dipolar charge-separated species called a betaine and a four-membered heterocyclic structure referred to as an oxaphosphatane. Cleavage of the oxaphosphatane to alkene and phosphine oxide products is exothermic and irreversible.
1) Nucleophillic attack on the carbonyl
2) Formation of a 4 membered ring
3) Formation of the alkene
Limitation of the Wittig reaction
If possible both E and Z isomer of the double bond will be formed. This should be considered when planning a synthesis involving a Wittig Reaction.
Problems
1) Please write the product of the following reactions.
2) Please indicate the starting material required to produce the product.
3) Please draw the structure of the oxaphosphetane which is made during the mechanism of the reaction given that produces product C.
4) Please draw the structure of the betaine which is made during the mechanism of the reaction given that produces product D.
5) Please give a detailed mechanism and the final product of this reaction
6) It has been shown that reacting and epoxide with triphenylphosphine forms an alkene. Please propose a mechanism for this reaction. Review the section on epoxide reactions if you need help.
Answers
1)
2)
3)
4)
5)
Nucleophillic attack on the carbonyl
Formation of a 4 membered ring
Formation of the alkene
6) Nucleophillic attack on the epoxide
Formation of a 4 membered ring
Formation of the alkene
19.02: Addition of 1 Amines
The reaction of aldehydes and ketones with ammonia or 1º-amines forms imine derivatives, also known as Schiff bases (compounds having a C=N function). Water is eliminated in the reaction, which is acid-catalyzed and reversible in the same sense as acetal formation. The pH for reactions which form imine compounds must be carefully controlled. The rate at which these imine compounds are formed is generally greatest near a pH of 5, and drops at higher and lower pH's. At high pH there will not be enough acid to protonate the OH in the intermediate to allow for removal as H2O. At low pH most of the amine reactant will be tied up as its ammonium conjugate acid and will become non-nucleophilic.
Converting reactants to products simply
Mechanism of imine formation
1) Nucleophilic attack
2) Proton transfer
3) Protonation of OH
4) Removal of water
5) Deprotonation
Reversibility of imine forming reactions
Imines can be hydrolyzed back to the corresponding primary amine under acidic conditons.
Reactions involving other reagents of the type Y-NH2
Imines are sometimes difficult to isolate and purify due to their sensitivity to hydrolysis. Consequently, other reagents of the type Y–NH2 have been studied, and found to give stable products (R2C=N–Y) useful in characterizing the aldehydes and ketones from which they are prepared. Some of these reagents are listed in the following table, together with the structures and names of their carbonyl reaction products. Hydrazones are used as part of the Wolff-Kishner reduction and will be discussed in more detail in another module.
With the exception of unsubstituted hydrazones, these derivatives are easily prepared and are often crystalline solids - even when the parent aldehyde or ketone is a liquid. Since melting points can be determined more quickly and precisely than boiling points, derivatives such as these are useful for comparison and identification of carbonyl compounds. It should be noted that although semicarbazide has two amino groups (–NH2) only one of them is a reactive amine. The other is amide-like and is deactivated by the adjacent carbonyl group.
Problems
1)Please draw the products of the following reactions.
2) Please draw the structure of the reactant needed to produce the indicated product.
3) Please draw the products of the following reactions.
1)
2)
3)
19.03: Addition of 2 Amines
Introduction
Most aldehydes and ketones react with 2º-amines to give products known as enamines. It should be noted that, like acetal formation, these are acid-catalyzed reversible reactions in which water is lost. Consequently, enamines are easily converted back to their carbonyl precursors by acid-catalyzed hydrolysis.
Example
Mechanism
1) Nuleophilic attack
2) Proton transfer
3) Protonation of OH
4) Removal of water
5) Deprotonation
Example
Problems
1) Please draw the products for the following reactions.
2) Please give the structure of the reactant needed to product the following product
1)
2) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/19%3A_Aldehydes_and_KetonesNucleophilic_Addition/19.01%3A_The_Wittig_Reaction.txt |
It has been demonstrated that water, in the presence of an acid or a base, adds rapidly to the carbonyl function of aldehydes and ketones establishing a reversible equilibrium with a hydrate (geminal-diol or gem-diol). The word germinal or gem comes from the Latin word for twin, geminus.
Example 1
Reversibility of the Reaction
Isolation of gem-diols is difficult because the reaction is reversibly. Removal of the water during a reaction can cause the conversion of a gem-diol back to the corresponding carbonyl.
Factors Affecting the Gem-diol Equilibrium
In most cases the resulting gem-diol is unstable relative to the reactants and cannot be isolated. Exceptions to this rule exist, one being formaldehyde where the weaker pi-component of the carbonyl double bond, relative to other aldehydes or ketones, and the small size of the hydrogen substituents favor addition. Thus, a solution of formaldehyde in water (formalin) is almost exclusively the hydrate, or polymers of the hydrate. The addition of electron donating alkyl groups stabilized the partial positive charge on the carbonyl carbon and decreases the amount of gem-diol product at equilibrium. Because of this ketones tend to form less than 1% of the hydrate at equilibrium. Likewise, the addition of strong electron-withdrawing groups destabilizes the carbonyl and tends to form stable gem-diols. Two examples of this are chloral, and 1,2,3-indantrione. It should be noted that chloral hydrate is a sedative and has been added to alcoholic beverages to make a “Knock-out” drink also called a Mickey Finn. Also, ninhydrin is commonly used by forensic investigators to resolve finger prints.
Mechanism of Gem-diol Formation
The mechanism is catalyzed by the addition of an acid or base. Note! This may speed up the reaction but is has not effect on the equilibriums discussed above. Basic conditions speed up the reaction because hydroxide is a better nucleophilic than water. Acidic conditions speed up the reaction because the protonated carbonyl is more electrophilic.
Basic conditions
1) Nucleophilic attack by hydroxide
2) Protonation of the alkoxide
Acidic conditions
1) Protonation of the carbonyl
2) Nucleophilic attack by water
3) Deprotonation
Problems
1) Draw the expected products of the following reactions.
2) Of the following pairs of molecules which would you expect to form a larger percentage of gem-diol at equilibrium? Please explain your answer.
3) Would you expect the following molecule to form appreciable amount of gem-diol in water? Please explain your answer.
Answers
1)
2) The compound on the left would. Fluorine is more electronegative than bromine and would remove more electron density from the carbonyl carbon. This would destabilize the carbonyl allowing for more gem-diol to form.
3) Although ketones tend to not form gem-diols this compound exists almost entirely in the gem-diol form when placed in water. Ketones tend to not form gem-diols because of the stabilizing effect of the electron donating alkyl group. However, in this case the electron donating effects of alkyl group is dominated by the presence of six highly electronegative fluorines.
19.05: Addition of AlcoholsAcetal Formation
In this organic chemistry topic, we shall see how alcohols (R-OH) add to carbonyl groups. Carbonyl groups are characterized by a carbon-oxygen double bond. The two main functional groups that consist of this carbon-oxygen double bond are Aldehydes and Ketones.
Introduction
It has been demonstrated that water adds rapidly to the carbonyl function of aldehydes and ketones to form geminal-diol. In a similar reaction alcohols add reversibly to aldehydes and ketones to form hemiacetals (hemi, Greek, half). This reaction can continue by adding another alcohol to form an acetal. Hemiacetals and acetals are important functional groups because they appear in sugars.
To achieve effective hemiacetal or acetal formation, two additional features must be implemented. First, an acid catalyst must be used because alcohol is a weak nucleophile; and second, the water produced with the acetal must be removed from the reaction by a process such as a molecular sieves or a Dean-Stark trap. The latter is important, since acetal formation is reversible. Indeed, once pure hemiacetal or acetals are obtained they may be hydrolyzed back to their starting components by treatment with aqueous acid and an excess of water.
Formation of Hemiacetals
Example 1: Formation of Hemiacetals
Example 2: Hemiacetal Reversibility
Formation of Acetals
Acetals are geminal-diether derivatives of aldehydes or ketones, formed by reaction with two equivalents (or an excess amount) of an alcohol and elimination of water. Ketone derivatives of this kind were once called ketals, but modern usage has dropped that term. It is important to note that a hemiacetal is formed as an intermediate during the formation of an acetal.
Example 3: Formation of Acetals
Example 4: Acetal Reversibility
Mechanism for Hemiacetal and Acetal Formation
The mechanism shown here applies to both acetal and hemiacetal formation
1) Protonation of the carbonyl
2) Nucleophilic attack by the alcohol
3) Deprotonation to form a hemiacetal
4) Protonation of the alcohol
5) Removal of water
6) Nucleophilic attack by the alcohol
7) Deprotonation by water | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/19%3A_Aldehydes_and_KetonesNucleophilic_Addition/19.04%3A_Addition_of_%28H_2O%29Hydration.txt |
Acetals as Protecting Groups
The importance of acetals as carbonyl derivatives lies chiefly in their stability and lack of reactivity in neutral to strongly basic environments. As long as they are not treated by acids, especially aqueous acid, acetals exhibit all the lack of reactivity associated with ethers in general. Among the most useful and characteristic reactions of aldehydes and ketones is their reactivity toward strongly nucleophilic (and basic) metallo-hydride, alkyl and aryl reagents. If the carbonyl functional group is converted to an acetal these powerful reagents have no effect; thus, acetals are excellent protective groups, when these irreversible addition reactions must be prevented.
In the following example we would like a Grignard reagent to react with the ester and not the ketone. This cannot be done without a protecting group because Grignard reagents react with esters and ketones.
19.07: Cyclic Hemiacetals
Formation of Cyclic Hemiacetal and Acetals
Molecules which have an alcohol and a carbonyl can undergo an intramolecular reaction to form a cyclic hemiacetal.
Intramolecular Hemiacetal formation is common in sugar chemistry. For example, the common sugar glucose exists in the cylcic manner more than 99% of the time in a mixture of aqueous solution.
Carbonyls reacting with diol produce a cyclic acetal. A common diol used to form cyclic acetals is ethylene glycol.
19.08: An Introduction to Carbohydrates
Carbohydrates are the most abundant class of organic compounds found in living organisms. They originate as products of photosynthesis, an endothermic reductive condensation of carbon dioxide requiring light energy and the pigment chlorophyll.
$nCO_2 + n H_2O + \text{Energy} \rightarrow C_nH_{2n}O_n + nO_2$
As noted here, the formulas of many carbohydrates can be written as carbon hydrates, $C_n(H_2O)_n$, hence their name. The carbohydrates are a major source of metabolic energy, both for plants and for animals that depend on plants for food. Aside from the sugars and starches that meet this vital nutritional role, carbohydrates also serve as a structural material (cellulose), a component of the energy transport compound ATP/ADP, recognition sites on cell surfaces, and one of three essential components of DNA and RNA.
The most useful carbohydrate classification scheme divides the carbohydrates into groups according to the number of individual simple sugar units. Monosaccharides contain a single unit; disaccharides contain two sugar units; and polysaccharides contain many sugar units as in polymers - most contain glucose as the monosaccharide unit.
Some sugars can undergo a intermolecular cyclization to form a hemi-acetal. The hemiacetal carbon atom (C-1) becomes a new stereogenic center, commonly referred to as the anomeric carbon, and the α and β-isomers are called anomers.
Disaccharides made up of other sugars are known, but glucose is often one of the components. Two important examples of such mixed disaccharides are displayed above. Lactose, also known as milk sugar, is a galactose-glucose compound joined as a beta-glycoside. It is a reducing sugar because of the hemiacetal function remaining in the glucose moiety. Many adults, particularly those from regions where milk is not a dietary staple, have a metabolic intolerance for lactose. Infants have a digestive enzyme which cleaves the beta-glycoside bond in lactose, but production of this enzyme stops with weaning. Sucrose, or cane sugar, is our most commonly used sweetening agent. It is a non-reducing disaccharide composed of glucose and fructose joined at the anomeric carbon of each by glycoside bonds (one alpha and one beta). In the formula shown here the fructose ring has been rotated 180º from its conventional perspective.
19.09: Introduction
A carbonyl group is a chemically organic functional group composed of a carbon atom double-bonded to an oxygen atom --> [C=O] The simplest carbonyl groups are aldehydes and ketones usually attached to another carbon compound. These structures can be found in many aromatic compounds contributing to smell and taste.
The Carbonyl Group
C=O is prone to additions and nucleophillic attack because or carbon's positive charge and oxygen's negative charge. The resonance of the carbon partial positive charge allows the negative charge on the nucleophile to attack the Carbonyl group and become a part of the structure and a positive charge (usually a proton hydrogen) attacks the oxygen. Just a reminder, the nucleophile is a good acid therefore "likes protons" so it will attack the side with a positive charge.
Before we consider in detail the reactivity of aldehydes and ketones, we need to look back and remind ourselves of what the bonding picture looks like in a carbonyl. Carbonyl carbons are sp2 hybridized, with the three sp2 orbitals forming soverlaps with orbitals on the oxygen and on the two carbon or hydrogen atoms. These three bonds adopt trigonal planar geometry. The remaining unhybridized 2p orbital on the central carbonyl carbon is perpendicular to this plane, and forms a ‘side-by-side’ pbond with a 2p orbital on the oxygen.
The carbon-oxygen double bond is polar: oxygen is more electronegative than carbon, so electron density is higher on the oxygen side of the bond and lower on the carbon side. Recall that bond polarity can be depicted with a dipole arrow, or by showing the oxygen as holding a partial negative charge and the carbonyl carbon a partial positive charge.
A third way to illustrate the carbon-oxygen dipole is to consider the two main resonance contributors of a carbonyl group: the major form, which is what you typically see drawn in Lewis structures, and a minor but very important contributor in which both electrons in the pbond are localized on the oxygen, giving it a full negative charge. The latter depiction shows the carbon with an empty 2p orbital and a full positive charge.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/19%3A_Aldehydes_and_KetonesNucleophilic_Addition/19.06%3A_Acetals_as_Protecting_Groups.txt |
he most potent and varied odors are aldehydes. Ketones are widely used as industrial solvents. Aldehydes and ketones contain the carbonyl group. Aldehydes are considered the most important functional group. They are often called the formyl or methanoyl group. Aldehydes derive their name from the dehydration of alcohols. Aldehydes contain the carbonyl group bonded to at least one hydrogen atom. Ketones contain the carbonyl group bonded to two carbon atoms.
Introduction
Aldehydes and ketones are organic compounds which incorporate a carbonyl functional group, C=O. The carbon atom of this group has two remaining bonds that may be occupied by hydrogen, alkyl or aryl substituents. If at least one of these substituents is hydrogen, the compound is an aldehyde. If neither is hydrogen, the compound is a ketone.
Naming Aldehydes
The IUPAC system of nomenclature assigns a characteristic suffix -al to aldehydes. For example, H2C=O is methanal, more commonly called formaldehyde. Since an aldehyde carbonyl group must always lie at the end of a carbon chain, it is always is given the #1 location position in numbering and it is not necessary to include it in the name. There are several simple carbonyl containing compounds which have common names which are retained by IUPAC.
Also, there is a common method for naming aldehydes and ketones. For aldehydes common parent chain names, similar to those used for carboxylic acids, are used and the suffix –aldehyde is added to the end. In common names of aldehydes, carbon atoms near the carbonyl group are often designated by Greek letters. The atom adjacent to the carbonyl function is alpha, the next removed is beta and so on.
If the aldehyde moiety (-CHO) is attached to a ring the suffix –carbaldehyde is added to the name of the ring. The carbon attached to this moiety will get the #1 location number in naming the ring.
Summary of Aldehyde Nomenclature rules
1. Aldehydes take their name from their parent alkane chains. The -e is removed from the end and is replaced with -al.
2. The aldehyde funtional group is given the #1 numbering location and this number is not included in the name.
3. For the common name of aldehydes start with the common parent chain name and add the suffix -aldehyde. Substituent positions are shown with Greek letters.
4. When the -CHO functional group is attached to a ring the suffix -carbaldehyde is added, and the carbon attached to that group is C1.
Example 1
The IUPAC system names are given on top while the common name is given on the bottom in parentheses.
Naming Ketones
The IUPAC system of nomenclature assigns a characteristic suffix of -one to ketones. A ketone carbonyl function may be located anywhere within a chain or ring, and its position is usually given by a location number. Chain numbering normally starts from the end nearest the carbonyl group. Very simple ketones, such as propanone and phenylethanone do not require a locator number, since there is only one possible site for a ketone carbonyl function
The common names for ketones are formed by naming both alkyl groups attached to the carbonyl then adding the suffix -ketone. The attached alkyl groups are arranged in the name alphabetically.
Summary of Ketone Nomenclature rules
1. Ketones take their name from their parent alkane chains. The ending -e is removed and replaced with -one.
2. The common name for ketones are simply the substituent groups listed alphabetically + ketone.
3. Some common ketones are known by their generic names. Such as the fact that propanone is commonly referred to as acetone.
Example 2
The IUPAC system names are given on top while the common name is given on the bottom in parentheses.
Naming Aldehydes and Ketones in the Same Molecule
As with many molecules with two or more functional groups, one is given priority while the other is named as a substituent. Because aldehydes have a higher priority than ketones, molecules which contain both functional groups are named as aldehydes and the ketone is named as an "oxo" substituent. It is not necessary to give the aldehyde functional group a location number, however, it is usually necessary to give a location number to the ketone.
Example 3
Naming Dialdehydes and Diketones
For dialdehydes the location numbers for both carbonyls are omitted because the aldehyde functional groups are expected to occupy the ends of the parent chain. The ending –dial is added to the end of the parent chain name.
Example 4
For diketones both carbonyls require a location number. The ending -dione or -dial is added to the end of the parent chain.
Example 5
Naming Cyclic Ketones and Diketones
In cyclic ketones the carbonyl group is assigned location position #1, and this number is not included in the name, unless more than one carbonyl group is present. The rest of the ring is numbered to give substituents the lowest possible location numbers. Remember the prefix cyclo is included before the parent chain name to indicate that it is in a ring. As with other ketones the –e ending is replaced with the –one to indicate the presence of a ketone.
With cycloalkanes which contain two ketones both carbonyls need to be given a location numbers. Also, an –e is not removed from the end but the suffix –dione is added.
Example 6
Naming Carbonyls and Hydroxyls in the Same Molecule
When and aldehyde or ketone is present in a molecule which also contains an alcohol functional group the carbonyl is given nomenclature priority by the IUPAC system. This means that the carbonyl is given the lowest possible location number and the appropriate nomenclature suffix is included. In the case of alcohols the OH is named as a hydroxyl substituent. However, the l in hydroxyl is generally removed.
Example 7
Naming Carbonyls and Alkenes in the Same Molecule
When and aldehyde or ketone is present in a molecule which also contains analkene functional group the carbonyl is given nomenclature priority by the IUPAC system. This means that the carbonyl is given the lowest possible location number and the appropriate nomenclature suffix is included.
When carbonyls are included with an alkene the following order is followed:
(Location number of the alkene)-(Prefix name for the longest carbon chain minus the -ane ending)-(an -en ending to indicate the presence of an alkene)-(the location number of the carbonyl if a ketone is present)-(either an –one or and -anal ending).
Remember that the carbonyl has priority so it should get the lowest possible location number. Also, remember that cis/tran or E/Z nomenclature for the alkene needs to be included if necessary.
Example 8
Aldehydes and Ketones as Fragments
• Alkanoyl is the common name of the fragment, though the older naming, acyl, is still widely used.
• Formyl is the common name of the fragment.
• Acety is the common name of the CH3-C=O- fragment.
Example 9
Additional Examples of Carbonyl Nomenclature
1) Please give the IUPAC name for each compound:
Answers for Question 1
1. 3,4-Dimethylhexanal
2. 5-Bromo-2-pentanone
3. 2,4-Hexanedione
4. cis-3-Penenal
5. 6-methyl-5-Hepten-3-one
6. 3-hydroxy-2,4-Pentanedione
7. 1,2-Cyclobutanedione
8. 2-methyl-Propanedial
9. 3-methyl-5-oxo-Hexanal
10. cis-2,3-dihydroxycyclohexanone
11. 2-methylcyclopentanecarboaldehyde
12. 3-bromo-2-methylpropanal
2) Please give the structure corresponding to each name:
A) Butanal
B) 2-Hydroxycyclopentanone
C) 2,3-Pentanedione
D) 1,3-Cyclohexanedione
E) 3,4-Dihydoxy-2-butanone
F) (E) 3-methyl-2-Hepten-4-one
G) 3-Oxobutanal
H) cis-3-Bromocyclohexanecarboaldehyde
I) Butanedial
J) trans-2-methyl-3-Hexenal
Answers to question 2: | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/19%3A_Aldehydes_and_KetonesNucleophilic_Addition/19.10%3A_Nomenclature.txt |
Introduction
A comparison of the properties and reactivity of aldehydes and ketones with those of the alkenes is warranted, since both have a double bond functional group. Because of the greater electronegativity of oxygen, the carbonyl group is polar, and aldehydes and ketones have larger molecular dipole moments (D) than do alkenes. The resonance structures in Figure 1 illustrate this polarity, and the relative dipole moments of formaldehyde, other aldehydes and ketones confirm the stabilizing influence that alkyl substituents have on carbocations (the larger the dipole moment the greater the polar character of the carbonyl group). We expect, therefore, that aldehydes and ketones will have higher boiling points than similar sized alkenes. Furthermore, the presence of oxygen with its non-bonding electron pairs makes aldehydes and ketones hydrogen-bond acceptors, and should increase their water solubility relative to hydrocarbons. Specific examples of these relationships are provided in the following table.
Figure 1: Resonance structures
Compound
Mol. Wt.
Boiling Point
Water
Solubility
(CH3)2C=CH2
56
-7.0 ºC
0.04 g/100
(CH3)2C=O
58
56.5 ºC
infinite
CH3CH2CH2CH=CH2
70
30.0 ºC
0.03 g/100
CH3CH2CH2CH=O
72
76.0 ºC
7 g/100
96
103.0 ºC
insoluble
98
155.6 ºC
5 g/100
The polarity of the carbonyl group also has a profound effect on its chemical reactivity, compared with the non-polar double bonds of alkenes. Thus, reversible addition of water to the carbonyl function is fast, whereas water addition to alkenes is immeasurably slow in the absence of a strong acid catalyst. Curiously, relative bond energies influence the thermodynamics of such addition reactions in the opposite sense.
The C=C of alkenes has an average bond energy of 146 kcal/mole. Since a C–C σ-bond has a bond energy of 83 kcal/mole, the π-bond energy may be estimated at 63 kcal/mole (i.e. less than the energy of the sigma bond). The C=O bond energy of a carbonyl group, on the other hand, varies with its location, as follows:
H2C=O 170 kcal/mole
RCH=O 175 kcal/mole
R2C=O 180 kcal/mole
The C–O σ-bond is found to have an average bond energy of 86 kcal/mole. Consequently, with the exception of formaldehyde, the carbonyl function of aldehydes and ketones has a π-bond energy greater than that of the sigma-bond, in contrast to the pi-sigma relationship in C=C. This suggests that addition reactions to carbonyl groups should be thermodynamically disfavored, as is the case for the addition of water. All of this is summarized in the following diagram (ΔHº values are for the addition reaction).
Although the addition of water to an alkene is exothermic and gives a stable product (an alcohol), the uncatalyzed reaction is extremely slow due to a high activation energy. The reverse reaction (dehydration of an alcohol) is even slower, and because of the kinetic barrier, both reactions are practical only in the presence of a strong acid.
In contrast, both the endothermic addition of water to a carbonyl function, and the exothermic elimination of water from the resulting geminal-diol are fast. The inherent polarity of the carbonyl group, together with its increased basicity (compared with alkenes), lowers the transition state energy for both reactions, with a resulting increase in rate. Acids and bases catalyze both the addition and elimination of water. Proof that rapid and reversible addition of water to carbonyl compounds occurs is provided by experiments using isotopically labeled water. If a carbonyl reactant composed of 16O (colored blue above) is treated with water incorporating the 18O isotope (colored red above), a rapid exchange of the oxygen isotope occurs. This can only be explained by the addition-elimination mechanism shown here.
Contributors
• William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
19.12: Spectroscopic Properties
IR Spectra
The carbonyl stretching vibration band C=O of saturated aliphatic ketones appears:
• C=O stretch - aliphatic ketones 1715 cm-1
- ?, ?-unsaturated ketones 1685-1666 cm-1
Figure 8. shows the spectrum of 2-butanone. This is a saturated ketone, and the C=O band appears at 1715.
Figure 8. Infrared Spectrum of 2-Butanone
If a compound is suspected to be an aldehyde, a peak always appears around 2720 cm-1 which often appears as a shoulder-type peak just to the right of the alkyl C–H stretches.
• H–C=O stretch 2830-2695 cm-1
• C=O stretch:
• aliphatic aldehydes 1740-1720 cm-1
• alpha, beta-unsaturated aldehydes 1710-1685 cm-1
Figure 9. shows the spectrum of butyraldehyde.
Figure 9. Infrared Spectrum of Butyraldehyde
NMR Spectra
Hydrogens attached to carbon adjacent to the sp2 hybridized carbon in aldehydes and ketones usually show up 2.0-2.5 ppm.
.
Aldehyde hydrogens are highly deshielded and appear far downfield as 9-10 ppm.
Example
Chemical shift of each protons is predicted by 1H chemical shift ranges (Ha): chemical shift of methyl groups (1.1 ppm). (Hb) The chemical shift of the -CH- group move downfield due to effect an adjacent aldehyde group: (2.4 ppm). The chemical shift of aldehyde hydrogen is highly deshielded (9.6 ppm).
4) Splitting pattern is determined by (N+1) rule: Ha is split into two peaks by Hb(#of proton=1). Hb has the septet pattern by Ha (#of proton=6). Hc has one peak.(Note that Hc has doublet pattern by Hb due to vicinal proton-proton coupling.) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/19%3A_Aldehydes_and_KetonesNucleophilic_Addition/19.11%3A_Physical_Properties.txt |
Aldehydes and ketones are widespread in nature and are often combined with other functional groups. Examples of naturally occurring molecules which contain a aldehyde or ketone functional group are shown in the following two figures. The compounds in the figure 1 are found chiefly in plants or microorganisms and those in the figure 2 have animal origins. Many of these molecular structures are chiral.
When chiral compounds are found in nature they are usually enantiomerically pure, although different sources may yield different enantiomers. For example, carvone is found as its levorotatory (R)-enantiomer in spearmint oil, whereas, caraway seeds contain the dextrorotatory (S)-enantiomer. In this case the change of the stereochemistry causes a drastic change in the perceived scent. Aldehydes and ketones are known for their sweet and sometimes pungent odors. The odor from vanilla extract comes from the molecule vanillin. Likewise, benzaldehyde provides a strong scent of almonds and is this author’s favorite chemical smell. Because of their pleasant fragrances aldehyde and ketone containing molecules are often found in perfumes. However, not all of the fragrances are pleasing. In particular, 2-Heptanone provides part of the sharp scent from blue cheese and (R)-Muscone is part of the musky smell from the Himalayan musk deer. Lastly, ketones show up in many important hormones such as progesterone (a female sex hormone) and testosterone (a male sex hormone). Notice how subtle differences in structure can cause drastic changes in biological activity. The ketone functionality also shows up in the anti-inflammatory steroid, Cortisone.
Figure 1. Aldehyde and ketone containing molecules isolated from plant sources.
Figure 2. Aldehyde and ketone containing molecules isolated from animal sources.
19.14: Preparation of Aldehydes and Ketones
Aldehydes and ketones can be prepared using a wide variety of reactions. Although these reactions are discussed in greater detail in other sections, they are listed here as a summary and to help with planning multistep synthetic pathways. Please use the appropriate links to see more details about the reactions.
Hydration of an alkyne to form aldehydes
Anti-Markovnikov addition of a hydroxyl group to an alkyne forms an aldehyde. The addition of a hydroxyl group to an alkyne causes tautomerization which subsequently forms a carbonyl.
Oxidation of 2o alcohols to form ketones
Typically uses Jones reagent (CrO3 in H2SO4) but many other reagents can be used
Hydration of an alkyne to form ketones
The addition of a hydroxyl group to an alkyne causes tautomerization which subsequently forms a carbonyl. Markovnikov addition of a hydroxyl group to an alkyne forms a ketone.
Alkenes can be cleaved using ozone (O3) to form aldehydes and/or ketones
This is an example of a Ozonolysis reaction.
19.15: Reactions of Aldehydes and KetonesGeneral Considerations
Reaction at the Carbonyl Carbon
Under neutral or basic conditions, nucleophilic attack of the electrophilic carbon takes place. As the nucleophile approaches the electrophilic carbon, two valence electrons from the nucleophile form a covalent bond to the carbon. As this occurs, the electron pair from the pie bond transfers completely over to the oxygen which produces the intermediate alkoxide ion. This alkoxide ion, with a negative charge on oxygen is susceptible to protonation from a protic solvent like water or alcohol, giving the final addition reaction.
Electrophilic Addition-Protonation
Under acidic conditions, electrophilic attack of the carbonyl oxygen takes place. Initially, protonation of the carbonyl group at the oxygen takes place because of excess H+ all around. Once protonation has occurred, nucleophilic attack by the nucleophile finishes the addition reaction. It should be noted that electrophilic attack is extremely unlikely, however, a few carbonyl groups do become protonated initially to initiate addition through electrophilic attack. This type of reaction works best when the reagent being used is a very mildly basic nucleophile.
Mechanism
1) Protonation of the carbonyl
2) Nuc addition
3) Deprotonation | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/19%3A_Aldehydes_and_KetonesNucleophilic_Addition/19.13%3A_Interesting_Aldehydes_and_Ketones.txt |
Addition of a hydride anion (H:-) to an aldehyde or ketone gives an alkoxide anion, which on protonation yields the corresponding alcohol. Aldehydes produce 1º-alcohols and ketones produce 2º-alcohols.
In metal hydrides reductions the resulting alkoxide salts are insoluble and need to be hydrolyzed (with care) before the alcohol product can be isolated. In the sodium borohydride reduction the methanol solvent system achieves this hydrolysis automatically. In the lithium aluminum hydride reduction water is usually added in a second step. The lithium, sodium, boron and aluminum end up as soluble inorganic salts at the end of either reaction. Note! LiAlH4 and NaBH4 are both capable of reducing aldehydes and ketones to the corresponding alcohol.
Mechanism
This mechanism is for a LiAlH4 reduction. The mechanism for a NaBH4 reduction is the same except methanol is the proton source used in the second step.
1) Nucleopilic attack by the hydride anion
2) The alkoxide is protonated
Addition of a organometallic reagent to an aldehyde or ketone gives an alkoxide anion, which on protonation yields the corresponding alcohol. Aldehydes produce 2º-alcohols and ketones produce 3º-alcohols.
Addition to formaldehyde gives 1o alcohols
Addition to aldehydes gives 2o alcohols
Addition to ketones gives 3o alcohols
Mechanism
1) Nucleophilic attack
2) Protonation
19.17: Nucleophilic Addition of CN
Cyanohydrins have the structural formula of R2C(OH)CN. The “R” on the formula represents an alkyl, aryl, or hydrogen. In order to form a cyanohydrin, a hydrogen cyanide adds reversibly to the carbonyl group of an organic compound thus forming a hydroxyalkanenitrile adducts (commonly known and called as cyanohydrins).
Introduction
Cyanohydrin reactions occurs when an aldehyde or ketone gets treated by a cyanide anion (such as HCN) or a nitrile forming a cyanohydrin product. This special reaction is a nucleophilic addition, where the nucleophilic CN- attacks the electrophilic carbonyl carbon on the ketone, following a protonation by HCN, thereby the cyanide anion being regenerated. This reaction is also reversible.
Cyanohydrins are also intermediates for the Strecker amino acid synthesis. The preparation of displacements of sulfite by cyanide salts are also followed under cyanohydrins.
Mechanism of Cyanohydrin Formation
Acid-catalysed hydrolysis of silylated cyanohydrins has recently been shown to give cyanohydrins instead of ketones; thus an efficient synthesis of cyanohydrins has been found which works with even highly hindered ketones.
Acetone Cyanohydrins
Acetone cyanohydrins (ACH) have the structural formula of (CH3)2C(OH)CN. It is an organic compound serves in the production of methyl methacrylate (also known as acrylic). It is classified as an extremely hazardous substance, since it rapidly decomposes when it's in contact with water. In ACH, sulfuric acid is treated to give the sulfate ester of the methacrylamid. Preparations of other cyanohydrins are also used from ACH: for HACN to Michael acceptors and for the formylation of arenas. The treatment with lithium hydride affords anhydrous lithium cyanide.
Other Cyanohydrins
Other cyanohydrins, excluding acetone cyanohydrins, are: mandelonitrile and glycolonitrile.
Mandelonitrile have a structural formula of C6H5CH(OH)CN and occur in pits of some fruits. Glycolonitrile is an organic compound with the structural formula of HOCH2CN, which is the simplest cyanohydrin that is derived by formaldehydes.
Contributors
• Kathy Wong (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/19%3A_Aldehydes_and_KetonesNucleophilic_Addition/19.16%3A_Nucleophilic_Addition_of_H_and_RA_Review.txt |
Carboxylic acids react with Thionyl Chloride (\(SOCl_2\)) to form acid chlorides.
During the reaction the hydroxyl group of the carboxylic acid is converted to a chlorosulfite intermediate making it a better leaving group. The chloride anion produced during the reaction acts a nucleophile.
Mechanism
1) Nucleophilic attack on Thionyl Chloride
2) Removal of Cl leaving group
3) Nucleophilic attack on the carbonyl
4) Leaving group removal
5) Deprotonation
Carboxylic acids can react with alcohols to form esters in a process called Fischer esterification.
Usually the alcohol is used as the reaction solvent. An acid catalyst is required.
Mechanism
1) Protonation of the carbonyl by the acid. The carbonyl is now activated toward nucleophilic attack.
2) Nucleophilic attack on the carbonyl
3) Proton transfer
4) Water leaves
5) Deprotonation
Conversion of Carboxylic Acids to Amides
The direct reaction of a carboxylic acid with an amine would be expected to be difficult because the basic amine would deprotonate the carboxylic acid to form a highly unreactive carboxylate. However when the ammonium carboxylate salt is heated to a temperature above 100 oC water is driven off and an amide is formed.
Conversion of Carboxylic acids to amide using DCC as an activating agent
The direct conversion of a carboxylic acid to an amide is difficult because amines are basic and tend to convert carboxylic acids to their highly unreactive carboxylates. In this reaction the carboxylic acid adds to the DCC molecule to form a good leaving group which can then be displaced by an amine during nucleophilic substitution. DCC induced coupling to form an amide linkage is an important reaction in the synthesis of peptides.
Mechanism
1) Deprotonation
2) Nucleophilic attack by the carboxylate
3) Nucleophilic attack by the amine
4) Proton transfer
5) Leaving group removal
20.02: Reactions of Esters
Example 1:
Mechanism
1) Protonation of the Carbonyl
2) Nucleophilic attack by water
3) Proton transfer
4) Leaving group removal
Esters can be cleaved back into a carboxylic acid and an alcohol by reaction with water and a base
The reaction is called a saponification from the Latin sapo which means soap. The name comes from the fact that soap used to me made by the ester hydrolysis of fats. Due to the basic conditions a carboxylate ion is made rather than a carboxylic acid.
Example 1:
Mechanism
1) Nucleophilic attack by hydroxide
2) Leaving group removal
3) Deprotonation
20.03: Application- Lipid Hydrolysis
Soap is a mixture of sodium salts of various naturally occurring fatty acids. Air bubbles added to a molten soap will decrease the density of the soap and thus it will float on water. If the fatty acid salt has potassium rather than sodium, a softer lather is the result. Soap is produced by a saponification or basic hydrolysis reaction of a fat or oil. Currently, sodium carbonate or sodium hydroxide is used to neutralize the fatty acid and convert it to the salt.
Introduction
General overall hydrolysis reaction:
fat + NaOH glycerol + sodium salt of fatty acid
Although the reaction is shown as a one step reaction, it is in fact two steps. The net effect as that the ester bonds are broken. The glycerol turns back into an alcohol (addition of the green H's). The fatty acid portion is turned into a salt because of the presence of a basic solution of the NaOH. In the carboxyl group, one oxygen (red) now has a negative charge that attracts the positive sodium ion.
Types of Soap
The type of fatty acid and length of the carbon chain determines the unique properties of various soaps. Tallow or animal fats give primarily sodium stearate (18 carbons) a very hard, insoluble soap. Fatty acids with longer chains are even more insoluble. As a matter of fact, zinc stearate is used in talcum powders because it is water repellent.
Coconut oil is a source of lauric acid (12 carbons) which can be made into sodium laurate. This soap is very soluble and will lather easily even in sea water. Fatty acids with only 10 or fewer carbons are not used in soaps because they irritate the skin and have objectionable odors.
Contributors
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook
20.04: Reactions of Amides
This page describes the hydrolysis of amides under both acidic and alkaline conditions. It also describes the use of alkaline hydrolysis in testing for amides.
What is hydrolysis?
Technically, hydrolysis is a reaction with water. That is exactly what happens when amides are hydrolyzed in the presence of dilute acids such as dilute hydrochloric acid. The acid acts as a catalyst for the reaction between the amide and water. The alkaline hydrolysis of amides actually involves reaction with hydroxide ions, but the result is similar enough that it is still classed as hydrolysis.
Hydrolysis under acidic conditions
Taking ethanamide as a typical amide. If ethanamide is heated with a dilute acid (such as dilute hydrochloric acid), ethanoic acid is formed together with ammonium ions. So, if you were using hydrochloric acid, the final solution would contain ammonium chloride and ethanoic acid.
\[ CH_3CONH_2 + H_2O + HCl \ rightarrow CH_3COOH + NH_4^+Cl^-\]
Hydrolysis under alkaline conditions
Also, if ethanamide is heated with sodium hydroxide solution, ammonia gas is given off and you are left with a solution containing sodium ethanoate.
\[ CH_3CONH_2 + NaOH \ rightarrow CH_3COONa + NH_3\]
Using alkaline hydrolysis to test for an amide
If you add sodium hydroxide solution to an unknown organic compound, and it gives off ammonia on heating (but not immediately in the cold), then it is an amide. You can recognize the ammonia by smell and because it turns red litmus paper blue.
The possible confusion using this test is with ammonium salts. Ammonium salts also produce ammonia with sodium hydroxide solution, but in this case there is always enough ammonia produced in the cold for the smell to be immediately obvious.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/20%3A_Carboxylic_Acids_and_Their_Derivatives_Nucleophilic_Acyl_Substitution/20.01%3A_Reactions_of_Carboxylic_Acids.txt |
Antibiotics are specific chemical substances derived from or produced by living organisms that are capable of inhibiting the life processes of other organisms. The first antibiotics were isolated from microorganisms but some are now obtained from higher plants and animals. Over 3,000 antibiotics have been identified but only a few dozen are used in medicine. Antibiotics are the most widely prescribed class of drugs comprising 12% of the prescriptions in the United States. The penicillins were the first antibiotics discovered as natural products from the mold Penicillium.
Introduction
In 1928, Sir Alexander Fleming, professor of bacteriology at St. Mary's Hospital in London, was culturing Staphylococcus aureus. He noticed zones of inhibition where mold spores were growing. He named the mold Penicillium rubrum. It was determined that a secretion of the mold was effective against Gram-positive bacteria.
Penicillins as well as cephalosporins are called beta-lactam antibiotics and are characterized by three fundamental structural requirements: the fused beta-lactam structure (shown in the blue and red rings, a free carboxyl acid group (shown in red bottom right), and one or more substituted amino acid side chains (shown in black). The lactam structure can also be viewed as the covalent bonding of pieces of two amino acids - cysteine (blue) and valine (red).
Penicillin-G where R = an ethyl pheny group, is the most potent of all penicillin derivatives. It has several shortcomings and is effective only against gram-positive bacteria. It may be broken down in the stomach by gastric acids and is poorly and irregularly absorbed into the blood stream. In addition many disease producing staphylococci are able to produce an enzyme capable of inactivating penicillin-G. Various semisynthetic derivatives have been produced which overcome these shortcomings.
Powerful electron-attracting groups attached to the amino acid side chain such as in phenethicillin prevent acid attack. A bulky group attached to the amino acid side chain provides steric hindrance which interferes with the enzyme attachment which would deactivate the pencillins i.e. methicillin. Refer to Table 2 for the structures. Finally if the polar character is increased as in ampicillin or carbenicillin, there is a greater activity against Gram-negative bacteria.
Penicillin Mode of Action
All penicillin derivatives produce their bacteriocidal effects by inhibition of bacterial cell wall synthesis. Specifically, the cross linking of peptides on the mucosaccharide chains is prevented. If cell walls are improperly made cell walls allow water to flow into the cell causing it to burst. Resemblances between a segment of penicillin structure and the backbone of a peptide chain have been used to explain the mechanism of action of beta-lactam antibiotics. The structures of a beta-lactam antibiotic and a peptide are shown on the left for comparison. Follow the trace of the red oxygens and blue nitrogen atoms.
Gram-positive bacteria possess a thick cell wall composed of a cellulose-like structural sugar polymer covalently bound to short peptide units in layers.The polysaccharide portion of the peptidoglycan structure is made of repeating units of N-acetylglucosamine linked b-1,4 to N-acetylmuramic acid (NAG-NAM). The peptide varies, but begins with L-Ala and ends with D-Ala. In the middle is a dibasic amino acid, diaminopimelate (DAP). DAP (orange) provides a linkage to the D-Ala (gray) residue on an adjacent peptide.
The bacterial cell wall synthesis is completed when a cross link between two peptide chains attached to polysaccharide backbones is formed. The cross linking is catalyzed by the enzyme transpeptidase. First the terminal alanine from each peptide is hydrolyzed and secondly one alanine is joined to lysine through an amide bond.
Penicillin binds at the active site of the transpeptidase enzyme that cross-links the peptidoglycan strands. It does this by mimicking the D-alanyl-D-alanine residues that would normally bind to this site. Penicillin irreversibly inhibits the enzyme transpeptidase by reacting with a serine residue in the transpeptidase. This reaction is irreversible and so the growth of the bacterial cell wall is inhibited. Since mammal cells do not have the same type of cell walls, penicillin specifically inhibits only bacterial cell wall synthesis.
Bacterial Resistance
As early as the 1940s, bacteria began to combat the effectiveness of penicillin. Penicillinases (or beta-lactamases) are enzymes produced by structurally susceptable bacteria which renders penicillin useless by hydrolysing the peptide bond in the beta-lactam ring of the nucleus. Penicillinase is a response of bacterial adaptation to its adverse environment, namely the presence of a substance which inhibits its growth. Many other antibiotics are also rendered ineffective because of this same type of resistance.
Severe Allergic Shock
It is estimated that between 300-500 people die each year from penicillin-induced anaphylaxis, a severe allergic shock reaction to penicillin. In afflicted individuals, the beta-lactam ring binds to serum proteins, initiating an IgE-mediated inflammatory response. Penicillin and ala-ala peptide - Chime in new window
Cephalosporins
Cephalosporins are the second major group of beta-lactam antibiotics. They differ from penicillins by having the beta-lactam ring as a 6 member ring. The other difference, which is more significant from a medicinal chemistry stand point, is the existence of a functional group (R) at position 3 of the fused ring system. This now allows for molecular variations to effect changes in properties by diversifying the groups at position 3.
The first member of the newer series of beta-lactams was isolated in 1956 from extracts of Cephalosporium acremonium, a sewer fungus. Like penicillin, cephalosporins are valuable because of their low toxicity and their broad spectrum of action against various diseases. In this way, cephalosporin is very similar to penicillin. Cephalosporins are one of the most widely used antibiotics, and economically speaking, has about 29% of the antibiotic market. The cephalosporins are possibly the single most important group of antibiotics today and are equal in importance to penicillin.
The structure and mode of action of the cephalosporins are similar to that of penicillin. They affect bacterial growth by inhibiting cell wall synthesis, in Gram-positive and negative bacteria. Some brand names include: cefachlor, cefadroxil, cefoxitin, ceftriaxone. Cephalexin - Chime in new window
Contributors
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/20%3A_Carboxylic_Acids_and_Their_Derivatives_Nucleophilic_Acyl_Substitution/20.05%3A_Application-_The_Mechanism_of_Action_of_-Lactam_Antibiotics.txt |
This is probably the single most important reaction of carboxylic acid derivatives. The overall transformation is defined by the following equation, and may be classified either as nucleophilic substitution at an acyl group or as acylation of a nucleophile. For certain nucleophilic reagents the reaction may assume other names as well. If Nuc-H is water the reaction is often called hydrolysis, if Nuc–H is an alcohol the reaction is called alcoholysis, and for ammonia and amines it is called aminolysis.
Different carboxylic acid derivatives have very different reactivities, acyl chlorides and bromides being the most reactive and amides the least reactive, as noted in the following qualitatively ordered list. The change in reactivity is dramatic. In homogeneous solvent systems, reaction of acyl chlorides with water occurs rapidly, and does not require heating or catalysts. Amides, on the other hand, react with water only in the presence of strong acid or base catalysts and external heating.
Reactivity: acyl halides > anhydrides >> esters ≈ acids >> amides
Because of these differences, the conversion of one type of acid derivative into another is generally restricted to those outlined in the following diagram. Methods for converting carboxylic acids into these derivatives were shown in a previous section, but the amide and anhydride preparations were not general and required strong heating. A better and more general anhydride synthesis can be achieved from acyl chlorides, and amides are easily made from any of the more reactive derivatives. Specific examples of these conversions will be displayed by clicking on the product formula. The carboxylic acids themselves are not an essential part of this diagram, although all the derivatives shown can be hydrolyzed to the carboxylic acid state (light blue formulas and reaction arrows). Base catalyzed hydrolysis produces carboxylate salts.
20.07: Natural and Synthetic Fibers
Polyamides
Just as the reaction of a diol and a diacid forms a polyester (Section 15.8 "Preparation of Esters"), the reaction of a diacid and a diamine yields a polyamide. The two difunctional monomers often employed are adipic acid and 1,6-hexanediamine. The monomers condense by splitting out water to form a new product, which is still difunctional and thus can react further to yield a polyamide polymer.
Some polyamides are known as nylons. Nylons are among the most widely used synthetic fibers—for example, they are used in ropes, sails, carpets, clothing, tires, brushes, and parachutes. They also can be molded into blocks for use in electrical equipment, gears, bearings, and valves.
Polyesters
A commercially important esterification reaction is condensation polymerization, in which a reaction occurs between a dicarboxylic acid and a dihydric alcohol (diol), with the elimination of water. Such a reaction yields an ester that contains a free (unreacted) carboxyl group at one end and a free alcohol group at the other end. Further condensation reactions then occur, producing polyester polymers.
The most important polyester, polyethylene terephthalate (PET), is made from terephthalic acid and ethylene glycol monomers:
Polyester molecules make excellent fibers and are used in many fabrics. A knitted polyester tube, which is biologically inert, can be used in surgery to repair or replace diseased sections of blood vessels. PET is used to make bottles for soda pop and other beverages. It is also formed into films called Mylar. When magnetically coated, Mylar tape is used in audio- and videocassettes. Synthetic arteries can be made from PET, polytetrafluoroethylene, and other polymers.
Contributors
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/20%3A_Carboxylic_Acids_and_Their_Derivatives_Nucleophilic_Acyl_Substitution/20.06%3A_Summary_of_Nucleophilic_Acyl_Substitution_Reactions.txt |
Glutamine synthetase
You have already learned that the carboxylate functional group is a very unreactive substrate for an enzyme-catalyzed acyl substitution reactions. How, then, does a living system accomplish an ‘uphill’ reaction such as the one shown below, where glutamate (a carboxylate) is converted to glutamine (an amide)?
It turns out that this conversion is not carried out directly. Rather, the first conversion is from a carboxylate (the least reactive acyl transfer substrate) to an acyl phosphate (the most reactive acyl transfer substrate). This transformation requires a reaction that we are familiar with from chapter 10: phosphorylation of a carboxylate oxygen with ATP as the phosphate donor.
Note that this is just one of the many ways that ATP is used as a energy storage unit: in order to make a high energy acyl phosphate molecule from a low energy carboxylate, the cell must ‘spend’ the energy of one ATP molecule.
The acyl phosphate version of glutamate is now ready to be converted directly to an amide (glutamine) via a nucleophilic acyl substitution reaction, as an ammonia molecule attacks the carbonyl and the phosphate is expelled.
Overall, this reaction can be written as:
Asparagine synthetase
Another common form of activated carboxylate group is an acyl adenosine phosphate. Consider another amino acid reaction, the conversion of aspartate to asparagine. In the first step, the carboxylate group of aspartate must be activated:
Once again, ATP provides the energy for driving the uphill reaction. This time, however, the activated carboxylate takes the form of an acyl adenosine (mono)phosphate. All that has happened is that the carboxylate oxygen has attacked the a-phosphate of ATP rather than the g-phosphate.
The reactive acyl-AMP version of aspartate is now ready to be converted to an amide (asparagine) via nucleophilic attack by ammonia. In the case of glutamine synthase, the source of ammonia was free ammonium ion in solution. In the case of asparagine synthase, the NH3 is derived from the hydrolysis of glutamine (this is simply another acyl substitution reaction):
The hydrolysis reaction is happening in the same enzyme active site – as the NH3 is expelled in the hydrolysis of glutamine, it immediately turns around and acts as the nucleophile in the conversion of aspartyl-AMP to asparagine:
Keep in mind that the same enzyme is also binding ATP and using it to activate aspartate – this is a busy construction zone!
Overall, this reaction can be written in condensed form as:
The use of glutamine as a ‘carrier’ for ammonia is a fairly common strategy in metabolic pathways. This strategy makes sense, as it allows cells to maintain a constant source of NH3 for reactions that require it, without the need for high solution concentrations of free ammonia.
Glycinamide ribonucleotide synthetase
One of the early steps in the construction of purine bases (the adenine and guanine bases in DNA and RNA) involves an acyl substitution reaction with an acyl phosphate intermediate. In this case, the attacking nucleophile is not ammonia but a primary amine. The strategy, however, is similar to that of glutamine synthase. The carboxylate group on glycine is converted to an acyl phosphate, at the cost of one ATP molecule. The acyl group is then transferred to 5-phosphoribosylamine, resulting in an amide product.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
20.09: Nitriles
Synthesis of Nitriles
Nitriles are formed by an SN2 reaction between a bromide and sodium cyanide
Reactivity of Nitriles
The carbon in a nitrile is electrophilic because a resonance structure can be drawn which places a positive charge on it. Because of this the triple bond of a nitrile accepts a nucleophile in a manner similar to a carbonyl.
Hydrolysis of Nitriles
Nitriles can be converted to carboxylic acid with heating in sulfuric acid. During the reaction an amide intermediate is formed.
Reduction of Nitriles
Nitriles can be converted to 1° amines by reaction with LiAlH4. During this reaction the hydride nucleophile attacks the electrophilic carbon in the nitrile to form an imine anion. Once stabilized by a Lewis acid-base complexation the imine salt can accept a second hydride to form a dianion. The dianion can then be converted to an amine by addition of water.
Mechanism
1) Nucleophilic Attack by the Hydride
2) Second nucleophilic attack by the hydride.
3) Protonation by addition of water to give an amine
Addition of Grignard Reagents
Grignard reagents can attack the electophillic carbon in a nitrile to form an imine salt. This salt can then be hydrolyzed to become a ketone.
Mechanism
1) Nucleophilic Attack by the Grignard Reagent
2) Protonation
3) Protonation
4) Nucleophilic attack by water
5) Proton Transfer
6) Leaving group removal
7) Deprotonation | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/20%3A_Carboxylic_Acids_and_Their_Derivatives_Nucleophilic_Acyl_Substitution/20.08%3A_Biological_Acylation_Reactions.txt |
Background and Properties
The important classes of organic compounds known as alcohols, phenols, ethers, amines and halides consist of alkyl and/or aryl groups bonded to hydroxyl, alkoxyl, amino and halo substituents respectively. If these same functional groups are attached to an acyl group (RCO–) their properties are substantially changed, and they are designated as carboxylic acid derivatives. Carboxylic acids have a hydroxyl group bonded to an acyl group, and their functional derivatives are prepared by replacement of the hydroxyl group with substituents, such as halo, alkoxyl, amino and acyloxy. Some examples of these functional derivatives were displayed earlier.
The following table lists some representative derivatives and their boiling points. An aldehyde and ketone of equivalent molecular weight are also listed for comparison. Boiling points are given for 760 torr (atmospheric pressure), and those listed as a range are estimated from values obtained at lower pressures. As noted earlier, the relatively high boiling point of carboxylic acids is due to extensive hydrogen bonded dimerization. Similar hydrogen bonding occurs between molecules of 1º and 2º-amides (amides having at least one N–H bond), and the first three compounds in the table serve as hydrogen bonding examples.
Other functional group combinations with the carbonyl group can be prepared from carboxylic acids, and are usually treated as related derivatives. Five common classes of these carboxylic acid derivatives are listed in the following table. Although nitriles do not have a carbonyl group, they are included here because the functional carbon atoms all have the same oxidation state. The top row (yellow shaded) shows the general formula for each class, and the bottom row (light blue) gives a specific example of each. As in the case of amines, amides are classified as 1º, 2º or 3º, depending on the number of alkyl groups bonded to the nitrogen.
Acyl Group Substitution
This is probably the single most important reaction of carboxylic acid derivatives. The overall transformation is defined by the following equation, and may be classified either as nucleophilic substitution at an acyl group or as acylation of a nucleophile. For certain nucleophilic reagents the reaction may assume other names as well. If Nuc-H is water the reaction is often called hydrolysis, if Nuc–H is an alcohol the reaction is called alcoholysis, and for ammonia and amines it is called aminolysis.
Different carboxylic acid derivatives have very different reactivities, acyl chlorides and bromides being the most reactive and amides the least reactive, as noted in the following qualitatively ordered list. The change in reactivity is dramatic. In homogeneous solvent systems, reaction of acyl chlorides with water occurs rapidly, and does not require heating or catalysts. Amides, on the other hand, react with water only in the presence of strong acid or base catalysts and external heating.
Reactivity: acyl halides > anhydrides >> esters ≈ acids >> amides
Because of these differences, the conversion of one type of acid derivative into another is generally restricted to those outlined in the following diagram. Methods for converting carboxylic acids into these derivatives were shown in a previous section, but the amide and anhydride preparations were not general and required strong heating. A better and more general anhydride synthesis can be achieved from acyl chlorides, and amides are easily made from any of the more reactive derivatives. Specific examples of these conversions will be displayed by clicking on the product formula. The carboxylic acids themselves are not an essential part of this diagram, although all the derivatives shown can be hydrolyzed to the carboxylic acid state (light blue formulas and reaction arrows). Base catalyzed hydrolysis produces carboxylate salts.
Before proceeding further, it is important to review the general mechanism by means of which all these acyl transfer or acylation reactions take place. Indeed, an alert reader may well be puzzled by the facility of these nucleophilic substitution reactions. After all, it was previously noted that halogens bonded to sp2 or sp hybridized carbon atoms do not usually undergo substitution reactions with nucleophilic reagents. Furthermore, such substitution reactions of alcohols and ethers are rare, except in the presence of strong mineral acids. Clearly, the mechanism by which acylation reactions occur must be different from the SN1 and SN2 procedures described earlier.
In any substitution reaction two things must happen. The bond from the substrate to the leaving group must be broken, and a bond to the replacement group must be formed. The timing of these events may vary with the reacting system. In nucleophilic substitution reactions of alkyl compounds examples of bond-breaking preceding bond-making (the SN1 mechanism), and of bond-breaking and bond-making occurring simultaneously (the SN2 mechanism) were observed. On the other hand, for most cases of electrophilic aromatic substitution bond-making preceded bond-breaking.
As illustrated in the following diagram, acylation reactions generally take place by an addition-elimination process in which a nucleophilic reactant bonds to the electrophilic carbonyl carbon atom to create a tetrahedral intermediate. This tetrahedral intermediate then undergoes an elimination to yield the products. In this two-stage mechanism bond formation occurs before bond cleavage, and the carbonyl carbon atom undergoes a hybridization change from sp2 to sp3 and back again. The facility with which nucleophilic reagents add to a carbonyl group was noted earlier for aldehydes and ketones.
Acid and base-catalyzed variations of this mechanism will be displayed in turn as the "Mechanism Toggle" button is clicked. Also, a specific example of acyl chloride formation from the reaction of a carboxylic acid with thionyl chloride will be shown. The number of individual steps in these mechanisms vary, but the essential characteristic of the overall transformation is that of addition followed by elimination. Acid catalysts act to increase the electrophilicity of the acyl reactant; whereas, base catalysts act on the nucleophilic reactant to increase its reactivity. In principle all steps are reversible, but in practice many reactions of this kind are irreversible unless changes in the reactants and conditions are made. The acid-catalyzed formation of esters from carboxylic acids and alcohols, described earlier, is a good example of a reversible acylation reaction, the products being determined by the addition or removal of water from the system. The reaction of an acyl chloride with an alcohol also gives an ester, but this conversion cannot be reversed by adding HCl to the reaction mixture. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/20%3A_Carboxylic_Acids_and_Their_Derivatives_Nucleophilic_Acyl_Substitution/20.10%3A_Introduction.txt |
Introduction
Carboxylic acid derivatives are functional groups whose chemistry is closely related. The main difference is the presence of an electronegative substituent that can act as a leaving group during a nucleophile substitution reaction. Although there are many types of carboxylic acid derivatives known, this article focuses on four: acid halides, acid anhydrides, esters, and amides.
General mechanism
1) Nucleophilic attack on the carbonyl
2) Leaving group is removed
Although aldehydes and ketones also contain carbonyls, their chemistry is distinctly different because they do not contain suitable leaving groups. Once a tetrahedral intermediate is formed, aldehydes and ketones cannot reform their carbonyls. Because of this, aldehydes and ketones typically undergo nucleophilic additions and not substitutions.
The relative reactivity of carboxylic acid derivatives toward nucleophile substitutions is related to the electronegative leaving group’s ability to activate the carbonyl. The more electronegative leaving groups withdraw electron density from the carbonyl, thereby increasing its electrophilicity.
20.12: Nomenclature
Nomenclature of acid halides
The nomenclature of acid halides starts with the name of the corresponding carboxylic acid. The –ic acid ending is removed and replaced with the ending -yl followed by the name of the halogen with an –ide ending. This is true for both common and IUPAC nomenclature. The carbonyl carbon is given the #1 location number. It is not necessary to include the location number in the name because it is assumed that the functional group will be on the end of the parent chain.
Example 1:
The acid anhydride functional group results when two carboxylic acids combine and lose water (anhydride = without water). Symmetrical acid anhydrides are named like carboxylic acids except the ending -acid is replaced with -anhydride. This is true for both the IUPAC and Common nomenclature.
Unsymmetrical acid anhydrides
Unsymmetrical acid anhydrides are named by first naming each component carboxylic acid alphabetically arranged (without the word acid) followed by spaces and then the word anhydride.
propanoic anhydride
ethanoic propanoic anhydride
Try to name the following compound
Try to draw a structure for the following compound�
• 1,2-benzenedicarboxylic anhydride J
Common names that you should know
acetic anhydride (Try to name this anhydride by the proper name. J )
succinic anhydride (Try to name this anhydride by the proper name. J )
Nomenclature of Esters
Esters are made from a carboxylic acid and an alcohol.
Esters are named as if the alkyl chain from the alcohol is a substituent. No number is assigned to this alkyl chain. This is followed by the name of the parent chain form the carboxylic acid part of the ester with an –e remove and replaced with the ending –oate.
Example 1:
Primary amides
Primary amides are named by changing the name of the acid by dropping the -oic acid or -ic acid endings and adding -amide. The carbonyl carbon is given the #1 location number. It is not necessary to include the location number in the name because it is assumed that the functional group will be on the end of the parent chain.
methanamide or formamide (left), ethanamide or acetamide (center) , benzamide (right)
Try to draw a structure for the following compound
• 3-chlorobenzamide J
Try to name the following compound
Secondary amides
Secondary amides are named by using an upper case N to designate that the alkyl group is on the nitrogen atom. Alkyl groups attached to the nitrogen are named as substituents. The letter N is used to indicate they are attached to the nitrogen. Tertiary amides are named in the same way.
N-methylpropanamide
Try to draw a structure for the following compound
• N,N-dimethylformamide J
Try to name the following compound
Name the parent alkane (include the carbon atom of the nitrile as part of the parent) followed with the word -nitrile. The carbon in the nitrile is given the #1 location position. It is not necessary to include the location number in the name because it is assumed that the functional group will be on the end of the parent chain.
Nomenclature of nitriles
Cycloalkanes are followed by the word -carbonitrile. The substituent name is cyano.
• 1-butanenitrile or 1-cyanopropane
Try to name the following compounds using these conventions�
Try to draw structures for the following compounds�
• butanedinitrile J
• 2-methycyclohexanecarbonitrile J
Some common names that you should know are...
acetonitrile
benzonitrile
Try to draw a structure for the following compound�
• 2-methoxybenzonitrile J | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/20%3A_Carboxylic_Acids_and_Their_Derivatives_Nucleophilic_Acyl_Substitution/20.11%3A_Structure_and_Bonding.txt |
Physical Properties of Some Carboxylic Acid Derivatives
Formula
IUPAC Name
Molecular Weight
Boiling Point
Water Solubility
CH3(CH2)2CO2H butanoic acid 88 164 ºC very soluble
CH3(CH2)2CONH2 butanamide 87 216-220 ºC soluble
CH3CH2CONHCH3 N-methylpropanamide 87 205 -210 ºC soluble
CH3CON(CH3)2 N,N-dimethylethanamide 87 166 ºC very soluble
HCON(CH3)CH2CH3 N-ethyl, N-methylmethanamide 87 170-180 ºC very soluble
CH3(CH2)3CN pentanenitrile 83 141 ºC slightly soluble
CH3CO2CHO ethanoic methanoic
anhydride
88 105-112 ºC reacts with water
CH3CH2CO2CH3 methyl propanoate 88 80 ºC slightly soluble
CH3CO2C2H5 ethyl ethanoate 88 77 ºC moderately soluble
CH3CH2COCl propanoyl chloride 92.5 80 ºC reacts with water
CH3(CH2)3CHO pentanal 86 103 ºC slightly soluble
CH3(CH2)2COCH3 2-pentanone 86 102 ºC slightly soluble
The last nine entries in the above table cannot function as hydrogen bond donors, so hydrogen bonded dimers and aggregates are not possible. The relatively high boiling points of equivalent 3º-amides and nitriles are probably due to the high polarity of these functions. Indeed, if hydrogen bonding is not present, the boiling points of comparable sized compounds correlate reasonably well with their dipole moments.
20.14: Spectroscopic Properties
The influence of heteroatom substituents on the reactivity of carbonyl functions toward nucleophiles was discussed earlier with respect to carboxylic acid derivatives. A useful relationship exists between the reactivity of these derivatives and their carbonyl stretching frequencies. Thus, the very reactive acyl halides and anhydrides absorb at frequencies significantly higher than ketones, whereas the relatively unreactive amides absorb at lower frequencies. These characteristics are listed below.
Infrared spectra of many carboxylic acid derivatives will be displayed in the figure below the table by clicking the appropriate buttons presented there.
Carbonyl Derivative
Carbonyl Absorption
Comments
Acyl Halides (RCOX)
X = F
X = Cl
X = Br
C=O stretch
1860 ± 20 cm-1
1800 ± 15
1800 ± 15
Conjugation lowers the C=O frequencies reported here, as with aldehydes & ketones.
In acyl chlorides a lower intensity shoulder or peak near 1740 cm-1 is due to an overtone interaction.
Acid Anhydride, (RCO)2O
acyclic
6-membered ring
5-membered ring
C=O stretch (2 bands)
1750 & 1820 cm-1
1750 &1820
1785 & 1865
Conjugation lowers the C=O frequencies reported here, as with aldehydes & ketones.
The two stretching bands are separated by 60 ± 30 cm-1, and for acyclic anhydrides the higher frequency (asymmetric stretching) band is stronger than the lower frequency (symmetric) absorption.
Cyclic anhydrides also display two carbonyl stretching absorptions, but the lower frequency band is the strongest.
One or two -CO-O-CO- stretching bands are observed in the 1000 to 1300 cm-1 region.
Esters & Lactones (RCOOR')
esters
6-membered lactone
5-membered lactone
4-membered lactone
C=O stretch
1740 cm ± 10 cm-1
1740 cm ± 10
1765 cm± 5
1840 cm ± 5
Conjugation lowers the C=O frequencies reported here, as with aldehydes & ketones
Strong CO-O stretching absorptions (one ot two) are found from 1150 to 1250 cm-1
Amides & Lactams (RCONR2)
1° & 2°-amides
3°-amides
6-membered lactams
5-membered lactams
4-membered lactams
C=O bands
1510 to 1700 cm-1 (2 bands)
1650± 15 (one band)
1670 ± 10 (one band)
1700 ± 15
1745 ± 15
The effect of conjugation is much less than for aldehydes & ketones.
The higher frequency absorption (1665± 30) is called the Amide I band. The lower frequency Amide II band (1620± 30 in 1° amides & 1530± 30 in 2° amides) is largely due to N-H bending trans to the carbonyl oxygen. In concentrated samples this absorption is often obscured by the stronger amide I absorption. Hydrogen bonded association shifts some of these absorptions, as well as the prominent N-H stretching absorptions.
N-H stretch: 3170 to 3500 cm-1. Two bands for 1°-amides, one for 2°-amides.
NMR Spectra
Protons on carbons adjacent to carbonyls absorb at ~2.0-2.5 ppm.
The N-H protons attached to primary and secondary amines absorb at ~7.5-8.5.
The carbonyl carbon in carboxylic acid derivatives show up at ~ 160-180 ppm.
The carbon in a nitrile appear ~ 115-120 ppm in their 13C NMR spectrum. This is because of their sp hybridization. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/20%3A_Carboxylic_Acids_and_Their_Derivatives_Nucleophilic_Acyl_Substitution/20.13%3A_Physical_Properties.txt |
Functional groups of this kind are found in many kinds of natural products. Some examples are shown below with the functional group colored red. Most of the functions are amides or esters, cantharidin being a rare example of a natural anhydride. Cyclic esters are called lactones, and cyclic amides are referred to as lactams. Penicillin G has two amide functions, one of which is a β-lactam. The Greek letter locates the nitrogen relative to the carbonyl group of the amide.
Contributors
• William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
20.16: Introduction to Nucleophilic Acyl Substitution
Introduction
Carboxylic acid derivatives are functional groups whose chemistry is closely related. The main difference is the presence of an electronegative substituent that can act as a leaving group during a nucleophile substitution reaction. Although there are many types of carboxylic acid derivatives known, this article focuses on four: acid halides, acid anhydrides, esters, and amides.
General mechanism
1) Nucleophilic attack on the carbonyl
2) Leaving group is removed
Although aldehydes and ketones also contain carbonyls, their chemistry is distinctly different because they do not contain suitable leaving groups. Once a tetrahedral intermediate is formed, aldehydes and ketones cannot reform their carbonyls. Because of this, aldehydes and ketones typically undergo nucleophilic additions and not substitutions.
The relative reactivity of carboxylic acid derivatives toward nucleophile substitutions is related to the electronegative leaving group’s ability to activate the carbonyl. The more electronegative leaving groups withdraw electron density from the carbonyl, thereby increasing its electrophilicity.
20.17: Reactions of Acid Chlorides
Acyl chlorides (also known as acid chlorides) are one of a number of types of compounds known as "acid derivatives". This is ethanoic acid:
If you remove the -OH group and replace it by a -Cl, you have produced an acyl chloride.
This molecule is known as ethanoyl chloride and for the rest of this topic will be taken as typical of acyl chlorides in general. Acyl chlorides are extremely reactive. They are open to attack by nucleophiles - with the overall result being a replacement of the chlorine by something else.
Why are acyl chlorides attacked by nucleophiles?
The carbon atom in the -COCl group has both an oxygen atom and a chlorine atom attached to it. Both of these are very electronegative. They both pull electrons towards themselves, leaving the carbon atom quite positively charged.
The Overall Reaction
We are going to generalize this for the moment by writing the reacting molecule as "Nu-H". Nu is the bit of the molecule which contains the nucleophilic oxygen or nitrogen atom. The attached hydrogen turns out to be essential to the reaction. The general equation for the reaction is:
In each case, the net effect is that you replace the -Cl by -Nu, and hydrogen chloride is formed as well.
Since the initial attack is by a nucleophile, and the overall result is substitution, it would seem reasonable to describe the reaction as nucleophilic substitution. However, the reaction happens in two distinct stages. The first involves an addition reaction, which is followed by an elimination reaction where HCl is produced. So the mechanism is also known as nucleophilic addition / eliminatio
Example 1:
Mechanism
1) Nucleophilic attack by the alcohol
2) Leaving group is removed
3) Deprotonation
Example 1:
Mechanism
1) Nucleophilic attack by water
2) Leaving group is removed
3) Deprotonation
Example 1:
Mechanism
1) Nucleophilic attack by the alcohol
2) Leaving group is removed
3) Deprotonation
Examples:
Mechanism
1) Nucleophilic attack by the amine
2) Leaving group is removed
3) Deprotonation
20.18: Reactions of Anhydrides
This page explains what acid anhydrides are and looks at their simple physical properties such as boiling points. It introduces their chemical reactivity in a general way. A carboxylic acid such as ethanoic acid has the structure:
If you took two ethanoic acid molecules and removed a molecule of water between them you would get the acid anhydride, ethanoic anhydride (old name: acetic anhydride).
You can actually make ethanoic anhydride by dehydrating ethanoic acid, but it is normally made in a more efficient, round-about way
Example 1:
Mechanism
1) Nucleophilic Attack by the water molcule
2) Deprotonation by pyridine
3) Leaving group removal
4) Protonation of the carboxylate
Acid Anhydrides react with alcohols to form esters
Reactions of anhydrides use Pyridine as a solvent
Example 1:
Mechanism
1) Nucleophilic Attack by the Alcohol
2) Deprotonation by pyridine
3) Leaving group removal
4) Protonation of the carboxylate
Example 1:
Mechanism
1) Nucleophilic Attack by the Amine
2) Deprotonation by the amine
3) Leaving group removal | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/20%3A_Carboxylic_Acids_and_Their_Derivatives_Nucleophilic_Acyl_Substitution/20.15%3A_Interesting_Esters_and_Amides.txt |
The acetoacetic ester synthesis allows for the conversion of ethyl acetoacetate into a methyl ketone with one or two alkyl groups on the alpha carbon.
Steps
1) Deprotonation with ethoxide
2) Alkylation via and SN2 Reaction
3) Hydrolysis and decarboxylation
Addition of a second alky group
After the first step and additional alkyl group can be added prior to the decarboxylation step. Overall this allows for the addition of two different alkyl groups.
21.02: Introduction
Nucleophilic Addition to Aldehydes and Ketones
The result of carbonyl bond polarization, however it is depicted, is straightforward to predict. The carbon, because it is electron-poor, is an electrophile: it is a great target for attack by an electron-rich nucleophilic group. Because the oxygen end of the carbonyl double bond bears a partial negative charge, anything that can help to stabilize this charge by accepting some of the electron density will increase the bond’s polarity and make the carbon more electrophilic. Very often a general acid group serves this purpose, donating a proton to the carbonyl oxygen.
After the carbonyl is attacked by the nucleophile, the negatively charged oxygen has the capacity to act as a nucleophile. However, most commonly the oxygen acts instead as a base, abstracting a proton from a nearby acid group in the solvent or enzyme active site.
Reactions at The Alpha Carbon
Now we will investigate reactions which occur a the carbon alpha to the carbonyl groups. These reactions involve two new neucleophilic species the enol and the enolate.
Note! The electrophile replaces the hydrogen on the alpha carbon.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
21.03: Enols
Acidity of Alpha Hydrogens
Functional groups, such as aldehydes, ketones and esters, contain a carbonyl group which is made up of a sp2 hybridized carbon and oxygen. Because they are sp2 hybridized the carbon and oxygen both have unhybridized p orbitals which can overlap to form the C=O $\pi$ bond.
Keto-enol Tautomerism
Because of the acidity of α hydrogens carbonyls undergo keto-enol tautomerism. Tautomers are rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom and a differently located double bond. The equilibrium between tautomers is not only rapid under normal conditions, but it often strongly favors one of the isomers (acetone, for example, is 99.999% keto tautomer). Even in such one-sided equilibria, evidence for the presence of the minor tautomer comes from the chemical behavior of the compound. Tautomeric equilibria are catalyzed by traces of acids or bases that are generally present in most chemical samples.
Mechanism for Enol Formation
Acid conditions
1) Protonation of the Carbonyl
2) Enol formation
Basic conditions
1) Enolate formation
2) Protonation
Contributors
• William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
• Prof. Steven Farmer (Sonoma State University)
21.04: Enolates
For alkylation reactions of enolate anions to be useful, these intermediates must be generated in high concentration in the absence of other strong nucleophiles and bases. The aqueous base conditions used for the aldol condensation are not suitable because the enolate anions of simple carbonyl compounds are formed in very low concentration, and hydroxide or alkoxide bases induce competing SN2 and E2 reactions of alkyl halides. It is necessary, therefore, to achieve complete conversion of aldehyde or ketone reactants to their enolate conjugate bases by treatment with a very strong base (pKa > 25) in a non-hydroxylic solvent before any alkyl halides are added to the reaction system. Some bases that have been used for enolate anion formation are: NaH (sodium hydride, pKa > 45), NaNH2 (sodium amide, pKa = 34), and LiN[CH(CH3)2]2 (lithium diisopropylamide, LDA, pKa 36). Ether solvents like tetrahydrofuran (THF) are commonly used for enolate anion formation. With the exception of sodium hydride and sodium amide, most of these bases are soluble in THF. Certain other strong bases, such as alkyl lithium and Grignard reagents, cannot be used to make enolate anions because they rapidly and irreversibly add to carbonyl groups. Nevertheless, these very strong bases are useful in making soluble amide bases. In the preparation of lithium diisopropylamide (LDA), for example, the only other product is the gaseous alkane butane.
Because of its solubility in THF, LDA is a widely used base for enolate anion formation. In this application, one equivalent of diisopropylamine is produced along with the lithium enolate, but this normally does not interfere with the enolate reactions and is easily removed from the products by washing with aqueous acid. Although the reaction of carbonyl compounds with sodium hydride is heterogeneous and slow, sodium enolates are formed with the loss of hydrogen, and no other organic compounds are produced.
The presence of these overlapping p orbitals gives $\alpha$ hydrogens (Hydrogens on carbons adjacent to carbonyls) special properties. In particular, $\alpha$ hydrogens are weakly acidic because the conjugate base, called an enolate, is stabilized though conjugation with the $\pi$ orbitals of the carbonyl. The effect of the carbonyl is seen when comparing the pKa for the $\alpha$ hydrogens of aldehydes (~16-18), ketones (~19-21), and esters (~23-25) to the pKa of an alkane (~50).
Of the two resonance structures of the enolate ion the one which places the negative charge on the oxygen is the most stable. This is because the negative change will be better stabilized by the greater electronegativity of the oxygen.
Acidity of α-Hydrogens in Some Activated Compounds
Compound RCH2–NO2 RCH2–COR RCH2–C≡N RCH2–SO2R
pKa 9 20 25 25
Examples
If the formed enolate is stabilized by more than one carbonyl it is possible to use a weaker base such as sodium ethoxide.
NaOCH2CH3 = Na+ -OCH2CH3 = NaOEt
Because of the acidity of α hydrogens, carbonyls undergo keto-enol tautomerism. Tautomers are rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom and a differently located double bond. The equilibrium between tautomers is not only rapid under normal conditions, but it often strongly favors one of the isomers (acetone, for example, is 99.999% keto tautomer). Even in such one-sided equilibria, evidence for the presence of the minor tautomer comes from the chemical behavior of the compound. Tautomeric equilibria are catalyzed by traces of acids or bases that are generally present in most chemical samples.
The Ambident Character of Enolate Anions
Since the negative charge of an enolate anion is delocalized over the alpha-carbon and the oxygen, as shown earlier, electrophiles may bond to either atom. Reactants having two or more reactive sites are called ambident, so this term is properly applied to enolate anions. Modestly electrophilic reactants such as alkyl halides are not sufficiently reactive to combine with neutral enol tautomers, but the increased nucleophilicity of the enolate anion conjugate base permits such reactions to take place. Because alkylations are usually irreversible, their products should reflect the inherent (kinetic) reactivity of the different nucleophilic sites.
If an alkyl halide undergoes an SN2 reaction at the carbon atom of an enolate anion the product is an alkylated aldehyde or ketone. On the other hand, if the SN2 reaction occurs at oxygen the product is an ether derivative of the enol tautomer; such compounds are stable in the absence of acid and may be isolated and characterized. These alkylations (shown above) are irreversible under the conditions normally used for SN2 reactions, so the product composition should provide a measure of the relative rates of substitution at carbon versus oxygen. It has been found that this competition is sensitive to a number of factors, including negative charge density, solvation, cation coordination and product stability. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/21%3A_Substitution_Reactions_of_Carbonyl_Compounds_at_the_Alpha_Carbon/21.01%3A_Acetoacetic_Ester_Synthesis.txt |
Enolate of Unsymmetrical Carbonyl Compounds
Now let’s consider what happens when an unsymmetrical carbonyl is treated with a base. In the case displayed below there are two possible enolates which can form. The removal of the 2o hydrogen forms the kinetic enolate and is formed faster because it is less substituted and thereby less sterically hindered. The removal of the 3o hydrogen forms the thermodynamic enolate which is more stable because it is more substituted.
Kinetic Enolates
Kinetic enolates are formed when a strong bulky base like LDA is used. The bulky base finds the 2o hydrogen less sterically hindered and preferable removes it.
Low temperature are typically used when forming the kinetic enolate to prevent equilibration to the more stable thermodynamic enolate. Typically a temperature of -78 oC is used.
Thermodynamic Enolates
The thermodynamic enolate is favored by conditions which allow for equilibration. The thermodynamic enolate is usually formed by using a strong base at room temperature. At equilibrium the lower energy of the thermodynamic enolate is preferred, so that the more stable, more stubstituted enolate is formed.
21.08: Halogenation at the Carbon
A carbonyl containing compound with α hydrogens can undergo a substitution reaction with halogens. This reaction comes about because of the tendency of carbonyl compounds to form enolates in basic condition and enols in acidic condition. In these cases even weak bases, such as the hydroxide anion, is sufficient enough to cause the reaction to occur because it is not necessary for a complete conversion to the enolate. For this reaction Cl2, Br2 or I2 can be used as the halogens.
General reaction
Example
Acid Catalyzed Mechanism
Under acidic conditions the reaction occurs thought the formation of an enol which then reacts with the halogen.
1) Protonation of the carbonyl
2) Enol formation
3) SN2 attack
4) Deprotonation
Base Catalyzed Mechanism
Under basic conditions the enolate forms and then reacts with the halogen. Note! This is base promoted and not base catalyzed because an entire equivalent of base is required.
1) Enolate formation
2) SN2 attack
Overreaction during base promoted α halogenation
The fact that an electronegative halogen is placed on an α carbon means that the product of a base promoted α halogenation is actually more reactive than the starting material. The electron withdrawing effect of the halogen makes the α carbon even more acidic and therefor promotes further reaction. Because of this multiple halogenations can occur. This effect is exploited in the haloform reaction discussed later. If a monohalo product is required then acidic conditions are usually used.
The Haloform Reaction
Methyl ketones typically undergo halogenation three times to give a trihalo ketone due to the increased reactivity of the halogenated product as discussed above. This trihalomethyl group is an effective leaving group due to the three electron withdrawing halogens and can be cleaved by a hydroxide anion to effect the haloform reaction. The product of this reaction is a carboxylate and a haloform molecule (CHCl3, CHBr3, CHI3). Overall the haloform reaction represents an effective method for the conversion of methyl ketones to carboxylic acids. Typically, this reaction is performed using iodine because the subsequent iodoform (CHI3) is a bright yellow solid which is easily filtered off.
General reaction
Example: The Haloform Reaction
Mechanism
1) Formation of the trihalo species
2) Nulceophilic attack on the carbonyl carbon
3) Removal of the leaving group
4) Deprotonation
Problems
1) Please draw the products of the following reactions
Answers
1)
Contributors
• Prof. Steven Farmer (Sonoma State University) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/21%3A_Substitution_Reactions_of_Carbonyl_Compounds_at_the_Alpha_Carbon/21.05%3A_Enolates_of_Unsymmetrical_Carbonyl_Compounds.txt |
Enolates can act as a nucleophile in SN2 type reactions. Overall an α hydrogen is replaced with an alkyl group. This reaction is one of the more important for enolates because a carbon-carbon bond is formed. These alkylations are affected by the same limitations as SN2 reactions previously discussed. A good leaving group, Chloride, Bromide, Iodide, Tosylate, should be used. Also, secondary and tertiary leaving groups should not be used because of poor reactivity and possible competition with elimination reactions. Lastly, it is important to use a strong base, such as LDA or sodium amide, for this reaction. Using a weaker base such as hydroxide or an alkoxide leaves the possibility of multiple alkylation’s occurring.
Example 1: Alpha Alkylation
Mechanism
1) Enolate formation
2) Sn2 attack
Alkylation of Unsymmetrical Ketones
Unsymmetrical ketones can be regioselctively alkylated to form one major product depending on the reagents.
Treatment with LDA in THF at -78oC tends to form the less substituted kinetic enolate.
Using sodium ethoxide in ethanol at room temperature forms the more substituted thermodynamic enolate.
Problems
1) Please write the structure of the product for the following reactions.
Answers
1)
21.10: Malonic Ester Synthesis
Malonic ester is a reagent specifically used in a reaction which converts alkyl halides to carboxylic acids called the Malonic Ester Synthesis. Malonic ester synthesis is a synthetic procedure used to convert a compound that has the general structural formula 1 into a carboxylic acid that has the general structural formula 2.
• R1 = alkyl group
• L = leaving group
The group —CH2CO2H in 2 is contributed by a malonic ester, hence the term malonic ester synthesis.
• R2 = alkyl, aryl
Mechanism
Malonic ester synthesis consists of four consecutive reactions that can be carried out in the same pot.
• reaction 1: acid-base reaction
• reaction 2: nucleophilic substitution
• reaction 3: ester hydrolysis (using saponification)
• reaction 4: decarboxylation
eg:
reaction 1:
reaction 2:
reaction 3:
reaction 4:
A more direct method to convert 3 into 4 is the reaction of 3 with the enolate ion (5) of ethyl acetate followed by hydrolysis of the resultant ester.
However, the generation of 5 from ethyl acetate quantitatively in high yield is not an easy task because the reaction requires a very strong base, such as LDA, and must be carried out at very low temperature under strictly anhydrous conditions.
Malonic ester synthesis provides a more convenient alternative to convert 3 to 4.
Malonic ester synthesis can be adapted to synthesize compounds that have the general structural formula 6.
R3, R4 = identical or different alkyl groups
eg:
reaction 1:
reaction 2:
reaction 1 (repeat):
reaction 2 (repeat):
reaction 3:
reaction 4:
Malonic Ester Synthesis
Due to the fact that Malonic ester’s α hydrogens are adjacent to two carbonyls, they can be deprotonated by sodium ethoxide (NaOEt) to form Sodio Malonic Ester.
Because Sodio Malonic Ester is an enolate, it can then be alkylated with alkyl halides.
After alkylation the product can be converted to a dicarboxylic acid through saponification and subsequently one of the carboxylic acids can be removed through a decarboxylation step.
Mechanism
1) Saponification
2) Decarboxylation
3) Tautomerization
All of the steps together form the Malonic ester synthesis.
$RX \rightarrow RCH_2CO_2H$
Example
• Bernard E. Hoogenboom , Phillip J. Ihrig , Arne N. Langsjoen , Carol J. Linn and Stephen D. Mulder, The malonic ester synthesis in the undergraduate laboratory, J. Chem. Educ., 1991, 68 (8), p 689 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/21%3A_Substitution_Reactions_of_Carbonyl_Compounds_at_the_Alpha_Carbon/21.09%3A_Direct_Enolate_Alkylation.txt |
A useful carbon-carbon bond-forming reaction known as the Aldol Reaction is yet another example of electrophilic substitution at the alpha carbon in enolate anions. The fundamental transformation in this reaction is a dimerization of an aldehyde (or ketone) to a beta-hydroxy aldehyde (or ketone) by alpha C–H addition of one reactant molecule to the carbonyl group of a second reactant molecule. Due to the carbanion like nature of enolates they can add to carbonyls in a similar manner as Grignard reagents. For this reaction to occur at least one of the reactants must have α hydrogens.
Going from reactants to products simply
Example 1: Aldol Reactions
Aldol Reaction Mechanism
Step 1: Enolate formation
Step 2: Nucleophilic attack by the enolate
Step 3: Protonation
Aldol Condensation: the dehydration of Aldol products to synthesize α, β unsaturated carbonyl (enones)
The products of aldol reactions often undergo a subsequent elimination of water, made up of an alpha-hydrogen and the beta-hydroxyl group. The product of this $\beta$-elimination reaction is an α,β-unsaturated aldehyde or ketone. Base-catalyzed elimination occurs with heating. The additional stability provided by the conjugated carbonyl system of the product makes some aldol reactions thermodynamically and mixtures of stereoisomers (E & Z) are obtained from some reactions. Reactions in which a larger molecule is formed from smaller components, with the elimination of a very small by-product such as water, are termed Condensations. Hence, the following examples are properly referred to as aldol condensations. Overall the general reaction involves a dehydration of an aldol product to form an alkene:
Figure: General reaction for an aldol condensation
Going from reactants to products simply
Figure: The aldol condensatio example
Example 2: Aldol Condensation
Aldol Condensation Mechanism
1) Form enolate
2) Form enone
When performing both reactions together always consider the aldol product first then convert to the enone. Note! The double bond always forms in conjugation with the carbonyl.
Example
22.02: Crossed Aldol Reactions
Mixed Aldol Reaction and Condensations
The previous examples of aldol reactions and condensations used a common reactant as both the enolic donor and the electrophilic acceptor. The product in such cases is always a dimer of the reactant carbonyl compound. Aldol condensations between different carbonyl reactants are called crossed or mixed reactions, and under certain conditions such crossed aldol condensations can be effective.
Example 3: Mixed Aldol Reaction
The success of these mixed aldol reactions is due to two factors. First, aldehydes are more reactive acceptor electrophiles than ketones, and formaldehyde is more reactive than other aldehydes. Second, aldehydes lacking alpha-hydrogens can only function as acceptor reactants, and this reduces the number of possible products by half. Mixed aldols in which both reactants can serve as donors and acceptors generally give complex mixtures of both dimeric (homo) aldols and crossed aldols. Because of this most mixed aldol reactions are usually not performed unless one reactant has no alpha hydrogens.
The following abbreviated formulas illustrate the possible products in such a case, red letters representing the acceptor component and blue the donor. If all the reactions occurred at the same rate, equal quantities of the four products would be obtained. Separation and purification of the components of such a mixture would be difficult.
AACH2CHO + BCH2CHO + NaOH → AA + BB + AB + BA
The aldol condensation of ketones with aryl aldehydes to form α,β-unsaturated derivatives is called the Claisen-Schmidt reaction.
Example 4: Claisen-Schmidt Reaction | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/22%3A_Carbonyl_Condensation_Reactions/22.01%3A_The_Aldol_Reaction.txt |
Directed aldol reactions are a variation of the crossed aldol reaction. The enolate is prepared with one carbonyl compound using LDA. This causes the other carbonyl compound to be the electrophile. Even though both components have alpha hydrogens only one acts as an enolate because it is formed with LDA. Then an unsymmetrical ketone is use the LDA will selectively form the less substituted enolate.
22.04: Intramolecular Aldol Reactions
Intramolecular aldol reaction
Molecules which contain two carbonyl functionalities have the possibility of forming a ring through an intramolecular aldol reaction. In most cases two sets of $\alpha$ hydrogens need to be considered. As with most ring forming reaction five and six membered rings are preferred.
As with other aldol reaction the addition of heat causes an aldol condensation to occur.
22.05: The Claisen Reaction
Because esters can contain α hydrogens they can undergo a condensation reaction similar to the aldol reaction called a Claisen Condensation. In a fashion similar to the aldol, one ester acts as a nucleophile while a second ester acts as the electrophile. During the reaction a new carbon-carbon bond is formed. The product is a β-keto ester. A major difference with the aldol reaction is the fact that hydroxide cannot be used as a base because it could possibly react with the ester. Instead, an alkoxide version of the alcohol used to synthesize the ester is used to prevent transesterification side products.
Claisen Condensation
Basic reaction
Going from reactants to products simply
Example 1: Claisen Condensation
Claisen Condensation Mechanism
1) Enolate formation
2) Nucleophilic attack
3) Removal of leaving group
22.06: The Crossed Claisen and Related Reactions
Crossed Claisen Condensation
Claisen condensations between different ester reactants are called Crossed Claisen reactions. Crossed Claisen reactions in which both reactants can serve as donors and acceptors generally give complex mixtures. Because of this most Crossed Claisen reactions are usually not performed unless one reactant has no alpha hydrogens.
Example 3: Crossed Claisen Condensation
22.07: The Dieckmann Reaction
Dieckmann Condensation
A diester can undergo an intramolecular reaction called a Dieckmann condensation.
Example 2: Dieckman Condensation
22.08: The Michael Reaction
Basic reaction of 1,4 addition
In 1,4 addition the Nucleophile is added to the carbon β to the carbonyl while the hydrogen is added to the carbon α to the carbonyl.
Mechanism for 1,4 addition
1) Nucleophilic attack on the carbon β to the carbonyl
2) Proton Transfer
Here we can see why this addition is called 1,4. The nucleophile bonds to the carbon in the one position and the hydrogen adds to the oxygen in the four position.
3) Tautomerization
Enolates undergo 1,4 addition to α, β-unsaturated carbonyl compounds is a process called a Michael addition. The reaction is named after American chemist Arthur Michael (1853-1942).
22.09: The Robinson Annulation
Many times the product of a Michael addition produces a dicarbonyl which can then undergo an intramolecular aldol reaction. These two processes together in one reaction creates two new carbon-carbon bonds and also creates a ring. Ring-forming reactions are called annulations after the Latin work for ring annulus. The reaction is named after English chemist Sir Robert Robinson (1886-1975) who developed it. He received the Nobel prize in chemistry in 1947. Remember that during annulations five and six membered rings are preferred. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/22%3A_Carbonyl_Condensation_Reactions/22.03%3A_Directed_Aldol_Reactions.txt |
Basicity of nitrogen groups
In this section we consider the relative basicity of several nitrogen-containing functional groups: amines, amides, anilines, imines, and nitriles.
When evaluating the basicity of a nitrogen-containing organic functional group, the central question we need to ask ourselves is: how reactive (and thus how basic) is the lone pair on the nitrogen? In other words, how much does that lone pair want to break away from the nitrogen nucleus and form a new bond with a hydrogen?
Comparing the basicity of alkyl amines to ammonia
Because alkyl groups donate electrons to the more electronegative nitrogen. The inductive effect makes the electron density on the alkylamine's nitrogen greater than the nitrogen of ammonium. Correspondingly, primary, secondary, and tertiary alkyl amines are more basic than ammonia.
Comparing the basicity of alkylamines to amides
With an alkyl amine the lone pair electron is localized on the nitrogen. However, the lone pair electron on an amide are delocalized between the nitrogen and the oxygen through resonance. This makes amides much less basic compared to alkylamines.
In fact,when and amide is reacted with an acid, the protonation occurs at the carbonyl oxygen and not the nitrogen. This is becase the cation resulting from oxygen protonation is resonance stabilised. The cation resulting for the protonation of nitrogen is not resonance stabilized.
Basicity of aniline
Aniline is substantially less basic than methylamine, as is evident by looking at the pKa values for their respective ammonium conjugate acids (remember that the lower the pKa of the conjugate acid, the weaker the base).
This difference is basicity can be explained by the observation that, in aniline, the basic lone pair on the nitrogen is to some extent tied up in – and stabilized by – the aromatic p system.
This effect is accentuated by the addition of an electron-withdrawing group such as a carbonyl, and reversed to some extent by the addition of an electron-donating group such as methoxide.
In the case of 4-methoxy aniline (the molecule on the left side of the figure above), the lone pair on the methoxy group donates electron density to the aromatic system, and a resonance contributor can be drawn in which a negative charge is placed on the carbon adjacent to the nitrogen, which makes the lone pair of the nitrogen more reactive. In effect, the methoxy group is 'pushing' electron density towards the nitrogen. Conversely, the aldehyde group on the right-side molecule is 'pulling' electron density away from the nitrogen, decreasing its basicity.
At this point, you should draw resonance structures to convince yourself that these resonance effects are possible when the substituent in question (methoxy or carbonyl) is located at the ortho or para position, but not at the meta position.an imine functional group is characterized by an sp2-hybridized nitrogen double-bonded to a carbon. Imines are somewhat basic, with pKa values for the protonated forms ranging around 7. Notice that this is significantly less basic than amine groups (eg. pKa = 10.6 for methylammonium), in which the nitrogen is sp3-hybridized. This phenomenon can be explained using orbital theory and the inductive effect: the sp2 orbitals of an imine nitrogen are one part s and two parts p, meaning that they have about 67% s character. The sp3 orbitals of an amine nitrogen, conversely, are only 25% s character (one part s, three parts p). Because the s atomic orbital holds electrons in a spherical shape, closer to the nucleus than a p orbital, sp2hybridization implies greater electronegative than sp3 hybridization. Finally, recall the inductive effect from section 7.3C: more electronegative atoms absorb electron density more easily, and thus are more acidic. Moral of the story: protonated imine nitrogens are more acidic than protonated amines, thus imines are less basic than amines.
Basicity of heterocyclic amines
When a nitrogen atom is incorporated directly into an aromatic ring, its basicity depends on the bonding context. In a pyridine ring, for example, the nitrogen lone pair occupies an sp2-hybrid orbital, and is not part of the aromatic sextet - it is essentially an imine nitrogen. Its electron pair is available for forming a bond to a proton, and thus the pyridine nitrogen atom is somewhat basic.
In a pyrrole ring, in contrast, the nitrogen lone pair is part of the aromatic sextet. This means that these electrons are very stable right where they are (in the aromatic system), and are much less available for bonding to a proton (and if they do pick up a proton, the aromic system is destroyed). For these reasons, pyrrole nitrogens are not strongly basic.
The aniline, pyridine, and pyrrole examples are good models for predicting the reactivity of nitrogen atoms in more complex ring systems (a huge diversity of which are found in nature). The tryptophan side chain, for example, contains a non-basic 'pyrrole-like' nitrogen, while adenine (a DNA/RNA base) contains all three types.
The lone pair electrons on the nitrogen of a nitrile are contained in a sp hybrid orbital. The 50% s character of an sp hybrid orbital means that the electrons are close to the nucleus and therefore not significantly basic.
A review of basic acid-base concepts should be helpful to the following discussion. Like ammonia, most amines are Brønsted and Lewis bases, but their base strength can be changed enormously by substituents. It is common to compare basicity's quantitatively by using the pKa's of their conjugate acids rather than their pKb's. Since pKa + pKb = 14, the higher the pKa the stronger the base, in contrast to the usual inverse relationship of pKa with acidity. Most simple alkyl amines have pKa's in the range 9.5 to 11.0, and their water solutions are basic (have a pH of 11 to 12, depending on concentration). The first four compounds in the following table, including ammonia, fall into that category.
The last five compounds (colored cells) are significantly weaker bases as a consequence of three factors. The first of these is the hybridization of the nitrogen. In pyridine the nitrogen is sp2 hybridized, and in nitriles (last entry) an sp hybrid nitrogen is part of the triple bond. In each of these compounds (shaded red) the non-bonding electron pair is localized on the nitrogen atom, but increasing s-character brings it closer to the nitrogen nucleus, reducing its tendency to bond to a proton.
Compound
NH3 CH3C≡N
pKa 11.0 10.7 10.7 9.3 5.2 4.6 1.0 0.0 -1.0 -10.0
Secondly, aniline and p-nitroaniline (first two green shaded structures) are weaker bases due to delocalization of the nitrogen non-bonding electron pair into the aromatic ring (and the nitro substituent). This is the same delocalization that results in activation of a benzene ring toward electrophilic substitution. The following resonance equations, which are similar to those used to explain the enhanced acidity of ortho and para-nitrophenols illustrate electron pair delocalization in p-nitroaniline. Indeed, aniline is a weaker base than cyclohexyl amine by roughly a million fold, the same factor by which phenol is a stronger acid than cyclohexanol. This electron pair delocalization is accompanied by a degree of rehybridization of the amino nitrogen atom, but the electron pair delocalization is probably the major factor in the reduced basicity of these compounds. A similar electron pair delocalization is responsible for the very low basicity (and nucleophilic reactivity) of amide nitrogen atoms (last green shaded structure). This feature was instrumental in moderating the influence of amine substituents on aromatic ring substitution, and will be discussed further in the section devoted to carboxylic acid derivatives.
Conjugated amine groups influence the basicity of an existing amine. Although 4-dimethylaminopyridine (DMAP) might appear to be a base similar in strength to pyridine or N,N-dimethylaniline, it is actually more than ten thousand times stronger, thanks to charge delocalization in its conjugate acid. The structure in the gray box shows the locations over which positive charge (colored red) is delocalized in the conjugate acid. This compound is often used as a catalyst for acyl transfer reactions.
Finally, the very low basicity of pyrrole (shaded blue) reflects the exceptional delocalization of the nitrogen electron pair associated with its incorporation in an aromatic ring. Indole (pKa = -2) and imidazole (pKa = 7.0), see above, also have similar heterocyclic aromatic rings. Imidazole is over a million times more basic than pyrrole because the sp2 nitrogen that is part of one double bond is structurally similar to pyridine, and has a comparable basicity. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/23%3A_Amines/23.01%3A_Relative_Basicity_of_Amines_and_Other_Compounds.txt |
The reaction of aldehydes and ketones with ammonia or 1º-amines forms imine derivatives, also known as Schiff bases (compounds having a C=N function). Water is eliminated in the reaction, which is acid-catalyzed and reversible in the same sense as acetal formation. The pH for reactions which form imine compounds must be carefully controlled. The rate at which these imine compounds are formed is generally greatest near a pH of 5, and drops at higher and lower pH's. At high pH there will not be enough acid to protonate the OH in the intermediate to allow for removal as H2O. At low pH most of the amine reactant will be tied up as its ammonium conjugate acid and will become non-nucleophilic.
Most aldehydes and ketones react with 2º-amines to give products known as enamines. It should be noted that, like acetal formation, these are acid-catalyzed reversible reactions in which water is lost. Consequently, enamines are easily converted back to their carbonyl precursors by acid-catalyzed hydrolysis.
Acid chlorides react with ammonia, 1o amines and 2o amines to form amides.
Examples:
23.03: Hofmann Elimination
Amine functions seldom serve as leaving groups in nucleophilic substitution or base-catalyzed elimination reactions. Indeed, they are even less effective in this role than are hydroxyl and alkoxyl groups. In the case of alcohols and ethers, a useful technique for enhancing the reactivity of the oxygen function was to modify the leaving group (OH(–) or OR(–)) to improve its stability as an anion (or equivalent). This stability is conveniently estimated from the strength of the corresponding conjugate acids.
As noted earlier, 1º and 2º-amines are much weaker acids than alcohols, so it is not surprising that it is difficult to force the nitrogen function to assume the role of a nucleophilic leaving group. For example, heating an amine with HBr or HI does not normally convert it to the corresponding alkyl halide, as in the case of alcohols and ethers. In this context we note that the acidity of the putative ammonium leaving group is at least ten powers of ten less than that of an analogous oxonium species. The loss of nitrogen from diazonium intermediates is a notable exception in this comparison, due to the extreme stability of this leaving group (the conjugate acid of N2 would be an extraordinarily strong acid).
One group of amine derivatives that have proven useful in SN2 and E2 reactions is that composed of the tetraalkyl (4º-) ammonium salts. Most applications involving this class of compounds are eliminations, but a few examples of SN2 substitution have been reported.
C6H5–N(CH3)3(+) Br(–) + R-S(–) Na(+)
acetone & heat
R-S-CH3 + C6H5–N(CH3)2 + NaBr
(CH3)4N(+) OH(–)
heat
CH3–OH + (CH3)3N
Hofmann Elimination
Elimination reactions of 4º-ammonium salts are termed Hofmann eliminations. Since the counter anion in most 4º-ammonium salts is halide, this is often replaced by the more basic hydroxide ion through reaction with silver hydroxide (or silver oxide). The resulting hydroxide salt must then be heated (100 - 200 ºC) to effect the E2-like elimination of a 3º-amine. Example #1 below shows a typical Hofmann elimination. Obviously, for an elimination to occur one of the alkyl substituents on nitrogen must have one or more beta-hydrogens, as noted earlier in examining elimination reactions of alkyl halides.
In example #2 above, two of the alkyl substituents on nitrogen have beta-hydrogens, all of which are on methyl groups (colored orange & magenta). The chief product from the elimination is the alkene having the more highly substituted double bond, reflecting not only the 3:1 numerical advantage of those beta-hydrogens, but also the greater stability of the double bond.
Example #3 illustrates two important features of the Hofmann elimination:
1. Simple amines are easily converted to the necessary 4º-ammonium salts by exhaustive alkylation, usually with methyl iodide (methyl has no beta-hydrogens and cannot compete in the elimination reaction). Exhaustive methylation is shown again in example #4.
2. When a given alkyl group has two different sets of beta-hydrogens available to the elimination process (colored orange & magenta here), the major product is often the alkene isomer having the less substituted double bond.
The tendency of Hofmann eliminations to give the less-substituted double bond isomer is commonly referred to as the Hofmann Rule, and contrasts strikingly with the Zaitsev Rule formulated for dehydrohalogenations and dehydrations. In cases where other activating groups, such as phenyl or carbonyl, are present, the Hofmann Rule may not apply. Thus, if 2-amino-1-phenylpropane is treated in the manner of example #3, the product consists largely of 1-phenylpropene (E & Z-isomers).
To understand why the base-induced elimination of 4º-ammonium salts behaves differently from that of alkyl halides it is necessary to reexamine the nature of the E2 transition state, first described for dehydrohalogenation. The energy diagram shown earlier for a single-step bimolecular E2 mechanism is repeated below.
The E2 transition state is less well defined than is that of SN2 reactions. More bonds are being broken and formed, with the possibility of a continuum of states in which the extent of C–H and C–X bond-breaking and C=C bond-making varies. For example, if the bond to the leaving group (X) is substantially broken relative to the other bond changes, the transition state approaches that for an E1 reaction (initial ionization followed by a fast second step). At the other extreme, if the acidity of the beta-hydrogens is enhanced, then substantial breaking of C–H may occur before the other bonds begin to be affected. For most simple alkyl halides it was proper to envision a balanced transition state, in which there was a synchronous change in all the bonds. Such a model was consistent with the Zaitsev Rule.
When the leaving group X carries a positive charge, as do the 4º-ammonium compounds discussed here, the inductive influence of this charge will increase the acidity of both the alpha and the beta-hydrogens. Furthermore, the 4º-ammonium substituent is much larger than a halide or hydroxyl group and may perturb the conformations available to substituted beta-carbons. It seems that a combination of these factors acts to favor base attack at the least substituted (least hindered and most acidic) set of beta-hydrogens. The favored anti orientation of the leaving group and beta-hydrogen, noted for dehydrohalogenation, is found for many Hofmann eliminations; but syn-elimination is also common, possibly because the attraction of opposite charges orients the hydroxide base near the 4º-ammonium leaving group.
Three additional examples of the Hofmann elimination are shown in the following diagram. Example #1 is interesting in two respects. First, it generates a 4º-ammonium halide salt in a manner different from exhaustive methylation. Second, this salt is not converted to its hydroxide analog prior to elimination. A concentrated aqueous solution of the halide salt is simply dropped into a refluxing sodium hydroxide solution, and the volatile hydrocarbon product is isolated by distillation.
Example #2 illustrates an important aspect of the Hofmann elimination. If the nitrogen atom is part of a ring, then a single application of this elimination procedure does not remove the nitrogen as a separate 3º-amine product. In order to sever the nitrogen function from the molecule, a second Hofmann elimination must be carried out. Indeed, if the nitrogen atom was a member of two rings (fused or spiro), then three repetitions of the Hofmann elimination would be required to sever the nitrogen from the remaining molecular framework.
Example #3 is noteworthy because the less stable trans-cyclooctene is the chief product, accompanied by the cis-isomer. An anti-E2-transition state would necessarily give the cis-cycloalkene, so the trans-isomer must be generated by a syn-elimination. The cis-cyclooctene produced in this reaction could also be formed by a syn-elimination. Cyclooctane is a conformationally complex structure. Several puckered conformations that avoid angle strain are possible, and one of the most stable of these is shown on the right. Some eclipsed bonds occur in all these conformers, and transannular hydrogen crowding is unavoidable. Since the trimethylammonium substituent is large (about the size of tert-butyl) it will probably assume an equatorial-like orientation to avoid steric crowding. An anti-E2 transition state is likely to require an axial-like orientation of this bulky group, making this an unfavorable path. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/23%3A_Amines/23.02%3A_Amines_as_Nucleophiles.txt |
Nitrous acid ($HNO_2$ or $HONO$) reacts with aliphatic amines in a fashion that provides a useful test for distinguishing primary, secondary and tertiary amines.
• 1°-Amines + HONO (cold acidic solution) $\rightarrow$ Nitrogen Gas Evolution from a Clear Solution
• 2°-Amines + HONO (cold acidic solution) $\rightarrow$ An Insoluble Oil (N-Nitrosamine)
• 3°-Amines + HONO (cold acidic solution) $\rightarrow$ A Clear Solution (Ammonium Salt Formation)
Nitrous acid is a Brønsted acid of moderate strength (pKa = 3.3). Because it is unstable, it is prepared immediately before use in the following manner:
Under the acidic conditions of this reaction, all amines undergo reversible salt formation:
This happens with 3º-amines, and the salts are usually soluble in water. The reactions of nitrous acid with 1°- and 2°- aliphatic amines may be explained by considering their behavior with the nitrosonium cation, NO(+), an electrophilic species present in acidic nitrous acid solutions.
Secondary Amines
The distinct behavior of 1º, 2º & 3º-aliphatic amines is an instructive challenge to our understanding of their chemistry, but is of little importance as a synthetic tool. The SN1 product mixtures from 1º-amines are difficult to control, and rearrangement is common when branched primary alkyl groups are involved. The N-nitrosamines formed from 2º-amines are carcinogenic, and are not generally useful as intermediates for subsequent reactions.
Aryl Amines
Nitrous acid reactions of 1º-aryl amines generate relatively stable diazonium species that serve as intermediates for a variety of aromatic substitution reactions. Diazonium cations may be described by resonance contributors, as in the bracketed formulas shown below. The left-hand contributor is dominant because it has greater bonding. Loss of nitrogen is slower than in aliphatic 1º-amines because the C-N bond is stronger, and aryl carbocations are comparatively unstable.
Aqueous solutions of these diazonium ions have sufficient stability at 0º to 10 ºC that they may be used as intermediates in a variety of nucleophilic substitution reactions. For example, if water is the only nucleophile available for reaction, phenols are formed in good yield.
2º-Aryl Amines:
2º-Aryl amines give N-nitrosamine derivatives on reaction with nitrous acid, and thus behave identically to their aliphatic counterparts.
3º-Aryl Amines:
Depending on ring substitution, 3º-Aryl amines may undergo aromatic ring nitrosation at sites ortho or para to the amine substituent. The nitrosonium cation is not sufficiently electrophilic to react with benzene itself, or even toluene, but highly activated aromatic rings such as amines and phenols are capable of substitution. Of course, the rate of reaction of NO(+) directly at nitrogen is greater than that of ring substitution, as shown in the previous example. Once nitrosated, the activating character of the amine nitrogen is greatly diminished; and N-nitrosoaniline derivatives, or indeed any amide derivatives, do not undergo ring nitrosation.
23.05: Substitution Reactions of Aryl Diazonium Salts
Aryl diazonium salts are important intermediates. They are prepared in cold (0 º to 10 ºC) aqueous solution, and generally react with nucleophiles with loss of nitrogen. Some of the more commonly used substitution reactions are shown in the following diagram. Since the leaving group (N2) is thermodynamically very stable, these reactions are energetically favored. Those substitution reactions that are catalyzed by cuprous salts are known as Sandmeyer reactions. Fluoride substitution occurs on treatment with BF4(–), a reaction known as the Schiemann reaction. Stable diazonium tetrafluoroborate salts may be isolated, and on heating these lose nitrogen to give an arylfluoride product. The top reaction with hypophosphorus acid, H3PO2, is noteworthy because it achieves the reductive removal of an amino (or nitro) group. Unlike the nucleophilic substitution reactions, this reduction probably proceeds by a radical mechanism.
These aryl diazonium substitution reactions significantly expand the tactics available for the synthesis of polysubstituted benzene derivatives. Consider the following options:
1. The usual precursor to an aryl amine is the corresponding nitro compound. A nitro substituent deactivates an aromatic ring and directs electrophilic substitution to meta locations.
2. Reduction of a nitro group to an amine may be achieved in several ways. The resulting amine substituent strongly activates an aromatic ring and directs electrophilic substitution to ortho & para locations.
3. The activating character of an amine substituent may be attenuated by formation of an amide derivative (reversible), or even changed to deactivating and meta-directing by formation of a quaternary-ammonium salt (irreversible).
4. Conversion of an aryl amine to a diazonium ion intermediate allows it to be replaced by a variety of different groups (including hydrogen), which may in turn be used in subsequent reactions.
The following examples illustrate some combined applications of these options to specific cases. You should try to conceive a plausible reaction sequence for each. Once you have done so, you may check suggested answers by clicking on the question mark for each.
23.06: Coupling Reactions of Aryl Diazonium Salts
Bonding to Nitrogen
A resonance description of diazonium ions shows that the positive charge is delocalized over the two nitrogen atoms. It is not possible for nucleophiles to bond to the inner nitrogen, but bonding (or coupling) of negative nucleophiles to the terminal nitrogen gives neutral azo compounds. As shown in the following equation, this coupling to the terminal nitrogen should be relatively fast and reversible. The azo products may exist as E / Z stereoisomers. In practice it is found that the E-isomer predominates at equilibrium.
Unless these azo products are trapped or stabilized in some manner, reversal to the diazonium ion and slow nucleophilic substitution at carbon (with irreversible nitrogen loss) will be the ultimate course of reaction, as described in the previous section. For example, if phenyldiazonium bisufate is added rapidly to a cold solution of sodium hydroxide a relatively stable solution of sodium phenyldiazoate (the conjugate base of the initially formed diazoic acid) is obtained. Lowering the pH of this solution regenerates phenyldiazoic acid (pKa ca. 7), which disassociates back to the diazonium ion and eventually undergoes substitution, generating phenol.
C6H5N2(+) HSO4(–) + NaOH (cold solution) C6H5N2–OH + NaOH (cold) C6H5N2–O(–) Na(+)
phenyldiazonium bisulfate phenyldiazoic acid sodium phenyldiazoate
Aryl diazonium salts may be reduced to the corresponding hydrazines by mild reducing agents such as sodium bisulfite, stannous chloride or zinc dust. The bisulfite reduction may proceed by an initial sulfur-nitrogen coupling, as shown in the following equation.
Ar-N2(+) X(–)
NaHSO3
Ar-N=N-SO3H
NaHSO3
Ar-NH-NH-SO3H
H2O
Ar-NH-NH2 + H2SO4
The most important application of diazo coupling reactions is electrophilic aromatic substitution of activated benzene derivatives by diazonium electrophiles. The products of such reactions are highly colored aromatic azo compounds that find use as synthetic dyestuffs, commonly referred to as azo dyes. Azobenzene (Y=Z=H) is light orange; however, the color of other azo compounds may range from red to deep blue depending on the nature of the aromatic rings and the substituents they carry. Azo compounds may exist as cis/trans isomer pairs, but most of the well-characterized and stable compounds are trans.
Some examples of azo coupling reactions are shown below. A few simple rules are helpful in predicting the course of such reactions:
1. At acid pH (< 6) an amino group is a stronger activating substituent than a hydroxyl group (i.e. a phenol). At alkaline pH (> 7.5) phenolic functions are stronger activators, due to increased phenoxide base concentration.
2. Coupling to an activated benzene ring occurs preferentially para to the activating group if that location is free. Otherwise ortho-coupling will occur.
3. Naphthalene normally undergoes electrophilic substitution at an alpha-location more rapidly than at beta-sites; however, ortho-coupling is preferred. See the diagram for examples of α / β notation in naphthalenes.
You should try to conceive a plausible product structure for each of the following couplings. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/23%3A_Amines/23.04%3A_Reaction_of_Amines_with_Nitrous_Acid.txt |
This page looks at some typical reactions of diazonium ions, including examples of both substitution reactions and coupling reactions. If you have come straight to this page from a search engine and want to know about the preparation of the diazonium ions, you will find a link at the bottom of the page.
Substitution reactions of diazonium ions
Diazonium ions are present in solutions such as benzenediazonium chloride solution. They contain an -N2+ group. In the case of benzenediazonium chloride, this is attached to a benzene ring. Benzenediazonium chloride looks like this:
In this set of reactions of the diazonium ion, the -N2+ group is replaced by something else. The nitrogen is released as nitrogen gas.
Substitution by an -OH group
To get this reaction, all you need to do is warm the benzenediazonium chloride solution. The diazonium ion reacts with the water in the solution and phenol is formed - either in solution or as a black oily liquid (depending on how much is formed). Nitrogen gas is evolved.
This is the same reaction that you get if you react phenylamine with nitrous acid in the warm. The diazonium ion is formed first and then immediately reacts with the water in the solution to give phenol.
Substitution by an iodine atom
This is a good example of the use of diazonium salts to substitute things into a benzene ring which are otherwise quite difficult to attach. (That's equally true of the previous reaction, by the way.) If you add potassium iodide solution to the benzenediazonium chloride solution in the cold, nitrogen gas is given off, and you get oily droplets of iodobenzene formed. There is a simple reaction between the diazonium ions and the iodide ions from the potassium iodide solution.
Coupling reactions of diazonium ions
In the substitution reactions above, the nitrogen in the diazonium ion is lost. In the rest of the reactions on this page, the nitrogen is retained and used to make a bridge between two benzene rings.
The reaction with phenol
Phenol is dissolved in sodium hydroxide solution to give a solution of sodium phenoxide.
The solution is cooled in ice, and cold benzenediazonium chloride solution is added. There is a reaction between the diazonium ion and the phenoxide ion and a yellow-orange solution or precipitate is formed. The product is one of the simplest of what are known as azo compounds, in which two benzene rings are linked by a nitrogen bridge.
The reaction with naphthalen-2-ol
Naphthalen-2-ol is also known as 2-naphthol or beta-naphthol. It contains an -OH group attached to a naphthalene molecule rather than to a simple benzene ring. Naphthalene has two benzene rings fused together.
The reaction is done under exactly the same conditions as with phenol. The naphthalen-2-ol is dissolved in sodium hydroxide solution to produce an ion just like the phenol one. This solution is cooled and mixed with the benzenediazonium chloride solution. An intense orange-red precipitate is formed - another azo compound.
The reaction with phenylamine (aniline)
Some liquid phenylamine is added to a cold solution of benzenediazonium chloride, and the mixture is shaken vigorously. A yellow solid is produced.
These strongly colored azo compounds are frequently used as dyes known as azo dyes. The one made from phenylamine (aniline) is known as "aniline yellow" (amongst many other things - see note above). Azo compounds account for more than half of modern dyes.
The use of an azo dye as an indicator - methyl orange
Azo compounds contain a highly delocalised system of electrons which takes in both benzene rings and the two nitrogen atoms bridging the rings. The delocalisation can also extend to things attached to the benzene rings as well.
If white light falls on one of these molecules, some wavelengths are absorbed by these delocalised electrons. The color you see is the result of the non-absorbed wavelengths. The groups which contribute to the delocalisation (and so to the absorption of light) are known as a chromophore.
Modifying the groups present in the molecule can have an effect on the light absorbed, and so on the color you see. You can take advantage of this in indicators. Methyl orange is an azo dye which exists in two forms depending on the pH:
As the hydrogen ion is lost or gained there is a shift in the exact nature of the delocalization in the molecule, and that causes a shift in the wavelength of light absorbed. Obviously that means that you see a different color.
When you add acid to methyl orange, a hydrogen ion attaches to give the red form. Methyl orange is red in acidic solutions (in fact solutions of pH less than 3.1). If you add an alkali, hydrogen ions are removed and you get the yellow form. Methyl orange is yellow at pH's greater than 4.4. In between, at some point there will be equal amounts of the red and yellow forms and so methyl orange looks orange.
23.08: Application- Sulfa Drugs
Sulfonamides are synthetic antimicrobial agents with a wide spectrum encompassing most gram-positive and many gram-negative organisms. These drugs were the first efficient treatment to be employed systematically for the prevention and cure of bacterial infections.
Introduction
Their use introduced and substantiated the concept of metabolic antagonism. Sulfonamides, as antimetabolites, compete with para-aminobenzoic acid (PABA) for incorporation into folic acid. The action of sulfonamides illustrates the principle of selective toxicity where some difference between mammal cells and bacterial cells is exploited. All cells require folic acid for growth. Folic acid (as a vitamin is in food) diffuses or is transported into human cells. However, folic acid cannot cross bacterial cell walls by diffusion or active transport. For this reason bacteria must synthesize folic acid from p-aminobenzoic acid. Sulfonamides or sulfa drugs have the following general structures as shown below.
Sulfanilamide which was the first compound used of this type has H's at R1 and R4. To date about 15,000 sulfonamide derivatives, analogues, and related compounds have been synthesized. This has lead to the discovery of many useful drugs which are effective for diuretics, antimalerial and leprosy agents, and antithyroid agents. The basic structure of sulfonamide cannot be modified if it is to be an effective competitive "mimic" for p-aminobenzoic acid. Essential structural features are the benzene ring with two substituents para to each other; an amino group in the fourth position; and the singly substituted 1-sulfonamido group.
Mechanism for Action
Normally folic acid is synthesized in two steps in bacteria by the top reaction on the left. If A sulfa drug is used, the first enzyme is not to specific and can use the sulfonamide in the first reaction. This reaction produces the product containing pteridine and the sulfa drug. The next and final step is the reaction PABA + with glutamic acid to make folic acid. If the sulfa drug has been substituted for the PABA, then the final enzyme is inhibited and no folic acid is produced.
Recent studies indicate that substituents on the N(1) nitrogen may play the role of competing for a site on the enzyme surface reserved for the glutamate residue in p-aminobenzoic acid-glutamate through one of the following two ways:
1. Direct competition in the linking of PABA-glutamate with the pteridine derivative.
2. Indirect interference with the coupling of glutamate to dihydropteroic acid.
Questions
1. In your own words explain how the sulfa drug works including enzyme inhibition, folic acid, and antimetabolite.
Contributors
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/23%3A_Amines/23.07%3A_Application-_Synthetic_Dyes.txt |
Amines are derivatives of ammonia in which one or more of the hydrogens has been replaced by an alkyl or aryl group.
The basic properties of amines
We are going to have to use two different definitions of the term "base" in this page. A base is
• a substance which combines with hydrogen ions. This is the Bronsted-Lowry theory.
• an electron pair donor. This is the Lewis theory.
The easiest way of looking at the basic properties of amines is to think of an amine as a modified ammonia molecule. In an amine, one or more of the hydrogen atoms in ammonia has been replaced by a hydrocarbon group. Replacing the hydrogens still leaves the lone pair on the nitrogen unchanged - and it is the lone pair on the nitrogen that gives ammonia its basic properties. Amines will therefore behave much the same as ammonia in all cases where the lone pair is involved.
The reactions of amines with acids
These are most easily considered using the Bronsted-Lowry theory of acids and bases - the base is a hydrogen ion acceptor. We'll do a straight comparison between amines and the familiar ammonia reactions. Ammonia reacts with acids to produce ammonium ions. The ammonia molecule picks up a hydrogen ion from the acid and attaches it to the lone pair on the nitrogen.
If the reaction is in solution in water (using a dilute acid), the ammonia takes a hydrogen ion (a proton) from a hydroxonium ion. (Remember that hydrogen ions present in solutions of acids in water are carried on water molecules as hydroxonium ions, H3O+.)
NH3(aq)+H3O+(aq)NH+4+H2O(l)
If the acid was hydrochloric acid, for example, you would end up with a solution containing ammonium chloride - the chloride ions, of course, coming from the hydrochloric acid. You could also write this last equation as:
NH3(aq)+H+NH+4(aq)
. . . but if you do it this way, you must include the state symbols. If you write H+ on its own, it implies an unattached hydrogen ion - a proton. Such things don't exist on their own in solution in water. If the reaction is happening in the gas state, the ammonia accepts a proton directly from the hydrogen chloride:
NH3(aq)+HCl(g)NH+4(s)+Cl(s)
This time you produce clouds of white solid ammonium chloride.
The designation 1o, 2o and 3o is determined by the number of alkyl groups attached to the nitrogen.
23.10: Structure and Bonding
Amines typically have three bonds and one pair of lone pair electrons. This makes the nitrogen sp3 hybridized, trigonal pyramidal, with a bond angle of roughly 109.5o.
Stereogenic Nitrogen
Single-bonded nitrogen is pyramidal in shape, with the non-bonding electron pair pointing to the unoccupied corner of a tetrahedral region. Since the nitrogen in these compounds is bonded to three different groups, its configuration is chiral. The non-identical mirror-image configurations are illustrated in the following diagram (the remainder of the molecule is represented by R, and the electron pair is colored yellow). If these configurations were stable, there would be four additional stereoisomers of ephedrine and pseudoephedrine. However, pyramidal nitrogen is normally not configurationally stable. It rapidly inverts its configuration (equilibrium arrows) by passing through a planar, sp2-hybridized transition state, leading to a mixture of interconverting R and S configurations. If the nitrogen atom were the only chiral center in the molecule, a 50:50 (racemic) mixture of R and S configurations would exist at equilibrium. If other chiral centers are present, as in the ephedrin isomers, a mixture of diastereomers will result. The take-home message is that nitrogen does not contribute to isolable stereoisomers.
23.11: Nomenclature
In the IUPAC system of nomenclature, functional groups are normally designated in one of two ways. The presence of the function may be indicated by a characteristic suffix and a location number. This is common for the carbon-carbon double and triple bonds which have the respective suffixes ene and yne. Halogens, on the other hand, do not have a suffix and are named as substituents, for example: (CH3)2C=CHCHClCH3 is 4-chloro-2-methyl-2-pentene. If you are uncertain about the IUPAC rules for nomenclature you should review them now.
Amines are derivatives of ammonia in which one or more of the hydrogens has been replaced by an alkyl or aryl group. The nomenclature of amines is complicated by the fact that several different nomenclature systems exist, and there is no clear preference for one over the others. Furthermore, the terms primary (1º), secondary (2º) & tertiary (3º) are used to classify amines in a completely different manner than they were used for alcohols or alkyl halides. When applied to amines these terms refer to the number of alkyl (or aryl) substituents bonded to the nitrogen atom, whereas in other cases they refer to the nature of an alkyl group. The four compounds shown in the top row of the following diagram are all C4H11N isomers. The first two are classified as 1º-amines, since only one alkyl group is bonded to the nitrogen; however, the alkyl group is primary in the first example and tertiary in the second. The third and fourth compounds in the row are 2º and 3º-amines respectively. A nitrogen bonded to four alkyl groups will necessarily be positively charged, and is called a 4º-ammonium cation. For example, (CH3)4N(+) Br(–) is tetramethylammonium bromide.
• The IUPAC names are listed first and colored blue. This system names amine functions as substituents on the largest alkyl group. The simple -NH substituent found in 1º-amines is called an amino group. For 2º and 3º-amines a compound prefix (e.g. dimethylamino in the fourth example) includes the names of all but the root alkyl group.
• The Chemical Abstract Service has adopted a nomenclature system in which the suffix -amine is attached to the root alkyl name. For 1º-amines such as butanamine (first example) this is analogous to IUPAC alcohol nomenclature (-ol suffix). The additional nitrogen substituents in 2º and 3º-amines are designated by the prefix N- before the group name. These CA names are colored magenta in the diagram.
• Finally, a common system for simple amines names each alkyl substituent on nitrogen in alphabetical order, followed by the suffix -amine. These are the names given in the last row (colored black).
Many aromatic and heterocyclic amines are known by unique common names, the origins of which are often unknown to the chemists that use them frequently. Since these names are not based on a rational system, it is necessary to memorize them. There is a systematic nomenclature of heterocyclic compounds, but it will not be discussed here. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/23%3A_Amines/23.09%3A_Introduction.txt |
Boiling Point and Water Solubility
It is instructive to compare the boiling points and water solubility of amines with those of corresponding alcohols and ethers. The dominant factor here is hydrogen bonding, and the first table below documents the powerful intermolecular attraction that results from -O-H---O- hydrogen bonding in alcohols (light blue columns). Corresponding -N-H---N- hydrogen bonding is weaker, as the lower boiling points of similarly sized amines (light green columns) demonstrate. Alkanes provide reference compounds in which hydrogen bonding is not possible, and the increase in boiling point for equivalent 1º-amines is roughly half the increase observed for equivalent alcohols.
Compound CH3CH3 CH3OH CH3NH2 CH3CH2CH3 CH3CH2OH CH3CH2NH2
Mol.Wt. 30 32 31 44 46 45
Boiling
Point ºC
-88.6º 65º -6.0º -42º 78.5º 16.6º
The second table illustrates differences associated with isomeric 1º, 2º & 3º-amines, as well as the influence of chain branching. Since 1º-amines have two hydrogens available for hydrogen bonding, we expect them to have higher boiling points than isomeric 2º-amines, which in turn should boil higher than isomeric 3º-amines (no hydrogen bonding). Indeed, 3º-amines have boiling points similar to equivalent sized ethers; and in all but the smallest compounds, corresponding ethers, 3º-amines and alkanes have similar boiling points. In the examples shown here, it is further demonstrated that chain branching reduces boiling points by 10 to 15 ºC.
Compound CH3(CH2)2CH3 CH3(CH2)2OH CH3(CH2)2NH2 CH3CH2NHCH3 (CH3)3CH (CH3)2CHOH (CH3)2CHNH2 (CH3)3N
Mol.Wt. 58 60 59 59 58 60 59 59
Boiling
Point ºC
-0.5º 97º 48º 37º -12º 82º 34º
The water solubility of 1º and 2º-amines is similar to that of comparable alcohols. As expected, the water solubility of 3º-amines and ethers is also similar. These comparisons, however, are valid only for pure compounds in neutral water. The basicity of amines (next section) allows them to be dissolved in dilute mineral acid solutions, and this property facilitates their separation from neutral compounds such as alcohols and hydrocarbons by partitioning between the phases of non-miscible solvents.
23.13: Spectroscopic Properties
IR
The infrared spectrum of aniline is shown beneath the following table. Some of the characteristic absorptions for C-H stretching and aromatic ring substitution are also marked, but not colored.
Amine Class
Stretching Vibrations
Bending Vibrations
Primary (1°)
The N-H stretching absorption is less sensitive to hydrogen bonding than are O-H absorptions. In the gas phase and in dilute CCl4 solution free N-H absorption is observed in the 3400 to 3500 cm-1 region. Primary aliphatic amines display two well-defined peaks due to asymmetric (higher frequency) and symmetric N-H stretching, separated by 80 to 100 cm-1. In aromatic amines these absorptions are usually 40 to 70 cm-1 higher in frequency. A smaller absorption near 3200 cm-1 (shaded orange in the spectra) is considered to be the result of interaction between an overtone of the 1600 cm-1 band with the symmetric N-H stretching band.
C-N stretching absorptions are found at 1200 to 1350 cm-1 for aromatic amines, and at 1000 to 1250 cm-1 for aliphatic amines.
Strong in-plane NH2 scissoring absorptions at 1550 to 1650 cm-1, and out-of-plane wagging at 650 to 900 cm-1 (usually broad) are characteristic of 1°-amines.
Secondary (2°)
Secondary amines exhibit only one absorption near 3420 cm-1. Hydrogen bonding in concentrated liquids shifts these absorptions to lower frequencies by about 100 cm-1. Again, this absorption appears at slightly higher frequency when the nitrogen atom is bonded to an aromatic ring.
The C-N absorptions are found in the same range, 1200 to 1350 cm-1(aromatic) and 1000 to 1250 cm-1 (aliphatic) as for 1°-amines.
A weak N-H bending absorption is sometimes visible at 1500 to 1600 cm-1. A broad wagging absorption at 650 to 900 cm-1 may be discerned in liquid film samples.
Tertiary (3°)
No N-H absorptions. The C-N absorptions are found in the same range, 1200 to 1350 cm-1 (aromatic) and 1000 to 1250 cm-1 (aliphatic) as for 1°-amines.
Aside from the C-N stretch noted on the left, these compounds have spectra characteristic of their alkyl and aryl substituents.
NMR
The hydrogens attached to an amine show up ~ 0.5-5.0 ppm. The location is dependent on the amount of hydrogen bonding and the sample's concentration.
The hydrogens on carbons directly bonded to an amine typically appear ~2.3-3.0 ppm. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/23%3A_Amines/23.12%3A_Physical_Properties.txt |
Nature abounds with nitrogen compounds, many of which occur in plants and are referred to as alkaloids. Structural formulas for some representative alkaloids and other nitrogen containing natural products are displayed below, and we can recognize many of the basic structural features listed above in their formulas. Thus, Serotonin and Thiamine are 1º-amines, Coniine is a 2º-amine, Atropine, Morphine and Quinine are 3º-amines, and Muscarine is a 4º-ammonium salt.
The reader should be able to recognize indole, imidazole, piperidine, pyridine, pyrimidine & pyrrolidine moieties among these structures. These will be identified by pressing the "Show Structures" button under the diagram.
Nitrogen atoms that are part of aromatic rings , such as pyridine, pyrrole & imidazole, have planar configurations (sp2 hybridization), and are not stereogenic centers. Nitrogen atoms bonded to carbonyl groups, as in caffeine, also tend to be planar. In contrast, atropine, coniine, morphine, nicotine and quinine have stereogenic pyramidal nitrogen atoms in their structural formulas (think of the non-bonding electron pair as a fourth substituent on a sp3 hybridized nitrogen). In quinine this nitrogen is restricted to one configuration by the bridged ring system. The other stereogenic nitrogens are free to assume two pyramidal configurations, but these are in rapid equilibrium so that distinct stereoisomers reflecting these sites cannot be easily isolated.
It should be noted that structural factors may serve to permit the resolution of pyramidal chiral amines. Two examples of such 3º-amines, compared with similar non-resolvable analogs, are shown in the following diagram. The two nitrogen atoms in Trögers base are the only stereogenic centers in the molecule. Because of the molecule's bridged structure, the nitrogens have the same configuration and cannot undergo inversion. The chloro aziridine can invert, but requires a higher activation energy to do so, compared with larger heterocyclic amines. It has in fact been resolved, and pure enantiomers isolated. An increase in angle strain in the sp2-hybridized planar transition state is responsible for the greater stability of the pyramidal configuration. The rough estimate of angle strain is made using a C-N-C angle of 60º as an arbitrary value for the three-membered heterocycle.
Of course, quaternary ammonium salts, such as that in muscarine, have a tetrahedral configuration that is incapable of inversion. With four different substituents, such a nitrogen would be a stable stereogenic center.
Amines as Antidepressants
Antidepressant drugs act by one or more of the following stimulation type mechanisms:
1. Increase release of norepinephrine:Amphetamines and electroconvulsive therapy act by this mechanism. Amphetamines mimic norepinephrine.
2. Prevent inactivation of norepinephrine:Monoamine oxidase (MAO) inhibitors are thought to act as antidepressant agents in part by preventing the breakdown and inactivation of norepinephrine.
3. Prevent the re uptake of norepinephrine:The action of norepinephrine at the receptor site is terminated by the re uptake of norepinephrine by the neuron from which it was originally released.
Tricyclic Antidepressants
The tricyclic antidepressants are the most effective drugs presently available for the treatment of depression. These act by increasing the release of norepinephrine. Amphetamine and cocaine can also act in this manner. Imipramine, amitriptylin, and other closely related drugs are among the drugs currently most widely used for the treatment of major depression.
• imipramine (Tofranil)
• desipramine (Norpramin)
The activity of the tricyclic drugs depends on the central ring of seven or eight atoms which confers an angled or twisted conformation. The side chain must have at least 2 carbons although 3 appear to be better. The amine group may be either tertiary or secondary. All tricyclic antidepressants block the re-uptake of norepinephrine at nerve terminals. However, the potency and selectivity for the inhibition of the uptake of norepinephrine, serotonin, and dopamine vary greatly among the agents. The tertiary amine tricyclics seem to inhibit the serotonin uptake pump, whereas the secondary amine ones seem better in switching off the NE pump. For instance, imipramine is a potent and selective blocker of serotonin transport, while desipramine inhibits the uptake of norepinephrine.
Serotonin
Serotonin (5-hydroxytryptamine or 5-HT) is a monoamine neurotransmitter found in cardiovascular tissue, in endothelial cells, in blood cells, and in the central nervous system. The role of serotonin in neurological function is diverse, and there is little doubt that serotonin is an important CNS neurotransmitter. Although some of the serotonin is metabolized by monoamine oxidase, most of the serotonin released into the post-synaptic space is removed by the neuron through a re uptake mechanism inhibited by the tricyclic antidepressants and the newer, more selective antidepressant re uptake inhibitors such as fluoxetine and sertraline.
Selective Serotonin Reuptake Inhibitors
In recent years, selective serotonin reuptake inhibitors have been introduced for the treatment of depression. Prozac is the most famous drug in this class. Clomiprimine, fluoxetine (Prozac), sertraline and paroxetine selectively block the re uptake of serotonin, thereby increasing the levels of serotonin in the central nervous system. Note the similarities and differences between the tricyclic antidepressants and the selective serotonin re uptake inhibitors. Clomipramine has been useful in the treatment of obsessive-compulsive disorders.
Monoamine Oxidase Inhibitors
Monoamine oxidase (MAO) causes the oxidative deamination of norephinephrine, serotonin, and other amines. This oxidation is the method of reducing the concentration of the neurotransmitter after it has sent the signal at the receptor site. A drug which inhibits this enzyme has the effect of increasing the concentration of the norepinephrine which in turn causes a stimulation effect. Most MAO inhibitors are hydrazine derivatives. Hydrazine is highly reactive and may form a strong covalent bond with MAO with consequent inhibition for up to 5 days.
These drugs are less effective and produce more side effects than the tricyclic antidepressants. For example, they lower blood pressure and were at one time used to treat hypertension. Their use in psychiatry has also become very limited as the tricyclic antidepressants have come to dominate the treatment of depression and allied conditions. Thus, MAOIs are used most often when tricyclic antidepressants give unsatisfactory results.
Phenelzine is the hydrazine analog of phenylethylamine, a substrate of MAO. This and several other MAOIs, such as isocarboxazide, are structurally related to amphetamine and were synthesized in an attempt to enhance central stimulant properties.
• phenelzine (Nardil)
• isocarboxazid (Marplan)
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/23%3A_Amines/23.14%3A_Interesting_and_Useful_Amines.txt |
This page summarises the reactions of amines as nucleophiles. This includes their reactions with halogenoalkanes (haloalkanes or alkyl halides), with acyl chlorides (acid chlorides) and with acid anhydrides.
Amines by direct nucleophilic substitution
A nucleophile is something which is attracted to, and then attacks, a positive or slightly positive part of another molecule or ion. All amines contain an active lone pair of electrons on the very electronegative nitrogen atom. It is these electrons which are attracted to positive parts of other molecules or ions.
The reactions of primary amines with halogenoalkanes
You get a complicated series of reactions on heating to give a mixture of products - probably one of the most confusing sets of reactions you will meet at this level. The products of the reactions include secondary and tertiary amines and their salts, and quaternary ammonium salts.
Making secondary amines and their salts
In the first stage of the reaction, you get the salt of a secondary amine formed. For example if you started with ethylamine and bromoethane, you would get diethylammonium bromide
In the presence of excess ethylamine in the mixture, there is the possibility of a reversible reaction. The ethylamine removes a hydrogen from the diethylammonium ion to give free diethylamine - a secondary amine.
Making tertiary amines and their salts
But it doesn't stop here! The diethylamine also reacts with bromoethane - in the same two stages as before. This is where the reaction would start if you reacted a secondary amine with a halogenoalkane.
In the first stage, you get triethylammonium bromide.
There is again the possibility of a reversible reaction between this salt and excess ethylamine in the mixture.
The ethylamine removes a hydrogen ion from the triethylammonium ion to leave a tertiary amine - triethylamine.
Making a quaternary ammonium salt
The final stage! The triethylamine reacts with bromoethane to give tetraethylammonium bromide - a quaternary ammonium salt (one in which all four hydrogens have been replaced by alkyl groups).
This time there isn't any hydrogen left on the nitrogen to be removed. The reaction stops here.
Preparation of Primary Amines
Although direct alkylation of ammonia by alkyl halides leads to 1º-amines, alternative procedures are preferred in many cases. These methods require two steps, but they provide pure product, usually in good yield. The general strategy is to first form a carbon-nitrogen bond by reacting a nitrogen nucleophile with a carbon electrophile. The following table lists several general examples of this strategy in the rough order of decreasing nucleophilicity of the nitrogen reagent. In the second step, extraneous nitrogen substituents that may have facilitated this bonding are removed to give the amine product.
Nitrogen
Reactant
Carbon
Reactant
1st Reaction
Type
Initial Product
2nd Reaction
Conditions
2nd Reaction
Type
Final Product
N3(–) RCH2-X or
R2CH-X
SN2 RCH2-N3 or
R2CH-N3
LiAlH4 or
4 H2 & Pd
Hydrogenolysis RCH2-NH2 or
R2CH-NH2
C6H5SO2NH(–) RCH2-X or
R2CH-X
SN2 RCH2-NHSO2C6H5 or
R2CH-NHSO2C6H5
Na in NH3 (liq) Hydrogenolysis RCH2-NH2 or
R2CH-NH2
CN(–) RCH2-X or
R2CH-X
SN2 RCH2-CN or
R2CH-CN
LiAlH4 Reduction RCH2-CH2NH2 or
R2CH-CH2NH2
NH3 RCH=O or
R2C=O
Addition /
Elimination
RCH=NH or
R2C=NH
H2 & Ni
or NaBH3CN
Reduction RCH2-NH2 or
R2CH-NH2
NH3 RCOX Addition /
Elimination
RCO-NH2 LiAlH4 Reduction RCH2-NH2
NH2CONH2
(urea)
R3C(+) SN1 R3C-NHCONH2 NaOH soln. Hydrolysis R3C-NH2
A specific example of each general class is provided in the diagram below. In the first two, an anionic nitrogen species undergoes an SN2 reaction with a modestly electrophilic alkyl halide reactant. For example #2 an acidic phthalimide derivative of ammonia has been substituted for the sulfonamide analog listed in the table. The principle is the same for the two cases, as will be noted later. Example #3 is similar in nature, but extends the carbon system by a methylene group (CH2). In all three of these methods 3º-alkyl halides cannot be used because the major reaction path is an E2 elimination.
The methods illustrated by examples #4 and #5 proceed by attack of ammonia, or equivalent nitrogen nucleophiles, at the electrophilic carbon of a carbonyl group. A full discussion of carbonyl chemistry is presented later, but for present purposes it is sufficient to recognize that the C=O double bond is polarized so that the carbon atom is electrophilic. Nucleophile addition to aldehydes and ketones is often catalyzed by acids. Acid halides and anhydrides are even more electrophilic, and do not normally require catalysts to react with nucleophiles. The reaction of ammonia with aldehydes or ketones occurs by a reversible addition-elimination pathway to give imines (compounds having a C=N function). These intermediates are not usually isolated, but are reduced as they are formed (i.e. in situ). Acid chlorides react with ammonia to give amides, also by an addition-elimination path, and these are reduced to amines by LiAlH4.
The 6th example is a specialized procedure for bonding an amino group to a 3º-alkyl group (none of the previous methods accomplishes this). Since a carbocation is the electrophilic species, rather poorly nucleophilic nitrogen reactants can be used. Urea, the diamide of carbonic acid, fits this requirement nicely. The resulting 3º-alkyl-substituted urea is then hydrolyzed to give the amine.
One important method of preparing 1º-amines, especially aryl amines, uses a reverse strategy. Here a strongly electrophilic nitrogen species (NO2(+)) bonds to a nucleophilic carbon compound. This nitration reaction gives a nitro group that can be reduced to a 1º-amine by any of several reduction procedures.
The Hofmann rearrangement of 1º-amides provides an additional synthesis of 1º-amines. To learn about this useful procedure Click Here.
Reduction of Nitro Groups
Several methods for reducing nitro groups to amines are known. These include catalytic hydrogenation (H2 + catalyst), zinc or tin in dilute mineral acid, and sodium sulfide in ammonium hydroxide solution. The procedures described above are sufficient for most case
Nitriles can be converted to 1° amines by reaction with LiAlH4
During this reaction the hydride nucleophile attacks the electrophilic carbon in the nitrile to form an imine anion. Once stabilized by a Lewis acid-base complexation the imine salt can accept a second hydride to form a dianion. The dianion can then be converted to an amine by addition of water.
Reductive amination
Aldehydes and ketones can be converted into 1o, 2o and 3o amines using reductive amination. The reaction takes place in two parts. The first step is the nucleophiic addition of the carbonyl group to form an imine. The second step is the reduction of the imine to an amine using an reducing agent. A reducing agent commonly used for this reaction is sodium cyanoborohydride (NaBH3CN). | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/23%3A_Amines/23.15%3A_Preparation_of_Amines.txt |
The basic properties of amines
We are going to have to use two different definitions of the term "base" in this page. A base is
• a substance which combines with hydrogen ions. This is the Bronsted-Lowry theory.
• an electron pair donor. This is the Lewis theory.
The easiest way of looking at the basic properties of amines is to think of an amine as a modified ammonia molecule. In an amine, one or more of the hydrogen atoms in ammonia has been replaced by a hydrocarbon group. Replacing the hydrogens still leaves the lone pair on the nitrogen unchanged - and it is the lone pair on the nitrogen that gives ammonia its basic properties. Amines will therefore behave much the same as ammonia in all cases where the lone pair is involved.
The reactions of amines with acids
These are most easily considered using the Bronsted-Lowry theory of acids and bases - the base is a hydrogen ion acceptor. We'll do a straight comparison between amines and the familiar ammonia reactions. Ammonia reacts with acids to produce ammonium ions. The ammonia molecule picks up a hydrogen ion from the acid and attaches it to the lone pair on the nitrogen.
If the reaction is in solution in water (using a dilute acid), the ammonia takes a hydrogen ion (a proton) from a hydroxonium ion. (Remember that hydrogen ions present in solutions of acids in water are carried on water molecules as hydroxonium ions, H3O+.)
NH3(aq)+H3O+(aq)NH+4+H2O(l)
If the acid was hydrochloric acid, for example, you would end up with a solution containing ammonium chloride - the chloride ions, of course, coming from the hydrochloric acid. You could also write this last equation as:
NH3(aq)+H+NH+4(aq)
. . . but if you do it this way, you must include the state symbols. If you write H+ on its own, it implies an unattached hydrogen ion - a proton. Such things don't exist on their own in solution in water. If the reaction is happening in the gas state, the ammonia accepts a proton directly from the hydrogen chloride:
NH3(aq)+HCl(g)NH+4(s)+Cl(s)
Electrophilic Substitution at Nitrogen
Ammonia and many amines are not only bases in the Brønsted sense, they are also nucleophiles that bond to and form products with a variety of electrophiles. A general equation for such electrophilic substitution of nitrogen is:
2 R2ÑH + E(+) R2NHE(+) R2ÑE + H(+) (bonded to a base)
A list of some electrophiles that are known to react with amines is shown here. In each case the electrophilic atom or site is colored red.
Electrophile
RCH2–X RCH2–OSO2R R2C=O R(C=O)X RSO2–Cl HO–N=O
Name
Alkyl Halide Alkyl Sulfonate Aldehyde
or Ketone
Acid Halide
or Anhydride
Sulfonyl Chloride Nitrous Acid
23.17: Amines as Bases
Basicity of Amines
A review of basic acid-base concepts should be helpful to the following discussion. Like ammonia, most amines are Brønsted and Lewis bases, but their base strength can be changed enormously by substituents. It is common to compare basicity's quantitatively by using the pKa's of their conjugate acids rather than their pKb's. Since pKa + pKb = 14, the higher the pKa the stronger the base, in contrast to the usual inverse relationship of pKa with acidity. Most simple alkyl amines have pKa's in the range 9.5 to 11.0, and their water solutions are basic (have a pH of 11 to 12, depending on concentration). The first four compounds in the following table, including ammonia, fall into that category.
The last five compounds (colored cells) are significantly weaker bases as a consequence of three factors. The first of these is the hybridization of the nitrogen. In pyridine the nitrogen is sp2 hybridized, and in nitriles (last entry) an sp hybrid nitrogen is part of the triple bond. In each of these compounds (shaded red) the non-bonding electron pair is localized on the nitrogen atom, but increasing s-character brings it closer to the nitrogen nucleus, reducing its tendency to bond to a proton.
Compound
NH3 CH3C≡N
pKa 11.0 10.7 10.7 9.3 5.2 4.6 1.0 0.0 -1.0 -10.0
Secondly, aniline and p-nitroaniline (first two green shaded structures) are weaker bases due to delocalization of the nitrogen non-bonding electron pair into the aromatic ring (and the nitro substituent). This is the same delocalization that results in activation of a benzene ring toward electrophilic substitution. The following resonance equations, which are similar to those used to explain the enhanced acidity of ortho and para-nitrophenols illustrate electron pair delocalization in p-nitroaniline. Indeed, aniline is a weaker base than cyclohexyl amine by roughly a million fold, the same factor by which phenol is a stronger acid than cyclohexanol. This electron pair delocalization is accompanied by a degree of rehybridization of the amino nitrogen atom, but the electron pair delocalization is probably the major factor in the reduced basicity of these compounds. A similar electron pair delocalization is responsible for the very low basicity (and nucleophilic reactivity) of amide nitrogen atoms (last green shaded structure). This feature was instrumental in moderating the influence of amine substituents on aromatic ring substitution, and will be discussed further in the section devoted to carboxylic acid derivatives.
Conjugated amine groups influence the basicity of an existing amine. Although 4-dimethylaminopyridine (DMAP) might appear to be a base similar in strength to pyridine or N,N-dimethylaniline, it is actually more than ten thousand times stronger, thanks to charge delocalization in its conjugate acid. The structure in the gray box shows the locations over which positive charge (colored red) is delocalized in the conjugate acid. This compound is often used as a catalyst for acyl transfer reactions.
Finally, the very low basicity of pyrrole (shaded blue) reflects the exceptional delocalization of the nitrogen electron pair associated with its incorporation in an aromatic ring. Indole (pKa = -2) and imidazole (pKa = 7.0), see above, also have similar heterocyclic aromatic rings. Imidazole is over a million times more basic than pyrrole because the sp2 nitrogen that is part of one double bond is structurally similar to pyridine, and has a comparable basicity.
Although resonance delocalization generally reduces the basicity of amines, a dramatic example of the reverse effect is found in the compound guanidine (pKa = 13.6). Here, as shown below, resonance stabilization of the base is small, due to charge separation, while the conjugate acid is stabilized strongly by charge delocalization. Consequently, aqueous solutions of guanidine are nearly as basic as are solutions of sodium hydroxide.
The relationship of amine basicity to the acidity of the corresponding conjugate acids may be summarized in a fashion analogous to that noted earlier for acids:
Strong bases have weak conjugate acids, and weak bases have strong conjugate acids.
Important Reagent Bases
The significance of all these acid-base relationships to practical organic chemistry lies in the need for organic bases of varying strength, as reagents tailored to the requirements of specific reactions. The common base sodium hydroxide is not soluble in many organic solvents, and is therefore not widely used as a reagent in organic reactions. Most base reagents are alkoxide salts, amines or amide salts. Since alcohols are much stronger acids than amines, their conjugate bases are weaker than amide bases, and fill the gap in base strength between amines and amide salts. In the following table, pKa again refers to the conjugate acid of the base drawn above it.
Base Name Pyridine Triethyl
Amine
Hünig's Base Barton's
Base
Potassium
t-Butoxide
Sodium HMDS LDA
Formula (C2H5)3N (CH3)3CO(–) K(+) [(CH3)3Si]2N(–) Na(+) [(CH3)2CH]2N(–) Li(+)
pKa 5.3 10.7 11.4 14 19 26 35.7
Pyridine is commonly used as an acid scavenger in reactions that produce mineral acid co-products. Its basicity and nucleophilicity may be modified by steric hindrance, as in the case of 2,6-dimethylpyridine (pKa=6.7), or resonance stabilization, as in the case of 4-dimethylaminopyridine (pKa=9.7). Hünig's base is relatively non-nucleophilic (due to steric hindrance), and like DBU is often used as the base in E2 elimination reactions conducted in non-polar solvents. Barton's base is a strong, poorly-nucleophilic, neutral base that serves in cases where electrophilic substitution of DBU or other amine bases is a problem. The alkoxides are stronger bases that are often used in the corresponding alcohol as solvent, or for greater reactivity in DMSO. Finally, the two amide bases see widespread use in generating enolate bases from carbonyl compounds and other weak carbon acids. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/23%3A_Amines/23.16%3A_Reactions_of_AminesGeneral_Features.txt |
Polymers are long chain, giant organic molecules are assembled from many smaller molecules called monomers. Polymers consist of many repeating monomer units in long chains, sometimes with branching or cross-linking between the chains. A polymer is analogous to a necklace made from many small beads (monomers). A chemical reaction forming polymers from monomers is called polymerization, of which there are many types. A common name for many synthetic polymer materials is plastic, which comes from the Greek word "plastikos", suitable for molding or shaping.
In the following illustrated example, many monomers called styrene are polymerized into a long chain polymer called polystyrene. The squiggly lines indicate that the polymer molecule extends further at both the left and right ends. In fact, polymer molecules are often hundreds or thousands of monomer units long.
The repeating structural unit of most simple polymers not only reflects the monomer(s) from which the polymers are constructed, but also provides a concise means for drawing structures to represent these macromolecules. For polyethylene, arguably the simplest polymer, this is demonstrated by the following equation. Here ethylene (ethene) is the monomer, and the corresponding linear polymer is called high-density polyethylene (HDPE). HDPE is composed of macromolecules in which n ranges from 10,000 to 100,000 (molecular weight $2 \times 10^5$ to $3 \times10^6$ ).
If Y and Z represent moles of monomer and polymer respectively, Z is approximately $10^{-5}$ Y. This polymer is called polyethylene rather than polymethylene, $\ce{(-CH_2-)_{n}}$, because ethylene is a stable compound (methylene is not), and it also serves as the synthetic precursor of the polymer. The two open bonds remaining at the ends of the long chain of carbons (colored magenta) are normally not specified, because the atoms or groups found there depend on the chemical process used for polymerization. The synthetic methods used to prepare this and other polymers will be described later in this chapter
Introduction
Many objects in daily use from packing, wrapping, and building materials include half of all polymers synthesized. Other uses include textiles, many electronic appliance casings, CD's, automobile parts, and many others are made from polymers. A quarter of the solid waste from homes is plastic materials - some of which may be recycled as shown in the table below.
Some products, such as adhesives, are made to include monomers which can be polymerized by the user in their application.
Types of Polymers
There are many types of polymers including synthetic and natural polymers.
Synthetic polymers
• Plastics
• Elastomers - solids with rubber-like qualities
• Rubber (carbon backbone often from hydrocarbon monomers)
• silicones (backbone of alternating silicon and oxygen atoms).
• Fibers
• Solid materials of intermediate characteristics
• Gels or viscous liquids
Classification of Polymers
• Homopolymers: These consist of chains with identical bonding linkages to each monomer unit. This usually implies that the polymer is made from all identical monomer molecules. These may be represented as : -[A-A-A-A-A-A]- Homopolymers are commonly named by placing the prefix poly in front of the constituent monomer name. For example, polystyrene is the name for the polymer made from the monomer styrene (vinylbenzene).
• Copolymers: These consist of chains with two or more linkages usually implying two or more different types of monomer units. These may be represented as : -[A-B-A-B-A-B]-
Polymers classified by mode of polymerization
• Addition Polymers: The monomer molecules bond to each other without the loss of any other atoms. Addition polymers from alkene monomers or substituted alkene monomers are the biggest groups of polymers in this class. Ring opening polymerization can occur without the loss of any small molecules.
• Condensation Polymers: Usually two different monomer combine with the loss of a small molecule, usually water. Most polyesters and polyamides (nylon) are in this class of polymers. Polyurethane Foam in graphic above.
Polymers classified by Physical Response to Heating
Thermoplastics
Plastics that soften when heated and become firm again when cooled. This is the more popular type of plastic because the heating and cooling may be repeated and the thermoplastic may be reformed.
Thermosets
These are plastics that soften when heated and can be molded, but harden permanently. They will decompose when reheated. An example is Bakelite, which is used in toasters, handles for pots and pans, dishes, electrical outlets and billiard balls.
Contributors
• Charles Ophardt, Professor Emeritus, Elmhurst College; Virtual Chembook | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/24%3A_Synthetic_Polymers/24.01%3A_Introduction.txt |
Polymers are long chain giant organic molecules are assembled from many smaller molecules called monomers. Polymers consist of many repeating monomer units in long chains. A polymer is analogous to a necklace made from many small beads (monomers). Many monomers are alkenes or other molecules with double bonds which react by addition to their unsaturated double bonds.
Introduction
The electrons in the double bond are used to bond two monomer molecules together. This is represented by the red arrows moving from one molecule to the space between two molecules where a new bond is to form. The formation of polyethylene from ethylene (ethene) may be illustrated in the graphic on the left as follows. In the complete polymer, all of the double bonds have been turned into single bonds. No atoms have been lost and you can see that the monomers have just been joined in the process of addition. A simple representation is -[A-A-A-A-A]-. Polyethylene is used in plastic bags, bottles, toys, and electrical insulation.
• LDPE - Low Density Polyethylene: The first commercial polyethylene process used peroxide catalysts at a temperature of 500 C and 1000 atmospheres of pressure. This yields a transparent polymer with highly branched chains which do not pack together well and is low in density. LDPE makes a flexible plastic. Today most LDPE is used for blow-molding of films for packaging and trash bags and flexible snap-on lids. LDPE is recyclable plastic #4.
• HDPE - High Density Polyethylene: An alternate method is to use Ziegler-Natta aluminum titanium catalysts to make HDPE which has very little branching, allows the strands to pack closely, and thus is high density. It is three times stronger than LDPE and more opaque. About 45% of the HDPE is blow molded into milk and disposable consumer bottles. HDPE is also used for crinkly plastic bags to pack groceries at grocery stores. HDPE is recyclable plastic #2.
Other Addition Polymers
• PVC (polyvinyl chloride), which is found in plastic wrap, simulated leather, water pipes, and garden hoses, is formed from vinyl chloride (H2C=CHCl). The reaction is shown in the graphic on the left. Notice how every other carbon must have a chlorine attached.
• Polypropylene: The reaction to make polypropylene (H2C=CHCH3) is illustrated in the middle reaction of the graphic. Notice that the polymer bonds are always through the carbons of the double bond. Carbon #3 already has saturated bonds and cannot participate in any new bonds. A methyl group is on every other carbon.
• Polystyrene: The reaction is the same for polystrene where every other carbon has a benzene ring attached. Polystyrene (PS) is recyclable plastic #6. In the following illustrated example, many styrene monomers are polymerized into a long chain polystyrene molecule. The squiggly lines indicate that the polystyrene molecule extends further at both the left and right ends.
• Blowing fine gas bubbles into liquid polystyrene and letting it solidify produces expanded polystyrene, called Styrofoam by the Dow Chemical Company.
• Polystyrene with DVB: Cross-linking between polymer chains can be introduced into polystyrene by copolymerizing with p-divinylbenzene (DVB). DVB has vinyl groups (-CH=CH2) at each end of its molecule, each of which can be polymerized into a polymer chain like any other vinyl group on a styrene monomer.
Other addition polymers
Table 1: Links to various polymers with Chime molecule - Macrogalleria at U. Southern Mississippi
Monomer Polymer Name Trade Name Uses
F2C=CF2 polytetrafluoroethylene Teflon
Non-stick coating for cooking utensils, chemically-resistant specialty plastic parts, Gore-Tex
H2C=CCl2 polyvinylidene dichloride Saran Clinging food wrap
H2C=CH(CN) polyacrylonitrile Orlon, Acrilan, Creslan Fibers for textiles, carpets, upholstery
H2C=CH(OCOCH3) polyvinyl acetate Elmer's glue - Silly Putty Demo
H2C=CH(OH) polyvinyl alcohol Ghostbusters Demo
H2C=C(CH3)COOCH3 polymethyl methacrylate Plexiglass, Lucite Stiff, clear, plastic sheets, blocks, tubing, and other shapes
Addition polymers from conjugated dienes
Polymers from conjugated dienes usually give elastomer polymers having rubber-like properties.
Table 2. Addition homopolymers from conjugated dienes
Monomer Polymer name Trade name Uses
H2C=CH-C(CH3)=CH2 polyisoprene natural or some synthetic rubber applications similar to natural rubber
H2C=CH-CH=CH2 polybutadiene polybutadiene synthetic rubber select synthetic rubber applications
H2C=CH-CCl=CH2 polychloroprene Neoprene chemically-resistant rubber
All the monomers from which addition polymers are made are alkenes or functionally substituted alkenes. The most common and thermodynamically favored chemical transformations of alkenes are addition reactions. Many of these addition reactions are known to proceed in a stepwise fashion by way of reactive intermediates, and this is the mechanism followed by most polymerizations. A general diagram illustrating this assembly of linear macromolecules, which supports the name chain growth polymers, is presented here. Since a pi-bond in the monomer is converted to a sigma-bond in the polymer, the polymerization reaction is usually exothermic by 8 to 20 kcal/mol. Indeed, cases of explosively uncontrolled polymerizations have been reported.
It is useful to distinguish four polymerization procedures fitting this general description.
• Radical Polymerization The initiator is a radical, and the propagating site of reactivity (*) is a carbon radical.
• Cationic Polymerization The initiator is an acid, and the propagating site of reactivity (*) is a carbocation.
• Anionic Polymerization The initiator is a nucleophile, and the propagating site of reactivity (*) is a carbanion.
• Coordination Catalytic Polymerization The initiator is a transition metal complex, and the propagating site of reactivity (*) is a terminal catalytic complex.
Radical Chain-Growth Polymerization
Virtually all of the monomers described above are subject to radical polymerization. Since this can be initiated by traces of oxygen or other minor impurities, pure samples of these compounds are often "stabilized" by small amounts of radical inhibitors to avoid unwanted reaction. When radical polymerization is desired, it must be started by using a radical initiator, such as a peroxide or certain azo compounds. The formulas of some common initiators, and equations showing the formation of radical species from these initiators are presented below.
By using small amounts of initiators, a wide variety of monomers can be polymerized. One example of this radical polymerization is the conversion of styrene to polystyrene, shown in the following diagram. The first two equations illustrate the initiation process, and the last two equations are examples of chain propagation. Each monomer unit adds to the growing chain in a manner that generates the most stable radical. Since carbon radicals are stabilized by substituents of many kinds, the preference for head-to-tail regioselectivity in most addition polymerizations is understandable. Because radicals are tolerant of many functional groups and solvents (including water), radical polymerizations are widely used in the chemical industry.
In principle, once started a radical polymerization might be expected to continue unchecked, producing a few extremely long chain polymers. In practice, larger numbers of moderately sized chains are formed, indicating that chain-terminating reactions must be taking place. The most common termination processes are Radical Combination and Disproportionation. These reactions are illustrated by the following equations. The growing polymer chains are colored blue and red, and the hydrogen atom transferred in disproportionation is colored green. Note that in both types of termination two reactive radical sites are removed by simultaneous conversion to stable product(s). Since the concentration of radical species in a polymerization reaction is small relative to other reactants (e.g. monomers, solvents and terminated chains), the rate at which these radical-radical termination reactions occurs is very small, and most growing chains achieve moderate length before termination.
The relative importance of these terminations varies with the nature of the monomer undergoing polymerization. For acrylonitrile and styrene combination is the major process. However, methyl methacrylate and vinyl acetate are terminated chiefly by disproportionation.
Another reaction that diverts radical chain-growth polymerizations from producing linear macromolecules is called chain transfer. As the name implies, this reaction moves a carbon radical from one location to another by an intermolecular or intramolecular hydrogen atom transfer (colored green). These possibilities are demonstrated by the following equations
Chain transfer reactions are especially prevalent in the high pressure radical polymerization of ethylene, which is the method used to make LDPE (low density polyethylene). The 1º-radical at the end of a growing chain is converted to a more stable 2º-radical by hydrogen atom transfer. Further polymerization at the new radical site generates a side chain radical, and this may in turn lead to creation of other side chains by chain transfer reactions. As a result, the morphology of LDPE is an amorphous network of highly branched macromolecules.
Chain topology
Polymers may also be classified as straight-chained or branched, leading to forms such as these:
The monomers can be joined end-to-end, and they can also be cross-linked to provide a harder material:
If the cross-links are fairly long and flexible, adjacent chains can move with respect to each other, producing an elastic polymer or
Cationic Chain-Growth Polymerization
Polymerization of isobutylene (2-methylpropene) by traces of strong acids is an example of cationic polymerization. The polyisobutylene product is a soft rubbery solid, Tg = _70º C, which is used for inner tubes. This process is similar to radical polymerization, as demonstrated by the following equations. Chain growth ceases when the terminal carbocation combines with a nucleophile or loses a proton, giving a terminal alkene (as shown here).
Monomers bearing cation stabilizing groups, such as alkyl, phenyl or vinyl can be polymerized by cationic processes. These are normally initiated at low temperature in methylene chloride solution. Strong acids, such as HClO4 , or Lewis acids containing traces of water (as shown above) serve as initiating reagents. At low temperatures, chain transfer reactions are rare in such polymerizations, so the resulting polymers are cleanly linear (unbranched).
Anionic Chain-Growth Polymerization
Treatment of a cold THF solution of styrene with 0.001 equivalents of n-butyllithium causes an immediate polymerization. This is an example of anionic polymerization, the course of which is described by the following equations. Chain growth may be terminated by water or carbon dioxide, and chain transfer seldom occurs. Only monomers having anion stabilizing substituents, such as phenyl, cyano or carbonyl are good substrates for this polymerization technique. Many of the resulting polymers are largely isotactic in configuration, and have high degrees of crystallinity.
Species that have been used to initiate anionic polymerization include alkali metals, alkali amides, alkyl lithiums and various electron sources. A practical application of anionic polymerization occurs in the use of superglue. This material is methyl 2-cyanoacrylate, CH2=C(CN)CO2CH3. When exposed to water, amines or other nucleophiles, a rapid polymerization of this monomer takes place.
Ring opening polymerization
In this kind of polymerization, molecular rings are opened in the formation of a polymer. Here epsilon-caprolactam, a 6-carbon cyclic monomer, undergoes ring opening to form a Nylon 6 homopolymer, which is somewhat similar to but not the same as Nylon 6,6 alternating copolymer.
Addition Copolymerization
Most direct copolymerizations of equimolar mixtures of different monomers give statistical copolymers, or if one monomer is much more reactive a nearly homopolymer of that monomer. The copolymerization of styrene with methyl methacrylate, for example, proceeds differently depending on the mechanism. Radical polymerization gives a statistical copolymer. However, the product of cationic polymerization is largely polystyrene, and anionic polymerization favors formation of poly(methyl methacrylate). In cases where the relative reactivities are different, the copolymer composition can sometimes be controlled by continuous introduction of a biased mixture of monomers into the reaction.
Formation of alternating copolymers is favored when the monomers have different polar substituents (e.g. one electron withdrawing and the other electron donating), and both have similar reactivities toward radicals. For example, styrene and acrylonitrile copolymerize in a largely alternating fashion.
Some Useful Copolymers
Monomer A
Monomer B
Copolymer
Uses
H2C=CHCl
H2C=CCl2
Saran
films & fibers
H2C=CHC6H5
H2C=C-CH=CH2
SBR
styrene butadiene rubber
tires
H2C=CHCN
H2C=C-CH=CH2
Nitrile Rubber
adhesives
hoses
H2C=C(CH3)2
H2C=C-CH=CH2
Butyl Rubber
inner tubes
F2C=CF(CF3)
H2C=CHF
Viton
gaskets
A terpolymer of acrylonitrile, butadiene and styrene, called ABS rubber, is used for high-impact containers, pipes and gaskets.
Block Copolymerization
Several different techniques for preparing block copolymers have been developed, many of which use condensation reactions (next section). At this point, our discussion will be limited to an application of anionic polymerization. In the anionic polymerization of styrene described above, a reactive site remains at the end of the chain until it is quenched. The unquenched polymer has been termed a living polymer, and if additional styrene or a different suitable monomer is added a block polymer will form. This is illustrated for methyl methacrylate in the following diagram.
Contributors
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/24%3A_Synthetic_Polymers/24.02%3A_Chain-Growth_PolymersAddition_Polymers.txt |
Another type of anionic polymerization involves an epoxide and forms a polyether. The strained three-memebered epoxide ring is easily opened by nuleophiles, typically a hydroxide or an alkoxy group. Note! This polymerization is different to the other chain-growth polymerizations discussed because the process forms a C-O bond in the polymer chain.
24.04: ZieglerNatta Catalysts and Polymer Stereochemistry
Symmetrical monomers such as ethylene and tetrafluoroethylene can join together in only one way. Monosubstituted monomers, on the other hand, may join together in two organized ways, described in the following diagram, or in a third random manner. Most monomers of this kind, including propylene, vinyl chloride, styrene, acrylonitrile and acrylic esters, prefer to join in a head-to-tail fashion, with some randomness occurring from time to time. The reasons for this regioselectivity will be discussed in the synthetic methods section.
If the polymer chain is drawn in a zig-zag fashion, as shown above, each of the substituent groups (Z) will necessarily be located above or below the plane defined by the carbon chain. Consequently we can identify three configurational isomers of such polymers. If all the substituents lie on one side of the chain the configuration is called isotactic. If the substituents alternate from one side to another in a regular manner the configuration is termed syndiotactic. Finally, a random arrangement of substituent groups is referred to as atactic. Examples of these configurations are shown here.
Many common and useful polymers, such as polystyrene, polyacrylonitrile and poly(vinyl chloride) are atactic as normally prepared. Customized catalysts that effect stereoregular polymerization of polypropylene and some other monomers have been developed, and the improved properties associated with the increased crystallinity of these products has made this an important field of investigation. The following values of Tg have been reported.
Polymer
Tg atactic
Tg isotactic
Tg syndiotactic
PP
–20 ºC
0 ºC
–8 ºC
PMMA
100 ºC
130 ºC
120 ºC
The properties of a given polymer will vary considerably with its tacticity. Thus, atactic polypropylene is useless as a solid construction material, and is employed mainly as a component of adhesives or as a soft matrix for composite materials. In contrast, isotactic polypropylene is a high-melting solid (ca. 170 ºC) which can be molded or machined into structural components.
Ziegler-Natta Catalytic Polymerization
An efficient and stereospecific catalytic polymerization procedure was developed by Karl Ziegler (Germany) and Giulio Natta (Italy) in the 1950's. Their findings permitted, for the first time, the synthesis of unbranched, high molecular weight polyethylene (HDPE), laboratory synthesis of natural rubber from isoprene, and configurational control of polymers from terminal alkenes like propene (e.g. pure isotactic and syndiotactic polymers). In the case of ethylene, rapid polymerization occurred at atmospheric pressure and moderate to low temperature, giving a stronger (more crystalline) product (HDPE) than that from radical polymerization (LDPE). For this important discovery these chemists received the 1963 Nobel Prize in chemistry.
Ziegler-Natta catalysts are prepared by reacting certain transition metal halides with organometallic reagents such as alkyl aluminum, lithium and zinc reagents. The catalyst formed by reaction of triethylaluminum with titanium tetrachloride has been widely studied, but other metals (e.g. V & Zr) have also proven effective. The following diagram presents one mechanism for this useful reaction. Others have been suggested, with changes to accommodate the heterogeneity or homogeneity of the catalyst. Polymerization of propylene through action of the titanium catalyst gives an isotactic product; whereas, a vanadium based catalyst gives a syndiotactic product. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/24%3A_Synthetic_Polymers/24.03%3A_Anionic_Polymerization_of_Epoxides.txt |
Rubber is an example of an elastomer type polymer, where the polymer has the ability to return to its original shape after being stretched or deformed. The rubber polymer is coiled when in the resting state. The elastic properties arise from the its ability to stretch the chains apart, but when the tension is released the chains snap back to the original position. The majority of rubber polymer molecules contain at least some units derived from conjugated diene monomers (see Polymerization of Conjugated Dienes). Such conjugated diene monomers have a constructive backbone of at least four carbon atoms with a double-single-double bond reactive core (C=C-C=C ). Most if not practically all such dienes undergo 1,4-addition to the polymer chain, where 1 and 4 refer to the 1st and 4th carbons of the backbone unit, which become single-bonded to the rest of the polymer chain. The diene's double bonds turn into single bonds, and the single bond between them turns into a Z or E configured double bond, depending on the polymerization conditions. The unit's backbone thus becomes like this (-C-C=C-C-). Rubber gets its elasticity when the formed double bond gets the Z configuration. For 1,3-butadiene, Z is equivalent to a cis and E is equivalent to a trans configuration.
Natural Rubber
Natural rubber is an addition polymer that is obtained as a milky white fluid known as latex from a tropical rubber tree. Natural rubber is from the monomer isoprene (2-methyl-1,3-butadiene), which is a conjugated diene hydrocarbon as mentioned above. In natural rubber, most of the double fonds formed in the polymer chain have the Z configuration, resulting in natural rubber's elastomer qualities.
Charles Goodyear accidentally discovered that by mixing sulfur and rubber, the properties of the rubber improved in being tougher, resistant to heat and cold, and increased in elasticity. This process was later called vulcanization after the Roman god of fire. Vulcanization causes shorter chains to cross link through the sulfur to longer chains. The development of vulcanized rubber for automobile tires greatly aided this industry.
Synthetic Rubber
Important conjugated dienes used in synthetic rubbers include isoprene (2-methyl-1,3-butadiene), 1,3-butadiene, and chloroprene (2-chloro-1,3-butadiene). Polymerized 1,3-butadiene is mostly referred to simply as polybutadiene. Polymerized chloroprene was developed by DuPont and given the trade name Neoprene.
In a number of cases, monomers which are not dienes are also used for certain types of synthetic rubber, often copolymerized with dienes. Some of the most commercially important addition polymers are the copolymers. These are polymers made by polymerizing a mixture of two or more monomers. An example is styrene-butadiene rubber (SBR) - which is a copolymer of 1,3-butadiene and styrene which is mixed in a 3 to 1 ratio, respectively.
SBR rubber was developed during World War II when important supplies of natural rubber were cut off. SBR is more resistant to abrasion and oxidation than natural rubber and can also be vulcanized. More than 40% of the synthetic rubber production is SBR and is used in tire production. A tiny amount is used for bubble-gum in the unvulcanized form.
Nitrile rubber is copolymerized from butadiene and acrylonitrile (H2C=CH-CN). Butyl rubber is copolymerized from isobutylene [which is methylpropene H2C=C(CH3)2 ] and a small percentage of isoprene. Silicone rubber and other compounds, chemically called polysiloxanes, are not from conjugated dienes but have repeating units like -O-SiR2- where R is some organic radical group like methyl. There is a separate page on Silicone Polymers.
Conjugated dienes (alkenes with two double bonds and a single bond in between) can be polymerized to form important compounds like rubber. This takes place, in different forms, both in nature and in the laboratory. Interactions between double bonds on multiple chains leads to cross-linkage which creates elasticity within the compound.
Polymerization of 1,3-Butadiene
For rubber compounds to be synthesized, 1,3-butadiene must be polymerized. Below is a simple illustration of how this compound is formed into a chain. The 1,4 polymerization is much more useful to polymerization reactions.
Above, the green structures represent the base units of the polymers that are synthesized and the red represents the bonds between these units which form these polymers. Whether the 1,3 product or the 1,4 product is formed depends on whether the reaction is thermally or kinetically controlled.
Synthetic Rubber
The most important synthetic rubber is Neoprene which is produced by the polymerization of 2-chloro-1,3-butadiene.
In this illustration, the dashed lines represent repetition of the same base units, so both the products and reactants are polymers. The reaction proceeds with a mechanism similar to the Friedel-Crafts mechanism. Cross-linkage between the chlorine atom of one chain and the double bond of another contributes to the overall elasticity of neoprene. This cross-linkage occurs as the chains lie next to each other at random angles, and the attractions between double bonds prevent them from sliding back and forth.
Natural Rubber
The synthesis of rubber in nature in somewhat similar the artificial synthesis of rubber except that it takes place within a plant. Instead of the 2-chloro-1,3-butadiene used in the synthesis of neoprene, natural rubber is synthesized from 2-methyl-1,3-butadiene. As an electrophile, the plant synthesizes the pyrophosphate 3-methyl-3-butenyl pyrophosphate is from phosphoric acid and 3-methyl-3-buten-1-ol. This pyrophosphate then catalyzes the reaction that leads to natural rubber.
The 3-methyl-3-butenyl pyrophosphate (OPP) is then used in the polymerization of natural rubber as it pulls electrons off 2-methyl-1,3-butadiene (see questions section for this process.)
Problem
Draw out the mechanism for the natural synthesis of rubber from 3-methyl-3-butenyl pyrophosphate and 2-methyl-1,3-butadiene. Show the movement of electrons with arrows.
Contributors
• Charles Ophardt, Professor Emeritus, Elmhurst College; Virtual Chembook
• User:H Padleckas | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/24%3A_Synthetic_Polymers/24.05%3A_Natural_and_Synthetic_Rubbers.txt |
A large number of important and useful polymeric materials are not formed by chain-growth processes involving reactive species such as radicals, but proceed instead by conventional functional group transformations of polyfunctional reactants. These polymerizations often (but not always) occur with loss of a small byproduct, such as water, and generally (but not always) combine two different components in an alternating structure. The polyester Dacron and the polyamide Nylon 66, shown here, are two examples of synthetic condensation polymers, also known as step-growth polymers. In contrast to chain-growth polymers, most of which grow by carbon-carbon bond formation, step-growth polymers generally grow by carbon-heteroatom bond formation (C-O & C-N in Dacron & Nylon respectively). Although polymers of this kind might be considered to be alternating copolymers, the repeating monomeric unit is usually defined as a combined moiety.
Examples of naturally occurring condensation polymers are cellulose, the polypeptide chains of proteins, and poly(β-hydroxybutyric acid), a polyester synthesized in large quantity by certain soil and water bacteria. Formulas for these will be displayed below by clicking on the diagram.
Characteristics of Condensation Polymers
Condensation polymers form more slowly than addition polymers, often requiring heat, and they are generally lower in molecular weight. The terminal functional groups on a chain remain active, so that groups of shorter chains combine into longer chains in the late stages of polymerization. The presence of polar functional groups on the chains often enhances chain-chain attractions, particularly if these involve hydrogen bonding, and thereby crystallinity and tensile strength. The following examples of condensation polymers are illustrative.
Note that for commercial synthesis the carboxylic acid components may actually be employed in the form of derivatives such as simple esters. Also, the polymerization reactions for Nylon 6 and Spandex do not proceed by elimination of water or other small molecules. Nevertheless, the polymer clearly forms by a step-growth process. Some Condensation Polymers
The difference in Tg and Tm between the first polyester (completely aliphatic) and the two nylon polyamides (5th & 6th entries) shows the effect of intra-chain hydrogen bonding on crystallinity. The replacement of flexible alkylidene links with rigid benzene rings also stiffens the polymer chain, leading to increased crystalline character, as demonstrated for polyesters (entries 1, 2 &3) and polyamides (entries 5, 6, 7 & 8). The high Tg and Tm values for the amorphous polymer Lexan are consistent with its brilliant transparency and glass-like rigidity. Kevlar and Nomex are extremely tough and resistant materials, which find use in bullet-proof vests and fire resistant clothing.
Many polymers, both addition and condensation, are used as fibers The chief methods of spinning synthetic polymers into fibers are from melts or viscous solutions. Polyesters, polyamides and polyolefins are usually spun from melts, provided the Tm is not too high. Polyacrylates suffer thermal degradation and are therefore spun from solution in a volatile solvent. Cold-drawing is an important physical treatment that improves the strength and appearance of these polymer fibers. At temperatures above Tg, a thicker than desired fiber can be forcibly stretched to many times its length; and in so doing the polymer chains become untangled, and tend to align in a parallel fashion. This cold-drawing procedure organizes randomly oriented crystalline domains, and also aligns amorphous domains so they become more crystalline. In these cases, the physically oriented morphology is stabilized and retained in the final product. This contrasts with elastomeric polymers, for which the stretched or aligned morphology is unstable relative to the amorphous random coil morphology.
This cold-drawing treatment may also be used to treat polymer films (e.g. Mylar & Saran) as well as fibers.
Step-growth polymerization is also used for preparing a class of adhesives and amorphous solids called epoxy resins. Here the covalent bonding occurs by an SN2 reaction between a nucleophile, usually an amine, and a terminal epoxide. In the following example, the same bisphenol A intermediate used as a monomer for Lexan serves as a difunctional scaffold to which the epoxide rings are attached. Bisphenol A is prepared by the acid-catalyzed condensation of acetone with phenol.
What are polyamides?
Polyamides are polymers where the repeating units are held together by amide links. An amide group has the formula - CONH2. An amide link has this structure:
In an amide itself, of course, the bond on the right is attached to a hydrogen atom.
Nylon
In nylon, the repeating units contain chains of carbon atoms. (That is different from Kevlar, where the repeating units contain benzene rings - see below.) There are various different types of nylon depending on the nature of those chains.
Nylon-6,6
Nylon-6,6 is made from two monomers each of which contain 6 carbon atoms - hence its name. One of the monomers is a 6 carbon acid with a -COOH group at each end - hexanedioic acid. The other monomer is a 6 carbon chain with an amino group, -NH2, at each end. This is 1,6-diaminohexane (also known as hexane-1,6-diamine).
When these two compounds polymerise, the amine and acid groups combine, each time with the loss of a molecule of water. This is known as condensation polymerization. Condensation polymerization is the formation of a polymer involving the loss of a small molecule. In this case, the molecule is water, but in other cases different small molecules might be lost.
The diagram shows the loss of water between two of the monomers:
This keeps on happening, and so you get a chain which looks like this:
Nylon-6
Iit is possible to get a polyamide from a single monomer. Nylon-6 is made from a monomer called caprolactam.
Notice that this already contains an amide link. When this molecule polymerizes, the ring opens, and the molecules join up in a continuous chain.
Kevlar
Kevlar is similar in structure to nylon-6,6 except that instead of the amide links joining chains of carbon atoms together, they join benzene rings. The two monomers are benzene-1,4-dicarboxylic acid and 1,4-diaminobenzene.
If you line these up and remove water between the -COOH and -NH2 groups in the same way as we did with nylon-6,6, you get the structure of Kevlar:
What is a polyester?
A polyester is a polymer (a chain of repeating units) where the individual units are held together by ester linkages.
The diagram shows a very small bit of the polymer chain and looks pretty complicated. But it is not very difficult to work out - and that's the best thing to do: work it out, not try to remember it. You will see how to do that in a moment.
The usual name of this common polyester is poly(ethylene terephthalate). The everyday name depends on whether it is being used as a fibre or as a material for making things like bottles for soft drinks. When it is being used as a fiber to make clothes, it is often just called polyester. It may sometimes be known by a brand name like Terylene. When it is being used to make bottles, for example, it is usually called PET.
Making polyesters as an example of condensation polymerisation
In condensation polymerisation, when the monomers join together a small molecule gets lost. That's different from addition polymerisation which produces polymers like poly(ethene) - in that case, nothing is lost when the monomers join together. A polyester is made by a reaction involving an acid with two -COOH groups, and an alcohol with two -OH groups. In the common polyester drawn below.
Figure: The acid is benzene-1,4-dicarboxylic acid (old name: terephthalic acid) and the alcohol is ethane-1,2-diol (old name: ethylene glycol).
Now imagine lining these up alternately and making esters with each acid group and each alcohol group, losing a molecule of water every time an ester linkage is made.
That would produce the chain shown above (although this time written without separating out the carbon-oxygen double bond - write it whichever way you like). | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/24%3A_Synthetic_Polymers/24.06%3A_Step-Growth_PolymersCondensation_Polymers.txt |
A comparison of the properties of polyethylene (both LDPE & HDPE) with the natural polymers rubber and cellulose is instructive. As noted above, synthetic HDPE macromolecules have masses ranging from 105 to 106 amu (LDPE molecules are more than a hundred times smaller). Rubber and cellulose molecules have similar mass ranges, but fewer monomer units because of the monomer's larger size. The physical properties of these three polymeric substances differ from each other, and of course from their monomers.
• HDPE is a rigid translucent solid which softens on heating above 100º C, and can be fashioned into various forms including films. It is not as easily stretched and deformed as is LDPE. HDPE is insoluble in water and most organic solvents, although some swelling may occur on immersion in the latter. HDPE is an excellent electrical insulator.
• LDPE is a soft translucent solid which deforms badly above 75º C. Films made from LDPE stretch easily and are commonly used for wrapping. LDPE is insoluble in water, but softens and swells on exposure to hydrocarbon solvents. Both LDPE and HDPE become brittle at very low temperatures (below -80º C). Ethylene, the common monomer for these polymers, is a low boiling (-104º C) gas.
• Natural (latex) rubber is an opaque, soft, easily deformable solid that becomes sticky when heated (above. 60º C), and brittle when cooled below -50º C. It swells to more than double its size in nonpolar organic solvents like toluene, eventually dissolving, but is impermeable to water. The C5H8 monomer isoprene is a volatile liquid (b.p. 34º C).
• Pure cellulose, in the form of cotton, is a soft flexible fiber, essentially unchanged by variations in temperature ranging from -70 to 80º C. Cotton absorbs water readily, but is unaffected by immersion in toluene or most other organic solvents. Cellulose fibers may be bent and twisted, but do not stretch much before breaking. The monomer of cellulose is the C6H12O6 aldohexose D-glucose. Glucose is a water soluble solid melting below 150º C.
To account for the differences noted here we need to consider the nature of the aggregate macromolecular structure, or morphology, of each substance. Because polymer molecules are so large, they generally pack together in a non-uniform fashion, with ordered or crystalline-like regions mixed together with disordered or amorphous domains. In some cases the entire solid may be amorphous, composed entirely of coiled and tangled macromolecular chains. Crystallinity occurs when linear polymer chains are structurally oriented in a uniform three-dimensional matrix. In the diagram on the right, crystalline domains are colored blue.
Increased crystallinity is associated with an increase in rigidity, tensile strength and opacity (due to light scattering). Amorphous polymers are usually less rigid, weaker and more easily deformed. They are often transparent.
Three factors that influence the degree of crystallinity are:
i) Chain length
ii) Chain branching
iii) Interchain bonding
The importance of the first two factors is nicely illustrated by the differences between LDPE and HDPE. As noted earlier, HDPE is composed of very long unbranched hydrocarbon chains. These pack together easily in crystalline domains that alternate with amorphous segments, and the resulting material, while relatively strong and stiff, retains a degree of flexibility. In contrast, LDPE is composed of smaller and more highly branched chains which do not easily adopt crystalline structures. This material is therefore softer, weaker, less dense and more easily deformed than HDPE. As a rule, mechanical properties such as ductility, tensile strength, and hardness rise and eventually level off with increasing chain length.
The nature of cellulose supports the above analysis and demonstrates the importance of the third factor (iii). To begin with, cellulose chains easily adopt a stable rod-like conformation. These molecules align themselves side by side into fibers that are stabilized by inter-chain hydrogen bonding between the three hydroxyl groups on each monomer unit. Consequently, crystallinity is high and the cellulose molecules do not move or slip relative to each other. The high concentration of hydroxyl groups also accounts for the facile absorption of water that is characteristic of cotton.
Natural rubber is a completely amorphous polymer. Unfortunately, the potentially useful properties of raw latex rubber are limited by temperature dependence; however, these properties can be modified by chemical change. The cis-double bonds in the hydrocarbon chain provide planar segments that stiffen, but do not straighten the chain. If these rigid segments are completely removed by hydrogenation (H2 & Pt catalyst), the chains lose all constrainment, and the product is a low melting paraffin-like semisolid of little value. If instead, the chains of rubber molecules are slightly cross-linked by sulfur atoms, a process called vulcanization which was discovered by Charles Goodyear in 1839, the desirable elastomeric properties of rubber are substantially improved. At 2 to 3% crosslinking a useful soft rubber, that no longer suffers stickiness and brittleness problems on heating and cooling, is obtained. At 25 to 35% crosslinking a rigid hard rubber product is formed. The following illustration shows a cross-linked section of amorphous rubber. By clicking on the diagram it will change to a display of the corresponding stretched section. The more highly-ordered chains in the stretched conformation are entropically unstable and return to their original coiled state when allowed to relax | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/24%3A_Synthetic_Polymers/24.07%3A_Polymer_Structure_and_Properties.txt |
Most plastics crumble into ever-tinier fragments as they are exposed to sunlight and the elements. Except for the small amount that's been incinerated–and it's a very small amount–every bit of plastic ever made still exists, unless the material's molecular structure is designed to favor biodegradation. Unfortunately, cleaning up the garbage patch is not a realistic option, and unless we change our disposal and recycling habits, it will undoubtedly get bigger. One sensible solution would require manufacturers to use natural biodegradable packaging materials whenever possible, and consumers to conscientiously dispose of their plastic waste. Thus, instead of consigning all plastic trash to a land fill, some of it may provide energy by direct combustion, and some converted for reuse as a substitute for virgin plastics. The latter is particularly attractive since a majority of plastics are made from petroleum, a diminishing resource with a volatile price.
The energy potential of plastic waste is relatively significant, ranging from 10.2 to 30.7MJ kg Ã, suggesting application as an energy source and temperature stabilizer in municipal incinerators, thermal power plants and cement kilns. The use of plastic waste as a fuel source would be an effective means of reducing landfill requirements while recovering energy. This, however, depends on using appropriate materials. Inadequate control of combustion, especially for plastics containing chlorine, fluorine and bromine, constitutes a risk of emitting toxic pollutants.
Whether used as fuels or a source of recycled plastic, plastic waste must be separated into different categories. To this end, an identification coding system was developed by the Society of the Plastics Industry (SPI) in 1988, and is used internationally. This code, shown on the right, is a set of symbols placed on plastics to identify the polymer type, for the purpose of allowing efficient separation of different polymer types for recycling. The abbreviations of the code are explained in the following table.
PETE
HDPE
V
LDPE
polyethylene
terephthalate
high density
polyethylene
polyvinyl chloride
low density
polyethylene
PP
PS
OTHER
polypropylene
polystyrene
polyesters, acrylics
polyamides, teflon etc.
Despite use of the recycling symbol in the coding of plastics, there is consumer confusion about which plastics are readily recyclable. In most communities throughout the United States, PETE and HDPE are the only plastics collected in municipal recycling programs. However, some regions are expanding the range of plastics collected as markets become available. (Los Angeles, for example, recycles all clean plastics numbered 1 through 7) In theory, most plastics are recyclable and some types can be used in combination with others. In many instances, however, there is an incompatibility between different types that necessitates their effective separation. Since the plastics utilized in a given manufacturing sector (e.g. electronics, automotive, etc.) is generally limited to a few types, effective recycling is often best achieved with targeted waste streams.
Recycled Plastics
Recycle Code
Abbreviation and Chemical Name of Plastic
Types of Uses and Examples
1
PET - polyethylene terephthalate
Many types of clear plastic consumer bottles, including clear, 2-liter beverage bottles
2
HDPE - High density polyethylene
Milk jugs, detergent bottles, some water bottles, some grocery plastic bags
3
PVC - Polyvinyl chloride
Plastic drain pipe, shower curtains, some water bottles
4
LDPE - Low density polyethylene
Plastic garbage and other bags, garment bags, snap-on lids such as coffee can lids
5
PP - Polypropylene
Many translucent (or opaque) plastic containers; containers for some products such as yogurt, soft butter, or margarine; aerosol can tops; rigid bottle caps; candy wrappers; bottoms of bottles
6
PS - Polystyrene
Hard clear plastic cups, foam cups, eating utensils, deli food containers, toy model kits, some packing popcorn
7
Other
Polycarbonate is a common type, Biodegradable, Some packing popcorn | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/24%3A_Synthetic_Polymers/24.09%3A_Polymer_Recycling_and_Disposal.txt |
Wöhler synthesis of Urea in 1828 heralded the birth of modern chemistry. The Art of synthesis is as old as Organic chemistry itself. Natural product chemistry is firmly rooted in the science of degrading a molecule to known smaller molecules using known chemical reactions and conforming the assigned structure by chemical synthesis from small, well known molecules using well established synthetic chemistry techniques. Once this art of synthesizing a molecule was mastered, chemists attempted to modify bioactive molecules in an attempt to develop new drugs and also to unravel the mystery of biomolecular interactions. Until the middle of the 20th Century, organic chemists approached the task of synthesis of molecules as independent tailor made projects, guided mainly by chemical intuition and a sound knowledge of chemical reactions. During this period, a strong foundation was laid for the development of mechanistic principles of organic reactions, new reactions and reagents. More than a century of such intensive studies on the chemistry of carbohydrates, alkaloids, terpenes and steroids laid the foundation for the development of logical approaches for the synthesis of molecules.
Organic chemistry encompasses a very large number of compounds (many millions), and our previous discussion and illustrations have focused on their structural characteristics. Now that we can recognize these actors (compounds), we turn to the roles they are inclined to play in the scientific drama staged by the multitude of chemical reactions that define organic chemistry.
If you scan any organic textbook you will encounter what appears to be a very large, often intimidating, number of reactions. These are the "tools" of a chemist, and to use these tools effectively, we must organize them in a sensible manner and look for patterns of reactivity that permit us make plausible predictions. Most of these reactions occur at special sites of reactivity known as functional groups, and these constitute one organizational scheme that helps us catalog and remember reactions.
Ultimately, the best way to achieve proficiency in organic chemistry is to understand how reactions take place, and to recognize the various factors that influence their course.
First, we identify four broad classes of reactions based solely on the structural change occurring in the reactant molecules. This classification does not require knowledge or speculation concerning reaction paths or mechanisms. The four main reaction classes are additions, eliminations, substitutions, and rearrangements.
The job of a synthetic chemist is akin to that of an architect (or civil engineer). While the architect could actually see the building he is constructing, a molecular architect called Chemist is handicapped by the fact that the molecule he is synthesizing is too small to be seen even through the most powerful microscope developed to date. With such a limitation, how does he ‘see’ the developing structure? For this purpose, a chemist makes use of spectroscopic tools. How does this chemist cut, tailor and glue the components on a molecule that cannot be seen with the naked eye? For this purpose chemists have developed molecular level tools called Reagents and Reactions. How does he clean the debris and produce pure molecules? This feat is achieved by crystallization, distillation and extensive use of chromatography techniques. A mastery over several such techniques enables the molecular architect (popularly known as organic chemist) to achieve the challenging task of synthesizing the mirade molecular structures encountered in Natural Products Chemistry, Drug Chemistry and modern Molecular Materials. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/01._Structure_and_Bonding_in_Organic_Molecules/1.1%3A_The__Scope_of__Organic_Chemistry.txt |
There are many types of chemical bonds and forces that bind molecules together. The two most basic types of bonds are characterized as either ionic or covalent. In ionic bonding, atoms transfer electrons to each other. Ionic bonds require at least one electron donor and one electron acceptor. In contrast, atoms with the same electronegativity share electrons in covalent bonds, because neither atom preferentially attracts or repels the shared electrons.
Introduction
Ionic bonding is the complete transfer of valence electron(s) between atoms. It is a type of chemical bond that generates two oppositely charged ions. In ionic bonds, the metal loses electrons to become a positively charged cation, whereas the nonmetal accepts those electrons to become a negatively charged anion. Ionic bonds require an electron donor, often a metal, and an electron acceptor, a nonmetal.
Ionic bonding is observed because metals have few electrons in their outer-most orbitals. By losing those electrons, these metals can achieve noble gas configuration and satisfy the octet rule. Similarly, nonmetals that have close to 8 electrons in their valence shells tend to readily accept electrons to achieve noble gas configuration. In ionic bonding, more than 1 electron can be donated or received to satisfy the octet rule. The charges on the anion and cation correspond to the number of electrons donated or received. In ionic bonds, the net charge of the compound must be zero.
This sodium molecule donates the lone electron in its valence orbital in order to achieve octet configuration. This creates a positively charged cation due to the loss of electron.
This chlorine atom receives one electron to achieve its octet configuration, which creates a negatively charged anion.
The predicted overall energy of the ionic bonding process, which includes the ionization energy of the metal and electron affinity of the nonmetal, is usually positive, indicating that the reaction is endothermic and unfavorable. However, this reaction is highly favorable because of the electrostatic attraction between the particles. At the ideal interatomic distance, attraction between these particles releases enough energy to facilitate the reaction. Most ionic compounds tend to dissociate in polar solvents because they are often polar. This phenomenon is due to the opposite charges on each ion.
Example \(1\): Chloride Salts
In this example, the sodium atom is donating its 1 valence electron to the chlorine atom. This creates a sodium cation and a chlorine anion. Notice that the net charge of the resulting compound is 0.
In this example, the magnesium atom is donating both of its valence electrons to chlorine atoms. Each chlorine atom can only accept 1 electron before it can achieve its noble gas configuration; therefore, 2 atoms of chlorine are required to accept the 2 electrons donated by the magnesium. Notice that the net charge of the compound is 0.
Covalent Bonding
Covalent bonding is the sharing of electrons between atoms. This type of bonding occurs between two atoms of the same element or of elements close to each other in the periodic table. This bonding occurs primarily between nonmetals; however, it can also be observed between nonmetals and metals.
If atoms have similar electronegativities (the same affinity for electrons), covalent bonds are most likely to occur. Because both atoms have the same affinity for electrons and neither has a tendency to donate them, they share electrons in order to achieve octet configuration and become more stable. In addition, the ionization energy of the atom is too large and the electron affinity of the atom is too small for ionic bonding to occur. For example: carbon does not form ionic bonds because it has 4 valence electrons, half of an octet. To form ionic bonds, Carbon molecules must either gain or lose 4 electrons. This is highly unfavorable; therefore, carbon molecules share their 4 valence electrons through single, double, and triple bonds so that each atom can achieve noble gas configurations. Covalent bonds include interactions of the sigma and pi orbitals; therefore, covalent bonds lead to formation of single, double, triple, and quadruple bonds.
Example \(2\): \(PCl_3\)
In this example, a phosphorous atom is sharing its three unpaired electrons with three chlorine atoms. In the end product, all four of these molecules have 8 valence electrons and satisfy the octet rule.
Bonding in Organic Chemistry
Ionic and covalent bonds are the two extremes of bonding. Polar covalent is the intermediate type of bonding between the two extremes. Some ionic bonds contain covalent characteristics and some covalent bonds are partially ionic. For example, most carbon-based compounds are covalently bonded but can also be partially ionic. Polarity is a measure of the separation of charge in a compound. A compound's polarity is dependent on the symmetry of the compound and on differences in electronegativity between atoms. Polarity occurs when the electron pushing elements, found on the left side of the periodic table, exchanges electrons with the electron pulling elements, on the right side of the table. This creates a spectrum of polarity, with ionic (polar) at one extreme, covalent (nonpolar) at another, and polar covalent in the middle.
Both of these bonds are important in organic chemistry. Ionic bonds are important because they allow the synthesis of specific organic compounds. Scientists can manipulate ionic properties and these interactions in order to form desired products. Covalent bonds are especially important since most carbon molecules interact primarily through covalent bonding. Covalent bonding allows molecules to share electrons with other molecules, creating long chains of compounds and allowing more complexity in life.
Problems
1. Are these compounds ionic or covalent?
2. In the following reactions, indicate whether the reactants and products are ionic or covalently bonded.
a)
b) Clarification: What is the nature of the bond between sodium and amide? What kind of bond forms between the anion carbon chain and sodium?
c)
Solutions
• 1) From left to right: Covalent, Ionic, Ionic, Covalent, Covalent, Covalent, Ionic.
• 2a) All products and reactants are ionic.
• 2b) From left to right: Covalent, Ionic, Ionic, Covalent, Ionic, Covalent, Covalent, Ionic.
• 2c) All products and reactants are covalent.
Further Reading
Michigan State Virtual Textbook of Organic Chemistry
MasterOrganicChemistry
Understanding Periodic Trends
Carey 4th Edition On-Line Activity
Khan Academy
Leah4Sci
Cliffs Notes
Slide Presentations
Web Pages
Videos | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/01._Structure_and_Bonding_in_Organic_Molecules/1.3%3A_Ionic__and__Covalent_Bonds_-_The__Octet__Rule.txt |
Using Lewis Dot Symbols to Describe Covalent Bonding
This sharing of electrons allowing atoms to "stick" together is the basis of covalent bonding. There is some intermediate distant, generally a bit longer than 0.1 nm, or if you prefer 100 pm, at which the attractive forces significantly outweigh the repulsive forces and a bond will be formed if both atoms can achieve a completen s2np6 configuration. It is this behavior that Lewis captured in his octet rule. The valence electron configurations of the constituent atoms of a covalent compound are important factors in determining its structure, stoichiometry, and properties. For example, chlorine, with seven valence electrons, is one electron short of an octet. If two chlorine atoms share their unpaired electrons by making a covalent bond and forming Cl2, they can each complete their valence shell:
Each chlorine atom now has an octet. The electron pair being shared by the atoms is called a bonding pair ; the other three pairs of electrons on each chlorine atom are called lone pairs. Lone pairs are not involved in covalent bonding. If both electrons in a covalent bond come from the same atom, the bond is called a coordinate covalent bond.
We can illustrate the formation of a water molecule from two hydrogen atoms and an oxygen atom using Lewis dot symbols:
The structure on the right is the Lewis electron structure, or Lewis structure, for H2O. With two bonding pairs and two lone pairs, the oxygen atom has now completed its octet. Moreover, by sharing a bonding pair with oxygen, each hydrogen atom now has a full valence shell of two electrons. Chemists usually indicate a bonding pair by a single line, as shown here for our two examples:
The following procedure can be used to construct Lewis electron structures for more complex molecules and ions:
1. Arrange the atoms to show specific connections. When there is a central atom, it is usually the least electronegative element in the compound. Chemists usually list this central atom first in the chemical formula (as in CCl4 and CO32−, which both have C as the central atom), which is another clue to the compound’s structure. Hydrogen and the halogens are almost always connected to only one other atom, so they are usually terminal rather than central.
Note
The central atom is usually the least electronegative element in the molecule or ion; hydrogen and the halogens are usually terminal.
1. Determine the total number of valence electrons in the molecule or ion. Add together the valence electrons from each atom. (Recall from Chapter 2 that the number of valence electrons is indicated by the position of the element in the periodic table.) If the species is a polyatomic ion, remember to add or subtract the number of electrons necessary to give the total charge on the ion. For CO32−, for example, we add two electrons to the total because of the −2 charge.
2. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. In $H_2O$, for example, there is a bonding pair of electrons between oxygen and each hydrogen.
3. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). These electrons will usually be lone pairs.
4. If any electrons are left over, place them on the central atom. Some atoms are able to accommodate more than eight electrons.
5. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. This will not change the number of electrons on the terminal atoms.
Now let’s apply this procedure to some particular compounds, beginning with one we have already discussed.
H2O
1. Because H atoms are almost always terminal, the arrangement within the molecule must be HOH.
2. Each H atom (group 1) has 1 valence electron, and the O atom (group 16) has 6 valence electrons, for a total of 8 valence electrons.
3. Placing one bonding pair of electrons between the O atom and each H atom gives H:O:H, with 4 electrons left over.
4. Each H atom has a full valence shell of 2 electrons.
5. Adding the remaining 4 electrons to the oxygen (as two lone pairs) gives the following structure:
This is the Lewis structure we drew earlier. Because it gives oxygen an octet and each hydrogen two electrons, we do not need to use step 6.
OCl−
1. With only two atoms in the molecule, there is no central atom.
2. Oxygen (group 16) has 6 valence electrons, and chlorine (group 17) has 7 valence electrons; we must add one more for the negative charge on the ion, giving a total of 14 valence electrons.
3. Placing a bonding pair of electrons between O and Cl gives O:Cl, with 12 electrons left over.
4. If we place six electrons (as three lone pairs) on each atom, we obtain the following structure:
Each atom now has an octet of electrons, so steps 5 and 6 are not needed. The Lewis electron structure is drawn within brackets as is customary for an ion, with the overall charge indicated outside the brackets, and the bonding pair of electrons is indicated by a solid line. OCl is the hypochlorite ion, the active ingredient in chlorine laundry bleach and swimming pool disinfectant.
CH2O
1. Because carbon is less electronegative than oxygen and hydrogen is normally terminal, C must be the central atom. One possible arrangement is as follows:
2. Each hydrogen atom (group 1) has one valence electron, carbon (group 14) has 4 valence electrons, and oxygen (group 16) has 6 valence electrons, for a total of [(2)(1) + 4 + 6] = 12 valence electrons.
3. Placing a bonding pair of electrons between each pair of bonded atoms gives the following:
Six electrons are used, and 6 are left over.
4. Adding all 6 remaining electrons to oxygen (as three lone pairs) gives the following:
Although oxygen now has an octet and each hydrogen has 2 electrons, carbon has only 6 electrons.
5. There are no electrons left to place on the central atom.
6. To give carbon an octet of electrons, we use one of the lone pairs of electrons on oxygen to form a carbon–oxygen double bond:
Both the oxygen and the carbon now have an octet of electrons, so this is an acceptable Lewis electron structure. The O has two bonding pairs and two lone pairs, and C has four bonding pairs. This is the structure of formaldehyde, which is used in embalming fluid.
An alternative structure can be drawn with one H bonded to O. Formal charges, discussed later in this section, suggest that such a structure is less stable than that shown previously.
Example
Write the Lewis electron structure for each species.
1. NCl3
2. S22−
3. NOCl
Given: chemical species
Asked for: Lewis electron structures
Strategy:
Use the six-step procedure to write the Lewis electron structure for each species.
Solution:
Nitrogen is less electronegative than chlorine, and halogen atoms are usually terminal, so nitrogen is the central atom. The nitrogen atom (group 15) has 5 valence electrons and each chlorine atom (group 17) has 7 valence electrons, for a total of 26 valence electrons. Using 2 electrons for each N–Cl bond and adding three lone pairs to each Cl account for (3 × 2) + (3 × 2 × 3) = 24 electrons. Rule 5 leads us to place the remaining 2 electrons on the central N:
Nitrogen trichloride is an unstable oily liquid once used to bleach flour; this use is now prohibited in the United States.
1. In a diatomic molecule or ion, we do not need to worry about a central atom. Each sulfur atom (group 16) contains 6 valence electrons, and we need to add 2 electrons for the −2 charge, giving a total of 14 valence electrons. Using 2 electrons for the S–S bond, we arrange the remaining 12 electrons as three lone pairs on each sulfur, giving each S atom an octet of electrons:
2. Because nitrogen is less electronegative than oxygen or chlorine, it is the central atom. The N atom (group 15) has 5 valence electrons, the O atom (group 16) has 6 valence electrons, and the Cl atom (group 17) has 7 valence electrons, giving a total of 18 valence electrons. Placing one bonding pair of electrons between each pair of bonded atoms uses 4 electrons and gives the following:
Adding three lone pairs each to oxygen and to chlorine uses 12 more electrons, leaving 2 electrons to place as a lone pair on nitrogen:
3. Because this Lewis structure has only 6 electrons around the central nitrogen, a lone pair of electrons on a terminal atom must be used to form a bonding pair. We could use a lone pair on either O or Cl. Because we have seen many structures in which O forms a double bond but none with a double bond to Cl, it is reasonable to select a lone pair from O to give the following:
All atoms now have octet configurations. This is the Lewis electron structure of nitrosyl chloride, a highly corrosive, reddish-orange gas.
Exercise
Write Lewis electron structures for CO2 and SCl2, a vile-smelling, unstable red liquid that is used in the manufacture of rubber.
Answer:
Formal Charges
It is sometimes possible to write more than one Lewis structure for a substance that does not violate the octet rule, as we saw for CH2O, but not every Lewis structure may be equally reasonable. In these situations, we can choose the most stable Lewis structure by considering the formal charge on the atoms, which is the difference between the number of valence electrons in the free atom and the number assigned to it in the Lewis electron structure. The formal charge is a way of computing the charge distribution within a Lewis structure; the sum of the formal charges on the atoms within a molecule or an ion must equal the overall charge on the molecule or ion. A formal charge does not represent a true charge on an atom in a covalent bond but is simply used to predict the most likely structure when a compound has more than one valid Lewis structure.
To calculate formal charges, we assign electrons in the molecule to individual atoms according to these rules:
• Nonbonding electrons are assigned to the atom on which they are located.
• Bonding electrons are divided equally between the bonded atoms.
For each atom, we then compute a formal charge:
To illustrate this method, let’s calculate the formal charge on the atoms in ammonia (NH3) whose Lewis electron structure is as follows:
A neutral nitrogen atom has five valence electrons (it is in group 15). From its Lewis electron structure, the nitrogen atom in ammonia has one lone pair and shares three bonding pairs with hydrogen atoms, so nitrogen itself is assigned a total of five electrons [2 nonbonding e + (6 bonding e/2)]. Substituting into Equation 5.3.1, we obtain
A neutral hydrogen atom has one valence electron. Each hydrogen atom in the molecule shares one pair of bonding electrons and is therefore assigned one electron [0 nonbonding e + (2 bonding e/2)]. Using Equation 4.4.1 to calculate the formal charge on hydrogen, we obtain
The hydrogen atoms in ammonia have the same number of electrons as neutral hydrogen atoms, and so their formal charge is also zero. Adding together the formal charges should give us the overall charge on the molecule or ion. In this example, the nitrogen and each hydrogen has a formal charge of zero. When summed the overall charge is zero, which is consistent with the overall charge on the NH3 molecule.
Typically, the structure with the most charges on the atoms closest to zero is the more stable Lewis structure. In cases where there are positive or negative formal charges on various atoms, stable structures generally have negative formal charges on the more electronegative atoms and positive formal charges on the less electronegative atoms. The next example further demonstrates how to calculate formal charges.
Example
Calculate the formal charges on each atom in the NH4+ ion.
Given: chemical species
Asked for: formal charges
Strategy:
Identify the number of valence electrons in each atom in the NH4+ ion. Use the Lewis electron structure of NH4+ to identify the number of bonding and nonbonding electrons associated with each atom and then use Equation 4.4.1 to calculate the formal charge on each atom.
Solution:
The Lewis electron structure for the NH4+ ion is as follows:
The nitrogen atom shares four bonding pairs of electrons, and a neutral nitrogen atom has five valence electrons. Using Equation 4.4.1, the formal charge on the nitrogen atom is therefore
formalcharge(N)=5(0+82)=0
Each hydrogen atom in has one bonding pair. The formal charge on each hydrogen atom is therefore
formalcharge(H)=1(0+22)=0
The formal charges on the atoms in the NH4+ ion are thus
Adding together the formal charges on the atoms should give us the total charge on the molecule or ion. In this case, the sum of the formal charges is 0 + 1 + 0 + 0 + 0 = +1.
Exercise
Write the formal charges on all atoms in BH4.
Answer:
If an atom in a molecule or ion has the number of bonds that is typical for that atom (e.g., four bonds for carbon), its formal charge is zero.
Using Formal Charges to Distinguish between Lewis Structures
As an example of how formal charges can be used to determine the most stable Lewis structure for a substance, we can compare two possible structures for CO2. Both structures conform to the rules for Lewis electron structures.
CO2
1. C is less electronegative than O, so it is the central atom.
2. C has 4 valence electrons and each O has 6 valence electrons, for a total of 16 valence electrons.
3. Placing one electron pair between the C and each O gives O–C–O, with 12 electrons left over.
4. Dividing the remaining electrons between the O atoms gives three lone pairs on each atom:
This structure has an octet of electrons around each O atom but only 4 electrons around the C atom.
5. No electrons are left for the central atom.
6. To give the carbon atom an octet of electrons, we can convert two of the lone pairs on the oxygen atoms to bonding electron pairs. There are, however, two ways to do this. We can either take one electron pair from each oxygen to form a symmetrical structure or take both electron pairs from a single oxygen atom to give an asymmetrical structure:
Both Lewis electron structures give all three atoms an octet. How do we decide between these two possibilities? The formal charges for the two Lewis electron structures of CO2 are as follows:
Both Lewis structures have a net formal charge of zero, but the structure on the right has a +1 charge on the more electronegative atom (O). Thus the symmetrical Lewis structure on the left is predicted to be more stable, and it is, in fact, the structure observed experimentally. Remember, though, that formal charges do not represent the actual charges on atoms in a molecule or ion. They are used simply as a bookkeeping method for predicting the most stable Lewis structure for a compound.
Note
The Lewis structure with the set of formal charges closest to zero is usually the most stable.
Example
The thiocyanate ion (SCN), which is used in printing and as a corrosion inhibitor against acidic gases, has at least two possible Lewis electron structures. Draw two possible structures, assign formal charges on all atoms in both, and decide which is the preferred arrangement of electrons.
Given: chemical species
Asked for: Lewis electron structures, formal charges, and preferred arrangement
Strategy:
A Use the step-by-step procedure to write two plausible Lewis electron structures for SCN.
B Calculate the formal charge on each atom using Equation 4.4.1.
C Predict which structure is preferred based on the formal charge on each atom and its electronegativity relative to the other atoms present.
Solution:
A Possible Lewis structures for the SCN ion are as follows:
B We must calculate the formal charges on each atom to identify the more stable structure. If we begin with carbon, we notice that the carbon atom in each of these structures shares four bonding pairs, the number of bonds typical for carbon, so it has a formal charge of zero. Continuing with sulfur, we observe that in (a) the sulfur atom shares one bonding pair and has three lone pairs and has a total of six valence electrons. The formal charge on the sulfur atom is therefore 6(6+22)=1.5(4+42)=1 In (c), nitrogen has a formal charge of −2.
C Which structure is preferred? Structure (b) is preferred because the negative charge is on the more electronegative atom (N), and it has lower formal charges on each atom as compared to structure (c): 0, −1 versus +1, −2.
Exercise
Salts containing the fulminate ion (CNO) are used in explosive detonators. Draw three Lewis electron structures for CNO and use formal charges to predict which is more stable. (Note: N is the central atom.)
Answer:
The second structure is predicted to be more stable.
Contributors
• Anonymous
Layne Morsch (University of Illinois Springfield)
Three cases can be constructed that do not follow the octet rule, and as such, they are known as the exceptions to the octet rule. Following the Octet Rule for Lewis Dot Structures leads to the most accurate depictions of stable molecular and atomic structures and because of this we always want to use the octet rule when drawing Lewis Dot Structures. However, it is hard to imagine that one rule could be followed by all molecules. There is always an exception, and in this case, three exceptions. The octet rule is violated in these three scenarios:
1. When there are an odd number of valence electrons
2. When there are too few valence electrons
3. When there are too many valence electrons
Exception 1: Species with Odd Numbers of Electrons
The first exception to the Octet Rule is when there are an odd number of valence electrons. An example of this would be Nitrogen (II) Oxide (NO ,refer to figure one). Nitrogen has 5 valence electrons while Oxygen has 6. The total would be 11 valence electrons to be used. The Octet Rule for this molecule is fulfilled in the above example, however that is with 10 valence electrons. The last one does not know where to go. The lone electron is called an unpaired electron. But where should the unpaired electron go? The unpaired electron is usually placed in the Lewis Dot Structure so that each element in the structure will have the lowest formal charge possible. The formal charge is the perceived charge on an individual atom in a molecule when atoms do not contribute equal numbers of electrons to the bonds they participate in. The formula to find a formal charge is:
Formal Charge= [# of valence e- the atom would have on its own] - [# of lone pair electrons on that atom]
- [# of bonds that atom participates in]
No formal charge at all is the most ideal situation. An example of a stable molecule with an odd number of valence electrons would be nitrogen monoxide. Nitrogen monoxide has 11 valence electrons. If you need more information about formal charges, see Lewis Structures. If we were to imagine nitrogen monoxide had ten valence electrons we would come up with the Lewis Structure (Figure 8.7.1):
Figure 8.7.1. This is if Nitrogen monoxide has only ten valence electrons, which it does not.
Let's look at the formal charges of Figure 8.7.2 based on this Lewis structure. Nitrogen normally has five valence electrons. In Figure 8.7.1, it has two lone pair electrons and it participates in two bonds (a double bond) with oxygen. This results in nitrogen having a formal charge of +1. Oxygen normally has six valence electrons. In Figure 8.7.1, oxygen has four lone pair electrons and it participates in two bonds with nitrogen. Oxygen therefore has a formal charge of 0. The overall molecule here has a formal charge of +1 (+1 for nitrogen, 0 for oxygen. +1 + 0 = +1). However, if we add the eleventh electron to nitrogen (because we want the molecule to have the lowest total formal charge), it will bring both the nitrogen and the molecule's overall charges to zero, the most ideal formal charge situation. That is exactly what is done to get the correct Lewis structure for nitrogen monoxide (Figure 8.7.2):
Figure 8.7.2. The proper Lewis structure for NO molecule
Free Radicals
There are actually very few stable molecules with odd numbers of electrons that exist, since that unpaired electron is willing to react with other unpaired electrons. Most odd electron species are highly reactive, which we call Free Radicals. Because of their instability, free radicals bond to atoms in which they can take an electron from in order to become stable, making them very chemically reactive. Radicals are found as both reactants and products, but generally react to form more stable molecules as soon as they can. In order to emphasize the existence of the unpaired electron, radicals are denoted with a dot in front of their chemical symbol as with OH, the hydroxyl radical. An example of a radical you may by familiar with already is the gaseous chlorine atom, denoted Cl. Interestingly, odd Number of Valence Electrons will result in the molecule being paramagnetic.
Exception 2: Incomplete Octets
The second exception to the Octet Rule is when there are too few valence electrons that results in an incomplete Octet. There are even more occasions where the octet rule does not give the most correct depiction of a molecule or ion. This is also the case with incomplete octets. Species with incomplete octets are pretty rare and generally are only found in some beryllium, aluminum, and boron compounds including the boron hydrides. Let's take a look at one such hydride, BH3 (Borane).
If one was to make a Lewis structure for BH3 following the basic strategies for drawing Lewis structures, one would probably come up with this structure (Figure 8.7.3):
Figure 8.7.3
The problem with this structure is that boron has an incomplete octet; it only has six electrons around it. Hydrogen atoms can naturally only have only 2 electrons in their outermost shell (their version of an octet), and as such there are no spare electrons to form a double bond with boron. One might surmise that the failure of this structure to form complete octets must mean that this bond should be ionic instead of covalent. However, boron has an electronegativity that is very similar to hydrogen, meaning there is likely very little ionic character in the hydrogen to boron bonds, and as such this Lewis structure, though it does not fulfill the octet rule, is likely the best structure possible for depicting BH3 with Lewis theory. One of the things that may account for BH3's incomplete octet is that it is commonly a transitory species, formed temporarily in reactions that involve multiple steps.
Let's take a look at another incomplete octet situation dealing with boron, BF3 (Boron trifluorine). Like with BH3, the initial drawing of a Lewis structure of BF3 will form a structure where boron has only six electrons around it (Figure 8.7.4).
Figure 8.7.4
If you look Figure 8.7.4, you can see that the fluorine atoms possess extra lone pairs that they can use to make additional bonds with boron, and you might think that all you have to do is make one lone pair into a bond and the structure will be correct. If we add one double bond between boron and one of the fluorines we get the following Lewis Structure (Figure 8.7.5):
Figure 8.7.5
Each fluorine has eight electrons, and the boron atom has eight as well! Each atom has a perfect octet, right? Not so fast. We must examine the formal charges of this structure. The fluorine that shares a double bond with boron has six electrons around it (four from its two lone pairs of electrons and one each from its two bonds with boron). This is one less electron than the number of valence electrons it would have naturally (Group Seven elements have seven valence electrons), so it has a formal charge of +1. The two flourines that share single bonds with boron have seven electrons around them (six from their three lone pairs and one from their single bonds with boron). This is the same amount as the number of valence electrons they would have on their own, so they both have a formal charge of zero. Finally, boron has four electrons around it (one from each of its four bonds shared with fluorine). This is one more electron than the number of valence electrons that boron would have on its own, and as such boron has a formal charge of -1.
This structure is supported by the fact that the experimentally determined bond length of the boron to fluorine bonds in BF3 is less than what would be typical for a single bond (see Bond Order and Lengths). However, this structure contradicts one of the major rules of formal charges: Negative formal charges are supposed to be found on the more electronegative atom(s) in a bond, but in the structure depicted in Figure 8.7.5, a positive formal charge is found on fluorine, which not only is the most electronegative element in the structure, but the most electronegative element in the entire periodic table (χ=4.0). Boron on the other hand, with the much lower electronegativity of 2.0, has the negative formal charge in this structure. This formal charge-electronegativity disagreement makes this double-bonded structure impossible.
However the large electronegativity difference here, as opposed to in BH3, signifies significant polar bonds between boron and fluorine, which means there is a high ionic character to this molecule. This suggests the possibility of a semi-ionic structure such as seen in Figure 8.7.6:
Figure 8.7.6
None of these three structures is the "correct" structure in this instance. The most "correct" structure is most likely a resonance of all three structures: the one with the incomplete octet (Figure 8.7.4), the one with the double bond (Figure 8.7.5), and the one with the ionic bond (Figure 8.7.6). The most contributing structure is probably the incomplete octet structure (due to Figure 8.7.5 being basically impossible and Figure 8.7.6 not matching up with the behavior and properties of BF3). As you can see even when other possibilities exist, incomplete octets may best portray a molecular structure.
As a side note, it is important to note that BF3 frequently bonds with a F- ion in order to form BF4- rather than staying as BF3. This structure completes boron's octet and it is more common in nature. This exemplifies the fact that incomplete octets are rare, and other configurations are typically more favorable, including bonding with additional ions as in the case of BF3 .
Example 8.7.1: NF3
Draw the Lewis structure for boron trifluoride (BF3).
SOLUTION
1. Add electrons (3*7) + 3 = 24
2. Draw connectivities:
3. Add octets to outer atoms:
4. Add extra electrons (24-24=0) to central atom:
5. Does central electron have octet?
• NO. It has 6 electrons
• Add a multiple bond (double bond) to see if central atom can achieve an octet:
6. The central Boron now has an octet (there would be three resonance Lewis structures)
However...
• In this structure with a double bond the fluorine atom is sharing extra electrons with the boron.
• The fluorine would have a '+' partial charge, and the boron a '-' partial charge, this is inconsistent with the electronegativities of fluorine and boron.
• Thus, the structure of BF3, with single bonds, and 6 valence electrons around the central boron is the most likely structure
BF3 reacts strongly with compounds which have an unshared pair of electrons which can be used to form a bond with the boron:
Exception 3: Expanded Valence Shells
More common than incomplete octets are expanded octets where the central atom in a Lewis structure has more than eight electrons in its valence shell. In expanded octets, the central atom can have ten electrons, or even twelve. Molecules with expanded octets involve highly electronegative terminal atoms, and a nonmetal central atom found in the third period or below, which those terminal atoms bond to. For example, PCl5 is a legitimate compound (whereas NCl5) is not:
Note
Expanded valence shells are observed only for elements in period 3 (i.e. n=3) and beyond
The 'octet' rule is based upon available ns and np orbitals for valence electrons (2 electrons in the s orbitals, and 6 in the p orbitals). Beginning with the n=3 principle quantum number, the d orbitals become available (l=2). The orbital diagram for the valence shell of phosphorous is:
Hence, the third period elements occasionally exceed the octet rule by using their empty d orbitals to accommodate additional electrons. Size is also an important consideration:
• The larger the central atom, the larger the number of electrons which can surround it
• Expanded valence shells occur most often when the central atom is bonded to small electronegative atoms, such as F, Cl and O.
There is currently much scientific exploration and inquiry into the reason why expanded valence shells are found. The top area of interest is figuring out where the extra pair(s) of electrons are found. Many chemists think that there is not a very large energy difference between the 3p and 3d orbitals, and as such it is plausible for extra electrons to easily fill the 3d orbital when an expanded octet is more favorable than having a complete octet. This matter is still under hot debate, however and there is even debate as to what makes an expanded octet more favorable than a configuration that follows the octet rule.
One of the situations where expanded octet structures are treated as more favorable than Lewis structures that follow the octet rule is when the formal charges in the expanded octet structure are smaller than in a structure that adheres to the octet rule, or when there are less formal charges in the expanded octet than in the structure a structure that adheres to the octet rule.
Example 8.7.2: The SO24 ion
Such is the case for the sulfate ion, SO4-2. A strict adherence to the octet rule forms the following Lewis structure:
Figure 8.7.12
If we look at the formal charges on this molecule, we can see that all of the oxygen atoms have seven electrons around them (six from the three lone pairs and one from the bond with sulfur). This is one more electron than the number of valence electrons then they would have normally, and as such each of the oxygens in this structure has a formal charge of -1. Sulfur has four electrons around it in this structure (one from each of its four bonds) which is two electrons more than the number of valence electrons it would have normally, and as such it carries a formal charge of +2.
If instead we made a structure for the sulfate ion with an expanded octet, it would look like this:
Figure 8.7.13
Looking at the formal charges for this structure, the sulfur ion has six electrons around it (one from each of its bonds). This is the same amount as the number of valence electrons it would have naturally. This leaves sulfur with a formal charge of zero. The two oxygens that have double bonds to sulfur have six electrons each around them (four from the two lone pairs and one each from the two bonds with sulfur). This is the same amount of electrons as the number of valence electrons that oxygen atoms have on their own, and as such both of these oxygen atoms have a formal charge of zero. The two oxygens with the single bonds to sulfur have seven electrons around them in this structure (six from the three lone pairs and one from the bond to sulfur). That is one electron more than the number of valence electrons that oxygen would have on its own, and as such those two oxygens carry a formal charge of -1. Remember that with formal charges, the goal is to keep the formal charges (or the difference between the formal charges of each atom) as small as possible. The number of and values of the formal charges on this structure (-1 and 0 (difference of 1) in Figure 8.7.12, as opposed to +2 and -1 (difference of 3) in Figure 8.7.12) is significantly lower than on the structure that follows the octet rule, and as such an expanded octet is plausible, and even preferred to a normal octet, in this case.
Example 8.7.3: The ICl4 Ion
Draw the Lewis structure for ICl4 ion.
SOLUTION
1. Count up the valence electrons: 7+(4*7)+1 = 36 electrons
2. Draw the connectivities:
3. Add octet of electrons to outer atoms:
4. Add extra electrons (36-32=4) to central atom:
5. The ICl4- ion thus has 12 valence electrons around the central Iodine (in the 5d orbitals)
Expanded Lewis structures are also plausible depictions of molecules when experimentally determined bond lengths suggest partial double bond characters even when single bonds would already fully fill the octet of the central atom. Despite the cases for expanded octets, as mentioned for incomplete octets, it is important to keep in mind that, in general, the octet rule applies. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/01._Structure_and_Bonding_in_Organic_Molecules/1.4%3A_Electron-Dot_Model_of__Bonding_-_Lewis__Structures.txt |
Resonance Structures
Sometimes, even when formal charges are considered, the bonding in some molecules or ions cannot be described by a single Lewis structure. Such is the case for ozone (O3), an allotrope of oxygen with a V-shaped structure and an O–O–O angle of 117.5°.
O3
1. We know that ozone has a V-shaped structure, so one O atom is central:
2. Each O atom has 6 valence electrons, for a total of 18 valence electrons.
3. Assigning one bonding pair of electrons to each oxygen–oxygen bond gives
with 14 electrons left over.
4. If we place three lone pairs of electrons on each terminal oxygen, we obtain
and have 2 electrons left over.
5. At this point, both terminal oxygen atoms have octets of electrons. We therefore place the last 2 electrons on the central atom:
6. The central oxygen has only 6 electrons. We must convert one lone pair on a terminal oxygen atom to a bonding pair of electrons—but which one? Depending on which one we choose, we obtain either
Which is correct? In fact, neither is correct. Both predict one O–O single bond and one O=O double bond. As you will learn in Section 4.8, if the bonds were of different types (one single and one double, for example), they would have different lengths. It turns out, however, that both O–O bond distances are identical, 127.2 pm, which is shorter than a typical O–O single bond (148 pm) and longer than the O=O double bond in O2 (120.7 pm).
Equivalent Lewis dot structures, such as those of ozone, are called resonance structures . The position of the atoms is the same in the various resonance structures of a compound, but the position of the electrons is different. Double-headed arrows link the different resonance structures of a compound:
Before the development of quantum chemistry it was thought that the double-headed arrow indicates that the actual electronic structure is an average of those shown, or that the molecule oscillates between the two structures. Today we know that the electrons involved in the double bonds occupy an orbital that extends over all three oxygen molecules, combining p orbitals on all three.
Figure 5.3.4 The resonance structure of ozone involves a molecular orbital extending all three oxygen atoms. In ozone, a molecular orbital extending over all three oxygen atoms is formed from three atom centered pz orbitals. Similar molecular orbitals are found in every resonance structure.
We will discuss the formation of these molecular orbitals in the next chapter but it is important to understand that resonance structures are based on molecular orbitals not averages of different bonds between atoms. We describe the electrons in such molecular orbitals as being delocalized, that is they cannot be assigned to a bond between two atoms.
Note the Pattern
When it is possible to write more than one equivalent resonance structure for a molecule or ion, the actual structure involves a molecular orbital which is a linear combination of atomic orbitals from each of the atoms.
CO32−
Like ozone, the electronic structure of the carbonate ion cannot be described by a single Lewis electron structure. Unlike O3, though, the Lewis structures describing CO32 has three equivalent representations.
1. Because carbon is the least electronegative element, we place it in the central position:
2. Carbon has 4 valence electrons, each oxygen has 6 valence electrons, and there are 2 more for the −2 charge. This gives 4 + (3 × 6) + 2 = 24 valence electrons.
3. Six electrons are used to form three bonding pairs between the oxygen atoms and the carbon:
4. We divide the remaining 18 electrons equally among the three oxygen atoms by placing three lone pairs on each and indicating the −2 charge:
5. No electrons are left for the central atom.
6. At this point, the carbon atom has only 6 valence electrons, so we must take one lone pair from an oxygen and use it to form a carbon–oxygen double bond. In this case, however, there are three possible choices:
As with ozone, none of these structures describes the bonding exactly. Each predicts one carbon–oxygen double bond and two carbon–oxygen single bonds, but experimentally all C–O bond lengths are identical. We can write resonance structures (in this case, three of them) for the carbonate ion:
As the case for ozone, the actual structure involves the formation of a molecular orbital from pz orbitals centered on each atom and sitting above and below the plane of the CO32 ion.
Example
Benzene is a common organic solvent that was previously used in gasoline; it is no longer used for this purpose, however, because it is now known to be a carcinogen. The benzene molecule (C6H6) consists of a regular hexagon of carbon atoms, each of which is also bonded to a hydrogen atom. Use resonance structures to describe the bonding in benzene.
Given: molecular formula and molecular geometry
Asked for: resonance structures
Strategy:
A Draw a structure for benzene illustrating the bonded atoms. Then calculate the number of valence electrons used in this drawing.
B Subtract this number from the total number of valence electrons in benzene and then locate the remaining electrons such that each atom in the structure reaches an octet.
C Draw the resonance structures for benzene.
Solution:
A Each hydrogen atom contributes 1 valence electron, and each carbon atom contributes 4 valence electrons, for a total of (6 × 1) + (6 × 4) = 30 valence electrons. If we place a single bonding electron pair between each pair of carbon atoms and between each carbon and a hydrogen atom, we obtain the following:
Each carbon atom in this structure has only 6 electrons and has a formal charge of +1, but we have used only 24 of the 30 valence electrons.
B If the 6 remaining electrons are uniformly distributed pairwise on alternate carbon atoms, we obtain the following:
Three carbon atoms now have an octet configuration and a formal charge of −1, while three carbon atoms have only 6 electrons and a formal charge of +1. We can convert each lone pair to a bonding electron pair, which gives each atom an octet of electrons and a formal charge of 0, by making three C=C double bonds.
C There are, however, two ways to do this:
Each structure has alternating double and single bonds, but experimentation shows that each carbon–carbon bond in benzene is identical, with bond lengths (139.9 pm) intermediate between those typically found for a C–C single bond (154 pm) and a C=C double bond (134 pm). We can describe the bonding in benzene using the two resonance structures, but the actual electronic structure is an average of the two. The existence of multiple resonance structures for aromatic hydrocarbons like benzene is often indicated by drawing either a circle or dashed lines inside the hexagon:
This combination of p orbitals for benzene can be visualized as a ring with a node in the plane of the carbon atoms.
Exercise
The sodium salt of nitrite is used to relieve muscle spasms. Draw two resonance structures for the nitrite ion (NO2).
Answer:
Resonance structures are particularly common in oxoanions of the p-block elements, such as sulfate and phosphate, and in aromatic hydrocarbons, such as benzene and naphthalene.
Rules for estimating stability of resonance structures
1. The greater the number of covalent bonds, the greater the stability since more atoms will have complete octets
2. The structure with the least number of formal charges is more stable
3. The structure with the least separation of formal charge is more stable
4. A structure with a negative charge on the more electronegative atom will be more stable
5. Positive charges on the least electronegative atom (most electropositive) is more stable
6. Resonance forms that are equivalent have no difference in stability and contribute equally. (eg. benzene)
The above resonance structures show that the electrons are delocalized within the molecule and through this process the molecule gains extra stability. Ozone with both of its opposite charges creates a neutral molecule and through resonance it is a stable molecule. The extra electron that created the negative charge on either terminal oxygen can be delocalized by resonance through the terminal oxygens.
Benzene is an extremely stable molecule and it is accounted for its geometry and molecular orbital interaction, but most importantly it's due to its resonance structures. The delocalized electrons in the benzene ring make the molecule very stable and with its characteristics of a nucleophile, it will react with a strong electrophile only and after the first reactivity, the substituted benzene will depend on its resonance to direct the next position for the reaction to add a second substituent.
The next molecule, the Amide, is a very stable molecule that is present in most biological systems, mainly in proteins. By studies of NMR spectroscopy and X-Ray crystallography it is confirmed that the stability of the amide is due to resonance which through molecular orbital interaction creates almost a double bond between the Nitrogen and the carbon.
Example: Multiple Resonance of other Molecules
Molecules with more than one resonance form
Some structural resonance conformations are the major contributor or the dominant forms that the molecule exists. For example, if we look at the above rules for estimating the stability of a molecule, we see that for the third molecule the first and second forms are the major contributors for the overall stability of the molecule. The nitrogen is more electronegative than carbon so, it can handle the negative charge more than carbon. A carbon with a negative charge is the least favorable conformation for the molecule to exist, so the last resonance form contributes very little for the stability of the Ion.
The Hybrid Resonance forms show the different Lewis structures with the electron been delocalized. This is very important for the reactivity of chloro-benzene because in the presence of an electrophile it will react and the formation of another bond will be directed and determine by resonance. The long pair of electrons delocalized in the aromatic substituted ring is where it can potentially form a new bond with an electrophile, as it is shown there are three possible places that reactivity can take place, the first to react will take place at the para position with respect to the chloro substituent and then to either ortho position.
Contributors
• Sharon Wei (UCD), Liza Chu (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/01._Structure_and_Bonding_in_Organic_Molecules/1.5%3A_Resonance_Forms.txt |
After completing this section, you should be able to
1. describe the physical significance of an orbital.
2. list the atomic orbitals from 1s to 3d in order of increasing energy.
3. sketch the shapes of s and p orbitals.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• nodal plane
• node
• orbital
• quantum mechanics
• wave function
Atomic Orbitals
An orbital is the quantum mechanical refinement of Bohr’s orbit. In contrast to his concept of a simple circular orbit with a fixed radius, orbitals are mathematically derived regions of space with different probabilities of having an electron.
One way of representing electron probability distributions was illustrated in Figure 6.5.2 for the 1s orbital of hydrogen. Because Ψ2 gives the probability of finding an electron in a given volume of space (such as a cubic picometer), a plot of Ψ2 versus distance from the nucleus (r) is a plot of the probability density. The 1s orbital is spherically symmetrical, so the probability of finding a 1s electron at any given point depends only on its distance from the nucleus. The probability density is greatest at r = 0 (at the nucleus) and decreases steadily with increasing distance. At very large values of r, the electron probability density is very small but not zero.
In contrast, we can calculate the radial probability (the probability of finding a 1s electron at a distance r from the nucleus) by adding together the probabilities of an electron being at all points on a series of x spherical shells of radius r1, r2, r3,…, rx − 1, rx. In effect, we are dividing the atom into very thin concentric shells, much like the layers of an onion (part (a) in Figure 1.2.1), and calculating the probability of finding an electron on each spherical shell. Recall that the electron probability density is greatest at r = 0 (part (b) in Figure 1.2.1), so the density of dots is greatest for the smallest spherical shells in part (a) in Figure 1.2.1. In contrast, the surface area of each spherical shell is equal to 4πr2, which increases very rapidly with increasing r (part (c) in Figure 1.2.1). Because the surface area of the spherical shells increases more rapidly with increasing r than the electron probability density decreases, the plot of radial probability has a maximum at a particular distance (part (d) in Figure 1.2.1). Most important, when r is very small, the surface area of a spherical shell is so small that the total probability of finding an electron close to the nucleus is very low; at the nucleus, the electron probability vanishes (part (d) in Figure 1.2.1).
Figure 1.2.1 Most Probable Radius for the Electron in the Ground State of the Hydrogen Atom. (a) Imagine dividing the atom’s total volume into very thin concentric shells as shown in the onion drawing. (b) A plot of electron probability density Ψ2 versus r shows that the electron probability density is greatest at r = 0 and falls off smoothly with increasing r. The density of the dots is therefore greatest in the innermost shells of the onion. (c) The surface area of each shell, given by 4πr2, increases rapidly with increasing r. (d) If we count the number of dots in each spherical shell, we obtain the total probability of finding the electron at a given value of r. Because the surface area of each shell increases more rapidly with increasing r than the electron probability density decreases, a plot of electron probability versus r (the radial probability) shows a peak. This peak corresponds to the most probable radius for the electron, 52.9 pm, which is exactly the radius predicted by Bohr’s model of the hydrogen atom.
For the hydrogen atom, the peak in the radial probability plot occurs at r = 0.529 Å (52.9 pm), which is exactly the radius calculated by Bohr for the n = 1 orbit. Thus the most probable radius obtained from quantum mechanics is identical to the radius calculated by classical mechanics. In Bohr’s model, however, the electron was assumed to be at this distance 100% of the time, whereas in the Schrödinger model, it is at this distance only some of the time. The difference between the two models is attributable to the wavelike behavior of the electron and the Heisenberg uncertainty principle.
Figure 1.2.2 compares the electron probability densities for the hydrogen 1s, 2s, and 3s orbitals. Note that all three are spherically symmetrical. For the 2s and 3s orbitals, however (and for all other s orbitals as well), the electron probability density does not fall off smoothly with increasing r. Instead, a series of minima and maxima are observed in the radial probability plots (part (c) in Figure 1.2.2). The minima correspond to spherical nodes (regions of zero electron probability), which alternate with spherical regions of nonzero electron probability.
Figure 1.2.2: Probability Densities for the 1s, 2s, and 3s Orbitals of the Hydrogen Atom. (a) The electron probability density in any plane that contains the nucleus is shown. Note the presence of circular regions, or nodes, where the probability density is zero. (b) Contour surfaces enclose 90% of the electron probability, which illustrates the different sizes of the 1s, 2s, and 3s orbitals. The cutaway drawings give partial views of the internal spherical nodes. The orange color corresponds to regions of space where the phase of the wave function is positive, and the blue color corresponds to regions of space where the phase of the wave function is negative. (c) In these plots of electron probability as a function of distance from the nucleus (r) in all directions (radial probability), the most probable radius increases as n increases, but the 2s and 3s orbitals have regions of significant electron probability at small values of r.
s Orbitals
Three things happen to s orbitals as n increases (Figure 1.2.2):
1. They become larger, extending farther from the nucleus.
2. They contain more nodes. This is similar to a standing wave that has regions of significant amplitude separated by nodes, points with zero amplitude.
3. For a given atom, the s orbitals also become higher in energy as n increases because of their increased distance from the nucleus.
Orbitals are generally drawn as three-dimensional surfaces that enclose 90% of the electron density, as was shown for the hydrogen 1s, 2s, and 3s orbitals in part (b) in Figure 1.2.2. Although such drawings show the relative sizes of the orbitals, they do not normally show the spherical nodes in the 2s and 3s orbitals because the spherical nodes lie inside the 90% surface. Fortunately, the positions of the spherical nodes are not important for chemical bonding.
p Orbitals
Only s orbitals are spherically symmetrical. As the value of l increases, the number of orbitals in a given subshell increases, and the shapes of the orbitals become more complex. Because the 2p subshell has l = 1, with three values of ml (−1, 0, and +1), there are three 2p orbitals.
Figure 1.2.3: Electron Probability Distribution for a Hydrogen 2p Orbital. The nodal plane of zero electron density separates the two lobes of the 2p orbital. As in Figure 1.2.2, the colors correspond to regions of space where the phase of the wave function is positive (orange) and negative (blue).
The electron probability distribution for one of the hydrogen 2p orbitals is shown in Figure 1.2.3. Because this orbital has two lobes of electron density arranged along the z axis, with an electron density of zero in the xy plane (i.e., the xy plane is a nodal plane), it is a 2pz orbital. As shown in Figure 1.2.4, the other two 2p orbitals have identical shapes, but they lie along the x axis (2px) and y axis (2py), respectively. Note that each p orbital has just one nodal plane. In each case, the phase of the wave function for each of the 2p orbitals is positive for the lobe that points along the positive axis and negative for the lobe that points along the negative axis. It is important to emphasize that these signs correspond to the phase of the wave that describes the electron motion, not to positive or negative charges.
Figure 1.2.4 The Three Equivalent 2p Orbitals of the Hydrogen Atom
The surfaces shown enclose 90% of the total electron probability for the 2px, 2py, and 2pz orbitals. Each orbital is oriented along the axis indicated by the subscript and a nodal plane that is perpendicular to that axis bisects each 2p orbital. The phase of the wave function is positive (orange) in the region of space where x, y, or z is positive and negative (blue) where x, y, or z is negative.
Just as with the s orbitals, the size and complexity of the p orbitals for any atom increase as the principal quantum number n increases. The shapes of the 90% probability surfaces of the 3p, 4p, and higher-energy p orbitals are, however, essentially the same as those shown in Figure 1.2.4.
he electron configuration of an atom is the representation of the arrangement of electrons distributed among the orbital shells and subshells. Commonly, the electron configuration is used to describe the orbitals of an atom in its ground state, but it can also be used to represent an atom that has ionized into a cation or anion by compensating with the loss of or gain of electrons in their subsequent orbitals. Many of the physical and chemical properties of elements can be correlated to their unique electron configurations. The valence electrons, electrons in the outermost shell, are the determining factor for the unique chemistry of the element.
Electron Configurations
Before assigning the electrons of an atom into orbitals, one must become familiar with the basic concepts of electron configurations. Every element on the periodic table consists of atoms, which are composed of protons, neutrons, and electrons. Electrons exhibit a negative charge and are found around the nucleus of the atom in electron orbitals, defined as the volume of space in which the electron can be found within 95% probability. The four different types of orbitals (s,p,d, and f) have different shapes, and one orbital can hold a maximum of two electrons. The p, d, and f orbitals have different sublevels, thus can hold more electrons.
As stated, the electron configuration of each element is unique to its position on the periodic table. The energy level is determined by the period and the number of electrons is given by the atomic number of the element. Orbitals on different energy levels are similar to each other, but they occupy different areas in space. The 1s orbital and 2s orbital both have the characteristics of an s orbital (radial nodes, spherical volume probabilities, can only hold two electrons, etc.) but, as they are found in different energy levels, they occupy different spaces around the nucleus. Each orbital can be represented by specific blocks on the periodic table. The s-block is the region of the alkali metals including helium (Groups 1 & 2), the d-block are the transition metals (Groups 3 to 12), the p-block are the main group elements from Groups 13 to 18, and the f-block are the lanthanides and actinides series.
Using the periodic table to determine the electron configurations of atoms is key, but also keep in mind that there are certain rules to follow when assigning electrons to different orbitals. The periodic table is an incredibly helpful tool in writing electron configurations. For more information on how electron configurations and the periodic table are linked, visit the Connecting Electrons to the Periodic Table module.
Rules for Assigning Electron Orbitals
Occupation of Orbitals
Electrons fill orbitals in a way to minimize the energy of the atom. Therefore, the electrons in an atom fill the principal energy levels in order of increasing energy (the electrons are getting farther from the nucleus). The order of levels filled looks like this:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, and 7p
One way to remember this pattern, probably the easiest, is to refer to the periodic table and remember where each orbital block falls to logically deduce this pattern. Another way is to make a table like the one below and use vertical lines to determine which subshells correspond with each other.
The number of valence electrons
The number of valence electrons of an element can be determined by the periodic table group (vertical column) in which the element is categorized. With the exception of groups 3–12 (the transition metals), the units digit of the group number identifies how many valence electrons are associated with a neutral atom of an element listed under that particular column.
The periodic table of the chemical elements
Periodic table group Valence electrons
Group 1: alkali metals 1
Group 2: alkaline earth metals 2
Groups 3-12: transition metals 2* (The 4s shell is complete and cannot hold any more electrons)
Group 13: boron group 3
Group 14: carbon group 4
Group 15: pnictogens 5
Group 16: chalcogens 6
Group 17: halogens 7
Group 18: noble gases 8**
* The general method for counting valence electrons is generally not useful for transition metals. Instead the modified d electron count method is used.
** Except for helium, which has only two valence electrons. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/01._Structure_and_Bonding_in_Organic_Molecules/1.6%3A_Atomic_Orbitals%3A_A_Quantum_Mechanical_Description_of__Electrons_Around_the__Nucleus.txt |
Learning Objectives
• To use molecular orbital theory to predict bond order
None of the approaches we have described so far can adequately explain why some compounds are colored and others are not, why some substances with unpaired electrons are stable, and why others are effective semiconductors. These approaches also cannot describe the nature of resonance. Such limitations led to the development of a new approach to bonding in which electrons are not viewed as being localized between the nuclei of bonded atoms but are instead delocalized throughout the entire molecule. Just as with the valence bond theory, the approach we are about to discuss is based on a quantum mechanical model.
Previously, we described the electrons in isolated atoms as having certain spatial distributions, called orbitals, each with a particular orbital energy. Just as the positions and energies of electrons in atoms can be described in terms of atomic orbitals (AOs), the positions and energies of electrons in molecules can be described in terms of molecular orbitals (MOs) A particular spatial distribution of electrons in a molecule that is associated with a particular orbital energy.—a spatial distribution of electrons in a molecule that is associated with a particular orbital energy. As the name suggests, molecular orbitals are not localized on a single atom but extend over the entire molecule. Consequently, the molecular orbital approach, called molecular orbital theory is a delocalized approach to bonding.
Molecular Orbital Theory: A Delocalized Bonding Approach
Although the molecular orbital theory is computationally demanding, the principles on which it is based are similar to those we used to determine electron configurations for atoms. The key difference is that in molecular orbitals, the electrons are allowed to interact with more than one atomic nucleus at a time. Just as with atomic orbitals, we create an energy-level diagram by listing the molecular orbitals in order of increasing energy. We then fill the orbitals with the required number of valence electrons according to the Pauli principle. This means that each molecular orbital can accommodate a maximum of two electrons with opposite spins.
Molecular Orbitals Involving Only ns Atomic Orbitals
We begin our discussion of molecular orbitals with the simplest molecule, H2, formed from two isolated hydrogen atoms, each with a 1s1 electron configuration. As we explained in Chapter 9, electrons can behave like waves. In the molecular orbital approach, the overlapping atomic orbitals are described by mathematical equations called wave functions. The 1s atomic orbitals on the two hydrogen atoms interact to form two new molecular orbitals, one produced by taking the sum of the two H 1s wave functions, and the other produced by taking their difference:
$\begin{matrix} MO(1)= & AO(atom\; A) & +& AO(atomB) \ MO(1)= & AO(atom\; A) & -&AO(atomB) \end{matrix} \label{9.7.1}$
The molecular orbitals created from Equation $\ref{9.7.1}$ are called linear combinations of atomic orbitals (LCAOs) Molecular orbitals created from the sum and the difference of two wave functions (atomic orbitals). A molecule must have as many molecular orbitals as there are atomic orbitals.
A molecule must have as many molecular orbitals as there are atomic orbitals.
Adding two atomic orbitals corresponds to constructive interference between two waves, thus reinforcing their intensity; the internuclear electron probability density is increased. The molecular orbital corresponding to the sum of the two H 1s orbitals is called a σ1s combination (pronounced “sigma one ess”) (part (a) and part (b) in Figure $1$). In a sigma (σ) orbital, (i.e., a bonding molecular orbital in which the electron density along the internuclear axis and between the nuclei has cylindrical symmetry), the electron density along the internuclear axis and between the nuclei has cylindrical symmetry; that is, all cross-sections perpendicular to the internuclear axis are circles. The subscript 1s denotes the atomic orbitals from which the molecular orbital was derived: The ≈ sign is used rather than an = sign because we are ignoring certain constants that are not important to our argument.
$\sigma _{1s} \approx 1s\left ( A \right ) + 1s\left ( B \right ) \label{9.7.2}$
Conversely, subtracting one atomic orbital from another corresponds to destructive interference between two waves, which reduces their intensity and causes a decrease in the internuclear electron probability density (part (c) and part (d) in Figure $1$). The resulting pattern contains a node where the electron density is zero. The molecular orbital corresponding to the difference is called $\sigma _{1s}^{*}$ (“sigma one ess star”). In a sigma star (σ*) orbital An antibonding molecular orbital in which there is a region of zero electron probability (a nodal plane) perpendicular to the internuclear axis., there is a region of zero electron probability, a nodal plane, perpendicular to the internuclear axis:
$\sigma _{1s}^{\star } \approx 1s\left ( A \right ) - 1s\left ( B \right ) \label{9.7.3}$
The electron density in the σ1s molecular orbital is greatest between the two positively charged nuclei, and the resulting electron–nucleus electrostatic attractions reduce repulsions between the nuclei. Thus the σ1s orbital represents a bonding molecular orbital. A molecular orbital that forms when atomic orbitals or orbital lobes with the same sign interact to give increased electron probability between the nuclei due to constructive reinforcement of the wave functions. In contrast, electrons in the $\sigma _{1s}^{\star }$ orbital are generally found in the space outside the internuclear region. Because this allows the positively charged nuclei to repel one another, the $\sigma _{1s}^{\star }$ orbital is an antibonding molecular orbital (a molecular orbital that forms when atomic orbitals or orbital lobes of opposite sign interact to give decreased electron probability between the nuclei due to destructive reinforcement of the wave functions).
Antibonding orbitals contain a node perpendicular to the internuclear axis; bonding orbitals do not.
Energy-Level Diagrams
Because electrons in the σ1s orbital interact simultaneously with both nuclei, they have a lower energy than electrons that interact with only one nucleus. This means that the σ1s molecular orbital has a lower energy than either of the hydrogen 1s atomic orbitals. Conversely, electrons in the $\sigma _{1s}^{\star }$ orbital interact with only one hydrogen nucleus at a time. In addition, they are farther away from the nucleus than they were in the parent hydrogen 1s atomic orbitals. Consequently, the $\sigma _{1s}^{\star }$ molecular orbital has a higher energy than either of the hydrogen 1s atomic orbitals. The σ1s (bonding) molecular orbital is stabilized relative to the 1s atomic orbitals, and the $\sigma _{1s}^{\star }$ (antibonding) molecular orbital is destabilized. The relative energy levels of these orbitals are shown in the energy-level diagram (a schematic drawing that compares the energies of the molecular orbitals (bonding, antibonding, and nonbonding) with the energies of the parent atomic orbitals) in Figure $2$
A bonding molecular orbital is always lower in energy (more stable) than the component atomic orbitals, whereas an antibonding molecular orbital is always higher in energy (less stable).
To describe the bonding in a homonuclear diatomic molecule (a molecule that consists of two atoms of the same element) such as H2, we use molecular orbitals; that is, for a molecule in which two identical atoms interact, we insert the total number of valence electrons into the energy-level diagram (Figure $2$). We fill the orbitals according to the Pauli principle and Hund’s rule: each orbital can accommodate a maximum of two electrons with opposite spins, and the orbitals are filled in order of increasing energy. Because each H atom contributes one valence electron, the resulting two electrons are exactly enough to fill the σ1s bonding molecular orbital. The two electrons enter an orbital whose energy is lower than that of the parent atomic orbitals, so the H2 molecule is more stable than the two isolated hydrogen atoms. Thus molecular orbital theory correctly predicts that H2 is a stable molecule. Because bonds form when electrons are concentrated in the space between nuclei, this approach is also consistent with our earlier discussion of electron-pair bonds.
Bond Order in Molecular Orbital Theory
In the Lewis electron structures, the number of electron pairs holding two atoms together was called the bond order. In the molecular orbital approach, bond order One-half the net number of bonding electrons in a molecule. is defined as one-half the net number of bonding electrons:
$bond\; order=\dfrac{number\; of \; bonding\; electrons-number\; of \; antibonding\; electrons}{2} \label{9.7.4}$
To calculate the bond order of H2, we see from Figure $2$ that the σ1s (bonding) molecular orbital contains two electrons, while the $\sigma _{1s}^{\star }$ (antibonding) molecular orbital is empty. The bond order of H2 is therefore
$\dfrac{2-0}{2}=1 \label{9.7.5}$
This result corresponds to the single covalent bond predicted by Lewis dot symbols. Thus molecular orbital theory and the Lewis electron-pair approach agree that a single bond containing two electrons has a bond order of 1. Double and triple bonds contain four or six electrons, respectively, and correspond to bond orders of 2 and 3. We can use energy-level diagrams such as the one in Figure $2$ to describe the bonding in other pairs of atoms and ions where n = 1, such as the H2+ ion, the He2+ ion, and the He2 molecule. Again, we fill the lowest-energy molecular orbitals first while being sure not to violate the Pauli principle or Hund’s rule.
Figure $\PageIndex{3a}$ shows the energy-level diagram for the H2+ ion, which contains two protons and only one electron. The single electron occupies the σ1s bonding molecular orbital, giving a (σ1s)1 electron configuration. The number of electrons in an orbital is indicated by a superscript. In this case, the bond order is
$\dfrac{1-0}{2}=1/2 \nonumber$
Because the bond order is greater than zero, the H2+ ion should be more stable than an isolated H atom and a proton. We can therefore use a molecular orbital energy-level diagram and the calculated bond order to predict the relative stability of species such as H2+. With a bond order of only 1/2 the bond in H2+ should be weaker than in the H2 molecule, and the H–H bond should be longer. As shown in Table $1$, these predictions agree with the experimental data.
Figure $\PageIndex{3b}$ is the molecular orbital energy-level diagram for He2+. This ion has a total of three valence electrons. Because the first two electrons completely fill the σ1s molecular orbital, the Pauli principle states that the third electron must be in the $\sigma _{1s}^{\star}$ antibonding orbital, giving a $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{1}$ electron configuration. This electron configuration gives a bond order of
$\dfrac{2-1}{2}=1/2 \nonumber$
As with H2+, the He2+ ion should be stable, but the He–He bond should be weaker and longer than in H2. In fact, the He2+ ion can be prepared, and its properties are consistent with our predictions (Table $1$).
Table $1$: Molecular Orbital Electron Configurations, Bond Orders, Bond Lengths, and Bond Energies for some Simple Homonuclear Diatomic Molecules and Ions
Molecule or Ion Electron Configuration Bond Order Bond Length (pm) Bond Energy (kJ/mol)
H2+ 1s)1 1/2 106 269
H2 1s)2 1 74 436
He2+ $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{1}$ 1/2 108 251
He2 $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{2}$ 0 not observed not observed
Finally, we examine the He2 molecule, formed from two He atoms with 1s2 electron configurations. Figure $\PageIndex{3c}$ is the molecular orbital energy-level diagram for He2. With a total of four valence electrons, both the σ1s bonding and $\sigma _{1s}^{\star }$ antibonding orbitals must contain two electrons. This gives a $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{2}$ electron configuration, with a predicted bond order of (2 − 2) ÷ 2 = 0, which indicates that the He2 molecule has no net bond and is not a stable species. Experiments show that the He2 molecule is actually less stable than two isolated He atoms due to unfavorable electron–electron and nucleus–nucleus interactions.
In molecular orbital theory, electrons in antibonding orbitals effectively cancel the stabilization resulting from electrons in bonding orbitals. Consequently, any system that has equal numbers of bonding and antibonding electrons will have a bond order of 0, and it is predicted to be unstable and therefore not to exist in nature. In contrast to Lewis electron structures and the valence bond approach, molecular orbital theory is able to accommodate systems with an odd number of electrons, such as the H2+ ion.
In contrast to Lewis electron structures and the valence bond approach, molecular orbital theory can accommodate systems with an odd number of electrons.
Example $1$
Use a molecular orbital energy-level diagram, such as those in Figure $2$, to predict the bond order in the He22+ ion. Is this a stable species?
Given: chemical species
Asked for: molecular orbital energy-level diagram, bond order, and stability
Strategy:
1. Combine the two He valence atomic orbitals to produce bonding and antibonding molecular orbital
2. s. Draw the molecular orbital energy-level diagram for the system.
3. Determine the total number of valence electrons in the He22+ ion. Fill the molecular orbitals in the energy-level diagram beginning with the orbital with the lowest energy. Be sure to obey the Pauli principle and Hund’s rule while doing so.
4. Calculate the bond order and predict whether the species is stable.
Solution:
A Two He 1s atomic orbitals combine to give two molecular orbitals: a σ1s bonding orbital at lower energy than the atomic orbitals and a $\sigma _{1s}^{\star }$ antibonding orbital at higher energy. The bonding in any diatomic molecule with two He atoms can be described using the following molecular orbital diagram:
B The He22+ ion has only two valence electrons (two from each He atom minus two for the +2 charge). We can also view He22+ as being formed from two He+ ions, each of which has a single valence electron in the 1s atomic orbital. We can now fill the molecular orbital diagram:
The two electrons occupy the lowest-energy molecular orbital, which is the bonding (σ1s) orbital, giving a (σ1s)2 electron configuration. To avoid violating the Pauli principle, the electron spins must be paired. C So the bond order is
$\dfrac{2-0}{2} =1 \nonumber$
He22+ is therefore predicted to contain a single He–He bond. Thus it should be a stable species.
Exercise $1$
Use a molecular orbital energy-level diagram to predict the valence-electron configuration and bond order of the H22 ion. Is this a stable species?
Answer
H22 has a valence electron configuration of $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{2}$ with a bond order of 0. It is therefore predicted to be unstable.
So far, our discussion of molecular orbitals has been confined to the interaction of valence orbitals, which tend to lie farthest from the nucleus. When two atoms are close enough for their valence orbitals to overlap significantly, the filled inner electron shells are largely unperturbed; hence they do not need to be considered in a molecular orbital scheme. Also, when the inner orbitals are completely filled, they contain exactly enough electrons to completely fill both the bonding and antibonding molecular orbitals that arise from their interaction. Thus the interaction of filled shells always gives a bond order of 0, so filled shells are not a factor when predicting the stability of a species. This means that we can focus our attention on the molecular orbitals derived from valence atomic orbitals.
A molecular orbital diagram that can be applied to any homonuclear diatomic molecule with two identical alkali metal atoms (Li2 and Cs2, for example) is shown in part (a) in Figure $4$, where M represents the metal atom. Only two energy levels are important for describing the valence electron molecular orbitals of these species: a σns bonding molecular orbital and a σ*ns antibonding molecular orbital. Because each alkali metal (M) has an ns1 valence electron configuration, the M2 molecule has two valence electrons that fill the σns bonding orbital. As a result, a bond order of 1 is predicted for all homonuclear diatomic species formed from the alkali metals (Li2, Na2, K2, Rb2, and Cs2). The general features of these M2 diagrams are identical to the diagram for the H2 molecule in Figure $4$. Experimentally, all are found to be stable in the gas phase, and some are even stable in solution.
Similarly, the molecular orbital diagrams for homonuclear diatomic compounds of the alkaline earth metals (such as Be2), in which each metal atom has an ns2 valence electron configuration, resemble the diagram for the He2 molecule in part (c) in Figure $2$. As shown in part (b) in Figure $4$, this is indeed the case. All the homonuclear alkaline earth diatomic molecules have four valence electrons, which fill both the σns bonding orbital and the σns* antibonding orbital and give a bond order of 0. Thus Be2, Mg2, Ca2, Sr2, and Ba2 are all expected to be unstable, in agreement with experimental data.In the solid state, however, all the alkali metals and the alkaline earth metals exist as extended lattices held together by metallic bonding. At low temperatures, $Be_2$ is stable.
Example $2$
Use a qualitative molecular orbital energy-level diagram to predict the valence electron configuration, bond order, and likely existence of the Na2 ion.
Given: chemical species
Asked for: molecular orbital energy-level diagram, valence electron configuration, bond order, and stability
Strategy:
1. Combine the two sodium valence atomic orbitals to produce bonding and antibonding molecular orbitals. Draw the molecular orbital energy-level diagram for this system.
2. Determine the total number of valence electrons in the Na2 ion. Fill the molecular orbitals in the energy-level diagram beginning with the orbital with the lowest energy. Be sure to obey the Pauli principle and Hund’s rule while doing so.
3. Calculate the bond order and predict whether the species is stable.
Solution:
A Because sodium has a [Ne]3s1 electron configuration, the molecular orbital energy-level diagram is qualitatively identical to the diagram for the interaction of two 1s atomic orbitals. B The Na2 ion has a total of three valence electrons (one from each Na atom and one for the negative charge), resulting in a filled σ3s molecular orbital, a half-filled σ3s* and a $\left ( \sigma _{3s} \right )^{2}\left ( \sigma _{3s}^{\star } \right )^{1}$ electron configuration.
C The bond order is (2-1)÷2=1/2 With a fractional bond order, we predict that the Na2 ion exists but is highly reactive.
Exercise $2$
Use a qualitative molecular orbital energy-level diagram to predict the valence electron configuration, bond order, and likely existence of the Ca2+ ion.
Answer
Ca2+ has a $\left ( \sigma _{4s} \right )^{2}\left ( \sigma _{4s}^{\star } \right )^{1}$ electron configurations and a bond order of 1/2 and should exist.
Molecular Orbitals Formed from ns and np Atomic Orbitals
Atomic orbitals other than ns orbitals can also interact to form molecular orbitals. Because individual p, d, and f orbitals are not spherically symmetrical, however, we need to define a coordinate system so we know which lobes are interacting in three-dimensional space. Recall that for each np subshell, for example, there are npx, npy, and npz orbitals. All have the same energy and are therefore degenerate, but they have different spatial orientations.
$\sigma _{np_{z}}=np_{z}\left ( A \right )-np_{z}\left ( B \right ) \label{9.7.6}$
Just as with ns orbitals, we can form molecular orbitals from np orbitals by taking their mathematical sum and difference. When two positive lobes with the appropriate spatial orientation overlap, as illustrated for two npz atomic orbitals in part (a) in Figure $5$, it is the mathematical difference of their wave functions that results in constructive interference, which in turn increases the electron probability density between the two atoms. The difference therefore corresponds to a molecular orbital called a $\sigma _{np_{z}}$ bonding molecular orbital because, just as with the σ orbitals discussed previously, it is symmetrical about the internuclear axis (in this case, the z-axis):
$\sigma _{np_{z}}=np_{z}\left ( A \right )-np_{z}\left ( B \right ) \label{9.7.7A}$
The other possible combination of the two npz orbitals is the mathematical sum:
$\sigma _{np_{z}}=np_{z}\left ( A \right )+np_{z}\left ( B \right ) \label{9.7.7}$
In this combination, shown in part (b) in Figure $5$, the positive lobe of one npz atomic orbital overlaps the negative lobe of the other, leading to destructive interference of the two waves and creating a node between the two atoms. Hence this is an antibonding molecular orbital. Because it, too, is symmetrical about the internuclear axis, this molecular orbital is called a $\sigma _{np_{z}}=np_{z}\left ( A \right )-np_{z}\left ( B \right )$ antibonding molecular orbital. Whenever orbitals combine, the bonding combination is always lower in energy (more stable) than the atomic orbitals from which it was derived, and the antibonding combination is higher in energy (less stable).
Overlap of atomic orbital lobes with the same sign produces a bonding molecular orbital, regardless of whether it corresponds to the sum or the difference of the atomic orbitals.
The remaining p orbitals on each of the two atoms, npx and npy, do not point directly toward each other. Instead, they are perpendicular to the internuclear axis. If we arbitrarily label the axes as shown in Figure $6$, we see that we have two pairs of np orbitals: the two npx orbitals lying in the plane of the page, and two npy orbitals perpendicular to the plane. Although these two pairs are equivalent in energy, the npx orbital on one atom can interact with only the npx orbital on the other, and the npy orbital on one atom can interact with only the npy on the other. These interactions are side-to-side rather than the head-to-head interactions characteristic of σ orbitals. Each pair of overlapping atomic orbitals again forms two molecular orbitals: one corresponds to the arithmetic sum of the two atomic orbitals and one to the difference. The sum of these side-to-side interactions increases the electron probability in the region above and below a line connecting the nuclei, so it is a bonding molecular orbital that is called a pi (π) orbital (a bonding molecular orbital formed from the side-to-side interactions of two or more parallel np atomic orbitals). The difference results in the overlap of orbital lobes with opposite signs, which produces a nodal plane perpendicular to the internuclear axis; hence it is an antibonding molecular orbital, called a pi star (π*) orbital An antibonding molecular orbital formed from the difference of the side-to-side interactions of two or more parallel np atomic orbitals, creating a nodal plane perpendicular to the internuclear axis..
$\pi _{np_{x}}=np_{x}\left ( A \right )+np_{x}\left ( B \right ) \label{9.7.8}$
$\pi ^{\star }_{np_{x}}=np_{x}\left ( A \right )-np_{x}\left ( B \right ) \label{9.7.9}$
The two npy orbitals can also combine using side-to-side interactions to produce a bonding $\pi _{np_{y}}$ molecular orbital and an antibonding $\pi _{np_{y}}^{\star }$ molecular orbital. Because the npx and npy atomic orbitals interact in the same way (side-to-side) and have the same energy, the $\pi _{np_{x}}$ and $\pi _{np_{y}}$molecular orbitals are a degenerate pair, as are the $\pi _{np_{x}}^{\star }$ and $\pi _{np_{y}}^{\star }$ molecular orbitals.
Figure $7$ is an energy-level diagram that can be applied to two identical interacting atoms that have three np atomic orbitals each. There are six degenerate p atomic orbitals (three from each atom) that combine to form six molecular orbitals, three bonding and three antibonding. The bonding molecular orbitals are lower in energy than the atomic orbitals because of the increased stability associated with the formation of a bond. Conversely, the antibonding molecular orbitals are higher in energy, as shown. The energy difference between the σ and σ* molecular orbitals is significantly greater than the difference between the two π and π* sets. The reason for this is that the atomic orbital overlap and thus the strength of the interaction are greater for a σ bond than a π bond, which means that the σ molecular orbital is more stable (lower in energy) than the π molecular orbitals.
Although many combinations of atomic orbitals form molecular orbitals, we will discuss only one other interaction: an ns atomic orbital on one atom with an npz atomic orbital on another. As shown in Figure $8$, the sum of the two atomic wave functions (ns + npz) produces a σ bonding molecular orbital. Their difference (nsnpz) produces a σ* antibonding molecular orbital, which has a nodal plane of zero probability density perpendicular to the internuclear axis.
Summary
Molecular orbital theory, a delocalized approach to bonding, can often explain a compound’s color, why a compound with unpaired electrons is stable, semiconductor behavior, and resonance, none of which can be explained using a localized approach. A molecular orbital (MO) is an allowed spatial distribution of electrons in a molecule that is associated with a particular orbital energy. Unlike an atomic orbital (AO), which is centered on a single atom, a molecular orbital extends over all the atoms in a molecule or ion. Hence the molecular orbital theory of bonding is a delocalized approach. Molecular orbitals are constructed using linear combinations of atomic orbitals (LCAOs), which are usually the mathematical sums and differences of wave functions that describe overlapping atomic orbitals. Atomic orbitals interact to form three types of molecular orbitals.
A completely bonding molecular orbital contains no nodes (regions of zero electron probability) perpendicular to the internuclear axis, whereas a completely antibonding molecular orbital contains at least one node perpendicular to the internuclear axis. A sigma (σ) orbital (bonding) or a sigma star (σ*) orbital (antibonding) is symmetrical about the internuclear axis. Hence all cross-sections perpendicular to that axis are circular. Both a pi (π) orbital (bonding) and a pi star (π*) orbital (antibonding) possess a nodal plane that contains the nuclei, with electron density localized on both sides of the plane.
The energies of the molecular orbitals versus those of the parent atomic orbitals can be shown schematically in an energy-level diagram. The electron configuration of a molecule is shown by placing the correct number of electrons in the appropriate energy-level diagram, starting with the lowest-energy orbital and obeying the Pauli principle; that is, placing only two electrons with opposite spin in each orbital. From the completed energy-level diagram, we can calculate the bond order, defined as one-half the net number of bonding electrons. In bond orders, electrons in antibonding molecular orbitals cancel electrons in bonding molecular orbitals, while electrons in nonbonding orbitals have no effect and are not counted. Bond orders of 1, 2, and 3 correspond to single, double, and triple bonds, respectively. Molecules with predicted bond orders of 0 are generally less stable than the isolated atoms and do not normally exist.
Contributors and Attributions
1. Orbitals or orbital lobes with the same sign interact to give increased electron probability along the plane of the internuclear axis because of constructive reinforcement of the wave functions. Consequently, electrons in such molecular orbitals help to hold the positively charged nuclei together. Such orbitals are bonding molecular orbitals, and they are always lower in energy than the parent atomic orbitals.
2. Orbitals or orbital lobes with opposite signs interact to give decreased electron probability density between the nuclei because of destructive interference of the wave functions. Consequently, electrons in such molecular orbitals are primarily located outside the internuclear region, leading to increased repulsions between the positively charged nuclei. These orbitals are called antibonding molecular orbitals, and they are always higher in energy than the parent atomic orbitals.
3. Some atomic orbitals interact only very weakly, and the resulting molecular orbitals give essentially no change in the electron probability density between the nuclei. Hence electrons in such orbitals have no effect on the bonding in a molecule or ion. These orbitals are nonbonding molecular orbitals, and they have approximately the same energy as the parent atomic orbitals. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/01._Structure_and_Bonding_in_Organic_Molecules/1.7%3A_Molecular_Orbitals_and__Covalent_Bonding.txt |
Formation of sigma bonds: the H2 molecule
The simplest case to consider is the hydrogen molecule, H2. When we say that the two electrons from each of the hydrogen atoms are shared to form a covalent bond between the two atoms, what we mean in valence bond theory terms is that the two spherical 1s orbitals overlap, allowing the two electrons to form a pair within the two overlapping orbitals.
These two electrons are now attracted to the positive charge of both of the hydrogen nuclei, with the result that they serve as a sort of ‘chemical glue’ holding the two nuclei together.
Bonding in Methane
Now let’s turn to methane, the simplest organic molecule. Recall the valence electron configuration of the central carbon:
This picture, however, is problematic. How does the carbon form four bonds if it has only two half-filled p orbitals available for bonding? A hint comes from the experimental observation that the four C-H bonds in methane are arranged with tetrahedral geometry about the central carbon, and that each bond has the same length and strength. In order to explain this observation, valence bond theory relies on a concept called orbital hybridization. In this picture, the four valence orbitals of the carbon (one 2s and three 2p orbitals) combine mathematically (remember: orbitals are described by equations) to form four equivalent hybrid orbitals, which are named sp3 orbitals because they are formed from mixing one s and three p orbitals. In the new electron configuration, each of the four valence electrons on the carbon occupies a single sp3 orbital.
The sp3 hybrid orbitals, like the p orbitals of which they are partially composed, are oblong in shape, and have two lobes of opposite sign. Unlike the p orbitals, however, the two lobes are of very different size. The larger lobes of the sp3 hybrids are directed towards the four corners of a tetrahedron, meaning that the angle between any two orbitals is 109.5o.
This geometric arrangement makes perfect sense if you consider that it is precisely this angle that allows the four orbitals (and the electrons in them) to be as far apart from each other as possible.This is simply a restatement of the Valence Shell Electron Pair Repulsion (VSEPR) theory that you learned in General Chemistry: electron pairs (in orbitals) will arrange themselves in such a way as to remain as far apart as possible, due to negative-negative electrostatic repulsion.
Each C-H bond in methane, then, can be described as an overlap between a half-filled 1s orbital in a hydrogen atom and the larger lobe of one of the four half-filled sp3 hybrid orbitals in the central carbon. The length of the carbon-hydrogen bonds in methane is 1.09 Å (1.09 x 10-10 m).
While previously we drew a Lewis structure of methane in two dimensions using lines to denote each covalent bond, we can now draw a more accurate structure in three dimensions, showing the tetrahedral bonding geometry. To do this on a two-dimensional page, though, we need to introduce a new drawing convention: the solid / dashed wedge system. In this convention, a solid wedge simply represents a bond that is meant to be pictured emerging from the plane of the page. A dashed wedge represents a bond that is meant to be pictured pointing into, or behind, the plane of the page. Normal lines imply bonds that lie in the plane of the page.
This system takes a little bit of getting used to, but with practice your eye will learn to immediately ‘see’ the third dimension being depicted.
Example
Imagine that you could distinguish between the four hydrogens in a methane molecule, and labeled them Ha through Hd. In the images below, the exact same methane molecule is rotated and flipped in various positions. Draw the missing hydrogen atom labels. (It will be much easier to do this if you make a model.)
Example
Describe, with a picture and with words, the bonding in chloroform, CHCl3.
Solutions
The bonding arrangement here is also tetrahedral: the three N-H bonds of ammonia can be pictured as forming the base of a trigonal pyramid, with the fourth orbital, containing the lone pair, forming the top of the pyramid.
Recall from your study of VSEPR theory in General Chemistry that the lone pair, with its slightly greater repulsive effect, ‘pushes’ the three N-H sbonds away from the top of the pyramid, meaning that the H-N-H bond angles are slightly less than tetrahedral, at 107.3˚ rather than 109.5˚.
VSEPR theory also predicts, accurately, that a water molecule is ‘bent’ at an angle of approximately 104.5˚. It would seem logical, then, to describe the bonding in water as occurring through the overlap of sp3-hybrid orbitals on oxygen with 1sorbitals on the two hydrogen atoms. In this model, the two nonbonding lone pairs on oxygen would be located in sp3 orbitals.
Some experimental evidence, however, suggests that the bonding orbitals on the oxygen are actually unhybridized 2p orbitals rather than sp3 hybrids. Although this would seem to imply that the H-O-H bond angle should be 90˚ (remember that p orbitals are oriented perpendicular to one another), it appears that electrostatic repulsion has the effect of distorting this p-orbital angle to 104.5˚. Both the hybrid orbital and the nonhybrid orbital models present reasonable explanations for the observed bonding arrangement in water, so we will not concern ourselves any further with the distinction.
Example
Draw, in the same style as the figures above, an orbital picture for the bonding in methylamine.
Solution
Formation of $\pi$ bonds - $sp^2$ and $sp$ hybridization
The valence bond theory, along with the hybrid orbital concept, does a very good job of describing double-bonded compounds such as ethene. Three experimentally observable characteristics of the ethene molecule need to be accounted for by a bonding model:
1. Ethene is a planar (flat) molecule.
2. Bond angles in ethene are approximately 120o, and the carbon-carbon bond length is 1.34 Å, significantly shorter than the 1.54 Å single carbon-carbon bond in ethane.
3. There is a significant barrier to rotation about the carbon-carbon double bond.
Clearly, these characteristics are not consistent with an sp3 hybrid bonding picture for the two carbon atoms. Instead, the bonding in ethene is described by a model involving the participation of a different kind of hybrid orbital. Three atomic orbitals on each carbon – the 2s, 2px and 2py orbitals – combine to form three sp2 hybrids, leaving the 2pz orbital unhybridized.
The three sp2 hybrids are arranged with trigonal planar geometry, pointing to the three corners of an equilateral triangle, with angles of 120°between them. The unhybridized 2pz orbital is perpendicular to this plane (in the next several figures, sp2 orbitals and the sigma bonds to which they contribute are represented by lines and wedges; only the 2pz orbitals are shown in the 'space-filling' mode).
The carbon-carbon double bond in ethene consists of one sbond, formed by the overlap of two sp2 orbitals, and a second bond, calleda π (pi) bond, which is formed by the side-by-side overlap of the two unhybridized 2pz orbitals from each carbon.
spacefilling image of bonding in ethene
The pi bond does not have symmetrical symmetry. Because they are the result of side-by-side overlap (rather then end-to-end overlap like a sigma bond), pi bonds are not free to rotate. If rotation about this bond were to occur, it would involve disrupting the side-by-side overlap between the two 2pz orbitals that make up the pi bond. The presence of the pi bond thus ‘locks’ the six atoms of ethene into the same plane. This argument extends to larger alkene groups: in each case, the six atoms of the group form a single plane.
Conversely, sbonds such as the carbon-carbon single bond in ethane (CH3CH3) exhibit free rotation, and can assume many different conformations, or shapes - this is one of the main subjects of Chapter 3.
Example 1.20
Circle the six atoms in the molecule below that are ‘locked’ into the same plane.
Example
What kinds of orbitals are overlapping in bonds a-d indicated below?
Example
What is wrong with the way the following structure is drawn?
Solutions
A similar picture can be drawn for the bonding in carbonyl groups, such as formaldehyde. In this molecule, the carbon is sp2-hybridized, and we will assume that the oxygen atom is also sp2 hybridized. The carbon has three sigma bonds: two are formed by overlap between two of its sp2 orbitals with the 1sorbital from each of the hydrogens, and the third sigma bond is formed by overlap between the remaining carbon sp2 orbital and an sp2 orbital on the oxygen. The two lone pairs on oxygen occupy its other two sp2 orbitals.
The pi bond is formed by side-by-side overlap of the unhybridized 2pz orbitals on the carbon and the oxygen. Just like in alkenes, the 2pz orbitals that form the pi bond are perpendicular to the plane formed by the sigma bonds.
Example
Describe and draw the bonding picture for the imine group shown below. Use the drawing of formaldehyde above as your guide.
Solution | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/01._Structure_and_Bonding_in_Organic_Molecules/1.8%3A_Hybrid_Orbitals%3A_Bonding_in_Complex_Molecules_and_Practice_Problems.txt |
It is necessary to draw structural formulas for organic compounds because in most cases a molecular formula does not uniquely represent a single compound. Different compounds having the same molecular formula are called isomers, and the prevalence of organic isomers reflects the extraordinary versatility of carbon in forming strong bonds to itself and to other elements. When the group of atoms that make up the molecules of different isomers are bonded together in fundamentally different ways, we refer to such compounds as constitutional isomers. There are seven constitutional isomers of C4H10O, and structural formulas for these are drawn in the following table. These formulas represent all known and possible C4H10O compounds, and display a common structural feature. There are no double or triple bonds and no rings in any of these structures.
Structural Formulas for C4H10O isomers
Kekulé Formula Condensed Formula Shorthand Formula
Simplification of structural formulas may be achieved without any loss of the information they convey. In condensed structural formulas the bonds to each carbon are omitted, but each distinct structural unit (group) is written with subscript numbers designating multiple substituents, including the hydrogens. Shorthand (line) formulas omit the symbols for carbon and hydrogen entirely. Each straight line segment represents a bond, the ends and intersections of the lines are carbon atoms, and the correct number of hydrogens is calculated from the tetravalency of carbon. Non-bonding valence shell electrons are omitted in these formulas.
Developing the ability to visualize a three-dimensional structure from two-dimensional formulas requires practice, and in most cases the aid of molecular models. As noted earlier, many kinds of model kits are available to students and professional chemists, and the beginning student is encouraged to obtain one.
Kekulé Formula
A structural formula displays the atoms of the molecule in the order they are bonded. It also depicts how the atoms are bonded to one another, for example single, double, and triple covalent bond. Covalent bonds are shown using lines. The number of dashes indicate whether the bond is a single, double, or triple covalent bond. Structural formulas are helpful because they explain the properties and structure of the compound which empirical and molecular formulas cannot always represent.
Ex. Kekulé Formula for Ethanol:
Condensed Formula
Condensed structural formulas show the order of atoms like a structural formula but are written in a single line to save space and make it more convenient and faster to write out. Condensed structural formulas are also helpful when showing that a group of atoms is connected to a single atom in a compound. When this happens, parenthesis are used around the group of atoms to show they are together.
Ex. Condensed Structural Formula for Ethanol: CH3CH2OH (Molecular Formula for Ethanol C2H6O).
Shorthand Formula
Because organic compounds can be complex at times, line-angle formulas are used to write carbon and hydrogen atoms more efficiently by replacing the letters with lines. A carbon atom is present wherever a line intersects another line. Hydrogen atoms are then assumed to complete each of carbon's four bonds. All other atoms that are connected to carbon atoms are written out. Line angle formulas help show structure and order of the atoms in a compound making the advantages and disadvantages similar to structural formulas.
Ex.Shorthand Formula for Ethanol:
Contributors
• Jean Kim (UCD), Kristina Bonnett (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/01._Structure_and_Bonding_in_Organic_Molecules/1.9%3A_Structures_and__Formulas_of__Organic_Molecules.txt |
Equilibrium Constant
For the hypothetical chemical reaction:
$aA + bB \rightleftharpoons cC + dD \tag{6.5.1}$
the equilibrium constant is defined as:
$K_C = \dfrac{[C]^c[D]^d}{[A]^a[B]^b} \tag{6.5.2}$
The notation [A] signifies the molar concentration of species A. An alternative expression for the equilibrium constant involves partial pressures:
$K_P = \dfrac{P_C^c P_D^d}{P_A^aP_B^b} \tag{6.5.3}$
Note that the expression for the equilibrium constant includes only solutes and gases; pure solids and liquids do not appear in the expression. For example, the equilibrium expression for the reaction
$CaH_2 (s) + 2H_2O (g) \rightleftharpoons Ca(OH)_2 (s) + 2H_2 (g) \tag{6.5.4}$
is the following:
$K_C = \dfrac{[H_2]^2}{[H_2O]^2} \tag{6.5.5}$
Observe that the gas-phase species H2O and H2 appear in the expression but the solids CaH2 and Ca(OH)2 do not appear.
The equilibrium constant is most readily determined by allowing a reaction to reach equilibrium, measuring the concentrations of the various solution-phase or gas-phase reactants and products, and substituting these values into the Law of Mass Action.
Gibbs Energy
The interaction between enthalpy and entropy changes in chemical reactions is best observed by studying their influence on the equilibrium constants of reversible reactions. To this end a new thermodynamic function called Free Energy (or Gibbs Free Energy), symbol ΔG, is defined as shown in the first equation below. Two things should be apparent from this equation. First, in cases where the entropy change is small, ΔG ≅ ΔH. Second, the importance of ΔS in determining ΔG increases with increasing temperature.
$ΔGº = ΔHº – TΔSº \tag{6.5.6}$
where $T$ is the absolute temperature measured in kelvin.
The free energy function provides improved insight into the thermodynamic driving forces that influence reactions. A negative ΔGº is characteristic of an exergonic reaction, one which is thermodynamically favorable and often spontaneous, as is the melting of ice at 1 ºC. Likewise a positive ΔGº is characteristic of an endergonic reaction, one which requires an input of energy from the surroundings.
Example 6.5.1: Decomposition of Cyclobutane
For an example of the relationship of free energy to enthalpy consider the decomposition of cyclobutane to ethene, shown in the following equation. The standard state for all the compounds is gaseous.
This reaction is endothermic, but the increase in number of molecules from one (reactants) to two (products) results in a large positive ΔSº.
At 25 ºC (298 ºK), ΔGº = 19 kcal/mol – 298(43.6) cal/mole = 19 – 13 kcal/mole = +6 kcal/mole. Thus, the entropy change opposes the enthalpy change, but is not sufficient to change the sign of the resulting free energy change, which is endergonic. Indeed, cyclobutane is perfectly stable when kept at room temperature.
Because the entropy contribution increases with temperature, this energetically unfavorable transformation can be made favorable by raising the temperature. At 200 ºC (473 ºK), ΔGº = 19 kcal/mol – 473(43.6) cal/mole = 19 – 20.6 kcal/mole = –1.6 kcal/mole. This is now an exergonic reaction, and the thermal cracking of cyclobutane to ethene is known to occur at higher temperatures.
$ΔGº = –RT \ln K = –2.303RT \log K$
with
• R = 1.987 cal/ ºK mole
• T = temperature in ºK
• K = equilibrium constant
A second equation, shown above, is important because it demonstrates the fundamental relationship of ΔGº to the equilibrium constant, K. Because of the negative logarithmic relationship between these variables, a negative ΔGº generates a K>1, whereas a positive ΔGº generates a K<1. When ΔGº = 0, K = 1. Furthermore, small changes in ΔGº produce large changes in K. A change of 1.4 kcal/mole in ΔGº changes K by approximately a factor of 10. This interrelationship may be explored with the calculator on the right. Entering free energies outside the range -8 to 8 kcal/mole or equilibrium constants outside the range 10-6 to 900,000 will trigger an alert, indicating the large imbalance such numbers imply.
Substituted Cyclohexanes
A Values
Substituents on a cyclohexane prefer to be in the equatorial position. When a substituent is in the axial position, there are two gauche butane interactions more than when a substituent is in the equatorial position. We quantify the energy difference between the axial and equatorial conformations as the A-value, which is equivalent to the negative of the ∆G°, for the equilibrium shown below. Therefore the A-value, or -∆G°, is the preference for the substituent to sit in the equatorial position.
Recall that the equilibrium constant is related to the change in Gibbs Energies for the reaction:
$\Delta{G^o} = -RT\ln {K_{eq}}$
The balance between reactants and products in a reaction will be determined by the free energy difference between the two sides of the reaction. The greater the free energy difference, the more the reaction will favor one side or the other (Table 6.5.1).
Table 6.5.1: Below is a table of A-values for some common substituents.
Substituent ∆G° (kcal/mol) A-value
-F
-0.28-0.24 0.24-0.28
-Cl -0.53 0.53
-Br -0.48 0.48
-I -0.47 0.47
-CH3 (-Me) -1.8 1.8
-CH2CH3 (-Et) -1.8 1.8
-CH(CH3)2 (-i-Pr) -2.1 2.1
-C(CH3)3 (-t-Bu) <-4.5 >4.5
-CHCH2 -1.7 1.7
-CCH -0.5 0.5
-CN -0.25-0.15 0.15-0.25
-C6H5 (-Ph) -2.9 2.9
Polysubstituted Cyclohexanes
1,4-disubstitution
The A-values of the substituents are roughly additive in either the cis- or trans-diastereomers.
1,3-disubstitution
A-values are only additive in the trans-diastereomer:
When there are cis-substituents on the chair, there is a new interaction in the di-axial conformation:
In the above example, each methyl group has one 1,3-diaxial interaction with a hydrogen. The methyl groups also interact with each other. This new diaxial interaction is extremely unfavorable based on their steric interaction (see: double-gauche pentane conformation).
Contributors
You may recall from general chemistry that it is often convenient to describe chemical reactions with energy diagrams. In an energy diagram, the vertical axis represents the overall energy of the reactants, while the horizontal axis is the ‘reaction coordinate’, tracing from left to right the progress of the reaction from starting compounds to final products. The energy diagram for a typical one-step reaction might look like this:
Despite its apparent simplicity, this energy diagram conveys some very important ideas about the thermodynamics and kinetics of the reaction. Recall that when we talk about the thermodynamics of a reaction, we are concerned with the difference in energy between reactants and products, and whether a reaction is ‘downhill’ (exergonic, energy releasing) or ‘uphill (endergonic, energy absorbing). When we talk about kinetics, on the other hand, we are concerned with the rate of the reaction, regardless of whether it is uphill or downhill thermodynamically.
First, let’s review what this energy diagram tells us about the thermodynamics of the reaction illustrated by the energy diagram above. The energy level of the products is lower than that of the reactants. This tells us that the change in standard Gibbs Free Energy for the reaction (Δrnx) is negative. In other words, the reaction is exergonic, or ‘downhill’. Recall that the Δrnx term encapsulates both Δrnx, the change in enthalpy (heat) and Δrnx , the change in entropy (disorder):
$ΔG˚ = ΔH˚- TΔS˚$
where T is the absolute temperature in Kelvin. For chemical processes where the entropy change is small (~0), the enthalpy change is essentially the same as the change in Gibbs Free Energy. Energy diagrams for these processes will often plot the enthalpy (H) instead of Free Energy for simplicity.
The standard Gibbs Free Energy change for a reaction can be related to the reaction's equilibrium constant ($K_{eq}\_) by a simple equation: $ΔG˚ = -RT \ln K_{eq}$ where: • Keq = [product] / [reactant] at equilibrium • R = 8.314 J×K-1×mol-1 or 1.987 cal× K-1×mol-1 • T = temperature in Kelvin (K) If you do the math, you see that a negative value for Δrnx (an exergonic reaction) corresponds - as it should by intuition - to Keq being greater than 1, an equilibrium constant which favors product formation. In a hypothetical endergonic (energy-absorbing) reaction the products would have a higher energy than reactants and thus Δrnx would be positive and Keq would be less than 1, favoring reactants. Now, let's move to kinetics. Look again at the energy diagram for exergonic reaction: although it is ‘downhill’ overall, it isn’t a straight downhill run. First, an ‘energy barrier’ must be overcome to get to the product side. The height of this energy barrier, you may recall, is called the ‘activation energy’ (ΔG). The activation energy is what determines the kinetics of a reaction: the higher the energy hill, the slower the reaction. At the very top of the energy barrier, the reaction is at its transition state (TS), which is the point at which the bonds are in the process of breaking and forming. The transition state is an ‘activated complex’: a transient and dynamic state that, unlike more stable species, does not have any definable lifetime. It may help to imagine a transition state as being analogous to the exact moment that a baseball is struck by a bat. Transition states are drawn with dotted lines representing bonds that are in the process of breaking or forming, and the drawing is often enclosed by brackets. Here is a picture of a likely transition state for a substitution reaction between hydroxide and chloromethane: $CH_3Cl + HO^- \rightarrow CH_3OH + Cl^-$ This reaction involves a collision between two molecules: for this reason, we say that it has second order kinetics. The rate expression for this type of reaction is: rate = k[reactant 1][reactant 2] . . . which tells us that the rate of the reaction depends on the rate constant k as well as on the concentration of both reactants. The rate constant can be determined experimentally by measuring the rate of the reaction with different starting reactant concentrations. The rate constant depends on the activation energy, of course, but also on temperature: a higher temperature means a higher k and a faster reaction, all else being equal. This should make intuitive sense: when there is more heat energy in the system, more of the reactant molecules are able to get over the energy barrier. Here is one more interesting and useful expression. Consider a simple reaction where the reactants are A and B, and the product is AB (this is referred to as a condensation reaction, because two molecules are coming together, or condensing). If we know the rate constant k for the forward reaction and the rate constant kreverse for the reverse reaction (where AB splits apart into A and B), we can simply take the quotient to find our equilibrium constant \(K_{eq}$:
This too should make some intuitive sense; if the forward rate constant is higher than the reverse rate constant, equilibrium should lie towards products.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
Further Reading
MasterOrganicChemistry
Equilibria
Websites
Reversible and irreversible reactions
Activation Energy
Since exothermic reactions are energetically (thermodynamically) favored, a careless thinker might conclude that all such reactions will proceed spontaneously to their products. Were this true, no life would exist on Earth, because the numerous carbon compounds that are present in and essential to all living organisms would spontaneously combust in the presence of oxygen to give carbon dioxide-a more stable carbon compound. The combustion of methane (eq.1), for example, does not occur spontaneously, but requires an initiating energy in the form of a spark or flame. The flaw in this careless reasoning is that we have focused only on the initial (reactant) and final (product) states of reactions. To understand why some reactions occur readily (almost spontaneously), whereas other reactions are slow, even to the point of being unobservable, we need to consider the intermediate stages of reactions.
Exothermic
Single Step Reaction
Endothermic
Single Step Reaction
Exothermic
Two Step Reaction
Every reaction in which bonds are broken will have a high energy transition state that must be reached before products can form. In order for the reactants to reach this transition state, energy must be supplied and reactant molecules must orient themselves in a suitable fashion. The energy needed to raise the reactants to the transition state energy level is called the activation energy, ΔE. An example of a single-step exothermic reaction profile is shown on the left above, and a similar single-step profile for an endothermic reaction is in the center. The activation energy is drawn in red in each case, and the overall energy change (ΔE) is in green.
The profile becomes more complex when a multi-step reaction path is described. An example of a two-step reaction proceeding by way of a high energy intermediate is shown on the right above. Here there are two transition states, each with its own activation energy. The overall activation energy is the difference in energy between the reactant state and the highest energy transition state. We see now why the rate of a reaction may not correlate with its overall energy change. In the exothermic diagram on the left, a significant activation energy must be provided to initiate the reaction. Since the reaction is strongly exothermic, it will probably generate enough heat to keep going as long as reactants remain. The endothermic reaction in the center has a similar activation energy, but this will have to be supplied continuously for the reaction to proceed to completion.
What is the source of the activation energy that enables a chemical reaction to occur? Often it is heat, as noted above in reference to the flame or spark that initiates methane combustion. At room temperature, indeed at any temperature above absolute zero, the molecules of a compound have a total energy that is a combination of translational (kinetic) energy, internal vibrational and rotational energies, as well as electronic and nuclear energies. The temperature of a system is a measure of the average kinetic energy of all the atoms and molecules present in the system. As shown in the following diagram, the average kinetic energy increases and the distribution of energies broadens as the temperature is raised from T1 to T2. Portions of this thermal or kinetic energy provide the activation energy for many reactions, the concentration of suitably activated reactant molecules increasing with temperature, e.g. orange area for T1 and yellow plus orange for T2. (Note that the area under a curve or a part of a curve is proportional to the number of molecules represented.)
Distribution of Molecular Kinetic Energy
at Two Different Temperatures, T1 & T2
Reaction Rates and Kinetics
Chemical reactivity is the focus of chemistry, and the study of reaction rates provides essential information about this subject. Some reactions proceed so rapidly they seem to be instantaneous, whereas other reactions are so slow they are nearly unobservable. Most of the reactions described in this text take place in from 0.2 to 12 hours at 25 ºC. Temperature is important, since fast reactions may be slowed or stopped by cooling, and slow reactions are accelerated by heating. When a reaction occurs between two reactant species, it proceeds faster at higher concentrations of the reactants. These facts lead to the following general analysis of reaction rates.
Reaction Rate =
Number of Collisions
between Reactant Molecules
per Unit of Time
•
Fraction of Collisions
with Sufficient Energy
to pass the Transition State
•
an Orientational
or Probability
Factor
Since reacting molecules must collide to interact, and the necessary activation energy must come from the kinetic energy of the colliding molecules, the first two factors are obvious. The third (probability) factor incorporates the orientational requirements of the reaction. For example, the addition of bromine to a double bond at the end of a six-carbon chain (1-hexene) could only occur if the colliding molecules came together in a way that allowed the bromine molecule to interact with the pi-electrons of the double bond.
The collision frequency of reactant molecules will be proportional to their concentration in the reaction system. This aspect of a reaction rate may be incorporated in a rate equation, which may take several forms depending on the number of reactants. Three general examples are presented in the following table.
Reaction Type
Rate Equation
Reaction Order
A ——> B
Reaction Rate = k•[A]
First Order Reaction (no collision needed)
A + B ——> C + D
Reaction Rate = k•[A]*[B]
Second Order Reaction
A + A ——> D
Reaction Rate = k•[A]2
Second Order Reaction
These rate equations take the form Reaction Rate = k[X] n[Y] m, where the proportionality constant k reflects the unique characteristics of a specific reaction, and is called the rate constant. The concentrations of reactants X and Y are [X] and [Y] respectively, and n & m are exponential numbers used to fit the rate equation to the experimental data. The sum n + m is termed the kinetic order of a reaction. The first example is a simple first order process. The next two examples are second order reactions, since n + m = 2. The kinetic order of a reaction is usually used to determine its molecularity.
In writing a rate equation we have disconnected the collision frequency term from the activation energy and probability factors defined above, which are necessarily incorporated in the rate constant k. This is demonstrated by the following equation.
The complex parameter A incorporates the probability factor. Because of the exponential relationship of k and the activation energy small changes in ΔE will cause relatively large changes in reaction rate. An increase in temperature clearly acts to increase k, but of greater importance is the increase in average molecular kinetic energy such an increase produces. This was illustrated in a previous diagram, increase in temperature from T1 to T2 producing a larger proportion of reactant molecules having energies equal or greater than the activation energy (designated by the red line. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/02._Structure_and_Reactivity%3A_Acids_and_Bases_Polar_and_Nonpolar_Molecules/2.1%3A_Kinetics_and__Thermodynamics_of__Simple_Chemical_Processes.txt |
The Arrow Notation in Mechanisms
Since chemical reactions involve the breaking and making of bonds, a consideration of the movement of bonding ( and non-bonding ) valence shell electrons is essential to this understanding. It is now common practice to show the movement of electrons with curved arrows, and a sequence of equations depicting the consequences of such electron shifts is termed a mechanism. In general, two kinds of curved arrows are used in drawing mechanisms:
A full head on the arrow indicates the movement or shift of an electron pair:
A partial head (fishhook) on the arrow indicates the shift of a single electron:
The use of these symbols in bond-breaking and bond-making reactions is illustrated below. If a covalent single bond is broken so that one electron of the shared pair remains with each fragment, as in the first example, this bond-breaking is called homolysis. If the bond breaks with both electrons of the shared pair remaining with one fragment, as in the second and third examples, this is called heterolysis.
Bond-Breaking Bond-Making
Other Arrow Symbols
Chemists also use arrow symbols for other purposes, and it is essential to use them correctly.
The Reaction Arrow
The Equilibrium Arrow
The Resonance Arrow
The following equations illustrate the proper use of these symbols:
Reactive Intermediates
The products of bond breaking, shown above, are not stable in the usual sense, and cannot be isolated for prolonged study. Such species are referred to as reactive intermediates, and are believed to be transient intermediates in many reactions. The general structures and names of four such intermediates are given below.
A pair of widely used terms, related to the Lewis acid-base notation, should also be introduced here.
Electrophile: An electron deficient atom, ion or molecule that has an affinity for an electron pair, and will bond to a base or nucleophile.
Nucleophile: An atom, ion or molecule that has an electron pair that may be donated in bonding to an electrophile (or Lewis acid).
Using these definitions, it is clear that carbocations ( called carbonium ions in the older literature ) are electrophiles and carbanions are nucleophiles. Carbenes have only a valence shell sextet of electrons and are therefore electron deficient. In this sense they are electrophiles, but the non-bonding electron pair also gives carbenes nucleophilic character. As a rule, the electrophilic character dominates carbene reactivity. Carbon radicals have only seven valence electrons, and may be considered electron deficient; however, they do not in general bond to nucleophilic electron pairs, so their chemistry exhibits unique differences from that of conventional electrophiles. Radical intermediates are often called free radicals.
The importance of electrophile / nucleophile terminology comes from the fact that many organic reactions involve at some stage the bonding of a nucleophile to an electrophile, a process that generally leads to a stable intermediate or product. Reactions of this kind are sometimes called ionic reactions, since ionic reactants or products are often involved. Some common examples of ionic reactions and their mechanisms may be examined by Clicking Here
The shapes ideally assumed by these intermediates becomes important when considering the stereochemistry of reactions in which they play a role. A simple tetravalent compound like methane, CH4, has a tetrahedral configuration. Carbocations have only three bonds to the charge bearing carbon, so it adopts a planar trigonal configuration. Carbanions are pyramidal in shape ( tetrahedral if the electron pair is viewed as a substituent ), but these species invert rapidly at room temperature, passing through a higher energy planar form in which the electron pair occupies a p-orbital. Radicals are intermediate in configuration, the energy difference between pyramidal and planar forms being very small. Since three points determine a plane, the shape of carbenes must be planar; however, the valence electron distribution varies. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/02._Structure_and_Reactivity%3A_Acids_and_Bases_Polar_and_Nonpolar_Molecules/2.2%3A_Keys_to_Success%3A_Using_Curved_Electron_Pushing_Arrows_to_Describe_Chem.txt |
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