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Please Note: The terms "acid halide" and "acyl halide" are synonymous and are both used in this text. In biochemistry, the term "acyl" is used more frequently.
Acid Halide Synthesis
Carboxylic acids react with thionyl chloride (SOCl2) or oxalyl chloride (C2O2Cl2) to form acid chlorides. Typically the reactions occur in the presence of a proton scavanger like pyridine to minimize unwanted side reactions. 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.
Analogous to the reactions of primary and secondary alcohols with PBr3 to produce the corresponding alkyl bromide, acid bromides can be formed from the reaction of phosphorous tribromide with carboxylic acids.
Acyl Halide Reactivity
Acyl halides can be hydrolyzed to carboxylic acids and converted to carboxylic acid derivatives. Acid halides can also undergo reduction reactions and reactions with Grignard reagents and organolithium cuprates along with Fridel-Crafts acylation of benzene. The reaction map below summarizes the reactivity of acyl halides.
Acid Halide Hydrolysis
The hydrolysis reaction of acid chlorides is shown below.
The hydrolysis of butonyl chloride is shown below as an example.
Example: Acyl Chloride Hydrolysis
Carboxylic Acid Derivative Synthesis from Acyl Chlorides
Carboxylic acid derivatives can be synthesized from acyl chlorides via the nucleophilic acyl substitution mechanism previously discussed.
Anhydride Synthesis
Acid chlorides react with carboxylic acids to form acid anhydrides as shown in the reaction below.
The synthesis of benzoic anhydride from benzoyl chloride and benzoic acid is shown as an example.
Example: Anhydride Synthesis from Acyl Chlorides
Ester Synthesis
Acid chlorides react with alcohols for form esters are shown in the reaction below.
The synthesis of ethyl benzoate from benzoyl chloride and ethanol is shown as an example.
Example: Ester Synthesis from Acid Chlorides
Amide Synthesis
Acid chlorides react with "ammonia, 1o amines and 2o amines" to form amides as shown in the reaction below.
The reaction requires 1 equivalent of the "ammonia/1o or 2o amine" with a proton scavenger (aka base) like pyridine OR two equivalents of "ammonia/1o or 2o amine". The additional equivalent the nucleophile or base is needed to maintain the nucleophilic character of "ammonia/1o or 2o amine". As shown in the mechanism below, the amide is briefly protonated after the carbonyl reforms from the tetrahedral complex. Since amides are considered neutral with no significant basicity, the "ammonia/1o or 2o amine" quickly accepts their proton and is no longer a nucleophile. The second equivalent of "ammonia/1o or 2o amine" restores the concentration of the nucleophile. For clarity, the mechanism below is shown with ammonia as the nucleophile.
Mechanism
The syntheses of acetamide and N-ethylacetamide from acetyl chloride are shown as examples.
Example: Amide Synthesis from Acyl Chlorides
Acid Chloride Reduction
Acid chlorides can be fully reduced to primary alcohols using either sodium borohydride or lithium aluminum hydride. Acid chlorides can be partially reduced to aldehydes using the lithium tri-tert-butoxyaluminum hydride (LiAlH(O-t-Bu)3. These reactions are summarized in the reaction map below.
The syntheses of 1-hexanol and hexanal from hexanoyl chloride are shown as examples.
Example: Reduction Acyl Chlorides
Acid Chloride Reactions with Organometallic Compounds
Grignard reagents
Acid chlorides react with Grignard reagents to produce tertiary alcohols. Two equivalents of the Grignard reagent are needed because the first equivalent reacts to form a ketone which then reacts with the second equivalent. Because of the high reactivity of the Grigard reagent, the reaction can NOT be stopped at the ketone.
The syntheses of 3-phenylpentan-3-ol from benzoyl chloride is shown as an example.
Example: Acyl Chloride Reactions with Grignard Reagents
Organolithium Cuprates
Organolithium cuprate reagents are less reactive than Grignard reagents and can convert acid chlorides to ketones as shown below.
The synthesis of 1-phenylpropan-1-one from benzoyl chloride is shown as an example.
Example: Acyl Chloride Reactions with Organolithium Cuprates
Friedel-Crafts Acylation of Benzene
Benzene rings can be acylated via the Friedel-Crafts acylation reaction with acid chlorides in the presence of aluminum chloride followed by an aqueous work-up as shown below.
The synthesis of 1-(4-tert-butylphenyl)butan-1-one from t-butylbenzene and butonyl chloride is shown as an example.
Example: Friedel-Crafts Acylation
Exercise
4. Draw the mechanism for the following reaction
5. Propose a synthesis of the following molecules from an acid chloride and an amide.
(a)
(b)
(c)
Answer
4.
5.
a) Acetyl chloride and dimethylamine
b) Benzoyl chloride and ethylamine
c) Acetyl chloride and ammonia
22.05: Acid Anhydride Chemistry
Synthesis of Acid Anhydrides
Acid chlorides react with carboxylic acids to form anhydrides as shown in the reaction below.
Some cyclic anhydrides can be synthesized from the corresponding dicarboxylic acid with gentle heating. The example below shows the reaction of glutaric acid to form a cyclic anhydride.
Example: Acid Anhydride Synthesis
Acid Anhydride Reactivity
Acid anhydrides undergo hydrolysis and nucleophilic acyl substitution reactions.
Acid Anhydride Hydrolysis
Acid anhydrides readily hydrolyze to carboxylic acids. In many cases, this reaction is an unwanted side reaction and steps will be taken in the lab to keep the system "dry" (aka water free). The presence of pyridine facilitates proton transfers during the reaction. The hydrolysis reaction for benzoic anhydride is shown below.
The mechanism is analogous to the mechanism for ester synthesis from acid anhydrides and is shown below is detail.
Nuclephilic Acyl Substitution Reactions from Acid Anhydrides
Carboxylic acid derivatives can be synthesized from acid anhydrides via the nucleophilic acyl substitution mechanism previously discussed.
Ester Synthesis
Acid anhydrides react with alcohols to produce esters as shown in the reaction below. The reactions of anhydrides frequently use pyridine as a solvent.
A carboxylic acid is also produced, but is not considered a synthetic product. The ester is considered the "product of interest".
The synthesis of methyl benzoate from benzoic anhydride and methanol is shown in the example.
Example: Ester Sythesis
The mechanism follows the nucleophililc acyl substitution mechanism as previously discussed and reviewed below.
1) Nucleophilic Alcohol reacts with Electrophilic Carbonyl
2) Deprotonation by pyridine
3) Leaving group removal
4) Protonation of the carboxylate
Amide Synthesis
Acid Anhydrides react with amines to form amides. As seen with acid halide reactions, a second equivalent of the amine must be present for the reaction to proceed.
Example: Amide Synthesis
The mechanism for amide synthesis is analogous to the mechanism for ester formation. The only minor difference is that a second equivalent of the amine or ammonia is used instead of the pyridine.
1) Nucleophilic Amine reacts with Electrophilic Carbonyl
2) Deprotonation by the amine
3) Leaving group removal
Exercise
6. Draw out the mechanism for the following reaction.
7. Draw the product of the reaction between these two molecules.
Answer
6.
7. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/22%3A_Carboxylic_Acid_Derivatives_and_Nitriles/22.04%3A_Acid_Halide_Chemistry.txt |
Introduction
Esters are readily synthesized and naturally abundant. Esters are frequently the source of flavors and aromas in many fruits and flowers.
Esters also make up the bulk of animal fats and vegetable oils—glycerides (fatty acid esters of glycerol). Soap is produced by a saponification (basic hydrolysis) reaction of a fat or oil.
Esters are also present in a number of important biological molecules and have several commercial and synthetic application. For example, 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.
The most important polyester, polyethylene terephthalate (PET), is made from terephthalic acid and ethylene glycol monomers:
Synthesis of Esters
Carboxylic acids can react with alcohols to form esters in the presence of an acid catalyst as shown in the reaction below.
Acid chlorides react with alcohols to form esters as shown in the reaction below.
Acid anhydrides also react with alcohols to form esters as shown in the reaction below.
As an example, the synthesis of banana oil (isoamyl acetate) is an example of these two reactions.
Example: Ester Synthesis
Esters can be also be synthesized from trans-esterification reactions. Trans-esterification is discussed in the next section on Reactivity of Esters
Reactivity of Esters
Esters can be hydrolyzed to carboxylic acids under acidic or basic conditions. Basic hydrolysis can be used to convert fats and oils into soap and is called a saponification reaction. Esters can be converted to amides via an aminolysis reaction. Esters can undergo trans-esterification reactions to form different esters by applying LeChatlier's principle to this equilibrium reaction. Esters can be reduced to form alcohols or aldehydes depending on the reducing agent. Esters also react with organometallic compounds to form tertiary alcohols. The reaction map for esters is shown below.
Ester Hydrolysis
Ester hydrolysis requires an acid catalyst or base promotion to occur. Esters are less reactive than acyl halides and acid anhydrides because the alkoxide group is a poor leaving group with its negative charge fully localized on a single oxygen atom.
Ester Hydrolysis - Acid Catalyzed
Esters can be cleaved back into a carboxylic acid and an alcohol by reaction with water and a catalytic amount of acid as shown in the reaction below.
The acid catalyzed hydrolysis of ethyl benzoate is shown below as an example.
Example: Acid Catalyzed Ester Hydrolysis
The mechanism for the acid catalyzed hydrolysis reaction begins with protonation of the carbonyl oxygen to increase the reactivity of the ester. The nucleophilic water reacts with the electrophilic carbonyl carbon atom to form the tetrahedral intermediate. Proton transfer reactions occur to create a good leaving group when the carbonyl reforms. The complete mechanism is shown below.
1) Protonation of the Carbonyl
2) Nucleophilic reaction by water
3) Proton transfer
4) Leaving group removal
Ester Hydrolysis - Base promoted
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. The hydroxide ions are consumed in the reaction so it is described as "base promoted".
The base promoted hydrolysis of ethyl benzoate is shown below as an example.
Example: Base Promoted Hydrolysis of Esters
The mechanism for the base promoted hydrolysis reaction begins with the nucleophilic hydroxide reacting with the electrophilic carbonyl carbon atom to form the tetrahedral intermediate. The carbonyl reforms with the loss of the alkoxide leaving group. The alkoxide then deprotonates the resulting carboxylic acid. The complete mechanism is shown below.
1) Nucleophilic reaction by hydroxide
2) Leaving group removal
3) Deprotonation
Nuclephilic Acyl Substitution Reactions from Esters
Carboxylic acid derivatives can be synthesized from esters via the nucleophilic acyl substitution mechanism previously discussed.
Ester Synthesis: Trans-Esterification
Trans-esterification is the conversion of a carboxylic acid ester into a different carboxylic acid ester. When an ester is placed in a large excess of an alcohol along with presence of either an acid or a base there can be an exchange of alkoxy groups. The large excess of alcohol is used to drive the reaction forward. The most common method of trans-esterification is the reaction of the ester with an alcohol in the presence of an acid catalyst as shown below.
The trans-esterification of ethyl acetate to methyl acetate and methyl benzoate to ethyl benzoate are shown below as examples.
Example: Trans-esterification Reactions
Under acidic conditions, the reaction mechanism begins with protonation of the carbonyl oxygen which increases the reactivity of the ester. An alcohol then reactions with protonated ester to form the tetrahedral intermediate. After several proton transfers, the carbonyl reforms to produce a new ester. The complete mechanism is shown below for the trans-esterification of ethyl actetae to methyl acetate.
Since both the reactants and the products are an ester and an alcohol, the reaction is reversible and the equilibrium constant is close to one. Consequently, the Le Chatelier’s principle has to be exploited to drive the reaction to completion. The simplest way to do so is to use the alcohol as the solvent as well.
Under basic conditions, the mechanism begins with the nucleophilic reaction of the alkoxide with the carbonyl carbon to produce the tetrahedral intermediate. The carbonyl reforms with the loss of the leaving group to produce a new ester.
1) Nucleophilic reaction by an alkoxide
2) Leaving group removal
Aminolysis: Conversion of Esters into Amides
Esters react with ammonia and 1o or 2o alkyl amines to yield amides in a reaction called aminolysis.
The aminolysis of ethyl benzoate is shown below as an example. The mechanism for this reaction is analogous to the base promoted hydrolysis reaction of esters shown above.
Example: Aminolysis of Esters
Ester Reduction Reactions
Ester Reduction to a 1o Alcohol
Esters can be converted to 1o alcohols using LiAlH4, while sodium borohydride (NaBH4) is not a strong enough reducing agent to perform this reaction.
The reduction of ethyl benzoate to benzyl alcohol and ethanol is shown as an example.
Example: Ester Reduction to a 1o Alcohol
The mechanism begins with a hydride nucleophile reacting with the ester carbonyl carbon to form the tetrahedral intermediate. The carbonyl reforms to produce an aldehyde with the loss of the alkoxide ion. The resulting aldehyde undergoes a subsequent reaction with a hydride nucleophile to form another tetrahedral intermediate. The carbonyl is not able to reform, because there are no stable leaving groups. Therefore, the alkoxide (tetrahedral intermediate) is protonated to produce a primary alcohol. The complete mechanism is shown below.
1) Nucleophilic reaction by the hydride
2) Leaving group removal
3) Nucleopilic reaction by the hydride anion
4) The alkoxide is protonated
Ester Reduction to an Aldehyde
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.
The reduction of methyl benzoate to form benzaldehyde is shown as an example. The mechanism is analogous to the LiAlH4 mechanism shown above with the important difference that the reaction stops after the aldehyde is produced because the DIBAH reducing agent is not strong enough to reduce the aldehyde at low temperatures.
Example: Ester Reduction to an Aldehyde
Ester Reactions with Organometallic Compounds
Grignard reagents
Esters react with Grignard reagents to form tertiary alcohols. This reaction is analogous to the reaction discussed for acid chlorides with Grignard reagents. The first equivalent of the Grignard reagent produces a ketone which reacts with the second equivalent of the Grignard reagent to produce a tertiary alcohol. In effect, the Grignard reagent adds twice as shown in the reaction below.
The reaction of methyl benzoate with a Grignard reagent to produce 3-phenyl-3-pentanol.
Example: Ester Reaction with a Grignard Reagent
The mechanism begins with a carbide nucleophile from the Grignard reagent reacting with the ester carbonyl carbon to form the tetrahedral intermediate. The carbonyl reforms to produce a ketone with the loss of the alkoxide ion. The resulting ketone undergoes a subsequent reaction with a carbide nucleophile from the Grignard reagent to form another tetrahedral intermediate. The carbonyl is not able to reform, because there are no stable leaving groups. Therefore, the alkoxide (tetrahedral intermediate) is protonated to produce a tertiary alcohol. The complete mechanism is shown below.
1) Nucleophilic reaction
2) Leaving group removal
3) Nucleophilic reaction
4) Protonation
Exercise
8. Why is the alkaline hydrolysis of an ester not a reversible process? Why doesn't the reaction with a hydroxide ion and a carboxylic acid produce an ester?
9. Draw the product of the reaction between the following molecule and LiAlH4, and the product of the reaction between the following molecule and DIBAL.
10. Prepare the following molecules from esters and Grignards?
(a)
(b)
(c)
Answer
8. The reaction between a carboxylic acid and a hydroxide ion is an acid base reaction, which produces water and a carboxylate anion.
9.
10.
(a)
(b)
(c)
Contributors and Attributions
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/22%3A_Carboxylic_Acid_Derivatives_and_Nitriles/22.06%3A_Ester_Chemistry.txt |
Synthesis of Amides
There are five synthetic routes to produce amides: nitrile conversion and the acyl nucleophilic substitution reactions of acid halides, acid anhydrides, and carboxylic acids.
Nitriles can be converted to amides. This reaction can be acid or base catalyzed
Carboxylic acid can be converted to amides by using DCC as an activating agent
Direct conversion of a carboxylic acid to an amide by reaction with an amine.
Acid chlorides react with ammonia, 1o amines and 2o amines to form amides
Acid Anhydrides react with ammonia, 1o amines and 2o amines to form amides
Hydrolysis of Amides
Hydrolysis under acidic conditions
Taking acetamide (ethanamide) as a typical amide. If acetamide is heated with a dilute acid (such as dilute hydrochloric acid), acetic acid is formed together with ammonium ions. So, if you were using hydrochloric acid, the final solution would contain ammonium chloride and acetic acid.
Hydrolysis under alkaline conditions
Also, if acetamide is heated with sodium hydroxide solution, ammonia gas is given off and you are left with a solution containing sodium acetate.
Peptide hydrolysis
Peptide hydrolysis of proteins is amide hydrolysis. What biologists and biochemists call a peptide link (in proteins, for example) is what chemists call an amide link. Apply either hydrolysis reaction above to the dipeptide below to produce two amino acids. The amines in the products are shown in their protonated form because this hydrolysis reaction was performed under acidic conditions.
Reduction of Amides into Amines
Amides can be converted to 1°, 2° or 3° amines using LiAlH4 followed by an aqueous work-up. Alkyl groups attached to the amide nitrogen do not affect the reaction. The amine classification correlates with the amide as shown in the reaction summary below.
The reductions of propanamide and N,N-dimethylpropanamide are shown as examples.
Example: Amide Reductions
The mechanism begins with nucleophilic hydride reacting with the carbonyl carbon to produce the tetrahedral intermediate. An imine forms in concert with the loss on the leaving group. A second hydride nucleophile reacts with the imine carbon to produce the final product.
1) Nucleophilic reaction by the hydride
2) Imine formation with loss of leaving group
3) Nucleophilic reaction by the hydride
Exercise
11. How would you prepare the following compounds from N-Propypl benzamide?
(a)
(b)
(c)
12.
Propose a synthesis for the following.
Answer
11.
a) NaOH, H2O
b) NaOH, H2O, then LiAlH4
c) LiAlH4
12.
22.08: Nitrile Chemistry
Interesting Nitriles
One of the most common occurrences of nitriles is nitrile rubber. Nitrile rubber is a synthetic copolymer of acrylonitrile and butadiene. This form of rubber is highly resistant to chemicals and is used to make protective gloves, hoses and seals.
Synthesis of Nitriles
Nitriles can be synthesized from the reaction of nucleophilic cyanide with electrophilic groups, such as the carbonyls (aldehydes and ketones) and alkyl halides that are suitable for SN2 reactions. Amides can react with thionyl chloride to produce nitriles.
Addition of cyanide (-:C≡N) to an aldehyde or ketone forms a cyanohydrin.
Nitriles are formed by an SN2 reaction between an alkyl bromide and sodium cyanide
Primary (1o) amides can be converted to nitriles by dehydration with thionyl chloride (or other dehydrating agents like P2O5, or POCl3).
For the reaction of primary amides with thionyl chloride, the mechanism begins with the lone pair of the nitrogen atom formimg a protonated imine and pushing the pi electrons of the carbonyl to form a sigma bond with the sulfur of thionyl chloride. The sulfonyl bond reforms in concert with the loss of the leaving group (Cl-). The protonated imine is neutralized by any base. The nitrile is produced by one last deprotonation reaction with a loss of sulfur dioxide and chloride as the leaving groups.
The complete mechanism is shown below.
1) Protonated imine formation with Nucleophilic reaction of carbonyl pi bond
2) Leaving group removal
3) Deprotonation
4) Deprotonation & Leaving group removal
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.
Nitriles can be converted to carboxylic acid with heating in sulfuric acid. During the reaction an amide intermediate is formed.
The hydrolysis of cyclopentanecarbonitrile is shown below as an example.
Example: Nitrile Hydrolysis
Note that the presence of water is understood.
The mechanism begins with the protonation of the nitrile to make it more electrophilic to nucleophilic water. Once the water has reacted with the nitrile carbon, proton transfers occur to produce a resonance stabilized intermediate. Water acts as a weak base to deprotonate the carbonyl to form the amide which is hydrolyzed to the carboxylic acid.
1) Protonation
2) Nucleophilic reaction by water
3) Proton Transfer
4) Resonance
5) Deprotonation
6) Further hydrolysis of the amide shown in amide section of this chapter.
Nitrile Reduction
Nitriles can be reduced to primary amines with lithium aluminum hydride followed by an aqueous work-up. During this reaction the hydride nucleophile reacts with 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 to neutralize the reaction environment. The general reaction is shown below.
The reduction of cyclopentanecarbonitrile is shown below as an example.
Example: Nitrile Reduction
The mechanism begins with the nucleophilic hydride reacting with the electrophilc carbon of the nitrile to form an anionic aluminum complex. A second hydride nucleophile reacts with the same electrophilic carbon to form a tetrahedral complex. Protonation by addition of water produces the primary amine in its neutral form.
1) Nucleophilic reaction by the Hydride
2) Second nucleophilic reaction by the hydride.
3) Protonation by addition of water to give an amine
Organometallic Reaction with Nitriles
Grignard reagents can react with nitriles to form an imine salt that can be hydrolyzed to form a ketone as shown in the reaction below.
The reaction of benzonitrile with the methyl-Grignard reagent to form acetophenone is shown below as an example.
Example: Nitrile reaction with a Grignard Reagent
The mechanism begins with the nucleophilic Grignard Reagent reacting with the electrophilic carbon of the nitrile to form an imine salt. The imine salt is hydrolyzed to produce a ketone through a series of nucleophilic and proton transfer reactions. The complete mechanism is shown below for those who are curious.
1) Nucleophilic reaction by the Grignard Reagent
2) Protonation
3) Protonation
4) Nucleophilic reaction by water
5) Proton Transfer
6) Leaving group removal
7) Deprotonation
Exercise
13. Propose two different synthetic routes to convert benzyl bromide into 2-phenyl acetic acid.
Answer
13. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/22%3A_Carboxylic_Acid_Derivatives_and_Nitriles/22.07%3A_Amide_Chemistry.txt |
Introduction to thioesters and Coenzyme A
In the metabolism of lipids (fats and oils), thioesters are the principal form of activated carboxylate groups. They are employed as acyl carriers, assisting with the transfer of acyl groups such as fatty acids from one acyl X substrate to another.
The ‘acyl X group’ in a thioester is a thiol. The most important thiol compound used to make thioesters is called coenzyme A, which has the following structure:
Coenzyme A is often abbreviated HSCoA, in order to emphasize that it is the thiol sulfur that provides the critical thioester linkage to acyl groups. When fuel (carbohydrate and fat) is broken down in your body, it is eventually converted to a simple two-carbon unit called acetyl CoA, which is essentially a thioester derivative of acetic acid:
Activation of fatty acids by coenzyme A: a thioesterification reaction
In the biologically active form of fatty acids, the carboxylate groups have been converted to thioesters using coenzyme A. For example, the activated form of the C16 fatty acid palmitate is:
Let’s take a look at how this activation takes place, in a reaction catalyzed by an enzyme called acyl CoA synthetase. You already know that carboxylates are not themselves good substrates for acyl substitution reactions, and must be activated. Thus, you might predict that the first step of this reaction requires ATP to make a high-energy acyl phosphate intermediate. In fact, the activated carboxylate in this case is an acyl-AMP, formed in the same way as the acyl-AMP intermediate in the asparagine synthetase reaction.
The activated acyl-AMP intermediate is then attacked by the thiol sulfur of coenzyme A, and the AMP group is expelled to form the fatty acyl CoA.
Transfer of fatty acyl groups to glycerol: a thioester to ester substitution
The -SCoA thioester form of the fatty acid is a good substrate for a number of metabolic transformations. This is the form of fatty acid, for example, that is oxidized and broken down for energy in the mitochondria of your cells. Fatty acyl CoA also serves as substrate for the construction of triacylglycerol, which is the fat molecule that your body uses to store energy in fat cells. Recall that triacylglycerol is composed of a glycerol ‘backbone’ connected to three fatty acid groups through ester linkages.
The reaction in which a fatty acid acyl group is linked to glycerol represents the conversion of a thioester (fatty acyl CoA) to an ester. First, however, a transthioesterification reaction occurs. A transthioesterification is merely the conversion of one thioester to another. In the case of monoacylglycerolacyltransferase, the fatty acyl group first trades its thioester link to coenzyme A for another thioester link to a cysteine residue in the active site of the enzyme. It is a common strategy for enzymes to first form a covalent link to one substrate before catalyzing the principle chemical reaction.
The fatty acyl group is now ready to be transferred to glycerol, trading its thioester linkage to the cysteine for a new ester linkage to one of the alcohol groups on glycerol. The attacking nucelophile in this reaction is of course the alcohol oxygen of monoacylglycerol.
Because esters are more stable than thioesters, this is an energetically downhill reaction.
Transthioesterification reactions
In the previous section we saw one example of a transthioesterification. Another important transthioesterification reaction involves acetyl CoA, the activated form of acetic acid and the basic two-carbon building block for fats and oils. Before it can be incorporated into a growing fatty acid molecule, acetyl CoA must first be linked to a so-called ‘acyl carrier protein’ (ACP). The acetyl group is linked to the acyl carrier protein via a thiol group on a carrier molecule that is covalently attached to the protein.
Notice that the structure of this carrier group (called phosphopantetheine) is identical to the region of coenzyme A (structure shown earlier in this section) near the thiol group. Once attached to the ACP, the two-carbon acetyl group condenses with another acyl group (which is also attached to its own ACP), and the fatty acid chain begins to grow.
Finally, a transthioesterification is the final step in one of the most important and well-studied reactions in animal metabolism: the conversion of pyruvate to acetyl CoA by a cluster of enzymes called the pyruvate dehydrogenase complex.
The overall reaction looks simple, but is actually quite complex and involves several intermediate species. The final step in the process is a transthioesterification, involving a dithiol molecule called lipoamide:
We will look more closely at the complete biochemical transformation catalyzed by the pyruvate dehydrogenase complex in section 16.12B.
Hydrolysis of thioesters
The acyl group of a thioester can be transferred to a water molecule in a hydrolysis reaction, resulting in a carboxylate. An example of thioester hydrolysis is the conversion of (S)-citryl CoA to citrate in the citric acid cycle (also known as the Krebs cycle).
Reactivity of thioesters and acyl phosphates
Thioesters are reactive among the biologically relevant acyl groups. However, thioesters are not as reactive as an acid chlorides or acid anhydrides.
Relative reactivity of biologically relevant acyl groups | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/22%3A_Carboxylic_Acid_Derivatives_and_Nitriles/22.09%3A_Thioesters-_Biological_Carboxylic_Acid_Derivative.txt |
Polyamides
Just as the reaction of a diol and a diacid forms a polyester, 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.
Condensation Polymers
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
Contributors and Attributions
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_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/22%3A_Carboxylic_Acid_Derivatives_and_Nitriles/22.10%3A_Polyamides_and_Polyesters-__Step-Growth_Polymers.txt |
The Mechanism of Action of β-Lactam Antibiotics
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 and Attributions
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/22%3A_Carboxylic_Acid_Derivatives_and_Nitriles/22.11%3A__Beta-Lactams-_An_Application.txt |
Glutamine synthetase
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)
22.13: Additional Exercises
General Review
22-1 Suggest a carboxylic acid and an acid derivative that could be reacted together to form the following molecule.
22-2 For the following reaction, predict the product if:
• only one equivalent of the Grignard reagent was used (and the product could be isolated)
• if two equivalents were used (followed by an acid workup)
22-3 Provide the final products of the following reactions.
22-4 Predict the final product of the following reaction.
Interconversion of Acid Derivatives by Nucleophilic Acyl Substitution
22-5 Predict the interconverted acid derivatives of the following reactions.
22-6 Predict the structure of the product and give its IUPAC nomenclature.
22-7 Choose the correct answer for the product of the following reaction.
Transesterification
22-8 Choose the correct IUPAC nomenclature of the product of the following reaction.
a) ethyl butanoate
b) propan-2-yl butanoate
c) dipropyl carbonate
d) propyl butanoate
22-9 Explain why transesterification can be done under acidic or basic conditions.
22-10 Give the products of the following transesterification reactions.
Hydrolysis of Carboxylic Acid Derivatives
22-11 Provide the correct structure of the product of the following hydrolysis reaction.
22-12 Provide the structure of all the products resulting from the hydrolysis of the following triglyceride.
22-13 Provide the structures and IUPAC nomenclature of the products.
Reduction of Acid Derivatives
22-14 Give the structure of the product of the following reaction.
22-15 Provide the structure of all the products (including leaving groups) formed in the following reaction.
22-16 Choose the correct answer that gives the products of a fully reduced 1,3-dimethyl-1,3-diazinan-2-one.
Reactions of Acid Derivatives with Organometallic Reagents
22-17 Predict the product of the following reaction.
22-18 Choose the correct IUPAC nomenclature of the product of the following reaction.
a) 6-amino-5-chlorohexan-2-one
b) 6-amino-5-chloro-2-methylhexan-2-ol
c) 6-amino-4-chloro-2-methylhexan-2-ol
d) 4-chloro-2-methylpiperidin-2-ol
22-19 Decide whether or not the following reaction is the best way to obtain the final product. If not, suggest a better route of synthesis.
22.14: Solutions to Additional Exercises
General Review
22-1
Possible set of reactants:
22-2
22-3
22-4
Interconversion of Acid Derivatives by Nucleophilic Acyl Substitution
22-5:
22-6:
22-7:
Answer: B
Transesterification
22-8:
Answer: D
22-9:
Under acidic conditions, the carbonyl oxygen atom is protonated, making it a better electrophile for the reaction to occur. Under basic conditions, the alcohol we are trying to add is deprotonated, making it a better nucleophile.
22-10:
Hydrolysis of Carboxylic Acid Derivatives
22-11:
22-12:
22-13:
Reduction of Acid Derivatives
22-14:
22-15:
22-16:
Answer: C
Reactions of Acid Derivatives with Organometallic Reagents
22-17:
22-18:
Answer: C
22-19:
A possibly better route of synthesis: | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/22%3A_Carboxylic_Acid_Derivatives_and_Nitriles/22.12%3A__Biological_Acylation_Reactions.txt |
learning objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• predict the relative acidity of the a-hydrogens on various carbonyl compounds (section 23.1)
• explain or predict the equilibrium of enol-keto tautomers (section 23.2)
• predict the products and specify the reagents for the following reactions
• Halogenation of the a-carbon of aldehydes and ketones (section 23.3 and 23.4)
• Halogenation of the a-carbon of carboxylic acids (Hell-Vollhard-Zelinski) (section 23.3 and 23.5)
• Alkylation of the a-carbon of carbonyl compounds via the LDA pathway (section 23.3 and 23.6)
• Alkylation of the a-carbon of aldehydes and ketones via the enamine intermediate (section 23.3 and 3.7)
• Aldol addition and condensation reactions – 2 aldehydes, 2 ketones, 1 aldehyde with 1 ketone (section 23.3 and 23.8)
• Claisen condensation reactions – 2 esters or 1 ester with 1 ketone (section 23.3 and 23.9)
• Diekmann condensation reactions (intramolecular Claisen) - (section 23.9)
• Conjugate Addition a.k.a. Michael reaction (section 23.3 and 23.10)
• Robinson annulation (section 23.10)
• Decarboxylation of 3-oxocarboxylic acids (section 23.3 and 23.12)
• Malonic ester synthesis of carboxylic acids
• Acetoacetic ester synthesis of methyl ketones
Designing synthesis using all of the reactions through this chapter with an emphasis on increasing the size of the carbon backbone by forming new carbon-carbon bonds
23: Alpha Substitutions and Condensations of Carbonyl Compounds
Acidity of Alpha Hydrogens
Alkyl hydrogen atoms bonded to a carbon atom in a a (alpha) position relative to a carbonyl group display unusual acidity. While the pKa values for alkyl C-H bonds is typically on the order of 40-50, pKa values for these alpha hydrogens is more on the order of 19-20. This can most easily be explained by resonance stabilization of the product carbanion, as illustrated in the diagram below.
In the presence of a proton source, the product can either revert back into the starting ketone or aldehyde or can form a new product, the enol. The equilibrium reaction between the ketone or aldehyde and the enol form is commonly referred to as "keto-enol tautomerism". The ketone or aldehyde is generally strongly favored in this reaction.
Because carbonyl groups are sp2 hybridized the carbon and oxygen both have unhybridized p orbitals which can overlap to form the C=O $\pi$ bond.
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.
Relative Acidity of Alpha Hydrogens
The acidity of alpha hydrogens varies by carbonyl functional group as shown in the table below. Evaluating the stability of the conjugate bases can explain the differences in the relative acidity of the alpha hydrogens.
Exercise
1. Draw the bond line structure for each compound in the table above including all relevant resonance forms to explain the relative acidity.
Answer
1.
Contributors and Attributions
Prof. Steven Farmer (Sonoma State University)
• Clarke Earley (Department of Chemistry, Kent State University Stark Campus) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/23%3A_Alpha_Substitutions_and_Condensations_of_Carbonyl_Compounds/23.01%3A__Relative_Acidity_of_alpha-Hyd.txt |
Introduction
Because of the acidity of the alpha-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 and Enolate Formation
Under acidic conditions, the enol tautomer forms. Under basic conditions, the enolate tautomer forms. Both the enol and enolate are nucleophiles that can undergo subsequent reactions. The mechanism for both acidic and basic reaction conditions are shown below.
Acid conditions
1) Protonation of the Carbonyl
2) Enol formation
Basic conditions
1) Enolate formation
2) Protonation
For 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.
Functional Group Structure pKa
carboxylic acid HO–(C=O)R 5
nitro RCH2–NO2 9
β-diketone * R(O=C)–CH2–(C=O)R 9
β-ketoester * R(O=C)–CH2–(C=O)OR 11
β-diester * RO(O=C)–CH2–(C=O)OR 13
amide RNH–(C=O)R 15
alcohol RCH2–OH 16
aldehyde RCH2–(C=O)H 17
ketone RCH2–(C=O)R 20
thioester RCH2–(C=O)SR 21
ester RCH2–(C=O)OR 25
nitrile RCH2–C≡N 25
sulfone RCH2–SO2R 25
amide RCH2–(C=O)N(CH3)2 30
alkane CH3–R 50
* Note methylene groups bridging between two electron withdrawing groups are more acidic than alpha protons next to only one carbonyl group.
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.
Exercises
2. Draw the enol forms of the following molecules
1. 4-methylcyclohexanone
2. Ethyl thioactetate
3. Methyl acetate
4. Butanal
5. Propionic Acid
6. 1-phenyl-2-butanone
3. How many acid protons do each of the molecules from the previous question have? Label them.
4. Draw all of the monoenol forms for the following molecule. Which ones are most stable? Why?
Answers
2.
(a)
(b)
(c)
(d)
(e)
(f)
3.
(a)
(b)
(c)
(d)
(e)
(f)
4.
The ability to resonate stabilizes this enol form.
This enol has no resonance forms and is therefore less stable. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/23%3A_Alpha_Substitutions_and_Condensations_of_Carbonyl_Compounds/23.02%3A_Enols_Enolate_Ions_and_Tautome.txt |
Overview
The reactivity of the alpha-carbon can be grouped into three main categories: alpha-substitution, condensation, and decarboxylation. Alpha,beta-unsaturated carbonyls can undergo conjugate addition reactions that are called Michael Additions when the nucleophile is an alpha-carbon. Because the reactants, reagents and products can contain multiple functional groups, it is helpful to initially focus on the overall conversions between functional groups BEFORE digging into the details of each reaction pathway. It is also useful to label the alpha and beta carbons to help follow the reactivity. An overview of the reactions that will be studied in this chapter are shown below.
23.04: Alpha Halogenation of Carbonyl
Ketones with alpha hydrogens can undergo a substitution reaction with halogens. This reaction occurs 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 1
Aldehydes are oxidized by the halogens so this reaction pathway is not synthetically useful. For example, when benzaldehyde is added to either set of reagents described above for ketones, the majore product is benzoic acid.
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 reaction
4) Deprotonation
Kinetic studies provide some evidence for the mechanism shown above. The rate law for the alpha-halogenation of a ketone can be given by:
rate = [ketone][H+]
The implication is that the rate determining step is dependent on the concentrations of the ketone and acid catalyst and therefore associated with the enol formation part of the mechanism. The halogen does not even appear in the rate law. Indeed, the overall rate is completely independent of the concentration of the halogen and suggests the halogenation step occurs rapidly.
Base Promoted 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.
It is difficult to stop the base promoted reaction after a single substitution, so acidic conditons are used when a monohalo product is required.
1) Enolate formation
2) SN2 reaction
The Haloform Qualitative Reaction to Identify Methyl Ketones
The overreaction during base promotion of alpha halogenation is used as a qualitative test called the haloform reaction to identify methyl ketones. Under basic conditions, subsequent halogenation reactions occur because the halogenated product is more reactive than the starting material due to the electron withdrawing effect of the halogen. The halogen inductively stabilizes the conjugate base and increases the relative acidity of the remaining alpha-carbons. Halogenations occur at the alpha-carbon until the haloform becomes a leaving group and is observed as a precipitate as shown in the example below.
Deuterium Exchange
Due to the acidic nature of α hydrogens they can be exchanged with deuterium by reaction with D2O (heavy water). The process is accelerated by the addition of an acid or base; an excess of D2O is required. The end result is the complete exchange of all α hydrogens with deuteriums.
Example 2
Mechanism in basic conditions
1) Enolate Formation
2) Deuteration
Example Question
Draw the product for the following reactions.
Solutions to example question
Exercises
5. Draw the products of the following reactions
6. Draw out the mechanism for the following reaction.
7. How might you form 2-hepten-4-one from 4-heptanone?
Answer
5.
6.
7. 1) Br2, H3O+; 2) Pyridine, Heat | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/23%3A_Alpha_Substitutions_and_Condensations_of_Carbonyl_Compounds/23.03%3A_Reaction_Overview.txt |
Although the alpha bromination of some carbonyl compounds, such as aldehydes and ketones, can be accomplished with Br2 under acidic conditions, the reaction will generally not occur with acids, esters, and amides. This is because only aldehydes and ketones enolize to a sufficient extent to allow the reaction to occur. However, carboxylic acids, can be brominated in the alpha position with a mixture of Br2 and PBr3 in a reaction called the Hell-Volhard-Zelinskii (HVZ) reaction.
For example, pentanoic acid can be converted to 2-bromopentanoic acid as shown in the example below.
The mechanism of this reaction involves an acid bromide enol instead of the expected carboxylic acid enol. The reaction starts with the reaction of the carboxylic acid with PBr3 to form the acid bromide and HBr. The HBr then catalyzes the formation of the acid bromide enol which subsequently reacts with Br2 to give alpha bromination. Lastly, the acid bromide reacts with water to reform the carboxylic acid.
Exercise
8. Draw the bond-line structure for the product of each reaction below.
Answer
8.
23.06: Alkylation of the alpha-Carbon
Alpha Alkylation
A strong base, such as lithium diisopropyl amide (LDA), sodium hydride, or sodium amide, creates the nucleophilic enolate ion which reacts with an alkyl halide suitable for the SN2 reactivity to form an alpha-alkylated product.
Example 1: Alpha Alkylation
Mechanism
The mechanism begins with enolate formation. The resulting enolate is the nucleophile in an SN2 reaction with a suitable alkyl halide.
1) Enolate formation
2) SN2 reaction
Example Question
Write the structure of the product for the following reactions.
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.
Exercises
9. How might you prepare the following compounds from an alkylation reaction?
(a)
(b)
(c)
(d)
(e)
(f)
Answer
9.
(a)
(b)
(c)
(d)
(e)
(f)
23.07: Alkylation of the Alpha-Carbo
Overview of the Stork Enamine Reaction
The reaction conditions for the direct alkylation of the alpha carbon with LDA or other very strong base are quite harsh. Many organic compounds cannot withstand the reaction environment at synthetically useful amounts. Therefore, an alternate synthetic pathway was developed by Gilbert Stork of Columbia University. Some of the advantages of using an enamine over an enolate are that enamines are neutral, easier to prepare, and usually prevent the overreaction problems plagued by enolates. As shown in the example below, the aldehyde or ketone can be recovered from the enamine via a hydrolysis reaction.
Example
Reversible
Enamines act as nucleophiles in a fashion similar to enolates. Because of this, enamines can be used as synthetic equivalents as enolates in many reactions. This process requires a three steps: 1) Formation of the enamine, 2) Reaction with an eletrophile to form an iminium salt, 3) Hydrolysis of the iminium salt to reform the aldehyde or ketone.
Typically we use the following 2o amines for enamine reactions
Alkylation of an Enamine
Enamined undergo an SN2 reaction with reactive alkyl halides to give the iminium salt. The iminium salt can be hydrolyzed back into the carbonyl.
Individual steps
1) Formation of an enamine
2) SN2 Alkylation
3) Reform the carbonyl by hydrolysis
All three steps together:
Exercises
10. Draw the product of the reaction with the enamine prepared from cyclopentanone and pyrrolidine, and the following molecules.
(a)
(b)
(c)
11. Propose a synthesis for the following compounds via an enamine.
(a)
(b)
Answers
10.
(a)
(b)
(c)
11.
(a) cyclopentanone enamine + 2-cyanopropene
(b) cyclohexanone enamine + ethyl acrylate | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/23%3A_Alpha_Substitutions_and_Condensations_of_Carbonyl_Compounds/23.05%3A_Bromination_of_Acids-_The_HVZ_.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 alpha hydrogens.
The aldol reactions for acetaldehyde and acetone are shown as examples.
Example: Aldol Reactions
Aldol Reaction Mechanism
Step 1: Enolate formation
Step 2: Nucleophilic reaction by the enolate
Step 3: Protonation
Aldol Condensation: the dehydration of aldol products to synthesize α, β unsaturated carbonyls (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 driven 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:
Going from reactants to products simply
Example: Aldol Condensation from an Aldol Reaction Product
1) Form enolate
2) Form enone
Aldol Condensation Acid Catalyzed Mechanism
Under acidic conditions an enol is formed and the hydroxy group is protonated. Water is expelled by either and E1 or E2 reaction.
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: Aldol Condensation Directly from the Ketones or Aldehydes
Aldol Reactions in Multiple Step Synthesis
Aldol reactions are excellent methods for the synthesis of many enones or beta hydroxy carbonyls. Because of this, being able to predict when an aldol reaction might be used in a synthesis in an important skill. This accomplished by mentally breaking apart the target molecule and then considering what the starting materials might be.
Fragments which are easily made by an aldol reaction
Steps to 'reverse' the aldol reaction (from the final aldol product towards identifying the starting compounds).
1) From an enone break the double bond and form two single bonds. Place and OH on the bond furthest from the carbonyl and an H on the bond closest to the carbonyl.
2) From the aldol product break the C-C bond between the alpha carbon and the carbon attached to the OH. Then turn the OH into a carbonyl and add an hydrogen to the other carbon.
Example: Determining the Reactant when given the Aldol Condensation Product
What reactant must be used to make the following molecule using an aldol condensation?
Solution
Mixed Aldol Reactions 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: Mixed Aldol Reaction (One Product)
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.
ACH2CHO + BCH2CHO + NaOH AA + BB + AB + BA
Example: Products of a Mixed Aldol Reaction
The aldol condensation of ketones with aryl aldehydes to form α,β-unsaturated derivatives is called the Claisen-Schmidt reaction.
Example: Claisen-Schmidt Reaction
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 (less ring strain).
As with other aldol reaction the addition of heat causes an aldol condensation to occur.
Exercise
12. Draw the bond-line structures for the products of the reactions below. Note: One of the reactions is a poorly designed aldol condensation producing four different products.
Answer
12.
23.09: The Claisen Condensation React
Because esters can contain alpha 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
The Claisen condensation reactions of methyl acetate and methyl propanoate are shown as examples.
Example: Claisen Condensation
Claisen Condensation Mechanism
1) Enolate formation
2) Nucleophilic reaction
3) Removal of leaving group
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: Crossed Claisen Condensation
Dieckmann Condensation
A diester can undergo an intramolecular reaction called a Dieckmann condensation.
Example: Dieckman Condensation
Mechanism
1. Dieckmann, W. Ber. Dtsch. Chem. Ges. 1894, 27, 102–103.
Exercise
13. Draw the bond-line structures for the products of the following reactions.
Answer
13. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/23%3A_Alpha_Substitutions_and_Condensations_of_Carbonyl_Compounds/23.08%3A_The_Aldol_Reaction_and_Condens.txt |
1,2 (Direct) versus 1,4 (Conjugate) Addition
There are two electrophilic sites in alpha,beta-unsaturated carbonyls: the carbonyl carbon and the beta carbon. The second electrophilic site is created through resonance as shown below.
It is the strength of the nucleophile that determines the dominant reaction pathway. Strong nucleophiles like Grignard reagents and hydrides will react directly at the carbonyl carbon following the reactivity previously studied. The strong nucleophile reacts with the carbonyl carbon to produce a tetrahedral intermediate that is protonated to form an alcohol. The 1,2 direct addition reaction of prop-2-enal with a Grignard reagent is shown as an example below.
In 1,4 conjugate addition, the nucleophile reacts with the carbon β to the carbonyl driving the formation of an enolate ion that tautermerizes back to the carbonyl upon protonation.while the hydrogen is added to the carbon alpha to the carbonyl.
Mechanism for 1,4 conjugate addition
1) Nucleophilic reaction at 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
Michael Additions
Enolates are weak nucleophiles and undergo 1,4 addition to α, β-unsaturated carbonyl compounds in a process called a Michael addition. The reaction is named after American chemist Arthur Michael (1853-1942). Two examples are shown below.
In combination with alkylations and condensations, the Michael reaction may be used to construct a wide variety of complex molecules from relatively simple starting materials. The nucleophile is called the Michael Donor and the electrophile (the alpha,beta-unsaturated carbonyl) is called the Michael Acceptor. The table below shows common reagents used for Michael addition reactions.
There are several examples shown below including cyanide as another potential Michael donor. These anions are sufficiently stable that their addition reactions may be presumed reversible. If this is so, the thermodynamic argument used for hetero-nucleophile additions would apply here as well, and would indicate preferential formation of 1,4-addition products. Cyanide addition does not always follow this rule, and aldehydes often give 1,2-products (cyanohydrins). In each case the initial reaction is a Michael addition, and the new carbon-carbon bond is colored magenta. Any subsequent bonds that are formed by other reactions are colored orange.
Exercise2
14. Draw the bond-line structures for the products of the reactions below.
15. Specify the reagents needed to perform the following chemical transformation.
Answers
14.
15.
Robinson Annulations
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.
Exercise
16. Draw the product of the reaction with the enamine prepared from cyclopentanone and pyrrolidine, and the following molecules.
(a)
(b)
(c)
17. Draw the product of the following reaction.
Answer
16.
(a)
(b)
(c)
17.
23.11: Decarboxylation Reactions
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, X= 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.
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.
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
The Acetoacetic Ester Synthesis
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.
Exercise
18. Propose a synthesis for each of the following molecules from this malonic ester.
(a)
(b)
(c)
19. Why can't we prepare tri substituted acetic acids from a malonic ester?
20. Propose a synthesis for the following molecule via a malonic ester.
Answer
18.
(a) 1) Malonic Ester, NaOEt, 2) 4-Methylbenzyl Bromide, 3) Base, 4) Acid, Heat
(b) 1) Malonic Ester, NaOEt, 2) 3-bromohexane, 3) Base, 4) Acid, Eat
(c) 1) Malonic Ester, NaOEt, 2) 1-Bromo-2,3,3-trimethylbutane, 3) Base, 4) Acid, Heat
19. Malonic esters only contain two acid protons.
20. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/23%3A_Alpha_Substitutions_and_Condensations_of_Carbonyl_Compounds/23.10%3A_Conjugate_Additions-_The_Micha.txt |
General Review
23-1 For each of the following reactions, predict the product. Then identify which one gives the kinetic product and which one gives the thermodynamic product. Explain your reasoning.
23-2 Predict the final product of the following reaction.
23-3 Explain why we obtain the products shown below as a result of the base catalyzed α-halogenation reaction, instead of a single or double halogenated product.
23-4 Identify the starting molecule of the aldol addition-condensation reaction that resulted in the final product below.
23-5 The following reaction is an example of an intramolecular aldol addition-condensation reaction. Given the aldol product, identify the starting molecule and the condensation product.
23-6 Provide the final product of the following reaction (for now, ignore any stereochemistry).
23-7 Predict the final product of the following reaction.
Enols and Enolate Ions
23-8 Choose the correct IUPAC nomenclature of the product of the following reaction and provide its structure.
a) dibromo(4-methylphenyl)methanol
b) 4-methylbenzoic acid
c) 4-methylbenzoyl bromide
d) bromo(4-methylphenyl)methanol
23-9 Draw the mechanism for the following self-catalyzed halogenation reaction.
23-10 Predict the product of the following reaction.
Formation and Alkylation of Enamines
23-11 Choose the correct IUPAC nomenclature for the product of the following reaction.
a) (2Z)-N,N-diethylpent-2-en-3-amine
b) N,N-diethylpentan-3-amine
c) 3-(diethylamino)pentan-3-ol
d) Tetraethylhydrazine
23-12 Predict the major products of the following reactions.
23-13 Predict the product of the following reaction.
Alpha Halogenation of Ketones
23-14 Predict the product of the following reaction.
23-15 Draw the mechanism for the α-halogenation step of the ketone in problem 23-14.
23-16 Provide the structures of the products of the following reactions.
Alpha Bromination of Acids: The HVZ Reaction
23-17 Give the product of the following reaction.
23-18 Choose the correct IUPAC nomenclature of the product of the following reaction.
a) 1-chlorocyclohexane-1-carboxylic acid
b) 1-bromocyclohexane-1-carbonyl chloride
c) 1-chlorocyclohexane-1-carbonyl chloride
d) 1-hydroxycyclohexane-1-carbonyl bromide
23-19 Provide the structure and IUPAC nomenclature of the product of the following reaction.
The Aldol Condensation of Ketones and Aldehydes
23-20 Give the structure of the product of the following aldol condensation reaction.
23-21 Predict the structure of the product of the following reaction.
23-22 Provide the structure of the product of the following reaction.
Dehydration of Aldol Products
23-23 Choose the correct IUPAC nomenclature for the product of the following aldol condensation reaction.
a) 5-ethyl-4-methylheptan-3-one
b) 5-ethyl-4-methylhept-4-en-3-one
c) 5-ethyl-5-hydroxy-4-methylheptan-3-one
d) (5Z)-5-ethyl-4-methylhept-5-en-3-one
23-24 Provide the mechanism for the reaction to the answer for the previous question.
23-25 For the following compounds, draw a possible product of the condensation step.
Crossed Aldol Condensations
23-26 Provide the structure of the product of the following reaction.
23-27 Suggest the structures of the starting compounds that were reacted to create this final crossed aldol condensation product.
23-28 Provide the structures of all possible products of the following crossed aldol condensation reaction.
Aldol Cyclizations
23-29 Provide the IUPAC name for the product of the following intramolecular aldol condensation reaction.
23-30 Identify the starting material used to create the following products of aldol cyclization.
23-31 Starting with a single diketone molecule, propose a method of synthesis for the following compound that includes an aldol cyclization step.
Claisen Condensations
23-32 Predict the product of the following Claisen condensation reaction.
23-33 Predict the final product of the following reaction.
23-34 Provide the structure of the products of the following crossed Claisen condensation reactions.
Syntheses Using β-Dicarbonyl Compounds
23-35 Give the structure and IUPAC nomenclature of the product of the following reaction.
23-36 Choose the correct IUPAC nomenclature of the product of the following reaction.
a_ 2,2,4,4-tetramethylpentan-3-one
b) 2,2,4-trimethylpentan-3-one
c) 2,4,4-trimethylpent-1-en-3-one
d) 2,4,4-trimethyl-3-oxopentanoic acid
23-37 Suggest a way to make (5-methylhexyl)benzene from 3-hydroxy-5-methylhexanoic acid.
Conjugate Additions: The Michael Reaction
23-38 For the following pairs of compounds, identify the Michael acceptor and the Michael donor.
23-39 Predict the product of the following reaction chain.
23-40 Pick the answer that correctly names the product of the following Michael addition reaction.
a) 3-(ethylamino)butan-2-one
b) 2-(ethylamino)butan-2-ol
c) 4-(ethylamino)butan-2-one
d) 4-(methylamino)butan-2-one
The Robinson Annulation
23-41 Predict a possible product of the following reactions.
23-42 Given the following compound, predict the Michael acceptor and donor that initially reacted to allow for the aldol condensation to occur.
23-43 Given the following Robinson annulation product, identify the intermediate compound that results from the Michael addition and exists before the aldol condensation.
23.13: Solutions to Additional Exerci
General Review
23-1
When a bulky base, such as LDA, is used, it will almost always deprotonate the least hindered position. In this case, the reaction is performed under low temperatures (to prevent the thermodynamic product from forming) and we get the kinetic product. When we use a strong base that is not bulky, it will favor deprotonating the position that will provide the most stable product at equilibrium. As a result, the enolate that forms when using NaH or NaOEt is more substituted and more stable.
23-2
23-3 When acetophenone is halogenated at the α-position, the α-carbon becomes more acidic as a result of the electron-withdrawing halogen. This makes it more likely to go through α-halogenation again, until it no longer has any α-protons. When the base performs a nucleophilic attack on the ketone, the triiodomethyl group becomes a good leaving group. The resulting products are sodium benzoate and triiodomethane.
23-4
23-5
23-6
23-7
Enols and Enolate Ions
23-8:
Answer: B
23-9:
23-10:
Formation and Alkylation of Enamines
23-11:
Answer: A
23-12:
23-13:
Alpha Halogenation of Ketones
23-14:
23-15:
23-16:
Alpha Bromination of Acids: The HVZ Reaction
23-17:
23-18:
Answer: B
23-19:
The Aldol Condensation of Ketones and Aldehydes
23-20:
23-21:
23-22:
Dehydration of Aldol Products
23-23:
Answer: B
23-24:
23-25:
Crossed Aldol Condensations
23-26:
23-27:
23-28:
Aldol Cyclizations
23-29:
23-30:
23-31:
Several answers possible. One plausible method of synthesis:
Claisen Condensations
23-32:
23-33:
23-34:
Syntheses Using β-Dicarbonyl Compounds
23-35:
23-36:
Answer: B
23-37:
Possible route of synthesis:
Conjugate Additions: The Michael Reaction
23-38:
Michael acceptors are generally α,β-unsaturated ketones; however, aldehydes or acid derivatives can also function as acceptors. Michael donors are weak bases/strong nucleophiles.
23-39:
23-40
Answer: C
The Robinson Annulation
23-41:
23-42:
23-43: | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/23%3A_Alpha_Substitutions_and_Condensations_of_Carbonyl_Compounds/23.12%3A_Additional_Exercises.txt |
Objectives
After completing this section, you should be able to
1. identify carbohydrates (sugars) as being polyhydroxylated aldehydes and ketones.
2. describe, briefly, the process of photosynthesis, and identify the role played by carbohydrates as an energy source for living organisms.
Key Terms
Make certain that you can define, and use in context, the key term below.
• carbohydrate
Introduction
All carbohydrates consist of carbon, hydrogen, and oxygen atoms and are polyhydroxy aldehydes or ketones or are compounds that can be broken down to form such compounds. Examples of carbohydrates include starch, fiber, the sweet-tasting compounds called sugars, and structural materials such as cellulose. The term carbohydrate had its origin in a misinterpretation of the molecular formulas of many of these substances. For example, because its formula is C6H12O6, glucose was once thought to be a “carbon hydrate” with the structure C6·6H2O.
Example 1
Which compounds would be classified as carbohydrates?
Solution
1. This is a carbohydrate because the molecule contains an aldehyde functional group with OH groups on the other two carbon atoms.
2. This is not a carbohydrate because the molecule does not contain an aldehyde or a ketone functional group.
3. This is a carbohydrate because the molecule contains a ketone functional group with OH groups on the other two carbon atoms.
4. This is not a carbohydrate; although it has a ketone functional group, one of the other carbons atoms does not have an OH group attached.
Exercise 1
Which compounds would be classified as carbohydrates?
Green plants are capable of synthesizing glucose (C6H12O6) from carbon dioxide (CO2) and water (H2O) by using solar energy in the process known as photosynthesis:
$6CO_2 + 6H_2O + \text{2870 kJ} \rightarrow C_6H_{12}O_6 + 6O_2 \tag{${1}$}$
(The 2870 kJ comes from solar energy.) Plants can use the glucose for energy or convert it to larger carbohydrates, such as starch or cellulose. Starch provides energy for later use, perhaps as nourishment for a plant’s seeds, while cellulose is the structural material of plants. We can gather and eat the parts of a plant that store energy—seeds, roots, tubers, and fruits—and use some of that energy ourselves. Carbohydrates are also needed for the synthesis of nucleic acids and many proteins and lipids.
Animals, including humans, cannot synthesize carbohydrates from carbon dioxide and water and are therefore dependent on the plant kingdom to provide these vital compounds. We use carbohydrates not only for food (about 60%–65% by mass of the average diet) but also for clothing (cotton, linen, rayon), shelter (wood), fuel (wood), and paper (wood).
24.02: Classification of Carbohydrates
Objectives
After completing this section, you should be able to
1. classify a specific carbohydrate as being a monosaccharide, disaccharide, trisaccharide, etc., given the structure of the carbohydrate or sufficient information about its structure.
2. classify a monosaccharide according to the number of carbon atoms present and whether it contains an aldehyde or ketone group.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• aldose
• disaccharide
• ketose
• monosaccharide (simple sugar)
• polysaccharide
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.
$nCO_2 + n H_2O + Energy \rightarrow C_nH_{2n}O_n + nO_2$
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.
Carbohydrates are called saccharides or, if they are relatively small, sugars. Several classifications of carbohydrates have proven useful, and are outlined in the following table.
Complexity
Simple Carbohydrates
monosaccharides
Complex Carbohydrates
disaccharides, oligosaccharides
& polysaccharides
Size
Tetrose
C4 sugars
Pentose
C5 sugars
Hexose
C6 sugars
Heptose
C7 sugars
etc.
C=O Function
Aldose
sugars having an aldehyde function or an acetal equivalent.
Ketose
sugars having a ketone function or an acetal equivalent.
Reactivity
Reducing
sugars oxidized by Tollens' reagent (or Benedict's or Fehling's reagents).
Non-reducing
sugars not oxidized by Tollens' or other reagents. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/24%3A_Carbohydrates/24.01%3A_Introduction.txt |
Objectives
After completing this section, you should be able to
1. draw the Fischer projection of a monosaccharide, given its wedge‑and‑broken‑line structure or a molecular model.
2. draw the wedge‑and‑broken‑line structure of a monosaccharide, given its Fischer projection or a molecular model.
3. construct a molecular model of a monosaccharide, given its Fischer projection or wedge‑and‑broken‑line structure.
Key Terms
Make certain that you can define, and use in context, the key term below.
• Fischer projection
Study Notes
When studying this section, use your molecular model set to assist you in visualizing the structures of the compounds that are discussed. It is important that you be able to determine whether two apparently different Fischer projections represent two different structures or one single structure. Often the simplest way to check is to construct a molecular model corresponding to each projection formula, and then compare the two models.
The problem of drawing three-dimensional configurations on a two-dimensional surface, such as a piece of paper, has been a long-standing concern of chemists. The wedge and hatched line notations we have been using are effective, but can be troublesome when applied to compounds having many chiral centers. As part of his Nobel Prize-winning research on carbohydrates, the great German chemist Emil Fischer, devised a simple notation that is still widely used. In a Fischer projection drawing, the four bonds to a chiral carbon make a cross with the carbon atom at the intersection of the horizontal and vertical lines. The two horizontal bonds are directed toward the viewer (forward of the stereogenic carbon). The two vertical bonds are directed behind the central carbon (away from the viewer). Since this is not the usual way in which we have viewed such structures, the following diagram shows how a stereogenic carbon positioned in the common two-bonds-in-a-plane orientation ( x–C–y define the reference plane ) is rotated into the Fischer projection orientation (the far right formula). When writing Fischer projection formulas it is important to remember these conventions. Since the vertical bonds extend away from the viewer and the horizontal bonds toward the viewer, a Fischer structure may only be turned by 180º within the plane, thus maintaining this relationship. The structure must not be flipped over or rotated by 90º.
In the above diagram, if x = CO2H, y = CH3, a = H & b = OH, the resulting formula describes (R)-(–)-lactic acid. The mirror-image formula, where x = CO2H, y = CH3, a = OH & b = H, would, of course, represent (S)-(+)-lactic acid.
The Fischer Projection consists of both horizontal and vertical lines, where the horizontal lines represent the atoms that are pointed toward the viewer while the vertical line represents atoms that are pointed away from the viewer. The point of intersection between the horizontal and vertical lines represents the central carbon.
Using the Fischer projection notation, the stereoisomers of 2-methylamino-1-phenylpropanol are drawn in the following manner. Note that it is customary to set the longest carbon chain as the vertical bond assembly.
The usefulness of this notation to Fischer, in his carbohydrate studies, is evident in the following diagram. There are eight stereoisomers of 2,3,4,5-tetrahydroxypentanal, a group of compounds referred to as the aldopentoses. Since there are three chiral centers in this constitution, we should expect a maximum of 23 stereoisomers. These eight stereoisomers consist of four sets of enantiomers. If the configuration at C-4 is kept constant (R in the examples shown here), the four stereoisomers that result will be diastereomers. Fischer formulas for these isomers, which Fischer designated as the "D"-family, are shown in the diagram. Each of these compounds has an enantiomer, which is a member of the "L"-family so, as expected, there are eight stereoisomers in all. Determining whether a chiral carbon is R or S may seem difficult when using Fischer projections, but it is actually quite simple. If the lowest priority group (often a hydrogen) is on a vertical bond, the configuration is given directly from the relative positions of the three higher-ranked substituents. If the lowest priority group is on a horizontal bond, the positions of the remaining groups give the wrong answer (you are in looking at the configuration from the wrong side), so you simply reverse it.
The aldopentose structures drawn above are all diastereomers. A more selective term, epimer, is used to designate diastereomers that differ in configuration at only one chiral center. Thus, ribose and arabinose are epimers at C-2, and arabinose and lyxose are epimers at C-3. However, arabinose and xylose are not epimers, since their configurations differ at both C-2 and C-3.
How to make Fischer Projections
To make a Fischer Projection, it is easier to show through examples than through words. Lets start with the first example, turning a 3D structure of ethane into a 2D Fischer Projection.
Example 25.2.1
Start by mentally converting a 3D structure into a Dashed-Wedged Line Structure. Remember, the atoms that are pointed toward the viewer would be designated with a wedged lines and the ones pointed away from the viewer are designated with dashed lines.
Notice the red balls (atoms) in Figure A above are pointed away from the screen. These atoms will be designated with dashed lines like those in Figure B by number 2 and 6. The green balls (atoms) are pointed toward the screen. These atoms will be designated with wedged lines like those in Figure B by number 3 and 5. The blue atoms are in the plane of the screen so they are designated with straight lines.
Now that we have our Dashed- Wedged Line Structure, we can convert it to a Fischer Projection. However, before we can convert this Dashed-Wedged Line Structure into a Fischer Projection, we must first convert it to a “flat” Dashed-Wedged Line Structure. Then from there we can draw our Fischer Projection. Lets start with a more simpler example. Instead of using the ethane shown in Figure A and B, we will start with a methane. The reason being is that it allows us to only focus on one central carbon, which make things a little bit easier.
Lets start with this 3D image and work our way to a dashed-wedged image. Start by imagining yourself looking directly at the central carbon from the left side as shown in Figure C. It should look something like Figure D. Now take this Figure D and flatten it out on the surface of the paper and you should get an image of a cross.
As a reminder, the horizontal line represents atoms that are coming out of the paper and the vertical line represents atoms that are going into the paper. The cross image to the right of the arrow is a Fischer projection. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/24%3A_Carbohydrates/24.03%3A_Fischer_Projections.txt |
Objectives
After completing this section, you should be able to
1. identify a specific enantioner of a monosaccharide as being D or L, given its Fischer projection.
2. identify the limitations of the D, L system of nomenclature for carbohydrates.
3. assign an R or S configuration to each of the chiral carbon atoms present in a monosaccharide, given its Fischer projection.
4. draw the Fischer projection formula for a monosaccharide, given its systematic name, complete with the configuration of each chiral carbon atom.
5. construct a molecular model of a monosaccharide, given its systematic name, complete with the configuration of each chiral carbon atom.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• D sugar
• L sugar
Study Notes
If you find that you have forgotten the meanings of terms such as dextrorotatory and polarimeter, refer back to Section 5.3 in which the fundamentals of optical activity were introduced.
How would you set about the task of deciding whether each chiral carbon has an R or an S configuration? True, you could use molecular models, but suppose that a model set had not been available—what would you have done then?
One approach is to focus on the carbon atom of interest and sketch a three-dimensional representation of the configuration around that atom, remembering the convention used in Fischer projections: vertical lines represent bonds going into the page, and horizontal lines represent bonds coming out of the page. Thus, the configuration around carbon atom 2 in structure a can be represented as follows:
In your mind, you should be able to imagine how this molecule would look if it was rotated so that the bonds that are shown as coming out of the page are now in the plane of the page. [One possible way of doing this is to try and imagine how the molecule would look if it was viewed from a point at the bottom of the page.] What you should see in your mind is a representation similar to the one drawn below.
To determine whether the configuration about the central carbon atom is R or S, we must rotate the molecule so that the group with the lowest priority (H), is directed away from the viewer. This effect can be achieved by keeping the hydroxyl group in its present position and moving each of the other three groups one position clockwise.
The Cahn-Ingold-Prelog order of priority for the three remaining groups is OH > CHO > CH(OH)CH2OH; thus, we see that we could trace out a counterclockwise path going from the highest-priority group to the second- and third-highest, and we conclude that the central carbon atom has an S configuration.
The Configuration of Glucose
The four chiral centers in glucose indicate there may be as many as sixteen (24) stereoisomers having this constitution. These would exist as eight diastereomeric pairs of enantiomers, and the initial challenge was to determine which of the eight corresponded to glucose. This challenge was accepted and met in 1891 by the German chemist Emil Fischer. His successful negotiation of the stereochemical maze presented by the aldohexoses was a logical tour de force, and it is fitting that he received the 1902 Nobel Prize for chemistry for this accomplishment. One of the first tasks faced by Fischer was to devise a method of representing the configuration of each chiral center in an unambiguous manner. To this end, he invented a simple technique for drawing chains of chiral centers, that we now call the Fischer projection formula. Click on this link for a review.
At the time Fischer undertook the glucose project it was not possible to establish the absolute configuration of an enantiomer. Consequently, Fischer made an arbitrary choice for (+)-glucose and established a network of related aldose configurations that he called the D-family. The mirror images of these configurations were then designated the L-family of aldoses. To illustrate using present day knowledge, Fischer projection formulas and names for the D-aldose family (three to six-carbon atoms) are shown below, with the asymmetric carbon atoms (chiral centers) colored red. The last chiral center in an aldose chain (farthest from the aldehyde group) was chosen by Fischer as the D / L designator site. If the hydroxyl group in the projection formula pointed to the right, it was defined as a member of the D-family. A left directed hydroxyl group (the mirror image) then represented the L-family. Fischer's initial assignment of the D-configuration had a 50:50 chance of being right, but all his subsequent conclusions concerning the relative configurations of various aldoses were soundly based. In 1951 x-ray fluorescence studies of (+)-tartaric acid, carried out in the Netherlands by Johannes Martin Bijvoet, proved that Fischer's choice was correct.
It is important to recognize that the sign of a compound's specific rotation (an experimental number) does not correlate with its configuration (D or L). It is a simple matter to measure an optical rotation with a polarimeter. Determining an absolute configuration usually requires chemical interconversion with known compounds by stereospecific reaction paths.
24.05: Configuration of Aldoses
Objectives
After completing this section, you should be able to
1. draw the structures of all possible aldotetroses, aldopentoses, and aldohexoses, without necessarily being able to assign names to the individual compounds.
2. draw the Fischer projection of D‑glyceraldehyde, D‑ribose and D‑glucose from memory.
The four chiral centers in glucose indicate there may be as many as sixteen (24) stereoisomers having this constitution. These would exist as eight diastereomeric pairs of enantiomers, and the initial challenge was to determine which of the eight corresponded to glucose. This challenge was accepted and met in 1891 by the German chemist Emil Fischer. His successful negotiation of the stereochemical maze presented by the aldohexoses was a logical tour de force, and it is fitting that he received the 1902 Nobel Prize for chemistry for this accomplishment. One of the first tasks faced by Fischer was to devise a method of representing the configuration of each chiral center in an unambiguous manner. To this end, he invented a simple technique for drawing chains of chiral centers, that we now call the Fischer projection formula.
At the time Fischer undertook the glucose project it was not possible to establish the absolute configuration of an enantiomer. Consequently, Fischer made an arbitrary choice for (+)-glucose and established a network of related aldose configurations that he called the D-family. The mirror images of these configurations were then designated the L-family of aldoses. To illustrate using present day knowledge, Fischer projection formulas and names for the D-aldose family (three to six-carbon atoms) are shown below, with the asymmetric carbon atoms (chiral centers) colored red.
The last chiral center in an aldose chain (farthest from the aldehyde group) was chosen by Fischer as the D / L designator site. If the hydroxyl group in the projection formula pointed to the right, it was defined as a member of the D-family. A left directed hydroxyl group (the mirror image) then represented the L-family. Fischer's initial assignment of the D-configuration had a 50:50 chance of being right, but all his subsequent conclusions concerning the relative configurations of various aldoses were soundly based. In 1951 x-ray fluorescence studies of (+)-tartaric acid, carried out in the Netherlands by Johannes Martin Bijvoet (pronounced "buy foot"), proved that Fischer's choice was correct.
It is important to recognize that the sign of a compound's specific rotation (an experimental number) does not correlate with its configuration (D or L). It is a simple matter to measure an optical rotation with a polarimeter. Determining an absolute configuration usually requires chemical interconversion with known compounds by stereospecific reaction paths. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/24%3A_Carbohydrates/24.04%3A_D_and_L_Sugars.txt |
Objectives
After completing this section, you should be able to
1. determine whether a given monosaccharide will exist as a pyranose or furanose.
2. draw the cyclic pyranose form of a monosaccharide, given its Fischer projection.
3. draw the Fischer projection of a monosaccharide, given its cyclic pyranose form.
4. draw, from memory, the cyclic pyranose form of D‑glucose.
5. determine whether a given cyclic pyranose form represents the D or L form of the monosaccharide concerned.
6. describe the phenomenon known as mutarotation.
7. explain, through the use of chemical equations, exactly what happens at the molecular level during the mutarotation process.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• alpha anomer
• anomer
• anomeric centre
• beta anomer
• furanose
• mutarotation
• pyranose
Study Notes
If necessary, before you attempt to study this section, review the formation of hemiacetals discussed in Section 19.10.
As noted above, the preferred structural form of many monosaccharides may be that of a cyclic hemiacetal. Five and six-membered rings are favored over other ring sizes because of their low angle and eclipsing strain. Cyclic structures of this kind are termed furanose (five-membered) or pyranose (six-membered), reflecting the ring size relationship to the common heterocyclic compounds furan and pyran shown on the right. Ribose, an important aldopentose, commonly adopts a furanose structure, as shown in the following illustration. By convention for the D-family, the five-membered furanose ring is drawn in an edgewise projection with the ring oxygen positioned away from the viewer. The anomeric carbon atom (colored red here) is placed on the right. The upper bond to this carbon is defined as beta, the lower bond then is alpha.
The cyclic pyranose forms of various monosaccharides are often drawn in a flat projection known as a Haworth formula, after the British chemist, Norman Haworth. As with the furanose ring, the anomeric carbon is placed on the right with the ring oxygen to the back of the edgewise view. In the D-family, the alpha and beta bonds have the same orientation defined for the furanose ring (beta is up & alpha is down). These Haworth formulas are convenient for displaying stereochemical relationships, but do not represent the true shape of the molecules. We know that these molecules are actually puckered in a fashion we call a chair conformation. Examples of four typical pyranose structures are shown below, both as Haworth projections and as the more representative chair conformers. The anomeric carbons are colored red.
The size of the cyclic hemiacetal ring adopted by a given sugar is not constant, but may vary with substituents and other structural features. Aldolhexoses usually form pyranose rings and their pentose homologs tend to prefer the furanose form, but there are many counter examples. The formation of acetal derivatives illustrates how subtle changes may alter this selectivity. A pyranose structure for D-glucose is drawn in the rose-shaded box on the left. Acetal derivatives have been prepared by acid-catalyzed reactions with benzaldehyde and acetone. As a rule, benzaldehyde forms six-membered cyclic acetals, whereas acetone prefers to form five-membered acetals. The top equation shows the formation and some reactions of the 4,6-O-benzylidene acetal, a commonly employed protective group. A methyl glycoside derivative of this compound (see below) leaves the C-2 and C-3 hydroxyl groups exposed to reactions such as the periodic acid cleavage, shown as the last step. The formation of an isopropylidene acetal at C-1 and C-2, center structure, leaves the C-3 hydroxyl as the only unprotected function. Selective oxidation to a ketone is then possible. Finally, direct di-O-isopropylidene derivatization of glucose by reaction with excess acetone results in a change to a furanose structure in which the C-3 hydroxyl is again unprotected. However, the same reaction with D-galactose, shown in the blue-shaded box, produces a pyranose product in which the C-6 hydroxyl is unprotected. Both derivatives do not react with Tollens' reagent. This difference in behavior is attributed to the cis-orientation of the C-3 and C-4 hydroxyl groups in galactose, which permits formation of a less strained five-membered cyclic acetal, compared with the trans-C-3 and C-4 hydroxyl groups in glucose. Derivatizations of this kind permit selective reactions to be conducted at different locations in these highly functionalized molecules.
Anomers of Simple Sugars: Mutarotation of Glucose
When a straight-chain monosaccharide, such as any of the structures shown in Figure 1, forms a cyclic structure, the carbonyl oxygen atom may be pushed either up or down, giving rise to two stereoisomers, as shown in Figure 2. The structure shown on the left side of Figure 2, with the OH group on the first carbon atom projected downward, represent what is called the alpha (α) form. The structures on the right side, with the OH group on the first carbon atom pointed upward, is the beta (β) form. These two stereoisomers of a cyclic monosaccharide are known as anomers; they differ in structure around the anomeric carbon—that is, the carbon atom that was the carbonyl carbon atom in the straight-chain form.
It is possible to obtain a sample of crystalline glucose in which all the molecules have the α structure or all have the β structure. The α form melts at 146°C and has a specific rotation of +112°, while the β form melts at 150°C and has a specific rotation of +18.7°. When the sample is dissolved in water, however, a mixture is soon produced containing both anomers as well as the straight-chain form, in dynamic equilibrium (part (a) of Figure 2). You can start with a pure crystalline sample of glucose consisting entirely of either anomer, but as soon as the molecules dissolve in water, they open to form the carbonyl group and then reclose to form either the α or the β anomer. The opening and closing repeats continuously in an ongoing interconversion between anomeric forms and is referred to as mutarotation (Latin mutare, meaning “to change”). At equilibrium, the mixture consists of about 36% α-D-glucose, 64% β-D-glucose, and less than 0.02% of the open-chain aldehyde form. The observed rotation of this solution is +52.7°.
Even though only a small percentage of the molecules are in the open-chain aldehyde form at any time, the solution will nevertheless exhibit the characteristic reactions of an aldehyde. As the small amount of free aldehyde is used up in a reaction, there is a shift in the equilibrium to yield more aldehyde. Thus, all the molecules may eventually react, even though very little free aldehyde is present at a time.
Commonly, (e.g., in Figures 1 and 2) the cyclic forms of sugars are depicted using a convention first suggested by Walter N. Haworth, an English chemist. The molecules are drawn as planar hexagons with a darkened edge representing the side facing toward the viewer. The structure is simplified to show only the functional groups attached to the carbon atoms. Any group written to the right in a Fischer projection appears below the plane of the ring in a Haworth projection, and any group written to the left in a Fischer projection appears above the plane in a Haworth projection.
The difference between the α and the β forms of sugars may seem trivial, but such structural differences are often crucial in biochemical reactions. This explains why we can get energy from the starch in potatoes and other plants but not from cellulose, even though both starch and cellulose are polysaccharides composed of glucose molecules linked together.
Summary
Monosaccharides that contain five or more carbons atoms form cyclic structures in aqueous solution. Two cyclic stereoisomers can form from each straight-chain monosaccharide; these are known as anomers. In an aqueous solution, an equilibrium mixture forms between the two anomers and the straight-chain structure of a monosaccharide in a process known as mutarotation.
Exercises
1. Draw the cyclic structure for β-D-glucose. Identify the anomeric carbon.
2. Given that the aldohexose D-mannose differs from D-glucose only in the configuration at the second carbon atom, draw the cyclic structure for α-D-mannose.
2. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/24%3A_Carbohydrates/24.06%3A_Cyclic_Structures_of_Monosaccharides.txt |
Objectives
After completing this section, you should be able to
1. write equations to illustrate that the hydroxyl groups of carbohydrates can react to form esters and ethers.
2. identify the product formed when a given monosaccharide is reacted with acetic anhydride or with silver oxide and an alkyl halide.
3. identify the reagents required to convert a given monosaccharide to its ester or ether.
1. write an equation to show how a monosaccharide can be converted to a glycoside using an alcohol and an acid catalyst.
2. identify the product formed when a given monosaccharide is treated with an alcohol and an acid catalyst.
3. write a detailed mechanism for the formation of a glycoside by the reaction of the cyclic form of a monosaccharide with an alcohol and an acid catalyst.
1. identify the ester formed by phosphorylation in biologically important compounds.
1. identify the product formed when a given monosaccharide is reduced with sodium borohydride.
2. identify the monosaccharide which should be reduced in order to form a given polyalcohol (alditol).
1. explain that a sugar with an aldehyde or hemiacetal can be oxidized to the corresponding carboxylic acid (also known as aldonic acid). Note: The sugar is able to reduce an oxidizing agent, and is thus called a reducing sugar. Tests for reducing sugars include the use of Tollens’ reagent, Fehling’s reagent and Benedict’s reagent.
2. explain why certain ketoses, such as fructose, behave as reducing sugars even though they do not contain an aldehyde group.
3. identify warm HNO3 as the reagent needed to form dicarboxylic acid (an aldaric acid).
1. describe the chain‑lengthening effect of the Kiliani‑Fischer synthesis.
2. predict the product that would be produced by the Kiliani‑Fischer synthesis of a given aldose.
3. identify the aldose that would yield a given product following Kiliani‑Fischer synthesis.
1. describe the chain‑shortening effect of the Wohl degradation.
2. predict the product that would be produced by the Wohl degradation of a given aldose.
3. identify the aldose or aldoses that would yield a given product following Wohl degradation.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• aldaric acid
• aldonic acid
• alditol
• aldonic acid
• glycoside
• Kiliani-Fischer synthesis
• neighbouring group effect
• reducing sugar
• Wohl degradation
Study Notes
While several reactions are covered in this section, keep in mind that you have encountered them in previous sections. The active functional groups on monosaccharides are essentially carbonyls and hydroxyls. Although they now are a part of much larger molecules, their chemistry should be familiar.
The formation of esters and ethers is quite straightforward and should not require further clarification.
Note that glycosides are in fact acetals, and that glycoside formation is therefore analogous to acetal formation. To refresh your memory about the chemistry of acetals, quickly review Section 19.10
Ester and Ether Formation
The -OH groups on a monosaccharide can be readily converted to esters and ethers. Esterfication can be done with an acid chloride (Section 21.4) or acid anhydride (Section 21.5), while treatment with an alkyl halide by a Williamson ether synthesis (Section 18.2) leads to the ether.
Glycoside Formation
Acetal derivatives formed when a monosaccharide reacts with an alcohol in the presence of an acid catalyst are called glycosides. This reaction is illustrated for glucose and methanol in the diagram below. In naming of glycosides, the "ose" suffix of the sugar name is replaced by "oside", and the alcohol group name is placed first. As is generally true for most acetals, glycoside formation involves the loss of an equivalent of water. The diether product is stable to base and alkaline oxidants such as Tollen's reagent. Since acid-catalyzed aldolization is reversible, glycosides may be hydrolyzed back to their alcohol and sugar components by aqueous acid.
The anomeric methyl glucosides are formed in an equilibrium ratio of 66% alpha to 34% beta. From the structures in the previous diagram, we see that pyranose rings prefer chair conformations in which the largest number of substituents are equatorial. In the case of glucose, the substituents on the beta-anomer are all equatorial, whereas the C-1 substituent in the alpha-anomer changes to axial. Since substituents on cyclohexane rings prefer an equatorial location over axial (methoxycyclohexane is 75% equatorial), the preference for alpha-glycopyranoside formation is unexpected, and is referred to as the anomeric effect.
Glycosides abound in biological systems. By attaching a sugar moiety to a lipid or benzenoid structure, the solubility and other properties of the compound may be changed substantially. Because of the important modifying influence of such derivatization, numerous enzyme systems, known as glycosidases, have evolved for the attachment and removal of sugars from alcohols, phenols and amines. Chemists refer to the sugar component of natural glycosides as the glycon and the alcohol component as the aglycon.
Two examples of naturally occurring glycosides and one example of an amino derivative are displayed above. Salicin, one of the oldest herbal remedies known, was the model for the synthetic analgesic aspirin. A large class of hydroxylated, aromatic oxonium cations called anthocyanins provide the red, purple and blue colors of many flowers, fruits and some vegetables. Peonin is one example of this class of natural pigments, which exhibit a pronounced pH color dependence. The oxonium moiety is only stable in acidic environments, and the color changes or disappears when base is added. The complex changes that occur when wine is fermented and stored are in part associated with glycosides of anthocyanins. Finally, amino derivatives of ribose, such as cytidine play important roles in biological phosphorylating agents, coenzymes and information transport and storage materials.
Biological Ester Formation: Phosphorylation
Recall that almost all biomolecules are charged species, which 1) keeps them water soluble, and 2) prevents them from diffusing across lipid bilayer membranes. Although many biomolecules are ionized by virtue of negatively charged carboxylate and positively charged amino groups, the most common ionic group in biologically important organic compounds is phosphate - thus the phosphorylation of alcohol groups is a critical metabolic step. In alcohol phosphorylations, ATP is almost always the phosphate donor, and the mechanism is very consistent: the alcohol oxygen acts as a nucleophile, attacking the gamma-phosphorus of ATP and expelling ADP (look again, for example, at the glucose kinase reaction that we first saw in section 10.1D).
Oxidation
As noted above, sugars may be classified as reducing or non-reducing based on their reactivity with Tollens', Benedict's or Fehling's reagents. If a sugar is oxidized by these reagents it is called reducing, since the oxidant (Ag(+) or Cu(+2)) is reduced in the reaction, as evidenced by formation of a silver mirror or precipitation of cuprous oxide. The Tollens' test is commonly used to detect aldehyde functions; and because of the facile interconversion of ketoses and aldoses under the basic conditions of this test, ketoses such as fructose also react and are classified as reducing sugars.
When the aldehyde function of an aldose is oxidized to a carboxylic acid the product is called an aldonic acid. Because of the 2º hydroxyl functions that are also present in these compounds, a mild oxidizing agent such as hypobromite must be used for this conversion (equation 1). If both ends of an aldose chain are oxidized to carboxylic acids the product is called an aldaric acid. By converting an aldose to its corresponding aldaric acid derivative, the ends of the chain become identical (this could also be accomplished by reducing the aldehyde to CH2OH, as noted below). Such an operation will disclose any latent symmetry in the remaining molecule. Thus, ribose, xylose, allose and galactose yield achiral aldaric acids which are, of course, not optically active. The ribose oxidation is shown in equation 2 below.
1.
2.
3.
Other aldose sugars may give identical chiral aldaric acid products, implying a unique configurational relationship. The examples of arabinose and lyxose shown in equation 3 above illustrate this result. Remember, a Fischer projection formula may be rotated by 180º in the plane of projection without changing its configuration.
Reduction
Sodium borohydride reduction of an aldose makes the ends of the resulting alditol chain identical, HOCH2(CHOH)nCH2OH, thereby accomplishing the same configurational change produced by oxidation to an aldaric acid. Thus, allitol and galactitol from reduction of allose and galactose are achiral, and altrose and talose are reduced to the same chiral alditol. A summary of these redox reactions, and derivative nomenclature is given in the following table.
Table: Derivatives of \(HOCH_2(CHOH)_nCHO\)
\(HOBr\) Oxidation
\(\longrightarrow\)
\(HOCH_2(CHOH)_nCO_2H\)
an Aldonic Acid
\(HNO_3\) Oxidation
\(\longrightarrow\)
\(H_2OC(CHOH)_nCO_2H\)
an Aldaric Acid
\(NaBH_4\) Reduction
\(\longrightarrow\)
\(HOCH_2(CHOH)_nCH_2OH\)
an Alditol
Chain Shortening and Lengthening
1. Ruff Degradation
2. Kiliani-Fischer Synthesis
These two procedures permit an aldose of a given size to be related to homologous smaller and larger aldoses. The importance of these relationships may be seen in the array of aldose structures presented earlier, where the structural connections are given by the dashed blue lines. Thus Ruff degradation of the pentose arabinose gives the tetrose erythrose. Working in the opposite direction, a Kiliani-Fischer synthesis applied to arabinose gives a mixture of glucose and mannose. An alternative chain shortening procedure known as the Wohl degradation is essentially the reverse of the Kiliani-Fischer synthesis.
Note that in the Kiliani-Fischer synthesis the first step is to generate a cyanohydrin intermediate, which is then has its nitrile group hydrolyzed to the carboxylic acid. From there the cyclic ester (lactone) is formed and reduced to the final products (epimers).
Wohl Degradation
The ability to shorten (degrade) an aldose chain by one carbon was an important tool in the structure elucidation of carbohydrates. This was commonly accomplished by the Ruff procedure. An interesting alternative technique, known as the Wohl degradation has also been used. The following equation illustrates the application of this procedure to the aldopentose, arabinose. Based on your knowledge of carbonyl chemistry, and considering that the Wohl degradation is in essence the reverse of the Kiliani-Fischer synthesis. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/24%3A_Carbohydrates/24.07%3A_Reactions_of_Monosaccharides.txt |
Objectives
After completing this section, you should be able to
1. identify disaccharides as compounds consisting of two monosaccharide units joined by a glycoside link between the C1 of one sugar and one of the hydroxyl groups of a second sugar.
2. identify the two monosaccharide units in a given disaccharide.
3. identify the type of glycoside link (e.g., 1,4′‑β) present in a given disaccharide structure.
4. draw the structure of a specific disaccharide, given the structure of the monosaccharide units and the type of glycoside link involved.
Note: If α‑ or β‑D‑glucose were one of the monosaccharide units, its structure would not be provided.
5. identify the structural feature that determines whether or not a given disaccharide behaves as a reducing sugar and undergoes mutarotation, and write equations to illustrate these phenomena.
6. identify the products formed from the hydrolysis of a given disaccharide.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• 1,4′ link
• disaccharide (see Section 25.1)
• invert sugar
Study Notes
Notice that most of the disaccharides discussed in this section contain one unit of D-glucose. You are not expected to remember the detailed structures of maltose, lactose and sucrose. Similarly, we do not expect you to remember the systematic names of these substances.
Previously, you learned that monosaccharides can form cyclic structures by the reaction of the carbonyl group with an OH group. These cyclic molecules can in turn react with another alcohol. Disaccharides (C12H22O11) are sugars composed of two monosaccharide units that are joined by a carbon–oxygen-carbon linkage known as a glycosidic linkage. This linkage is formed from the reaction of the anomeric carbon of one cyclic monosaccharide with the OH group of a second monosaccharide.
The disaccharides differ from one another in their monosaccharide constituents and in the specific type of glycosidic linkage connecting them. There are three common disaccharides: maltose, lactose, and sucrose. All three are white crystalline solids at room temperature and are soluble in water. We’ll consider each sugar in more detail.
Maltose
Maltose occurs to a limited extent in sprouting grain. It is formed most often by the partial hydrolysis of starch and glycogen. In the manufacture of beer, maltose is liberated by the action of malt (germinating barley) on starch; for this reason, it is often referred to as malt sugar. Maltose is about 30% as sweet as sucrose. The human body is unable to metabolize maltose or any other disaccharide directly from the diet because the molecules are too large to pass through the cell membranes of the intestinal wall. Therefore, an ingested disaccharide must first be broken down by hydrolysis into its two constituent monosaccharide units. In the body, such hydrolysis reactions are catalyzed by enzymes such as maltase. The same reactions can be carried out in the laboratory with dilute acid as a catalyst, although in that case the rate is much slower, and high temperatures are required. Whether it occurs in the body or a glass beaker, the hydrolysis of maltose produces two molecules of D-glucose.
$\mathrm{maltose \xrightarrow{H^+\: or\: maltase} \textrm{2 D-glucose}}$
Maltose is a reducing sugar. Thus, its two glucose molecules must be linked in such a way as to leave one anomeric carbon that can open to form an aldehyde group. The glucose units in maltose are joined in a head-to-tail fashion through an α-linkage from the first carbon atom of one glucose molecule to the fourth carbon atom of the second glucose molecule (that is, an α-1,4-glycosidic linkage; see Figure 1). The bond from the anomeric carbon of the first monosaccharide unit is directed downward, which is why this is known as an α-glycosidic linkage. The OH group on the anomeric carbon of the second glucose can be in either the α or the β position, as shown in Figure 1.
• Lactose
Lactose is known as milk sugar because it occurs in the milk of humans, cows, and other mammals. In fact, the natural synthesis of lactose occurs only in mammary tissue, whereas most other carbohydrates are plant products. Human milk contains about 7.5% lactose, and cow’s milk contains about 4.5%. This sugar is one of the lowest ranking in terms of sweetness, being about one-sixth as sweet as sucrose. Lactose is produced commercially from whey, a by-product in the manufacture of cheese. It is important as an infant food and in the production of penicillin.
Lactose is a reducing sugar composed of one molecule of D-galactose and one molecule of D-glucose joined by a β-1,4-glycosidic bond (the bond from the anomeric carbon of the first monosaccharide unit being directed upward). The two monosaccharides are obtained from lactose by acid hydrolysis or the catalytic action of the enzyme lactase:
Many adults and some children suffer from a deficiency of lactase. These individuals are said to be lactose intolerant because they cannot digest the lactose found in milk. A more serious problem is the genetic disease galactosemia, which results from the absence of an enzyme needed to convert galactose to glucose. Certain bacteria can metabolize lactose, forming lactic acid as one of the products. This reaction is responsible for the “souring” of milk.
Example 1
For this trisaccharide, indicate whether each glycosidic linkage is α or β.
Solution
The glycosidic linkage between sugars 1 and 2 is β because the bond is directed up from the anomeric carbon. The glycosidic linkage between sugars 2 and 3 is α because the bond is directed down from the anomeric carbon.
To Your Health: Lactose Intolerance and Galactosemia
Lactose makes up about 40% of an infant’s diet during the first year of life. Infants and small children have one form of the enzyme lactase in their small intestines and can digest the sugar easily; however, adults usually have a less active form of the enzyme, and about 70% of the world’s adult population has some deficiency in its production. As a result, many adults experience a reduction in the ability to hydrolyze lactose to galactose and glucose in their small intestine. For some people the inability to synthesize sufficient enzyme increases with age. Up to 20% of the US population suffers some degree of lactose intolerance.
In people with lactose intolerance, some of the unhydrolyzed lactose passes into the colon, where it tends to draw water from the interstitial fluid into the intestinal lumen by osmosis. At the same time, intestinal bacteria may act on the lactose to produce organic acids and gases. The buildup of water and bacterial decay products leads to abdominal distention, cramps, and diarrhea, which are symptoms of the condition.
The symptoms disappear if milk or other sources of lactose are excluded from the diet or consumed only sparingly. Alternatively, many food stores now carry special brands of milk that have been pretreated with lactase to hydrolyze the lactose. Cooking or fermenting milk causes at least partial hydrolysis of the lactose, so some people with lactose intolerance are still able to enjoy cheese, yogurt, or cooked foods containing milk. The most common treatment for lactose intolerance, however, is the use of lactase preparations (e.g., Lactaid), which are available in liquid and tablet form at drugstores and grocery stores. These are taken orally with dairy foods—or may be added to them directly—to assist in their digestion.
Galactosemia is a condition in which one of the enzymes needed to convert galactose to glucose is missing. Consequently, the blood galactose level is markedly elevated, and galactose is found in the urine. An infant with galactosemia experiences a lack of appetite, weight loss, diarrhea, and jaundice. The disease may result in impaired liver function, cataracts, mental retardation, and even death. If galactosemia is recognized in early infancy, its effects can be prevented by the exclusion of milk and all other sources of galactose from the diet. As a child with galactosemia grows older, he or she usually develops an alternate pathway for metabolizing galactose, so the need to restrict milk is not permanent. The incidence of galactosemia in the United States is 1 in every 65,000 newborn babies.
Sucrose
Sucrose, probably the largest-selling pure organic compound in the world, is known as beet sugar, cane sugar, table sugar, or simply sugar. Most of the sucrose sold commercially is obtained from sugar cane and sugar beets (whose juices are 14%–20% sucrose) by evaporation of the water and recrystallization. The dark brown liquid that remains after the recrystallization of sugar is sold as molasses.
The sucrose molecule is unique among the common disaccharides in having an α-1,β-2-glycosidic (head-to-head) linkage. Because this glycosidic linkage is formed by the OH group on the anomeric carbon of α-D-glucose and the OH group on the anomeric carbon of β-D-fructose, it ties up the anomeric carbons of both glucose and fructose.
This linkage gives sucrose certain properties that are quite different from those of maltose and lactose. As long as the sucrose molecule remains intact, neither monosaccharide “uncyclizes” to form an open-chain structure. Thus, sucrose is incapable of mutarotation and exists in only one form both in the solid state and in solution. In addition, sucrose does not undergo reactions that are typical of aldehydes and ketones. Therefore, sucrose is a nonreducing sugar.
The hydrolysis of sucrose in dilute acid or through the action of the enzyme sucrase (also known as invertase) gives an equimolar mixture of glucose and fructose. This 1:1 mixture is referred to as invert sugar because it rotates plane-polarized light in the opposite direction than sucrose. The hydrolysis reaction has several practical applications. Sucrose readily recrystallizes from a solution, but invert sugar has a much greater tendency to remain in solution. In the manufacture of jelly and candy and in the canning of fruit, the recrystallization of sugar is undesirable. Therefore, conditions leading to the hydrolysis of sucrose are employed in these processes. Moreover, because fructose is sweeter than sucrose, the hydrolysis adds to the sweetening effect. Bees carry out this reaction when they make honey.
The average American consumes more than 100 lb of sucrose every year. About two-thirds of this amount is ingested in soft drinks, presweetened cereals, and other highly processed foods. The widespread use of sucrose is a contributing factor to obesity and tooth decay. Carbohydrates such as sucrose, are converted to fat when the caloric intake exceeds the body’s requirements, and sucrose causes tooth decay by promoting the formation of plaque that sticks to teeth.
Summary
Maltose is composed of two molecules of glucose joined by an α-1,4-glycosidic linkage. It is a reducing sugar that is found in sprouting grain. Lactose is composed of a molecule of galactose joined to a molecule of glucose by a β-1,4-glycosidic linkage. It is a reducing sugar that is found in milk. Sucrose is composed of a molecule of glucose joined to a molecule of fructose by an α-1,β-2-glycosidic linkage. It is a nonreducing sugar that is found in sugar cane and sugar beets.
Concept Review Exercise
1. What monosaccharides are obtained by the hydrolysis of each disaccharide?
1. sucrose
2. maltose
3. lactose
Answer
1. D-glucose and D-fructose
2. two molecules of D-glucose
3. D-glucose and D-galactose
Exercises
1. Identify each sugar by its common chemical name.
1. milk sugar
2. table sugar
2. For each disaccharide, indicate whether the glycosidic linkage is α or β.
3. Identify each disaccharide in Exercise 2 as a reducing or nonreducing sugar. If it is a reducing sugar, draw its structure and circle the anomeric carbon. State if the OH group at the anomeric carbon is in the α or the β position
4. Melibiose is a disaccharide that occurs in some plant juices. Its structure is as follows:
1. What monosaccharide units are incorporated into melibiose?
2. What type of linkage (α or β) joins the two monosaccharide units of melibiose?
3. Melibiose has a free anomeric carbon and is thus a reducing sugar. Circle the anomeric carbon and indicate whether the OH group is α or β
Answers
1. lactose
2. sucrose
2. a.
2. b.
3. a. nonreducing
3. b. reducing
4. a. galactose and glucose
4. b.α-glycosidic linkage
4. c. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/24%3A_Carbohydrates/24.08%3A_Disaccharides_and_Glycosidic_Bonds.txt |
Objectives
After completing this section, you should be able to
1. identify the structural difference between cellulose and the cold‑water‑insoluble fraction of starch (amylose), and identify both of these substances as containing many glucose molecules joined by 1,4′‑glycoside links.
2. identify the cold‑water‑soluble fraction of starch (amylopectin) as having a more complex structure than amylose because of the existence of 1,6′‑glycoside links in addition to the 1,4′‑links.
3. compare and contrast the structures and uses of starch, glycogen and cellulose.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• amylopectin
• amylose
• polysaccharide
Learning Objectives
• To compare and contrast the structures and uses of starch, glycogen, and cellulose.
The polysaccharides are the most abundant carbohydrates in nature and serve a variety of functions, such as energy storage or as components of plant cell walls. Polysaccharides are very large polymers composed of tens to thousands of monosaccharides joined together by glycosidic linkages. The three most abundant polysaccharides are starch, glycogen, and cellulose. These three are referred to as homopolymers because each yields only one type of monosaccharide (glucose) after complete hydrolysis. Heteropolymers may contain sugar acids, amino sugars, or noncarbohydrate substances in addition to monosaccharides. Heteropolymers are common in nature (gums, pectins, and other substances) but will not be discussed further in this textbook. The polysaccharides are nonreducing carbohydrates, are not sweet tasting, and do not undergo mutarotation.
Starch
Starch is the most important source of carbohydrates in the human diet and accounts for more than 50% of our carbohydrate intake. It occurs in plants in the form of granules, and these are particularly abundant in seeds (especially the cereal grains) and tubers, where they serve as a storage form of carbohydrates. The breakdown of starch to glucose nourishes the plant during periods of reduced photosynthetic activity. We often think of potatoes as a “starchy” food, yet other plants contain a much greater percentage of starch (potatoes 15%, wheat 55%, corn 65%, and rice 75%). Commercial starch is a white powder.
Starch is a mixture of two polymers: amylose and amylopectin. Natural starches consist of about 10%–30% amylose and 70%–90% amylopectin. Amylose is a linear polysaccharide composed entirely of D-glucose units joined by the α-1,4-glycosidic linkages we saw in maltose (part (a) of Figure \(1\)). Experimental evidence indicates that amylose is not a straight chain of glucose units but instead is coiled like a spring, with six glucose monomers per turn (part (b) of Figure \(1\)). When coiled in this fashion, amylose has just enough room in its core to accommodate an iodine molecule. The characteristic blue-violet color that appears when starch is treated with iodine is due to the formation of the amylose-iodine complex. This color test is sensitive enough to detect even minute amounts of starch in solution.
Amylopectin is a branched-chain polysaccharide composed of glucose units linked primarily by α-1,4-glycosidic bonds but with occasional α-1,6-glycosidic bonds, which are responsible for the branching. A molecule of amylopectin may contain many thousands of glucose units with branch points occurring about every 25–30 units (Figure \(2\)). The helical structure of amylopectin is disrupted by the branching of the chain, so instead of the deep blue-violet color amylose gives with iodine, amylopectin produces a less intense reddish brown.
Dextrins are glucose polysaccharides of intermediate size. The shine and stiffness imparted to clothing by starch are due to the presence of dextrins formed when clothing is ironed. Because of their characteristic stickiness with wetting, dextrins are used as adhesives on stamps, envelopes, and labels; as binders to hold pills and tablets together; and as pastes. Dextrins are more easily digested than starch and are therefore used extensively in the commercial preparation of infant foods.
The complete hydrolysis of starch yields, in successive stages, glucose:
starch → dextrins → maltose → glucose
In the human body, several enzymes known collectively as amylases degrade starch sequentially into usable glucose units.
Glycogen
Glycogen is the energy reserve carbohydrate of animals. Practically all mammalian cells contain some stored carbohydrates in the form of glycogen, but it is especially abundant in the liver (4%–8% by weight of tissue) and in skeletal muscle cells (0.5%–1.0%). Like starch in plants, glycogen is found as granules in liver and muscle cells. When fasting, animals draw on these glycogen reserves during the first day without food to obtain the glucose needed to maintain metabolic balance.
Glycogen is structurally quite similar to amylopectin, although glycogen is more highly branched (8–12 glucose units between branches) and the branches are shorter. When treated with iodine, glycogen gives a reddish brown color. Glycogen can be broken down into its D-glucose subunits by acid hydrolysis or by the same enzymes that catalyze the breakdown of starch. In animals, the enzyme phosphorylase catalyzes the breakdown of glycogen to phosphate esters of glucose.
About 70% of the total glycogen in the body is stored in muscle cells. Although the percentage of glycogen (by weight) is higher in the liver, the much greater mass of skeletal muscle stores a greater total amount of glycogen.
Cellulose
Cellulose, a fibrous carbohydrate found in all plants, is the structural component of plant cell walls. Because the earth is covered with vegetation, cellulose is the most abundant of all carbohydrates, accounting for over 50% of all the carbon found in the vegetable kingdom. Cotton fibrils and filter paper are almost entirely cellulose (about 95%), wood is about 50% cellulose, and the dry weight of leaves is about 10%–20% cellulose. The largest use of cellulose is in the manufacture of paper and paper products. Although the use of noncellulose synthetic fibers is increasing, rayon (made from cellulose) and cotton still account for over 70% of textile production.
Like amylose, cellulose is a linear polymer of glucose. It differs, however, in that the glucose units are joined by β-1,4-glycosidic linkages, producing a more extended structure than amylose (part (a) of Figure \(3\)). This extreme linearity allows a great deal of hydrogen bonding between OH groups on adjacent chains, causing them to pack closely into fibers (part (b) of Figure \(3\)). As a result, cellulose exhibits little interaction with water or any other solvent. Cotton and wood, for example, are completely insoluble in water and have considerable mechanical strength. Because cellulose does not have a helical structure, it does not bind to iodine to form a colored product.
Cellulose yields D-glucose after complete acid hydrolysis, yet humans are unable to metabolize cellulose as a source of glucose. Our digestive juices lack enzymes that can hydrolyze the β-glycosidic linkages found in cellulose, so although we can eat potatoes, we cannot eat grass. However, certain microorganisms can digest cellulose because they make the enzyme cellulase, which catalyzes the hydrolysis of cellulose. The presence of these microorganisms in the digestive tracts of herbivorous animals (such as cows, horses, and sheep) allows these animals to degrade the cellulose from plant material into glucose for energy. Termites also contain cellulase-secreting microorganisms and thus can subsist on a wood diet. This example once again demonstrates the extreme stereospecificity of biochemical processes.
Career Focus: Certified Diabetes Educator
Certified diabetes educators come from a variety of health professions, such as nursing and dietetics, and specialize in the education and treatment of patients with diabetes. A diabetes educator will work with patients to manage their diabetes. This involves teaching the patient to monitor blood sugar levels, make good food choices, develop and maintain an exercise program, and take medication, if required.
Summary
Starch is a storage form of energy in plants. It contains two polymers composed of glucose units: amylose (linear) and amylopectin (branched). Glycogen is a storage form of energy in animals. It is a branched polymer composed of glucose units. It is more highly branched than amylopectin. Cellulose is a structural polymer of glucose units found in plants. It is a linear polymer with the glucose units linked through β-1,4-glycosidic bonds. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/24%3A_Carbohydrates/24.09%3A_Polysaccharides.txt |
Objectives
After completing this section, you should be able to identify deoxy and amino sugars, given their structures.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• amino sugar
• deoxy sugar
The sugars in the backbone of DNA
The backbone of DNA is based on a repeated pattern of a sugar group and a phosphate group. The full name of DNA, deoxyribonucleic acid, gives you the name of the sugar present - deoxyribose. Deoxyribose is a modified form of another sugar called ribose. I'm going to give you the structure of that first, because you will need it later anyway. Ribose is the sugar in the backbone of RNA, ribonucleic acid.
This diagram misses out the carbon atoms in the ring for clarity. Each of the four corners where there isn't an atom shown has a carbon atom. The heavier lines are coming out of the screen or paper towards you. In other words, you are looking at the molecule from a bit above the plane of the ring.
So that's ribose. Deoxyribose, as the name might suggest, is ribose which has lost an oxygen atom - "de-oxy".
The only other thing you need to know about deoxyribose (or ribose, for that matter) is how the carbon atoms in the ring are numbered. The carbon atom to the right of the oxygen as we have drawn the ring is given the number 1, and then you work around to the carbon on the CH2OH side group which is number 5.
You will notice that each of the numbers has a small dash by it - 3' or 5', for example. If you just had ribose or deoxyribose on its own, that wouldn't be necessary, but in DNA and RNA these sugars are attached to other ring compounds. The carbons in the sugars are given the little dashes so that they can be distinguished from any numbers given to atoms in the other rings. You read 3' or 5' as "3-prime" or "5-prime".
Amino sugar
An amino sugar (or more technically a 2-amino-2-deoxysugar) is a sugar molecule in which a hydroxyl group has been replaced with an amine group. More than 60 amino sugars are known, with one of the most abundant being N-acetylglucosamine, which is the main component of chitin.
Structure of the chitin molecule, showing two of the N-acetylglucosamine units that repeat to form long chains in β-(1→4)-linkage.
Chitin is is a polymer of N-acetylglucosamine and is found in many places throughout the natural world. It is a characteristic component of the cell walls of fungi, the exoskeletons of arthropods such as crustaceans (e.g., crabs, lobsters and shrimps) and insects, the radulae of molluscs, and the beaks and internal shells of cephalopods, including squid and octopuses and on the scales and other soft tissues of fish and lissamphibians.
24.11: Cell Surface Carbohydrates and Influenza Viruses
Objectives
You may omit Section 25.11.
Carbohydrates are covalently attached to many different biomolecules, including lipids, to form glycolipids, and proteins, to form glycoproteins. Glycoproteins and glycolipids are often found in biological membranes, to which they are anchored by through nonpolar interactions. A special kind of glycoprotein, a proteoglycan, actually has more carbohydrate mass than protein. What is the function of these carbohydrates? Two are apparent. First, glycosylation of proteins helps protect the protein from degradation by enzyme catalysts within the body. However, there main functions arises from the fact that covalently attached carbohydrates that "decorate" the surface of glycoproteins or glycolipids provide new binding site interactions that allow interactions with other biomolecules. Hence glycosylation allows for cell:cell, cell:protein, or protein:protein interactions. Unfortunately, bacteria and viruses often recognize glycosylated molecules on cell membranes, allowing for their import into the cell.
Here are some "cartoon" examples of carbohydrates covalently linked to the amino acid asparagine (Asn) on a glycoprotein.
Here are some examples of biomolecular interactions promoted by IMFs involving carbohydrates.
Influenza Virus binding to Cell Surface Glycoproteins with Neu5Ac - A protein on the surface of influenza virus, hemagluttinin, bind to sialic acid (Sia), which is covalently attached to many cell membrane glycoproteins on host cells. The sialic acid is usually connected through an alpha (2,3) or alpha (2,6) link to galactose on N-linked glycoproteins. The subtypes found in avian (and equine) influenza isolates bind preferentially to Sia (alpha 2,3) Gal which predominates in avian GI tract where viruses replicate. Human virus of H1, H2, and H3 subtype (cause of the 1918, 1957, and 1968 pandemics) recognize Sia (alpha 2,6) Gal, the major form in human respiratory tract. The swine influenza HA bind to Sia (alpha 2,6) Gal and some Sia (alpha 2,3)both of which found in swine.
Leukocyte: Cell Wall binding - During inflammation, circulating leukocytes (a type of white blood cell) tether and roll on the walls of blood vessels where they become active. E-, L- and P-selectin proteins are the primary proteins responsible for the tethering and rolling of these leukocytes. P-selectin binds, in part, to a tetrasaccharide, sialyl-Lewisx (SLEX) on the cell surface.. The interaction between P-selectin and the cell mediates the initial binding/rolling of the leukocyte on the vessel wall. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/24%3A_Carbohydrates/24.10%3A_Other_Important_Carbohydrates.txt |
Objectives
After completing this section, you should be able to
1. give examples of the various biological roles played by proteins.
2. identify amino acids as being the building blocks from which all proteins are made.
3. show, in a general way, how the joining together of a number of amino acids through the formation of peptide bonds results in the formation of proteins.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• amino acid
• enzyme
• peptide bond
• protein
Study Notes
The “peptide bond” or “peptide linkage” that is formed between the amino group of one amino acid and the carboxyl group of a second amino acid is identical to the C$\ce{-}$N bond present in amides (see Section 21.7). We shall review the nature of such bonds in Section 26.4.
Proteins are polymers of amino acids, linked by amide groups known as peptide bonds. An amino acid can be thought of as having two components: a 'backbone', or 'main chain', composed of an ammonium group, an 'alpha-carbon', and a carboxylate, and a variable 'side chain' (in green below) bonded to the alpha-carbon.
There are twenty different side chains in naturally occurring amino acids, and it is the identity of the side chain that determines the identity of the amino acid: for example, if the side chain is a -CH3 group, the amino acid is alanine, and if the side chain is a -CH2OH group, the amino acid is serine. Many amino acid side chains contain a functional group (the side chain of serine, for example, contains a primary alcohol), while others, like alanine, lack a functional group, and contain only a simple alkane.
The two 'hooks' on an amino acid monomer are the amine and carboxylate groups. Proteins (polymers of ~50 amino acids or more) and peptides (shorter polymers) are formed when the amino group of one amino acid monomer reacts with the carboxylate carbon of another amino acid to form an amide linkage, which in protein terminology is a peptide bond. Which amino acids are linked, and in what order - the protein sequence - is what distinguishes one protein from another, and is coded for by an organism's DNA. Protein sequences are written in the amino terminal (N-terminal) to carboxylate terminal (C-terminal) direction, with either three-letter or single-letter abbreviations for the amino acids (see amino acid table). Below is a four amino acid peptide with the sequence "cysteine - histidine - glutamate - methionine". Using the single-letter code, the sequence is abbreviated CHEM.
When an amino acid is incorporated into a protein it loses a molecule of water and what remains is called a residue of the original amino acid. Thus we might refer to the 'glutamate residue' at position 3 of the CHEM peptide above.
Once a protein polymer is constructed, it in many cases folds up very specifically into a three-dimensional structure, which often includes one or more 'binding pockets' in which other molecules can be bound. It is this shape of this folded structure, and the precise arrangement of the functional groups within the structure (especially in the area of the binding pocket) that determines the function of the protein.
Enzymes are proteins which catalyze biochemical reactions. One or more reacting molecules - often called substrates - become bound in the active site pocket of an enzyme, where the actual reaction takes place. Receptors are proteins that bind specifically to one or more molecules - referred to as ligands - to initiate a biochemical process. For example, we saw in the introduction to this chapter that the TrpVI receptor in mammalian tissues binds capsaicin (from hot chili peppers) in its binding pocket and initiates a heat/pain signal which is sent to the brain.
Shown below is an image of the glycolytic enzyme fructose-1,6-bisphosphate aldolase (in grey), with the substrate molecule bound inside the active site pocket.
(x-ray crystallographic data are from Protein Science 1999, 8, 291; pdb code 4ALD. Image produced with JMol First Glance)
Intro to nucleic acids ⇒
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/25%3A_Amino_Acids_Peptides_and_Proteins/25.01%3A_Introduction.txt |
Objectives
After completing this section, you should be able to
1. identify the structural features present in the 20 amino acids commonly found in proteins.
Note: You are not expected to remember the detailed structures of all these amino acids, but you should be prepared to draw the structures of the two simplest members, glycine and alanine.
2. draw the Fischer projection formula of a specified enantiomer of a given amino acid.
Note: To do so, you must remember that in the S enantiomer, the carboxyl group appears at the top of the projection formula and the amino group is on the left.
3. classify an amino acid as being acidic, basic or neutral, given its Kekulé, condensed or shorthand structure.
4. draw the zwitterion form of a given amino acid.
5. account for some of the typical properties of amino acids (e.g., high melting points, solubility in water) in terms of zwitterion formation.
6. write appropriate equations to illustrate the amphoteric nature of amino acids.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• α‑amino acids
• amphoteric
• essential amino acids
• zwitterion
Study Notes
This is a good point at which to review some of the principles of stereochemistry presented in Chapter 5. Be sure to make full use of molecular models when any stereochemical issues arise.
You should recognize that a three‑letter shorthand code is often used to represent individual amino acids. You need not memorize this code.
The distinction between essential and nonessential amino acids is not as clear‑cut as one might suppose. For example, arginine is often regarded as being nonessential.
Introduction to Amino Acids
Amino acids form polymers through a condensation reaction by the amino group of an amino acid with the carboxyl group of another amino acid. The carboxyl group of the amino acid must first be activated to provide a better leaving group than OH-. (We will discuss this activation by ATP later in the course.) The resulting link between the amino acids is an amide link which biochemists call a peptide bond. In this reaction, water is released. In a reverse reaction, the peptide bond can be cleaved by water (hydrolysis).
When two amino acids link together to form an amide link, the resulting structure is called a dipeptide. Likewise, we can have tripeptides, tetrapeptides, and other polypeptides. At some point, when the structure is long enough, it is called a protein. There are many different ways to represent the structure of a polypeptide or protein, each showing differing amounts of information.
Figure: Different Representations of a Polypeptide (Heptapeptide)
(Note: above picture represents the amino acid in an unlikely protonation state with the weak acid protonated and the weak base deprotonated for simplicity in showing removal of water on peptide bond formation and the hydrolysis reaction.) Proteins are polymers of twenty naturally occurring amino acids. In contrast, nucleic acids are polymers of just 4 different monomeric nucleotides. Both the sequence of a protein and its total length differentiate one protein from another. Just for an octapeptide, there are over 25 billion different possible arrangements of amino acids. Compare this to just 65536 different oligonucleotides of 8 monomeric units (8mer). Hence the diversity of possible proteins is enormous.
Stereochemistry
The amino acids are all chiral, with the exception of glycine, whose side chain is H. As with lipids, biochemists use the L and D nomenclature. All naturally occuring proteins from all living organisms consist of L amino acids. The absolute stereochemistry is related to L-glyceraldehyde, as was the case for triacylglycerides and phospholipids. Most naturally occurring chiral amino acids are S, with the exception of cysteine. As the diagram below shows, the absolute configuration of the amino acids can be shown with the H pointed to the rear, the COOH groups pointing out to the left, the R group to the right, and the NH3 group upwards. You can remember this with the anagram CORN.
Figure: Stereochemistry of Amino Acids.
Why do biochemists still use D and L for sugars and amino acids? This explanation (taken from the link below) seems reasonable.
"In addition, however, chemists often need to define a configuration unambiguously in the absence of any reference compound, and for this purpose the alternative (R,S) system is ideal, as it uses priority rules to specify configurations. These rules sometimes lead to absurd results when they are applied to biochemical molecules. For example, as we have seen, all of the common amino acids are L, because they all have exactly the same structure, including the position of the R group if we just write the R group as R. However, they do not all have the same configuration in the (R,S) system: L-cysteine is also (R)-cysteine, but all the other L-amino acids are (S), but this just reflects the human decision to give a sulphur atom higher priority than a carbon atom, and does not reflect a real difference in configuration. Worse problems can sometimes arise in substitution reactions: sometimes inversion of configuration can result in no change in the (R) or (S) prefix; and sometimes retention of configuration can result in a change of prefix.
It follows that it is not just conservatism or failure to understand the (R,S) system that causes biochemists to continue with D and L: it is just that the DL system fulfils their needs much better. As mentioned, chemists also use D and L when they are appropriate to their needs. The explanation given above of why the (R,S) system is little used in biochemistry is thus almost the exact opposite of reality. This system is actually the only practical way of unambiguously representing the stereochemistry of complicated molecules with several asymmetric centres, but it is inconvenient with regular series of molecules like amino acids and simple sugars. "
Natural α-Amino Acids
Hydrolysis of proteins by boiling aqueous acid or base yields an assortment of small molecules identified as α-aminocarboxylic acids. More than twenty such components have been isolated, and the most common of these are listed in the following table. Those amino acids having green colored names are essential diet components, since they are not synthesized by human metabolic processes. The best food source of these nutrients is protein, but it is important to recognize that not all proteins have equal nutritional value. For example, peanuts have a higher weight content of protein than fish or eggs, but the proportion of essential amino acids in peanut protein is only a third of that from the two other sources. For reasons that will become evident when discussing the structures of proteins and peptides, each amino acid is assigned a one or three letter abbreviation.
Natural α-Amino Acids
Some common features of these amino acids should be noted. With the exception of proline, they are all 1º-amines; and with the exception of glycine, they are all chiral. The configurations of the chiral amino acids are the same when written as a Fischer projection formula, as in the drawing on the right, and this was defined as the L-configuration by Fischer. The R-substituent in this structure is the remaining structural component that varies from one amino acid to another, and in proline R is a three-carbon chain that joins the nitrogen to the alpha-carbon in a five-membered ring. Applying the Cahn-Ingold-Prelog notation, all these natural chiral amino acids, with the exception of cysteine, have an S-configuration. For the first seven compounds in the left column the R-substituent is a hydrocarbon. The last three entries in the left column have hydroxyl functional groups, and the first two amino acids in the right column incorporate thiol and sulfide groups respectively. Lysine and arginine have basic amine functions in their side-chains; histidine and tryptophan have less basic nitrogen heterocyclic rings as substituents. Finally, carboxylic acid side-chains are substituents on aspartic and glutamic acid, and the last two compounds in the right column are their corresponding amides.
The formulas for the amino acids written above are simple covalent bond representations based upon previous understanding of mono-functional analogs. The formulas are in fact incorrect. This is evident from a comparison of the physical properties listed in the following table. All four compounds in the table are roughly the same size, and all have moderate to excellent water solubility. The first two are simple carboxylic acids, and the third is an amino alcohol. All three compounds are soluble in organic solvents (e.g. ether) and have relatively low melting points. The carboxylic acids have pKa's near 4.5, and the conjugate acid of the amine has a pKa of 10. The simple amino acid alanine is the last entry. By contrast, it is very high melting (with decomposition), insoluble in organic solvents, and a million times weaker as an acid than ordinary carboxylic acids.
Physical Properties of Selected Acids and Amines
Compound
Formula
Mol.Wt.
Solubility in Water
Solubility in Ether
Melting Point
pKa
isobutyric acid (CH3)2CHCO2H 88 20g/100mL complete -47 ºC 5.0
lactic acid CH3CH(OH)CO2H 90 complete complete 53 ºC 3.9
3-amino-2-butanol CH3CH(NH2)CH(OH)CH3 89 complete complete 9 ºC 10.0
alanine CH3CH(NH2)CO2H 89 18g/100mL insoluble ca. 300 ºC 9.8
Zwitterion
These differences above all point to internal salt formation by a proton transfer from the acidic carboxyl function to the basic amino group. The resulting ammonium carboxylate structure, commonly referred to as a zwitterion, is also supported by the spectroscopic characteristics of alanine.
CH3CH(NH2)CO2H CH3CH(NH3)(+)CO2(–)
As expected from its ionic character, the alanine zwitterion is high melting, insoluble in nonpolar solvents and has the acid strength of a 1º-ammonium ion. Examples of a few specific amino acids may also be viewed in their favored neutral zwitterionic form. Note that in lysine the amine function farthest from the carboxyl group is more basic than the alpha-amine. Consequently, the positively charged ammonium moiety formed at the chain terminus is attracted to the negative carboxylate, resulting in a coiled conformation.
The structure of an amino acid allows it to act as both an acid and a base. An amino acid has this ability because at a certain pH value (different for each amino acid) nearly all the amino acid molecules exist as zwitterions. If acid is added to a solution containing the zwitterion, the carboxylate group captures a hydrogen (H+) ion, and the amino acid becomes positively charged. If base is added, ion removal of the H+ ion from the amino group of the zwitterion produces a negatively charged amino acid. In both circumstances, the amino acid acts to maintain the pH of the system—that is, to remove the added acid (H+) or base (OH) from solution.
Example 26.1
1. Draw the structure for the anion formed when glycine (at neutral pH) reacts with a base.
2. Draw the structure for the cation formed when glycine (at neutral pH) reacts with an acid.
Solution
1. The base removes H+ from the protonated amine group.
2. The acid adds H+ to the carboxylate group.
Other Natural Amino Acids
The twenty alpha-amino acids listed above are the primary components of proteins, their incorporation being governed by the genetic code. Many other naturally occurring amino acids exist, and the structures of a few of these are displayed below. Some, such as hydroxylysine and hydroxyproline, are simply functionalized derivatives of a previously described compound. These two amino acids are found only in collagen, a common structural protein. Homoserine and homocysteine are higher homologs of their namesakes. The amino group in beta-alanine has moved to the end of the three-carbon chain. It is a component of pantothenic acid, HOCH2C(CH3)2CH(OH)CONHCH2CH2CO2H, a member of the vitamin B complex and an essential nutrient. Acetyl coenzyme A is a pyrophosphorylated derivative of a pantothenic acid amide. The gamma-amino homolog GABA is a neurotransmitter inhibitor and antihypertensive agent.
Many unusual amino acids, including D-enantiomers of some common acids, are produced by microorganisms. These include ornithine, which is a component of the antibiotic bacitracin A, and statin, found as part of a pentapeptide that inhibits the action of the digestive enzyme pepsin. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/25%3A_Amino_Acids_Peptides_and_Proteins/25.02%3A_Structure_and_Stereochemistry_of_the_Amino_Acids.txt |
Objectives
After completing this section, you should be able to
1. draw the predominant form of a given amino acid in a solution of known pH, given the isoelectric point of the amino acid.
2. describe, briefly, how a mixture of amino acids may be separated by paper electrophoresis.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• electrophoresis
• isoelectric point
Since amino acids, as well as peptides and proteins, incorporate both acidic and basic functional groups, the predominant molecular species present in an aqueous solution will depend on the pH of the solution. In order to determine the nature of the molecular and ionic species that are present in aqueous solutions at different pH's, we make use of the Henderson-Hasselbalch Equation, written below. Here, the pKa represents the acidity of a specific conjugate acid function (HA). When the pH of the solution equals pKa, the concentrations of HA and A(-) must be equal (log 1 = 0).
\[ pK_a = pH + \log_{10} \dfrac{[HA]}{A^-]}\]
The titration curve for alanine in Figure \(2\) demonstrates this relationship. At a pH lower than 2, both the carboxylate and amine functions are protonated, so the alanine molecule has a net positive charge. At a pH greater than 10, the amine exists as a neutral base and the carboxyl as its conjugate base, so the alanine molecule has a net negative charge. At intermediate pH's the zwitterion concentration increases, and at a characteristic pH, called the isoelectric point (pI), the negatively and positively charged molecular species are present in equal concentration. This behavior is general for simple (difunctional) amino acids. Starting from a fully protonated state, the pKa's of the acidic functions range from 1.8 to 2.4 for -CO2H, and 8.8 to 9.7 for -NH3(+). The isoelectric points range from 5.5 to 6.2. Titration curves show the neutralization of these acids by added base, and the change in pH during the titration.
The distribution of charged species in a sample can be shown experimentally by observing the movement of solute molecules in an electric field, using the technique of electrophoresis (Figure \(2\)). For such experiments an ionic buffer solution is incorporated in a solid matrix layer, composed of paper or a crosslinked gelatin-like substance. A small amount of the amino acid, peptide or protein sample is placed near the center of the matrix strip and an electric potential is applied at the ends of the strip, as shown in the following diagram. The solid structure of the matrix retards the diffusion of the solute molecules, which will remain where they are inserted, unless acted upon by the electrostatic potential.
At pH 6.00 alanine and isoleucine exist on average as neutral zwitterionic molecules, and are not influenced by the electric field. Arginine is a basic amino acid. Both base functions exist as "onium" conjugate acids in the pH 6.00 matrix. The solute molecules of arginine therefore carry an excess positive charge, and they move toward the cathode. The two carboxyl functions in aspartic acid are both ionized at pH 6.00, and the negatively charged solute molecules move toward the anode in the electric field. Structures for all these species are shown to the right of the display.
It should be clear that the result of this experiment is critically dependent on the pH of the matrix buffer. If we were to repeat the electrophoresis of these compounds at a pH of 3.80, the aspartic acid would remain at its point of origin, and the other amino acids would move toward the cathode. Ignoring differences in molecular size and shape, the arginine would move twice as fast as the alanine and isoleucine because its solute molecules on average would carry a double positive charge.
As noted earlier, the titration curves of simple amino acids display two inflection points, one due to the strongly acidic carboxyl group (pKa1 = 1.8 to 2.4), and the other for the less acidic ammonium function (pKa2 = 8.8 to 9.7). For the 2º-amino acid proline, pKa2 is 10.6, reflecting the greater basicity of 2º-amines.
Table \(1\): pKa Values of Polyfunctional Amino Acids
Amino Acid α-CO2H pKa1 α-NH3 pKa2 Side Chain pKa3 pI
Arginine 2.1 9.0 12.5 10.8
Aspartic Acid 2.1 9.8 3.9 3.0
Cysteine 1.7 10.4 8.3 5.0
Glutamic Acid 2.2 9.7 4.3 3.2
Histidine 1.8 9.2 6.0 7.6
Lysine 2.2 9.0 10.5 9.8
Tyrosine 2.2 9.1 10.1 5.7
Some amino acids have additional acidic or basic functions in their side chains. These compounds are listed in Table \(1\). A third pKa, representing the acidity or basicity of the extra function, is listed in the fourth column of the table. The pI's of these amino acids (last column) are often very different from those noted above for the simpler members. As expected, such compounds display three inflection points in their titration curves, illustrated by the titrations of arginine and aspartic acid (Figure\ (3\)). For each of these compounds four possible charged species are possible, one of which has no overall charge. Formulas for these species are written to the right of the titration curves, together with the pH at which each is expected to predominate. The very high pH required to remove the last acidic proton from arginine reflects the exceptionally high basicity of the guanidine moiety at the end of the side chain.
The Isoelectric Point
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 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/25%3A_Amino_Acids_Peptides_and_Proteins/25.03%3A_Isoelectric_Points_and_Electrophoresis.txt |
Objectives
After completing this section, you should be able to
1. outline, by means of equations, how a racemic mixture of given amino acid can be prepared from a carboxylic acid using reactions you studied earlier in the course.
1. outline, by means of equations, the preparation of a given amino acid by the amidomalonate synthesis.
2. identify the amino acid formed from using a given alkyl halide in an amidomalonate synthesis.
3. identify the alkyl halide needed to produce a given amino acid by the amidomalonate synthesis.
2. describe, by means of equations, how an α‑keto acid can be transformed to an amino acid by reductive amination.
1. describe a general method for resolving a racemic mixture of a given amino acid.
2. provide a brief example of how a biological method may be employed to resolve a racemic mixture of a given amino acid.
3. show the enantioselective preparation of an amino acid from the corresponding Z enamido acid.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• amidomalonate synthesis
• enantioselective synthesis
• racemic mixture
Study Notes
Do not be alarmed by the number of methods to synthesize amino acids described in this section.You have seen many of these reactions in previous sections and should already be familiar with the approaches discussed here.
To fulfill the requirements of Objective 1, review the Hell‑Volhard‑Zelinskii reaction (Section 22.4) and the Gabriel phthalimide synthesis (Section 24.6).
The amidomalonate synthesis is a simple variation of the malonic ester synthesis (Section 22.7). A base abstracts a proton from the alpha carbon, which is then alkylated with an alkyl halide. Then both the hydrolysis of the esters and the amide protecting group under aqueous acidic conditions generates the α‑amino acid.
Another method of getting to the α‑amino acid is by reductive amination of the α‑keto acid which you have also previously encountered (Section 24.6).
Synthesis of α-Amino Acids
1) Amination of alpha-bromocarboxylic acids, illustrated by the following equation, provides a straightforward method for preparing alpha-aminocarboxylic acids. The bromoacids, in turn, are conveniently prepared from carboxylic acids by reaction with Br2 + PCl3. Although this direct approach gave mediocre results when used to prepare simple amines from alkyl halides, it is more effective for making amino acids, thanks to the reduced nucleophilicity of the nitrogen atom in the product. Nevertheless, more complex procedures that give good yields of pure compounds are often chosen for amino acid synthesis.
2) By modifying the nitrogen as a phthalimide salt, the propensity of amines to undergo multiple substitutions is removed, and a single clean substitution reaction of 1º- and many 2º-alkylhalides takes place. This procedure, known as the Gabriel synthesis, can be used to advantage in aminating bromomalonic esters, as shown in the upper equation of the following scheme. Since the phthalimide substituted malonic ester has an acidic hydrogen (colored orange), activated by the two ester groups, this intermediate may be converted to an ambident anion and alkylated. Finally, base catalyzed hydrolysis of the phthalimide moiety and the esters, followed by acidification and thermal decarboxylation, produces an amino acid and phthalic acid (not shown).
3) An elegant procedure, known as the Strecker synthesis, assembles an alpha-amino acid from ammonia (the amine precursor), cyanide (the carboxyl precursor), and an aldehyde. This reaction (shown below) is essentially an imino analog of cyanohydrin formation. The alpha-amino nitrile formed in this way can then be hydrolyzed to an amino acid by either acid or base catalysis.
4) Resolution The three synthetic procedures described above, and many others that can be conceived, give racemic amino acid products. If pure L or D enantiomers are desired, it is necessary to resolve these racemic mixtures. A common method of resolving racemates is by diastereomeric salt formation with a pure chiral acid or base. This is illustrated for a generic amino acid in the following diagram. Be careful to distinguish charge symbols, shown in colored circles, from optical rotation signs, shown in parenthesis.
In the initial display, the carboxylic acid function contributes to diastereomeric salt formation. The racemic amino acid is first converted to a benzamide derivative to remove the basic character of the amino group. Next, an ammonium salt is formed by combining the carboxylic acid with an optically pure amine, such as brucine (a relative of strychnine). The structure of this amine is not shown, because it is not a critical factor in the logical progression of steps. Since the amino acid moiety is racemic and the base is a single enantiomer (levorotatory in this example), an equimolar mixture of diastereomeric salts is formed (drawn in the green shaded box). Diastereomers may be separated by crystallization, chromatography or other physical manipulation, and in this way one of the isomers may be isolated for further treatment, in this illustration it is the (+):(-) diastereomer. Finally the salt is broken by acid treatment, giving the resolved (+)-amino acid derivative together with the recovered resolving agent (the optically active amine). Of course, the same procedure could be used to obtain the (-)-enantiomer of the amino acid.
Since amino acids are amphoteric, resolution could also be achieved by using the basic character of the amine function. For this approach we would need an enantiomerically pure chiral acid such as tartaric acid to use as the resolving agent. This alternative resolution strategy will be illustrated. Note that the carboxylic acid function is first esterified, so that it will not compete with the resolving acid.
Resolution of aminoacid derivatives may also be achieved by enzymatic discrimination in the hydrolysis of amides. For example, an aminoacylase enzyme from pig kidneys cleaves an amide derivative of a natural L-amino acid much faster than it does the D-enantiomer. If the racemic mixture of amides shown in the green shaded box above is treated with this enzyme, the L-enantiomer (whatever its rotation) will be rapidly converted to its free zwitterionic form, whereas the D-enantiomer will remain largely unchanged. Here, the diastereomeric species are transition states rather than isolable intermediates. This separation of enantiomers, based on very different rates of reaction, is called kinetic resolution.
Enantioselective Synthesis
Till now all of the synthetic routes to α-amino acids we have discussed yield a racemic mixture. Once produced one could resolve the mixture to obtain pure L or D enantiomers. However, enantioselective synthetic methods to produce pure compounds directly are being developed. For instance, several catalysts are now available for reduction of C=C to expose enantiopure amino acids. A good example is the industrial synthesis of L-DOPA, a drug used in the treatment of Parkinson’s disease. W.S. Knowles shared the 2001 Nobel Price with R. Noyori and K.B. Sharpless for their contributions in the area of asymmetric catalytic reductions. Knowles developed several chiral phosphine–metal catalysts for asymmetric reductions. The rhodium(I) catalyst shown, which is complexed by large organic ligands, facilitates production of almost pure L-DOPA. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/25%3A_Amino_Acids_Peptides_and_Proteins/25.04%3A_Synthesis_of_Amino_Acids.txt |
Objectives
After completing this section, you should be able to
1. show, by means of a diagram, how two different amino acid residues can be combined to give two different dipeptides.
2. draw the structure of a relatively simple peptide, given its full or abbreviated name and the structures of the appropriate amino acids.
3. draw, or name, the six possible isomeric tripeptides that can be formed by combining three different amino acid residues (amino acid units) of given structure.
1. account for the fact that there is restricted rotation about the C$\ce{-}$N bonds in peptides.
2. illustrate the formation of a disulfide linkage between two cysteine residues, and show how such bonds can link together two separate peptide chains or can provide a bridge between two cysteine residues present in a single peptide molecule.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• C‑terminal amino acid
• N‑terminal amino acid
• peptides
• residues
Study Notes
If necessary, review the discussion of the delocalization of the nitrogen lone‑pair electrons in amides that was presented in Section 24.3. Similarly, you may wish to refer back to Section 18.8 to review the interconversion of thiols and disulfides.
Peptide Bond Formation or Amide Synthesis
The formation of peptides is nothing more than the application of the amide synthesis reaction. By convention, the amide bond in the peptides should be made in the order that the amino acids are written. The amine end (N terminal) of an amino acid is always on the left, while the acid end (C terminal) is on the right. The reaction of glycine with alanine to form the dipeptide glyclalanine is written as shown in the graphic on the left. Oxygen (red) from the acid and hydrogens (red) on the amine form a water molecule. The carboxyl oxygen (green) and the amine nitrogen (green) join to form the amide bond.
If the order of listing the amino acids is reversed, a different dipeptide is formed such as alaninylglycine.
Exercise $1$
Write the reactions for:
1. ala + gly ---> Answer graphic
2. phe + ser ----> Answer graphic
Resonance contributors for the peptide bonds
A consideration of resonance contributors is crucial to any discussion of the amide functional group. One of the most important examples of amide groups in nature is the ‘peptide bond’ that links amino acids to form polypeptides and proteins.
Critical to the structure of proteins is the fact that, although it is conventionally drawn as a single bond, the C-N bond in a peptide linkage has a significant barrier to rotation, almost as if it were a double bond.
This, along with the observation that the bonding around the peptide nitrogen has trigonal planar geometry, strongly suggests that the nitrogen is sp2-hybridized. An important resonance contributor has a C=N double bond and a C-O single bond, with a separation of charge between the oxygen and the nitrogen.
Although B is a minor contributor due to the separation of charges, it is still very relevant in terms of peptide and protein structure – our proteins would simply not fold up properly if there was free rotation about the peptide C-N bond.
Backbone Peptide or Protein Structure
The structure of a peptide can be written fairly easily without showing the complete amide synthesis reaction by learning the structure of the "backbone" for peptides and proteins.
The peptide backbone consists of repeating units of "N-H 2, CH, C double bond O; N-H 2, CH, C double bond O; etc. See the graphic on the left .
After the backbone is written, go back and write the specific structure for the side chains as represented by the "R" as gly-ala-leu for this example. The amine end (N terminal) of an amino acid is always on the left (gly), while the acid end (C terminal) is on the right (leu).
Exercise $2$
Write the tripeptide structure for val-ser-cys. First write the "backbone" and then add the specific side chains.
Solution
Answer graphic
QUES. Write the structure for the tripeptide:
2 a ) glu-cys-gly ---> Answer graphic
2 b) phe-tyr-asn ---> Answer graphic
Disulfide Bridges and Oxidation-Reduction
The amino acid cysteine undergoes oxidation and reduction reactions involving the -SH (sulfhydryl group). The oxidation of two sulfhydryl groups results in the formation of a disulfide bond by the removal of two hydrogens. The oxidation of two cysteine amino acids is shown in the graphic. An unspecified oxidizing agent (O) provides an oxygen which reacts with the hydrogen (red) on the -SH group to form water. The sulfurs (yellow) join to make the disulfide bridge. This is an important bond to recognize in protein tertiary structure. The reduction of a disulfide bond is the opposite reaction which again leads to two separate cysteine molecules. Remember that reduction is the addition of hydrogen.
Cysteine residues in the the peptide chain can form a loop buy forming the disulfide bond (—S—S—), while cysteine residues in different peptide chains can actually link what were otherwise separate chains. Insulin was the first protein whose amino acid sequence was determined. This pioneering work, completed in 1953 after some 10 years of effort, earned a Nobel Prize for British biochemist Frederick Sanger (born 1918). He found the primary structure to comprise of two chains linked by two cysteine disulfide bridges. Also note the first peptide chain possesses an internal loop.
Insulin
Contributors and Attributions
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/25%3A_Amino_Acids_Peptides_and_Proteins/25.05%3A_Peptides_and_Proteins.txt |
Objectives
After completing this section, you should be able to describe, briefly, how the identity and amounts of each amino acid residue present in a peptide of unknown structure may be determined.
Key Terms
Make certain that you can define, and use in context, the key term below.
• amino acid analyzer
Study Notes
You need not memorize the reaction between ninhydrin and an α‑amino acid.
Ion-exchange chromatography
When a protein is to be analyzed, it is first heated with acid to hydrolyse all the peptide bonds. When such a mixture of amino acids is to be purified and estimated quantitatively, ion-exchange chromatography is the technique of choice. Fully automated amino acid analyzers are now available, which are equipped with a solvent pump to deliver the required buffer(s) in a programmed manner. There is a column, filled with Dowex 50 resin (Fig 26.5.1). This solid support is made up of polymeric beads. Chemically speaking they are polymers bearing arylsulfonic acid groups. The cation exchange resin helps in the separation of amino acids. In a typical run (Fig 26.5.2), the eluent is a buffer. The pH value of the buffer could be varied as step elution or as gradient elution. The chromatogram shown in Fig 26.5.2 is a chromatogram run with gradient elution technique, using ninhydrin as the post column treatment. The detector is a UV detector scanning the wavelengths 570 nm and 440 nm.
Fig 26.5.1: A Cation Rasin like Dowex 50 is a polymeric bead bearing aryl sulfonic acid groups
Fig 26.5.2: Some typical chromatograms from an amino acid analyzer
The Ninhydrin Reaction
Alpha-amino acids show reactivity at their the carboxylic acid and amine sites typical of those functional groups. In addition to these common reactions of amines and carboxylic acids, common alpha-amino acids, except proline, undergo a unique reaction with the triketohydrindene hydrate known as ninhydrin. Among the products of this unusual reaction (shown on the left below) is a purple colored imino derivative, which provides as a useful color test for these amino acids, most of which are colorless. A common application of the ninhydrin test is the visualization of amino acids in paper chromatography. As shown in the graphic on the right, samples of amino acids or mixtures thereof are applied along a line near the bottom of a rectangular sheet of paper (the baseline). The bottom edge of the paper is immersed in an aqueous buffer, and this liquid climbs slowly toward the top edge. As the solvent front passes the sample spots, the compounds in each sample are carried along at a rate which is characteristic of their functionality, size and interaction with the cellulose matrix of the paper. Some compounds move rapidly up the paper, while others may scarcely move at all. The ratio of the distance a compound moves from the baseline to the distance of the solvent front from the baseline is defined as the retardation (or retention) factor Rf. Different amino acids usually have different Rf's under suitable conditions. In the example on the right, the three sample compounds (1, 2 & 3) have respective Rf values of 0.54, 0.36 & 0.78. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/25%3A_Amino_Acids_Peptides_and_Proteins/25.06%3A_Amino_Acid_Analysis_of_Peptides.txt |
Objectives
After completing this section, you should be able to
1. describe how an Edman degradation is used to determine the sequence of the amino acid residues in peptides containing up to 20 such residues.
2. describe, briefly, how the procedure is modified to deal with peptides and proteins containing more than 20 amino acid residues.
3. write a detailed mechanism for the Edman degradation.
4. determine the structure of a peptide, given a list of the fragments that are produced by a partial acid hydrolysis.
5. determine the structure of a peptide, given a list of the fragments that are produced when the peptide is cleaved by a specific enzyme and the details of the types of bonds cleaved by that enzyme.
6. predict the fragments that would be produced when a peptide of known structure is cleaved by a specific enzyme, given sufficient information about the types of bonds that are cleaved by the enzyme in question.
Key Terms
Make certain that you can define, and use in context, the key term below.
• Edman degradation
Study Notes
The reagent used in the Edman degradation is phenyl isothiocyanate. You may find it helpful to review the relationship between cyanates, isocyanates, thiocyanates and isothiocyanates.
You need not memorize the specific peptide bonds that are broken by the enzymes trypsin and chymotrypsin.
Edman degradation is the process of purifying protein by sequentially removing one residue at a time from the amino end of a peptide. To solve the problem of damaging the protein by hydrolyzing conditions, Pehr Edman created a new way of labeling and cleaving the peptide. Edman thought of a way of removing only one residue at a time, which did not damage the overall sequencing. This was done by adding Phenyl isothiocyanate, which creates a phenylthiocarbamoyl derivative with the N-terminal. The N-terminal is then cleaved under less harsh acidic conditions, creating a cyclic compound of phenylthiohydantoin PTH-amino acid. This does not damage the protein and leaves two constituents of the peptide. This method can be repeated for the rest of the residues, separating one residue at a time.
Edman degradation is very useful because it does not damage the protein. This allows sequencing of the protein to be done in less time. Edman sequencing is done best if the composition of the amino acid is known. As we saw in Section 26.5, to determine the composition of the amino acid, the peptide must be hydrolyzed. This can be done by denaturing the protein and heating it and adding HCl for a long time. This causes the individual amino acids to be separated, and they can be separated by ion exchange chromatography. They are then dyed with ninhydrin and the amount of amino acid can be determined by the amount of optical absorbance. This way, the composition but not the sequence can be determined
Sequencing Larger Proteins
Larger proteins cannot be sequenced by the Edman sequencing because of the less than perfect efficiency of the method. A strategy called divide and conquer successfully cleaves the larger protein into smaller, practical amino acids. This is done by using a certain chemical or enzyme which can cleave the protein at specific amino acid residues. The separated peptides can be isolated by chromatography. Then they can be sequenced using the Edman method, because of their smaller size.
In order to put together all the sequences of the different peptides, a method of overlapping peptides is used. The strategy of divide and conquer followed by Edman sequencing is used again a second time, but using a different enzyme or chemical to cleave it into different residues. This allows two different sets of amino acid sequences of the same protein, but at different points. By comparing these two sequences and examining for any overlap between the two, the sequence can be known for the original protein.
For example, trypsin can be used on the initial peptide to cleave it at the carboxyl side of arginine and lysine residues. Using trypsin to cleave the protein and sequencing them individually with Edman degradation will yield many different individual results. Although the sequence of each individual cleaved amino acid segment is known, the order is scrambled. Chymotrypsin, which cleaves on the carboxyl side of aromatic and other bulky nonpolar residues, can be used. The sequence of these segments overlap with those of the trypsin. They can be overlapped to find the original sequence of the initial protein. However, this method is limited in analyzing larger sized proteins (more than 100 amino acids) because of secondary hydrogen bond interference. Other weak intermolecular bonding such as hydrophobic interactions cannot be properly predicted. Only the linear sequence of a protein can be properly predicted assuming the sequence is small enough. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/25%3A_Amino_Acids_Peptides_and_Proteins/25.07%3A_Peptide_Sequencing-_The_Edman_Degradation.txt |
Objectives
After completing this section, you should be able to
1. describe why it is necessary to protect certain amino and carboxyl groups during the synthesis of a peptide.
2. describe, using appropriate equations, how carboxyl groups are protected by ester formation and amino groups are protected by the formation of their tert‑butoxycarbonyl amide derivatives.
3. write a detailed mechanism for the formation of a peptide link between an amino acid with a protected amino group and an amino acid with a protected carboxyl group using dicyclohexylcarbodiimide.
4. outline the five steps required in order to form a dipeptide from two given amino acids.
In order to synthesize a peptide from its component amino acids, two obstacles must be overcome. The first of these is statistical in nature, and is illustrated by considering the dipeptide Ala-Gly as a proposed target. If we ignore the chemistry involved, a mixture of equal molar amounts of alanine and glycine would generate four different dipeptides. These are: Ala-Ala, Gly-Gly, Ala-Gly & Gly-Ala. In the case of tripeptides, the number of possible products from these two amino acids rises to eight. Clearly, some kind of selectivity must be exercised if complex mixtures are to be avoided.
The second difficulty arises from the fact that carboxylic acids and 1º or 2º-amines do not form amide bonds on mixing, but will generally react by proton transfer to give salts (the intermolecular equivalent of zwitterion formation).
From the perspective of an organic chemist, peptide synthesis requires selective acylation of a free amine. To accomplish the desired amide bond formation, we must first deactivate all extraneous amine functions so they do not compete for the acylation reagent. Then we must selectively activate the designated carboxyl function so that it will acylate the one remaining free amine. Fortunately, chemical reactions that permit us to accomplish these selections are well known.
First, the basicity and nucleophilicity of amines are substantially reduced by amide formation. Consequently, the acylation of amino acids by treatment with acyl chlorides or anhydrides at pH > 10, as described earlier, serves to protect their amino groups from further reaction.
Second, acyl halide or anhydride-like activation of a specific carboxyl reactant must occur as a prelude to peptide (amide) bond formation. This is possible, provided competing reactions involving other carboxyl functions that might be present are precluded by preliminary ester formation. Remember, esters are weaker acylating reagents than either anhydrides or acyl halides, as noted earlier.
Finally, dicyclohexylcarbodiimide (DCC) effects the dehydration of a carboxylic acid and amine mixture to the corresponding amide under relatively mild conditions. The structure of this reagent and the mechanism of its action have been described. Its application to peptide synthesis will become apparent in the following discussion.
The strategy for peptide synthesis, as outlined here, should now be apparent. The following example shows a selective synthesis of the dipeptide Ala-Gly.
An important issue remains to be addressed. Since the N-protective group is an amide, removal of this function might require conditions that would also cleave the just formed peptide bond. Furthermore, the harsh conditions often required for amide hydrolysis might cause extensive racemization of the amino acids in the resulting peptide. This problem strikes at the heart of our strategy, so it is important to give careful thought to the design of specific N-protective groups. In particular, three qualities are desired:
1. The protective amide should be easy to attach to amino acids.
2. The protected amino group should not react under peptide forming conditions.
3. The protective amide group should be easy to remove under mild conditions.
A number of protective groups that satisfy these conditions have been devised; and two of the most widely used, carbobenzoxy (Cbz) and t-butoxycarbonyl (BOC or t-BOC), are described here.
The reagents for introducing these N-protective groups are the acyl chlorides or anhydrides shown in the left portion of the above diagram. Reaction with a free amine function of an amino acid occurs rapidly to give the "protected" amino acid derivative shown in the center. This can then be used to form a peptide (amide) bond to a second amino acid. Once the desired peptide bond is created the protective group can be removed under relatively mild non-hydrolytic conditions. Equations showing the protective group removal will be displayed above by are shown above. Cleavage of the reactive benzyl or tert-butyl groups generates a common carbamic acid intermediate (HOCO-NHR) which spontaneously loses carbon dioxide, giving the corresponding amine. If the methyl ester at the C-terminus is left in place, this sequence of reactions may be repeated, using a different N-protected amino acid as the acylating reagent. Removal of the protective groups would then yield a specific tripeptide, determined by the nature of the reactants and order of the reactions.
25.09: Automated Peptide Synthesis- The Merrifield Solid-Phase
Objectives
After completing this section, you should be able to describe, briefly, the Merrifield solid‑phase technique for the synthesis of polypeptides.
Key Terms
Make certain that you can define, and use in context, the key term below.
• solid‑phase method (solid‑phase synthesis)
Study Notes
The solid‑phase used in this method is a polymer support. You will not be examined on the details of the Merrifield solid‑phase method; however, you should be prepared to write a couple of paragraphs describing this important process.
For his work on the synthesis of peptides, Bruce Merrifield was awarded the 1984 Nobel Prize in chemistry.
The synthesis of a peptide of significant length (e.g. ten residues) by this approach requires many steps, and the product must be carefully purified after each step to prevent unwanted cross-reactions. To facilitate the tedious and time consuming purifications, and reduce the material losses that occur in handling, a clever modification of this strategy has been developed. This procedure, known as the Merrifield Synthesis after its inventor R. Bruce Merrifield, involves attaching the C-terminus of the peptide chain to a polymeric solid, usually having the form of very small beads. Separation and purification is simply accomplished by filtering and washing the beads with appropriate solvents. The reagents for the next peptide bond addition are then added, and the purification steps repeated. The entire process can be automated, and peptide synthesis machines based on the Merrifield approach are commercially available. A series of equations illustrating the Merrifield synthesis may be viewed below. The final step, in which the completed peptide is released from the polymer support, is a simple benzyl ester cleavage. This is not shown in the display.
The Merrifield Peptide Synthesis
Two or more moderately sized peptides can be joined together by selective peptide bond formation, provided side-chain functions are protected and do not interfere. In this manner good sized peptides and small proteins may be synthesized in the laboratory. However, even if chemists assemble the primary structure of a natural protein in this or any other fashion, it may not immediately adopt its native secondary, tertiary and quaternary structure. Many factors, such as pH, temperature and inorganic ion concentration influence the conformational coiling of peptide chains. Indeed, scientists are still trying to understand how and why these higher structures are established in living organisms. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/25%3A_Amino_Acids_Peptides_and_Proteins/25.08%3A_Peptide_Synthesis.txt |
Objectives
After completing this section, you should be able to
1. discuss, with reference to a suitable example (either given or of your own choice), the structure of proteins, paying particular attention to distinguishing between the primary, secondary, tertiary and quaternary structure.
2. describe the α‑helical secondary structure displayed by many proteins.
3. describe the β‑pleated‑sheet structure displayed by many proteins.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• α helix
• β pleated sheet
• primary structure
• quaternary structure
• secondary structure
• tertiary structure
Study Notes
Note that in a diagram of the α‑helical structure of a protein, the C‑terminal of the protein is at the bottom of the diagram and the N‑terminal is at the top. In an α helix, such as the one shown in Figure 26.9.1, the bulky R groups are all found on the outside of the helix, where they have the most room.
The four levels of protein structure
Protein structure can be discussed at four distinct levels. A protein’s primary structure is two-dimensional - simply the sequence of amino acids in the peptide chain. Below is a Lewis structure of a short segment of a protein with the sequence CHEM (cysteine - histidine - glutamate - methionine)
Secondary structure is three-dimensional, but is a local phenomenon, confined to a relatively short stretch of amino acids. For the most part, there are three important elements of secondary structure: helices, beta-sheets, and loops. In a helix, the main chain of the protein adopts the shape of a clockwise spiral staircase, and the side chains point out laterally.
In a beta-sheet (or beta-strand) structure, two sections of protein chain are aligned side-by-side in an extended conformation. The figure below shows two different views of the same beta-sheet: in the left-side view, the two regions of protein chain are differentiated by color.
Loops are relatively disordered segments of protein chain, but often assume a very ordered structure when in contact with a second protein or a smaller organic compound.
Both helix and the beta-sheet structures are held together by very specific hydrogen-bonding interactions between the amide nitrogen on one amino acid and the carbonyl oxygen on another. The hydrogen bonding pattern in a section of a beta-strand is shown below.
Secondary structure refers to the shape of a folding protein due exclusively to hydrogen bonding between its backbone amide and carbonyl groups. Secondary structure does not include bonding between the R-groups of amino acids, hydrophobic interactions, or other interactions associated with tertiary structure. The two most commonly encountered secondary structures of a polypeptide chain are α-helices and beta-pleated sheets. These structures are the first major steps in the folding of a polypeptide chain, and they establish important topological motifs that dictate subsequent tertiary structure and the ultimate function of the protein.
α-Helices
An α-helix is a right-handed coil of amino-acid residues on a polypeptide chain, typically ranging between 4 and 40 residues. This coil is held together by hydrogen bonds between the oxygen of C=O on top coil and the hydrogen of N-H on the bottom coil. Such a hydrogen bond is formed exactly every 4 amino acid residues, and every complete turn of the helix is only 3.6 amino acid residues. This regular pattern gives the α-helix very definite features with regards to the thickness of the coil and the length of each complete turn along the helix axis.
The structural integrity of an α-helix is in part dependent on correct steric configuration. Amino acids whose R-groups are too large (tryptophan, tyrosine) or too small (glycine) destabilize α-helices. Proline also destabilizes α-helices because of its irregular geometry; its R-group bonds back to the nitrogen of the amide group, which causes steric hindrance. In addition, the lack of a hydrogen on Proline's nitrogen prevents it from participating in hydrogen bonding.
Another factor affecting α-helix stability is the total dipole moment of the entire helix due to individual dipoles of the C=O groups involved in hydrogen bonding. Stable α-helices typically end with a charged amino acid to neutralize the dipole moment.
BETA-PLEATED SHEETS
This structure occurs when two (or more, e.g. ψ-loop) segments of a polypeptide chain overlap one another and form a row of hydrogen bonds with each other. This can happen in a parallel arrangement:
Or in anti-parallel arrangement:
Parallel and anti-parallel arrangement is the direct consequence of the directionality of the polypeptide chain. In anti-parallel arrangement, the C-terminus end of one segment is on the same side as the N-terminus end of the other segment. In parallel arrangement, the C-terminus end and the N-terminus end are on the same sides for both segments. The "pleat" occurs because of the alternating planes of the peptide bonds between amino acids; the aligned amino and carbonyl group of each opposite segment alternate their orientation from facing towards each other to facing opposite directions.
The parallel arrangement is less stable because the geometry of the individual amino acid molecules forces the hydrogen bonds to occur at an angle, making them longer and thus weaker. Contrarily, in the anti-parallel arrangement the hydrogen bonds are aligned directly opposite each other, making for stronger and more stable bonds.
Commonly, an anti-parallel beta-pleated sheet forms when a polypeptide chain sharply reverses direction. This can occur in the presence of two consecutive proline residues, which create an angled kink in the polypeptide chain and bend it back upon itself. This is not necessary for distant segments of a polypeptide chain to form beta-pleated sheets, but for proximal segments it is a definite requirement. For short distances, the two segments of a beta-pleated sheet are separated by 4+2n amino acid residues, with 4 being the minimum number of residues.
α-PLEATED SHEETS
A similar structure to the beta-pleated sheet is the α-pleated sheet. This structure is energetically less favorable than the beta-pleated sheet, and is fairly uncommon in proteins. An α-pleated sheet is characterized by the alignment of its carbonyl and amino groups; the carbonyl groups are all aligned in one direction, while all the N-H groups are aligned in the opposite direction. The polarization of the amino and carbonyl groups results in a net dipole moment on the α-pleated sheet. The carbonyl side acquires a net negative charge, and the amino side acquires a net positive charge.
A protein’s tertiary structure is the shape in which the entire protein chain folds together in three-dimensional space, and it is this level of structure that provides protein scientists with the most information about a protein’s specific function.
While a protein's secondary and tertiary structure is defined by how the protein chain folds together, quaternary structure is defined by how two or more folded protein chains come together to form a 'superstructure'. Many proteins consist of only one protein chain, or subunit, and thus have no quaternary structure. Many other proteins consist of two identical subunits (these are called homodimers) or two non-identical subunits (these are called heterodimers).
Quaternary structures can be quite elaborate: below we see a protein whose quaternary structure is defined by ten identical subunits arranged in two five-membered rings, forming what can be visualized as a 'double donut' shape (this is fructose 1,6-bisphosphate aldolase):
The molecular forces that hold proteins together
The question of exactly how a protein ‘finds’ its specific folded structure out of the vast number of possible folding patterns is still an active area of research. What is known, however, is that the forces that cause a protein to fold properly and to remain folded are the same basic noncovalent forces that we talked about in chapter 2: ion-ion, ion-dipole, dipole-dipole, hydrogen bonding, and hydrophobic (van der Waals) interactions. One interesting type of hydrophobic interaction is called ‘aromatic stacking’, and occurs when two or more planar aromatic rings on the side chains of phenylalanine, tryptophan, or tyrosine stack together like plates, thus maximizing surface area contact.
Hydrogen bonding networks are extensive within proteins, with both side chain and main chain atoms participating. Ionic interactions often play a role in protein structure, especially on the protein surface, as negatively charged residues such as aspartate interact with positively-charged groups on lysine or arginine.
One of the most important ideas to understand regarding tertiary structure is that a protein, when properly folded, is polar on the surface and nonpolar in the interior. It is the protein's surface that is in contact with water, and therefore the surface must be hydrophilic in order for the whole structure to be soluble. If you examine a three dimensional protein structure you will see many charged side chains (e.g. lysine, arginine, aspartate, glutamate) and hydrogen-bonding side chains (e.g. serine, threonine, glutamine, asparagine) exposed on the surface, in direct contact with water. Inside the protein, out of contact with the surrounding water, there tend to be many more hydrophobic residues such as alanine, valine, phenylalanine, etc. If a protein chain is caused to come unfolded (through exposure to heat, for example, or extremes of pH), it will usually lose its solubility and form solid precipitates, as the hydrophobic residues from the interior come into contact with water. You can see this phenomenon for yourself if you pour a little bit of vinegar (acetic acid) into milk. The solid clumps that form in the milk are proteins that have come unfolded due to the sudden acidification, and precipitated out of solution.
In recent years, scientists have become increasing interested in the proteins of so-called ‘thermophilic’ (heat-loving) microorganisms that thrive in hot water environments such as geothermal hot springs. While the proteins in most organisms (including humans) will rapidly unfold and precipitate out of solution when put in hot water, the proteins of thermophilic microbes remain completely stable, sometimes even in water that is just below the boiling point. In fact, these proteins typically only gain full biological activity when in appropriately hot water - at room-temperature they act is if they are ‘frozen’. Is the chemical structure of these thermostable proteins somehow unique and exotic? As it turns out, the answer to this question is ‘no’: the overall three-dimensional structures of thermostable proteins look very much like those of ‘normal’ proteins. The critical difference seems to be simply that thermostable proteins have more extensive networks of noncovalent interactions, particularly ion-ion interactions on their surface, that provides them with a greater stability to heat. Interestingly, the proteins of ‘psychrophilic’ (cold-loving) microbes isolated from pockets of water in arctic ice show the opposite characteristic: they have far fewer ion-ion interactions, which gives them greater flexibility in cold temperatures but leads to their rapid unfolding in room temperature water. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/25%3A_Amino_Acids_Peptides_and_Proteins/25.10%3A_Levels_of_Protein_Structure.txt |
Objectives
After completing this section, you should be able to
1. describe the catalytic role of an enzyme in a biochemical reaction.
2. give an example of one fat‑soluble and one water‑soluble vitamin.
Key Terms
Make certain that you can define, and use in context, the key term below.
• coenzyme
• cofactor
• enzyme
• substrate
• vitamin
Study Notes
You should have a general knowledge of the function of enzymes, but you need not memorize specific names or the classification system.
A catalyst is any substance that increases the rate or speed of a chemical reaction without being changed or consumed in the reaction. Enzymes are biological catalysts, and nearly all of them are proteins. In addition, enzymes are highly specific in their action; that is, each enzyme catalyzes only one type of reaction in only one compound or a group of structurally related compounds. The compound or compounds on which an enzyme acts are known as its substrates. Enzymes are classified by reaction type into six categories show in Table $1$.
Table $1$: Classes of Enzymes
Class Type of Reaction Catalyzed Examples
oxidoreductases oxidation-reduction reactions Dehydrogenases catalyze oxidation-reduction reactions involving hydrogen and reductases catalyze reactions in which a substrate is reduced.
transferases transfer reactions of groups, such as methyl, amino, and acetyl Transaminases catalyze the transfer of amino group, and kinases catalyze the transfer of a phosphate group.
hydrolases hydrolysis reactions Lipases catalyze the hydrolysis of lipids, and proteases catalyze the hydrolysis of proteins
lyases reactions in which groups are removed without hydrolysis or addition of groups to a double bond Decarboxylases catalyze the removal of carboxyl groups.
isomerases reactions in which a compound is converted to its isomer Isomerases may catalyze the conversion of an aldose to a ketose, and mutases catalyze reactions in which a functional group is transferred from one atom in a substrate to another.
ligases reactions in which new bonds are formed between carbon and another atom; energy is required Synthetases catalyze reactions in which two smaller molecules are linked to form a larger one.
Enzyme-catalyzed reactions occur in at least two steps. In the first step, an enzyme molecule (E) and the substrate molecule or molecules (S) collide and react to form an intermediate compound called the enzyme-substrate (E–S) complex (Equation $\ref{step1}$). This step is reversible because the complex can break apart into the original substrate or substrates and the free enzyme. Once the E–S complex forms, the enzyme is able to catalyze the formation of product (P), which is then released from the enzyme surface (Equation $\ref{step2}$):
$S + E \rightleftharpoons E–S \label{step1}$
$E–S → P + E \label{step2}$
Hydrogen bonding and other electrostatic interactions hold the enzyme and substrate together in the complex. The structural features or functional groups on the enzyme that participate in these interactions are located in a cleft or pocket on the enzyme surface. This pocket, where the enzyme combines with the substrate and transforms the substrate to product is called the active site of the enzyme (Figure $1$).
Figure $1$: Substrate Binding to the Active Site of an Enzyme. The enzyme dihydrofolate reductase is shown with one of its substrates: NADP+ (a) unbound and (b) bound. The NADP+ (shown in red) binds to a pocket that is complementary to it in shape and ionic properties.
The active site possesses a unique conformation (including correctly positioned bonding groups) that is complementary to the structure of the substrate, so that the enzyme and substrate molecules fit together in much the same manner as a key fits into a tumbler lock. In fact, an early model describing the formation of the enzyme-substrate complex was called the lock-and-key model (Figure $2$). This model portrayed the enzyme as conformationally rigid and able to bond only to substrates that exactly fit the active site.
Working out the precise three-dimensional structures of numerous enzymes has enabled chemists to refine the original lock-and-key model of enzyme actions. They discovered that the binding of a substrate often leads to a large conformational change in the enzyme, as well as to changes in the structure of the substrate or substrates. The current theory, known as theinduced-fit model, says that enzymes can undergo a change in conformation when they bind substrate molecules, and the active site has a shape complementary to that of the substrate only after the substrate is bound, as shown for hexokinase in Figure $3$. After catalysis, the enzyme resumes its original structure.
The structural changes that occur when an enzyme and a substrate join together bring specific parts of a substrate into alignment with specific parts of the enzyme’s active site. Amino acid side chains in or near the binding site can then act as acid or base catalysts, provide binding sites for the transfer of functional groups from one substrate to another or aid in the rearrangement of a substrate. The participating amino acids, which are usually widely separated in the primary sequence of the protein, are brought close together in the active site as a result of the folding and bending of the polypeptide chain or chains when the protein acquires its tertiary and quaternary structure. Binding to enzymes brings reactants close to each other and aligns them properly, which has the same effect as increasing the concentration of the reacting compounds.
Example $1$
1. What type of interaction would occur between an OH group present on a substrate molecule and a functional group in the active site of an enzyme?
2. Suggest an amino acid whose side chain might be in the active site of an enzyme and form the type of interaction you just identified.
Solution
1. An OH group would most likely engage in hydrogen bonding with an appropriate functional group present in the active site of an enzyme.
2. Several amino acid side chains would be able to engage in hydrogen bonding with an OH group. One example would be asparagine, which has an amide functional group.
Exercise $1$
1. What type of interaction would occur between an COO group present on a substrate molecule and a functional group in the active site of an enzyme?
2. Suggest an amino acid whose side chain might be in the active site of an enzyme and form the type of interaction you just identified.
Enzyme Cofactors and Vitamins
Many enzymes are simple proteins consisting entirely of one or more amino acid chains. Other enzymes contain a nonprotein component called a cofactor that is necessary for the enzyme’s proper functioning. There are two types of cofactors: inorganic ions [e.g., zinc or Cu(I) ions] and organic molecules known as coenzymes. Most coenzymes are vitamins or are derived from vitamins.
Vitamins are organic compounds that are essential in very small (trace) amounts for the maintenance of normal metabolism. They generally cannot be synthesized at adequate levels by the body and must be obtained from the diet. The absence or shortage of a vitamin may result in a vitamin-deficiency disease. In the first half of the 20th century, a major focus of biochemistry was the identification, isolation, and characterization of vitamins. Despite accumulating evidence that people needed more than just carbohydrates, fats, and proteins in their diets for normal growth and health, it was not until the early 1900s that research established the need for trace nutrients in the diet.
Table $2$: Fat-Soluble Vitamins and Physiological Functions
Vitamin Physiological Function Effect of Deficiency
vitamin A (retinol) formation of vision pigments; differentiation of epithelial cells night blindness; continued deficiency leads to total blindness
vitamin D (cholecalciferol) increases the body’s ability to absorb calcium and phosphorus osteomalacia (softening of the bones); known as rickets in children
vitamin E (tocopherol) fat-soluble antioxidant damage to cell membranes
vitamin K (phylloquinone) formation of prothrombin, a key enzyme in the blood-clotting process increases the time required for blood to clot
Because organisms differ in their synthetic abilities, a substance that is a vitamin for one species may not be so for another. Over the past 100 years, scientists have identified and isolated 13 vitamins required in the human diet and have divided them into two broad categories: the fat-soluble vitamins (Table $2$), which include vitamins A, D, E, and K, and the water-soluble vitamins, which are the B complex vitamins and vitamin C (Table $3$). All fat-soluble vitamins contain a high proportion of hydrocarbon structural components. There are one or two oxygen atoms present, but the compounds as a whole are nonpolar. In contrast, water-soluble vitamins contain large numbers of electronegative oxygen and nitrogen atoms, which can engage in hydrogen bonding with water. Most water-soluble vitamins act as coenzymes or are required for the synthesis of coenzymes. The fat-soluble vitamins are important for a variety of physiological functions.
Table $3$: Water-Soluble Vitamins and Physiological Functions
Vitamin Coenzyme Coenzyme Function Deficiency Disease
vitamin B1 (thiamine) thiamine pyrophosphate decarboxylation reactions beri-beri
vitamin B2 (riboflavin) flavin mononucleotide or flavin adenine dinucleotide oxidation-reduction reactions involving two hydrogen atoms
vitamin B3 (niacin) nicotinamide adenine dinucleotide or nicotinamide adenine dinucleotide phosphate oxidation-reduction reactions involving the hydride ion (H) pellagra
vitamin B6 (pyridoxine) pyridoxal phosphate variety of reactions including the transfer of amino groups
vitamin B12 (cyanocobalamin) methylcobalamin or deoxyadenoxylcobalamin intramolecular rearrangement reactions pernicious anemia
biotin biotin carboxylation reactions
folic acid tetrahydrofolate carrier of one-carbon units such as the formyl group anemia
pantothenic Acid coenzyme A carrier of acyl groups
vitamin C (ascorbic acid) none antioxidant; formation of collagen, a protein found in tendons, ligaments, and bone scurvy
One characteristic that distinguishes an enzyme from all other types of catalysts is its substrate specificity. An inorganic acid such as sulfuric acid can be used to increase the reaction rates of many different reactions, such as the hydrolysis of disaccharides, polysaccharides, lipids, and proteins, with complete impartiality. In contrast, enzymes are much more specific. Some enzymes act on a single substrate, while other enzymes act on any of a group of related molecules containing a similar functional group or chemical bond. Some enzymes even distinguish between D- and L-stereoisomers, binding one stereoisomer but not the other. Urease, for example, is an enzyme that catalyzes the hydrolysis of a single substrate—urea—but not the closely related compounds methyl urea, thiourea, or biuret. The enzyme carboxypeptidase, on the other hand, is far less specific. It catalyzes the removal of nearly any amino acid from the carboxyl end of any peptide or protein.
Enzyme specificity results from the uniqueness of the active site in each different enzyme because of the identity, charge, and spatial orientation of the functional groups located there. It regulates cell chemistry so that the proper reactions occur in the proper place at the proper time. Clearly, it is crucial to the proper functioning of the living cell.
Concept Review Exercises
1. Distinguish between the lock-and-key model and induced-fit model of enzyme action.
2. Which enzyme has greater specificity—urease or carboxypeptidase? Explain.
Answers
1. The lock-and-key model portrays an enzyme as conformationally rigid and able to bond only to substrates that exactly fit the active site. The induced fit model portrays the enzyme structure as more flexible and is complementary to the substrate only after the substrate is bound.
2. Urease has the greater specificity because it can bind only to a single substrate. Carboxypeptidase, on the other hand, can catalyze the removal of nearly any amino acid from the carboxyl end of a peptide or protein.
Takeaways
• A substrate binds to a specific region on an enzyme known as the active site, where the substrate can be converted to product.
• The substrate binds to the enzyme primarily through hydrogen bonding and other electrostatic interactions.
• The induced-fit model says that an enzyme can undergo a conformational change when binding a substrate.
• Enzymes exhibit varying degrees of substrate specificity.
Exercises
1. What type of interaction would occur between each group present on a substrate molecule and a functional group of the active site in an enzyme?
1. COO-
2. NH3+
3. OH
4. CH(CH3)2
2. What type of interaction would occur between each group present on a substrate molecule and a functional group of the active site in an enzyme?
1. SH
2. NH2
3. C6H5
4. COO
3. For each functional group in Exercise 1, suggest an amino acid whose side chain might be in the active site of an enzyme and form the type of interaction you identified.
4. For each functional group in Exercise 2, suggest an amino acid whose side chain might be in the active site of an enzyme and form the type of interaction you identified.
Answers
1. ionic bonding (salt bridge)
2. ionic bonding (salt bridge)
3. hydrogen bonding
4. dispersion forces
2. a. disulfide bridge
b. H-bond because we can infer this amino group is part of an amide since it is not ionized
c. London dispersion
d. salt bridge (ionic bonding)
1. The amino acid has a positively charged side chain capable of forming salt bridges: histidine or arginine or lysine.
2. The amino acid has a negatively charged side chain capable of forming salt bridges; aspartic acid or glutamic acid.
3. The amino acid has a polar side chain capable of engaging in hydrogen bonding: serine or threonine or tyrosine or asparagine or glutamine or cysteine.
4. The amino acid has a nonpolar side chain with London dispersion forces: alanine or valine or phenylalanine or leucine or isoleucine or methionine or tryptophan or proline or glycine.
4. a. The amino acid can form a disulfide bridge: cysteine OR the amino acid has a polar side chain capable of engaging in hydrogen bonding: serine or threonine or tyrosine or asparagine or glutamine
b. The amino acid has a polar side chain capable of engaging in hydrogen bonding: serine or threonine or tyrosine or asparagine or glutamine or cysteine.
c. The amino acid has a nonpolar side chain with London dispersion forces: alanine or valine or phenylalanine or leucine or isoleucine or methionine or tryptophan or proline or glycine.
d. The amino acid has a positively charged side chain capable of forming salt bridges: histidine or arginine or lysine.
25.12: How do enzymes work
Objectives
After completing this section, you should be able to
1. describe and explain the general function of an enzyme like citrate synthase in a reaction.
2. identify the structures of ten common coenzymes.
Oxaloacetate to Citrate Catalyzed by Citrate Synthase
Citrate synthase is a protein with 433 amino acids with various functional groups that can react with substrates. This enzyme catalyzes oxaloacetate to eventually produce citrate as part of the citric acid (Krebs) cycle. In the first step of the citric acid (Krebs) cycle, acetyl CoA condenses with oxaloacetate to form (S)-citryl CoA. The carboxylate group of an aspartic acid (B:) on citrate synthase removes the acidic alpha proton on acetyl CoA, while a histidine site (H-A) donates a proton to form the enol. Then a second histidine site (H-A) protonates the carbonyl oxygen of oxaloacetate, while the carbon of the carbonyl is attacked by the enol. Simultaeously, that first histidine (:A-) deprotonates the acetyl CoA enol. (S)-citryl CoA is generated.
The acyl group of a thioester of (S)-citryl CoA can be transferred to a water molecule in a hydrolysis reaction to converting (S)-citryl CoA to citrate. Again histidine sites on citrate synthase are an integral part of the mechanism and assist with removal and addition of protons. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/25%3A_Amino_Acids_Peptides_and_Proteins/25.11%3A_Enzymes_and_Coenzymes.txt |
• 26.1: Introduction
Although many lipids have complex structures, they are all insoluble in water. The major lipid classifications are: fatty acids, steroids, waxes, triglycerides (fats and oils), phospholipids, sphingolipids, glycolipids, terpenes, and prostaglandins (eicosanoids).
• 26.2: Waxes, Fats, and Oils
Fats play an important role in human nutrition, and most people are aware of the desirability of limiting their dietary intake of saturated fats, as these compounds have been associated with heart disease. Unsaturated fats are generally considered to be much more desirable from the point of view of good health. Notice that all the fatty acids derived from naturally occurring fats have a Z (i.e., cis) configuration.
• 26.3: Saponification of Fats and Oils; Soaps and Detergents
Soaps are the carboxylate salts of fatty acids, while detergents are sulfonate salts with long hydrocarbon tails.
• 26.4: Phospholipids
Phospholipids are the main constituents of cell membranes.
• 26.5: Prostaglandins and other Eicosanoids
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.
• 26.6: Terpenes and Terpenoids
The terpenoids are an estimated 60% of known natural products and a diverse group of lipids derived from five-carbon isoprene units assembled in thousands of combinations. Technically a terpenoid contains oxygen, while a terpene is a hydrocarbon, however, the two terms are commonly used to refer collectively to both groups.
• 26.7: Steroids
Steroids may be recognized by their tetracyclic skeleton consisting of three fused six-membered and one five-membered ring.
• 26.8: Biosynthesis of Steroids
A series of cation-like cyclizations and rearrangements, known as the Stork-Eschenmoser hypothesis, were identified in the biosynthesis of the triterpene lanosterol. Lanosterol is a precursor in the biosynthesis of steroids.
26: Lipids
Objectives
After completing this section, you should be able to identify fats and steroids as being examples of lipids.
Key Terms
Make certain that you can define, and use in context, the key term below.
• lipid
Lipids are not defined by the presence of specific functional groups, as carbohydrates are, but by a physical property—solubility. Compounds isolated from body tissues are classified as lipids if they are more soluble in organic solvents, such as dichloromethane, than in water. By this criterion, the lipid category includes not only fats and oils, which are esters of the trihydroxy alcohol glycerol and fatty acids, but also compounds that incorporate functional groups derived from phosphoric acid, carbohydrates, or amino alcohols, as well as steroid compounds such as cholesterol. The diagram below presents one scheme for classifying the various kinds of lipids.
26.02: Waxes Fats and Oils
Objectives
After completing this section, you should be able to
1. identify waxes as being mixtures of long‑chain esters, and write the general structure for such compounds.
2. identify fats and oils as being triacylglycerols, and write a general structure for such compounds.
3. relate the physical properties of animal fats and vegetable oils to their structures.
4. predict the behaviour of a given fat or oil when it is subjected to some of the more common reactions discussed in previous units; examples would include hydrolysis, reduction and ozonolysis.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• fat
• fatty acids
• triacylglycerols
Study Notes
You are not expected to memorize the trivial names or formulas of the fatty acids listed in Table 27.1. The systematic names for these compounds are shown in the tables below.
Saturated Fatty Acids
Table 27.1 Systematic names of some common saturated fatty acids
Trivial Name Systematic Name
lauric acid dodecanoic acid
myristic acid tetradecanoic acid
palmitic acid hexadecanoic acid
stearic acid octadecanoic acid
arachidic acid eicosanoic acid
Unsaturated Fatty Acids
Table 27.2 Systematic names of some common unsaturated fatty acids
Trivial Name Systematic Name
palmitoleic acid (Z)‑9‑hexadecenoic acid
oleic acid (Z)‑9‑octadecenoic acid
ricinoleic acid (Z)‑12‑hydroxy‑9‑octadecenoic acid
linoleic acid (Z,Z)‑9,12‑octadecadienoic acid
linolenic acid (Z,Z,Z)‑9,12,15‑octadecatrienoic acid
arachidonic acid (Z,Z,Z,Z)‑5,8,11,14‑eicosatetraenoic acid
The systematic names in these tables may be somewhat unfamiliar, but they are derived in exactly the same way as the names of the simpler carboxylic acids that we discussed in Chapter 20. You may wish to review the alkane names (Section 3.2) and the E,Z system (Section 7.5) to satisfy yourself that you understand how these names originate.
In many older textbooks, the term “fatty acid” is used to describe all carboxylic acids, not only those that are obtained from the hydrolysis of triacylglycerols.
Fats play an important role in human nutrition, and most people are aware of the desirability of limiting their dietary intake of saturated fats, as these compounds have been associated with heart disease. Unsaturated fats are generally considered to be much more desirable from the point of view of good health. Notice that all the fatty acids derived from naturally occurring fats have a Z (i.e., cis) configuration.
Linoleic acid is an “essential” nutrient; that is, it cannot be made by the body in sufficient quantity to meet our physiological needs, and must be obtained from food. A deficiency in linoleic acid results in skin problems and liver abnormalities. Historically, linolenic and arachidonic acids were also thought to be essential nutrients, but recent research suggests that they can be synthesized in the body if sufficient linoleic acid is present.
Waxes are esters of fatty acids with long chain monohydric alcohols (one hydroxyl group). Natural waxes are often mixtures of such esters, and may also contain hydrocarbons. The formulas for three well known waxes are given below, with the carboxylic acid moiety colored red and the alcohol colored blue.
Spermaceti
Beeswax
Carnuba wax
CH3(CH2)14CO2-(CH2)15CH3
CH3(CH2)24CO2-(CH2)29CH3
CH3(CH2)30CO2-(CH2)33CH3
Waxes are widely distributed in nature. The leaves and fruits of many plants have waxy coatings, which may protect them from dehydration and small predators. The feathers of birds and the fur of some animals have similar coatings which serve as a water repellent. Carnuba wax is valued for its toughness and water resistance.
Triglycerides are esters of fatty acids and a trifunctional alcohol - glycerol (IUPAC name is 1,2,3-propantriol). The properties of fats and oils follow the same general principles as already described for the fatty acids. The important properties to be considered are: melting points and degree of unsaturation from component fatty acids. Since glycerol has three alcohol functional groups, three fatty acids must react to make three ester functional groups. The three fatty acids may or may not be identical. In fact, three different fatty acids may be present. The synthesis of a triglyceride is another application of the ester synthesis reaction. To write the structure of the triglyceride you must know the structure of glycerol and be given or look up the structure of the fatty acid in the table.
The common fats and oils including fatty acid content are listed below.
glycerides
Fat or Oil Saturated Unsaturated
Palmitic Stearic Oleic Linoleic Other
Animal Origin
Butter 29 9 27 4 31
Lard 30 18 41 6 5
Beef 32 25 38 3 2
Vegetable Origin
Corn oil 10 4 34 48 4
Soybean 7 3 25 56 9
Peanut 7 5 60 21 7
Olive 6 4 83 7 -
The higher melting points of the saturated fatty acids reflect the uniform rod-like shape of their molecules. The cis-double bond(s) in the unsaturated fatty acids introduce a kink in their shape, which makes it more difficult to pack their molecules together in a stable repeating array or crystalline lattice. The trans-double bond isomer of oleic acid, known as elaidic acid, has a linear shape and a melting point of 45 ºC (32 ºC higher than its cis isomer). The shapes of stearic and oleic acids are displayed in the models below.
Stearic acid
Oleic acid
Two polyunsaturated fatty acids, linoleic and linolenic, are designated "essential" because their absence in the human diet has been associated with health problems, such as scaley skin, stunted growth and increased dehydration. These acids are also precursors to the prostaglandins, a family of physiologically potent lipids present in minute amounts in most body tissues.
Synthesis of a Triglyceride
Since glycerol, (IUPAC name is 1,2,3-propantriol), has three alcohol functional groups, three fatty acids must react to make three ester functional groups. The three fatty acids may or may not be identical. In fact, three different fatty acids may be present. nThe synthesis of a triglyceride is another application of the ester synthesis reaction. To write the structure of the triglyceride you must know the structure of glycerol and be given or look up the structure of the fatty acid in the table - find lauric acid.
Glycerol
The simplified reaction reveals the process of breaking some bonds and forming the ester and the by product, water. Refer to the graphic on the left for the synthesis of trilauroylglycerol. First, the -OH (red) bond on the acid is broken and the -H (red) bond on the alcohol is also broken. Both join to make HOH, a water molecule. Secondly, the oxygen of the alcohol forms a bond (green) to the acid at the carbon with the double bond oxygen. This forms the ester functional group. This process is carried out three times to make three ester groups and three water molecules.
As might be expected from the properties of the fatty acids, fats have a predominance of saturated fatty acids, and oils are composed largely of unsaturated acids. Thus, the melting points of triglycerides reflect their composition, as shown by the following examples. Natural mixed triglycerides have somewhat lower melting points, the melting point of lard being near 30 º C, whereas olive oil melts near -6 º C. Since fats are valued over oils by some Northern European and North American populations, vegetable oils are extensively converted to solid triglycerides (e.g. Crisco) by partial hydrogenation of their unsaturated components. Some of the remaining double bonds are isomerized (to trans) in this operation. These saturated and trans-fatty acid glycerides in the diet have been linked to long-term health issues such as atherosclerosis.
Triglycerides having three identical acyl chains, such as tristearin and triolein (above), are called "simple", while those composed of different acyl chains are called "mixed". If the acyl chains at the end hydroxyl groups (1 & 3) of glycerol are different, the center carbon becomes a chiral center and enantiomeric configurations must be recognized.
The hydrogenation of vegetable oils to produce semisolid products has had unintended consequences. Although the hydrogenation imparts desirable features such as spreadability, texture, "mouth feel," and increased shelf life to naturally liquid vegetable oils, it introduces some serious health problems. These occur when the cis-double bonds in the fatty acid chains are not completely saturated in the hydrogenation process. The catalysts used to effect the addition of hydrogen isomerize the remaining double bonds to their trans configuration. These unnatural trans-fats appear to to be associated with increased heart disease, cancer, diabetes and obesity, as well as immune response and reproductive problems. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/26%3A_Lipids/26.01%3A_Introduction.txt |
Learning Objectives
After completing this section, you should be able to
1. write an equation to represent the formation of a soap.
2. identify the structure of the fat required to produce a given soap.
3. identify the structure of a soap, given the structure of the fat from which it is produced.
1. describe the mechanism by which soaps exert their cleansing action.
2. give a chemical explanation of the problems encountered when carboxylate soaps are used in hard‑water areas, and explain how they may be overcome by the use of sulphonate detergents.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• hydrophilic
• lipophilic (hydrophobic)
• amphiphilic
• micelles
Carboxylic acids and salts having alkyl chains longer than eight 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. Fatty acids made up of ten or more carbon atoms are nearly insoluble in water, and because of their lower density, float on the surface when mixed with water. Unlike paraffin or other alkanes, which tend to puddle on the waters surface, these fatty acids spread evenly over an extended water surface, eventually forming a monomolecular layer in which the polar carboxyl groups are hydrogen bonded at the water interface, and the hydrocarbon chains are aligned together away from the water. This behavior is illustrated in the diagram on the right. Substances that accumulate at water surfaces and change the surface properties are called surfactants.
Alkali metal salts of fatty acids are more soluble in water than the acids themselves, and the amphiphilic character of these substances also make them strong surfactants. The most common examples of such compounds are soaps and detergents, four of which are shown below. Note that each of these molecules has a nonpolar hydrocarbon chain, the "tail", and a polar (often ionic) "head group". The use of such compounds as cleaning agents is facilitated by their surfactant character, which lowers the surface tension of water, allowing it to penetrate and wet a variety of materials.
Very small amounts of these surfactants dissolve in water to give a random dispersion of solute molecules. However, when the concentration is increased an interesting change occurs. The surfactant molecules reversibly assemble into polymolecular aggregates called micelles. By gathering the hydrophobic chains together in the center of the micelle, disruption of the hydrogen bonded structure of liquid water is minimized, and the polar head groups extend into the surrounding water where they participate in hydrogen bonding. These micelles are often spherical in shape, but may also assume cylindrical and branched forms, as illustrated on the right. Here the polar head group is designated by a blue circle, and the nonpolar tail is a zig-zag black line.
The oldest amphiphilic cleaning agent known to humans is soap. Soap is manufactured by the base-catalyzed hydrolysis (saponification) of animal fat (see below). Before sodium hydroxide was commercially available, a boiling solution of potassium carbonate leached from wood ashes was used. Soft potassium soaps were then converted to the harder sodium soaps by washing with salt solution. The importance of soap to human civilization is documented by history, but some problems associated with its use have been recognized. One of these is caused by the weak acidity (pKa ca. 4.9) of the fatty acids. Solutions of alkali metal soaps are slightly alkaline (pH 8 to 9) due to hydrolysis. If the pH of a soap solution is lowered by acidic contaminants, insoluble fatty acids precipitate and form a scum. A second problem is caused by the presence of calcium and magnesium salts in the water supply (hard water). These divalent cations cause aggregation of the micelles, which then deposit as a dirty scum.
These problems have been alleviated by the development of synthetic amphiphiles called detergents (or syndets). By using a much stronger acid for the polar head group, water solutions of the amphiphile are less sensitive to pH changes. Also the sulfonate functions used for virtually all anionic detergents confer greater solubility on micelles incorporating the alkaline earth cations found in hard water. Variations on the amphiphile theme have led to the development of other classes, such as the cationic and nonionic detergents shown above. Cationic detergents often exhibit germicidal properties, and their ability to change surface pH has made them useful as fabric softeners and hair conditioners. These versatile chemical "tools" have dramatically transformed the household and personal care cleaning product markets over the past fifty years
Chemical Reactions of Fats and Oils
Fats and oils can participate in a variety of chemical reactions—for example, because triglycerides are esters, they can be hydrolyzed in the presence of an acid, a base, or specific enzymes known as lipases. The hydrolysis of fats and oils in the presence of a base is used to make soap and is called saponification. Today most soaps are prepared through the hydrolysis of triglycerides (often from tallow, coconut oil, or both) using water under high pressure and temperature [700 lb/in2 (∼50 atm or 5,000 kPa) and 200°C]. Sodium carbonate or sodium hydroxide is then used to convert the fatty acids to their sodium salts (soap molecules):
26.04: Phospholipids
Learning Objectives
After completing this section, you should be able to
1. draw the general structure of a phosphoglyceride.
2. describe the occurrence and importance of phosphoglycerides in plant and animal tissues.
Key Terms
Make certain that you can define, and use in context, the key terms below.
Phospholipids are the main constituents of cell membranes. They resemble the triglycerides in being ester or amide derivatives of glycerol or sphingosine with fatty acids and phosphoric acid. The phosphate moiety of the resulting phosphatidic acid is further esterified with ethanolamine, choline or serine in the phospholipid itself. The following diagram shows the structures of some of these components. Clicking on the diagram will change it to display structures for two representative phospholipids. Note that the fatty acid components (R & R') may be saturated or unsaturated.
As ionic amphiphiles, phospholipids aggregate or self-assemble when mixed with water, but in a different manner than the soaps and detergents. Because of the two pendant alkyl chains present in phospholipids and the unusual mixed charges in their head groups, micelle formation is unfavorable relative to a bilayer structure. If a phospholipid is smeared over a small hole in a thin piece of plastic immersed in water, a stable planar bilayer of phospholipid molecules is created at the hole. As shown in the following diagram, the polar head groups on the faces of the bilayer contact water, and the hydrophobic alkyl chains form a nonpolar interior. The phospholipid molecules can move about in their half the bilayer, but there is a significant energy barrier preventing migration to the other side of the bilayer.
This bilayer membrane structure is also found in aggregate structures called liposomes. Liposomes are microscopic vesicles consisting of an aqueous core enclosed in one or more phospholipid layers. They are formed when phospholipids are vigorously mixed with water. Unlike micelles, liposomes have both aqueous interiors and exteriors.
A cell may be considered a very complex liposome. The bilayer membrane that separates the interior of a cell from the surrounding fluids is largely composed of phospholipids, but it incorporates many other components, such as cholesterol, that contribute to its structural integrity. Protein channels that permit the transport of various kinds of chemical species in and out of the cell are also important components of cell membranes.
The interior of a cell contains a variety of structures (organelles) that conduct chemical operations vital to the cells existence. Molecules bonded to the surfaces of cells serve to identify specific cells and facilitate interaction with external chemical entities. The sphingomyelins are also membrane lipids. They are the major component of the myelin sheath surrounding nerve fibers. Multiple Sclerosis is a devastating disease in which the myelin sheath is lost, causing eventual paralysis. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/26%3A_Lipids/26.03%3A_Saponification_of_Fats_and_Oils_Soaps_and_Detergents.txt |
Learning Objectives
After completing this section, you should be able to
1. describe the general structure of the prostaglandins, and identify a prostaglandin from a given list of organic structures.
2. identify at least two important biological functions of prostaglandins.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• eicosanoid
• prostaglandin
Eicosanoids
The members of this group of structurally related natural hormones have an extraordinary range of biological effects. They can lower gastric secretions, stimulate uterine contractions, lower blood pressure, influence blood clotting and induce asthma-like allergic responses. Because their genesis in body tissues is tied to the metabolism of the essential fatty acid arachadonic acid (5,8,11,14-eicosatetraenoic acid) they are classified as eicosanoids. Many properties of the common drug aspirin result from its effect on the cascade of reactions associated with these hormones. Eicosanoids include prostaglandins, leukotrienes, and thromboxanes.
Eicosanoids are primarily named off the ring system present in the molecule prostaglandin (PG), thromboxane (TX), leukotriene (LT). Common substitution patterns on the ring system are indicated with a letter in the name. Also, the number of double bonds in the molecule is indicated with a subscript number.
Prostaglandin structures for PGA, PGD, PGE and PGF
Structures for prostaglandins PGG, PGH and PGI and thromboxanes TXA and TXB
Thus the molecule PGH2 means that it has the prostaglandin ring system with a type H Substitution pattern. The subscript 2 indicated that the molecule contains two double bonds.
Prostaglandin PGH2
Prostaglandins
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.
Prostaglandins are found in low concentrations distributed in a large number of organs, tissues, and body fluids of mammals. They exhibit a broad spectrum of physiological activity and are remarkably potent. Their precise biological role is not entirely clear, but they are known to induce strong contractions of smooth muscle tissue (lungs, uterus) and to lower blood pressure and sodium levels. Prostaglandins also have been implicated in the control of pituitary hormones released from the hypothalamus, and in the incidence of "pain" as a response to fever and inflammation. In fact, the analgesic property of aspirin possibly may result from the inhibition of prostaglandin biosynthesis
Prostaglandins are unsaturated carboxylic acids, consisting of of a 20 carbon skeleton that also contains a five member ring and are based upon prostanoic acid, which is a C20 fatty acid in which there is a cyclopentane ring formed by connecting the C8 and C12 positions.
They are biochemically synthesized from the fatty acid, arachidonic acid. 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 of the prostaglandin.
Arachidonic acid
There are a variety of functional groups present in prostaglandin structures. The can have one, two, or three double bonds. On the five member ring there may also be double bonds, a ketone, or alcohol groups. Some typical structures are shown below.
Prostaglandin E2
Prostaglandin F
Thromboxanes
Thromboxanes play roles in clot formation and named for their role in thrombosis. They are potent vasoconstrictors and facilitate platelet aggregation. They are synthesized in platelets, as well. The anti-clotting effects of aspirin have their roots in the inhibition of synthesis of PGH2, which is the precursor of the thromboxanes. The most common thromboxanes are A2 (Figure 2.217) and B2.
Thromboxane A22
Leukotrienes
Another group of eicosanoid compounds are the leukotrienes (Figure 2.219). Like prostaglandins, leukotrienes are made from arachidonic acid. The enzyme catalyzing their formation is a dioxygenase known as arachidonate 5-lipoxygenase. Leukotrienes are involved in regulating immune responses. They are found in leukocytes and other immunocompetent cells, such as neutrophils, monocytes, mast cells, eosinophils, and basophils. Leukotrienes are associated with production of histamines and prostaglandins, which act as mediators of inflammation. Leukotrienes also trigger contractions in the smooth muscles of the bronchioles. When overproduced, they may pay a role in asthma and allergic reactions. Some treatments for asthma aim at inhibiting production or action of leukotrienes.
Leukotriene A4 (LTA4)
Eicosanoid Biosynthesis
The biosynthesis of eicosanoids begins with the reaction of arachidonic acid with O2 which can be catalyzed by two different cyclooxygenase enzymes (COX-1 & COX-2).
In nature, prostaglandins arise by an oxidative cyclization of poly-unsaturated twenty-carbon fatty acids, which begins with enantiospecific removal of the L-hydrogen atom of the prochiral methylene group at C-13 coupled with enantiospecific introduction of oxygen at the allylic C-15 position. Subsequent cyclization and termination by addition of a second molecule of oxygen leads to a 15-hydroperoxy bicyclic peroxide (PGG), that is reduced to a 15-hydroxy bicyclic peroxide (PGH). These intermediates, known as prostaglandin endoperoxides, have been isolated and shown to yield prostaglandins. Reduction of the peroxy bridge gives PGF, while disproportionation gives β-hydroxy ketones PGE and PGD. The carbons in prostaglandins are numbered one to twenty starting at the carboxyl carbon and following the numbering system of the biosynthetic precursor fatty acids. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/26%3A_Lipids/26.05%3A_Prostaglandins_and_other_Eicosanoids.txt |
Objectives
After completing this section, you should be able to
1. identify a terpene from a given list of organic structures.
2. analyse the structure of a given terpene in terms of the isoprene rule.
3. classify a given terpene structure according to the number of isoprene units present; that is, determine whether a given terpene is a monoterpene, sesquiterpene, diterpene, etc.
4. identify and draw the structure of the precursors of isopentenyl diphosphate: (R) mevalonate or 1‑deoxy‑D‑xylulose 5‑phosphate.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• isoprene rule
• terpenoid
• terpene
Study Notes
You are not expected to memorize all the details of the synthetic mechanisms for terpenoids. However, you should know the overall general synthetic pathway illustrated under “Terpenoid Biosynthesis.” As you read through the details of these mechanisms, realize that they may be complex, but they are based on experimental evidence. You should also note the important role of enzymes in many natural systems transformations. Finally, you will recognize that essentially, individual steps are often reactions you have already encountered in previous sections.
The terpenoids (aka isoprenoids) are a large (estimated 60% of known natural products ) and diverse group of lipids derived from five-carbon isoprene units assembled in thousands of combinations. Technically a terpenoid contains oxygen, while a terpene is a hydrocarbon. Often the two terms are used to refer collectively to both groups.
Isoprene Rule
Compounds classified as terpenes constitute what is arguably the largest and most diverse class of natural products. A majority of these compounds are found only in plants, but some of the larger and more complex terpenes (e.g. squalene & lanosterol) occur in animals. Terpenes incorporating most of the common functional groups are known, so this does not provide a useful means of classification. Instead, the number and structural organization of carbons is a definitive characteristic. Terpenes may be considered to be made up of isoprene (more accurately isopentane) units, an empirical feature known as the isoprene rule. Because of this, terpenes usually have \(5n\) carbon atoms (\(n\) is an integer), and are subdivided as follows:
Classification Isoprene Units Carbon Atoms
monoterpenes 2 C10
sesquiterpenes 3 C15
diterpenes 4 C20
sesterterpenes 5 C25
triterpenes 6 C30
Isoprene itself, a C5H8 gaseous hydrocarbon, is emitted by the leaves of various plants as a natural byproduct of plant metabolism. Next to methane it is the most common volatile organic compound found in the atmosphere. Examples of C10 and higher terpenes, representing the four most common classes are shown in the following diagrams. Most terpenes may be structurally dissected into isopentane segments. How this is done can be seen in the diagram directly below.
The isopentane units in most of these terpenes are easy to discern, and are defined by the shaded areas. In the case of the monoterpene camphor, the units overlap to such a degree it is easier to distinguish them by coloring the carbon chains. This is also done for alpha-pinene. In the case of the triterpene lanosterol we see an interesting deviation from the isoprene rule. This thirty carbon compound is clearly a terpene, and four of the six isopentane units can be identified. However, the ten carbons in center of the molecule cannot be dissected in this manner. Evidence exists that the two methyl groups circled in magenta and light blue have moved from their original isoprenoid locations (marked by small circles of the same color) to their present location. This rearrangement is described in the biosynthesis section. Similar alkyl group rearrangements account for other terpenes that do not strictly follow the isoprene rule.
Polymeric isoprenoid hydrocarbons have also been identified. Rubber is undoubtedly the best known and most widely used compound of this kind. It occurs as a colloidal suspension called latex in a number of plants, ranging from the dandelion to the rubber tree (Hevea brasiliensis). Rubber is a polyene, and exhibits all the expected reactions of the C=C function. Bromine, hydrogen chloride and hydrogen all add with a stoichiometry of one molar equivalent per isoprene unit. Ozonolysis of rubber generates a mixture of levulinic acid ( \(CH_3COCH_2CH_2CO_2H\) ) and the corresponding aldehyde. Pyrolysis of rubber produces the diene isoprene along with other products.
The double bonds in rubber all have a Z-configuration, which causes this macromolecule to adopt a kinked or coiled conformation. This is reflected in the physical properties of rubber. Despite its high molecular weight (about one million), crude latex rubber is a soft, sticky, elastic substance. Chemical modification of this material is normal for commercial applications. Gutta-percha (structure above) is a naturally occurring E-isomer of rubber. Here the hydrocarbon chains adopt a uniform zig-zag or rod like conformation, which produces a more rigid and tough substance. Uses of gutta-percha include electrical insulation and the covering of golf balls.
Terpenoid Biosynthesis
While we can identify isoprene units within a terpenoid structure and use that in its classification, the building block for terpenoid synthesis in nature is isopentenyl diphosphate (formerly called isopentenyl pyrophosphate and abbreviated IPP). There are two major routes to the synthesis of IPP; namely (1) the mevalonate pathway and (2) the 1-deoxyxylulose pathway.
Mevalonate Pathway
Step 1 - Claisen Condensation
An early step in the biosynthesis of cholesterol and other ‘isoprenoid’ compounds is a Claisen condensation between two acetyl CoA molecules. An initial trans-thioesterase process transfers the acetyl group of the first acetyl CoA to an enzymatic cysteine (Reaction 1). In the Claisen condensation phase of the reaction, the alpha-carbon of a second acetyl CoA is deprotonated, forming an enolate (Reaction 2).
The enolate carbon attacks the electrophilic thioester carbon, forming a tetrahedral intermediate (Reaction 3) which quickly collapses to expel the cysteine thiol (Reaction 4) and produce acetoacetyl CoA.
Step 2 - Aldol Condensation
Acetyl CoA then reacts with the acetoacetyl CoA in an aldol-like addition. Subsequent hydrolysis produces (3S)-3-hydroxy-3-methylglutaryl CoA (HMG-CoA).
Generating HMG-CoA
Step 3 - Reduction of the Thioester
The thioester is reduced first to an aldehyde, then to a primary alcohol by two equivalents of NADPH producing (R)-mevalonate. The enzyme catalyzing this reaction is the target of the statin family of cholesterol-lowering drugs.
Generating (R)-Mevalonate
Step 4 - Mevalonate Phosphorylation
Two phsophorylations by adenosine triphosphate (ATP) occur at the terminal hydroxyl/phosphorus group through nucleophilic substitution, followed by a third ATP phosphorylation of the tertiary hydroxyl group.
Step 5 - Decarboxylation
Finally isopentenyl diphosphate (IPP), the 'building block' for all isoprenoid compounds, is formed from a decarboxylation-elimination reaction.
Conversion of IPP to Terpenoids
The electrophilic double bond isomerization catalyzed by IPP isomerase is a highly reversible reaction, with an equilibrium IPP:DMAPP ratio of about 6:1. In the next step of isoprenoid biosynthesis, the two five-carbon isomers condense to form a 10-carbon isoprenoid product called geranyl diphosphate (GPP).
This is a nice example of an electrophilic addition/elimination mechanism:
The first step is ionization of the electrophile - in other words, the leaving group departs and a carbocation intermediate is formed. In this case, the pyrophosphate group on DMAPP is the leaving group, and the electrophilic species is the resulting allylic carbocation.
In the condensation (addition) step, the C3-C4 double bond in IPP attacks the positively-charged C1 of DMAPP, resulting in a new carbon-carbon bond and a second carbocation intermediate, this time at a tertiary carbon. In the elimination phase, proton abstraction leads to re-establishment of a double bond in the GPP product. Notice that the enzyme specifically takes the pro-R proton in this step.
To continue the chain elongation process, another IPP molecule can then condense, in a very similar reaction, with C1 of geranyl diphosphate to form a 15-carbon product called farnesyl diphosphate (FPP).
How do we know that these are indeed SN1-like mechanisms with carbocation intermediates, rather than concerted SN2-like mechanisms? First of all, recall that the question of whether a substitution is dissociative (SN1-like) or associative (SN2-like) is not always clear-cut - it could be somewhere in between, like the protein prenyltransferase reaction. The protein prenyltransferase reaction and the isoprenoid chain elongation reactions are very similar: the electrophile is the same, but in the former the nucleophile is a thiolate, while in the latter the nucleophile is a pi bond.
This difference in the identity of the nucleophilic species would lead one to predict that the chain elongation reaction has more SN1-like character than the protein prenylation reaction. A thiolate is a very powerful nucleophile, and thus is able to push the pyrophosphate leaving group off, implying some degree of SN2 character. The electrons in a pi bond, in contrast, are only weakly nucleophilic, and thus need to be pulled in by a powerful electrophile - ie. a carbocation.
So it makes perfect sense that the chain elongation reaction should more SN1-like than SN2-like. Is this in fact the case? We know how to answer this question experimentally - just run the reaction with fluorinated DMAPP or GPP substrates and observe how much the fluorines slow things down.
If the reaction is SN1-like, the electron-withdrawing fluorines should destabilize the allylic carbocation intermediate and thus slow the reaction down considerably. If the mechanism is SN2-like, the fluorine substitutions should not have a noticeable effect, because a carbocation intermediate would not be formed. When this experiment was performed with FPP synthase, the results were dramatic: the presence of a single fluorine slowed down the rate of the reaction by a factor of about 60, while two and three fluorines resulted in a reaction that was 500,000 and 3 million times slower, respectively (J. Am. Chem. Soc. 1981, 103, 3926.) These results strongly suggest indicate the formation of a carbocation intermediate in an SN1-like displacement.
In this section, we will briefly examine the reaction catalyzed by an enzyme called squalene synthase, an important enzymatic transformation that involves some very interesting and unusual electrophilic additions, rearrangements, and reactive intermediates. This particular enzyme is also of interest because it represents a potential new target for cholesterol-lowering drugs.
Cholesterol, as we discussed earlier in this chapter, is derived from a 30-carbon isoprenoid molecule called squalene. Squalene, in turn, is derived from the condensation of two molecules of farnesyl diphosphate (FPP), a 15-carbon isoprenoid. You may recall that FPP is the product of the C4 to C1, or 'head to tail' electrophilic condensation of isoprenoid chains:
The condensation of two molecules of FPP to form squalene, however, is something different: this is a 'head to head' condensation, where C1 of the first molecule forms a bond to C1 of the second. The chemistry involved is quite a bit more complicated.
The first two steps are familiar: first, the pyrophosphate on one FPP molecule leaves (step 1), resulting in an allylic carbocation that is attacked by the C2-C3 π bond of the second molecule (step 2).
This results in a new carbon-carbon bond between the two FPP molecules, but with incorrect C1 to C2 connectivity (remember, the overall reaction is a C1 to C1 condensation). In step 3, a proton is abstracted and the electrons from the broken C-H bond bridge across a 2-carbon gap to form a cyclopropyl intermediate.
In the second stage of squalene synthesis, the second pyrophosphate group leaves, generating a cyclopropylcarbinyl cation (step 4). Because this is a primary carbocation, you probably are wondering about how stable it could be (and thus how likely an intermediate). As it turns out, such carbocations are remarkably stable, due to favorable interactions between the empty orbital and orbitals on the three-membered ring (the level of bonding theory needed to really understand this idea is beyond the scope of this text, but you may learn about it if you take a class in advanced organic chemistry). What occurs next is an alkyl shift leading to a tertiary carbocation (step 5).
Discussion of the final step (step 6) will need to be put off - this is a reduction with a hydride nucleophile derived from a coenzyme called NADPH. Although this may seem like an extremely convoluted (and perhaps unlikely!) mechanism, there is much experimental evidence to back it up. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/26%3A_Lipids/26.06%3A_Terpenes_and_Terpenoids.txt |
Objectives
After completing this section, you should be able to
1. draw the tetracyclic ring system on which the structure of all steroids is based.
2. identify the occurrence and biological roles of at least two common steroids.
3. sketch the stereochemical conformation of a steroid, given an adequate wedge‑and‑broken line structure, and determine whether the ring substituents in such a compound occupy axial or equatorial positions.
4. construct a molecular model of a steroid, given a suitable written description or a wedge‑and‑broken line structure from which to work.
Key Terms
Make certain that you can define, and use in context, the key term below.
• steroid
Study Notes
Since the 1988 Olympic Games in Seoul, even individuals who have no interest in chemistry or sport have heard the word “steroid” and are aware that some athletes use these substances to enhance their athletic abilities. Stanozolol, the substance which Canadian sprinter, Ben Johnson, was found to have used, has the structure shown below:
The important class of lipids called steroids are actually metabolic derivatives of terpenes, but they are customarily treated as a separate group. Steroids may be recognized by their tetracyclic skeleton, consisting of three fused six-membered and one five-membered ring, as shown in the diagram to the right. The four rings are designated A, B, C & D as noted, and the peculiar numbering of the ring carbon atoms (shown in red) is the result of an earlier misassignment of the structure. The substituents designated by R are often alkyl groups, but may also have functionality. The R group at the A:B ring fusion is most commonly methyl or hydrogen, that at the C:D fusion is usually methyl. The substituent at C-17 varies considerably, and is usually larger than methyl if it is not a functional group. The most common locations of functional groups are C-3, C-4, C-7, C-11, C-12 & C-17. Ring A is sometimes aromatic. Since a number of tetracyclic triterpenes also have this tetracyclic structure, it cannot be considered a unique identifier.
Steroids are widely distributed in animals, where they are associated with a number of physiological processes. Examples of some important steroids are shown in the following diagram. Norethindrone is a synthetic steroid, all the other examples occur naturally. A common strategy in pharmaceutical chemistry is to take a natural compound, having certain desired biological properties together with undesired side effects, and to modify its structure to enhance the desired characteristics and diminish the undesired. This is sometimes accomplished by trial and error.
The generic steroid structure drawn above has seven chiral stereocenters (carbons 5, 8, 9, 10, 13, 14 & 17), which means that it may have as many as 128 stereoisomers. With the exception of C-5, natural steroids generally have a single common configuration. This is shown in the last of the toggled displays, along with the preferred conformations of the rings.
Chemical studies of the steroids were very important to our present understanding of the configurations and conformations of six-membered rings. Substituent groups at different sites on the tetracyclic skeleton will have axial or equatorial orientations that are fixed because of the rigid structure of the trans-fused rings. This fixed orientation influences chemical reactivity, largely due to the greater steric hindrance of axial groups versus their equatorial isomers. Thus an equatorial hydroxyl group is esterified more rapidly than its axial isomer.
It is instructive to examine a simple bicyclic system as a model for the fused rings of the steroid molecule. Decalin, short for decahydronaphthalene, exists as cis and trans isomers at the ring fusion carbon atoms. Planar representations of these isomers are drawn at the top of the following diagram, with corresponding conformational formulas displayed underneath. The numbering shown for the ring carbons follows IUPAC rules, and is different from the unusual numbering used for steroids. For purposes of discussion, the left ring is labeled A (colored blue) and the right ring B (colored red). In the conformational drawings the ring fusion and the angular hydrogens are black.
The trans-isomer is the easiest to describe because the fusion of the A & B rings creates a rigid, roughly planar, structure made up of two chair conformations. Each chair is fused to the other by equatorial bonds, leaving the angular hydrogens (Ha) axial to both rings. Note that the bonds directed above the plane of the two rings alternate from axial to equatorial and back if we proceed around the rings from C-1 to C-10 in numerical order. The bonds directed below the rings also alternate in a complementary fashion.
Conformational descriptions of cis- decalin are complicated by the fact that two energetically equivalent fusions of chair cyclohexanes are possible, and are in rapid equilibrium as the rings flip from one chair conformation to the other. In each of these all chair conformations the rings are fused by one axial and one equatorial bond, and the overall structure is bent at the ring fusion. In the conformer on the left, the red ring (B) is attached to the blue ring (A) by an axial bond to C-1 and an equatorial bond to C-6 (these terms refer to ring A substituents). In the conformer on the right, the carbon bond to C-1 is equatorial and the bond to C-6 is axial. Each of the angular hydrogens (Hae or Hea) is oriented axial to one of the rings and equatorial to the other. This relationship reverses when double ring flipping converts one cis-conformer into the other.
Cis-decalin is less stable than trans-decalin by about 2.7 kcal/mol (from heats of combustion and heats of isomerization data). This is due to steric crowding (hindrance) of the axial hydrogens in the concave region of both cis-conformers, as may be seen in the model display activated by the following button. This difference is roughly three times the energy of a gauche butane conformer relative to its anti conformer. Indeed three gauche butane interactions may be identified in each of the cis-decalin conformations, as will be displayed by clicking on the above conformational diagram. These gauche interactions are also shown in the model.
Steroids in which rings A and B are fused cis, such as the example on the right, do not have the sameconformational mobility exhibited by cis-decalin. The fusion of ring C to ring B in a trans configuration prevents ring B from undergoing a conformational flip to another chair form. If this were to occur, ring C would have to be attached to ring B by two adjacent axial bonds directed 180º apart. This is too great a distance to be bridged by the four carbon atoms making up ring C. Consequently, the steroid molecule is locked in the all chair conformation shown here. Of course, all these steroids and decalins may have one or more six-membered rings in a boat conformation. However the high energy of boat conformers relative to chairs would make such structures minor components in the overall ensemble of conformations available to these molecules.
Steroid Hormones
Hormones are chemical messengers that are released in one tissue and transported through the circulatory system to one or more other tissues. One group of hormones is known as steroid hormones because these hormones are synthesized from cholesterol, which is also a steroid. There are two main groups of steroid hormones: adrenocortical hormones and sex hormones.
The adrenocortical hormones, such as aldosterone and cortisol (Table 17.3), are produced by the adrenal gland, which is located adjacent to each kidney. Aldosterone acts on most cells in the body, but it is particularly effective at enhancing the rate of reabsorption of sodium ions in the kidney tubules and increasing the secretion of potassium ions and/or hydrogen ions by the tubules. Because the concentration of sodium ions is the major factor influencing water retention in tissues, aldosterone promotes water retention and reduces urine output. Cortisol regulates several key metabolic reactions (for example, increasing glucose production and mobilizing fatty acids and amino acids). It also inhibits the inflammatory response of tissue to injury or stress. Cortisol and its analogs are therefore used pharmacologically as immunosuppressants after transplant operations and in the treatment of severe skin allergies and autoimmune diseases, such as rheumatoid arthritis.
Table 17.3 Representative Steroid Hormones and Their Physiological Effects
Hormone Effect
regulates salt metabolism; stimulates kidneys to retain sodium and excrete potassium
stimulates the conversion of proteins to carbohydrates
regulates the menstrual cycle; maintains pregnancy
stimulates female sex characteristics; regulates changes during the menstrual cycle
stimulates and maintains male sex characteristics
The sex hormones are a class of steroid hormones secreted by the gonads (ovaries or testes), the placenta, and the adrenal glands. Testosterone and androstenedione are the primary male sex hormones, or androgens, controlling the primary sexual characteristics of males, or the development of the male genital organs and the continuous production of sperm. Androgens are also responsible for the development of secondary male characteristics, such as facial hair, deep voice, and muscle strength. Two kinds of sex hormones are of particular importance in females: progesterone, which prepares the uterus for pregnancy and prevents the further release of eggs from the ovaries during pregnancy, and the estrogens, which are mainly responsible for the development of female secondary sexual characteristics, such as breast development and increased deposition of fat tissue in the breasts, the buttocks, and the thighs. Both males and females produce androgens and estrogens, differing in the amounts of secreted hormones rather than in the presence or absence of one or the other.
Sex hormones, both natural and synthetic, are sometimes used therapeutically. For example, a woman who has had her ovaries removed may be given female hormones to compensate. Some of the earliest chemical compounds employed in cancer chemotherapy were sex hormones. For example, estrogens are one treatment option for prostate cancer because they block the release and activity of testosterone. Testosterone enhances prostate cancer growth. Sex hormones are also administered in preparation for sex-change operations, to promote the development of the proper secondary sexual characteristics. Oral contraceptives are synthetic derivatives of the female sex hormones; they work by preventing ovulation.
Adrenocorticoid Hormones
The adrenocorticoid hormones are products of the adrenal glands ("adrenal" means adjacent to the renal (kidney). The most important mineralocrticoid is aldosterone, which regulates the reabsorption of sodium and chloride ions in the kidney tubules and increases the loss of potassium ions. Aldosterone is secreted when blood sodium ion levels are too low to cause the kidney to retain sodium ions. If sodium levels are elevated, aldosterone is not secreted, so that some sodium will be lost in the urine. Aldosterone also controls swelling in the tissues.
Cortisol, the most important glucocortinoid, has the function of increasing glucose and glycogen concentrations in the body. These reactions are completed in the liver by taking fatty acids from lipid storage cells and amino acids from body proteins to make glucose and glycogen.
In addition, cortisol and its ketone derivative, cortisone, have the ability to inflammatory effects. Cortisone or similar synthetic derivatives such as prednisolone are used to treat inflammatory diseases, rheumatoid arthritis, and bronchial asthma. There are many side effects with the use of cortisone drugs, so there use must be monitored carefully.
26.08: Biosynthesis of Steroids
On the diagram below, the series of cation-like cyclizations and rearrangements, known as the Stork-Eschenmoser hypothesis, is shown, which were identified in the biosynthesis of the triterpene lanosterol. Lanosterol is a precursor in the biosynthesis of steroids. This takes place by metabolic removal of three methyl groups and degradation of the side chain.
Rearrangements are particularly important in carbocation-intermediate reactions in which isoprenoid molecules cyclize to form complex multi-ring structures. One of the key steps in the biosynthesis of cholesterol is the electrophilic cyclization of oxidosqualene to form a steroid called lanosterol.
This fascinating reaction has two phases. The first phase, in which the actual cyclization takes place, is a series of electrophilic addition steps. The second phase is a series of hydride and methyl shifts. There is some argument about whether these processes occur in a stepwise fashion (with discreet carbocation intermediates) or in a concerted manner. For the sake of clarity, we will show the reaction proceeding stepwise.
The cyclization phase begins with attack by pi electrons on an epoxide electrophile (step 1 - review epoxide ring-opening reactions in section 8.6B).
Steps 2, 3, and 4 are simply successive attacks by pi electrons on the carbocation generated by the previous attack. The overall result of this electrophilic cascade is the opening of the epoxide ring, and closure of three six-membered and one five-membered ring. Next comes the rearrangement phase of the reaction. which is a series of two hydride shifts and two methyl shifts, followed by a proton abstraction which finally quenches the positive charge to form lanosterol. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/26%3A_Lipids/26.07%3A_Steroids.txt |
Objectives
After completing this section, you should be able to
1. outline the relationship between nucleic acids, nucleotides and nucleosides.
2. identify, in general terms, the enzymatic hydrolysis products of nucleosides.
3. explain the structural difference between the sugar components of DNA and RNA.
4. identify by name the four heterocyclic amine bases found in deoxyribonucleotides.
5. identify by name the four heterocyclic amine bases found in ribonucleotides.
6. draw the general structure of a nucleotide and a nucleoside.
7. indicate the nitrogen atom by which a given purine or pyrimidine base attaches to the sugar component in nucleotides and nucleosides.
8. sketch a section of nucleic acid to show how the nucleotide units are joined together.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• deoxyribonucleic acid (DNA)
• nucleosides nucleotides
• ribonucleic acid (RNA)
Study Notes
The five bases that are found in nucleotides are often represented by their initial letter: adenine, A; guanine, G; cytosine, C; thymine, T; and uracil, U. Note that A, G, C and T occur in DNA; A, G, C and U occur in RNA. You are not required to memorize the structures of these bases, but you must know how each one bonds to the sugar unit in a nucleotide.
To fulfill Objective 6, you should be able to reproduce the figure below.
The repeating, or monomer, units that are linked together to form nucleic acids are known as nucleotides. The deoxyribonucleic acid (DNA) of a typical mammalian cell contains about 3 × 109 nucleotides. Nucleotides can be further broken down to phosphoric acid (H3PO4), a pentose sugar (a sugar with five carbon atoms), and a nitrogenous base (a base containing nitrogen atoms).
$\mathrm{nucleic\: acids \underset{down\: into}{\xrightarrow{can\: be\: broken}} nucleotides \underset{down\: into}{\xrightarrow{can\: be\: broken}} H_3PO_4 + nitrogen\: base + pentose\: sugar} \tag{28.1.1}$
If the pentose sugar is ribose, the nucleotide is more specifically referred to as a ribonucleotide, and the resulting nucleic acid is ribonucleic acid (RNA). If the sugar is 2-deoxyribose, the nucleotide is a deoxyribonucleotide, and the nucleic acid is DNA.
The nitrogenous bases found in nucleotides are classified as pyrimidines or purines. Pyrimidines are heterocyclic amines with two nitrogen atoms in a six-member ring and include uracil, thymine, and cytosine. Purines are heterocyclic amines consisting of a pyrimidine ring fused to a five-member ring with two nitrogen atoms. Adenine and guanine are the major purines found in nucleic acids (Figure $1$).
The formation of a bond between C1′ of the pentose sugar and N1 of the pyrimidine base or N9 of the purine base joins the pentose sugar to the nitrogenous base. In the formation of this bond, a molecule of water is removed. Table 28.1.1 summarizes the similarities and differences in the composition of nucleotides in DNA and RNA.
Example $1$
The numbering convention is that primed numbers designate the atoms of the pentose ring, and unprimed numbers designate the atoms of the purine or pyrimidine ring.
Table $1$: Composition of Nucleotides in DNA and RNA
Composition DNA RNA
purine bases adenine and guanine adenine and guanine
pyrimidine bases cytosine and thymine cytosine and uracil
pentose sugar 2-deoxyribose ribose
inorganic acid phosphoric acid (H3PO4) H3PO4
The names and structures of the major ribonucleotides and one of the deoxyribonucleotides are given in Figure $2$.
Apart from being the monomer units of DNA and RNA, the nucleotides and some of their derivatives have other functions as well. Adenosine diphosphate (ADP) and adenosine triphosphate (ATP), shown in Figure $3$, have a role in cell metabolism. Moreover, a number of coenzymes, including flavin adenine dinucleotide (FAD), nicotinamide adenine dinucleotide (NAD+), and coenzyme A, contain adenine nucleotides as structural components.
Primary Structure of Nucleic Acids
Nucleotides are joined together through the phosphate group of one nucleotide connecting in an ester linkage to the OH group on the third carbon atom of the sugar unit of a second nucleotide. This unit joins to a third nucleotide, and the process is repeated to produce a long nucleic acid chain (Figure 28.1.4). The backbone of the chain consists of alternating phosphate and sugar units (2-deoxyribose in DNA and ribose in RNA). The purine and pyrimidine bases branch off this backbone.
Note
Each phosphate group has one acidic hydrogen atom that is ionized at physiological pH. This is why these compounds are known as nucleic acids.
Like proteins, nucleic acids have a primary structure that is defined as the sequence of their nucleotides. Unlike proteins, which have 20 different kinds of amino acids, there are only 4 different kinds of nucleotides in nucleic acids. For amino acid sequences in proteins, the convention is to write the amino acids in order starting with the N-terminal amino acid. In writing nucleotide sequences for nucleic acids, the convention is to write the nucleotides (usually using the one-letter abbreviations for the bases, shown in Figure 28.1.4) starting with the nucleotide having a free phosphate group, which is known as the 5′ end, and indicate the nucleotides in order. For DNA, a lowercase d is often written in front of the sequence to indicate that the monomers are deoxyribonucleotides. The final nucleotide has a free OH group on the 3′ carbon atom and is called the 3′ end. The sequence of nucleotides in the DNA segment shown in Figure 28.1.4 would be written 5′-dG-dT-dA-dC-3′, which is often further abbreviated to dGTAC or just GTAC.
Concept Review Exercises
1. Identify the three molecules needed to form the nucleotides in each nucleic acid.
1. DNA
2. RNA
2. Classify each compound as a pentose sugar, a purine, or a pyrimidine.
1. adenine
2. guanine
3. deoxyribose
4. thymine
5. ribose
6. cytosine
Answers
1. nitrogenous base (adenine, guanine, cytosine, and thymine), 2-deoxyribose, and H3PO4
2. nitrogenous base (adenine, guanine, cytosine, and uracil), ribose, and H3PO4
1. purine
2. purine
3. pentose sugar
4. pyrimidine
5. pentose sugar
6. pyrimidine
Key Takeaways
• Nucleotides are composed of phosphoric acid, a pentose sugar (ribose or deoxyribose), and a nitrogen-containing base (adenine, cytosine, guanine, thymine, or uracil).
• Ribonucleotides contain ribose, while deoxyribonucleotides contain deoxyribose.
Exercises
1. What is the sugar unit in each nucleic acid?
1. RNA
2. DNA
2. Identify the major nitrogenous bases in each nucleic acid.
1. DNA
2. RNA
3. For each structure, circle the sugar unit and identify the nucleotide as a ribonucleotide or a deoxyribonucleotide.
4. For each structure, circle the sugar unit and identify the nucleotide as a ribonucleotide or a deoxyribonucleotide.
5. For each structure, circle the nitrogenous base and identify it as a purine or pyrimidine.
6. For each structure, circle the nitrogenous base and identify it as a purine or pyrimidine.
Answers
1. ribose
2. deoxyribose | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/27%3A_Nucleic_Acids/27.01%3A_Nucleotides_and_Nucleic_Acids.txt |
Objectives
After completing this section, you should be able, given the necessary Kekulé structures, to show how hydrogen bonding can occur between thymine and adenine, and between guanine and cytosine; and to explain the significance of such interactions to the primary and secondary structures of DNA.
Study Notes
Watson and Crick received the Nobel Prize in 1962 for elucidating the structure of DNA and proposing the mechanism for gene reproduction. Their work rested heavily on X‑ray crystallographic work done on RNA and DNA by Franklin and Wilkins. Wilkins shared the Nobel Prize with Watson and Crick, but Franklin had been dead four years at the time of the award (you cannot be awarded the Nobel Prize posthumously).
The history of Watson and Crick’s proposed DNA model is controversial and a travesty of scientific ethics. Rosalind Franklin was deeply involved in the determination of the structure of DNA, and had collected numerous diffraction patterns. Watson attended a departmental colloquium at King’s College given by Franklin, and came into possession of an internal progress report she had written. Both departmental colloquia and progress reports are merely methods of discussion between colleagues; works presented in these fora are not considered by scientists to be “published” works, and therefore are not in the public domain. Watson and Crick not only were aware of Franklin’s work, but used her unpublished data, presented in confidence within her own college.
The final blow came about a year after the colloquium. Watson visited Wilkins at King’s College, and Wilkins inexplicably handed over Franklin’s diffraction photographs without her consent. Had Franklin’s work not been secretly taken from her, she might quite possibly have solved the DNA structure before Watson and Crick, who at the time did not yet have their own photographs. This is truly one of the sadder episodes of questionable scientific ethics and discovery that I have ever encountered.
References
Kass‑Simon, G., and P. Farnes. Women of Science: Righting the Record. Bloomington, IN: Indiana University Press, 1990.
Maddox, B. Rosalind Franklin: The Dark Lady of DNA. New York: HarperCollins, 2002.
Intermolecular Forces in Nucleic Acids
The nucleic acids RNA and DNA are involved in the storage and expression of genetic information in a cell. Both are polymers of monomeric nucleotides. DNA exists in the cell as double-stranded helices while RNA typically is a single-stranded molecule which can fold in 3D space to form complex secondary (double-stranded helices) and tertiary structures in a fashion similar to proteins. The complex 3D structures formed by RNA allow it to perform functions other than simple genetic information storage, such as catalysis. Hence most scientists believe that RNA preceded both DNA and proteins in evolution as it can both store genetic information and catalyze chemical reactions.
DNA
DNA is a polymer, consisting of monomers call deoxynucleotides. The monomer contains a simple sugar (deoxyribose, shown in black below), a phosphate group (in red), and a cyclic organic R group (in blue) that is analogous to the side chain of an amino acid.
Only four bases are used in DNA (in contrast to the 20 different side chains in proteins) which we will abbreviate, for simplicity, as A, G, C and T. They are bases since they contain amine groups that can accept protons. The polymer consists of a sugar - phosphate - sugar - phosphate backbone, with one base attached to each sugar molecule. As with proteins, the DNA backbone is polar but also charged. It is a polyanion. The bases, analogous to the side chains of amino acids, are predominately polar. Given the charged nature of the backbone, you might expect that DNA does not fold to a compact globular (spherical) shape, even if positively charged cations like Mg bind to and stabilize the charge on the polymer. Instead, DNA exists usually as a double-stranded (ds) structure with the sugar-phosphate backbones of the two different strands running in opposite directions (5'-3' and the other 3'-5'). The strands are held together by hydrogen bonds between bases on complementary strands. Hence like proteins, DNA has secondary structure but in this case, the hydrogen bonds are not within the backbone but between the "side chain" bases on opposing strands. It is actually a misnomer to call dsDNA a molecule, since it really consists of two different, complementary strands held together by hydrogen bonds. A structure of ds-DNA showing the opposite polarity of the strands is shown below.
In double stranded DNA, the guanine (G) base on one strand can form three H-bonds with a cytosine (C) base on another strand (this is called a GC base pair). The thymine (T) base on one strand can form two H-bonds with an adenine (A) base on the other strand (this is called an AT base pair). Double-stranded DNA has a regular geometric structure with a fixed distance between the two backbones. This requires the bases pairs to consists of one base with a two-ring (bicyclic) structure (these bases are called purines) and one with a single ring structure (these bases are called pyrimidines). Hence a G and A or a T and C are not possible base pair partners.
Double stranded DNA varies in length (number of sugar-phosphate units connected), base composition (how many of each set of bases) and sequence (the order of the bases in the backbone). The following links provide interactive Jmol models of dsDNA made by Angel Herráez, Univ. de Alcalá (Spain) and Eric Martz.
Chromosomes consist of one dsDNA with many different bound proteins. The human genome has about 3 billion base pairs of DNA. Therefore, on average, each single chromosome of a pair has about 150 million base pairs and lots of proteins bound to it. dsDNA is a highly charged molecule, and can be viewed, to a first approximation, as a long rod-like molecule with a large negative charge. It is a polyanion. This very large molecule must somehow be packed into a small nucleus of a tiny cell. In complex (eukaryotic) cells, this packing problem is solved by coiling DNA around a core complex of four different pairs (eight proteins total) of histone proteins (H2A, H2B, H3, and H4) which have net positive charges. The histone core complex with dsDNA wound around approximately 2.5 times is called the nucleosome.
Jmol model of the nucleosome
DNA can adopt two other types of double-helical forms. The one discovered by Watson and Crick and found in most textbooks is called B-DNA. Depending on the actual DNA sequence and the hydration state of the DNA, it can be coaxed to form two other types of double-stranded helices, Z and A DNA. The A form is much more open then the B form.
The 3.2 billion base pairs of DNA in humans contains about 24,000 short stretches (genes) that encode different proteins. These genes are interspersed among DNA that helps determining if the gene is decoded into RNA molecules (see below) and ultimately into proteins. For a particular gene to be activated (or "turned on"), specific proteins must bind to the region of a particular gene. How can binding proteins find specific binding targets among the vast number of base pairs that to a first approximation have a repetitive sugar-phosphate-base repeat? The Jmol below shows how specificity can be achieved. When DNA winds into a double helix through base-pairs between AT and GC, hydrogen bond donors (amide Hs) and acceptors (Os) on the bases that are not used in intrastrand base pairing,are still available in the major and minor grove of the ds-DNA helix (see Jmol below). Unique base pair sequences will display unique patterns of H bond donors and acceptors in the major grove. These donors/acceptors can be recognized by specific DNA binding proteins which on binding can lead to gene activation. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/27%3A_Nucleic_Acids/27.02%3A_DNA_Base_Pairs.txt |
Objectives
After completing this section, you should be able to describe, very briefly, the replication of DNA.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• replication
• semiconservative replication
Study Notes
Notice that the objective for this section requires only that you be able to describe the replication process briefly.
New cells are continuously forming in the body through the process of cell division. For this to happen, the DNA in a dividing cell must be copied in a process known as replication. The complementary base pairing of the double helix provides a ready model for how genetic replication occurs. If the two chains of the double helix are pulled apart, disrupting the hydrogen bonding between base pairs, each chain can act as a template, or pattern, for the synthesis of a new complementary DNA chain.
The nucleus contains all the necessary enzymes, proteins, and nucleotides required for this synthesis. A short segment of DNA is “unzipped,” so that the two strands in the segment are separated to serve as templates for new DNA. DNA polymerase, an enzyme, recognizes each base in a template strand and matches it to the complementary base in a free nucleotide. The enzyme then catalyzes the formation of an ester bond between the 5′ phosphate group of the nucleotide and the 3′ OH end of the new, growing DNA chain. In this way, each strand of the original DNA molecule is used to produce a duplicate of its former partner (Figure 28.3.1). Whatever information was encoded in the original DNA double helix is now contained in each replicate helix. When the cell divides, each daughter cell gets one of these replicates and thus all of the information that was originally possessed by the parent cell.
Example
A segment of one strand from a DNA molecule has the sequence 5′‑TCCATGAGTTGA‑3′. What is the sequence of nucleotides in the opposite, or complementary, DNA chain?
Solution
Knowing that the two strands are antiparallel and that T base pairs with A, while C base pairs with G, the sequence of the complementary strand will be 3′‑AGGTACTCAACT‑5′ (can also be written as TCAACTCATGGA).
Exercise
A segment of one strand from a DNA molecule has the sequence 5′‑CCAGTGAATTGCCTAT‑3′. What is the sequence of nucleotides in the opposite, or complementary, DNA chain?
What do we mean when we say information is encoded in the DNA molecule? An organism’s DNA can be compared to a book containing directions for assembling a model airplane or for knitting a sweater. Letters of the alphabet are arranged into words, and these words direct the individual to perform certain operations with specific materials. If all the directions are followed correctly, a model airplane or sweater is produced.
In DNA, the particular sequences of nucleotides along the chains encode the directions for building an organism. Just as saw means one thing in English and was means another, the sequence of bases CGT means one thing, and TGC means something different. Although there are only four letters—the four nucleotides—in the genetic code of DNA, their sequencing along the DNA strands can vary so widely that information storage is essentially unlimited. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/27%3A_Nucleic_Acids/27.03%3A_DNA_Replication.txt |
Objectives
After completing this section, you should be able to
1. describe, very briefly, how RNA is synthesized in the nucleus of the cell by transcription of DNA.
2. identify the important structural differences between DNA and DNA.
3. given the appropriate Kekulé structures, show how uracil can form strong hydrogen bonds to adenine.
4. identify the base sequence in RNA that would be complementary to a given base sequence in DNA.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• messenger RNA
• RNA polymerase
• ribosomal RNA
• transcription
• transfer RNA
Study Notes
“Messenger RNA” (mRNA) carries the genetic information from the DNA in the nucleus to the cytoplasm where protein synthesis occurs. The code carried by mRNA is read by “transfer RNA” (tRNA) in a process called translation (see Section 28.5).
“Ribosomal RNA” (rRNA) is the term used to describe the RNA molecules which, together with proteins, make up the ribosomes on which proteins are synthesized.
For the hereditary information in DNA to be useful, it must be “expressed,” that is, used to direct the growth and functioning of an organism. The first step in the processes that constitute DNA expression is the synthesis of RNA, by a template mechanism that is in many ways analogous to DNA replication. Because the RNA that is synthesized is a complementary copy of information contained in DNA, RNA synthesis is referred to as transcription.
There are three key differences between replication and transcription: (1) RNA molecules are much shorter than DNA molecules; only a portion of one DNA strand is copied or transcribed to make an RNA molecule. (2) RNA is built from ribonucleotides rather than deoxyribonucleotides. (3) The newly synthesized RNA strand does not remain associated with the DNA sequence it was transcribed from.
The DNA sequence that is transcribed to make RNA is called the template strand, while the complementary sequence on the other DNA strand is called the coding or informational strand. To initiate RNA synthesis, the two DNA strands unwind at specific sites along the DNA molecule. Ribonucleotides are attracted to the uncoiling region of the DNA molecule, beginning at the 3′ end of the template strand, according to the rules of base pairing. Thymine in DNA calls for adenine in RNA, cytosine specifies guanine, guanine calls for cytosine, and adenine requires uracil. RNA polymerase—an enzyme—binds the complementary ribonucleotide and catalyzes the formation of the ester linkage between ribonucleotides, a reaction very similar to that catalyzed by DNA polymerase (Figure 28.4.1). Synthesis of the RNA strand takes place in the 5′ to 3′ direction, antiparallel to the template strand. Only a short segment of the RNA molecule is hydrogen-bonded to the template strand at any time during transcription. When transcription is completed, the RNA is released, and the DNA helix reforms. The nucleotide sequence of the RNA strand formed during transcription is identical to that of the corresponding coding strand of the DNA, except that U replaces T.
Example
A portion of the template strand of a gene has the sequence 5′‑TCCATGAGTTGA‑3′. What is the sequence of nucleotides in the RNA that is formed from this template?
Solution
Four things must be remembered in answering this question: (1) the DNA strand and the RNA strand being synthesized are antiparallel; (2) RNA is synthesized in a 5′ to 3′ direction, so transcription begins at the 3′ end of the template strand; (3) ribonucleotides are used in place of deoxyribonucleotides; and (4) thymine (T) base pairs with adenine (A), A base pairs with uracil (U; in RNA), and cytosine (C) base pairs with guanine (G). The sequence is determined to be 3′‑AGGUACUCAACU‑5′ (can also be written as 5′‑UCAACUCAUGGA‑3′).
Exercise
A portion of the template strand of a gene has the sequence 5′‑CCAGTGAATTGCCTAT‑3′. What is the sequence of nucleotides in the RNA that is formed from this template?
Three types of RNA are formed during transcription: messenger RNA (mRNA), ribosomal RNA (rRNA), and transfer RNA (tRNA). These three types of RNA differ in function, size, and percentage of the total cell RNA (Table 28.4.1). mRNA makes up only a small percent of the total amount of RNA within the cell, primarily because each molecule of mRNA exists for a relatively short time; it is continuously being degraded and resynthesized. The molecular dimensions of the mRNA molecule vary according to the amount of genetic information a given molecule contains. After transcription, which takes place in the nucleus, the mRNA passes into the cytoplasm, carrying the genetic message from DNA to the ribosomes, the sites of protein synthesis. Elsewhere, we shall see how mRNA directly determines the sequence of amino acids during protein synthesis.
Table: Properties of Cellular RNA in Escherichia coli
Type Function Approximate Number of Nucleotides Percentage of Total Cell RNA
mRNA codes for proteins 100–6,000 ~3
rRNA component of ribosomes 120–2900 83
tRNA adapter molecule that brings the amino acid to the ribosome 75–90 14
Ribosomes are cellular substructures where proteins are synthesized. They contain about 65% rRNA and 35% protein, held together by numerous noncovalent interactions, such as hydrogen bonding, in an overall structure consisting of two globular particles of unequal size.
Molecules of tRNA, which bring amino acids (one at a time) to the ribosomes for the construction of proteins, differ from one another in the kinds of amino acid each is specifically designed to carry. A set of three nucleotides, known as a codon, on the mRNA determines which kind of tRNA will add its amino acid to the growing chain. Each of the 20 amino acids found in proteins has at least one corresponding kind of tRNA, and most amino acids have more than one.
The two-dimensional structure of a tRNA molecule has three distinctive loops, reminiscent of a cloverleaf (Figure 28.4.2). On one loop is a sequence of three nucleotides that varies for each kind of tRNA. This triplet, called the anticodon, is complementary to and pairs with the codon on the mRNA. At the opposite end of the molecule is the acceptor stem, where the amino acid is attached. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/27%3A_Nucleic_Acids/27.04%3A_Transcription_of_DNA.txt |
Objectives
After completing this section, you should be able to describe, very briefly, the roles of messenger RNA and transfer RNA in the biosynthesis of proteins.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• anticodon
• codon
• translation
Study Notes
As in the preceding section, you should not be too concerned about trying to memorize details. The objective requires you to have a general understanding of the roles played by mRNA and tRNA in the biosynthesis of proteins, and that you be able to describe this process.
One of the definitions of a gene is as follows: a segment of deoxyribonucleic acid (DNA) carrying the code for a specific polypeptide. Each molecule of messenger RNA (mRNA) is a transcribed copy of a gene that is used by a cell for synthesizing a polypeptide chain. If a protein contains two or more different polypeptide chains, each chain is coded by a different gene. We turn now to the question of how the sequence of nucleotides in a molecule of ribonucleic acid (RNA) is translated into an amino acid sequence.
How can a molecule containing just 4 different nucleotides specify the sequence of the 20 amino acids that occur in proteins? If each nucleotide coded for 1 amino acid, then obviously the nucleic acids could code for only 4 amino acids. What if amino acids were coded for by groups of 2 nucleotides? There are 42, or 16, different combinations of 2 nucleotides (AA, AU, AC, AG, UU, and so on). Such a code is more extensive but still not adequate to code for 20 amino acids. However, if the nucleotides are arranged in groups of 3, the number of different possible combinations is 43, or 64. Here we have a code that is extensive enough to direct the synthesis of the primary structure of a protein molecule.
The genetic code can therefore be described as the identification of each group of three nucleotides and its particular amino acid. The sequence of these triplet groups in the mRNA dictates the sequence of the amino acids in the protein. Each individual three-nucleotide coding unit, as we have seen, is called a codon.
Protein synthesis is accomplished by orderly interactions between mRNA and the other ribonucleic acids (transfer RNA [tRNA] and ribosomal RNA [rRNA]), the ribosome, and more than 100 enzymes. The mRNA formed in the nucleus during transcription is transported across the nuclear membrane into the cytoplasm to the ribosomes—carrying with it the genetic instructions. The process in which the information encoded in the mRNA is used to direct the sequencing of amino acids and thus ultimately to synthesize a protein is referred to as translation.
Before an amino acid can be incorporated into a polypeptide chain, it must be attached to its unique tRNA. Each tRNA molecule has an anticodon for the amino acid it carries. An anticodon is a sequence of 3 bases, and is complementary to the codon for an amino acid. For example, the amino acid lysine has the codon AAG, so the anticodon is UUC. Therefore, lysine would be carried by a tRNA molecule with the anticodon UUC. Wherever the codon AAG appears in mRNA, a UUC anticodon on a tRNA temporarily binds to the codon. This crucial process requires an enzyme known as aminoacyl-tRNA synthetase (Figure 28.5.1). There is a specific aminoacyl-tRNA synthetase for each amino acid. This high degree of specificity is vital to the incorporation of the correct amino acid into a protein. After the amino acid molecule has been bound to its tRNA carrier, protein synthesis can take place. Figure 28.5.2 depicts a schematic stepwise representation of this all-important process.
Early experimenters were faced with the task of determining which of the 64 possible codons stood for each of the 20 amino acids. The cracking of the genetic code was the joint accomplishment of several well-known geneticists—notably Har Khorana, Marshall Nirenberg, Philip Leder, and Severo Ochoa—from 1961 to 1964. The genetic dictionary they compiled, summarized in Figure 28.5.3, shows that 61 codons code for amino acids, and 3 codons serve as signals for the termination of polypeptide synthesis (much like the period at the end of a sentence). Notice that only methionine (AUG) and tryptophan (UGG) have single codons. All other amino acids have two or more codons.
Example
A portion of an mRNA molecule has the sequence 5′‑AUGCCACGAGUUGAC‑3′. What amino acid sequence does this code for?
Solution
Use the Genetic Code Figure above to determine what amino acid each set of three nucleotides (codon) codes for. Remember that the sequence is read starting from the 5′ end and that a protein is synthesized starting with the N-terminal amino acid. The sequence 5′‑AUGCCACGAGUUGAC‑3′ codes for met-pro-arg-val-asp.
Exercise
A portion of an RNA molecule has the sequence 5′‑AUGCUGAAUUGCGUAGGA‑3′. What amino acid sequence does this code for?
Further experimentation threw much light on the nature of the genetic code, as follows:
1. The code is virtually universal; animal, plant, and bacterial cells use the same codons to specify each amino acid (with a few exceptions).
2. The code is “degenerate”; in all but two cases (methionine and tryptophan), more than one triplet codes for a given amino acid.
3. The first two bases of each codon are most significant; the third base often varies. This suggests that a change in the third base by a mutation may still permit the correct incorporation of a given amino acid into a protein. The third base is sometimes called the “wobble” base.
4. The code is continuous and nonoverlapping; there are no nucleotides between codons, and adjacent codons do not overlap.
5. The three termination codons are read by special proteins called release factors, which signal the end of the translation process.
6. The codon AUG codes for methionine and is also the initiation codon. Thus methionine is the first amino acid in each newly synthesized polypeptide. This first amino acid is usually removed enzymatically before the polypeptide chain is completed; the vast majority of polypeptides do not begin with methionine.
Concept Review Exercises
1. What are the roles of mRNA and tRNA in protein synthesis?
2. What is the initiation codon?
3. What are the termination codons and how are they recognized?
Answers
1. mRNA provides the code that determines the order of amino acids in the protein; tRNA transports the amino acids to the ribosome to incorporate into the growing protein chain.
2. AUG
3. UAA, UAG, and UGA; they are recognized by special proteins called release factors, which signal the end of the translation process.
Key Takeaways
• In translation, the information in mRNA directs the order of amino acids in protein synthesis.
• A set of three nucleotides (codon) codes for a specific amino acid.
Exercises
1. Write the anticodon on tRNA that would pair with each mRNA codon.
1. 5′‑UUU‑3′
2. 5′‑CAU‑3′
3. 5′‑AGC‑3′
4. 5′‑CCG‑3′
2. Write the codon on mRNA that would pair with each tRNA anticodon.
1. 5′‑UUG‑3′
2. 5′‑GAA‑3′
3. 5′‑UCC‑3′
4. 5′‑CAC‑3′
3. The peptide hormone oxytocin contains 9 amino acid units. What is the minimum number of nucleotides needed to code for this peptide?
4. Myoglobin, a protein that stores oxygen in muscle cells, has been purified from a number of organisms. The protein from a sperm whale is composed of 153 amino acid units. What is the minimum number of nucleotides that must be present in the mRNA that codes for this protein?
5. Use Figure 28.5.3to identify the amino acids carried by each tRNA molecule in Exercise 1.
6. Use Figure 28.5.3to identify the amino acids carried by each tRNA molecule in Exercise 2.
7. Use Figure 28.5.3to determine the amino acid sequence produced from this mRNA sequence: 5′‑AUGAGCGACUUUGCGGGAUUA‑3′.
8. Use Figure 28.5.3to determine the amino acid sequence produced from this mRNA sequence: 5′‑AUGGCAAUCCUCAAACGCUGU‑3′
Answers
1. 3′‑AAA‑5′
2. 3′‑GUA‑5′
3. 3′‑UCG‑5′
4. 3′‑GGC‑5′
1. 27 nucleotides (3 nucleotides/codon)
1. 1a: phenyalanine; 1b: histidine; 1c: serine; 1d: proline
1. met-ser-asp-phe-ala-gly-leu | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/27%3A_Nucleic_Acids/27.05%3A_Translation_of_RNA-_Protein_Biosynthesis.txt |
We will discuss one method of reading the sequence of DNA. This method, developed by Sanger won him a second Nobel prize. To sequence a single stranded piece of DNA, the complementary strand is synthesized. Four different reaction mixtures are set up. Each contain all 4 radioactive deoxynucleotides (dATP, dCTP, dGTP, dTTP) required for the reaction and DNA polymerase. In addition, dideoxyATP (ddATP) is added to one reaction tube The dATP and ddATP attach randomly to the growing 3' end of the complementary stranded. If ddATP is added no further nucleotides can be added after since its 3' end has an H and not a OH. That's why they call it dideoxy. The new chain is terminated.. If dATP is added, the chain will continue to grow until another A needs to be added. Hence a whole series of discreet fragments of DNA chains will be made, all terminated when ddATP was added. The same scenario occurs for the other 3 tubes, which contain dCTP and ddCTP, dTTP and ddTTP, and dGTP and ddGTP respectively. All the fragments made in each tube will be placed in separate lanes for electrophoresis, where the fragments will separate by size.
Didexoynucleotides
Example
You will pretend to sequence a single stranded piece of DNA as shown below. The new nucleotides are added by the enzyme DNA polymerase to the primer, GACT, in the 5' to 3' direction. You will set up 4 reaction tubes, Each tube contains all the dXTP's. In addition, add ddATP to tube 1, ddTTP to tube 2, ddCTP to tube 3, and ddGTP to tube 4. For each separate reaction mixture, determine all the possible sequences made by writing the possible sequences on one of the unfinished complementary sequences below. Cut the completed sequences from the page, determine the size of the polynucleotide sequences made, and place them as they would migrate (based on size) in the appropriate lane of a imaginary gel which you have drawn on a piece of paper. Lane 1 will contain the nucleotides made in tube 1, etc. Then draw lines under the positions of the cutout nucleotides to represent DNA bands in the gel. Read the sequence of the complementary DNA synthesized. Then write the sequence of the ssDNA that was to be sequenced.
• 5' T C A A C G A T C T G A 3' (STAND TO SEQUENCE)
• 3' G A C T 5' (primer)
• 3' G A C T 5' (primer)
• 3' G A C T 5' (primer)
• 3' G A C T 5' (primer)
• 3' G A C T 5' (primer)
• 3' G A C T 5' (primer)
• 3' G A C T 5' (primer)
• 3' G A C T 5' (primer)
Since the DNA fragments have no detectable color, they can not be directly visualized in the gel. Alternative methods are used. In the one described above, radiolabeled ddXTP's where used. Once the sequencing gel is run, it can be dried and the bands visualized by radioautography (also called autoradiography). A place of x-ray film is placed over the dried gel in a dark environment. The radiolabeled bands will emit radiation which will expose the x-ray film directly over the bands. The film can be developed to detect the bands. In a newer technique, the primer can be labeled with a flourescent dye. If a different dye is used for each reaction mixture, all the reaction mixtures can be run in one lane of a gel. (Actually only one reaction mix containing all the ddXTP's together need be performed.) The gel can then be scanned by a laser, which detects fluorescence from the dyes, each at a different wavelength.
One recent advance in sequencing allows for real-time determination of a sequence. The four deoxynucleotides are each labeled with a different fluorphore on the 5' phosphate (not the base as above). A tethered DNA polymerase elongates the DNA on a template, releasing the fluorophore into solution (i.e. the fluorophore is not incorporated into the DNA chain). The reaction takes place in a visualization chamber called a zero mode waveguide which is a cylindrical metallic chamber with a width of 70 nm and a volume of 20 zeptoliters (20 x 10-21 L). It sits on a glass support through which laser illumination of the sample is achieved. Given the small volume, non-incorporated fluorescently tagged deoxynucleotides diffuse in and out in the microsecond timescale. When a deoxynucleotide is incorporated into the DNA, its residence time is in the millisecond time scale. This allows for prolonged detection of fluorescence which give a high signal to noise ratio. This method might bring the cost of sequencing the human genome down from the initial billion dollar range to \$100.
27.07: Polymerase Chain Reactions
In the mid 80's a new method was developed to copy (amplify) DNA in a test tube. It doesn't require a plasmid or a virus. It just requires a DNA fragment, some primers (small polynucleotides complementary to sections of DNA on each strand and straddling the section of DNA to be amplified. Just add to this mixture dATP, dCTP, dGTP, dTTP, and a heat stable DNA polymerase from the organism Thermophilus aquaticus (which lives in hot springs), and off you go. The mixture is first heated to a temperature which will cause the DsDNA strands to separate. The temperature is cooled allowing a large stoichimetric excess of the primers to anneal to the ssDNA. The heat stable Taq polymerase (from Thermophilus aquaticus) polymerizes DNA from the primers. The temperature is raised again, allowing dsDNA strand separation. On cooling the primers anneal again to the original and newly synthesized DNA from the last cycle and synthesis of DNA occurs again. This cycle is repeated as shown in the diagram. This chain reaction is called the polymerase chain reaction (PCR). The target DNA synthesized is amplified a million times in 20 cycles, or a billion times in 30 cycles, which can be done in a few hours.
Figure: Copying DNA in the test tube - the polymerase chain reaction (PCR)
Figure: Another View of PCR | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/27%3A_Nucleic_Acids/27.06%3A_DNA_Sequencing.txt |
Learning Objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• discuss the origins of organic chemistry - refer to section 1.1
• use and apply the language of Atomic Structure (atomic number, mass number, isotopes) - refer to section 1.2
• draw, interpret, and convert between Lewis (Kekule), Condensed, and Bond-line Structures - refer to sections 1.3, 1.4, 1.5, and 1.6
• apply bonding patterns and polarity to organic compounds - refer to section 1.7 and 1.8
• identify polar bonds and compounds - refer to section 1.9
• draw resonance forms and predict the relative contribution of each resonance form to the overall structure of the compound or ion - refer to section 1.10
• recognize acids and bases - refer to sections 1.11 and 1.12
• use the definition of Lewis Acids and Bases to recognize electron movement in reactions - refer to section 1.13
• predict reaction products of acid-base reactions - refer to sections 1.11, 1.12 ,and 1.13
• determine relative strengths of acids and bases from their pKa values - refer to section 1.14
• determine the form of an acid or base at a specified pH (given the pKa) - refer to section 1.14
• predict relative strengths of acids and bases from their structure, bonding and resonance - refer to section 1.15
• determine the empirical and molecular formulas from combustion data - refer to section 1.16
01: Introduction and Review
Learning Objective
• discuss the origins of organic chemistry
All living things on earth are formed mostly of carbon compounds. The prevalence of carbon compounds in living things has led to the epithet “carbon-based” life. The truth is we know of no other kind of life. Early chemists regarded substances isolated from organisms (plants and animals) as a different type of matter that could not be synthesized artificially, and these substances were thus known as organic compounds. The widespread belief called vitalism held that organic compounds were formed by a vital force present only in living organisms. The German chemist Friedrich Wöhler was one of the early chemists to refute this aspect of vitalism, when, in 1828, he reported the synthesis of urea, a component of many body fluids, from nonliving materials. Since then, it has been recognized that organic molecules obey the same natural laws as inorganic substances, and the category of organic compounds has evolved to include both natural and synthetic compounds that contain carbon. Some carbon-containing compounds are not classified as organic, for example, carbonates and cyanides, and simple oxides, such as \(\ce{CO}\) and \(\ce{CO2}\). Although a single, precise definition has yet to be identified by the chemistry community, most agree that a defining trait of organic molecules is the presence of carbon as the principal element, bonded to hydrogen and other carbon atoms.
Today, organic compounds are key components of plastics, soaps, perfumes, sweeteners, fabrics, pharmaceuticals, and many other substances that we use every day. The value to us of organic compounds ensures that organic chemistry is an important discipline within the general field of chemistry. In this chapter, we discuss why the element carbon gives rise to a vast number and variety of compounds, how those compounds are classified, and the role of organic compounds in representative biological and industrial settings. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/01%3A_Introduction_and_Review/1.01%3A_The_Origins_of_Organic_Chemistry.txt |
Learning Objective
• Use and apply the language of Atomic Structure (atomic number, mass number, isotopes)
The precise physical nature of atoms finally emerged from a series of elegant experiments carried out between 1895 and 1915. The most notable of these achievements was Ernest Rutherford's famous 1911 alpha-ray scattering experiment, which established that
• Almost all of the mass of an atom is contained within a tiny (and therefore extremely dense) nucleus which carries a positive electric charge whose value identifies each element and is known as the atomic number of the element.
• Almost all of the volume of an atom consists of empty space in which electrons, the fundamental carriers of negative electric charge, reside. The extremely small mass of the electron (1/1840 the mass of the hydrogen nucleus) causes it to behave as a quantum particle, which means that its location at any moment cannot be specified; the best we can do is describe its behavior in terms of the probability of its manifesting itself at any point in space. It is common (but somewhat misleading) to describe the volume of space in which the electrons of an atom have a significant probability of being found as the electron cloud. The latter has no definite outer boundary, so neither does the atom. The radius of an atom must be defined arbitrarily, such as the boundary in which the electron can be found with 95% probability. Atomic radii are typically 30-300 pm.
The nucleus is itself composed of two kinds of particles. Protons are the carriers of positive electric charge in the nucleus; the proton charge is exactly the same as the electron charge, but of opposite sign. This means that in any [electrically neutral] atom, the number of protons in the nucleus (often referred to as the nuclear charge) is balanced by the same number of electrons outside the nucleus. The other nuclear particle is the neutron. As its name implies, this particle carries no electrical charge. Its mass is almost the same as that of the proton. Most nuclei contain roughly equal numbers of neutrons and protons, so we can say that these two particles together account for almost all the mass of the atom.
Because the electrons of an atom are in contact with the outside world, it is possible for one or more electrons to be lost, or some new ones to be added. The resulting electrically-charged atom is called an ion.
Elements
To date, about 115 different elements have been discovered; by definition, each is chemically unique. To understand why they are unique, you need to understand the structure of the atom (the fundamental, individual particle of an element) and the characteristics of its components. Atoms consist of electrons, protons, and neutrons. Although this is an oversimplification that ignores the other subatomic particles that have been discovered, it is sufficient for discussion of chemical principles. Some properties of these subatomic particles are summarized in Table $1$, which illustrates three important points:
1. Electrons and protons have electrical charges that are identical in magnitude but opposite in sign. Relative charges of −1 and +1 are assigned to the electron and proton, respectively.
2. Neutrons have approximately the same mass as protons but no charge. They are electrically neutral.
3. The mass of a proton or a neutron is about 1836 times greater than the mass of an electron. Protons and neutrons constitute the bulk of the mass of atoms.
The discovery of the electron and the proton was crucial to the development of the modern model of the atom and provides an excellent case study in the application of the scientific method. In fact, the elucidation of the atom’s structure is one of the greatest detective stories in the history of science.
Table $1$: Properties of Subatomic Particles*
Particle Mass (g) Atomic Mass (amu) Electrical Charge (coulombs) Relative Charge
electron $9.109 \times 10^{-28}$ 0.0005486 −1.602 × 10−19 −1
proton $1.673 \times 10^{-24}$ 1.007276 +1.602 × 10−19 +1
neutron $1.675 \times 10^{-24}$ 1.008665 0 0
In most cases, the symbols for the elements are derived directly from each element’s name, such as C for carbon, U for uranium, Ca for calcium, and Po for polonium. Elements have also been named for their properties [such as radium (Ra) for its radioactivity], for the native country of the scientist(s) who discovered them [polonium (Po) for Poland], for eminent scientists [curium (Cm) for the Curies], for gods and goddesses [selenium (Se) for the Greek goddess of the moon, Selene], and for other poetic or historical reasons. Some of the symbols used for elements that have been known since antiquity are derived from historical names that are no longer in use; only the symbols remain to indicate their origin. Examples are Fe for iron, from the Latin ferrum; Na for sodium, from the Latin natrium; and W for tungsten, from the German wolfram. Examples are in Table $2$.
Table $2$: Element Symbols Based on Names No Longer in Use
Element Symbol Derivation Meaning
antimony Sb stibium Latin for “mark”
copper Cu cuprum from Cyprium, Latin name for the island of Cyprus, the major source of copper ore in the Roman Empire
gold Au aurum Latin for “gold”
iron Fe ferrum Latin for “iron”
lead Pb plumbum Latin for “heavy”
mercury Hg hydrargyrum Latin for “liquid silver”
potassium K kalium from the Arabic al-qili, “alkali”
silver Ag argentum Latin for “silver”
sodium Na natrium Latin for “sodium”
tin Sn stannum Latin for “tin”
tungsten W wolfram German for “wolf stone” because it interfered with the smelting of tin and was thought to devour the tin
Recall that the nuclei of most atoms contain neutrons as well as protons. Unlike protons, the number of neutrons is not absolutely fixed for most elements. Atoms that have the same number of protons, and hence the same atomic number, but different numbers of neutrons are called isotopes. All isotopes of an element have the same number of protons and electrons, which means they exhibit the same chemistry. The isotopes of an element differ only in their atomic mass, which is given by the mass number (A), the sum of the numbers of protons and neutrons.
The element carbon (C) has an atomic number of 6, which means that all neutral carbon atoms contain 6 protons and 6 electrons. In a typical sample of carbon-containing material, 98.89% of the carbon atoms also contain 6 neutrons, so each has a mass number of 12. An isotope of any element can be uniquely represented as $^A_Z X$, where X is the atomic symbol of the element. The isotope of carbon that has 6 neutrons is therefore $_6^{12} C$. The subscript indicating the atomic number is actually redundant because the atomic symbol already uniquely specifies Z. Consequently, $_6^{12} C$ is more often written as 12C, which is read as “carbon-12.” Nevertheless, the value of Z is commonly included in the notation for nuclear reactions because these reactions involve changes in Z.
Figure $2$: Formalism used for identifying specific nuclide (any particular kind of nucleus)
In addition to $^{12}C$, a typical sample of carbon contains 1.11% $_6^{13} C$ (13C), with 7 neutrons and 6 protons, and a trace of $_6^{14} C$ (14C), with 8 neutrons and 6 protons. The nucleus of 14C is not stable, however, but undergoes a slow radioactive decay that is the basis of the carbon-14 dating technique used in archaeology. Many elements other than carbon have more than one stable isotope; tin, for example, has 10 isotopes. The properties of some common isotopes are in Table $3$.
Table $3$: Properties of Selected Isotopes
Element Symbol Atomic Mass (amu) Isotope Mass Number Isotope Masses (amu) Percent Abundances (%)
hydrogen H 1.0079 1 1.007825 99.9855
2 2.014102 0.0115
boron B 10.81 10 10.012937 19.91
11 11.009305 80.09
carbon C 12.011 12 12 (defined) 99.89
13 13.003355 1.11
oxygen O 15.9994 16 15.994915 99.757
17 16.999132 0.0378
18 17.999161 0.205
iron Fe 55.845 54 53.939611 5.82
56 55.934938 91.66
57 56.935394 2.19
58 57.933276 0.33
uranium U 238.03 234 234.040952 0.0054
235 235.043930 0.7204
238 238.050788 99.274
Sources of isotope data: G. Audi et al., Nuclear Physics A 729 (2003): 337–676; J. C. Kotz and K. F. Purcell, Chemistry and Chemical Reactivity, 2nd ed., 1991.
Example $1$
An element with three stable isotopes has 82 protons. The separate isotopes contain 124, 125, and 126 neutrons. Identify the element and write symbols for the isotopes.
Given: number of protons and neutrons
Asked for: element and atomic symbol
Strategy:
1. Refer to the periodic table and use the number of protons to identify the element.
2. Calculate the mass number of each isotope by adding together the numbers of protons and neutrons.
3. Give the symbol of each isotope with the mass number as the superscript and the number of protons as the subscript, both written to the left of the symbol of the element.
Solution:
A The element with 82 protons (atomic number of 82) is lead: Pb.
B For the first isotope, A = 82 protons + 124 neutrons = 206. Similarly, A = 82 + 125 = 207 and A = 82 + 126 = 208 for the second and third isotopes, respectively. The symbols for these isotopes are $^{206}_{82}Pb$, $^{207}_{82}Pb$, and $^{208}_{82}Pb$, which are usually abbreviated as $^{206}Pb$, $^{207}Pb$, and $^{208}Pb$.
Exercise $1$
Identify the element with 35 protons and write the symbols for its isotopes with 44 and 46 neutrons.
Answer
$\ce{^{79}_{35}Br}$ and $\ce{^{81}_{35}Br}$ or, more commonly, $\ce{^{79}Br}$ and $\ce{^{81}Br}$.
Summary
The atom consists of discrete particles that govern its chemical and physical behavior. Each atom of an element contains the same number of protons, which is the atomic number (Z). Neutral atoms have the same number of electrons and protons. Atoms of an element that contain different numbers of neutrons are called isotopes. Each isotope of a given element has the same atomic number but a different mass number (A), which is the sum of the numbers of protons and neutrons. The relative masses of atoms are reported using the atomic mass unit (amu), which is defined as one-twelfth of the mass of one atom of carbon-12, with 6 protons, 6 neutrons, and 6 electrons. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/01%3A_Introduction_and_Review/1.02%3A_Principles_of_Atomic_Structure_%28Review%29.txt |
Learning Objective
Draw, interpret, and convert between Lewis (Kekule), Condensed, and Bond-line Structures
Note: The review of general chemistry in sections 1.3 - 1.6 is integrated into the above Learning Objective for organic chemistry in sections 1.7 and 1.8.
The primary skills needed are the ability to determine the electron configurations of elements following the concepts of "aufbau" or "build up", Hund's Rule, and the Pauli Exclusion Principle, as well as visualizing the orbitals, subshells, and shells spatially and energetically. Converting electron configurations to orbital diagrams is a useful skill as we transition to organic chemistry. These skills are practiced in the next section.
The Wave Nature of Light
A wave is a periodic oscillation by which energy is transmitted through space. All waves are periodic, repeating regularly in both space and time. Waves are characterized by several interrelated properties.
• Electronic structure: arrangement of electrons in atoms
• Electromagnetic radiation: aka radiant energy; form of energy that has wave characteristics and carries energy through space. All types of electromagnetic radiation move through a vacuum at a speed of 3.00 X 108 m/s (speed of light).
• Wavelength: the distance between identical points on successive waves
• Frequency: the number of complete wavelengths that pass a given point in 1s
$\nu λ = c$
where $\nu$ = frequency, $λ$= wavelength, and $c$ = speed of light
Wavelength is expressed in units of length.
• Electromagnetic spectrum: various types of electromagnetic radiations arranged in order of increasing wavelength.
Frequency is expressed in Hertz (Hz), also denoted by s-1 or /s
Quantum Effects and Photons
• Quantum: smallest quantity of energy that can be emitted or absorbed as electromagnetic radiation
$E = hv$
where $E$ = energy, $h$ = Planck’s constant, $\nu$ = frequency
Planck’s constant = 6.63 X 10-34 J/s
According to Planck’s theory, energy is always emitted or absorbed in whole-number multiples of hv, for example, hv, 2hv, 3hv, and so forth. We say that the allowed energies are quantized (that is, their values are restricted to certain quantities).
Quantized Energy and Photons
Blackbody radiation is the radiation emitted by hot objects and could not be explained with classical physics. Max Planck postulated that energy was quantized and may be emitted or absorbed only in integral multiples of a small unit of energy, known as a quantum. The energy of a quantum is proportional to the frequency of the radiation; the proportionality constant h is a fundamental constant (Planck’s constant). Albert Einstein used the quantization of energy to explain the photoelectric effect
• The photoelectric effect: when photons of sufficiently high energy strike a metal surface, electrons are emitted from the metal. The emitted electrons are drawn toward the other electrode, which is a positive terminal. As a result, current flows in a circuit.
• Photon: smallest increment (a quantum) of radiant energy; a photon of light with frequency v has an energy equal to hv.
• When a photon strikes the metal, its energy is transferred to an electron in the metal. A certain amount of energy is required for the electron to overcome the attractive forces that hold it within the metal. If the photons have less energy that this energy threshold, the electrons cannot escape from the metal surface. If a photon has sufficient energy, an electron is emitted. If the photon has more energy than necessary, the excess appears as kinetic energy of the emitted electron.
Line Spectra and the Bohr Model
There is an intimate connection between the atomic structure of an atom and its spectral characteristics. Most light is polychromatic and contains light of many wavelengths. Light that has only a single wavelength is monochromatic and is produced by devices called lasers, which use transitions between two atomic energy levels to produce light in a very narrow range of wavelengths. Atoms can also absorb light of certain energies, resulting in a transition from the ground state or a lower-energy e
Line Spectra
Radiation composed of a single wavelength is said to be monochromatic.
• Spectrum: distribution among various wavelengths of the radiant energy emitted or absorbed by an object
• Continuous spectrum: rainbow of colors, containing light of all wavelengths. Not all radiation sources produce a continuous spectrum
• Line spectrum: spectrum containing radiation of only specific wavelengths
$v = C \left(\dfrac{1}{2^2} – \dfrac{1}{n^2}\right)$
with $n$ = 3, 4, 5, 6, and $C = 3.29 \times 10^{15} s^{-1}$ (constant)
Bohr’s Model
Electrons in a permitted orbit have a specific energy and are said to be in an "allowed" energy state. An electron in an allowed energy state will not radiate energy and therefore will not spiral into the nucleus.
$E_n = -R_H \dfrac{1}{n^2}$
• $R_H$ = Rydberg constant: 2.18 X 10-18 J
• $n$ = principal quantum number, corresponds to the different allowed orbits for the electron
All energies given by this equation will be negative. The lower (more negative) the energy is, the more stable the atom is. The lowest energy state is that for which n=1.
• Ground state: lowest energy state of an atom, $n=1$
• Excited state: when the electron is in higher energy orbit (less negative), n=2 or higher
If n becomes infinitely large (∞), the electron is completely separated from the nucleus:
$E_∞ = (-2.18 \times 10^{-18} J) \left(\dfrac{1}{∞^2}\right) = 0$
Thus, the state in which the electron is removed from the nucleus is the reference, or zero-energy, state of the hydrogen atom. It is important to remember that this zero-energy state is higher in energy than the states with negative energies
Electrons can change from one energy state to another by absorbing or emitting radiant energy. Radiant energy must be absorbed for an electron to move to a higher energy state, but is emitted when the electron moves to a lower energy state. .
$\Delta E = E_f – E_i$
• If $n_f > n_i$, then ∆E is positive, radiant energy is absorbed
• If $n_f < n_i$, then ∆E is negative, radiant energy is emitted
The Wave Behavior of Matter
An electron possesses both particle and wave properties. Louis de Broglie showed that the wavelength of a particle is equal to Planck’s constant divided by the mass times the velocity of the particle. The electron in Bohr’s circular orbits could thus be described as a standing wave, one that does not move through space. Werner Heisenberg’s uncertainty principle states that it is impossible to precisely describe both the location and the speed of particles that exhibit wavelike behavior.
• Momentum: the product of the mass, m, and the velocity, v, of a particle
• Matter waves: term used to describe the wave characteristics of a particle
$λ = \dfrac{h}{mv}$
where $λ$ is the wavelength, $h$ is Planck’s constant, $m$ is the particle mass, and $v$ is the velocity
The Uncertainty Principle
• Uncertainty principle: theory first put forth by Heisenberg, states that is it impossible to determine both the exact momentum of the electron and its exact location.
Quantum Mechanics and Atomic Orbitals
There is a relationship between the motions of electrons in atoms and molecules and their energies that is described by quantum mechanics. Because of wave–particle duality, scientists must deal with the probability of an electron being at a particular point in space. To do so required the development of quantum mechanics, which uses wavefunctions to describe the mathematical relationship between the motion of electrons in atoms and molecules and their energies.
• Wave functions: represented by ψ, square of wave function, ψ2, provides information about an electron’s location when it is in an allowed energy state.
• Probability density: represented by ψ2, value that represents the probability that an electron will be found at a given point in space
• Electron density: the probability of finding and electron at any particular point in an atom. Equals ψ2.
Orbitals and Quantum Numbers
-Orbital: allowed energy state of an electron in the quantum-mechanical model of the atom; also used to describe the spatial distribution of an electron. Defined by the value of 3 quantum numbers; n, l, and ml.
Value of l
0
1
2
3
Letter used
s
p
d
f
1. The principal quantum number, n, can have integral values of 1, 2, 3 and so forth. As n increases, the orbital becomes larger; the electron has a higher energy and is farther away from the nucleus.
2. The second quantum number, l, can have integral values from 0 to n – 1 for each value of n. This quantum number defines the shape of the orbital. Generally designated by the letters s, p, d, and f. These correspond to values ranging from 0 to 3.
3. The magnetic quantum number, ml, can have integral values between l and –l, including zero. This quantum number describes the orientation of the orbital in space.
Electron shell: collection of orbitals with the same value of n
Subshell: one or more orbitals with the same set of n and l values
1. Each shell is divided into the number of subshells equal to the principal quantum number, n, for that shell. The first shell consists of only the 1s subshell; the second shell consists of two subshells, 2s and 2p; the third of three subshell, 3s, 3p and 3d, and so forth.
2. Each subshell is divided into orbitals. Each s subshell consists of one orbital; each p subshell of three orbitals, each d subshell of five, and each f subshell of seven orbitals.
3D Representation of Orbitals
Orbitals with l = 0 are s orbitals and are spherically symmetrical, with the greatest probability of finding the electron occurring at the nucleus. Orbitals with values of n > 1 and l = 0 contain one or more nodes. Orbitals with l = 1 are p orbitals and contain a nodal plane that includes the nucleus, giving rise to a dumbbell shape. Orbitals with l = 2 are d orbitals and have more complex shapes with at least two nodal surfaces. l = 3 orbitals are f orbitals, which are still more complex.
The s Orbitals: 1s orbital: most stable, spherically symmetric, figure indicates that the probability decreases as we move away from the nucleus. ALL s ORBITALS ARE SPHERICALLY SYMMETRIC.
• Nodal surfaces (nodes): intermediate regions where ψ2 goes to zero. The number of nodes increases with increasing value for the principal quantum number, n.
The p Orbitals: Electron density is concentrated on two sides of the nucleus, separated by a node at the nucleus. The orbitals of a given subshell have the same size and shape but differ from each other in orientation. The axis along which the orbital is oriented is not related to ml.
Many-Electron Atoms
In addition to the three quantum numbers ($n$, $l$, $m_l$) dictated by quantum mechanics, a fourth quantum number is required to explain certain properties of atoms. This is the electron spin quantum number ($m_s$), which can have values of +½ or −½ for any electron, corresponding to the two possible orientations of an electron in a magnetic field. This is important for chemistry because the Pauli exclusion principle implies that no orbital can contain more than two electrons (with opposite spin).
Although the shapes of the orbitals for many-electron atoms are the same as those for hydrogen, the presence of more than one electron greatly changes the energies of the orbitals. In hydrogen, the energy of an orbital depends only on its principal quantum number, however, in many-electron atoms, electron-electron repulsions cause different subshells to be at different energies
Effective Nuclear Charge
• Effective nuclear charge: net positive charge attracting electrons
$Z_{eff} = Z – S$
where $Z_{eff}$ is the effective nuclear charge, $Z$ is the number of protons in the nucleus, and $S$ is the average number of electrons between the nucleus and electron in question.
• Screening effect: effect of inner electrons in decreasing the nuclear charge experienced by outer electrons
Energies of Orbitals
The extent to which an electron will be screened by the other electrons depends on its electron distribution as we move outward from the nucleus.
• In a many-electron atom, for a given value of n, Zeff decreases with increasing value of l.
• In a many-electron atom, for a given value of n, the energy of an orbital increases with increasing value of l.
Degenerate: orbitals that have the same energy
Electron Spin and the Pauli Exclusion Principle
• Electron spin: property of the electron that makes it behave as though it were a tiny magnet. The electron behaves as if it were spinning on its axis; electron spin is quantized.
• Electron spin quantum number: denoted as ms. It can only have two possible values, +½ and –½, which we can interpret as indicating the two opposite directions in which the electron can spin.
• Pauli exclusion principle: states that no two electrons in an atom can have the same set of four quantum numbers n, l, ml, ms. This means that if we wish to put two electrons in an orbital and satisfy Pauli’s exclusion principle, our only choice is to assign different ms values to the electrons. Because there are only two values, we can conclude that an orbital can hold a maximum of two electrons and they must have opposite spins.
Electron Configurations
Based on the Pauli principle and a knowledge of orbital energies obtained using hydrogen-like orbitals, it is possible to construct the periodic table by filling up the available orbitals beginning with the lowest-energy orbitals (the aufbau principle), which gives rise to a particular arrangement of electrons for each element (its electron configuration). Hund’s rule says that the lowest-energy arrangement of electrons is the one that places them in degenerate orbitals with parallel spins.
• Electron configuration: the way in which the electrons are distributed among the various orbitals. The most stable, or ground, electron configuration of an atom is that in which the electrons are in the lowest possible energy level
• Orbital diagram: representation of electron configuration in which each orbital is represented by a box and each electron by a half-arrow. A half-arrow pointing upward represents an electron with positive spin; one pointing downward represents an electron with a negative spin.
Writing Electron Configurations
• Hund’s rule: rule stating that electrons occupy degenerate orbitals in such a way as to maximize the number of electrons with the same spin. In other words, each orbital has one electron placed in it before paring of electron in orbitals occurs. Note that this rule applies to orbitals that are degenerate, which means that they have the same energy.
• Valence electrons: electrons in the outer shells
• Core electrons: electrons in the inner shells
• Transition elements: aka Transition metals; elements of the d orbitals
• Lanthanide elements: aka Rare-earth elements; 14 elements of the 4f orbitals, # 58-71
• Actinide elements: 14 elements of 5f orbitals, # 90-103. Most are not found in nature.
Electron Configurations and the Periodic Table
The arrangement of atoms in the periodic table results in blocks corresponding to filling of the ns, np, nd, and nf orbitals to produce the distinctive chemical properties of the elements in the s block, p block, d block, and f block, respectively.
• Main group elements: aka Representatives; s and p block elements
• F-block metals: 28 elements located below the table, f block elements | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/01%3A_Introduction_and_Review/1.03%3A_Electronic_Structure_%28Review%29.txt |
Learning Objective
Draw, interpret, and convert between Lewis (Kekule), Condensed, and Bond-line Structures
Note: The review of general chemistry in sections 1.3 - 1.6 is integrated into the above Learning Objective for organic chemistry in sections 1.7 and 1.8.
The 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.
Introduction
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.
Pauli Exclusion Principle
The Pauli exclusion principle states that no two electrons can have the same four quantum numbers. The first three (n, l, and ml) may be the same, but the fourth quantum number must be different. A single orbital can hold a maximum of two electrons, which must have opposing spins; otherwise they would have the same four quantum numbers, which is forbidden. One electron is spin up (ms = +1/2) and the other would spin down (ms = -1/2). This tells us that each subshell has double the electrons per orbital. The s subshell has 1 orbital that can hold up to 2 electrons, the p subshell has 3 orbitals that can hold up to 6 electrons, the d subshell has 5 orbitals that hold up to 10 electrons, and the f subshell has 7 orbitals with 14 electrons.
Example 1: Hydrogen and Helium
The first three quantum numbers of an electron are n=1, l=0, ml=0. Only two electrons can correspond to these, which would be either ms = -1/2 or ms = +1/2. As we already know from our studies of quantum numbers and electron orbitals, we can conclude that these four quantum numbers refer to the 1s subshell. If only one of the ms values are given then we would have 1s1 (denoting hydrogen) if both are given we would have 1s2 (denoting helium). Visually, this is be represented as:
As shown, the 1s subshell can hold only two electrons and, when filled, the electrons have opposite spins.
Hund's Rule
When assigning electrons in orbitals, each electron will first fill all the orbitals with similar energy (also referred to as degenerate) before pairing with another electron in a half-filled orbital. Atoms at ground states tend to have as many unpaired electrons as possible. When visualizing this processes, think about how electrons are exhibiting the same behavior as the same poles on a magnet would if they came into contact; as the negatively charged electrons fill orbitals they first try to get as far as possible from each other before having to pair up.
Example 2: Oxygen and Nitrogen
If we look at the correct electron configuration of the Nitrogen (Z = 7) atom, a very important element in the biology of plants: 1s2 2s2 2p3
We can clearly see that p orbitals are half-filled as there are three electrons and three p orbitals. This is because Hund's Rule states that the three electrons in the 2p subshell will fill all the empty orbitals first before filling orbitals with electrons in them. If we look at the element after Nitrogen in the same period, Oxygen (Z = 8) its electron configuration is: 1s2 2s2 2p4 (for an atom).
Oxygen has one more electron than Nitrogen and as the orbitals are all half filled the electron must pair up.
The Aufbau Process
Aufbau comes from the German word "aufbauen" meaning "to build." When writing electron configurations, orbitals are built up from atom to atom. When writing the electron configuration for an atom, orbitals are filled in order of increasing atomic number. However, there are some exceptions to this rule.
Example 3: 3rd row elements
Following the pattern across a period from B (Z=5) to Ne (Z=10), the number of electrons increases and the subshells are filled. This example focuses on the p subshell, which fills from boron to neon.
• B (Z=5) configuration: 1s2 2s2 2p1
• C (Z=6) configuration:1s2 2s2 2p2
• N (Z=7) configuration:1s2 2s2 2p3
• O (Z=8) configuration:1s2 2s2 2p4
• F (Z=9) configuration:1s2 2s2 2p5
• Ne (Z=10) configuration:1s2 2s2 2p6
Example
The electron configuration for sulfur is 1s2 2s2 2p6 3s2 3p4 and can be represented using the orbital diagram below.
Exercises
Write the electron configuration for phosphorus and draw the orbital diagram.
Solution:
The electron configuration for phosphorus is 1s2 2s2 2p6 3s2 3p3 and the orbital diagram is drawn below. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/01%3A_Introduction_and_Review/1.04%3A_Electron_Configurations_and_Electronic_Orbital_Diagrams_%28Review%29.txt |
Learning Objective
Draw, interpret, and convert between Lewis (Kekule), Condensed, and Bond-line Structures
Note: The review of general chemistry in sections 1.3 - 1.6 is integrated into the above Learning Objective for organic chemistry in sections 1.7 and 1.8.
For organic chemistry, the emphasis is on the chemistry of carbon. The chemistry of carbon becomes more interesting when carbon is bonded to oxygen and/or nitrogen or other heteroatoms, atoms that are NOT carbon or hydrogen. Therefore, the octet rule is a strong factor in organic chemistry and is only violated by non-carbon elements like hydrogen, boron, aluminum, sulfur, and phosphorus.
Why are some substances chemically bonded molecules and others are an association of ions? The answer to this question depends upon the electronic structures of the atoms and nature of the chemical forces within the compounds. Although there are no sharply defined boundaries, chemical bonds are typically classified into three main types: ionic bonds, covalent bonds, and metallic bonds. In this chapter, each type of bond will be discussed and the general properties found in typical substances in which the bond type occurs
1. Ionic bonds results from electrostatic forces that exist between ions of opposite charge. These bonds typically involves a metal with a nonmetal
2. Covalent bonds result from the sharing of electrons between two atoms. The bonds typically involves one nonmetallic element with another
3. Metallic bonds These bonds are found in solid metals (copper, iron, aluminum) with each metal bonded to several neighboring groups and bonding electrons free to move throughout the 3-dimensional structure.
The Octet Rule
In 1904, Richard Abegg formulated what is now known as Abegg's rule, which states that the difference between the maximum positive and negative valences of an element is frequently eight. This rule was used later in 1916 when Gilbert N. Lewis formulated the "octet rule" in his cubical atom theory. The octet rule refers to the tendency of atoms to prefer to have eight electrons in the valence shell. When atoms have fewer than eight electrons, they tend to react and form more stable compounds. Atoms will react to get in the most stable state possible. A complete octet is very stable because all orbitals will be full. Atoms with greater stability have less energy, so a reaction that increases the stability of the atoms will release energy in the form of heat or light ;reactions that decrease stability must absorb energy, getting colder.
The Octet Rule: Atoms often gain, lose, or share electrons to achieve the same number of electrons as the noble gas closest to them in the periodic table.
When discussing the octet rule, we do not consider d or f electrons. Only the s and p electrons are involved in the octet rule, making it a useful rule for the main group elements (elements not in the transition metal or inner-transition metal blocks); an octet in these atoms corresponds to an electron configurations ending with s2p6.
Covalent Bonds
Covalent bonds form when atoms share electrons. Hydrogen is a first shell element with only one valence electron, so it can only form one bond creating a duet, an exception to the octet rule. With its four valence electrons, carbon can form four bonds to create an octet.
1. Normally two electrons pairs up and forms a bond, e.g., \(H_2\)
2. For most atoms there will be a maximum of eight electrons in the valence shell (octet structure), e.g., \(CH_4\)
The other tendency of atoms is to maintain a neutral charge. Only the noble gases (the elements on the right-most column of the periodic table) have zero charge with filled valence octets. All of the other elements have a charge when they have eight electrons all to themselves. The result of these two guiding principles is the explanation for much of the reactivity and bonding that is observed within atoms: atoms seek to share electrons in a way that minimizes charge while fulfilling an octet in the valence shell.
Ionic Bonds
Some atoms do not share electrons. Energetically, it is more favorable to fully gain or lose electrons to form ions. Ionic compounds form through the eletrostatic attraction of the ions to create a crystal lattice.
The formula for table salt is NaCl. It is the result of Na+ ions and Cl- ions bonding together. If sodium metal and chlorine gas mix under the right conditions, they will form salt. The sodium loses an electron, and the chlorine gains that electron. In the process, a great amount of light and heat is released. The resulting salt is mostly unreactive — it is stable. It will not undergo any explosive reactions, unlike the sodium and chlorine that it is made of. Why? Referring to the octet rule, atoms attempt to get a noble gas electron configuration, which is eight valence electrons. Sodium has one valence electron, so giving it up would result in the same electron configuration as neon. Chlorine has seven valence electrons, so if it takes one it will have eight (an octet). Chlorine has the electron configuration of argon when it gains an electron.
The octet rule could have been satisfied if chlorine gave up all seven of its valence electrons and sodium took them. In that case, both would have the electron configurations of noble gasses, with a full valence shell. However, their charges would be much higher. It would be Na7- and Cl7+, which is much less stable than Na+ and Cl-. Atoms are more stable when they have no charge, or a small charge.
Ionic Bonds Example
Lewis dot symbols can also be used to represent the ions in ionic compounds. The reaction of cesium with fluorine, for example, to produce the ionic compound CsF can be written as follows:
No dots are shown on Cs+ in the product because cesium has lost its single valence electron to fluorine. The transfer of this electron produces the Cs+ ion, which has the valence electron configuration of Xe, and the F ion, which has a total of eight valence electrons (an octet) and the Ne electron configuration. This description is consistent with the statement that among the main group elements, ions in simple binary ionic compounds generally have the electron configurations of the nearest noble gas. The charge of each ion is written in the product, and the anion and its electrons are enclosed in brackets. This notation emphasizes that the ions are associated electrostatically; no electrons are shared between the two elements.
Noble Gases
The noble gases rarely form compounds. They have the most stable configuration (full octet, no charge), so they have no reason to react and change their configuration. All other elements attempt to gain, lose, or share electrons to achieve a noble gas configuration.
Summary
Lewis dot symbols can be used to predict the number of bonds formed by most elements in their compounds. One convenient way to predict the number and basic arrangement of bonds in compounds is by using Lewis electron dot symbols, which consist of the chemical symbol for an element surrounded by dots that represent its valence electrons, grouped into pairs often placed above, below, and to the left and right of the symbol. The structures reflect the fact that the elements in period 2 and beyond tend to gain, lose, or share electrons to reach a total of eight valence electrons in their compounds, the so-called octet rule. Hydrogen, with only two valence electrons, does not obey the octet rule.
Lewis
Contributors and Attributions
• National Programme on Technology Enhanced Learning (India) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/01%3A_Introduction_and_Review/1.05%3A_Octet_Rule_-_Ionic_and_Covalent_Bonding_%28Review%29.txt |
Learning Objective
Draw, interpret, and convert between Lewis (Kekule), Condensed, and Bond-line Structures
Note: The review of general chemistry in sections 1.3 - 1.6 is integrated into the above Learning Objective for organic chemistry in sections 1.7 and 1.8.
Lewis Structures
Lewis structures, also known as Lewis-dot diagrams, show the bonding relationship between atoms of a molecule and the lone pairs of electrons in a molecule. While it can be helpful initially to write the individual shared electrons, this approach quickly becomes awkward.
A single line is used to represent one pair of shared electrons. Line representations are only used for shared electrons. Lone pair (unshared) electrons are still shown as individual electrons. Double and triple bonds can also be communicated with lines as shown below.
2 shared electrons form a single bond shown as ‘:’ or ‘–‘
4 shared electrons form a double bond shown as ‘::’ or ‘=’
6 shared electrons form at triple bond shown as ‘:::’ or
Unshared electrons are also called ‘Lone Pairs’ and are shown as ‘:’
Drawing Lone Pairs
Since the lone pair electrons are often NOT shown in chemical structures, it is important to mentally add the lone pairs. In the beginning, it can be helpful to physically add the lone pair electrons.
Bonding Patterns
For organic chemistry, the common bonding patterns of carbon, oxygen, and nitrogen have useful applications when evaluating chemical structures and reactivity.
Formal charges
Organic molecules can also have positive or negative charges associated with them. During chemical reactions, it is common to have charge reactant, intermediates, and/or products. Recognizing and distinguishing between neutral and charged bonding patterns will be helpful in learning reaction mechanisms. Consider the Lewis structure of methanol, CH3OH (methanol is the so-called ‘wood alcohol’ that unscrupulous bootleggers sometimes sold during the prohibition days in the 1920's, often causing the people who drank it to go blind). Methanol itself is a neutral molecule, but can lose a proton to become a molecular anion (CH3O-), or gain a proton to become a molecular cation (CH3OH2+).
The molecular anion and cation have overall charges of -1 and +1, respectively. But we can be more specific than that - we can also state for each molecular ion that a formal charge is located specifically on the oxygen atom, rather than on the carbon or any of the hydrogen atoms.
Figuring out the formal charge on different atoms of a molecule is a straightforward process - it’s simply a matter of adding up valence electrons.
A unbound oxygen atom has 6 valence electrons. When it is bound as part of a methanol molecule, however, an oxygen atom is surrounded by 8 valence electrons: 4 nonbonding electrons (two 'lone pairs') and 2 electrons in each of its two covalent bonds (one to carbon, one to hydrogen). In the formal charge convention, we say that the oxygen 'owns' all 4 nonbonding electrons. However, it only 'owns' one electron from each of the two covalent bonds, because covalent bonds involve the sharing of electrons between atoms. Therefore, the oxygen atom in methanol owns 2 + 2 + (½ x 4) = 6 valence electrons.
The formal charge on an atom is calculated as the number of valence electrons owned by the isolated atom minus the number of valence electrons owned by the bound atom in the molecule:
Determining formal charge on an atom
formal charge =
(number of valence electrons owned by the isolated atom)
- (number of valence electrons owned by the bound atom)
or . . .
formal charge =
(number of valence electrons owned by the isolated atom)
- (number of non-bonding electrons on the bound atom)
- ( ½ the number of bonding electrons on the bound atom)
Using this formula for the oxygen atom of methanol, we have:
formal charge on oxygen =
(6 valence electrons on isolated atom)
- (4 non-bonding electrons)
- (½ x 4 bonding electrons)
= 6 - 4 - 2 = 0. Thus, oxygen in methanol has a formal charge of zero (in other words, it has no formal charge).
How about the carbon atom in methanol? An isolated carbon owns 4 valence electrons. The bound carbon in methanol owns (½ x 8) = 4 valence electrons:
formal charge on carbon =
(4 valence electron on isolated atom)
- (0 nonbonding electrons)
- (½ x 8 bonding electrons)
= 4 - 0 - 4 = 0. So the formal charge on carbon is zero.
For each of the hydrogens in methanol, we also get a formal charge of zero:
formal charge on hydrogen =
(1 valence electron on isolated atom)
- (0 nonbonding electrons)
- (½ x 2 bonding electrons)
= 1 - 0 - 1 = 0
Now, let's look at the cationic form of methanol, CH3OH2+. The bonding picture has not changed for carbon or for any of the hydrogen atoms, so we will focus on the oxygen atom.
The oxygen owns 2 non-bonding electrons and 3 bonding elections, so the formal charge calculations becomes:
formal charge on oxygen =
(6 valence electrons in isolated atom)
- (2 non-bonding electrons)
- (½ x 6 bonding electrons)
= 6 - 2 - 3 = 1. A formal charge of +1 is located on the oxygen atom.
For methoxide, the anionic form of methanol, the calculation for the oxygen atom is:
formal charge on oxygen =
(6 valence electrons in isolated atom)
- (6 non-bonding electrons)
- (½ x 2 bonding electrons)
= 6 - 6 - 1 = -1. A formal charge of -1 is located on the oxygen atom.
A very important rule to keep in mind is that the sum of the formal charges on all atoms of a molecule must equal the net charge on the whole molecule.
When drawing the structures of organic molecules, it is very important to show all non-zero formal charges, being clear about where the charges are located. A structure that is missing non-zero formal charges is not correctly drawn, and will probably be marked as such on an exam!
At this point, thinking back to what you learned in general chemistry, you are probably asking “What about dipoles? Doesn’t an oxygen atom in an O-H bond ‘own’ more of the electron density than the hydrogen, because of its greater electronegativity?” This is absolutely correct, and we will be reviewing the concept of bond dipoles later on. For the purpose of calculating formal charges, however, bond dipoles don’t matter - we always consider the two electrons in a bond to be shared equally, even if that is not an accurate reflection of chemical reality. Formal charges are just that - a formality, a method of electron book-keeping that is tied into the Lewis system for drawing the structures of organic compounds and ions. Later, we will see how the concept of formal charge can help us to visualize how organic molecules react.
Finally, don't be lured into thinking that just because the net charge on a structure is zero there are no atoms with formal charges: one atom could have a positive formal charge and another a negative formal charge, and the net charge would still be zero. Zwitterions, such as amino acids, have both positive and negative formal charges on different atoms:
Even though the net charge on glycine is zero, it is still neccessary to show the location of the positive and negative formal charges.
Exercise 1.4
Fill in all missing lone pair electrons and formal charges in the structures below. Assume that all atoms have a complete valence shell of electrons. Net charges are shown outside the brackets.
Solutions to exercises | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/01%3A_Introduction_and_Review/1.06%3A_Lewis_Structures_and_Formal_Charges_%28Review%29.txt |
Learning Objective
Draw, interpret, and convert between Lewis (Kekule), Condensed, and Bond-line Structures
Note: The review of general chemistry in sections 1.3 - 1.6 is integrated into the above Learning Objective for organic chemistry in sections 1.7 and 1.8.
Common bonding patterns in organic structures
The Lewis structure below (of one of the four nucleoside building blocks that make up DNA) can appear complex and confusing at first glance. Fortunately, common bonding patterns occur that can allow for simplifications when drawing structures. The rules for the simplifies structures rely on the neutral bonding patterns for carbon, orygen, nitrogen, phosphorus, and sulfur primarily. Since organic compounds have a hydrocarbon backbone, the atoms that are NOT carbon and hydrogen are called heteroatoms. 2'-deoxycytidine contains seven heteroatoms: four oxygen atoms land three nitrogen atoms. Heteroatoms are a primary source of chemical reactivity for organic chemistry.
Heteroatoms: atoms in an organic compound that are NOT carbon or hydrogen,
typically oxygen, nitrogen, phosphorus, and sulfur
The ability to quickly and efficiently draw large structures and determine formal charges is not terribly hard to come by - all it takes is a few shortcuts and some practice at recognizing common bonding patterns.
Let’s start with carbon, the most important element for organic chemists. Carbon is said to be tetravalent, meaning that it tends to form four bonds. If you look at the simple structures of methane, methanol, ethane, ethene, and ethyne in the figures from the previous section, you should quickly recognize that in each molecule, the carbon atom has four bonds, and a formal charge of zero.
This is a pattern that holds throughout most of the organic molecules we will see, but there are also exceptions.
<style type="text/css"></style> In carbon dioxide, the carbon atom has double bonds to oxygen on both sides (O=C=O). Later on in this chapter and throughout this book we will see examples of organic ions called ‘carbocations’ and carbanions’, in which a carbon atom bears a positive or negative formal charge, respectively. If a carbon has only three bonds and an unfilled valence shell (in other words, if it does not fulfill the octet rule), it will have a positive formal charge.
If, on the other hand, it has three bonds plus a lone pair of electrons, it will have a formal charge of -1. Another possibility is a carbon with three bonds and a single, unpaired (free radical) electron: in this case, the carbon has a formal charge of zero. (One last possibility is a highly reactive species called a ‘carbene’, in which a carbon has two bonds and one lone pair of electrons, giving it a formal charge of zero. You may encounter carbenes in more advanced chemistry courses, but they will not be discussed any further in this book).
You should certainly use the methods you have learned to check that these formal charges are correct for the examples given above. More importantly, you will need, before you progress much further in your study of organic chemistry, to simply recognize these patterns (and the patterns described below for other atoms) and be able to identify carbons that bear positive and negative formal charges by a quick inspection.
The pattern for hydrogens is easy: hydrogen atoms have only one bond, and no formal charge. The exceptions to this rule are the proton, H+, and the hydride ion, H-, which is a proton plus two electrons. Because we are concentrating in this book on organic chemistry as applied to living things, however, we will not be seeing ‘naked’ protons and hydrides as such, because they are too reactive to be present in that form in aqueous solution. Nonetheless, the idea of a proton will be very important when we discuss acid-base chemistry, and the idea of a hydride ion will become very important much later in the book when we discuss organic oxidation and reduction reactions. As a rule, though, all hydrogen atoms in organic molecules have one bond, and no formal charge.
Let us next turn to oxygen atoms. Typically, you will see an oxygen bonding in three ways, all of which fulfill the octet rule.
If it has two bonds and two lone pairs, as in water, it will have a formal charge of zero. If it has one bond and three lone pairs, as in hydroxide ion, it will have a formal charge of-1. If it has three bonds and one lone pair, as in hydronium ion, it will have a formal charge of +1.
When we get to our discussion of free radical chemistry in chapter 17, we will see other possibilities, such as where an oxygen atom has one bond, one lone pair, and one unpaired (free radical) electron, giving it a formal charge of zero. For now, however, concentrate on the three main non-radical examples, as these will account for virtually everything we see until chapter 17.
Nitrogen has two major bonding patterns, both of which fulfill the octet rule:
If a nitrogen has three bonds and a lone pair, it has a formal charge of zero. If it has four bonds (and no lone pair), it has a formal charge of +1. In a fairly uncommon bonding pattern, negatively charged nitrogen has two bonds and two lone pairs.
Two third row elements are commonly found in biological organic molecules: sulfur and phosphorus. Although both of these elements have other bonding patterns that are relevant in laboratory chemistry, in a biological context sulfur almost always follows the same bonding/formal charge pattern as oxygen, while phosphorus is present in the form of phosphate ion (PO43-), where it has five bonds (almost always to oxygen), no lone pairs, and a formal charge of zero. Remember that atoms of elements in the third row and below in the periodic table have 'expanded valence shells' with d orbitals available for bonding, and the the octet rule does not apply.
Finally, the halogens (fluorine, chlorine, bromine, and iodine) are very important in laboratory and medicinal organic chemistry, but less common in naturally occurring organic molecules. Halogens in organic compounds usually are seen with one bond, three lone pairs, and a formal charge of zero. Sometimes, especially in the case of bromine, we will encounter reactive species in which the halogen has two bonds (usually in a three-membered ring), two lone pairs, and a formal charge of +1.
These rules, if learned and internalized so that you don’t even need to think about them, will allow you to draw large organic structures, complete with formal charges, quite quickly.
Once you have gotten the hang of drawing Lewis structures, it is not always necessary to draw lone pairs on heteroatoms, as you can assume that the proper number of electrons are present around each atom to match the indicated formal charge (or lack thereof). Occasionally, though, lone pairs are drawn if doing so helps to make an explanation more clear.
Exercise 1: Draw one structure that corresponds to each of the following molecular formulas, using the common bonding patters covered above. Be sure to include all lone pairs and formal charges where applicable, and assume that all atoms have a full valence shell of electrons. More than one correct answer is possible for each, so you will want to check your answers with your instructor or tutor.
a) C5H10O b) C5H8O c) C6H8NO+ d) C4H3O2-
Solutions to exercises | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/01%3A_Introduction_and_Review/1.07%3A_Common_Bonding_Patterns_for_Organic_Chemistry.txt |
Learning Objective
Draw, interpret, and convert between Lewis (Kekule), Condensed, and Bond-line Structures
Note: The review of general chemistry in sections 1.3 - 1.6 is integrated into the above Learning Objective for organic chemistry in sections 1.7 and 1.8.
Shorthand notations to represent organic molecules rely on our knowledge of common neutral bonding patterns. Knowing these patterns, we can fill in the missing structural information. Some of these shorthand ways of drawing molecules give us insight into the bond angles and relative positions of atoms in the molecule, while some notations eliminate the carbon and hydrogen atoms and only indicate the heteroatoms (the atoms that are NOT carbon or hydrogen).
There are three primary methods to communicate chemical structure of organic molecules:
Kekule: Lewis structures using lines to represent covalent bonds and showing all atoms and lone pair electrons
Bond-line (Skeletl-line): shows bonds between carbon atoms and heteroatoms) (with lone pair electrons when requested)
Condensed: all atoms are written to communicate structure without drawing any chemical bonds based on the carbon backbone
Introduction
Observe the following drawings of the structure of Retinol, the most common form of vitamin A. The first drawing follows the straight-line (a.k.a. Kekulé) structure which is helpful when you want to look at every single atom; however, showing all of the hydrogen atoms makes it difficult to compare the overall structure with other similar molecules and makes it difficult to focus in on the double bonds and OH group.
Retinol: Kekulé straight-line drawing
The following is a bond-line (a.k.a. zig-zag) formula for retinol. With this simiplified representation, one can easily see the carbon-carbon bonds, double bonds, OH group, and CH3 groups sticking off of the the main ring and chain. Also, it is much quicker to draw this than the one above. You will learn to appreciate this type of formula writing after drawing a countless number of organic molecules.
Retinol: Bond-line or zig-zag formula
Importance of Structure
Learning and practicing the basics of Organic Chemistry will help you immensely in the long run as you learn new concepts and reactions. Some people say that Organic Chemistry is like another language, and in some aspects, it is. At first it may seem difficult or overwhelming, but the more you practice looking at and drawing organic molecules, the more familiar you will become with the structures and formulas. Another good idea is to get a model kit and physically make the molecules that you have trouble picturing in your head.
Through general chemistry, you may have already experienced looking at molecular structure. The different ways to draw organic molecules include Keku (straight-line), Condensed Formulas, and Bond-Line Formulas (zig-zag). It will be more helpful if you become comfortable going from one style of drawing to another, and look at drawings and understanding what they mean, than knowing which kind of drawing is named what.
An example of a drawing that incorporates all three ways to draw organic molecules would be the following additional drawing of Retinol. The majority of the drawing is Bond-line (zig-zag) formula, but the -CH3 are written as condensed formulas, and the -OH group is written in Kekulé form.
A widely used way of showing the 3D structure of molecules is the use of dashes, wedges, and straight lines. This drawing method is essential because the placement of different atoms could yield different molecules even if the molecular formulas were exactly the same. Below are two drawings of a 4-carbon molecule with two chlorines and two bromines attached.
4-carbon molecule with 2 chlorines and 2 bromines 4-carbon molecule with 2 chlorines and 2 bromines
Both drawings look like they represent the same molecule; however, if we add dashes and wedged we will see that two different molecules could be depicted:
The two molecules above are different, prove this to yourself by building a model. An easier way to compare the two molecules is to rotate one of the bonds (here, it is the bond on the right):
Notice how the molecule on the right has both bromines on the same side and chlorines on the same side, whereas the first molecule is different. Read about Dashed-Wedged Line structures, bottom of page, to understand what has been introduced above. You will learn more about the importance of atomic connectivity in molecules as you continue on to learn about Stereochemistry.
Drawing the Structure of Organic Molecules
Although larger molecules may look complicated, they can be easily understood by breaking them down and looking at their smaller components.
All atoms want to have their valence shell full, a "closed shell." Hydrogen wants to have 2 e- whereas carbon, oxygen, and nitrogen want to have 8 e-. When looking at the different representations of molecules, keep in mind the Octet Rule. Also remember that hydrogen can bond one time, oxygen can bond up to two times, nitrogen can bond up to three times, and carbon can bond up to four times.
Kekulé (a.k.a. Lewis Structures)
Kekulé structures are similar to Lewis Structures, but instead of covalent bonds being represented by electron dots, the two shared electrons are shown by a line.
(A) (B)(C)
Lone pairs remain as two electron dots, or are sometimes left out even though they are still there. Notice how the three lone pairs of electrons were not draw in around chlorine in example B.
Condensed Formulas
A condensed formula is made up of the elemental symbols. The order of the atoms suggests the connectivity. Condensed formulas can be read from either direction and H3C is the same as CH3, although the latter is more common because Look at the examples below and match them with their identical molecule under Kekulé structures and bond-line formulas.
(A) CH3CH2OH (B) ClCH2CH2CH(OCH3)CH3 (C) H3CNHCH2COOH
Let's look closely at example B. As you go through a condensed formula, you want to focus on the carbons and other elements that aren't hydrogen. The hydrogen's are important, but are usually there to complete octets. Also, notice the -OCH3 is in written in parentheses which tell you that it not part of the main chain of carbons. As you read through a a condensed formula, if you reach an atom that doesn't have a complete octet by the time you reach the next hydrogen, then it's possible that there are double or triple bonds. In example C, the carbon is double bonded to oxygen and single bonded to another oxygen. Notice how COOH means C(=O)-O-H instead of CH3-C-O-O-H because carbon does not have a complete octet and oxygens.
Bond-Line (a.k.a. zig-zag) Formulas
The name gives away how this formula works. This formula is full of bonds and lines, and because of the typical (more stable) bonds that atoms tend to make in molecules, they often end up looking like zig-zag lines. If you work with a molecular model kit you will find it difficult to make stick straight molecules (unless they contain sp triple bonds) whereas zig-zag molecules and bonds are much more feasible.
(A) (B) (C)
These molecules correspond to the exact same molecules depicted for Kekulé structures and condensed formulas. Notice how the carbons are no longer drawn in and are replaced by the ends and bends of a lines. In addition, the hydrogens have been omitted, but could be easily drawn in (see practice problems). Although we do not usually draw in the H's that are bonded to carbon, we do draw them in if they are connected to other atoms besides carbon (example is the OH group above in example A) . This is done because it is not always clear if the non-carbon atom is surrounded by lone pairs or hydrogens. Also in example A, notice how the OH is drawn with a bond to the second carbon, but it does not mean that there is a third carbon at the end of that bond/ line.
Dashed-Wedged Line Structure
As you may have guessed, the Dashed-Wedged Line structure is all about lines, dashes, and wedges. At first it may seem confusing, but with practice, understanding dash-wedged line structures will become like second nature. The following are examples of each, and how they can be used together.
Above are 4-carbon chains with attached OH groups or Cl and Br atoms. Remember that each line represents a bond and that the carbons and hydrogens have been omitted. When you look at or draw these structures, the straight lines illustrate atoms and bonds that are in the same plane, the plane of the paper (in this case, computer screen). Dashed lines show atoms and bonds that go into the page, behind the plane, away from you. In the above example, the OH group is going into the plane, while at the same time a hydrogen comes out (wedged).
Blue bead= OH group; White bead=H
Wedged lines illustrate bonds and atoms that come out of the page, in front of the plane, toward you. In the 2D diagram above, the OH group is coming out of the plane of the paper, while a hydrogen goes in (dashed).
Blue bead= OH group; White bead=H
As stated before, straight lines illustrate atoms and bonds that are in the same plane as the paper, but in the 2D example, the straight line bond for OH means that it it unsure or irrelevant whether OH is going away or toward you. It is also assumed that hydrogen is also connected to the same carbon that OH is on.
Blue bead= OH group; H is not shown
Try using your model kit to see that the OH group cannot lie in the same plane at the carbon chain (don't forget your hydrogens!). In the final 2Dexample, both dashed and wedged lines are used because the attached atoms are not hydrogens (although dashed and wedged lines can be used for hydrogens).The chlorine is coming out the page while bromine is going into the page.
Blue bead=Cl; Red bead=Br
Example: Converting between Structural Formulas
Throughout the course, it will be helpful to convert compounds into different structural formulas (Kekule (Lewis Structures), Bond-line, and Condensed) depending on the type of question that is asked. Standardized exams frequently include a high percentage of condensed formulas because it is easier and cheaper to type letters and numbers than to import figures. Initially, it can be tricky writing a bond-line structure directly from a condensed formula. First write the Kekule structure from the condensed formula and then draw the bond-line structure from the Kekule. Practice will quickly allow you to convert directly between condensed and bond-line structures.
The condensed formula for propanal is CH3CH2CHO. Can you visualize the bond-line structure of propanal? If yes, excellent, If no, the following practice will help.
The Kekule structure for propanal is shown below.
The bond-line structure for propanal is shown below.
All three structures represent the same compound, propanal.
Exercises
1. How many carbons are in the following drawing? How many hydrogens?
2. How many carbons are in the following drawing? How many hydrogens?
3. How many carbons are in the following drawing? How many hydrogens?
4. Look at the following molecule of vitamin A and draw in the hidden hydrogens and electron pairs.
(hint: Do all of the carbons have 4 bonds? Do all the oxygens have a full octet?)
5. How many bonds can hydrogen make?
6. How many bonds can chlorine make?
7. Dashed lines means the atomic bond goes ___________(away/toward) you.
8. Draw ClCH2CH2CH(OCH3)CH3 in Kekuléand zig-zag form.
9. Extra practice problems can be found ______?
Solutions
1. Remember the octet rule and how many times carbons and hydrogens are able to bond to other atoms.
2. Electron pairs drawn in blue and hydrogens draw in red.
3. Hygrogen can make one bond.
4. Chlorine can make one bond.
5. Away
6. See (B) under Kekulé and Bond-line (zig-zag) formulas.
7. Extra practice problems can be found: in your textbook, homework, lecture notes, online, reference books, and more. Try making up some of your own molecules, they may exist!
Contributors and Attributions
• Choo, Ezen (2009, UCD '11)
The building block of structural organic chemistry is the tetravalent carbon atom. With few exceptions, carbon compounds can be formulated with four covalent bonds to each carbon, regardless of whether the combination is with carbon or some other element. The two-electron bond, which is illustrated by the carbon-hydrogen bonds in methane or ethane and the carbon-carbon bond in ethane, is called a single bond. In these and many related substances, each carbon is attached to four other atoms:
There exist, however, compounds such as ethene (ethylene), $C_2H_4$, in which two electrons from each of the carbon atoms are mutually shared, thereby producing two two-electron bonds, an arrangement which is called a double bond. Each carbon in ethene is attached to only three other atoms:
Similarly, in ethyne (acetylene), $C_2H_2$, three electrons from each carbon atom are mutually shared, producing three two-electron bonds, called a triple bond, in which each carbon is attached to only two other atoms:
Of course, in all cases each carbon has a full octet of electrons. Carbon also forms double and triple bonds with several other elements that can exhibit a covalence of two or three. The carbon-oxygen (or carbonyl) double bond appears in carbon dioxide and many important organic compounds such as methanal (formaldehyde) and ethanoic acid (acetic acid). Similarly, a carbon-nitrogen triple bond appears in methanenitrile (hydrogen cyanide) and ethanenitrile (acetonitrile).
By convention, a single straight line connecting the atomic symbols is used to represent a single (two-electron) bond, two such lines to represent a double (four-electron) bond, and three lines a triple (six-electron) bond. Representations of compounds by these symbols are called structural formulas; some examples are
A point worth noting is that structural formulas usually do not indicate the nonbonding electron pairs. This is perhaps unfortunate because they play as much a part in the chemistry of organic molecules as do the bonding electrons and their omission may lead the unwary reader to overlook them. However, when it is important to represent them, this can be done best with pairs of dots, although a few authors use lines:
To save space and time in the representation of organic structures, it is common practice to use "condensed formulas" in which the bonds are not shown explicitly. In using condensed formulas, normal atomic valences are understood throughout. Examples of condensed formulas are
Another type of abbreviation that often is used, particularly for ring compounds, dispenses with the symbols for carbon and hydrogen atoms and leaves only the lines in a structural formula. For instance, cyclopentane, $C_5H_{10}$, often is represented as a regular pentagon in which it is understood that each apex represents a carbon atom with the requisite number of hydrogens to satisfy the tetravalence of carbon:
Likewise, cyclopropane, $C_3H_6$; cyclobutane, $C_4H_8$; and cyclohexane, $C_6H_{12}$, are drawn as regular polygons:
Although this type of line drawing is employed most commonly for cyclic structures, its use for open chain (acyclic) structures is becoming increasingly widespread. There is no special merit to this abbreviation for simple structures such as butane, $C_4H_{10}$; 1-butene, $C_4H_8$; or 1,3-butadiene, $C_4H_6$, but it is of value in representing more complex molecules such as $\beta$-carotene, $C_{40}H_{56}$:
Line structures also can be modified to represent the three-dimensional shapes of molecules, and the way that this is done will be discussed in detail in Chapter 5. At the onset of you study of organic chemistry, you should write out the formulas rather completely until you are thoroughly familiar with what these abbreviations stand for.
Contributors and Attributions
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_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/01%3A_Introduction_and_Review/1.08%3A_Structural_Formulas_-_Lewis_Kekule_Bond-line_Condensed_and_Perspecti.txt |
Learning Objective
• Identify polar bonds and compounds
Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. The Pauling scale is the most commonly used. Fluorine (the most electronegative element) is assigned a value of 4.0, and values range down to cesium and francium which are the least electronegative at 0.7.
Patterns of electronegativity in the Periodic Table
Electronegativity is defined as the ability of an atom in a particular molecule to attract electrons to itself. The greater the value, the greater the attractiveness for electrons.
The positively charged protons in the nucleus attract the negatively charged electrons. As the number of protons in the nucleus increases, the electronegativity or attraction will increase. Therefore electronegativity increases from left to right in a row in the periodic table. This effect only holds true for a row in the periodic table because the attraction between charges falls off rapidly with distance. The chart shows electronegativities from sodium to chlorine (ignoring argon since it does not does not form bonds).
As you go down a group, electronegativity decreases. (If it increases up to fluorine, it must decrease as you go down.) The chart shows the patterns of electronegativity in Groups 1 and 7.
Explaining the patterns in electronegativity
The attraction that a bonding pair of electrons feels for a particular nucleus depends on:
• the number of protons in the nucleus;
• the distance from the nucleus;
• the amount of screening by inner electrons.
Why does electronegativity increase across a period?
Consider sodium at the beginning of period 3 and chlorine at the end (ignoring the noble gas, argon). Think of sodium chloride as if it were covalently bonded.
Both sodium and chlorine have their bonding electrons in the 3-level. The electron pair is screened from both nuclei by the 1s, 2s and 2p electrons, but the chlorine nucleus has 6 more protons in it. It is no wonder the electron pair gets dragged so far towards the chlorine that ions are formed. Electronegativity increases across a period because the number of charges on the nucleus increases. That attracts the bonding pair of electrons more strongly.
Why does electronegativity fall as you go down a group?
As you go down a group, electronegativity decreases because the bonding pair of electrons is increasingly distant from the attraction of the nucleus. Consider the hydrogen fluoride and hydrogen chloride molecules:
The bonding pair is shielded from the fluorine's nucleus only by the 1s2 electrons. In the chlorine case it is shielded by all the 1s22s22p6 electrons. In each case there is a net pull from the center of the fluorine or chlorine of +7. But fluorine has the bonding pair in the 2-level rather than the 3-level as it is in chlorine. If it is closer to the nucleus, the attraction is greater.
Dipole moments occur when there is a separation of charge. They can occur between two ions in an ionic bond or between atoms in a covalent bond; dipole moments arise from differences in electronegativity. The larger the difference in electronegativity, the larger the dipole moment. The distance between the charge separation is also a deciding factor into the size of the dipole moment. The dipole moment is a measure of the polarity of the molecule.
Bond Polarity & Dipole Moment
Atoms with differences in electronegativity will share electrons unequally. The shared electrons of the covalent bond are held more tightly at the more electronegative element creating a partial negative charge, while the less electronegative element has a partial positive charge, . The larger the difference in electronegativity between the two atoms, the more polar the bond. To be considered a polar bond, the difference in electronegativity must >0.4 on the Pauling scale. Since the two electrical partial charges have opposite sign and equal magnitude and are separated by a distance, a dipole is established. Dipole moment is measured in debye units, which is equal to the distance between the charges multiplied by the charge (1 debye equals 3.34 x 10-30 coulomb-meters).
Polarity and Structure of Molecules
The shape of a molecule AND the polarity of its bonds. A molecule that contains polar bonds, might not have any overall polarity, depending upon its shape. The simple definition of whether a complex molecule is polar or not depends upon whether its overall centers of positive and negative charges overlap. If these centers lie at the same point in space, then the molecule has no overall polarity (and is non polar).
If a molecule is completely symmetric, then the dipole moment vectors on each molecule will cancel each other out, making the molecule nonpolar. A molecule can only be polar if the structure of that molecule is not symmetric.
A good example of a nonpolar molecule that contains polar bonds is carbon dioxide. This is a linear molecule and the C=O bonds are, in fact, polar. The central carbon will have a net positive charge, and the two outer oxygens a net negative charge. However, since the molecule is linear, these two bond dipoles cancel each other out (i.e. vector addition of the dipoles equals zero). And the overall molecule has no dipole (μ=0.
Although a polar bond is a prerequisite for a molecule to have a dipole, not all molecules with polar bonds exhibit dipoles
Geometric Considerations
Example 1: Polar Bonds vs. Polar Molecules
In a simple diatomic molecule like HCl, if the bond is polar, then the whole molecule is polar. What about more complicated molecules?
Consider CCl4, (left panel in figure above), which as a molecule is not polar - in the sense that it doesn't have an end (or a side) which is slightly negative and one which is slightly positive. The whole of the outside of the molecule is somewhat negative, but there is no overall separation of charge from top to bottom, or from left to right.
In contrast, CHCl3 is a polar molecule (right panel in figure above). The hydrogen at the top of the molecule is less electronegative than carbon and so is slightly positive. This means that the molecule now has a slightly positive "top" and a slightly negative "bottom", and so is overall a polar molecule.
A polar molecule will need to be "lop-sided" in some way.
Example 2: C2Cl4
Although the C–Cl bonds are rather polar, the individual bond dipoles cancel one another in this symmetrical structure, and Cl2C=CCl2 does not have a net dipole moment.
Example 3: CH3Cl
C-Cl, the key polar bond, is 178 pm. Measurement reveals 1.87 D. From this data, % ionic character can be computed. If this bond were 100% ionic (based on proton & electron),
μ=178100(4.80D)=8.54D
Example 4: HCl
Since measurement 1.87 D,
% ionic = (1.7/8.54)x100 = 22%HCl
u = 1.03 D (measured) H-Cl bond length 127 pm
If 100% ionic,
μ=127100(4.80D)=6.09D
ionic = (1.03/6.09)x100 = 17%
A "spectrum" of bonds
The implication of all this is that there is no clear-cut division between covalent and ionic bonds. In a pure covalent bond, the electrons are held on average exactly half way between the atoms. In a polar bond, the electrons have been dragged slightly towards one end. How far does this dragging have to go before the bond counts as ionic? There is no real answer to that. Sodium chloride is typically considered an ionic solid, but even here the sodium has not completely lost control of its electron. Because of the properties of sodium chloride, however, we tend to count it as if it were purely ionic. Lithium iodide, on the other hand, would be described as being "ionic with some covalent character". In this case, the pair of electrons has not moved entirely over to the iodine end of the bond. Lithium iodide, for example, dissolves in organic solvents like ethanol - not something which ionic substances normally do.
Summary
• No electronegativity difference between two atoms leads to a pure non-polar covalent bond.
• A small electronegativity difference leads to a polar covalent bond.
• A large electronegativity difference leads to an ionic bond.
Example 1: Polar Bonds vs. Polar Molecules
In a simple diatomic molecule like HCl, if the bond is polar, then the whole molecule is polar. What about more complicated molecules?
Consider CCl4, (left panel in figure above), which as a molecule is not polar - in the sense that it doesn't have an end (or a side) which is slightly negative and one which is slightly positive. The whole of the outside of the molecule is somewhat negative, but there is no overall separation of charge from top to bottom, or from left to right.
In contrast, CHCl3 is a polar molecule (right panel in figure above). The hydrogen at the top of the molecule is less electronegative than carbon and so is slightly positive. This means that the molecule now has a slightly positive "top" and a slightly negative "bottom", and so is overall a polar molecule.
A polar molecule will need to be "lop-sided" in some way.
Exercises
For the following compounds,
a) add lone pairs of electrons to complete octets
b) add dipole moment arrows or partial +/- signs to indicate polar bonds
c) predict the molecular polarity (Remember to visualize each compound in three dimensions.)
Solutions
Contributors and Attributions
• Jim Clark (Chemguide.co.uk)
• Prof. Richard Bank, Boise State University, Emeritus, | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/01%3A_Introduction_and_Review/1.09%3A_Electronegativity_and_Bond_Polarity_%28Review%29.txt |
Learning Objective
• Draw resonance forms and predict the relative contribution of each resonance form to the overall structure of the compound or ion
Recognizing resonance
Resonance contributors involve the ‘imaginary movement’ of pi-bonded electrons or of lone-pair electrons that are adjacent to (i.e. conjugated to) pi bonds. You can never shift the location of electrons in sigma bonds – if you show a sigma bond forming or breaking, you are showing a chemical reaction taking place. Likewise, the positions of atoms in the molecule cannot change between two resonance contributors.
Because benzene will appear throughout this course, it is important to recognize the stability gained through the resonance delocalization of the six pi electrons throughout the six carbon atoms. Benzene also illustrates one way to recognize resonance - when it is possible to draw two or more equivalent Lewis strucutres. If we were to draw the structure of an aromatic molecule such as 1,2-dimethylbenzene, there are two ways that we could draw the double bonds:
Which way is correct? There are two simple answers to this question: 'both' and 'neither one'. Both ways of drawing the molecule are equally acceptable approximations of the bonding picture for the molecule, but neither one, by itself, is an accurate picture of the delocalized pi bonds. The two alternative drawings, however, when considered together, give a much more accurate picture than either one on its own. This is because they imply, together, that the carbon-carbon bonds are not double bonds, not single bonds, but about halfway in between.
When it is possible to draw more than one valid structure for a compound or ion, we have identified resonance contributors: two or more different Lewis structures depicting the same molecule or ion that, when considered together, do a better job of approximating delocalized pi-bonding than any single structure. By convention, resonance contributors are linked by a double-headed arrow, and are sometimes enclosed by brackets:
In order to make it easier to visualize the difference between two resonance contributors, small, curved arrows are often used. Each of these arrows depicts the ‘movement’ of two pi electrons. A few chapters from now when we begin to study organic reactions - a process in which electron density shifts and covalent bonds between atoms break and form - this ‘curved arrow notation’ will become extremely important in depicting electron movement. In the drawing of resonance contributors, however, this electron ‘movement’ occurs only in our minds, as we try to visualize delocalized pi bonds. Nevertheless, use of the curved arrow notation is an essential skill that you will need to develop in drawing resonance contributors.
The depiction of benzene using the two resonance contributors A and B in the figure above does not imply that the molecule at one moment looks like structure A, then at the next moment shifts to look like structure B. Rather, at all moments, the molecule is a combination, or resonance hybrid of both A and B.
Caution! It is very important to be clear that in drawing two (or more) resonance contributors, we are not drawing two different molecules: they are simply different depictions of the exact same molecule. Furthermore, the double-headed resonance arrow does NOT mean that a chemical reaction has taken place.
Usually, derivatives of benzene (and phenyl groups, when the benzene ring is incorporated into a larger organic structure) are depicted with only one resonance contributor, and it is assumed that the reader understands that resonance hybridization is implied. This is the convention that will be used for the most part in this book. In other books or articles, you may sometimes see benzene or a phenyl group drawn with a circle inside the hexagon, either solid or dashed, as a way of drawing a resonance hybrid.
Curved arrows communicate electron flow (movement)
Curved arrows indicate electron flow. The base of the curved arrow is placed at the source of the electrons that are moving. The head of the arrow is placed at the destination of the electrons. A single barbed arrow represents one electron and a double barb represents two electrons. Electrons move from regions of relative high density to regions of low density or toward electronegative elements. It is important to use accuracy and precision when drawing curved arrows.
It is also important to consciously use the correct type of arrow. There are four primary types of arrows used by chemists to communicate one of the following: completion reaction, equilibrium reaction, electron movement, resonance forms. The three other types of arrows are shown below to build discernment between them.
Resonance Delocalizes Charge to Increase Stability
Resonance is most useful wen it delocalizes charge to stabilize reactive intermediates and products. Recognizing, drawing, and evaluating the relative stability of resonance contributors is essential to understanding organic reaction mechanisms.
Guidelines for drawing and working with resonance contributors
Learning to draw and interpret resonance structures, there are a few basic guidelines to help avoid drawing nonsensical structures. All of these guidellines make perfect sense as long as we remember that resonance contributors are merely a human-invented convention for depicting the delocalization of pi electrons in conjugated systems. When we see two different resonance contributors, we are not seeing a chemical reaction! We are seeing the exact same molecule or ion depicted in two different ways. All resonance contributors must be drawn as proper Lewis structures, with correct formal charges. Never show curved 'electron movement' arrows that would lead to a situation where a second-row element (ie. carbon, nitrogen, or oxygen) has more than eight electrons: this would break the 'octet rule'. Sometimes, however, we will draw resonance contributors in which a carbon atom has only six electrons (ie. a carbocation). In general, all oxygen and nitrogen atoms should have a complete octet of valence electrons.
1. There is ONLY ONE STRUCTURE for each compound or ion. This structure takes its character from the sum of all the contributors, not all resonance structures contribute equally to the sum.
2. Atoms must maintain their same position.
3. Only e- move !
4. All resonance contributors for a molecule or ion must have the same net charge.
5. Recognize which electrons can participate in resonance
a) unshared e- pairs or radicals
b) pi bond electrons
6. Recognize electron receptors
a) atoms with a positive (+) charge
b) electronegative atoms that can tolerate a negative charge
c) atoms which possess delocalizable electrons - see #4 above
7. Common electron flow patterns
a) move pi e- toward positive (+) charge or other pi bonds
b) move non-bonding e- pairs toward pi bonds
c) move single non-bonding e- toward pi bonds
Evaluating Resonance Contributors
1. Identical structures are equally important.
2. Structures will a greater number of bonds are more important.
3. Structures with charge separation are less important.
4. Pay attention to electronegativities.
5. Neutral atoms need to have complete octets.
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 pi 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 sigma bonds, 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.
Exercise 2.13: There is a third resonance contributor for formate (which we will soon learn is considered a 'minor' contributor). Draw this resonance contributor.
Here's another example, this time with a carbocation. Recall from section 2.1 that carbocations are sp2-hybridized, with an empty 2p orbital oriented perpendicular to the plane formed by three sigma bonds. If a carbocation is adjacent to a double bond, then three 2p orbitals can overlap and share the two pi electrons - another kind of conjugated pi system in which the positive charge is shared over two carbons.
Exercise 2.14: Draw the resonance contributors that correspond to the curved, two-electron movement arrows in the resonance expressions below.
Exercise 2.15: In each resonance expression, draw curved two-electron movement arrows on the left-side contributor that shows how we get to the right-side contributor. Be sure to include formal charges.
Solutions to exercises
Guided Resonance Practice
Below are a few more examples of 'legal' resonance expressions. Confirm for yourself that the octet rule is not exceeded for any atoms, and that formal charges are correct.
Exercise 2.16: Each of the 'illegal' resonance expressions below contains one or more mistakes. Explain what is incorrect in each.
Solutions to exercises
Major vs minor resonance contributors
Different resonance contributors do not always make the same contribution to the overall structure of the hybrid - rather, in many cases one contributor comes closer to depicting the actual bonding picture than another. In the case of carboxylates, contributors A and B below are equivalent in terms of their relative contribution to the hybrid structure. However, there is also a third resonance contributor ‘C, in which the carbon bears a positive formal charge and both oxygens are single-bonded and bear negative charges.
Structure C makes a less important contribution to the overall bonding picture of the group relative to A and B. How do we know that structure C is the ‘minor’ contributor? There are four basic rules which you need to learn in order to evaluate the relative importance of different resonance contributors. We will number them 5-8 so that they may be added to in the 'rules for resonance' list earlier on this page.
Rules for determining major and minor resonance contributors:
1. 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.
2. 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.
3. In structure C, there are only three bonds, compared to four in A and B. In general, a resonance structure with a lower number of total bonds is relatively less important.
4. The resonance contributor in which a negative formal charge is located on a more electronegative atom, usually oxygen or nitrogen, is more stable than one in which the negative charge is located on a less electronegative atom such as carbon. An example is in the upper left expression in the next figure.
Below are some additional examples of major and minor resonance contributors:
Why do we worry about a resonance contributor if it is the minor one? We will see later that very often a minor contributor can still be extremely important to our understanding of how a molecule reacts.
Exercise 2.17:
1. Draw a minor resonance structure for acetone (IUPAC name 2-propanone). Explain why it is a minor contributor.
2. Are acetone and 2-propanol resonance contributors of each other? Explain.
Exercise 2.18: Draw four additional resonance contributors for the molecule below. Label each one as major or minor (the structure below is of a major contributor).
Exercise 2.19: Draw three resonance contributors of methyl acetate (IUPAC name methyl methanoate), and order them according to their relative importance to the bonding picture of the molecule. Explain your reasoning.
Solutions to exercises
Resonance and peptide bonds
What is the hybridization state of the nitrogen atom in an amide? At first glance, it would seem logical to say that it is sp3-hybridized, because, like the nitrogen in an amine, the Lewis structure shows three single bonds and a lone pair. The picture looks quite different, though, if we consider another resonance contributor in which the nitrogen has a double bond to the carbonyl carbon: in this case, we would have to say that applicable hybridization is sp2, and the bonding geometry trigonal planar rather than tetrahedral.
In fact, the latter picture is more accurate: the lone pair of electrons on an amide nitrogen are not localized in an sp3 orbital, rather, they are delocalized as part of a conjugated pi system, and the bonding geometry around the nitrogen is trigonal planar as expected for sp2 hybridization. This is a good illustration of an important point: conjugation and the corresponding delocalization of electron density is stabilizing, thus if conjugation can occur, it probably will.
One of the most important examples of amide groups in nature is the ‘peptide bond’ that links amino acids to form polypeptides and proteins.
Critical to the structure of proteins is the fact that, although it is conventionally drawn as a single bond, the C-N bond in a peptide linkage has a significant barrier to rotation, indicating that to some degree, C-N pi overlap is present - in other words, there is some double bond character, and the nitrogen is sp2 hybridized with trigonal planar geometry.
The barrier to rotation in peptide bonds is an integral part of protein structure, introducing more rigidity to the protein's backbone. If there were no barrier to rotation in a peptide bond, proteins would be much more 'floppy' and three dimensional folding would be very different.
Exercise 2.20: Draw two pictures showing the unhybridized 2p orbitals and the location of pi electrons in methyl amide. One picture should represent the major resonance contributor, the other the minor contributor. How many overlapping 2p orbitals are sharing how many pi-bonded electrons?
Exercise 2.21: Draw two pictures showing the unhybridized 2p orbitals and the location of pi electrons in the 'enolate' anion shown below. One picture should represent the major resonance contributor, the other the minor contributor. How many overlapping 2p orbitals are sharing how many pi-bonded electrons?
Exercise 2.22: Below is a minor resonance contributor of a species known as an 'enamine', which we will study more in chapter 12. Draw the major resonance contributor for the enamine, and explain why your contributor is the major one (refer to resonance rules #5-8 from this section).
Solutions to exercises
Solved example: Draw the major resonance contributor of the structure below. Include in your figure the appropriate curved arrows showing how you got from the given structure to your structure. Explain why your contributor is the major one. In what kind of orbitals are the two lone pairs on the oxygen?
Solution: In the structure above, the carbon with the positive formal charge does not have a complete octet of valence electrons. Using the curved arrow convention, a lone pair on the oxygen can be moved to the adjacent bond to the left, and the electrons in the double bond shifted over to the left (see the rules for drawing resonance contributors to convince yourself that these are 'legal' moves).
The resulting resonance contributor, in which the oxygen bears the formal charge, is the major one because all atoms have a complete octet, and there is one additional bond drawn (resonance rules #5 and #7 both apply). This system can be thought of as four parallel 2p orbitals (one each on C2, C3, and C4, plus one on oxygen) sharing four pi electrons. One lone pair on the oxygen is in an unhybridized 2p orbital and is part of the conjugated pi system, and the other is located in an sp2 orbital.
Also note that one additional contributor can be drawn, but it is also minor because it has a carbon with an incomplete octet:
Exercise 2.23:
a) Draw three additional resonance contributors for the carbocation below. Include in your figure the appropriate curved arrows showing how one contributor is converted to the next.
b) Fill in the blanks: the conjugated pi system in this carbocation is composed of ______ 2p orbitals sharing ________ delocalized pi electrons.
Exercise 2.24: Draw the major resonance contributor for each of the anions below.
c) Fill in the blanks: the conjugated pi system in part (a) is composed of ______ 2p orbitals containing ________ delocalized pi electrons.
Exercise 2.25: The figure below shows how the negative formal charge on the oxygen can be delocalized to the carbon indicated by an arrow. More resonance contributors can be drawn in which negative charge is delocalized to three other atoms on the molecule.
a) Circle these atoms.
b) Draw the two most important resonance contributors for the molecule.
Solutions to exercises
A word of advice
Becoming adept at drawing resonance contributors, using the curved arrow notation to show how one contributor can be converted to another, and understanding the concepts of conjugation and resonance delocalization are some of the most challenging but also most important jobs that you will have as a beginning student of organic chemistry. If you work hard now to gain a firm grasp of these ideas, you will have come a long way toward understanding much of what follows in your organic chemistry course. Conversely, if you fail to come to grips with these concepts now, a lot of what you see later in the course will seem like a bunch of mysterious and incomprehensible lines, dots, and arrows, and you will be in for a rough ride, to say the least. More so than many other topics in organic chemistry, understanding bonding, conjugation, and resonance is something that most students really need to work on 'in person' with an instructor or tutor, preferably using a molecular modeling kit. Keep working problems, keep asking questions, and keep at it until it all makes sense!
Kahn Academy video tutorials
Drawing resonance structures
More on resonance
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/01%3A_Introduction_and_Review/1.10%3A_Resonance.txt |
LEARNING objective
• recognize acids or bases
Introduction
In 1884, the Swedish chemist Svante Arrhenius proposed two specific classifications of compounds, termed acids and bases. When dissolved in an aqueous solution, certain ions were released into the solution. As defined by Arrhenius, acid-base reactions are characterized by acids, which dissociate in aqueous solution to form hydrogen ions (H+) and bases, which form hydroxide (OH) ions. Arrhenius received the lowest passing score for his doctoral thesis with these innovative ideas about acids and bases. Ten years later he was awarded the Nobel Prize for his insights.
Acids are defined as a compound or element that releases hydrogen (H+) ions into the solution (mainly water).
$\ce{NHO_3 (aq) + H_2O(l) \rightarrow H_3O^+ + NO_3^- (aq)}$
In this reaction nitric acid (HNO3) disassociates into hydrogen (H+) and nitrate (NO3-) ions when dissolved in water. Bases are defined as a compound or element that releases hydroxide (OH-) ions into the solution.
$\ce{LiOH(s) ->[\ce{H2O}] Li^{+}(aq) + OH^{-}(aq)}$
In this reaction lithium hydroxide ($\ce{LiOH}$) dissociates into lithium ($\ce{Li^{+}}$) and hydroxide ($\ce{OH^{-}}$) ions when dissolved in water.
One way to define a class of compounds is by describing the various characteristics its members have in common. In the case of the compounds known as acids, the common characteristics include a sour taste, the ability to change the color of the vegetable dye litmus to red, and the ability to dissolve certain metals and simultaneously produce hydrogen gas. For the compounds called bases, the common characteristics are a slippery texture, a bitter taste, and the ability to change the color of litmus to blue. Acids and bases also react with each other to form compounds generally known as salts.
Although we include their tastes among the common characteristics of acids and bases, we never advocate tasting an unknown chemical!
Chemists prefer, however, to have definitions for acids and bases in chemical terms. The Swedish chemist Svante Arrhenius developed the first chemical definitions of acids and bases in the late 1800s. Arrhenius defined an acid as a compound that increases the concentration of hydrogen ion (H+) in aqueous solution. Many acids are simple compounds that release a hydrogen cation into solution when they dissolve. Similarly, Arrhenius defined a base as a compound that increases the concentration of hydroxide ion (OH) in aqueous solution. Many bases are ionic compounds that have the hydroxide ion as their anion, which is released when the base dissolves in water.
Table $1$: Formulas and Names for Some Acids and Bases
Acids Bases
Formula Name Formula Name
HCl(aq) hydrochloric acid NaOH(aq) sodium hydroxide
HBr(aq) hydrobromic acid KOH(aq) potassium hydroxide
HI(aq) hydriodic acid Mg(OH)2(aq) magnesium hydroxide
H2S(aq) hydrosulfuric acid Ca(OH)2(aq) calcium hydroxide
HC2H3O2(aq) acetic acid NH3(aq) ammonia
HNO3(aq) nitric acid
HNO2(aq) nitrous acid
H2SO4(aq) sulfuric acid
H2SO3(aq) sulfurous acid
HClO3(aq) chloric acid
HClO4(aq) perchloric acid
HClO2(aq) chlorous acid
H3PO4(aq) phosphoric acid
H3PO3(aq) phosphorous acid
Many bases and their aqueous solutions are named using the normal rules of ionic compounds that were presented previously; that is, they are named as hydroxide compounds. For example, the base sodium hydroxide (NaOH) is both an ionic compound and an aqueous solution. However, aqueous solutions of acids have their own naming rules. The names of binary acids (compounds with hydrogen and one other element in their formula) are based on the root of the name of the other element preceded by the prefix hydro- and followed by the suffix -ic acid. Thus, an aqueous solution of HCl [designated “HCl(aq)”] is called hydrochloric acid, H2S(aq) is called hydrosulfuric acid, and so forth. Acids composed of more than two elements (typically hydrogen and oxygen and some other element) have names based on the name of the other element, followed by the suffix -ic acid or -ous acid, depending on the number of oxygen atoms in the acid’s formula. Other prefixes, like per- and hypo-, also appear in the names for some acids. Unfortunately, there is no strict rule for the number of oxygen atoms that are associated with the -ic acid suffix; the names of these acids are best memorized. Table $1$ lists some acids and bases and their names. Note that acids have hydrogen written first, as if it were the cation, while most bases have the negative hydroxide ion, if it appears in the formula, written last.
The name oxygen comes from the Latin meaning “acid producer” because its discoverer, Antoine Lavoisier, thought it was the essential element in acids. Lavoisier was wrong, but it is too late to change the name now.
Example $1$
Name each substance.
1. HF(aq)
2. Sr(OH)2(aq)
Solution
1. This acid has only two elements in its formula, so its name includes the hydro- prefix. The stem of the other element’s name, fluorine, is fluor, and we must also include the -ic acid ending. Its name is hydrofluoric acid.
2. This base is named as an ionic compound between the strontium ion and the hydroxide ion: strontium hydroxide.
Exercise $1$
Name each substance.
1. H2Se(aq)
2. Ba(OH)2(aq)
Notice that one base listed in Table $1$—ammonia—does not have hydroxide as part of its formula. How does this compound increase the amount of hydroxide ion in aqueous solution? Instead of dissociating into hydroxide ions, ammonia molecules react with water molecules by taking a hydrogen ion from the water molecule to produce an ammonium ion and a hydroxide ion:
$NH_{3(aq)} + H_2O_{(ℓ)} \rightarrow NH^+_{4(aq)} + OH^−_{(aq)} \label{Eq1}$
Because this reaction of ammonia with water causes an increase in the concentration of hydroxide ions in solution, ammonia satisfies the Arrhenius definition of a base. Many other nitrogen-containing compounds are bases because they too react with water to produce hydroxide ions in aqueous solution. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/01%3A_Introduction_and_Review/1.11%3A_Arrhenius_Acids_and_Bases_%28Review%29.txt |
Learning Objective
• Use the definition of Lewis Acids and Bases to recognize electron movement in reactions
Acids and bases are an important part of chemistry. One of the most applicable theories is the Lewis acid/base motif that extends the definition of an acid and base beyond H+ and OH- ions as described by Brønsted-Lowry acids and bases.
Introduction
The Brønsted acid-base theory has been used throughout the history of acid and base chemistry. However, this theory is very restrictive and focuses primarily on acids and bases acting as proton donors and acceptors. Sometimes conditions arise where the theory doesn't necessarily fit, such as in solids and gases. In 1923, G.N. Lewis from UC Berkeley proposed an alternate theory to describe acids and bases. His theory gave a generalized explanation of acids and bases based on structure and bonding. Through the use of the Lewis definition of acids and bases, chemists are now able to predict a wider variety of acid-base reactions. Lewis' theory used electrons instead of proton transfer and specifically stated that an acid is a species that accepts an electron pair while a base donates an electron pair.
The reaction of a Lewis acid and a Lewis base will produce a coordinate covalent bond, as shown in Figure $1$ above. A coordinate covalent bond is just a type of covalent bond in which one reactant gives it electron pair to another reactant. In this case the lewis base donates its electrons to the lewis acid. When they do react this way the resulting product is called an addition compound, or more commonly an adduct.
• Lewis Acid: a species that accepts an electron pair (i.e., an electrophile) and will have vacant orbitals
• Lewis Base: a species that donates an electron pair (i.e., a nucleophile) and will have lone-pair electrons
Lewis Acids
Lewis acids accept an electron pair. Lewis Acids are Electrophilic meaning that they are electron attracting. When bonding with a base the acid uses its lowest unoccupied molecular orbital or LUMO (Figure 2).
• Various species can act as Lewis acids. All cations are Lewis acids since they are able to accept electrons. (e.g., Cu2+, Fe2+, Fe3+)
• An atom, ion, or molecule with an incomplete octet of electrons can act as an Lewis acid (e.g., BF3, AlF3).
• Molecules where the central atom can have more than 8 valence shell electrons can be electron acceptors, and thus are classified as Lewis acids (e.g., SiBr4, SiF4).
• Molecules that have multiple bonds between two atoms of different electronegativities (e.g., CO2, SO2)
Lewis Bases
Lewis Bases donate an electron pair. Lewis Bases are Nucleophilic meaning that they “attack” a positive charge with their lone pair. They utilize the highest occupied molecular orbital or HOMO (Figure 2). An atom, ion, or molecule with a lone-pair of electrons can thus be a Lewis base. Each of the following anions can "give up" their electrons to an acid, e.g., $OH^-$, $CN^-$, $CH_3COO^-$, $:NH_3$, $H_2O:$, $CO:$. Lewis base's HOMO (highest occupied molecular orbital) interacts with the Lewis acid's LUMO (lowest unoccupied molecular orbital) to create bonded molecular orbitals. Both Lewis Acids and Bases contain HOMO and LUMOs but only the HOMO is considered for Bases and only the LUMO is considered for Acids (Figure $2$).
Complex Ion / Coordination Compounds
Complex ions are polyatomic ions, which are formed from a central metal ion that has other smaller ions joined around it. While Brønsted theory can't explain this reaction Lewis acid-base theory can help. A Lewis Base is often the ligand of a coordination compound with the metal acting as the Lewis Acid (see Oxidation States of Transition Metals).
$Al^{3+} + 6 H_2O \rightleftharpoons [Al(H_2O)_6]^{3+} \label{1}$
The aluminum ion is the metal and is a cation with an unfilled valence shell, and it is a Lewis Acid. Water has lone-pair electrons and is an anion, thus it is a Lewis Base.
The Lewis Acid accepts the electrons from the Lewis Base which donates the electrons. Another case where Lewis acid-base theory can explain the resulting compound is the reaction of ammonia with Zn2+.
$Zn^{2+} + 4NH_3 \rightarrow [Zn(NH_3)_4]^{4+} \label{2}$
Similarly, the Lewis Acid is the zinc Ion and the Lewis Base is NH3. Note how Brønsted Theory of Acids and Bases will not be able to explain how this reaction occurs because there are no $H^+$ or $OH^-$ ions involved. Thus, Lewis Acid and Base Theory allows us to explain the formation of other species and complex ions which do not ordinarily contain hydronium or hydroxide ions. One is able to expand the definition of an acid and a base via the Lewis Acid and Base Theory. The lack of $H^+$ or $OH^-$ ions in many complex ions can make it harder to identify which species is an acid and which is a base. Therefore, by defining a species that donates an electron pair and a species that accepts an electron pair, the definition of a acid and base is expanded.
Amphoterism
As of now you should know that acids and bases are distinguished as two separate things however some substances can be both an acid and a base. You may have noticed this with water, which can act as both an acid or a base. This ability of water to do this makes it an amphoteric molecule. Water can act as an acid by donating its proton to the base and thus becoming its conjugate acid, OH-. However, water can also act as a base by accepting a proton from an acid to become its conjugate base, H3O+.
• Water acting as an Acid:
$H_2O + NH_3 \rightarrow NH_4^+ + OH^- \label{3}$
• Water acting as a Base:
$H_2O + HCl \rightarrow Cl^- + H_3O^+ \label{4}$
You may have noticed that the degree to which a molecule acts depends on the medium in which the molecule has been placed in. Water does not act as an acid in an acid medium and does not act as a base in a basic medium. Thus, the medium which a molecule is placed in has an effect on the properties of that molecule. Other molecules can also act as either an acid or a base. For example,
$Al(OH)_3 + 3H^+ \rightarrow Al^{3+} + 3H_2O \label{5}$
• where Al(OH)3 is acting as a Lewis Base.
$Al(OH)_3 + OH^- \rightarrow Al(OH)_4^- \label{6}$
• where Al(OH)3 is acting as an Lewis Acid.
Note how the amphoteric properties of the Al(OH)3 depends on what type of environment that molecule has been placed in.
Lewis Bases & Acids as Nucleophiles & Electrophiles
The emphasis on electron flow in the Lewis Theory of acids and bases is an important foundation for learning and predicting reaction mechanisms. The electron rich Lewis base can be described as a nucleophile. Nucleophiles are attracted to and can react with compounds or ions that have full or partial positive charge (like the nucleus). The electron poor Lewis acids can be described as electrophiles. Electrophiles attract nucleophiles until orbital overlap occurs between them triggering a reaction. At this point in the course, we can indicate electron flow using curved arrows when both the reactant(s) and product(s) are given.
Example
Exercises
For the following reactions,
a) add curved arrows to indicate the electron flow
b) label each reactant as the Nu (nucleophile) or E+ (electrophile).
Solutions
• Very Detailed review of Lewis Acids and Bases, covering all topics of this type of chemistry
• Very Complex and Detailed "Lewis Acid and Base Interaction Matrix"
• Youtube Video about Lewis Acids/Bases
Contributors and Attributions
• Adam Abudra (UCD), Tajinder Badial (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/01%3A_Introduction_and_Review/1.12%3A_Lewis_Acids_and_Bases.txt |
Learning Objective
• Determine relative strengths of acids and bases from their pKa values
• Determine the form of an acid or base at a specified pH (given the pKa)
The Henderson-Hasselbach Equation - a Quantitative View
We will use the general reaction for a weak acid to write the Ka expression.
\[HA_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + A^-_{(aq)}\]
\[K_a = \dfrac{[H_3O^+][A^-]}{[HA]}\]
pKa = -log Ka
where each bracketed term represents the concentration of that substance in solution.
The stronger an acid, the greater the ionization, the lower the pKa, and the lower the pH the compound will produce in solution.
It is important to realize that pKa is not at all the same thing as pH: the former is an inherent property of a compound or functional group, while the latter is the measure of the hydronium ion concentration in a particular aqueous solution:
pH = -log [H3O+]
Additional reagents can be added to a reaction solution to change the pH of the reaction conditions beyond the effects of an individual compound.
Relative Acidity and pKa Values
An application of the Henderson-Hasselbach Equation is the ability to determine the relative acidity of compounds by comparing their pKa values. The stronger an acid, the greater the ionization, the lower the pKa, and the lower the pH the compound will produce in solution. Some selected pKa values for compounds in the study of organic chemistry are shown bellow. Since organic reactions can be performed in non-aqueous environments, the pH can exceed 14 and organic compounds can have pKa values above 16. It is a variation on that line from the Wizard of Oz, "We don't live in water anymore."
It is a very good idea to commit to memory the approximate pKa ranges of the compounds above. A word of caution: when using the pKa table, be absolutely sure that you are considering the correct conjugate acid/base pair. If you are asked to say something about the basicity of ammonia (NH3) compared to that of ethoxide ion (CH3CH2O-), for example, the relevant pKa values to consider are 9.2 (the pKa of ammonium ion) and 16 (the pKa of ethanol). From these numbers, you know that ethoxide is the stronger base. Do not make the mistake of using the pKa value of 38: this is the pKa of ammonia acting as an acid, and tells you how basic the NH2- ion is (very basic!)
* A note on the pKa of water: The pKa of water is 14. Biochemistry and organic chemistry texts often list the value as 15.7. These texts have incorrectly factored the molar value for the concentration of water into the equilibrium constant. The correct derivation of the equilibrium constant involves the activity of water, which has a value of 1.
Example
While this course begins with single functional groups, we will eventually work with interesting compounds containing multiple functional groups. Recognizing which hydrogens can be ionized as acidic protons and which hydrogens can NOT, is a useful skill. Notice in this example that we need to evaluate the potential acidity at four different locations on the molecule.
Aldehyde and aromatic protons are not at all acidic (pKavalues are above 40 – not on our table). The two protons on the carbon next to the carbonyl are slightly acidic, with pKa values around 19-20 according to the table. The most acidic proton is on the phenol group, so if the compound were to be subjected to a single molar equivalent of strong base, this is the proton that would be donated.
Acidic & Basic Environments - Everything is Relative in Reactivity
Because our goal is understanding dynamic chemical reactivity, we do NOT need to know the specific amount of the protonated and unprotonated forms of a compound. We simply need to know which form is predominate. When the pH of the environment is less than the pKa of the compound, the environment is considered acidic and the compound will exist predominately in its protonated form. When the pH of the environment is greater than the pKa of the compound, the environment is considered basic and the compound will exist predominately in its deprotonated form.
For example, the pKa of acetic acid is about 5. At a pH of 1, the environment is considered acidic and acetic acid exists predominately in its protonated form. At pH 8, the environment is considered basic, and acetic acid becomes deprotonated to form acetate (CH3CO2-). Conversely, the pKa of phenol is 10. At pH 8, the environment is considered acidic for phenol and it remains primarily protonated.
It is also important to remember that organic chemistry does NOT have to occur in water so pKa values can be as high as 50.
Exercise
1. Complete the table below to indicate whether each compound exists predominantly in its protonated (acidic environment) or deprotonated (basic environment) form.
compound (pKa) pH 1 environment pH 8 environment pH 13 environment
Answer
1.
compound (pKa) pH 1 environment pH 8 environment pH 13 environment | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/01%3A_Introduction_and_Review/1.13%3A_Distinguishing_between_pH_and_pKa.txt |
Learning Objective
• Predict relative strengths of acids and bases from their structure, bonding and resonance
Since compounds are neutral, it can be difficult to evaluate and compare their overall stability without going through the tedious process of performing bond energy calculations.
When acidic compounds donate hydrogen ions or accepts electrons, they become ionized. It is much easier to compare ions because we can evaluate the charge density. The lower the charge density, the more stable the ion. Conversely, the higher the charge density, the less stable the ion. Charge density is analogous to density of matter. We place charge in the numerator, instead of mass, and volume can still be found in the denominator.
$\text{charge density} =\dfrac{\text{charge}}{\text{volume}}$
The six strong acids (HCl, HBr, HI, HNO3, H2SO4, HClO4) fully ionize to form the highly stable anions (Cl-, Br-, I-, NO3-, SO42-, ClO4-) respectively.
For the remaining weak acids (HA), we can determine their relative acidity by comparing the relative electron densities of their conjugate bases (A-).
the lower the electron density, the more stable the conjugate base
Structural Effects on Electron Density - Four Considerations
There are four main considerations for evaluating electron density.
1. Identity of the element or atoms holding the charge
2. Can the charge be delocalized by resonance?
3. Are there any inductive effects?
4. Hybridization of orbital holding the charge
These considerations are listed in order of importance and are explained individually, but must be looked at collectively.
Identity of the Element
When comparing the identity of the elements, it depends on the positional relationship of the elements on the periodic table.
Within a Group (aka down a column) As we move down the periodic table, the electrons are occupying higher energy subshells creating a larger atomic size and volume. As the volume increases, the electron density decreases.
Figure $1$ shows spheres representing the atoms of the s and p blocks from the periodic table to scale, showing the two trends for the atomic radius.
This relationship of atomic size and electron density is illustrated when we compare the relative acidities of methanol, CH3OH, with methanethiol, CH3SH. The lower pKa value of 10.4 for methanethiol indicates that it is a stronger acid than methanol with a pKa value of 15.5. It is important to remember that neither compound is considered an acid. These relationships become useful when trying to deprotonate compounds to increase their chemical reactivity in non-aqueous reaction conditions.
Across a Period (aka across a row) As we move across a period of the main group elements, the valence electrons all occupy orbitals in the same shell. These electrons have comparable energy, so this factor does not help us discern differences relative stability. Differences in electronegativity are now the dominant factor. This trend is shown when comparing the pKa values of methane, ammonia, water, and hydrofluoric acid reflects the relative electonegativities of the C < N < O < F.
Compound pKa Reaction
methane 50 $CH_4 \rightleftharpoons CH_3^- + H^+$
ammonia 36 $NH_3 \rightleftharpoons NH_2^- + H^+$
water 14 $H_2O \rightleftharpoons OH^- + H^+$
hydrofluoric acid 3 $HF \rightleftharpoons F^- + H^+$
First, we will focus on individual atoms, and think about trends associated with the position of an element on the periodic table. We’ll use as our first models the simple organic compounds ethane, methylamine, and ethanol, but the concepts apply equally to more complex biomolecules, such as the side chains of alanine, lysine, and serine.
We can see a clear trend in acidity as we move from left to right along the second row of the periodic table from carbon to nitrogen to oxygen. The key to understanding this trend is to consider the hypothetical conjugate base in each case: the more stable (weaker) the conjugate base, the stronger the acid. Look at where the negative charge ends up in each conjugate base. In the ethyl anion, the negative charge is borne by carbon, while in the methylamine anion and ethoxide anion the charges are located on a nitrogen and an oxygen, respectively. Remember the periodic trend in electronegativity (section 2.3A): it also increases as we move from left to right along a row, meaning that oxygen is the most electronegative of the three, and carbon the least. The more electronegative an atom, the better it is able to bear a negative charge. Thus, the ethoxide anion is the most stab = 0.00le (lowest energy, least basic) of the three conjugate bases, and the ethyl anion is the least stable (highest energy, most basic).
We can use the same set of ideas to explain the difference in basicity between water and ammonia.
$pK_a = 0.00 \,\,\,\,\,\, H_3O^+ \rightleftharpoons H_2O + H^+$
$pK_a = 9.26 \,\,\,\,\,\, NH_4^+ \rightleftharpoons NH_3 + H^+$
By looking at the pKavalues for the appropriate conjugate acids, we know that ammonia is more basic than water. Oxygen, as the more electronegative element, holds more tightly to its lone pair than the nitrogen. The nitrogen lone pair, therefore, is more likely to break away and form a new bond to a proton - it is, in other words, more basic. Once again, a more reactive (stronger) conjugate base means a less reactive (weaker) conjugate acid.
When moving vertically within a given column of the periodic table, we again observe a clear periodic trend in acidity. This is best illustrated with the halides: basicity, like electronegativity, increases as we move up the column.
Conversely, acidity in the haloacids increases as we move down the column.
In order to make sense of this trend, we will once again consider the stability of the conjugate bases. Because fluorine is the most electronegative halogen element, we might expect fluoride to also be the least basic halogen ion. But in fact, it is the least stable, and the most basic! It turns out that when moving vertically in the periodic table, the size of the atom trumps its electronegativity with regard to basicity. The atomic radius of iodine is approximately twice that of fluorine, so in an iodine ion, the negative charge is spread out over a significantly larger volume:
This illustrates a fundamental concept in organic chemistry that is important enough to put in red:
Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ than when they are confined to one atom.
We will see this idea expressed again and again throughout our study of organic reactivity, in many different contexts. For now, the concept is applied only to the influence of atomic radius on anion stability. Because fluoride is the least stable (most basic) of the halide conjugate bases, HF is the least acidic of the haloacids, only slightly stronger than acetic acid. HI, with a pKa of about -9, is one the strongest acids known.
More importantly to the study of biological organic chemistry, this trend tells us that thiols are more acidic than alcohols. The pKa of the thiol group on the cysteine side chain, for example, is approximately 8.3, while the pKa for the hydroxyl on the serine side chain is on the order of 17.
To reiterate: acid strength increases as we move to the right along a row of the periodic table, and as we move down a column.
Example $1$:
Draw the structure of the conjugate base that would form if the compound below were to react with 1 molar equivalent of sodium hydroxide:
Solution
Is resonance possible to localize the charge?
In the previous section we focused our attention on periodic trends - the differences in acidity and basicity between groups where the exchangeable proton was bound to different elements. Now, it is time to think about how the structure of different organic groups contributes to their relative acidity or basicity, even when we are talking about the same element acting as the proton donor/acceptor. When evaluating conjugate bases for the presence of the resonance contributors, remember to look for movable electrons as described in section 1.10 of this chapter. Delocalizing electrons over two or more atoms lowers the electron density.
The first model pair we will consider is ethanol and acetic acid, but the conclusions we reach will be equally valid for all alcohol and carboxylic acid groups. Despite the fact that they are both oxygen acids, the pKa values of ethanol and acetic acid are very different. What makes a carboxylic acid so much more acidic than an alcohol? As before, 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, according to our ‘rules of resonance’ (section 2.2C). What this means, you may recall, 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.
Now is the time to think back to that statement from the previous section that was so important that it got printed in bold font in its own paragraph – in fact, it is so important that we’ll just say it again: "Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ than when they are confined to one atom." Now, we are seeing this concept in another context, where 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 (and remember, pKa is a log expression, so we are talking about a difference of over 1012 between the acidity constants for the two molecules). The acetate ion is that much more stable than the ethoxide ion, all due to the effects of resonance delocalization.
The resonance effect also nicely explains why a nitrogen atom is basic when it is in an amine, but not 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 p system, and is very ready to form a bond with any acidic proton that might be nearby.
Often it requires some careful thought to predict the most acidic proton on a molecule. Ascorbic acid, also known as Vitamin C, has a pKa of 4.1.
There are four hydroxyl groups on this molecule – which one is the most acidic? If we consider all four possible conjugate bases, we find that there is only one for which we can delocalized the negative charge over two oxygen atoms.
Example $1$:
Rank the compounds below from most acidic to least acidic, and explain your reasoning.
(CC-NC-SA; Timothy Soderberg via UMn Morris Digital Well)
Inductive Effects
The inductive effect is an experimentally observed effect of the transmission of charge through a chain of atoms in a molecule, resulting in a permanent dipole in a bond. Inductive effects decrease quickly with distance. The inductive effect can be electron donating which helps stabilize positive charge. Alkyl groups (hydrocarbons) are inductive electron donators. The inductive effect can also be electron withdrawing. Electronegativity indicates the strength of electron withdrawing induction. Halogens are inductive electron withdrawing groups.
The effects of induction on relative acidity can also be seen when comparing acetic acid with trifluoroacetic acid. The difference in acidity does not have to do with resonance delocalization because no additional resonance structures can be drawn for the fluorinated molecule. The fluorine atoms inductively pull some of the electron density away from the carboxylate ion to further delocalize the negative charge of the conjugate base.
Compare the pKa values of acetic acid and its mono-, di-, and tri-chlorinated derivatives:
(CC-NC-SA; Timothy Soderberg via UMn Morris Digital Well)
The presence of the chlorines clearly increases the acidity of the carboxylic acid group. A chlorine atom is more electronegative than a hydrogen, and thus is able to ‘induce’, or ‘pull’ electron density towards itself, away from the carboxylate group. In helping to further spread out the electron density of the conjugate base, which as we know has a stabilizing effect. In this context, the chlorine substituent is called an electron-withdrawing group. Notice that the pKa-lowering effect of each chlorine atom, while significant, is not as dramatic as the delocalizing resonance effect illustrated by the difference in pKa values between an alcohol and a carboxylic acid. In general, resonance effects are more powerful than inductive effects. The inductive electron-withdrawing effect of the chlorines takes place through covalent bonds, and its influence decreases markedly with distance – thus a chlorine two carbons away from a carboxylic acid group has a decreased effect compared to a chlorine just one carbon away.
Exercise
Rank the compounds below from most acidic to least acidic, and explain your reasoning.
Solution
Orbital Hybridization
The hybridization of an orbital affects its electronegativity. Within a shell, the s orbitals occupy the region closer to the nucleus than the p orbitals. Therefore, the spherical s orbitals are more electronegative than the lobed p orbitals. The relative electronegativity of hybridized orbitals is sp > sp2 > sp3. This trend indicates the sp hybridized orbitals are more stable with a -1 charge than sp3 hybridized orbitals. The table below shows how orbital hybridization compares with the identity of the atom when predicting relative acidity.
Guided Practice
Let's practice by comparing the relative acidity of phenol with acetic acid.
Which compound is the stronger acid - phenol or acetic acid?
To answer this question, we draw all the relevant resonance contributors for each conjugate base, phenoxide and acetate, respectively.
Phenoxide has four resonance contributors, but three of the contributors have a negative charge on a carbon atom while both resonance contributors for acetate have a negative charge on the more electronegative element oxygen. There are no inductive effects or orbital hybridization differences to consider in this example, so we would predict acetic acid to be the stronger acid. The acetate ion is more stable than the phenoxide ion, so we would expect acetic acid to be the stronger acid.
The pKa table below supports our prediction. Acetic acid has a pKa of 4.7 while phenol has a pKa of 9.9.
A word of caution: when using the pKa table, be absolutely sure that you are considering the correct conjugate acid/base pair. If you are asked to say something about the basicity of ammonia (NH3) compared to that of ethoxide ion (CH3CH2O-), for example, the relevant pKa values to consider are 9.2 (the pKa of ammonium ion) and 16 (the pKa of ethanol). From these numbers, you know that ethoxide is the stronger base. Do not make the mistake of using the pKa value of 38: this is the pKa of ammonia acting as an acid, and tells you how basic the NH2- ion is (very basic!)
Contributors and Attributions
• Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
• Layne A. Morsch (University of Illinois Springfield) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/01%3A_Introduction_and_Review/1.14%3A_Predicting_Relative_Acidity.txt |
Learning objective
• Determine the empirical and molecular formulas from combustion data
Molecular formulas tell you how many atoms of each element are in a compound, and empirical formulas tell you the simplest or most reduced ratio of elements in a compound. If a compound's molecular formula cannot be reduced any more, then the empirical formula is the same as the molecular formula. Combustion analysis can determine the empirical formula of a compound, but cannot determine the molecular formula (other techniques can though). Once known, the molecular formula can be calculated from the empirical formula.
Empirical Formulas
An empirical formula tells us the relative ratios of different atoms in a compound. The ratios hold true on the molar level as well. Thus, H2O is composed of two atoms of hydrogen and 1 atom of oxygen. Likewise, 1.0 mole of H2O is composed of 2.0 moles of hydrogen and 1.0 mole of oxygen. We can also work backwards from molar ratios since if we know the molar amounts of each element in a compound we can determine the empirical formula.
Example $1$: Mercury Chloride
Mercury forms a compound with chlorine that is 73.9% mercury and 26.1% chlorine by mass. What is the empirical formula?
Let's say we had a 100 gram sample of this compound. The sample would therefore contain 73.9 grams of mercury and 26.1 grams of chlorine. How many moles of each atom do the individual masses represent?
For Mercury:
$(73.9 \;g) \times \left(\dfrac{1\; mol}{200.59\; g}\right) = 0.368 \;moles$
For Chlorine:
$(26.1\; g) \times \left(\dfrac{1\; mol}{35.45\; g}\right) = 0.736\; mol$
What is the molar ratio between the two elements?
$\dfrac{0.736 \;mol \;Cl}{0.368\; mol\; Hg} = 2.0$
Thus, we have twice as many moles (i.e. atoms) of Cl as Hg. The empirical formula would thus be (remember to list cation first, anion last):
$HgCl_2$
Molecular Formula from Empirical Formula
The chemical formula for a compound obtained by composition analysis is always the empirical formula. We can obtain the chemical formula from the empirical formula if we know the molecular weight of the compound. The chemical formula will always be some integer multiple of the empirical formula (i.e. integer multiples of the subscripts of the empirical formula). The general flow for this approach is shown in Figure $1$ and demonstrated in Example $2$.
Example $2$: Ascorbic Acid
Vitamin C (ascorbic acid) contains 40.92 % C, 4.58 % H, and 54.50 % O, by mass. The experimentally determined molecular mass is 176 amu. What is the empirical and chemical formula for ascorbic acid?
Solution
Consider an arbitrary amount of 100 grams of ascorbic acid, so we would have:
• 40.92 grams C
• 4.58 grams H
• 54.50 grams O
This would give us how many moles of each element?
• Carbon
$(40.92\; \cancel{g\; C}) \times \left( \dfrac{1\; mol\; C}{12.011\; \cancel{g\; C}} \right) = 3.407\; mol \; C$
• Hydrogen
$(4.58\; \cancel{g\; H}) \times \left( \dfrac{1\; mol\; H}{1.008\; \cancel{g\; H}} \right) = 4.544\; mol \;H$
• Oxygen
$(54.50\; \cancel{g\; O}) \times \left( \dfrac{1\; mol\; O}{15.999\; \cancel{g\; O}} \right) = 3.406\; mol \; O$
Determine the simplest whole number ratio by dividing by the smallest molar amount (3.406 moles in this case - see oxygen):
• Carbon
$C= \dfrac{3.407\; mol}{3.406\; mol} \approx 1.0$
• Hydrogen
$C= \dfrac{4.5.44\; mol}{3.406\; mol} = 1.0$
• Oxygen
$C= \dfrac{3.406\; mol}{3.406\; mol} = 1.0$
The relative molar amounts of carbon and oxygen appear to be equal, but the relative molar amount of hydrogen is higher. Since we cannot have "fractional" atoms in a compound, we need to normalize the relative amount of hydrogen to be equal to an integer. 1.333 would appear to be 1 and 1/3, so if we multiply the relative amounts of each atom by '3', we should be able to get integer values for each atom.
C = (1.0)*3 = 3
H = (1.333)*3 = 4
O = (1.0)*3 = 3
or
C3H4O3
This is our empirical formula for ascorbic acid.
What about the chemical formula? We are told that the experimentally determined molecular mass is 176 amu. What is the molecular mass of our empirical formula?
(3*12.011) + (4*1.008) + (3*15.999) = 88.062 amu
The molecular mass from our empirical formula is significantly lower than the experimentally determined value. What is the ratio between the two values?
(176 amu/88.062 amu) = 2.0
Thus, it would appear that our empirical formula is essentially one half the mass of the actual molecular mass. If we multiplied our empirical formula by '2', then the molecular mass would be correct. Thus, the actual molecular formula is:
2* C3H4O3 = C6H8O6
Combustion Analysis
When a compound containing carbon and hydrogen is subject to combustion with oxygen in a special combustion apparatus all the carbon is converted to CO2 and the hydrogen to H2O (Figure $2$). The amount of carbon produced can be determined by measuring the amount of CO2 produced. This is trapped by the sodium hydroxide, and thus we can monitor the mass of CO2 produced by determining the increase in mass of the CO2 trap. Likewise, we can determine the amount of H produced by the amount of H2O trapped by the magnesium perchlorate.
One of the most common ways to determine the elemental composition of an unknown hydrocarbon is an analytical procedure called combustion analysis. A small, carefully weighed sample of an unknown compound that may contain carbon, hydrogen, nitrogen, and/or sulfur is burned in an oxygen atmosphere,Other elements, such as metals, can be determined by other methods. and the quantities of the resulting gaseous products (CO2, H2O, N2, and SO2, respectively) are determined by one of several possible methods. One procedure used in combustion analysis is outlined schematically in Figure $3$ and a typical combustion analysis is illustrated in Examples $3$ and $4$.
Example $3$: Combustion of Isopropyl Alcohol
What is the empirical formulate for isopropyl alcohol (which contains only C, H and O) if the combustion of a 0.255 grams isopropyl alcohol sample produces 0.561 grams of CO2 and 0.306 grams of H2O?
Solution
From this information quantitate the amount of C and H in the sample.
$(0.561\; \cancel{g\; CO_2}) \left( \dfrac{1 \;mol\; CO_2}{44.0\; \cancel{g\;CO_2}}\right)=0.0128\; mol \; CO_2$
Since one mole of CO2 is made up of one mole of C and two moles of O, if we have 0.0128 moles of CO2 in our sample, then we know we have 0.0128 moles of C in the sample. How many grams of C is this?
$(0.0128 \; \cancel{mol\; C}) \left( \dfrac{12.011\; g \; C}{1\; \cancel{mol\;C}}\right)=0.154\; g \; C$
How about the hydrogen?
$(0.306 \; \cancel{g\; H_2O}) \left( \dfrac{1\; mol \; H_2O}{18.0\; \cancel{g \;H_2O}}\right)=0.017\; mol \; H_2O$
Since one mole of H2O is made up of one mole of oxygen and two moles of hydrogen, if we have 0.017 moles of H2O, then we have 2*(0.017) = 0.034 moles of hydrogen. Since hydrogen is about 1 gram/mole, we must have 0.034 grams of hydrogen in our original sample.
When we add our carbon and hydrogen together we get:
0.154 grams (C) + 0.034 grams (H) = 0.188 grams
But we know we combusted 0.255 grams of isopropyl alcohol. The 'missing' mass must be from the oxygen atoms in the isopropyl alcohol:
0.255 grams - 0.188 grams = 0.067 grams oxygen
This much oxygen is how many moles?
$(0.067 \; \cancel{g\; O}) \left( \dfrac{1\; mol \; O}{15.994\; \cancel{g \;O}}\right)=0.0042\; mol \; O$
Overall therefore, we have:
• 0.0128 moles Carbon
• 0.0340 moles Hydrogen
• 0.0042 moles Oxygen
Divide by the smallest molar amount to normalize:
• C = 3.05 atoms
• H = 8.1 atoms
• O = 1 atom
Within experimental error, the most likely empirical formula for propanol would be $C_3H_8O$
Example $4$: Combustion of Naphalene
Naphthalene, the active ingredient in one variety of mothballs, is an organic compound that contains carbon and hydrogen only. Complete combustion of a 20.10 mg sample of naphthalene in oxygen yielded 69.00 mg of CO2 and 11.30 mg of H2O. Determine the empirical formula of naphthalene.
Given: mass of sample and mass of combustion products
Asked for: empirical formula
Strategy:
1. Use the masses and molar masses of the combustion products, CO2 and H2O, to calculate the masses of carbon and hydrogen present in the original sample of naphthalene.
2. Use those masses and the molar masses of the elements to calculate the empirical formula of naphthalene.
Solution:
A Upon combustion, 1 mol of $\ce{CO2}$ is produced for each mole of carbon atoms in the original sample. Similarly, 1 mol of H2O is produced for every 2 mol of hydrogen atoms present in the sample. The masses of carbon and hydrogen in the original sample can be calculated from these ratios, the masses of CO2 and H2O, and their molar masses. Because the units of molar mass are grams per mole, we must first convert the masses from milligrams to grams:
$mass \, of \, C = 69.00 \, mg \, CO_2 \times {1 \, g \over 1000 \, mg } \times {1 \, mol \, CO_2 \over 44.010 \, g \, CO_2} \times {1 \, mol C \over 1 \, mol \, CO_2 } \times {12.011 \,g \over 1 \, mol \, C}$
$= 1.883 \times 10^{-2} \, g \, C$
$mass \, of \, H = 11.30 \, mg \, H_2O \times {1 \, g \over 1000 \, mg } \times {1 \, mol \, H_2O \over 18.015 \, g \, H_2O} \times {2 \, mol H \over 1 \, mol \, H_2O } \times {1.0079 \,g \over 1 \, mol \, H}$
$= 1.264 \times 10^{-3} \, g \, H$
B To obtain the relative numbers of atoms of both elements present, we need to calculate the number of moles of each and divide by the number of moles of the element present in the smallest amount:
$moles \, C = 1.883 \times 10^{-2} \,g \, C \times {1 \, mol \, C \over 12.011 \, g \, C} = 1.568 \times 10^{-3} \, mol C$
$moles \, H = 1.264 \times 10^{-3} \,g \, H \times {1 \, mol \, H \over 1.0079 \, g \, H} = 1.254 \times 10^{-3} \, mol H$
Dividing each number by the number of moles of the element present in the smaller amount gives
$H: {1.254\times 10^{−3} \over 1.254 \times 10^{−3}} = 1.000 \, \, \, C: {1.568 \times 10^{−3} \over 1.254 \times 10^{−3}}= 1.250$
Thus naphthalene contains a 1.25:1 ratio of moles of carbon to moles of hydrogen: C1.25H1.0. Because the ratios of the elements in the empirical formula must be expressed as small whole numbers, multiply both subscripts by 4, which gives C5H4 as the empirical formula of naphthalene. In fact, the molecular formula of naphthalene is C10H8, which is consistent with our results.
Exercise 1 $4$
1. Xylene, an organic compound that is a major component of many gasoline blends, contains carbon and hydrogen only. Complete combustion of a 17.12 mg sample of xylene in oxygen yielded 56.77 mg of CO2 and 14.53 mg of H2O. Determine the empirical formula of xylene.
2. The empirical formula of benzene is CH (its molecular formula is C6H6). If 10.00 mg of benzene is subjected to combustion analysis, what mass of CO2 and H2O will be produced?
Answer a
The empirical formula is C4H5. (The molecular formula of xylene is actually C8H10.)
Answer b
33.81 mg of CO2; 6.92 mg of H2O
Exercise 2
Elemental analysis of an organic compound indicates its composition to be 37.82% carbon, 6.36% hydrogen, and 55.82% chlorine.
a. What is the empirical formula for this compound?
b. Mass spectral analysis indicates a molar mass of 129 g/mol. What is the molecular formula for this compound?
c. Draw all the possible bond-line structures with this molecular formula.
Solutions to Exercise 2
a. C2H5Cl with a molar mass of 64.5 g/mol
b. C4H10Cl2
c. There 8 possible structures with the molecular formula C4H10Cl2. It can help to start with the different carbon backbones and then systematically add any branches (substituents). | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/01%3A_Introduction_and_Review/1.15%3A_Molecular_Formulas_and_Empirical_Formulas_%28Review%29.txt |
Bond Formation: The Octet Rule
1-1 Identify the number of valence electrons for each of the following elements. Then, identify the maximum number of covalent bonds it can form with other atoms while keeping a neutral net charge.
a) Oxygen
b) Carbon
c) Chlorine
d) Sulfur
e) Hydrogen
f) Boron
1-2 Which of the following atoms can bond with Br - to satisfy the octet rule?
a) Mg+2
b) O-2
c) Cl-
d) K+
1-3 Draw the Lewis dot structure of the correct answer from the previous problem 1-2 (a) - (d).
1-4 Identify which of the following compounds could not form due to an unfilled octet.
a) NCl3
b) NaOH
c) PCl
d) CF4
Lewis Structures
1-5 Draw the Lewis structures for the following compounds.
a) H2O
b) O3
c) BH3
d) SOCl2
1-6 Name the element that corresponds to each electronic configuration and identify how many valence electrons it has.
a) 1s22s22p6
b) 1s22s22p63s2
c) 1s22s22p4
d) 1s22s22p63s23p64s23d104p5
1-7 Draw the Lewis structures for PF3 and PF5.
1-8 Draw the Lewis structure for furan.
1-9 Identify the correct Lewis structure for hydroperoxyl, HO2.
Electronegativity and Bond Polarity
1-10 For the indicated bond in each of the following compounds, identify which atom is more electronegative, if applicable.
1-11 For each of the compounds in the previous problem, add a dipole moment arrow.
Formal Charges
1-12 For the following compounds, draw the structural formula. Then calculate the formal charge on each atom other than hydrogen.
a) N(CH3)4+
b) HSO4-
c) CH3CC-
1-13 Identify the formal charge for the following compounds.
1-14 Identify the formal charges for the central carbon in each of the following compounds.
Ionic Structures
1-15 Identify the substituent ions that make up the following salts.
a) NaCl
b) MgBr2
c) KNO3
d) NaH2PO4
1-16 Identify the products of the following reactions.
1-17 Give the correct nomenclature or write the correct chemical formula for the following ionic compounds.
a) NaCN
b) calcium oxalate
c) Al(OH)3
d) tin (II) phosphate
e) potassium hypochlorite
Resonance
1-18 For the following structure, draw its resonance structure(s).
1-19 Which resonance form from the previous problem has the most stable carbocation? Explain your answer.
1-20 Draw the important resonance forms to show the delocalization of charges in the following compounds.
1-21 Explain how resonance contributes to the lower pKa of acetic acid CH3CO2H (pKa= 4.75) compared to the pKa of ethanol CH3CH2OH (pKa=15.9).
1-22 Draw the resonance structure(s) for fulminic acid (HCNO).
Structural, Molecular and Empirical Formulas
1-23 Identify the molecular and empirical formula for the following structures.
1-24 Draw all possible structural formulas for the following compounds.
a) C4H10
b) CHN
c) C4H9Cl
1-25 True or False: You can always calculate the exact molecular weight of a molecule from its empirical formula.
1-26 For the following molecular formulas, provide the empirical formula.
a) C4H4O2
b) C8H6N2
c) C9H21N3O3
Acids and Bases - Arrhenius, Bronsted-Lowry, and Lewis
1-27 Briefly explain the three different definitions of acids and bases.
1-28 Calculate the Ka of nitric acid (HNO3). pKa of nitric acid is -1.4.
1-29 Rank the following in order of decreasing acidity: NH4+ HF H3O+ H2O
1-30 Rank the following in order of decreasing basicity: HSO4- H2O CH3COO- NH2-
1-31 Identify which compound is the stronger base. Identify which compound is the stronger acid.
1.32 Identify which group is more likely to grab a H+.
1.17: Solutions to Additional Exercises
Bond Formation: The Octet Rule
1-1:
a) 6 v.e. / 2 covalent bonds
b) 4 v.e. / 4 covalent bonds
c) 7 v.e. / 1 covalent bond
d) 6 v.e. / 2 covalent bonds (can also expand the octet to make 6 covalent bonds)
e) 1 v.e. / 1 covalent bond
f) 3 v.e. / 3 covalent bonds
1-2:
(d) K +
1-3:
1-4:
(c) PCl
Lewis Structures
1-5:
1-6:
a) Neon, 8 valence electrons
b) Magnesium, 2 valence electrons
c) Oxygen, 6 valence electrons
d) Bromine, 7 valence electrons
1-7:
1-8:
1-9: C.
Electronegativity and Bond Polarity
1-10:
1-11:
a) no dipole moment arrow b/c non-polar
1-12:
1-13:
a) 0
b) -1
c) -2
1-14:
a) 0
b) -1
c) 0
d) +1
Ionic Structures
1-15:
a) Na+ and Cl-
b) Mg+2 and 2 Br-
c) K+ and NO3-
d) Na+ and H2PO4-
1-16:
1-17:
a) sodium cyanide
b) CaC2O4
c) aluminum hydroxide
d) Sn3(PO4)2
e) KClO
Resonance
1-18:
1-19:
The resonance structure that is most stable has the tertiary carbocation. This tertiary carbocation is stabilized by hyperconjugation as well as two possible directions for resonance (compared to one immediate resonance structure for the other two carbocations).
1-20:
1-21:
When comparing the deprotonated forms of acetic acid and ethanol, acetate and ethoxide respectively, you can observe that acetate delocalizes the negative charge over the entire carboxylate group. Ethoxide, however, can only hold the negative charge on the alkoxide, making it a better base, but worse as an acid.
1-22:
Structural, Molecular and Empirical Formulas
1-23:
1-24:
1-25:
False; empirical formulas are the simplest whole number ratios that are useful in calculating percent compositions of atoms in a molecule. However, as they do not give the absolute number of atoms in a molecule, they cannot be used to calculate the molecular weight of the molecule.
1-26:
a) C2H2O
b) C4H3N
c) C9H7NO
Acids and Bases - Arrhenius, Bronsted-Lowry, and Lewis
1-27:
Arrhenius: An Arrhenius acid is a species that will donate a H+ when dissolved in water. An Arrhenius base is a species that will break down to yield a OH- when dissolved in water.
Bronsted-Lowry: A Bronsted-Lowry acid is a species that will donate a H+ when dissolved in solution (not only in water). A Bronsted-Lowry base is a species that can accept a H+ in solution (not only in water).
Lewis: A Lewis acid is an electron pair acceptor. A Lewis base is an electron base donor.
1-28:
Ka = 2.4 x 101
1-29:
H3O+ > HF > NH4+ > H2O
1-30:
NH2- > H2O > CH3COO- > HSO4-
1-31:
Compound A is the stronger base. Compound B is the stronger acid.
1-32:
Group A will want to grab the H+ more than group B. Since C is less electronegative than N, it cannot stabilize the negative charge as well and will want to grab a H+ in order to get rid of the charge. (grab = react with) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/01%3A_Introduction_and_Review/1.16%3A_Additional_Exercises.txt |
learning objective
• recognize acids and bases
The Brønsted-Lowry Theory of Acids and Bases
In 1923, Danish chemist Johannes Brønsted and English chemist Thomas Lowry independently proposed new definitions for acids and bases, ones that focus on proton transfer. A Brønsted-Lowry acid is any species that can donate a proton (H+) to another molecule. A Brønsted-Lowry base is any species that can accept a proton from another molecule. In short, a Brønsted-Lowry acid is a proton donor (PD), while a Brønsted-Lowry base is a proton acceptor (PA).
A Brønsted-Lowry acid is a proton donor, while a Brønsted-Lowry base is a proton acceptor.
Let us use the reaction of ammonia in water to demonstrate the Brønsted-Lowry definitions of an acid and a base. Ammonia and water molecules are reactants, while the ammonium ion and the hydroxide ion are products:
$\ce{NH3(aq) + H2O (ℓ) <=> NH^{+}4(aq) + OH^{−}(aq) }\label{Eq1}$
What has happened in this reaction is that the original water molecule has donated a hydrogen ion to the original ammonia molecule, which in turn has accepted the hydrogen ion. We can illustrate this as follows:
Because the water molecule donates a hydrogen ion to the ammonia, it is the Brønsted-Lowry acid, while the ammonia molecule—which accepts the hydrogen ion—is the Brønsted-Lowry base. Thus, ammonia acts as a base in both the Arrhenius sense and the Brønsted-Lowry sense.
Is an Arrhenius acid like hydrochloric acid still an acid in the Brønsted-Lowry sense? Yes, but it requires us to understand what really happens when HCl is dissolved in water. Recall that the hydrogen atom is a single proton surrounded by a single electron. To make the hydrogen ion, we remove the electron, leaving a bare proton. Do we really have bare protons floating around in aqueous solution? No, we do not. What really happens is that the H+ ion attaches itself to H2O to make H3O+, which is called the hydronium ion. For most purposes, H+ and H3O+ represent the same species, but writing H3O+ instead of H+ shows that we understand that there are no bare protons floating around in solution. Rather, these protons are actually attached to solvent molecules.
The Hydronium IOn
A proton in aqueous solution may be surrounded by more than one water molecule, leading to formulas like $\ce{H5O2^{+}}$ or $\ce{H9O4^{+}}$ rather than $\ce{H3O^{+}}$. It is simpler, however, to use $\ce{H3O^{+}}$ to represent the hydronium ion.
With this in mind, how do we define HCl as an acid in the Brønsted-Lowry sense? Consider what happens when HCl is dissolved in H2O:
$\ce{HCl(g) + H_2O (ℓ) \rightarrow H_3O^{+}(aq) + Cl^{−}(aq) }\label{Eq2}$
We can depict this process using Lewis electron dot diagrams:
Now we see that a hydrogen ion is transferred from the HCl molecule to the H2O molecule to make chloride ions and hydronium ions. As the hydrogen ion donor, HCl acts as a Brønsted-Lowry acid; as a hydrogen ion acceptor, H2O is a Brønsted-Lowry base. So HCl is an acid not just in the Arrhenius sense but also in the Brønsted-Lowry sense. Moreover, by the Brønsted-Lowry definitions, H2O is a base in the formation of aqueous HCl. So the Brønsted-Lowry definitions of an acid and a base classify the dissolving of HCl in water as a reaction between an acid and a base—although the Arrhenius definition would not have labeled H2O a base in this circumstance.
• A Brønsted-Lowry acid is a proton (hydrogen ion) donor.
• A Brønsted-Lowry base is a proton (hydrogen ion) acceptor.
• All Arrhenius acids and bases are Brønsted-Lowry acids and bases as well. However, not all Brønsted-Lowry acids and bases are Arrhenius acids and bases.
Example $1$
Aniline (C6H5NH2) is slightly soluble in water. It has a nitrogen atom that can accept a hydrogen ion from a water molecule just like the nitrogen atom in ammonia does. Write the chemical equation for this reaction and identify the Brønsted-Lowry acid and base.
Solution
C6H5NH2 and H2O are the reactants. When C6H5NH2 accepts a proton from H2O, it gains an extra H and a positive charge and leaves an OH ion behind. The reaction is as follows:
$\ce{C6H5NH2(aq) + H2O(ℓ) <=> C6H5NH3^{+}(aq) + OH^{−}(aq)} \nonumber$
Because C6H5NH2 accepts a proton, it is the Brønsted-Lowry base. The H2O molecule, because it donates a proton, is the Brønsted-Lowry acid.
Exercise $1$
Identify the Brønsted-Lowry acid and the Brønsted-Lowry base in this chemical equation.
$\ce{H2PO4^{-} + H_2O <=> HPO4^{2-} + H3O^{+}}$
Answer:
Brønsted-Lowry acid: H2PO4-; Brønsted-Lowry base: H2O
Exercise $2$
Which of the following compounds is a Bronsted-Lowry base?
1. HCl
2. HPO42-
3. H3PO4
4. NH4+
5. CH3NH3+
Answer:
A Brønsted-Lowry Base is a proton acceptor, which means it will take in an H+. This eliminates $\ce{HCl}$, $\ce{H3PO4}$ , $\ce{NH4^{+}}$ and $\ce{CH_3NH_3^{+}}$ because they are Bronsted-Lowry acids. They all give away protons. In the case of $\ce{HPO4^{2-}}$, consider the following equation:
$\ce{HPO4^{2-} (aq) + H2O (l) \rightarrow PO4^{3-} (aq) + H3O^{+}(aq) } \nonumber$
Here, it is clear that HPO42- is the acid since it donates a proton to water to make H3O+ and PO43-. Now consider the following equation:
$\ce{ HPO4^{2-}(aq) + H2O(l) \rightarrow H2PO4^{-} + OH^{-}(aq)} \nonumber$
In this case, HPO42- is the base since it accepts a proton from water to form H2PO4- and OH-. Thus, HPO42- is an acid and base together, making it amphoteric.
Since HPO42- is the only compound from the options that can act as a base, the answer is (b) HPO42-.
Conjugate Acid-Base Pair
In reality, all acid-base reactions involve the transfer of protons between acids and bases. For example, consider the acid-base reaction that takes place when ammonia is dissolved in water. A water molecule (functioning as an acid) transfers a proton to an ammonia molecule (functioning as a base), yielding the conjugate base of water, $\ce{OH^-}$, and the conjugate acid of ammonia, $\ce{NH4+}$:
In the reaction of ammonia with water to give ammonium ions and hydroxide ions, ammonia acts as a base by accepting a proton from a water molecule, which in this case means that water is acting as an acid. In the reverse reaction, an ammonium ion acts as an acid by donating a proton to a hydroxide ion, and the hydroxide ion acts as a base. The conjugate acid–base pairs for this reaction are $NH_4^+/NH_3$ and $H_2O/OH^−$.
The strongest acids are at the bottom left, and the strongest bases are at the top right. The conjugate base of a strong acid is a very weak base, and, conversely, the conjugate acid of a strong base is a very weak acid.
Example $2$
Identify the conjugate acid-base pairs in this equilibrium.
$\ce{CH3CO2H + H2O <=> H3O^{+} + CH3CO2^{-}} \nonumber$
Solution
Similarly, in the reaction of acetic acid with water, acetic acid donates a proton to water, which acts as the base. In the reverse reaction, $H_3O^+$ is the acid that donates a proton to the acetate ion, which acts as the base.
Once again, we have two conjugate acid–base pairs:
• the parent acid and its conjugate base ($CH_3CO_2H/CH_3CO_2^−$) and
• the parent base and its conjugate acid ($H_3O^+/H_2O$).
Example $3$
Identify the conjugate acid-base pairs in this equilibrium.
$(CH_{3})_{3}N + H_{2}O\rightleftharpoons (CH_{3})_{3}NH^{+} + OH^{-} \nonumber$
Solution
One pair is H2O and OH, where H2O has one more H+ and is the conjugate acid, while OH has one less H+ and is the conjugate base.
The other pair consists of (CH3)3N and (CH3)3NH+, where (CH3)3NH+ is the conjugate acid (it has an additional proton) and (CH3)3N is the conjugate base.
Exercise $3$
Identify the conjugate acid-base pairs in this equilibrium.
$\ce{NH2^{-} + H2O\rightleftharpoons NH3 + OH^{-}} \nonumber$
Answer:
H2O (acid) and OH (base); NH2 (base) and NH3 (acid)
Contributors and Attributions
• Henry Agnew (UC Davis) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/01%3A_Introduction_and_Review/1.18%3A_Brnsted-Lowry_Acids_and_Bases_%28Review%29.txt |
Learning Objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• define the terms "sterics" and "electrostatics" - refer to section 2.1
• write and interpret molecular orbital (MO) diagrams - refer to section 2.2
• predict the hybridization and geometry of atoms in a molecule - refer to section 2.3
• draw accurate 3-D representations of molecules with approximate bond angles - refer to section 2.3
• recognize conjugated pi bond systems - refer to section 2.4
• recognize that benzene is aromatic - refer to section 2.4
• identify the orbitals occupied by lone pair electrons - refer to section 2.5
• distinguish between bonds that can rotate and those that cannot - refer to section 2.6
• recognize the relationships between constitutional (structural) isomers, conformational isomers, and geometric isomers - refer to section 2.7
• apply the homologous series to organic molecules with 1-10 carbons - refer to section 2.8
• classify hydrocarbons as saturated or unsaturated - refer to section 2.8
• classify hydrocarbons as alkanes, alkenes, alkynes, cycloalkanes, or aromatics (arenes) - refer to section 2.8
• recognize and classify the common functional groups of organic chemistry (alkanes, alkenes, alkynes, alkyl halides, alcohols, amines, ethers, aldehydes, ketones, carboxylic acids, esters, and amides - refer to section 2.9
• determine the dominant intermolecular forces (IMFs) of organic compounds - refer to section 2.10
• predict the relative boil points of organic compounds - refer to section 2.11
• predict whether a mixture of compounds will a form homogeneous or heterogeneous solution - refer to section 2.12
• distinguish between organic compounds that are H-bond donors versus H-bond acceptors - refer to section 2.13
• apply the terms sterics and electrostatics to organic compounds - refer to sections 2.1- 2.13
• 2.1: Pearls of Wisdom
A few "pearls of wisdom" about "sterics" and "electrostatics" to provide context when applying the concepts of general chemistry to organic compounds.
• 2.2: Molecular Orbital (MO) Theory (Review)
Molecular orbital (MO) theory describes the behavior of electrons in a molecule in terms of combinations of the atomic wavefunctions.
• 2.3: Hybridization and Molecular Shapes (Review)
Hybridization and the Valence Shell Electron Pair Repulsion Theory effectively predict the three-dimensional structure of organic molecules. Since carbon can only form four bonds, we can limit our study to the tetrahedral, trigonal planar, and linear electron geometries.
• 2.4: 2.4 Conjugated Pi Bond Systems
A conjugated system is a system of connected p-orbitals with delocalized electrons in compounds with alternating single and multiple bonds, which in general may lower the overall energy of the molecule and increase stability. Recognizing the conjugated systems is helpful in determining reaction pathways.
• 2.5: Lone Pair Electrons and Bonding Theories
The chemical reactivity of lone pair electrons can be determined from the identity of the orbital they occupy. This concept will be further refined when we study aromaticity.
• 2.6: Bond Rotation
Single bonds can rotate, while double and triple bonds are rigid.
• 2.7: Isomerism Introduction
Structural (constitutional) isomers have the same molecular formula but a different bonding arrangement among the atoms. Stereoisomers have identical molecular formulas and arrangements of atoms. They differ from each other only in the spatial orientation of groups in the molecule.
• 2.8: Hydrocarbons and the Homologous Series
Since we will be spending the rest of the course working with compounds with a carbon backbone, there is no time like the present to learn the homologous series, the names for simple, straight hydrocarbon chains and branches.
• 2.9: Organic Functional Groups
Functional groups are to organic chemistry what ions are to general chemistry. We simply must be able to recognize and distinguish between functional grouops to learn organic chemistry.
• 2.10: Intermolecular Forces (IMFs) - Review
Intermolecular forces (IMFs) have many useful applications in organic chemistry. For students interested in biochemistry, the concepts of IMFs are called non-covalent interactions when they occur within a large biological molecule creating secondary and tertiary structure.
• 2.11: Intermolecular Forces and Relative Boiling Points (bp)
The relative strength of the intermolecular forces (IMFs) can be used to predict the relative boiling points of pure substances.
• 2.12: Intermolecular Forces and Solubilities
Organic chemistry can perform reactions in non-aqueous solutions using organic solvents. It is important to start considering the solvent as a reaction parameter and the solubility of each reagent.
• 2.13: Additional Practice Problems
• 2.14: Organic Functional Groups- H-bond donors and H-bond acceptors
When evaluating organic compounds, we want to visualize the compounds in their three-dimensional shapes exerting intermolecular forces on their environment. Because reactions will occur in aqueous and non-aqueous (organic) solutions, it is important to recognize which functional groups are both H-bond donors and H-bond acceptors and which groups are only H-bond acceptors.
• 2.15: Solutions to Additional Exercises
This section has the solutions to the exercises in the previous section.
• 2.16: Additional Exercises
This section has additional exercises for the key learning objectives of this chapter.
02: Structure and Properties of Organic Molecules
Learning Objective
• define the terms "sterics" and "electrostatics"
Introduction
Functional groups are the common bonding patterns found in organic compounds. Organic compounds are classified by their functional groups.
To talk about organic chemistry, we need to be able to
a) recognize and name the major organic functional groups (see chapter 3 for nomenclature)
b) apply bonding theories to the structure of functional groups
c) visual functional groups in three dimensions
d) determine the polarity & intermolecular forces of organic compounds
Ultimately, all of the information above will be integrated at the end of this chapter to predict solubilities and relative boiling points of organic compounds. In future chapters, these skills will help elucidate reaction mechanisms and pathways.
Sterics & Electrostatics - all roads lead to one or other
Sterics and electrostatics are primary considerations when learning the reactions of organic chemistry.
Sterics is the spatial arrangement (3-dimensional structure) of atoms in a molecule or ion.
Electrostatics is the the charge distribution within a molecule or ion.
Depending on the reaction mechanism, either sterics or electrostatics (charge stabilization) will play a dominant role in the rate determining step. For concerted (one-step) reactions, sterics will strongly influence the orientation of reactants in the transition state. For two-step reactions, there is typically a charged intermediate that requires stabilization for the reaction to proceed. The intermediate with the lowest charge distribution is the most stable and reacts preferentially.
Sterics can be predicted using bonding theories. Molecular orbital (MO) theory uses the combination of atomic orbitals to yield molecular orbitals that are delocalized over the entire molecule. In valence bond theory (VB) theory, atomic orbitals can be hybridized. VB theory assumes that all bonds are localized bonds formed between two atoms by the donation of an electron from each atom. As discussed in chapter 1, this assumption is invalid because some atoms can bond using delocalized electrons through resonance. VB theory does a good job of qualitatively describing the shapes of covalent compounds which is important in determining the sterics of the reactions. While Molecular Orbital (MO) theory is good for understanding bonding in general and the electrostatics of a reactant, intermediate, or product.
Electrostatics are determined by applying the same concepts used to determine the relative acidity of compounds by evaluating the electron density of their conjugate bases. The less charge, the more stable an ion is. The stability of ions is determined by the identity of the element being ionized, charge delocalization via resonance, inductive effects, and orbital hybridization. Inductive effects can be electron withdrawing (aka electronegative) or electron donating, such alkyl group stabilization of carbocations. These parameters are listed in order of importance with the overall character of an ion must be evaluated to determine its relative stability. Refer to section 1.15 of this text for the full explanation. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.01%3A_Pearls_of_Wisdom.txt |
Learning Objective
• write and interpret molecular orbital (MO) diagrams
Overview
Molecular orbital (MO) theory describes the behavior of electrons in a molecule in terms of combinations of the atomic wavefunctions. The resulting molecular orbitals may extend over all the atoms in the molecule. Bonding molecular orbitals are formed by in-phase combinations of atomic wavefunctions, and electrons in these orbitals stabilize a molecule. Antibonding molecular orbitals result from out-of-phase combinations and electrons in these orbitals make a molecule less stable.
Molecular orbital theory describes the distribution of electrons in molecules in much the same way that the distribution of electrons in atoms is described using atomic orbitals. Using quantum mechanics, the behavior of an electron in a molecule is still described by a wave function, Ψ, analogous to the behavior in an atom. Just like electrons around isolated atoms, electrons around atoms in molecules are limited to discrete (quantized) energies. The region of space in which a valence electron in a molecule is likely to be found is called a molecular orbital (Ψ2). Like an atomic orbital, a molecular orbital is full when it contains two electrons with opposite spin.
We will consider the molecular orbitals in molecules composed of two identical atoms (H2 or Cl2, for example). Such molecules are called homonuclear diatomic molecules. In these diatomic molecules, several types of molecular orbitals occur.
The mathematical process of combining atomic orbitals to generate molecular orbitals is called the linear combination of atomic orbitals (LCAO). The wave function describes the wavelike properties of an electron. Molecular orbitals are combinations of atomic orbital wave functions. Combining waves can lead to constructive interference, in which peaks line up with peaks, or destructive interference, in which peaks line up with troughs (Figure $2$). In orbitals, the waves are three dimensional, and they combine with in-phase waves producing regions with a higher probability of electron density and out-of-phase waves producing nodes, or regions of no electron density.
There are two types of molecular orbitals that can form from the overlap of two atomic s orbitals on adjacent atoms. The two types are illustrated in Figure 8.4.3. The in-phase combination produces a lower energy σs molecular orbital (read as "sigma-s") in which most of the electron density is directly between the nuclei. The out-of-phase addition (which can also be thought of as subtracting the wave functions) produces a higher energy $σ^∗_s$ molecular orbital (read as "sigma-s-star") molecular orbital in which there is a node between the nuclei. The asterisk signifies that the orbital is an antibonding orbital. Electrons in a σs orbital are attracted by both nuclei at the same time and are more stable (of lower energy) than they would be in the isolated atoms. Adding electrons to these orbitals creates a force that holds the two nuclei together, so we call these orbitals bonding orbitals. Electrons in the $σ^∗_s$ orbitals are located well away from the region between the two nuclei. The attractive force between the nuclei and these electrons pulls the two nuclei apart. Hence, these orbitals are called antibonding orbitals. Electrons fill the lower-energy bonding orbital before the higher-energy antibonding orbital, just as they fill lower-energy atomic orbitals before they fill higher-energy atomic orbitals.
In p orbitals, the wave function gives rise to two lobes with opposite phases, analogous to how a two-dimensional wave has both parts above and below the average. We indicate the phases by shading the orbital lobes different colors. When orbital lobes of the same phase overlap, constructive wave interference increases the electron density. When regions of opposite phase overlap, the destructive wave interference decreases electron density and creates nodes. When p orbitals overlap end to end, they create σ and σ* orbitals (Figure $4$). If two atoms are located along the x-axis in a Cartesian coordinate system, the two px orbitals overlap end to end and form σpx (bonding) and $σ^∗_{px}$ (antibonding) (read as "sigma-p-x" and "sigma-p-x star," respectively). Just as with s-orbital overlap, the asterisk indicates the orbital with a node between the nuclei, which is a higher-energy, antibonding orbital.
The side-by-side overlap of two p orbitals gives rise to a pi (π) bonding molecular orbital and a π* antibonding molecular orbital, as shown in Figure $5$. In valence bond theory, we describe π bonds as containing a nodal plane containing the internuclear axis and perpendicular to the lobes of the p orbitals, with electron density on either side of the node. In molecular orbital theory, we describe the π orbital by this same shape, and a π bond exists when this orbital contains electrons. Electrons in this orbital interact with both nuclei and help hold the two atoms together, making it a bonding orbital. For the out-of-phase combination, there are two nodal planes created, one along the internuclear axis and a perpendicular one between the nuclei.
In the molecular orbitals of diatomic molecules, each atom also has two sets of p orbitals oriented side by side (py and pz), so these four atomic orbitals combine pairwise to create two π orbitals and two π* orbitals. The πpy and $π^∗_{py}$ orbitals are oriented at right angles to the πpz and $π^∗_{pz}$ orbitals. Except for their orientation, the πpy and πpz orbitals are identical and have the same energy; they are degenerate orbitals. The $π^∗_{py}$ and $π^∗_{pz}$ antibonding orbitals are also degenerate and identical except for their orientation. A total of six molecular orbitals results from the combination of the six atomic p orbitals in two atoms: σpx and $σ^∗_{px}$, πpy and $π^∗_{py}$, πpz and $π^∗_{pz}$.
Example $1$
Molecular Orbitals Predict what type (if any) of molecular orbital would result from adding the wave functions so each pair of orbitals shown overlap. The orbitals are all similar in energy.
Solution
1. This is an in-phase combination, resulting in a σ3p orbital
2. This will not result in a new orbital because the in-phase component (bottom) and out-of-phase component (top) cancel out. Only orbitals with the correct alignment can combine.
3. This is an out-of-phase combination, resulting in a $π^∗_{3p}$ orbital.
Exercise $1$
Label the molecular orbital shown as σ or π, bonding or antibonding and indicate where the node occurs.
Answer
The orbital is located along the internuclear axis, so it is a σ orbital. There is a node bisecting the internuclear axis, so it is an antibonding orbital.
Molecular Orbital Energy Diagrams
The relative energy levels of atomic and molecular orbitals are typically shown in a molecular orbital diagram (Figure $7$). For a diatomic molecule, the atomic orbitals of one atom are shown on the left, and those of the other atom are shown on the right. Each horizontal line represents one orbital that can hold two electrons. The molecular orbitals formed by the combination of the atomic orbitals are shown in the center. Dashed lines show which of the atomic orbitals combine to form the molecular orbitals. For each pair of atomic orbitals that combine, one lower-energy (bonding) molecular orbital and one higher-energy (antibonding) orbital result. Thus we can see that combining the six 2p atomic orbitals results in three bonding orbitals (one σ and two π) and three antibonding orbitals (one σ* and two π*).
We predict the distribution of electrons in these molecular orbitals by filling the orbitals in the same way that we fill atomic orbitals, by the Aufbau principle. Lower-energy orbitals fill first, electrons spread out among degenerate orbitals before pairing, and each orbital can hold a maximum of two electrons with opposite spins (Figure $7$). Just as we write electron configurations for atoms, we can write the molecular electronic configuration by listing the orbitals with superscripts indicating the number of electrons present. For clarity, we place parentheses around molecular orbitals with the same energy. In this case, each orbital is at a different energy, so parentheses separate each orbital. Thus we would expect a diatomic molecule or ion containing seven electrons (such as $\ce{Be2+}$) would have the molecular electron configuration $(σ_{1s})^2(σ^∗_{1s})^2(σ_{2s})^2(σ^∗_{2s})^1$. It is common to omit the core electrons from molecular orbital diagrams and configurations and include only the valence electrons.
Bond Order
The filled molecular orbital diagram shows the number of electrons in both bonding and antibonding molecular orbitals. The net contribution of the electrons to the bond strength of a molecule is identified by determining the bond order that results from the filling of the molecular orbitals by electrons.
When using Lewis structures to describe the distribution of electrons in molecules, we define bond order as the number of bonding pairs of electrons between two atoms. Thus a single bond has a bond order of 1, a double bond has a bond order of 2, and a triple bond has a bond order of 3. We define bond order differently when we use the molecular orbital description of the distribution of electrons, but the resulting bond order is usually the same. The MO technique is more accurate and can handle cases when the Lewis structure method fails, but both methods describe the same phenomenon.
In the molecular orbital model, an electron contributes to a bonding interaction if it occupies a bonding orbital and it contributes to an antibonding interaction if it occupies an antibonding orbital. The bond order is calculated by subtracting the destabilizing (antibonding) electrons from the stabilizing (bonding) electrons. Since a bond consists of two electrons, we divide by two to get the bond order. We can determine bond order with the following equation:
$\textrm{bond order}=\dfrac{(\textrm{number of bonding electrons})−(\textrm{number of antibonding electrons})}{2}$
The order of a covalent bond is a guide to its strength; a bond between two given atoms becomes stronger as the bond order increases. If the distribution of electrons in the molecular orbitals between two atoms is such that the resulting bond would have a bond order of zero, a stable bond does not form. We next look at some specific examples of MO diagrams and bond orders.
Bonding in Diatomic Molecules
A dihydrogen molecule (H2) forms from two hydrogen atoms. When the atomic orbitals of the two atoms combine, the electrons occupy the molecular orbital of lowest energy, the σ1s bonding orbital. A dihydrogen molecule, H2, readily forms because the energy of a H2 molecule is lower than that of two H atoms. The σ1s orbital that contains both electrons is lower in energy than either of the two 1s atomic orbitals.
A molecular orbital can hold two electrons, so both electrons in the H2 molecule are in the σ1s bonding orbital; the electron configuration is $(σ_{1s})^2$. We represent this configuration by a molecular orbital energy diagram (Figure $8$) in which a single upward arrow indicates one electron in an orbital, and two (upward and downward) arrows indicate two electrons of opposite spin.
A dihydrogen molecule contains two bonding electrons and no antibonding electrons so we have
$\ce{bond\: order\: in\: H2}=\dfrac{(2−0)}{2}=1$
Because the bond order for the H–H bond is equal to 1, the bond is a single bond.
A helium atom has two electrons, both of which are in its 1s orbital. Two helium atoms do not combine to form a dihelium molecule, He2, with four electrons, because the stabilizing effect of the two electrons in the lower-energy bonding orbital would be offset by the destabilizing effect of the two electrons in the higher-energy antibonding molecular orbital. We would write the hypothetical electron configuration of He2 as $(σ_{1s})^2(σ^∗_{1s})^2$ as in Figure $9$ . The net energy change would be zero, so there is no driving force for helium atoms to form the diatomic molecule. In fact, helium exists as discrete atoms rather than as diatomic molecules. The bond order in a hypothetical dihelium molecule would be zero.
$\ce{bond\: order\: in\: He2}=\dfrac{(2−2)}{2}=0$
A bond order of zero indicates that no bond is formed between two atoms.
C2 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.02%3A_Molecular_Orbital_%28MO%29_Theory_%28Review%29.txt |
Learning Objectives
• predict the hybridization and geometry of atoms in a molecule - refer to section 2.3
• draw accurate 3-D representations of molecules with approximate bond angles
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.)
Exercise
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.
Exercise
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.
Exercise
Circle the six atoms in the molecule below that are ‘locked’ into the same plane.
Exercise
What kinds of orbitals are overlapping in bonds a-d indicated below?
Exercise
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.
Exercise
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_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.03%3A_Hybridization_and_Molecular_Shapes_%28Review%2.txt |
Learning Objective
• recognize conjugated pi bond systems
• recognize that benzene is aromatic
Introduction
It is important to train our eye to recognize structural features that have stabilizing effects. Alternating single and double bonds create a conjugated pi bond system across multiple atoms that lowers the energy and stabilizes the molecule or ion. When we look at carbon-carbon double bonds (C=C), we need to look and see if they are isolated or conjugated.
To understand the source of this stabilization we will use molecular orbital (MO) theory. Valence bond theory does a remarkably good job at explaining the bonding geometry of many of the functional groups in organic compounds, however, it fails to adequately account for the stability contained in alternating double and single bonds. In order to understand these properties, we will use the ideas of MO theory.
Let’s go back and consider again the simplest possible covalent bond: the one in molecular hydrogen (H2). When we described the hydrogen molecule using valence bond theory, we said that the two 1s orbitals from each atom overlap, allowing the two electrons to be shared and thus forming a covalent bond. In molecular orbital theory, we make a further statement: we say that the two atomic 1s orbitals mathematically combine to form two new orbitals. Recall that an atomic orbital (such as the 1s orbital of a hydrogen atom) describes a region of space around a single atom inside which electrons are likely to be found. A molecular orbital describes a region of space around two or more atoms inside which electrons are likely to be found.
Mathematical principles tell us that when orbitals combine, the number of orbitals before the combination takes place must equal the number of new orbitals that result from the combination – orbitals don’t just disappear! We saw this previously when we discussed hybrid orbitals: one s and three p orbitals make four sp3 hybrids. When two atomic 1s orbitals combine in the formation of H2, the result is two sigma (σ) orbitals.
Molecular orbitals for H2
According to MO theory, one sigma orbital is lower in energy than either of the two isolated atomic 1s orbitals –this lower sigma orbital is referred to as a bonding molecular orbital. The second, 'sigma star' orbital is higher in energy than the two atomic 1s orbitals, and is referred to as an antibonding molecular orbital.
The bonding sigma orbital, which holds both electrons in the ground state of the molecule, is egg-shaped, encompassing the two nuclei, and with the highest likelihood of electrons being in the area between the two nuclei. The high-energy, antibonding sigma* orbital can be visualized as a pair of droplets, with areas of higher electron density near each nucleus and a ‘node’, (area of zero electron density) midway between the two nuclei.
Remember that we are thinking here about electron behavior as wave behavior. When two separate waves combine, they can do so with constructive interference, where the two amplitudes build up and reinforce one another, or destructive interference, where the two amplitudes cancel one another out. Bonding MOs are the consequence of constructive interference between two atomic orbitals, which results in an attractive interaction and an increase in electron density between the nuclei. Antibonding MO’s are the consequence of destructive interference which results in a repulsive interaction and a region of zero electron density between the nuclei (in other words, a node).
Following the same aufbau ('building up') principle you learned in General Chemistry for writing out electron configurations, we place the two electrons in the H2 molecule in the lowest energy molecular orbital, which is the (bonding) sigma orbital. The bonding (attracting) MO is full, and the antibonding (repulsing) MO is empty.
MO theory and conjugated pi bonds
The advantage of using MO theory to understand bonding in organic molecules becomes more apparent when we think about pi bonds. Let’s first consider the pi bond in ethene from an MO theory standpoint (in this example we will be disregarding the s bonds in the molecule, and thinking only about the π bond). We start with two atomic orbitals: one unhybridized 2p orbital from each carbon. Each contains a single electron. In MO theory, the two atomic combine mathematically to form two pi molecular orbitals, one a low-energy pi bonding orbital and one a high-energy pi* antibonding orbital.
Molecular orbitals for ethene (ethylene)
In the bonding pi orbital, the two shaded lobes of the p orbitals interact constructively with each other, as do the two unshaded lobes (remember, the arbitrary shading choice represents mathematical (+) and (-) signs for the mathematical wavefunction describing the orbital). There is increased electron density between the two carbon nuclei in the molecular orbital - it is a bonding interaction.
In the higher-energy antibonding pi* orbital, the shaded lobe of one p orbital interacts destructively with the unshaded lobe of the second p orbital, leading to a node between the two nuclei and overall repulsion between the carbon nuclei.
Again using the 'building up' principle, we place the two electrons in the lower-energy, bonding pi molecular orbital. The antibonding pi* orbital remains empty.
Next, we'll consider the 1,3-butadiene molecule. From valence orbital theory alone we might expect that the C2-C3 bond in this molecule, because it is a sigma bond, would be able to rotate freely.
Experimentally, however, it is observed that there is a significant barrier to rotation about the C2-C3 bond, and that the entire molecule is planar. In addition, the C2-C3 bond is 148 pm long, shorter than a typical carbon-carbon single bond (about 154 pm), though longer than a typical double bond (about 134 pm).
Molecular orbital theory accounts for these observations with the concept of delocalized pi bonds. In this picture, the four 2p atomic orbitals combine mathematically to form four pi molecular orbitals of increasing energy. Two of these - the bonding pi orbitals - are lower in energy than the p atomic orbitals from which they are formed, while two - the antibonding pi* orbitals - are higher in energy.
The lowest energy molecular orbital, pi1, has only constructive interaction and zero nodes. Higher in energy, but still lower than the isolated p orbitals, the pi2 orbital has one node but two constructive interactions - thus it is still a bonding orbital overall. Looking at the two antibonding orbitals, pi3* has two nodes and one constructive interaction, while pi4* has three nodes and zero constructive interactions.
By the aufbau principle, the four electrons from the isolated 2pz atomic orbitals are placed in the bonding pi1 and pi2 MO’s. Because pi1 includes constructive interaction between C2 and C3, there is a degree, in the 1,3-butadiene molecule, of pi-bonding interaction between these two carbons, which accounts for its shorter length and the barrier to rotation. The valence bond picture of 1,3-butadiene shows the two pi bonds as being isolated from one another, with each pair of pi electrons ‘stuck’ in its own pi bond. However, molecular orbital theory predicts (accurately) that the four pi electrons are to some extent delocalized, or ‘spread out’, over the whole pi system.
space-filling view
1,3-butadiene is the simplest example of a system of conjugated pi bonds. To be considered conjugated, two or more pi bonds must be separated by only one single bond – in other words, there cannot be an intervening sp3-hybridized carbon, because this would break up the overlapping system of parallel p orbitals. In the compound below, for example, the C1-C2 and C3-C4 double bonds are conjugated, while the C6-C7 double bond is isolated from the other two pi bonds by sp3-hybridized C5.
A very important concept to keep in mind is that there is an inherent thermodynamic stability associated with conjugation. This stability can be measured experimentally by comparing the heat of hydrogenation of two different dienes. (Hydrogenation is a reaction type that we will learn much more about in chapter 15: essentially, it is the process of adding a hydrogen molecule - two protons and two electrons - to a p bond). When the two conjugated double bonds of 1,3-pentadiene are 'hydrogenated' to produce pentane, about 225 kJ is released per mole of pentane formed. Compare that to the approximately 250 kJ/mol released when the two isolated double bonds in 1,4-pentadiene are hydrogenated, also forming pentane.
The conjugated diene is lower in energy: in other words, it is more stable. In general, conjugated pi bonds are more stable than isolated pi bonds.
Conjugated pi systems can involve oxygen and nitrogen atoms as well as carbon. In the metabolism of fat molecules, some of the key reactions involve alkenes that are conjugated to carbonyl groups.
In chapter 4, we will see that MO theory is very useful in explaining why organic molecules that contain extended systems of conjugated pi bonds often have distinctive colors. beta-carotene, the compound responsible for the orange color of carrots, has an extended system of 11 conjugated pi bonds.
Exercise: Identify all conjugated and isolated double bonds in the structures below. For each conjugated pi system, specify the number of overlapping p orbitals, and how many pi electrons are shared among them.
Exercise: Identify all isolated and conjugated pi bonds in lycopene, the red-colored compound in tomatoes. How many pi electrons are contained in the conjugated pi system?
Solutions to exercises
Aromaticity - The Ultimate Conjugated System
Molecular orbital theory is especially helpful in explaining the unique properties of aromatic compounds such as benzene:
Although benzene is most often drawn with three double bonds and three single bonds, in fact all of the carbon-carbon bonds iare exactly the same length (138 pm). In addition, the pi bonds in benzene are significantly less reactive than 'normal' pi bonds, either isolated or conjugated. Something about the structure of benzene makes its pi bonding arrangement especially stable. This ‘something’ has a name: it is called ‘aromaticity’.
What exactly is this ‘aromatic’ property that makes the pi bonds in benzene so stable? In 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 pi electrons to be delocalized in molecular orbitals that extend all the way around the ring, above and below the plane. For this to happen, of course, the ring must be planar – otherwise the p 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? The answer, in fact, is ‘no’. The eight-membered cyclooctatetraene ring shown below is not flat, and its π bonds react like 'normal' alkenes.
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 of the pi molecular orbitals in benzene.
Quantum mechanical calculations tell us that the six pi molecular orbitals in benzene, formed from six atomic p orbitals, occupy four separate energy levels. pi1 and pi6* have unique energy levels, while the pi2 - pi3 and pi4*- pi5* pairs are degenerate, meaning they are at the same energy level. When we use the aufbau principle to fill up these orbitals with the six pi 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 in the atomic orbitals of the stable noble gases 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 (pi1 and pi8*) have unique energy levels, while the other six form degenerate pairs. Notice that pi4 and pi5 are 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 pi 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.
For now, the important learning objective is to recognize conjugated pi bonds systems and understand the benzene is exceptionally stable exhibiting a property called aromaticity. Aromaticity and chemistry of aromatic compounds is relatively complex and is discussed in greater detail in subsequent chapters of this text.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.04%3A_2.4_Conjugated_Pi_Bond_Systems.txt |
Learning Objective
• identify the orbitals occupied by lone pair electrons
Valence Bond and Molecular Orbital Theories
The table below summarizes the main points of the two complementary bonding theories. Both theories provide different, useful ways of describing molecular structure. We will use both theories and often blend them to analyze and predict chemical structure and reactivity. The bonding theories are reviewed in greater detail in the next two sections.
Comparison of Bonding Theories
Valence Bond Theory Molecular Orbital Theory
considers bonds as localized between one pair of atoms considers electrons delocalized throughout the entire molecule
creates bonds from overlap of atomic orbitals (s, p, d…) and hybrid orbitals (sp, sp2, sp3…) combines atomic orbitals to form molecular orbitals (σ, σ*, π, π*)
forms σ or π bonds creates bonding and antibonding interactions based on which orbitals are filled
predicts molecular shape based on the number of regions of electron density predicts the arrangement of electrons in molecules
needs multiple structures to describe resonance
Orbitals of Lone Pair Electrons
There are situations in which we will want to integrate molecular orbital and valence bond theories. Identifying the orbitals of lone pair electrons is one situation. Hybridized orbitals create sigma bonds and hold lone pairs. The sigma bonds create the "framework" that holds all the atoms together as a molecule or ion. Un-hybridized p orbitals create pi bonds perpendicular to this sigma framework. In the future, we will learn that some lone pair electrons on heteroatoms of rings can occupy p orbitals to create aromaticity. Stay tuned for upcoming attractions. For the first ten chapters of this text, we will only focus on non-aromatic compounds.
To identify the orbitals of the lone pair electrons in non-aromatic compounds, we can follow a two-step approach.
Step 1: Add any missing lone pair electrons to the heteroatoms (atoms other than carbon and hydrogen).
Step 2: Determine the hybridization of any atoms with lone pairs (heteroatoms). Lone pairs occupy the hybridized orbitals.
Example
To identify the orbitals of the lone pair electrons in the compound below, we will follow the approach above.
Step 1: Add lone pairs.
Step 2: Determine the hybridization of any atom with lone pairs.
The lone pairs on each heteroatom occupy the indicated hybridized orbital.
NOTE: These guidelines only apply for non-aromatic compounds. There can be exceptions to these guidelines for some heterocyclic aromatic compounds. These exceptions are fully explained in a later chapter of this text.
Exercise
1. For the compound below:
a) add the lone pair electrons
b) label the hybridization and electron geometry for all non-hydrogen atoms
c) specify the hybridization of the orbital for each lone pair
d) What is the chemical formula of this compound?
Answer
1.
2.06: Bond Rotation
Learning Objective
• distinguish between bonds that can rotate and those that cannot
Sigma Bonds can Rotate
We learned in section 2.1 that single bonds in organic molecules are free to rotate, due to the 'end-to-end' (sigma) nature of their orbital overlap. Consider the carbon-oxygen bond in ethanol, for example: with a 180o rotation about this bond, the shape of the molecule would look quite different:
For ethane, rotation about the carbon-carbon sigma bond results in many different possible three-dimensional arrangements of the atoms.
Pi Bonds are rigid
Pi bonds are created from overlapping p orbitals. The lobes of the p orbitals prevent the atoms sharing pi bonds from rotating as shown in the diagram below.
Double and Triple Bonds cannot Rotate
The pi bonds in double and triple bonds prevent these bonds from rotating. This rigidity has an effect on the physical structure of compounds and can influence chemical reactivity. For now, we want to build the habit of looking at static drawings and diagrams of organic compounds and visualizing their dynamic nature.
For ethene, there is no rotation about the carbon-carbon double bond because of the pi bond.
Exercise
1. Label the selected bonds in the compound below as "Rotates" or "Rigid."
Answer
1. Arrows for (a) and (c) are pointing to single bonds that can rotate. Arrow (b) is pointing to a double bond that is rigid because of the pi bond. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.05%3A_Lone_Pair_Electrons_and_Bonding_Theories.txt |
Learning Objective
• recognize the relationships between constitutional (structural) isomers, conformational isomers, and geometric isomers
Isomers
Isomers always have the same chemical formula. When the chemical formulae are different, then the compounds are completely different. Important information can be gained from the chemical formulas when comparing compounds.
Structural (constitutional) isomers have the same molecular formula but a different bonding arrangement among the atoms. Stereoisomers have identical molecular formulas and arrangements of atoms. They differ from each other only in the spatial orientation of groups in the molecule. For organic chemistry, there are several types of stereoisomers: enantiomers, diasteriomers, geometric isomers, and conformers. These stereoisomers will be introduced and explained throughout several chapters.
Structural (Constitutional) Isomers
Because carbon forms four bonds, there can be multiple ways to form molecules that follow the octet rule. Even with only four carbon atoms, there are two possible structures for the carbon backbone. The carbon atoms can be bonded to make a four carbon chain (butane) or there can be a one carbon branch from a three carbon chain (2-methylpropane). Butane and 2-methylpropane are structural isomers because they both have the chemical formula C4H10.
Identical vs Conformer
The rotation about single bonds creates dynamic molecules. When drawing and discussing molecules, it is important to be aware that our drawing are static while the molecule themselves are rotating. Although there are seven sigma bonds in the ethane molecule, rotation about the six carbon-hydrogen bonds does not result in any change in the shape of the molecule because the hydrogen atoms are essentially spherical. Rotation about the carbon-carbon bond, however, results in many different possible molecular conformations. Conformers are the simplest example of stereoisomerism.
Identical compounds are the same compound shown with ALL atoms in the same spatial orientation.
Conformers are the same compound shown with different rotations about single bonds.
In the example below, we can compare two identical structures for ethane with two conformers of ethane.
Geometric Isomers - an example of stereoisomerism
The rigidity of the pi bonds in double bonds can create geometric isomerism. Without rotation, there are two different orientations possible across the carbon-carbon double bond (C=C). The rigidity of the double bond creates a line of reference for spatial orientation. The prefixes cis and trans are used to distinguish between geometric isomers. The cis-stereoisomer has both non-hydrogen atoms on the same side of the double bond. Whereas, the trans-stereoisomer has the non-hydrogen atoms across the double bond. In the same way, we cross the ocean an a trans-Atlantic journey. This small difference may seem insignificant, but geometric isomers are different chemical compounds with different physical properties as shown in the example below.
For now, it is important to distinguish between structural differences and spatial differences when comparing compounds. In the future, we will look more closely at isomerism.
Example
Let's look ate the bond-line structures below and determine the relationships between the following pairs of compounds: identical, conformers, structural isomers, geometric isomers, or completely different compounds.
The first important step (that is often skipped) is to determine the chemical formula of each compound. If the chemical formulas are different, then the compounds are completely different and there is NO isomeric relationship. If the chemical formulas are the same, then we identify the difference between to the compounds to determine their relatioship. If there are structural differences in the bonding patterns, then the compounds are constitutional (structural isomers). If the compounds have the same structural connections, but the spatial orientations are different, then the compounds are stereoisomers. For now, the possible stereoisomers are conformers showing the same compound with different carbon-carbon single bond rotations or geometric isomers of compounds with different orientations at the carbon-carbon double bonds.
Applying the logic above to our example, we determine the following.
Exercise
1. What is the relationship between the following pairs of compounds: identical, conformers, structural isomers, geometric isomers, or completely different compounds?
Answer
1.
a) geometric isomers
b) conformers
c) structural isomers
d) completely different compounds
e) identical | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.07%3A_Isomerism_Introduction.txt |
Learning Objective
• classify hydrocarbons as saturated or unsaturated
• classify hydrocarbons as alkanes, alkenes, alkynes, cycloalkanes, or aromatics (arenes)
• apply the homologous series to organic molecules with 1-10 carbons
Hydrocarbon Classifications
Hydrocarbons are organic compounds that contain only carbon and hydrogen. The inherent ability of hydrocarbons to bond to themselves is known as catenation, and allows hydrocarbon to form more complex molecules, such as cyclohexane and benzene. Catenation comes from the fact that the bond character between carbon atoms is entirely non-polar.
The four general classes of hydrocarbons are: alkanes, alkenes, alkynes and arenes. Aromatic compounds derive their names from the fact that many of these compounds in the early days of discovery were grouped because they were oils with fragrant odors. The classifications for hydrocarbons are summarized below.
Saturated hydrocarbons (alkanes) are the simplest of the hydrocarbon species. They are composed entirely of single bonds and are saturated with hydrogen. The general formula for saturated hydrocarbons is CnH2n+2 (assuming non-cyclic structures). Saturated hydrocarbons are the basis of petroleum fuels and are found as either linear or branched species.The simplest alkanes have their C atoms bonded in a straight chain; these are called normal alkanes. They are named according to the number of carbon atoms in the chain. The smallest alkane is methane:
Unsaturated hydrocarbons have double and/or triple bonds between carbon atoms. Those with double bond are called alkenes and have the general formula CnH2n (assuming non-cyclic structures). Those containing triple bonds are called alkynes and have general formula CnH2n-2. The smallest alkene—ethene—has two C atoms and is also known by its common name ethylene and the smallest alkyne is ethyne, also known as acetylene.
Cycloalkanes are hydrocarbons containing one or more carbon rings to which hydrogen atoms are attached. The prefix "cyclo" is added to the name to communicate the ring structure. The general formula for a saturated hydrocarbon containing one ring is CnH2n.
Aromatic hydrocarbons, also known as arenes, are hydrocarbons that have at least one aromatic ring. Aromatic compounds contain the benzene unit. Benzene itself is composed of six C atoms in a ring, with alternating single and double C–C bonds:
For most compounds, information beyond the chemical formula will be needed to elucidate their structure. However, the ratio of C:H in a chemical formula can provide insights into the chemical structure.
For example, let's look at some of the possible structures and chemical formulas for hydrocarbons containing six carbon atoms.
The saturated alkane has the highest ratio of hydrogen to carbon. The unsaturated alkene and the six membered alkane ring share the same chemical formula. It is important to remember this relationship. The unsaturated alkyne has a lower ratio of hydrogen to carbon than alkenes with a second pi bond. Benzene rings have the lowest hydrogen ratio to carbon at 1:1.
Exercise
1. Classify the following compounds are saturated or unsaturated. For unsaturated hydrocarbons, refine the classification by indicating whether the compound is an alkene, alkyne, or arene.
Answer
1.
The number of carbons continuously bonded together is an important structural feature and is described using the Homologous Series. In first year organic chemistry, the first ten names of the Homologous Series are usually all that need to be memorized. Of course, your professor will set the standard. Most of the prefixes are familiar from the Greek prefixes for binary covalent compounds. It is the prefixes for the first four carbon chain lengths that may be unfamiliar. Interestingly, three of these hydrocarbons frequently appear in every day life. Methane gas is a primary component of flatulence and is the ingredient that ignites when farts are lit - don't try this at home. Propane and butane are gases at room temperature. They are stored under pressure to create the liquid state. Propane is the fuel for bbqs, while butane is used in lighters. The suffix "ane" is used to distinguish between the longest continuous carbon chain, while the shorter carbon branches (substituents) are indicated with "yl" as the suffix.
Exercise
2. Complete the table below.
Condensed Structural Formula Chemical Name
propane
C6H6
CH3CH2CH2CH2CH2CH3
Answer
2.
Condensed Structural Formula Chemical Name
CH3CH2CH3 propane
C6H6 benzene
CH3CH2CH2CH2CH2CH3 hexane
• Wikipedia | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.08%3A_Hydrocarbons_and_the_Homologous_Series.txt |
Learning Objective
• recognize and classify the common functional groups of organic chemistry (alkanes, alkenes, alkynes, alkyl halides, alcohols, amines, ethers, aldehydes, ketones, carboxylic acids, esters, and amides
Functional groups in organic compounds
Functional groups are structural units within organic compounds that are defined by specific bonding arrangements between specific atoms. The structure of capsaicin, the compound responsible for the heat in peppers, incorporates several functional groups, labeled in the figure below and explained throughout this section.
As we progress in our study of organic chemistry, it will become extremely important to be able to quickly recognize the most common functional groups, because they are the key structural elements that define how organic molecules react. For now, we will only worry about drawing and recognizing each functional group, as depicted by Lewis and line structures. Much of the remainder of your study of organic chemistry will be taken up with learning about how the different functional groups behave in organic reactions.
The 'default' in organic chemistry (essentially, the lack of any functional groups) is given the term alkane, characterized by single bonds between carbon and carbon, or between carbon and hydrogen. Methane, CH4, is the natural gas you may burn in your furnace. Octane, C8H18, is a component of gasoline.
Alkanes
Alkenes (sometimes called olefins) have carbon-carbon double bonds, and alkynes have carbon-carbon triple bonds. Ethene, the simplest alkene example, is a gas that serves as a cellular signal in fruits to stimulate ripening. (If you want bananas to ripen quickly, put them in a paper bag along with an apple - the apple emits ethene gas, setting off the ripening process in the bananas). Ethyne, commonly called acetylene, is used as a fuel in welding blow torches.
Alkenes and alkynes
Alkenes have trigonal planar electron geometry while alkynes have linear geometry. Furthermore, many alkenes can take two geometric forms: cis or trans. The cis and trans forms of a given alkene are different molecules with different physical properties there is a very high energy barrier to rotation about a double bond. In the example below, the difference between cis and trans alkenes is readily apparent.
Alkanes, alkenes, and alkynes are all classified as hydrocarbons, because they are composed solely of carbon and hydrogen atoms. Alkanes are said to be saturated hydrocarbons, because the carbons are bonded to the maximum possible number of hydrogens - in other words, they are saturated with hydrogen atoms. The double and triple-bonded carbons in alkenes and alkynes have fewer hydrogen atoms bonded to them - they are thus referred to as unsaturated hydrocarbons. As we will see in chapter 15, hydrogen can be added to double and triple bonds, in a type of reaction called 'hydrogenation'.
The aromatic group is exemplified by benzene (which used to be a commonly used solvent on the organic lab, but which was shown to be carcinogenic), and naphthalene, a compound with a distinctive 'mothball' smell. Aromatic groups are planar (flat) ring structures, and are widespread in nature. We will learn more about the structure and reactions of aromatic groups in chapters 2 and 14.
Aromatics
When the carbon of an alkane is bonded to one or more halogens, the group is referred to as a alkyl halide or haloalkane. Chloroform is a useful solvent in the laboratory, and was one of the earlier anesthetic drugs used in surgery. Chlorodifluoromethane was used as a refrigerant and in aerosol sprays until the late twentieth century, but its use was discontinued after it was found to have harmful effects on the ozone layer. Bromoethane is a simple alkyl halide often used in organic synthesis. Alkyl halides groups are quite rare in biomolecules.
Haloalkanes
In the alcohol functional group, a carbon is single-bonded to an OH group (the OH group, by itself, is referred to as a hydroxyl). Except for methanol, all alcohols can be classified as primary, secondary, or tertiary. In a primary alcohol, the carbon bonded to the OH group is also bonded to only one other carbon. In a secondary alcohol and tertiary alcohol, the carbon is bonded to two or three other carbons, respectively. When the hydroxyl group is directly attached to an aromatic ring, the resulting group is called a phenol. The sulfur analog of an alcohol is called a thiol (from the Greek thio, for sulfur).
Alcohols, phenols, and thiols
Note that the definition of a phenol states that the hydroxyl oxygen must be directly attached to one of the carbons of the aromatic ring. The compound below, therefore, is not a phenol - it is a primary alcohol.
The distinction is important, because as we will see later, there is a significant difference in the reactivity of alcohols and phenols.
The deprotonated forms of alcohols, phenols, and thiols are called alkoxides, phenolates, and thiolates, respectively. A protonated alcohol is an oxonium ion.
In an ether functional group, a central oxygen is bonded to two carbons. Below is the structure of diethyl ether, a common laboratory solvent and also one of the first compounds to be used as an anesthetic during operations. The sulfur analog of an ether is called a thioether or sulfide.
Ethers and sulfides
Phosphate and its derivative functional groups are ubiquitous in biomolecules. Phosphate linked to a single organic group is called a phosphate ester; when it has two links to organic groups it is called a phosphate diester. A linkage between two phosphates creates a phosphate anhydride.
Organic phosphates
There are a number of functional groups that contain a carbon-oxygen double bond, which is commonly referred to as a carbonyl. Ketones and aldehydes are two closely related carbonyl-based functional groups that react in very similar ways. In a ketone, the carbon atom of a carbonyl is bonded to two other carbons. In an aldehyde, the carbonyl carbon is bonded on one side to a hydrogen, and on the other side to a carbon. The exception to this definition is formaldehyde, in which the carbonyl carbon has bonds to two hydrogens.
A group with a carbon-nitrogen double bond is called an imine, or sometimes a Schiff base (in this book we will use the term 'imine'). The chemistry of aldehydes, ketones, and imines will be covered in chapter 10.
Aldehydes, ketones, and imines
When a carbonyl carbon is bonded on one side to a carbon (or hydrogen) and on the other side to an oxygen, nitrogen, or sulfur, the functional group is considered to be one of the ‘carboxylic acid derivatives’, a designation that describes a set of related functional groups. The eponymous member of this family is the carboxylic acid functional group, in which the carbonyl is bonded to a hydroxyl group. The conjugate base of a carboxylic acid is a carboxylate. Other derivatives are carboxylic esters (usually just called 'esters'), thioesters, amides, acyl phosphates, acid chlorides, and acid anhydrides. With the exception of acid chlorides and acid anhydrides, the carboxylic acid derivatives are very common in biological molecules and/or metabolic pathways, and their structure and reactivity will be discussed in detail in chapter 11.
Carboxylic acid derivatives
A single compound often contains several functional groups, particularly in biological organic chemistry. The six-carbon sugar molecules glucose and fructose, for example, contain aldehyde and ketone groups, respectively, and both contain five alcohol groups (a compound with several alcohol groups is often referred to as a ‘polyol’).
The hormone testosterone, the amino acid phenylalanine, and the glycolysis metabolite dihydroxyacetone phosphate all contain multiple functional groups, as labeled below.
While not in any way a complete list, this section has covered most of the important functional groups that we will encounter in biological organic chemistry.
Exercise:
1. Identify the functional groups (other than alkanes) in the following organic compounds. State whether alcohols and amines are primary, secondary, or tertiary.
Solution
1.
Amines are characterized by nitrogen atoms with single bonds to hydrogen and carbon. Just as there are primary, secondary, and tertiary alcohols, there are primary, secondary, and tertiary amines. Ammonia is a special case with no carbon atoms.
One of the most important properties of amines is that they are basic, and are readily protonated to form ammonium cations. In the case where a nitrogen has four bonds to carbon (which is somewhat unusual in biomolecules), it is called a quaternary ammonium ion.
Amines
Note: Do not be confused by how the terms 'primary', 'secondary', and 'tertiary' are applied to alcohols and amines - the definitions are different. In alcohols, what matters is how many other carbons the alcohol carbon is bonded to, while in amines, what matters is how many carbons the nitrogen is bonded to.
Finally, a nitrile group is characterized by a carbon triple-bonded to a nitrogen.
Nitriles
Exercise
2. Draw one example of each compound that includes the specified structural features. Be sure to designate the location of all non-zero formal charges. All atoms should have complete octets (phosphorus may exceed the octet rule). There are many possible correct answers for these, so be sure to check your structures with your instructor or tutor.
a) a compound with molecular formula C6H11NO that includes alkene, secondary amine, and primary alcohol functional groups
b) an ion with molecular formula C3H5O6P 2- that includes aldehyde, secondary alcohol, and phosphate functional groups.
c) A compound with molecular formula C6H9NO that has an amide functional group, and does not have an alkene group.
Answer
2. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.09%3A_Organic_Functional_Groups.txt |
learning objective
• determine the dominant intermolecular forces (IMFs) of organic compounds
The properties of liquids are intermediate between those of gases and solids, but are more similar to solids. In contrast to intramolecular forces, such as the covalent bonds that hold atoms together in molecules and polyatomic ions, intermolecular forces hold molecules together in a liquid or solid. Intermolecular forces are generally much weaker than covalent bonds. For example, it requires 927 kJ to overcome the intramolecular forces and break both O–H bonds in 1 mol of water, but it takes only about 41 kJ to overcome the intermolecular attractions and convert 1 mol of liquid water to water vapor at 100°C. (Despite this seemingly low value, the intermolecular forces in liquid water are among the strongest such forces known!) Given the large difference in the strengths of intra- and intermolecular forces, changes between the solid, liquid, and gaseous states almost invariably occur for molecular substances without breaking covalent bonds.
The properties of liquids are intermediate between those of gases and solids but are more similar to solids.
Intermolecular forces determine bulk properties such as the melting points of solids and the boiling points of liquids. Liquids boil when the molecules have enough thermal energy to overcome the intermolecular attractive forces that hold them together, thereby forming bubbles of vapor within the liquid. Similarly, solids melt when the molecules acquire enough thermal energy to overcome the intermolecular forces that lock them into place in the solid.
Intermolecular forces are electrostatic in nature; that is, they arise from the interaction between positively and negatively charged species. Like covalent and ionic bonds, intermolecular interactions are the sum of both attractive and repulsive components. Because electrostatic interactions fall off rapidly with increasing distance between molecules, intermolecular interactions are most important for solids and liquids, where the molecules are close together. These interactions become important for gases only at very high pressures, where they are responsible for the observed deviations from the ideal gas law at high pressures. (For more information on the behavior of real gases and deviations from the ideal gas law,.)
In this section, we explicitly consider three kinds of intermolecular interactions: There are two additional types of electrostatic interaction that you are already familiar with: the ion–ion interactions that are responsible for ionic bonding and the ion–dipole interactions that occur when ionic substances dissolve in a polar substance such as water. The first two are often described collectively as van der Waals forces.
Dipole–Dipole Interactions
Polar covalent bonds behave as if the bonded atoms have localized fractional charges that are equal but opposite (i.e., the two bonded atoms generate a dipole). If the structure of a molecule is such that the individual bond dipoles do not cancel one another, then the molecule has a net dipole moment. Molecules with net dipole moments tend to align themselves so that the positive end of one dipole is near the negative end of another and vice versa, as shown in Figure \(\PageIndex{1a}\).
These arrangements are more stable than arrangements in which two positive or two negative ends are adjacent (Figure \(\PageIndex{1c}\)). Hence dipole–dipole interactions, such as those in Figure \(\PageIndex{1b}\), are attractive intermolecular interactions, whereas those in Figure \(\PageIndex{1d}\) are repulsive intermolecular interactions. Because molecules in a liquid move freely and continuously, molecules always experience both attractive and repulsive dipole–dipole interactions simultaneously, as shown in Figure \(2\). On average, however, the attractive interactions dominate.
Because each end of a dipole possesses only a fraction of the charge of an electron, dipole–dipole interactions are substantially weaker than the interactions between two ions, each of which has a charge of at least ±1, or between a dipole and an ion, in which one of the species has at least a full positive or negative charge. In addition, the attractive interaction between dipoles falls off much more rapidly with increasing distance than do the ion–ion interactions. Recall that the attractive energy between two ions is proportional to 1/r, where r is the distance between the ions. Doubling the distance (r → 2r) decreases the attractive energy by one-half. In contrast, the energy of the interaction of two dipoles is proportional to 1/r3, so doubling the distance between the dipoles decreases the strength of the interaction by 23, or 8-fold. Thus a substance such as \(\ce{HCl}\), which is partially held together by dipole–dipole interactions, is a gas at room temperature and 1 atm pressure, whereas \(\ce{NaCl}\), which is held together by interionic interactions, is a high-melting-point solid. Within a series of compounds of similar molar mass, the strength of the intermolecular interactions increases as the dipole moment of the molecules increases, as shown in Table \(1\).
Table \(1\): Relationships between the Dipole Moment and the Boiling Point for Organic Compounds of Similar Molar Mass
Compound Molar Mass (g/mol) Dipole Moment (D) Boiling Point (K)
C3H6 (cyclopropane) 42 0 240
CH3OCH3 (dimethyl ether) 46 1.30 248
CH3CN (acetonitrile) 41 3.9 355
The attractive energy between two ions is proportional to 1/r, whereas the attractive energy between two dipoles is proportional to 1/r6.
Example
Arrange ethyl methyl ether (CH3OCH2CH3), 2-methylpropane [isobutane, (CH3)2CHCH3], and acetone (CH3COCH3) in order of increasing boiling points. Their structures are as follows:
Given: compounds
Asked for: order of increasing boiling points
Strategy:
Compare the molar masses and the polarities of the compounds. Compounds with higher molar masses and that are polar will have the highest boiling points.
Solution:
The three compounds have essentially the same molar mass (58–60 g/mol), so we must look at differences in polarity to predict the strength of the intermolecular dipole–dipole interactions and thus the boiling points of the compounds.
The first compound, 2-methylpropane, contains only C–H bonds, which are not very polar because C and H have similar electronegativities. It should therefore have a very small (but nonzero) dipole moment and a very low boiling point.
Ethyl methyl ether has a structure similar to H2O; it contains two polar C–O single bonds oriented at about a 109° angle to each other, in addition to relatively nonpolar C–H bonds. As a result, the C–O bond dipoles partially reinforce one another and generate a significant dipole moment that should give a moderately high boiling point.
Acetone contains a polar C=O double bond oriented at about 120° to two methyl groups with nonpolar C–H bonds. The C–O bond dipole therefore corresponds to the molecular dipole, which should result in both a rather large dipole moment and a high boiling point.
Thus we predict the following order of boiling points:
2-methylpropane < ethyl methyl ether < acetone.
This result is in good agreement with the actual data: 2-methylpropane, boiling point = −11.7°C, and the dipole moment (μ) = 0.13 D; methyl ethyl ether, boiling point = 7.4°C and μ = 1.17 D; acetone, boiling point = 56.1°C and μ = 2.88 D.
Exercise
Arrange carbon tetrafluoride (CF4), ethyl methyl sulfide (CH3SC2H5), dimethyl sulfoxide [(CH3)2S=O], and 2-methylbutane [isopentane, (CH3)2CHCH2CH3] in order of decreasing boiling points.
Answer
dimethyl sulfoxide (boiling point = 189.9°C) > ethyl methyl sulfide (boiling point = 67°C) > 2-methylbutane (boiling point = 27.8°C) > carbon tetrafluoride (boiling point = −128°C)
London Dispersion Forces
Thus far we have considered only interactions between polar molecules, but other factors must be considered to explain why many nonpolar molecules, such as bromine, benzene, and hexane, are liquids at room temperature, and others, such as iodine and naphthalene, are solids. Even the noble gases can be liquefied or solidified at low temperatures, high pressures, or both (Table \(2\)).
What kind of attractive forces can exist between nonpolar molecules or atoms? This question was answered by Fritz London (1900–1954), a German physicist who later worked in the United States. In 1930, London proposed that temporary fluctuations in the electron distributions within atoms and nonpolar molecules could result in the formation of short-lived instantaneous dipole moments, which produce attractive forces called London dispersion forces between otherwise nonpolar substances.
Table \(2\): Normal Melting and Boiling Points of Some Elements and Nonpolar Compounds
Substance Molar Mass (g/mol) Melting Point (°C) Boiling Point (°C)
Ar 40 −189.4 −185.9
Xe 131 −111.8 −108.1
N2 28 −210 −195.8
O2 32 −218.8 −183.0
F2 38 −219.7 −188.1
I2 254 113.7 184.4
CH4 16 −182.5 −161.5
Consider a pair of adjacent He atoms, for example. On average, the two electrons in each He atom are uniformly distributed around the nucleus. Because the electrons are in constant motion, however, their distribution in one atom is likely to be asymmetrical at any given instant, resulting in an instantaneous dipole moment. As shown in part (a) in Figure \(3\), the instantaneous dipole moment on one atom can interact with the electrons in an adjacent atom, pulling them toward the positive end of the instantaneous dipole or repelling them from the negative end. The net effect is that the first atom causes the temporary formation of a dipole, called an induced dipole, in the second. Interactions between these temporary dipoles cause atoms to be attracted to one another. These attractive interactions are weak and fall off rapidly with increasing distance. London was able to show with quantum mechanics that the attractive energy between molecules due to temporary dipole–induced dipole interactions falls off as 1/r6. Doubling the distance therefore decreases the attractive energy by 26, or 64-fold.
Instantaneous dipole–induced dipole interactions between nonpolar molecules can produce intermolecular attractions just as they produce interatomic attractions in monatomic substances like Xe. This effect, illustrated for two H2 molecules in part (b) in Figure \(3\), tends to become more pronounced as atomic and molecular masses increase (Table \(2\)). For example, Xe boils at −108.1°C, whereas He boils at −269°C. The reason for this trend is that the strength of London dispersion forces is related to the ease with which the electron distribution in a given atom can be perturbed. In small atoms such as He, the two 1s electrons are held close to the nucleus in a very small volume, and electron–electron repulsions are strong enough to prevent significant asymmetry in their distribution. In larger atoms such as Xe, however, the outer electrons are much less strongly attracted to the nucleus because of filled intervening shells. As a result, it is relatively easy to temporarily deform the electron distribution to generate an instantaneous or induced dipole. The ease of deformation of the electron distribution in an atom or molecule is called its polarizability. Because the electron distribution is more easily perturbed in large, heavy species than in small, light species, we say that heavier substances tend to be much more polarizable than lighter ones.
For similar substances, London dispersion forces get stronger with increasing molecular size.
The polarizability of a substance also determines how it interacts with ions and species that possess permanent dipoles. Thus London dispersion forces are responsible for the general trend toward higher boiling points with increased molecular mass and greater surface area in a homologous series of compounds, such as the alkanes (part (a) in Figure \(4\)). The strengths of London dispersion forces also depend significantly on molecular shape because shape determines how much of one molecule can interact with its neighboring molecules at any given time. For example, part (b) in Figure \(4\) shows 2,2-dimethylpropane (neopentane) and n-pentane, both of which have the empirical formula C5H12. Neopentane is almost spherical, with a small surface area for intermolecular interactions, whereas n-pentane has an extended conformation that enables it to come into close contact with other n-pentane molecules. As a result, the boiling point of neopentane (9.5°C) is more than 25°C lower than the boiling point of n-pentane (36.1°C).
All molecules, whether polar or nonpolar, are attracted to one another by London dispersion forces in addition to any other attractive forces that may be present. In general, however, dipole–dipole interactions in small polar molecules are significantly stronger than London dispersion forces, so the former predominate.
Example
Arrange n-butane, propane, 2-methylpropane [isobutene, (CH3)2CHCH3], and n-pentane in order of increasing boiling points.
Given: compounds
Asked for: order of increasing boiling points
Strategy:
Determine the intermolecular forces in the compounds and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point.
Solution:
The four compounds are alkanes and nonpolar, so London dispersion forces are the only important intermolecular forces. These forces are generally stronger with increasing molecular mass, so propane should have the lowest boiling point and n-pentane should have the highest, with the two butane isomers falling in between. Of the two butane isomers, 2-methylpropane is more compact, and n-butane has the more extended shape. Consequently, we expect intermolecular interactions for n-butane to be stronger due to its larger surface area, resulting in a higher boiling point. The overall order is thus as follows, with actual boiling points in parentheses: propane (−42.1°C) < 2-methylpropane (−11.7°C) < n-butane (−0.5°C) < n-pentane (36.1°C).
Exercise
Arrange GeH4, SiCl4, SiH4, CH4, and GeCl4 in order of decreasing boiling points.
Answer
GeCl4 (87°C) > SiCl4 (57.6°C) > GeH4 (−88.5°C) > SiH4 (−111.8°C) > CH4 (−161°C)
Hydrogen Bonds
Molecules with hydrogen atoms bonded to electronegative atoms such as O, N, and F (and to a much lesser extent Cl and S) tend to exhibit unusually strong intermolecular interactions. These result in much higher boiling points than are observed for substances in which London dispersion forces dominate, as illustrated for the covalent hydrides of elements of groups 14–17 in Figure \(5\). Methane and its heavier congeners in group 14 form a series whose boiling points increase smoothly with increasing molar mass. This is the expected trend in nonpolar molecules, for which London dispersion forces are the exclusive intermolecular forces. In contrast, the hydrides of the lightest members of groups 15–17 have boiling points that are more than 100°C greater than predicted on the basis of their molar masses. The effect is most dramatic for water: if we extend the straight line connecting the points for H2Te and H2Se to the line for period 2, we obtain an estimated boiling point of −130°C for water! Imagine the implications for life on Earth if water boiled at −130°C rather than 100°C.
Why do strong intermolecular forces produce such anomalously high boiling points and other unusual properties, such as high enthalpies of vaporization and high melting points? The answer lies in the highly polar nature of the bonds between hydrogen and very electronegative elements such as O, N, and F. The large difference in electronegativity results in a large partial positive charge on hydrogen and a correspondingly large partial negative charge on the O, N, or F atom. Consequently, H–O, H–N, and H–F bonds have very large bond dipoles that can interact strongly with one another. Because a hydrogen atom is so small, these dipoles can also approach one another more closely than most other dipoles. The combination of large bond dipoles and short dipole–dipole distances results in very strong dipole–dipole interactions called hydrogen bonds, as shown for ice in Figure \(6\). A hydrogen bond is usually indicated by a dotted line between the hydrogen atom attached to O, N, or F (the hydrogen bond donor) and the atom that has the lone pair of electrons (the hydrogen bond acceptor). Because each water molecule contains two hydrogen atoms and two lone pairs, a tetrahedral arrangement maximizes the number of hydrogen bonds that can be formed. In the structure of ice, each oxygen atom is surrounded by a distorted tetrahedron of hydrogen atoms that form bridges to the oxygen atoms of adjacent water molecules. The bridging hydrogen atoms are not equidistant from the two oxygen atoms they connect, however. Instead, each hydrogen atom is 101 pm from one oxygen and 174 pm from the other. In contrast, each oxygen atom is bonded to two H atoms at the shorter distance and two at the longer distance, corresponding to two O–H covalent bonds and two O⋅⋅⋅H hydrogen bonds from adjacent water molecules, respectively. The resulting open, cagelike structure of ice means that the solid is actually slightly less dense than the liquid, which explains why ice floats on water rather than sinks.
Each water molecule accepts two hydrogen bonds from two other water molecules and donates two hydrogen atoms to form hydrogen bonds with two more water molecules, producing an open, cagelike structure. The structure of liquid water is very similar, but in the liquid, the hydrogen bonds are continually broken and formed because of rapid molecular motion.
Hydrogen bond formation requires both a hydrogen bond donor and a hydrogen bond acceptor.
Because ice is less dense than liquid water, rivers, lakes, and oceans freeze from the top down. In fact, the ice forms a protective surface layer that insulates the rest of the water, allowing fish and other organisms to survive in the lower levels of a frozen lake or sea. If ice were denser than the liquid, the ice formed at the surface in cold weather would sink as fast as it formed. Bodies of water would freeze from the bottom up, which would be lethal for most aquatic creatures. The expansion of water when freezing also explains why automobile or boat engines must be protected by “antifreeze” and why unprotected pipes in houses break if they are allowed to freeze.
Example
Considering CH3OH, C2H6, Xe, and (CH3)3N, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.
Given: compounds
Asked for: formation of hydrogen bonds and structure
Strategy:
1. Identify the compounds with a hydrogen atom attached to O, N, or F. These are likely to be able to act as hydrogen bond donors.
2. Of the compounds that can act as hydrogen bond donors, identify those that also contain lone pairs of electrons, which allow them to be hydrogen bond acceptors. If a substance is both a hydrogen donor and a hydrogen bond acceptor, draw a structure showing the hydrogen bonding.
Solution:
A Of the species listed, xenon (Xe), ethane (C2H6), and trimethylamine [(CH3)3N] do not contain a hydrogen atom attached to O, N, or F; hence they cannot act as hydrogen bond donors.
B The one compound that can act as a hydrogen bond donor, methanol (CH3OH), contains both a hydrogen atom attached to O (making it a hydrogen bond donor) and two lone pairs of electrons on O (making it a hydrogen bond acceptor); methanol can thus form hydrogen bonds by acting as either a hydrogen bond donor or a hydrogen bond acceptor. The hydrogen-bonded structure of methanol is as follows:
Exercise
Considering CH3CO2H, (CH3)3N, NH3, and CH3F, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.
Answer
CH3CO2H and NH3;
Although hydrogen bonds are significantly weaker than covalent bonds, with typical dissociation energies of only 15–25 kJ/mol, they have a significant influence on the physical properties of a compound. Compounds such as HF can form only two hydrogen bonds at a time as can, on average, pure liquid NH3. Consequently, even though their molecular masses are similar to that of water, their boiling points are significantly lower than the boiling point of water, which forms four hydrogen bonds at a time.
Example: Buckyballs
Arrange C60 (buckminsterfullerene, which has a cage structure), NaCl, He, Ar, and N2O in order of increasing boiling points.
Given: compounds
Asked for: order of increasing boiling points
Strategy:
Identify the intermolecular forces in each compound and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point.
Solution:
Electrostatic interactions are strongest for an ionic compound, so we expect NaCl to have the highest boiling point. To predict the relative boiling points of the other compounds, we must consider their polarity (for dipole–dipole interactions), their ability to form hydrogen bonds, and their molar mass (for London dispersion forces). Helium is nonpolar and by far the lightest, so it should have the lowest boiling point. Argon and N2O have very similar molar masses (40 and 44 g/mol, respectively), but N2O is polar while Ar is not. Consequently, N2O should have a higher boiling point. A C60 molecule is nonpolar, but its molar mass is 720 g/mol, much greater than that of Ar or N2O. Because the boiling points of nonpolar substances increase rapidly with molecular mass, C60 should boil at a higher temperature than the other nonionic substances. The predicted order is thus as follows, with actual boiling points in parentheses:
He (−269°C) < Ar (−185.7°C) < N2O (−88.5°C) < C60 (>280°C) < NaCl (1465°C).
Exercise
Arrange 2,4-dimethylheptane, Ne, CS2, Cl2, and KBr in order of decreasing boiling points.
Answer
KBr (1435°C) > 2,4-dimethylheptane (132.9°C) > CS2 (46.6°C) > Cl2 (−34.6°C) > Ne (−246°C)
Example
Identify the most significant intermolecular force in each substance.
1. C3H8
2. CH3OH
3. H2S
Solution
a. Although C–H bonds are polar, they are only minimally polar. The most significant intermolecular force for this substance would be dispersion forces.
b. This molecule has an H atom bonded to an O atom, so it will experience hydrogen bonding.
c. Although this molecule does not experience hydrogen bonding, the Lewis electron dot diagram and VSEPR indicate that it is bent, so it has a permanent dipole. The most significant force in this substance is dipole-dipole interaction.
Exercise
Identify the most significant intermolecular force in each substance.
1. HF
2. HCl
Answer a
hydrogen bonding
Answer b
dipole-dipole interactions
More complex examples of hydrogen bonding
The hydration of negative ions
When an ionic substance dissolves in water, water molecules cluster around the separated ions. This process is called hydration. Water frequently attaches to positive ions by co-ordinate (dative covalent) bonds. It bonds to negative ions using hydrogen bonds.
If you are interested in the bonding in hydrated positive ions, you could follow this link to co-ordinate (dative covalent) bonding.
The diagram shows the potential hydrogen bonds formed to a chloride ion, Cl-. Although the lone pairs in the chloride ion are at the 3-level and would not normally be active enough to form hydrogen bonds, in this case they are made more attractive by the full negative charge on the chlorine.
However complicated the negative ion, there will always be lone pairs that the hydrogen atoms from the water molecules can hydrogen bond to.
Hydrogen bonding in alcohols
An alcohol is an organic molecule containing an -OH group. Any molecule which has a hydrogen atom attached directly to an oxygen or a nitrogen is capable of hydrogen bonding. Such molecules will always have higher boiling points than similarly sized molecules which don't have an -O-H or an -N-H group. The hydrogen bonding makes the molecules "stickier", and more heat is necessary to separate them.
Ethanol, CH3CH2OH, and methoxymethane, CH3OCH3, are structural isomers with the same molecular formula, C2H6O.
They have the same number of electrons, and a similar length to the molecule. The van der Waals attractions (both dispersion forces and dipole-dipole attractions) in each will be much the same. However, ethanol has a hydrogen atom attached directly to an oxygen - and that oxygen still has exactly the same two lone pairs as in a water molecule. Hydrogen bonding can occur between ethanol molecules, although not as effectively as in water. The hydrogen bonding is limited by the fact that there is only one hydrogen in each ethanol molecule with sufficient + charge.
In methoxymethane, lone pairs on the oxygen are still there, but the hydrogens are not sufficiently + for hydrogen bonds to form. Except in some rather unusual cases, the hydrogen atom has to be attached directly to the very electronegative element for hydrogen bonding to occur. The boiling points of ethanol and methoxymethane show the dramatic effect that the hydrogen bonding has on the stickiness of the ethanol molecules:
ethanol (with hydrogen bonding) 78.5°C
methoxymethane (without hydrogen bonding) -24.8°C
The hydrogen bonding in the ethanol has lifted its boiling point about 100°C. It is important to realize that hydrogen bonding exists in addition to van der Waals attractions. For example, all the following molecules contain the same number of electrons, and the first two are much the same length. The higher boiling point of the butan-1-ol is due to the additional hydrogen bonding.
Comparing the two alcohols (containing -OH groups), both boiling points are high because of the additional hydrogen bonding due to the hydrogen attached directly to the oxygen - but they are not the same. The boiling point of the 2-methylpropan-1-ol isn't as high as the butan-1-ol because the branching in the molecule makes the van der Waals attractions less effective than in the longer butan-1-ol.
Hydrogen bonding in organic molecules containing nitrogen
Hydrogen bonding also occurs in organic molecules containing N-H groups - in the same sort of way that it occurs in ammonia. Examples range from simple molecules like CH3NH2 (methylamine) to large molecules like proteins and DNA. The two strands of the famous double helix in DNA are held together by hydrogen bonds between hydrogen atoms attached to nitrogen on one strand, and lone pairs on another nitrogen or an oxygen on the other one.
Donors and Acceptors
In order for a hydrogen bond to occur there must be both a hydrogen donor and an acceptor present. The donor in a hydrogen bond is the atom to which the hydrogen atom participating in the hydrogen bond is covalently bonded, and is usually a strongly electronegative atom such as N,O, or F. The hydrogen acceptor is the neighboring electronegative ion or molecule, and must posses a lone electron pair in order to form a hydrogen bond.
Why does a hydrogen bond occur?
Since the hydrogen donor is strongly electronegative, it pulls the covalently bonded electron pair closer to its nucleus, and away from the hydrogen atom. The hydrogen atom is then left with a partial positive charge, creating a dipole-dipole attraction between the hydrogen atom bonded to the donor, and the lone electron pair on the accepton. This results in a hydrogen bond.(see Interactions Between Molecules With Permanent Dipoles)
Types of hydrogen bonds
Hydrogen bonds can occur within one single molecule, between two like molecules, or between two unlike molecules.
Intramolecular hydrogen bonds
Intramolecular hydrogen bonds are those which occur within one single molecule. This occurs when two functional groups of a molecule can form hydrogen bonds with each other. In order for this to happen, both a hydrogen donor an acceptor must be present within one molecule, and they must be within close proximity of each other in the molecule. For example, intramolecular hydrogen bonding occurs in ethylene glycol (C2H4(OH)2) between its two hydroxyl groups due to the molecular geometry.
Intermolecular hydrogen bonds
Intermolecular hydrogen bonds occur between separate molecules in a substance. They can occur between any number of like or unlike molecules as long as hydrogen donors and acceptors are present an in positions in which they can interact.For example, intermolecular hydrogen bonds can occur between NH3 molecules alone, between H2O molecules alone, or between NH3 and H2O molecules.
Properties and effects of hydrogen bonds
On Boiling Point
When we consider the boiling points of molecules, we usually expect molecules with larger molar masses to have higher normal boiling points than molecules with smaller molar masses. This, without taking hydrogen bonds into account, is due to greater dispersion forces (see Interactions Between Nonpolar Molecules). Larger molecules have more space for electron distribution and thus more possibilities for an instantaneous dipole moment. However, when we consider the table below, we see that this is not always the case.
Compound Molar Mass Normal Boiling Point
\(H_2O\) 18 g/mol 373 K
\(HF\) 20 g/mol 292.5 K
\(NH_3\) 17 g/mol 239.8 K
\(H_2S\) 34 g/mol 212.9 K
\(HCl\) 36.4 g/mol 197.9 K
\(PH_3\) 34 g/mol 185.2 K
We see that H2O, HF, and NH3 each have higher boiling points than the same compound formed between hydrogen and the next element moving down its respective group, indicating that the former have greater intermolecular forces. This is because H2O, HF, and NH3 all exhibit hydrogen bonding, whereas the others do not. Furthermore, \(H_2O\) has a smaller molar mass than HF but partakes in more hydrogen bonds per molecule, so its boiling point is consequently higher.
On Viscosity
The same effect that is seen on boiling point as a result of hydrogen bonding can also be observed in the viscosity of certain substances. Those substances which are capable of forming hydrogen bonds tend to have a higher viscosity than those that do not. Substances which have the possibility for multiple hydrogen bonds exhibit even higher viscosities.
Factors preventing Hydrogen bonding
Electronegativity
Hydrogen bonding cannot occur without significant electronegativity differences between hydrogen and the atom it is bonded to. Thus, we see molecules such as PH3, which no not partake in hydrogen bonding. PH3 exhibits a trigonal pyramidal molecular geometry like that of ammmonia, but unlike NH3 it cannot hydrogen bond. This is due to the similarity in the electronegativities of phosphorous and hydrogen. Both atoms have an electronegativity of 2.1, and thus, no dipole moment occurs. This prevents the hydrogen bonding from acquiring the partial positive charge needed to hydrogen bond with the lone electron pair in another molecule. (see Polarizability)
Atom Size
The size of donors and acceptors can also effect the ability to hydrogen bond. This can account for the relatively low ability of Cl to form hydrogen bonds. When the radii of two atoms differ greatly or are large, their nuclei cannot achieve close proximity when they interact, resulting in a weak interaction.
Hydrogen Bonding in Nature
Hydrogen bonding plays a crucial role in many biological processes and can account for many natural phenomena such as the Unusual properties of Water. In addition to being present in water, hydrogen bonding is also important in the water transport system of plants, secondary and tertiary protein structure, and DNA base pairing.
Plants
The cohesion-adhesion theory of transport in vascular plants uses hydrogen bonding to explain many key components of water movement through the plant's xylem and other vessels. Within a vessel, water molecules hydrogen bond not only to each other, but also to the cellulose chain which comprises the wall of plant cells. This creates a sort of capillary tube which allows for capillary action to occur since the vessel is relatively small. This mechanism allows plants to pull water up into their roots. Furthermore,hydrogen bonding can create a long chain of water molecules which can overcome the force of gravity and travel up to the high altitudes of leaves.
Proteins
Hydrogen bonding is present abundantly in the secondary structure of proteins, and also sparingly in tertiary conformation. The secondary structure of a protein involves interactions (mainly hydrogen bonds) between neighboring polypeptide backbones which contain Nitrogen-Hydrogen bonded pairs and oxygen atoms. Since both N and O are strongly electronegative, the hydrogen atoms bonded to nitrogen in one polypeptide backbone can hydrogen bond to the oxygen atoms in another chain and visa-versa. Though they are relatively weak,these bonds offer great stability to secondary protein structure because they repeat a great number of times.
In tertiary protein structure,interactions are primarily between functional R groups of a polypeptide chain; one such interaction is called a hydrophobic interaction. These interactions occur because of hydrogen bonding between water molecules around the hydrophobe and further reinforce conformation.
• Jose Pietri | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.10%3A_Intermolecular_Forces_%28IMFs%29_-_Review.txt |
Learning Objective
• predict the relative boil points of organic compounds
Intermolecular forces (IMFs) can be used to predict relative boiling points. The stronger the IMFs, the lower the vapor pressure of the substance and the higher the boiling point. Therefore, we can compare the relative strengths of the IMFs of the compounds to predict their relative boiling points.
H-bonding > dipole-dipole > London dispersion (van der Waals)
When comparing compounds with the same IMFs, we use size and shape as tie breakers since the London dispersion forces increase as the surface area increases. Since all compounds exhibit some level of London dispersion forces and compounds capable of H-bonding also exhibit dipole-dipole, we will use the phrase "dominant IMF" to communicate the IMF most responsible for the physical properties of the compound.
In the table below, we see examples of these relationships. When comparing the structural isomers of pentane (pentane, isopentane, and neopentane), they all have the same molecular formula C5H12. However, as the carbon chain is shortened to create the carbon branches found in isopentane and neopentane the overall surface area of the molecules decreases. The visual image of MO theory can be helpful in seeing each compound as a cloud of electrons in an all encompassing MO system. Branching creates more spherical shapes noting that the sphere allows the maximum volume with the least surface area. The structural isomers with the chemical formula C2H6O have different dominant IMFs. The H-bonding of ethanol results in a liquid for cocktails at room temperature, while the weaker dipole-dipole of the dimethylether results in a gas a room temperature. In the last example, we see the three IMFs compared directly to illustrate the relative strength IMFs to boiling points.
Boiling points and melting points
The observable melting and boiling points of different organic molecules provides an additional illustration of the effects of noncovalent interactions. The overarching principle involved is simple: the stronger the noncovalent interactions between molecules, the more energy that is required, in the form of heat, to break them apart. Higher melting and boiling points signify stronger noncovalent intermolecular forces.
Consider the boiling points of increasingly larger hydrocarbons. More carbons means a greater surface area possible for hydrophobic interaction, and thus higher boiling points.
As you would expect, the strength of intermolecular hydrogen bonding and dipole-dipole interactions is reflected in higher boiling points. Just look at the trend for hexane (nonpolar London dispersion interactions only ), 3-hexanone (dipole-dipole interactions), and 3-hexanol (hydrogen bonding).
Of particular interest to biologists (and pretty much anything else that is alive in the universe) is the effect of hydrogen bonding in water. Because it is able to form tight networks of intermolecular hydrogen bonds, water remains in the liquid phase at temperatures up to 100 OC, (slightly lower at high altitude). The world would obviously be a very different place if water boiled at 30 OC.
Exercise
1. Based on their structures, rank phenol, benzene, benzaldehyde, and benzoic acid in terms of lowest to highest boiling point.
Solution
By thinking about noncovalent intermolecular interactions, we can also predict relative melting points. All of the same principles apply: stronger intermolecular interactions result in a higher melting point. Ionic compounds, as expected, usually have very high melting points due to the strength of ion-ion interactions (there are some ionic compounds, however, that are liquids at room temperature). The presence of polar and especially hydrogen-bonding groups on organic compounds generally leads to higher melting points. Molecular shape, and the ability of a molecule to pack tightly into a crystal lattice, has a very large effect on melting points. The flat shape of aromatic compounds such as napthalene and biphenyl allows them to stack together efficiently, and thus aromatics tend to have higher melting points compared to alkanes or alkenes with similar molecular weights.
Comparing the melting points of benzene and toluene, you can see that the extra methyl group on toluene disrupts the molecule's ability to stack, thus decreasing the cumulative strength of intermolecular London dispersion forces.
Note also that the boiling point for toluene is 111 oC, well above the boiling point of benzene (80 oC). The key factor for the boiling point trend in this case is size (toluene has one more carbon), whereas for the melting point trend, shape plays a much more important role. This makes sense when you consider that melting involves ‘unpacking’ the molecules from their ordered array, whereas boiling involves simply separating them from their already loose (liquid) association with each other.
If you are taking an organic lab course, you may have already learned that impurities in a crystalline substance will cause the observed melting point to be lower compared to a pure sample of the same substance. This is because impurities disrupt the ordered packing arrangement of the crystal, and make the cumulative intermolecular interactions weaker.
The melting behavior of lipid structures
An interesting biological example of the relationship between molecular structure and melting point is provided by the observable physical difference between animal fats like butter or lard, which are solid at room temperature, and vegetable oils, which are liquid. Both solid fats and liquid oils are based on a ‘triacylglycerol’ structure, where three hydrophobic hydrocarbon chains of varying length are attached to a glycerol backbone through an ester functional group (compare this structure to that of the membrane lipids discussed in section 2.4B).
Interactive 3D image of a saturated triacylglycerol (BioTopics)
Saturated vs mono-unsaturated fatty acid (BioTopics)
In vegetable oils, the hydrophobic chains are unsaturated, meaning that they contain one or more double bonds. Solid animal fat, in contrast, contains saturated hydrocarbon chains, with no double bonds. The double bonds in vegetable oils cause those hydrocarbon chains to be more rigid, and ‘bent’ at an angle (remember that rotation is restricted around double bonds), with the result that they don’t pack together as closely, and thus can be broken apart (ie. melted) more readily. Shown in the figure above is a polyunsaturated fatty acid chain (two double bonds), and you can click on the link to see interactive images of a saturated fatty acid compared to a monounsaturated fatty acid (one double bond).
Exercise
2. Arrange the following compounds in order of decreasing boiling point.
Answer
2.
Contributors
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.11%3A_Intermolecular_Forces_and_Relative_Boiling_Poi.txt |
Learning Objective
• predict whether a mixture of compounds will a form homogeneous or heterogeneous solution
The type of intermolecular forces (IMFs) exhibited by compounds can be used to predict whether two different compounds can be mixed to form a homogeneous solution (soluble or miscible). Because organic chemistry can perform reactions in non-aqueous solutions using organic solvents. It is important to consider the solvent as a reaction parameter and the solubility of each reagent. With this said, solvent effects are secondary to the sterics and electrostatics of the reactants. Make sure that you do not drown in the solvent.
Solubility
Virtually all of the organic chemistry that you will see in this course takes place in the solution phase. In the organic laboratory, reactions are often run in nonpolar or slightly polar solvents such as toluene (methylbenzene), hexane, dichloromethane, or diethylether. In recent years, much effort has been made to adapt reaction conditions to allow for the use of ‘greener’ (in other words, more environmentally friendly) solvents such as water or ethanol, which are polar and capable of hydrogen bonding. In organic reactions that occur in the cytosolic region of a cell, the solvent is of course water. It is critical for any organic chemist to understand the factors which are involved in the solubility of different molecules in different solvents.
You probably remember the rule you learned in general chemistry regarding solubility: ‘like dissolves like’ (and even before you took any chemistry at all, you probably observed at some point in your life that oil does not mix with water). Let’s revisit this old rule, and put our knowledge of covalent and noncovalent bonding to work.
Imagine that you have a flask filled with water, and a selection of substances that you will test to see how well they dissolve in the water. The first substance is table salt, or sodium chloride. As you would almost certainly predict, especially if you’ve ever inadvertently taken a mouthful of water while swimming in the ocean, this ionic compound dissolves readily in water. Why? Because water, as a very polar molecule, is able to form many ion-dipole interactions with both the sodium cation and the chloride anion, the energy from which is more than enough to make up for energy required to break up the ion-ion interactions in the salt crystal and some water-water hydrogen bonds.
The end result, then, is that in place of sodium chloride crystals, we have individual sodium cations and chloride anions surrounded by water molecules – the salt is now in solution. Charged species as a rule dissolve readily in water: in other words, they are very hydrophilic (water-loving).
Now, we’ll try a compound called biphenyl, which, like sodium chloride, is a colorless crystalline substance (the two compounds are readily distinguishable by sight, however – the crystals look quite different).
Biphenyl does not dissolve at all in water. Why is this? Because it is a very non-polar molecule, with only carbon-carbon and carbon-hydrogen bonds. It is able to bond to itself very well through nonpolar (London dispersion) interactions, but it is not able to form significant attractive interactions with the very polar solvent molecules. Thus, the energetic cost of breaking up the biphenyl-to-biphenyl interactions in the solid is high, and very little is gained in terms of new biphenyl-water interactions. Water is a terrible solvent for nonpolar hydrocarbon molecules: they are very hydrophobic ('water-fearing').
Next, you try a series of increasingly large alcohol compounds, starting with methanol (1 carbon) and ending with octanol (8 carbons).
You find that the smaller alcohols - methanol, ethanol, and propanol - dissolve easily in water. This is because the water is able to form hydrogen bonds with the hydroxyl group in these molecules, and the combined energy of formation of these water-alcohol hydrogen bonds is more than enough to make up for the energy that is lost when the alcohol-alcohol hydrogen bonds are broken up. When you try butanol, however, you begin to notice that, as you add more and more to the water, it starts to form its own layer on top of the water.
The longer-chain alcohols - pentanol, hexanol, heptanol, and octanol - are increasingly non-soluble. What is happening here? Clearly, the same favorable water-alcohol hydrogen bonds are still possible with these larger alcohols. The difference, of course, is that the larger alcohols have larger nonpolar, hydrophobic regions in addition to their hydrophilic hydroxyl group. At about four or five carbons, the hydrophobic effect begins to overcome the hydrophilic effect, and water solubility is lost.
Now, try dissolving glucose in the water – even though it has six carbons just like hexanol, it also has five hydrogen-bonding, hydrophilic hydroxyl groups in addition to a sixth oxygen that is capable of being a hydrogen bond acceptor.
We have tipped the scales to the hydrophilic side, and we find that glucose is quite soluble in water.
We saw that ethanol was very water-soluble (if it were not, drinking beer or vodka would be rather inconvenient!) How about dimethyl ether, which is a constitutional isomer of ethanol but with an ether rather than an alcohol functional group? We find that diethyl ether is much less soluble in water. Is it capable of forming hydrogen bonds with water? Yes, in fact, it is –the ether oxygen can act as a hydrogen-bond acceptor. The difference between the ether group and the alcohol group, however, is that the alcohol group is both a hydrogen bond donor and acceptor.
The result is that the alcohol is able to form more energetically favorable interactions with the solvent compared to the ether, and the alcohol is therefore more soluble.
Here is another easy experiment that can be done (with proper supervision) in an organic laboratory. Try dissolving benzoic acid crystals in room temperature water – you'll find that it is not soluble. As we will learn when we study acid-base chemistry in a later chapter, carboxylic acids such as benzoic acid are relatively weak acids, and thus exist mostly in the acidic (protonated) form when added to pure water.
Acetic acid, however, is quite soluble. This is easy to explain using the small alcohol vs large alcohol argument: the hydrogen-bonding, hydrophilic effect of the carboxylic acid group is powerful enough to overcome the hydrophobic effect of a single methyl group on acetic acid, but not the larger hydrophobic effect of the 6-carbon benzene group on benzoic acid.
Now, try slowly adding some aqueous sodium hydroxide to the flask containing undissolved benzoic acid. As the solvent becomes more and more basic, the benzoic acid begins to dissolve, until it is completely in solution.
What is happening here is that the benzoic acid is being converted to its conjugate base, benzoate. The neutral carboxylic acid group was not hydrophilic enough to make up for the hydrophobic benzene ring, but the carboxylate group, with its full negative charge, is much more hydrophilic. Now, the balance is tipped in favor of water solubility, as the powerfully hydrophilic anion part of the molecule drags the hydrophobic part, kicking and screaming, (if a benzene ring can kick and scream) into solution. If you want to precipitate the benzoic acid back out of solution, you can simply add enough hydrochloric acid to neutralize the solution and reprotonate the carboxylate.
If you are taking a lab component of your organic chemistry course, you will probably do at least one experiment in which you will use this phenomenon to separate an organic acid like benzoic acid from a hydrocarbon compound like biphenyl.
Similar arguments can be made to rationalize the solubility of different organic compounds in nonpolar or slightly polar solvents. In general, the greater the content of charged and polar groups in a molecule, the less soluble it tends to be in solvents such as hexane. The ionic and very hydrophilic sodium chloride, for example, is not at all soluble in hexane solvent, while the hydrophobic biphenyl is very soluble in hexane.
Exercise
1. Vitamins can be classified as water-soluble or fat-soluble (consider fat to be a very non-polar, hydrophobic 'solvent'. Decide on a classification for each of the vitamins shown below.
Solutions
Exercise
2. Both aniline and phenol are insoluble in pure water. Predict the solubility of these two compounds in 10% aqueous hydrochloric acid, and explain your reasoning. Hint – in this context, aniline is basic, phenol is not!
Solutions
Illustrations of solubility concepts: metabolic intermediates, lipid bilayer membranes, soaps and detergents
Because water is the biological solvent, most biological organic molecules, in order to maintain water-solubility, contain one or more charged functional groups. These are most often phosphate, ammonium or carboxylate, all of which are charged when dissolved in an aqueous solution buffered to pH 7.
Sugars often lack charged groups, but as we discussed in our ‘thought experiment’ with glucose, they are quite water-soluble due to the presence of multiple hydroxyl groups.
Some biomolecules, in contrast, contain distinctly nonpolar, hydrophobic components. The ‘lipid bilayer’ membranes of cells and subcellular organelles serve to enclose volumes of water and myriad biomolecules in solution. The lipid (fat) molecules that make up membranes are amphipathic: they have a charged, hydrophilic ‘head’ and a hydrophobic hydrocarbon ‘tail’.
interactive 3D image of a membrane phospholipid (BioTopics)
Notice that the entire molecule is built on a ‘backbone’ of glycerol, a simple 3-carbon molecule with three alcohol groups. In a biological membrane structure, lipid molecules are arranged in a spherical bilayer: hydrophobic tails point inward and bind together by London dispersion forces, while the hydrophilic head groups form the inner and outer surfaces in contact with water.
Interactive 3D Image of a lipid bilayer (BioTopics)
Because the interior of the bilayer is extremely hydrophobic, biomolecules (which as we know are generally charged species) are not able to diffuse through the membrane– they are simply not soluble in the hydrophobic interior. The transport of molecules across the membrane of a cell or organelle can therefore be accomplished in a controlled and specific manner by special transmembrane transport proteins, a fascinating topic that you will learn more about if you take a class in biochemistry.
A similar principle is the basis for the action of soaps and detergents. Soaps are composed of fatty acids, which are long (typically 18-carbon), hydrophobic hydrocarbon chains with a (charged) carboxylate group on one end,
Fatty acids are derived from animal and vegetable fats and oils. In aqueous solution, the fatty acid molecules in soaps will spontaneously form micelles, a spherical structure that allows the hydrophobic tails to avoid contact with water and simultaneously form favorable London dispersion contacts.
Interactive 3D images of a fatty acid soap molecule and a soap micelle (Edutopics)
Because the outside of the micelle is charged and hydrophilic, the structure as a whole is soluble in water. Micelles will form spontaneously around small particles of oil that normally would not dissolve in water (like that greasy spot on your shirt from the pepperoni slice that fell off your pizza), and will carry the particle away with it into solution. We will learn more about the chemistry of soap-making in a later chapter (section 12.4B).
Synthetic detergents are non-natural amphipathic molecules that work by the same principle as that described for soaps. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.12%3A_Intermolecular_Forces_and_Solubilities.txt |
Learning Objective
• distinguish between organic compounds that are H-bond donors versus H-bond acceptors
H-bond donors vs H-bond acceptors
Compounds with H-bonding as their dominant intermolecular force (IMF) are BOTH H-bond donors and H-bond acceptors. They are H-bond donors because they have a highly polar hydrogen atom bonded to a strongly electronegative atom, primarily nitrogen, oxygen, or fluorine (NOF). Because there is an equivalent partial negative charge on the atom bonded to hydrogen (mostly NOF), this atom can accept H-bonds from another atoms. Since H-bond donors are ALWAYS H-bond acceptors, we simplify communication to "H-bond donor". There are two H-bonding interactions for H-bond donors. The strongly electronegative elements (primarily nitrogen, oxygen, and fluorine) will always form a relatively large partial negative charge when bonded with carbon. These elements can accept H-bonds when they are part of the organic molecule. In this situation, there is only one H-bonding interaction. The diagram below illustrates the similarities and differences between H-bond donors and H-bond acceptors. Water and alcohols may serve as both donors and acceptors, whereas ethers, aldehydes, ketones and esters can function only as acceptors. Similarly, primary and secondary amines are both donors and acceptors, but tertiary amines function only as acceptors.
Exercise
1. Classify the compounds below as H-bond donors, H-bond acceptors, or neither.
Answer
1.
Comparing Physical Properties of H-bond Donors vs H-bond Acceptors
Once we are able to recognize compounds that can exhibit intermolecular hydrogen bonding, the relatively high boiling points they exhibit become understandable. The data in the following table serve to illustrate this point.
Compound Formula Mol. Wt. Boiling Point Melting Point
dimethyl ether CH3OCH3 46 –24ºC –138ºC
ethanol CH3CH2OH 46 78ºC –130ºC
propanol CH3(CH2)2OH 60 98ºC –127ºC
diethyl ether (CH3CH2)2O 74 34ºC –116ºC
propyl amine CH3(CH2)2NH2 59 48ºC –83ºC
methylaminoethane CH3CH2NHCH3 59 37ºC
trimethylamine (CH3)3N 59 3ºC –117ºC
ethylene glycol HOCH2CH2OH 62 197ºC –13ºC
acetic acid CH3CO2H 60 118ºC 17ºC
ethylene diamine H2NCH2CH2NH2 60 118ºC 8.5ºC
Alcohols boil cosiderably higher than comparably sized ethers (first two entries), and isomeric 1º, 2º & 3º-amines, respectively, show decreasing boiling points, with the two hydrogen bonding isomers being substantially higher boiling than the 3º-amine (entries 5 to 7). Also, O–H---O hydrogen bonds are clearly stronger than N–H---N hydrogen bonds, as we see by comparing propanol with the amines.
As expected, the presence of two hydrogen bonding functions in a compound raises the boiling point even further. Acetic acid (the ninth entry) is an interesting case. A dimeric species, shown on the right, held together by two hydrogen bonds is a major component of the liquid state. If this is an accurate representation of the composition of this compound then we would expect its boiling point to be equivalent to that of a C4H8O4 compound (formula weight = 120). A suitable approximation of such a compound is found in tetramethoxymethane, (CH3O)4C, which is actually a bit larger (formula weight = 136) and has a boiling point of 114ºC. Thus, the dimeric hydrogen bonded structure appears to be a good representation of acetic acid in the condensed state.
A related principle is worth noting at this point. Although the hydrogen bond is relatively weak (ca. 4 to 5 kcal per mole), when several such bonds exist the resulting structure can be quite robust. The hydrogen bonds between cellulose fibers confer great strength to wood and related materials. For additional information on this subject Click Here. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.14%3A_Organic_Functional_Groups-_H-bond_donors_and_H.txt |
Hybridization
2-1
2-2
Longest to shortest bond length: b > a > c
Strongest to weakest bond: c > a > b
2-3
Sigma bonds: 7
Pi bonds: 0
2-4
Sigma bonds: 5
Pi bonds: 1
2-5
Sigma bonds: 3
Pi bonds: 2
Hybridization, Electron Geometry, and Molecular Shape
2-6 Correct answer is (b) sp3, tetrahedral.
2-7
a) Tetrahedral
b) Trigonal bipyramidal
c) Tetrahedral
d) Trigonal planar
2-8
2-9
2-10 Boron has trigonal planar geometry. The hydrogen atoms are at a 120° angle from each other to be as far apart as possible.
Bond Rotation
2-11 This molecule can rotate freely around the middle bond as there are no major steric hindrance interactions.
2-12 This molecule cannot rotate freely around the middle bond as the large bromine substituents attached at the ortho positions of the benzene rings experience significant steric hindrance with each other.
2-13 No; the pi-bond prevents free rotation about the C=C bond.
Polarity of Bonds and Molecules
2-14
2-15
a) 2
b) No dipole moment
c) 1
2-16 True
Intermolecular Forces (IMFs)
2-17
a) Cannot H-bond
b) Can H-bond
c) Can H-bond
d) Can H-bond
e) Cannot H-bond
f) Cannot H-bond
2-18
2-19
a) London Dispersion Forces
b) Dipole-Dipole Interactions
c) Ionic Forces
d) Hydrogen bonding
IMFs and Solubility
2-20
a) Not miscible
b) Miscible
c) Not miscible
d) Soluble
e) Not soluble
f) Soluble
2-21 Caffeine will dissolve in dichloromethane (DCM) significantly more than in hexanes as DCM is a more polar solvent and caffeine is a polar molecule (like dissolves like).
Hydrocarbons and an Introduction to Isomerism
2-22
a) Alkene
b) Alkane
c) Alkyne
d) Alkane
e) Alkene
f) Alkene
2-23
2-24
2-25 It does not have cis/trans configuration, as the triple bond in the compound (CH3)2CHC☰CCH3 holds the four carbons in a straight line due to the sp hybridization of the middle two carbons (which have a linear geometric configuration).
Organic Compounds with Oxygen
2-26
a) Ether
b) Ketone
c) Carboxylic acid
d) Alcohol and Amine
e) Amide
f) Ether and Alkene
2-27
a) Alcohol and Amine (We will learn that the most correct classification for hydroxyl groups bonded to benzene rings is phenol)
b) Alcohol, Ether, Ketone, Amine and Alkene
c) Ester, Ether, Amine and Alkene
2-28
a) Aldehyde and carboxylic acid
b) Alcohol, Ketone, Amine
c) Alcohol, Ketone, Carboxylic acid
2-29
Organic Compounds with Nitrogen
2-30 Compound B has a slight dipole moment due to the cis configuration of the amine groups. Since it has a dipole moment, it experiences dipole-dipole interactions in addition to hydrogen bonding, thus increasing its boiling point.
2-31
2.16: Additional Exercises
Hybridization
2-1 For each of the following compounds, identify the hybridization of each carbon or nitrogen atom with an arrow pointed at it.
2-2 Rank the following bonds in order of decreasing bond length. Then rank the bonds in order from strongest to weakest.
2-3 How many sigma and pi bonds are in a molecule of ethane (C2H6)?
2-4 How many sigma and pi bonds are in a molecule of ethylene (C2H4)?
2-5 How many sigma and pi bonds are in a molecule of acetylene (C2H2)?
Hybridization, Electron Geometry, and Molecular Shape
2-6 What is the hybridization state and geometry of the carbon atom in methane (CH4)?
a) sp, linear
b) sp3, tetrahedral
c) sp2, trigonal planar
d) None of the above
2-7 Identify the electron geometry of the following compounds.
a) H2O
b) PF5
c) NH4+
d) The carbonyl carbon of acetone (CH3)2CO. (Note that double bonds between carbon and oxygen must be recognized.)
2-8 For the following compounds, identify which atoms have sp2 hybridization.
2-9 Draw the orbitals showing the geometric shape of ammonia (NH3). Identify its geometric shape.
2-10 What is the geometric shape of the boron atom in BH3? What is the bond angle of the hydrogen atoms?
Bond Rotation
2-11 Will the following compound experience free rotation around the middle bond? Explain why or why not.
2-12 Will the following compound experience free rotation around the middle bond? Explain why or why not.
2-13 Can a molecule of ethylene experience free rotation around the C=C bond?
Polarity of Bonds and Molecules
2-14 For the following compounds, draw an arrow to show the direction of the dipole moment (if any).
2-15 In the following pairs of compounds, identify the compound with the larger dipole moment (if any).
2-16 True or False: Generally, the larger the difference in electronegativity of connected atoms, the greater the dipole moment.
Intermolecular Forces (IMFs)
2-17 Identify which of the following compounds can form hydrogen bonds.
2-18 For the compounds in the previous problem ( 2-17 above) that can hydrogen bond, draw how they can form those bonds.
2-19 Identify what type of intermolecular force the following compounds experience.
IMFs and Solubility
2-20 Identify whether the following compounds are miscible or soluble in water.
2-21 Identify which solvent, hexanes or dichloromethane (DCM), would be the better solvent to dissolve 3.0 grams of caffeine. Explain your answer.
Hydrocarbons and an Introduction to Isomerism
2-22 Identify whether the following hydrocarbons are alkanes, alkenes, or alkynes.
2-23 For the following compounds, identify the hybridization state of each labeled carbon.
2-24 Draw all possible isomers for the following compounds.
a) C4H10
b) C6H14
c) C3H6
2-25 Does (CH3)2CHCCCH3 show cis/trans isomerism? Explain why or why not.
Organic Compounds with Oxygen
2-26 Identify the functional group(s) of each compound.
2-27 What functional groups are found in the following compounds?
2-28 What oxygen-containing functional groups are present in the following compounds. The nitrogen-containing group is a challenge question.
2-29 Identify whether the functional groups on the following compounds are classified correctly. If not, give the correct classification.
Organic Compounds with Nitrogen
2-30 First, identify which of the following compounds has a dipole moment. Then, predict which of the following compounds will have the higher boiling point and explain why.
2-31 Identify whether the functional groups of the following compounds are classified correctly. If not, give the correct classification. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.15%3A_Solutions_to_Additional_Exercises.txt |
learning objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• use R groups to draw generic functional groups - refer to section 3.1
• name alkanes, cycloalkanes, alkenes, alkynes, alkyl halides, ethers, alcohols, amines, benzene and its derivatives, aldehydes, ketones, amines, carboxylic acids, and carboxylic acid derivatives using IUPAC (systematic) and selected common name nomenclature - refer to sections 3.2 - 3.14
• draw the structure of alkanes, cycloalkanes, alkenes, alkynes, alkyl halides, ethers, alcohols, amines, benzene and its derivatives, aldehydes, ketones, amines, carboxylic acids, and carboxylic acid derivatives from the IUPAC (systematic) and selected common names - refer to sections 3.2 - 3.14
• classify alkyl halides, alcohols and amines - refer to sections 3.5, 3.8, and 3.12 respectively.
A Pearl of Wisdom: Most common names were derived from older systems of nomenclature that some may argue were "not systematic at all". However, it is helpful to note that the older systems of nomenclature were often based on shared structural features and/or chemical reactivity. Learning carefully selected common names can offer insights into chemical reactivity and structural patterns. Additionally, there are some common names that are so prevalent, they need to be memorized.
Please note: The nomenclature for organic compounds with sulfur and phosphorus are introduced so that students can interpret a given name and draw the correct structure. Derivation of names can be required by the professor and requires additional instruction.
• 3.1: Generic (Abbreviated) Structures (aka R Groups)
It is not always necessary to draw the entire structure of a compound. The correct use of the "R group" is explained.
• 3.2: Overview of the IUPAC Naming Strategy
The International Union of Pure and Applied Chemistry (IUPAC) names for organic compounds all follow the same set of rules and can have up to four parts. Recognizing the overall pattern can simplify the learning process.
• 3.3: Alkanes
Alkanes form the carbon backbone of all organic compounds.
• 3.4: Cycloalkanes
The rotational limits of cycloalkanes introduce stereochemistry to some compounds. Some disubstituted cycloalkanes can exist as geometric isomers (cis/trans).
• 3.5: Haloalkane - Classification and Nomenclature
The reactivity of the alkyl halides (haloalkanes) can be predicted using their structural classifications of primary, secondary, or tertiary.
• 3.6: Alkenes
The rigid, carbon-carbon double bond (C=C) can also introduce stereochemical considerations in naming.
• 3.7: Alkynes
The linear geometry of carbon-carbon triple bonds simplifies the names of this functional group.
• 3.8: 3.8 Alcohols - Classification and Nomenclature
Alcohols are organic compounds with hydroxyl groups as a unique structural feature. Alcohol classification will be useful as we learn patterns of chemical reactivity. The terms "vicinal" and "germinal" are also explained.
• 3.9: Ethers, Epoxides and Sulfides
Ethers, epoxides, and sulfides all have the heteroatom disrupting the continuous carbon chain. There is no IUPAC suffix for ethers. The alkoxy group is always a substituent.
• 3.10: Benzene and its Derivatives
This section focuses on naming benzene derivatives and the important distinction between a phenyl group and a benzyl group. Phenol nomenclature and with other important benzene derivatives are discussed.
• 3.11: Aldehydes and Ketones
While there are many functional groups that include the carbonyl structural feature, aldehydes and ketones are collectively referred to as "the carbonyls." With their highly similar chemical reactivity, their nomenclature is taught together.
• 3.12: Amines - Classification and Nomenclature
Amines are important weak bases, organic reactants. and biologically active compounds, such as alkaloids. Amine classification is helpful in recognizing patterns of chemical reactivity.
• 3.13: Carboxylic Acids
Carboxylic acids are important natural products and synthetic precursors. The common names are provided for carboxylic acids (1 to 10 carbons) and selected dicarboxylic acids.
• 3.14: The Carboxylic Acid Derivatives
All of these functional groups can be hydrolyzed to form carboxylic acids, so they are collective called the carboxylic acid derivatives.
• 3.15: Additional Exercises
This section has additional exercises for the key learning objectives of this chapter.
• 3.16: Solutions to Additional Exercises
This section has the solutions to the additional exercises from the previous section.
• 3.17: Appendix - IUPAC Nomenclature Rules
This appendix provides a link to the full text of the IUPAC rules of nomenclature for organic compounds.
03: Functional Groups and Nomenclature
learning objective
• use R groups to draw generic functional groups - refer to section 3.1
Drawing Generic (abbreviated) Organic Structures
In chapter 2, we learned to recognize and distinguish between organic functional groups. Often when drawing organic structures, chemists find it convenient to use the letter 'R' to designate part of a molecule outside of the region of interest. "R" represents the "Rest of the Molecule". If we just want to refer in general to a functional group without drawing a specific molecule, for example, we can use 'R groups' to focus attention on the group of interest:
The 'R' group is a convenient way to abbreviate the structures of large biological molecules, especially when we are interested in something that is occurring specifically at one location on the molecule. For example, in chapter 15 when we look at biochemical oxidation-reduction reactions involving the flavin molecule, we will abbreviate a large part of the flavin structure which does not change at all in the reactions of interest:
As an alternative, we can use a 'break' symbol to indicate that we are looking at a small piece or section of a larger molecule. This is used commonly in the context of drawing groups on large polymers such as proteins or DNA.
Finally, 'R' groups can be used to concisely illustrate a series of related compounds, such as the family of penicillin-based antibiotics.
Using abbreviations appropriately is a very important skill to develop when studying organic chemistry in a biological context, because although many biomolecules are very large and complex (and take forever to draw!), usually we are focusing on just one small part of the molecule where a change is taking place.
As a rule, you should never abbreviate any atom involved in a bond-breaking or bond-forming event that is being illustrated: only abbreviate that part of the molecule which is not involved in the reaction of interest.
For example, carbon #2 in the reactant/product below most definitely is involved in bonding changes, and therefore should not be included in the 'R' group.
If you are unsure whether to draw out part of a structure or abbreviate it, the safest thing to do is to draw it out.
Exercise
1. a) If you intend to draw out the chemical details of a reaction in which the methyl ester functional group of cocaine (see earlier figure) was converted to a carboxylate plus methanol, what would be an appropriate abbreviation to use for the cocaine structure (assuming that you only wanted to discuss the chemistry specifically occurring at the ester group)?
b) Below is the (somewhat complicated) reaction catalyzed by an enzyme known as 'Rubisco', by which plants 'fix' carbon dioxide. Carbon dioxide and the oxygen of water are colored red and blue respectively to help you see where those atoms are incorporated into the products. Propose an appropriate abbreviation for the starting compound (ribulose 1,5-bisphosphate), using two different 'R' groups, R1 and R2.
Solutions to exercises
3.02: Overview of the IUPAC Naming Strategy
learning objectives
• name alkanes, cycloalkanes, alkenes, alkynes, alkyl halides, ethers, alcohols, amines, benzene and its derivatives, aldehydes, ketones, amines, carboxylic acids, and carboxylic acid derivatives using IUPAC (systematic) and selected common name nomenclature
• draw the structure of alkanes, cycloalkanes, alkenes, alkynes, alkyl halides, ethers, alcohols, amines, benzene and its derivatives, aldehydes, ketones, amines, carboxylic acids, and carboxylic acid derivatives from the IUPAC (systematic) and selected common names
Overview of the IUPAC System for Naming Organic Compounds
The International Union of Pure and Applied Chemistry (IUPAC) has established the rules of nomenclature of all chemical compounds. IUPAC nomenclature can also be called "systematic" nomenclature because there is an overall system and structure to the names. This section provides an overview of the general naming strategy and structure for organic compounds.
Naming organic compounds according to the IU{AC system requires up to four pieces of information
1. recognize & prioritize the functional group(s) present
2. identify & number the longest continuous carbon chain to give the highest ranking group the lowest possible number
3. cite the substituents (branches) alphabetically using the numbering determined above
4. recognize & classify any stereochemistry (E/Z, R/S, cis/trans, etc)
With these four pieces of information, the IUPAC name is written using the format below. This same format applies to ALL the organic compounds.
Recognize & Prioritize the Functional Group(s) Present
The IUPAC Rules of Organic Nomenclature assume that the following table is understood and memorized.
Identify & Number the Longest Continuous Carbon Chain with the Highest Priority Group
The longest continuous carbon chain (parent) is named using the Homologous Series, as well as any carbon branches. The suffixes and location within the name distinguish between the parent and the branches.
When Alkenes and Alkynes have Lower Priority
The hydrocarbon suffixes differ in the letter preceeding the "n". When alkenes and alkynes occur in compounds with higher priority functional groups, then the distinction between hydrocarbons is communicated with a single letter: "e", or "y" for alkenes and alkynes, respectively, An example is shown below to illustrate the application of this rule.
substituents (branches) alphabetically
The major substituents are listed above and need to be memorized. There are a few additional substituents that will introduced later in the text.
Stereochemistry
Distinguishing spatial orientations of atoms (stereochemistry) are communicated at the beginning of the name using the appropriate symbols, such as E/Z, R/S, cis/trans, etc. This nomenclature will be discussed when it is possible for a functional group. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.01%3A_Generic_%28Abbreviated%29_Structures_%28aka_R_Groups%29.txt |
learning objectives
• name alkanes using IUPAC (systematic) and selected common name nomenclature
• draw the structure of alkanes from IUPAC (systematic) and selected common names
Alkanes are hydrocarbons that can be described by the general formula CnH2n+2. They consist only of carbon and hydrogen and contain only single bonds. Alkanes are also known as "saturated hydrocarbons."
The following table contains the systematic names for the first twenty straight chain alkanes. It will be important to familiarize yourself with these names because they will be the basis for naming many other organic molecules throughout your course of study.
Name Molecular Formula Condensed Structural Formula
Methane CH4 CH4
Ethane C2H6 CH3CH3
Propane C3H8 CH3CH2CH3
Butane C4H10 CH3(CH2)2CH3
Pentane C5H12 CH3(CH2)3CH3
Hexane C6H14 CH3(CH2)4CH3
Heptane C7H16 CH3(CH2)5CH3
Octane C8H18 CH3(CH2)6CH3
Nonane C9H20 CH3(CH2)7CH3
Decane C10H22 CH3(CH2)8CH3
Undecane C11H24 CH3(CH2)9CH3
Dodecane C12H26 CH3(CH2)10CH3
Tridecane C13H28 CH3(CH2)11CH3
Tetradecane C14H30 CH3(CH2)12CH3
Pentadecane C15H32 CH3(CH2)13CH3
Hexadecane C16H34 CH3(CH2)14CH3
Heptadecane C17H36 CH3(CH2)15CH3
Octadecane C18H38 CH3(CH2)16CH3
Nonadecane C19H40 CH3(CH2)17CH3
Eicosane C20H42 CH3(CH2)18CH3
Carbon Atom Classifications
To assign the prefixes sec-, which stands for secondary, and tert-, for tertiary, it is important that we first learn how to classify carbon atoms. If a carbon is attached to only one other carbon, it is called a primary carbon. If a carbon is attached to two other carbons, it is called a seconday carbon. A tertiary carbon is attached to three other carbons and last, a quaternary carbon is attached to four other carbons. These terms are summarixed with an example in the table below.
Classification Example
methyl CH4
primary
secondary
tertiary
Using Common Names with Branched Alkanes
Certain branched alkanes have common names that are still widely used today. These common names make use of prefixes, such as iso-, sec-, tert-, and neo-.
Isoalkanes
The prefix iso-, which stands for isomer, is commonly given to 2-methyl alkanes. In other words, if there is methyl group located on the second carbon of a carbon chain, we can use the prefix iso-. The prefix will be placed in front of the alkane name that indicates the total number of carbons as in isopentane which is the same as 2-methylbutane and isobutane which is the same as 2-methylpropane. The pattern is illustrated below
Sec- and Tert-alkanes
Secondary and tertiary alkanes can be further distinguished from their "iso-counter parts" by applying comparing the carbon classifications. The common names for three and four carbon branches are summarized below. Notice that the "iso" prefix is joined directly to the alkyl name. When alphabetizing branches, the "i" is considered. For "sec" and "tert", the prefix is separated from the alkyl name and is NOT considered when alphabetizing branches.
For some compounds, the common names bring a simple a simple elegance to the experience.
Neo-alkanes
A five carbon alkane and the corresponding five carbon branch can form a structural pattern commonly known as neopentane and neopentyl respectively. The prefix neo- can also be applied to larger alkanes as shown below.
Alkyl Groups
An alkyl group is formed by removing one hydrogen from the alkane chain and is described by the formula CnH2n+1. The removal of this hydrogen results in a stem change from -ane to -yl. Take a look at the following examples.
The same approach can be used with any of the alkanes in the table above and with common names.
Alkyl Group Common Names that sound like Alkenes
In long hydrocarbon chains there can be many -CH2- groups in a series. These internal -(CH2)n- groups are names using the homologous series stem with the suffice "ene" even though there are NO carbon-carbon double bonds present. For example, the common name for dichloromethane, CH2Cl2, is methylene chloride; and the common names for the major ingredients in antifreeze are ethylene glycol (CH2(OH)CH2OH) and propylene glycol (CH2(OH)CH2(OH)CH3), in spite of the fact that there are no carbon-carbon double bonds in any of these three compounds.
Three Basic Principles of Naming
1. Choose the longest, most substituted carbon chain containing a functional group.
2. A carbon bonded to a functional group must have the lowest possible carbon number. If there are no functional groups, then any substitute present must have the lowest possible number.
3. Take the alphabetical order into consideration; that is, after applying the first two rules given above, make sure that your substitutes and/or functional groups are written in alphabetical order.
Example
Solution
Rule #1 Choose the longest, most substituted carbon chain containing a functional group. This example does not contain any functional groups, so we only need to be concerned with choosing the longest, most substituted carbon chain. The longest carbon chain has been highlighted in red and consists of eight carbons.
Rule #2 Carbons bonded to a functional group must have the lowest possible carbon number. If there are no functional groups, then any substitute present must have the lowest possible number. Because this example does not contain any functional groups, we only need to be concerned with the two substitutes present, that is, the two methyl groups. If we begin numbering the chain from the left, the methyls would be assigned the numbers 4 and 7, respectively. If we begin numbering the chain from the right, the methyls would be assigned the numbers 2 and 5. Therefore, to satisfy the second rule, numbering begins on the right side of the carbon chain as shown below. This gives the methyl groups the lowest possible numbering.
In this example, there is no need to utilize the third rule. Because the two substitutes are identical, neither takes alphabetical precedence with respect to numbering the carbons. This concept will become clearer in the next example.
Example
Solution
Rule #1 Choose the longest, most substituted carbon chain containing a functional group. This example contains two functional groups, bromine and chlorine. The longest carbon chain has been highlighted in red and consists of seven carbons.
Rule #2 Carbons bonded to a functional group must have the lowest possible carbon number. If there are no functional groups, then any substitute present must have the lowest possible number. In this example, numbering the chain from the left or the right would satisfy this rule. If we number the chain from the left, bromine and chlorine would be assigned the second and sixth carbon positions, respectively. If we number the chain from the right, chlorine would be assigned the second position and bromine would be assigned the sixth position. In other words, whether we choose to number from the left or right, the functional groups occupy the second and sixth positions in the chain. To select the correct numbering scheme, we need to utilize the third rule.
Rule #3 After applying the first two rules, take the alphabetical order into consideration. Alphabetically, bromine comes before chlorine. Therefore, bromine is assigned the second carbon position, and chlorine is assigned the sixth carbon position.
Example
Solution
Rule #1 Choose the longest, most substituted carbon chain containing a functional group. This example contains two functional groups, bromine and chlorine, and one substitute, the methyl group. The longest carbon chain has been highlighted in red and consists of seven carbons.
Rule #2 Carbons bonded to a functional group must have the lowest possible carbon number. After taking functional groups into consideration, any substitutes present must have the lowest possible carbon number. This particular example illustrates the point of difference principle. If we number the chain from the left, bromine, the methyl group and chlorine would occupy the second, fifth and sixth positions, respectively. This concept is illustrated in the second drawing below. If we number the chain from the right, chlorine, the methyl group and bromine would occupy the second, third and sixth positions, respectively, which is illustrated in the first drawing below. The position of the methyl, therefore, becomes a point of difference. In the first drawing, the methyl occupies the third position. In the second drawing, the methyl occupies the fifth position. To satisfy the second rule, we want to choose the numbering scheme that provides the lowest possible numbering of this substitute. Therefore, the first of the two carbon chains shown below is correct.
Therefore, the first numbering scheme is the appropriate one to use.
Once you have determined the correct numbering of the carbons, it is often useful to make a list, including the functional groups, substitutes, and the name of the parent chain.
Parent chain: heptane 2-Chloro 3-Methyl 6-Bromo
6-bromo-2-chloro-3-methylheptane
Exercises
Write the IUPAC (systematic) name for each of the compounds below. The parent chains have numbered for the first two compounds to help you begin.
a)
b)
c)
Solutions
a) 9-chloro-7-ethyl-2,2,4-tromethyldecane
b) 3-chloro-5-ethyl-4,4-dimethylheptane
c) 2-bromo-6-ethyloctane | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.03%3A_Alkanes.txt |
learning objectives
• name cycloalkanes using IUPAC (systematic) and selected common name nomenclature
• draw the structure of cycloalkanes from IUPAC (systematic) and selected common names
Cycloalkanes are cyclic hydrocarbons, meaning that the carbons of the molecule are arranged in the form of a ring. Cycloalkanes are also saturated, meaning that all of the carbons atoms that make up the ring are single bonded to other atoms (no double or triple bonds). There are also polycyclic alkanes, which are molecules that contain two or more cycloalkanes that are joined, forming multiple rings.
Introduction
Many organic compounds found in nature or created in a laboratory contain rings of carbon atoms with distinguishing chemical properties; these compounds are known as cycloalkanes. Cycloalkanes only contain carbon-hydrogen bonds and carbon-carbon single bonds, but in cycloalkanes, the carbon atoms are joined in a ring. The smallest cycloalkane is cyclopropane.
If you count the carbons and hydrogens, you will see that they no longer fit the general formula \(C_nH_{2n+2}\). By joining the carbon atoms in a ring,two hydrogen atoms have been lost. The general formula for a cycloalkane is \(C_nH_{2n}\). Cyclic compounds are not all flat molecules. All of the cycloalkanes, from cyclopentane upwards, exist as "puckered rings". Cyclohexane, for example, has a ring structure that looks like this:
In addition to being saturated cyclic hydrocarbons, cycloalkanes may have multiple substituents or functional groups that further determine their unique chemical properties. The most common and useful cycloalkanes in organic chemistry are cyclopentane and cyclohexane, although other cycloalkanes varying in the number of carbons can be synthesized. Understanding cycloalkanes and their properties are crucial in that many of the biological processes that occur in most living things have cycloalkane-like structures.
Glucose (6 carbon sugar) Ribose (5 carbon sugar)
Cholesterol (polycyclic)
Although polycyclic compounds are important, they are highly complex and typically have common names accepted by IUPAC. However, the common names do not generally follow the basic IUPAC nomenclature rules. The general formula of the cycloalkanes is \(C_nH_{2n}\) where \(n\) is the number of carbons. The naming of cycloalkanes follows a simple set of rules that are built upon the same basic steps in naming alkanes. Cyclic hydrocarbons have the prefix "cyclo-".
Contents
For simplicity, cycloalkane molecules can be drawn in the form of skeletal structures in which each intersection between two lines is assumed to have a carbon atom with its corresponding number of hydrogens.
same as same as
Cycloalkane Molecular Formula Basic Structure
Cyclopropane C3H6
Cyclobutane C4H8
Cyclopentane C5H10
Cyclohexane C6H12
Cycloheptane C7H14
Cyclooctane C8H16
Cyclononane C9H18
Cyclodecane C10H20
IUPAC Rules for Nomenclature
1. Determine the cycloalkane to use as the parent chain. The parent chain is the one with the highest number of carbon atoms. If there are two cycloalkanes, use the cycloalkane with the higher number of carbons as the parent chain.
2. If there is an alkyl straight chain that has a greater number of carbons than the cycloalkane, then the alkyl chain must be used as the primary parent chain. Cycloalkane acting as a substituent to an alkyl chain has an ending "-yl" and, therefore, must be named as a cycloalkyl.
Cycloalkane Cycloalkyl
cyclopropane cyclopropyl
cyclobutane cyclobutyl
cyclopentane cyclopentyl
cyclohexane cyclohexyl
cycloheptane cycloheptyl
cyclooctane cyclooctyl
cyclononane cyclononanyl
cyclodecane cyclodecanyl
Example \(1\):
The longest straight chain contains 10 carbons, compared with cyclopropane, which only contains 3 carbons. Because cyclopropane is a substituent, it would be named a cyclopropyl-substituted alkane.
3) Determine any functional groups or other alkyl groups.
4) Number the carbons of the cycloalkane so that the carbons with functional groups or alkyl groups have the lowest possible number. A carbon with multiple substituents should have a lower number than a carbon with only one substituent or functional group. One way to make sure that the lowest number possible is assigned is to number the carbons so that when the numbers corresponding to the substituents are added, their sum is the lowest possible.
(1+3=4) NOT (1+5=6)
5) When naming the cycloalkane, the substituents and functional groups must be placed in alphabetical order.
(ex: 2-bromo-1-chloro-3-methylcyclopentane)
6) Indicate the carbon number with the functional group with the highest priority according to alphabetical order. A dash"-" must be placed between the numbers and the name of the substituent. After the carbon number and the dash, the name of the substituent can follow. When there is only one substituent on the parent chain, indicating the number of the carbon atoms with the substituent is not necessary.
(ex: 1-chlorocyclohexane or cholorocyclohexane is acceptable)
7) If there is more than one of the same functional group on one carbon, write the number of the carbon two, three, or four times, depending on how many of the same functional group is present on that carbon. The numbers must be separated by commas, and the name of the functional group that follows must be separated by a dash. When there are two of the same functional group, the name must have the prefix "di". When there are three of the same functional group, the name must have the prefix "tri". When there are four of the same functional group, the name must have the prefix "tetra". However, these prefixes cannot be used when determining the alphabetical priorities.
There must always be commas between the numbers and the dashes that are between the numbers and the names.
Example \(2\)
(2-bromo-1,1-dimethylcyclohexane)
Notice that "f" of fluoro alphabetically precedes the "m" of methyl. Although "di" alphabetically precedes "f", it is not used in determining the alphabetical order.
Example 3
(2-fluoro-1,1,-dimethylcyclohexane NOT 1,1-dimethyl-2-fluorocyclohexane)
8) If the substituents of the cycloalkane are related by the cis or trans configuration, then indicate the configuration by placing "cis-" or "trans-" in front of the name of the structure.
Blue=Carbon Yellow=Hydrogen Green=Chlorine
Notice that chlorine and the methyl group are both pointed in the same direction on the axis of the molecule; therefore, they are cis.
cis-1-chloro-2-methylcyclopentane
9) After all the functional groups and substituents have been mentioned with their corresponding numbers, the name of the cycloalkane can follow.
Summary
1. Determine the parent chain: the parent chain contains the most carbon atoms.
2. Number the substituents of the chain so that the sum of the numbers is the lowest possible.
3. Name the substituents and place them in alphabetical order.
4. If stereochemistry of the compound is shown, indicate the orientation as part of the nomenclature.
5. Cyclic hydrocarbons have the prefix "cyclo-" and have an "-alkane" ending unless there is an alcohol substituent present. When an alcohol substituent is present, the molecule has an "-ol" ending.
Glossary
• alkyl: A structure that is formed when a hydrogen atom is removed from an alkane.
• cyclic: Chemical compounds arranged in the form of a ring or a closed chain form.
• cycloalkanes: Cyclic saturated hydrocarbons with a general formula of CnH(2n). Cycloalkanes are alkanes with carbon atoms attached in the form of a closed ring.
• functional groups: An atom or groups of atoms that substitute for a hydrogen atom in an organic compound, giving the compound unique chemical properties and determining its reactivity.
• hydrocarbon: A chemical compound containing only carbon and hydrogen atoms.
• saturated: All of the atoms that make up a compound are single bonded to the other atoms, with no double or triple bonds.
• skeletal structure: A simplified structure in which each intersection between two lines is assumed to have a carbon atom with its corresponding number of hydrogens.
Exercises
1. Name the following structures. (Note: The structures are complex for practice purposes and may not be found in nature.)
a) b) c) d) e) f)
g)
2. Draw the following structures.
a) 1,1-dibromo-5-fluoro-3-butyl-7-methylcyclooctane
b) trans-1-bromo-2-chlorocyclopentane
c) 1,1-dibromo-2,3-dichloro-4-propylcyclobutane
d) 2-methyl-1-ethyl-1,3-dipropylcyclopentane
e) cycloheptane-1,3,5-triol
3. Name the following structures.
Blue=Carbon Yellow=Hydrogen Red=Oxygen Green=Chlorine
a) b) c) d) e)
f) g)
Solutions
1. a) cyclodecane
b) chlorocyclopentane or 1-chlorocyclopentane
c) trans-1-chloro-2-methylcycloheptane
d) 3-cyclopropyl-6-methyldecane
e) cyclopentylcyclodecane or 1-cyclopentylcyclodecane
f) 1,3-dibromo-1-chloro-2-fluorocycloheptane
g) 1-cyclobutyl-4-isopropylcyclohexane
2.
a) b) c) d) e)
3. a) cyclohexane
b) cyclohexanol
c) chlorocyclohexane
d) cyclopentylcyclohexane
e) 1-chloro-3-methylcyclobutane
f) 2,3-dimethylcyclohexanol
g) cis-1-propyl-2-methylcyclopentane
Contributors and Attributions
• Pwint Zin
• Jim Clark (ChemGuide) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.04%3A_Cycloalkanes.txt |
learning objectives
• classify alkyl halides as primary, secondary, or tertiary
• name alkyl halides using IUPAC (systematic) and selected common name nomenclature
• draw the structure of alkyl halides from IUPAC (systematic) and selected common names
The haloalkanes, also known as alkyl halides, are a group of chemical compounds comprised of an alkane with one or more hydrogens replaced by a halogen atom (fluorine, chlorine, bromine, or iodine). There is a fairly large distinction between the structural and physical properties of haloalkanes and the structural and physical properties of alkanes. As mentioned above, the structural differences are due to the replacement of one or more hydrogens with a halogen atom. The differences in physical properties are a result of factors such as electronegativity, bond length, bond strength, and molecular size. A few representative alklyl halides are shown below.
Alkyl halides are a versatile and useful functional group for multi-step organic synthesis. The reactivity of the alkyl halides can be predicted using their structural classifications. To communicate the three different structures, the terms primary, secondary, and tertiary are used. The classification is determined by the number of carbons bonded to the carbon bearing the halide. This classification strategy is analogous to the one used for alcohols and is explained in further detail below.
Classification of Alkyl Halides
Functional group classifications are based on the bonding patterns of the atoms involved. There is only one neutral bonding pattern for halogens (three lone pairs and a single bond) so the halogens cannot be used to determine their classification. To determine the classification of alkyl halides, the bonding pattern of the carbon bonded to the halogen is used as shown in the diagram below.
Primary alkyl halides
In a primary (1°) halogenoalkane, the carbon which carries the halogen atom is only attached to one other alkyl group.Some examples of primary alkyl halides include:
Notice that it doesn't matter how complicated the attached alkyl group is. In each case there is only one linkage to an alkyl group from the CH2 group holding the halogen. There is an exception to this: CH3Br and the other methyl halides are often counted as primary alkyl halides even though there are no alkyl groups attached to the carbon with the halogen on it.
Secondary alkyl halides
In a secondary (2°) halogenoalkane, the carbon with the halogen attached is joined directly to two other alkyl groups, which may be the same or different. Examples:
Tertiary alkyl halides
In a tertiary (3°) halogenoalkane, the carbon atom holding the halogen is attached directly to three alkyl groups, which may be any combination of same or different. Examples:
IUPAC and Common Nomenclature
The replacement of only one hydrogen atom gives an alkyl halide (or haloalkane) so the nomenclature system is closely related to the system for alkanes. The common names of alkyl halides consist of two parts: the name of the alkyl group plus the stem of the name of the halogen, with the ending -ide.
Common Name Format: alkyl name + halide name
The IUPAC system uses the name of the parent alkane with a prefix indicating the halogen substituents, preceded by number indicating the substituent’s location. The prefixes are fluoro-, chloro-, bromo-, and iodo-. Thus CH3CH2Cl has the common name ethyl chloride and the IUPAC name chloroethane. For simple halo alkanes, the IUPAC name includes the three parts shown below.
IUPAC Name Format: locator # + halo prefix + parent alkane
Alkyl halides with simple alkyl groups (one to four carbon atoms) are often called by common names. Those with a larger number of carbon atoms are usually given IUPAC names.
Example
Give the common and IUPAC names for each compound.
1. CH3CH2CH2Br
2. (CH3)2CHCl
Solution
1. The alkyl group (CH3CH2CH2–) is a propyl group, and the halogen is bromine (Br). The common name is therefore propyl bromide. For the IUPAC name, the prefix for bromine (bromo) is combined with the name for a three-carbon chain (propane), preceded by a number identifying the carbon atom to which the Br atom is attached, so the IUPAC name is 1-bromopropane.
2. The alkyl group [(CH3)2CH–] has three carbon atoms, with a chlorine (Cl) atom attached to the middle carbon atom. The alkyl group is therefore isopropyl, and the common name of the compound is isopropyl chloride. For the IUPAC name, the Cl atom (prefix chloro-) attached to the middle (second) carbon atom of a propane chain results in 2-chloropropane.
Exercise
Give common and IUPAC names for each compound.
1. CH3CH2I
• CH3CH2CH2CH2F
• Exercise
Give the IUPAC name for each compound.
3.
4.
Solution
3. The parent alkane has five carbon atoms in the longest continuous chain; it is pentane. A bromo (Br) group is attached to the second carbon atom of the chain. The IUPAC name is 2-bromopentane.
4. The parent alkane is hexane. Methyl (CH3) and bromo (Br) groups are attached to the second and fourth carbon atoms, respectively. Listing the substituents in alphabetical order gives the name 4-bromo-2-methylhexane.
Exercise
Give the IUPAC name for each compound.
5.
6.
Solutions
5. 2-choro-3-methylbutane
6. 1-bromo-2-chloro-4-methylpentane | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.05%3A_Haloalkane_-_Classification_and_Nomenclature.txt |
learning objectives
• name alkenes using IUPAC (systematic) and selected common name nomenclature
• draw the structure of alkenes from IUPAC (systematic) and selected common names
Alkenes contain carbon-carbon double bonds and are unsaturated hydrocarbons with the molecular formula is CnH2n. Be aware - this is also the same molecular formula ratio as cycloalkanes as shown in the example below.
Introduction
The parent structure is the longest chain containing both carbon atoms of the double bond. The two carbon atoms of a double bond and the four atoms attached to them lie in a plane, with bond angles of approximately 120° A double bond consists of one sigma bond formed by overlap of sp2 hybrid orbitals and one pi bond formed by overlap of parallel 2 p orbitals.
The carbon atoms sharing the double bond can be referred to as the "vinyl carbons". This common name arose because alkenes the source for vinyl polymers.
Pi Bond Rigidity & Geometric Isomers
The rigidity of the pi bond in alkenes creates the possibility of stereoisomers called geometric isomers. To name alkenes, it may be neccessary to communicate the stereochemistry of the structure using the cis/trans or E/Z systems.
cis Isomers
The two largest groups are on the same side of the double bond.
trans Isomers
The two largest groups are on opposite sides of the double bond.
E/Z nomenclature
When the cis/trans system is ambiguous, the E/Z system can be used where E = entgegan ("trans") and Z = zusamen ("cis"). The E/Z system is used to prioritize when there are 3 or 4 different non-hydrogen atoms are attached to the vinyl carbons (carbons sharing the double bond). This system bases priority on the atomic number (Z) and/or atomic mass (A) of the atoms bonded to the vinyl carbons. An atom attached by a multiple bond is counted once for each bond. If there is a tie in priority, then move to the next atom along each chain until a difference occurs. Atomic number has higher priority than atomic mass. Atomic mass is used to establish priority for isotopes, therefore deuterium (D) has higher priority than hydrogen (H).
For example, when comparing atoms bonded to the vinylic carbons in the the compound below,
we would rank the priority as
fluorine atom > propyl group > ethyl group > methyl
Z = 9 > 3 x C chain > 2 x C chain > 1 x C chain
and name the compound ((2E)-3-ethyl-2-fluorohex-2-ene.
The double bond of the allylic group creates higher priority over a simple propyl grup such that -CH2-CH=CH2 > -CH2CH2CH3.
For straight chain alkenes, it is the same basic rules as nomenclature of alkanes except change the suffix to "-ene."
1. Find the Longest Carbon Chain that Contains the Carbon Carbon double bond. If you have two ties for longest Carbon chain, and both chains contain a Carbon Carbon double bond, then identify the most substituted chain.
2. Give the lowest possible number to the Carbon Carbon double bond.
a) Do not need to number cycloalkenes because it is understood that the double bond is in the one position.
b) Alkenes that have the same molecular formula but the location of the double bonds are different means they are constitutional isomers.
c) Functional Groups with higher priority determine the suffix
3. Add substituents and their position to the alkene as prefixes. Of course remember to give the lowest numbers possible. And remember to name them in alphabetical order when writting them.
4. Next is identifying stereoisomers - cis/trans. when there are only two non hydrogen attachments to the alkene then use cis and trans to name the molecule.
For example, the compound below is a cis isomer. It has both the substituents going upward. This molecule would be called (cis) 5-chloro-3-heptene.
5. Next is identifying stereoisomers - E/Z if cis/trans is ambiguous or skip 4 above and jump straight to E/Z.
For example, if we look at the "trans" alternative lo the previous compound, the cis/trans system cannot be applied.
cis or trans?
This molecule would be called (3E)-4-chlorohept-3-ene. It is E because the ethyl group (-CH2CH3) has the higher priority for the vicinal carbon on the left and the chlorine atom has the higher priority for the vicinal carbon on the right and these two groups are on opposite sides of the double bond.
6. An example of functional group priorities in nomenclature is that the hydroxyl group gets precedence (has higher priority) over the double bond.
Therefore, alkenes containing alcohol groups are called alkenols with the sufffix --enol.
7. Lastly remember that alkene substituents are called alkenyl. Suffix --enyl.
Here is a chart containing the systemic name for the first twenty straight chain alkenes.
Name Molecular formula
Ethene C2H4
Propene C3H6
Butene C4H8
Pentene C5H10
Hexene C6H12
Heptene C7H14
Octene C8H16
Nonene C9H18
Decene C10H20
Undecene C11H22
Dodecene C12H24
Tridecene C13H26
Tetradecene C14H28
Pentadecene C15H30
Hexadecene C16H32
Heptadecene C17H34
Octadecene C18H36
Nonadecene C19H38
Eicosene C20H40
Did you notice how there is no methene? Because it is impossible for a carbon to have a double bond with nothing.
Common names
The carbon atoms sharing the double bond can be referred to as the "vinyl carbons". The carbon atoms adjacent to the vinyl carbon atoms are called "allylic carbons". These carbon atoms have unique reactivity because of the potential for interaction with the pi bond.
Overall, remove the -ane suffix and add -ylene.
There are a couple of unique ones like ethenyl's common name is vinyl and 2-propenyl's common name is allyl, that you should know are...
• vinyl substituent H2C=CH-
• allyl substituent H2C=CH-CH2-
• allene molecule H2C=C=CH2
• isoprene is shown below
Endocyclic & Exocyclic Alkenes
Endocyclic double bonds have both carbons in the ring and exocyclic double bonds have only one carbon as part of the ring.
Cyclopentene is an example of an endocyclic double bond.
Methylenecylopentane is an example of an exocyclic double bond.
For example, when naming the compound below, the methyl group is considered when numbering the double bond.
The compound can be named 1-methylcyclobutene or 1-methylcyclobut-1-ene.
When naming this next compound, the ethenyl group is considered when numbering the double bond.
The IUPAC name for the compound can be 1-ethenylcyclohexene or 1-ethenylcyclohex-1-ene. The common name would be 1-vinylcyclohexene. For the compound below, the name is 2-vinyl-1,3-cyclohexadiene.
Exercise
1. Give the IUPAC name for the following compounds. When stereochemistry is included, write the name using both the cis/trans and E/Z names if possible.
2. Draw the bond-line structures for the following compounds.
a) trans-2-pentene
b) (Z)-3-heptene
c) 4-methyl-2-pentene
d) (Z)-5-Chloro-3-ethly-4-hexen-2-ol.
Answer
1. a) 1-pentene or pent-1-ene
b) 2-ethyl-1-hexene or 2-ethylhex-1-ene (parent chain must include the double bond)
c) cis-2-hexene or (Z)-2-hexene or (2Z)-hex-2-ene
d) (2E)-3-methylpent-2-ene or (E)-3-methyl-2-pentene (cis/trans cannnt be applied)
e) trans-2-methyl-4-octene or (4E)-2-methyloct-4-ene or (E)-2-methyl-4-octene (branch breaks the tie in numbering the parent chain since both directions begin the double bond at carbon 4.
2.
Contributors and Attributions
• S. Devarajan (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.06%3A_Alkenes.txt |
learning objectives
• name alkynes using IUPAC (systematic) and selected common name nomenclature
• draw the structure of alkynes from IUPAC (systematic) and selected common names
Alkynes are organic molecules made of the functional group carbon-carbon triple bonds and are written in the empirical formula of \(C_nH_{2n-2}\). They are unsaturated hydrocarbons. Like alkenes have the suffix –ene, alkynes use the ending –yne; this suffix is used when there is only one alkyne in the molecule.
Straight Chain Alkynes
Here are the molecular formulas and names of the first ten carbon straight chain alkynes.
Name
Molecular Formula
Ethyne
C2H2
Propyne
C3H4
1-Butyne
C4H6
1-Pentyne
C5H8
1-Hexyne
C6H10
1-Heptyne
C7H12
1-Octyne
C8H14
1-Nonyne
C9H16
1-Decyne
C10H18
Naming Alkynes
Like previously mentioned, the IUPAC rules are used for the naming of alkynes.
Rule 1
Find the longest carbon chain that includes both carbons of the triple bond.
Rule 2
Number the longest chain starting at the end closest to the triple bond. A 1-alkyne is referred to as a terminal alkyne and alkynes at any other position are called internal alkynes.
For example:
4-chloro-6-diiodo-7-methyl-2-nonyne
Rule 3
After numbering the longest chain with the lowest number assigned to the alkyne, label each of the substituents at its corresponding carbon. While writing out the name of the molecule, arrange the substituents in alphabetical order. If there are more than one of the same substituent use the prefixes di, tri, and tetra for two, three, and four substituents respectively. These prefixes are not taken into account in the alphabetical order.
For example:
1-triiodo-4-dimethyl-2-nonyne
If there is an alcohol present in the molecule, number the longest chain starting at the end closest to it, and follow the same rules. However, the suffix would be –ynol, because the alcohol group takes priority over the triple bond.
5- methyl-7-octyn-3-ol
When there are two triple bonds in the molecule, find the longest carbon chain including both the triple bonds. Number the longest chain starting at the end closest to the triple bond that appears first. The suffix that would be used to name this molecule would be –diyne.
For example:
4-methyl-1,5-octadiyne
Rule 4
Substituents containing a triple bond are called alkynyl.
For example:
1-chloro-1-ethynyl-4-bromocyclohexane
Here is a table with a few of the alkynyl substituents:
Name
Molecule
Ethynyl
-C?CH
2- Propynyl
-CH2C?CH
2-Butynyl
-CH3C?CH2CH3
Rule 5
A molecule that contains both double and triple bonds is called an alkenyne. The chain can be numbered starting with the end closest to the functional group that appears first. For example:
6-ethyl-3-methyl-1,4-nonenyne
Common Names
The more commonly used name for ethyne is acetylene, which used industrially.
Similar to the allylic carbon position of alkenes, the carbons bonded to the alkyne carbons are called "propargyl" carbons and also have differences in chemical reactivity because of the interaction of the two pi bonds with the propargyl carbons.
HC≡CH–CH2 Propargyl group
Exercise
1. Briefly identify the important differences between an alkene and an alkyne. How are they similar?
2. The alkene (CH3)2CHCH2CH=CH2 is named 4-methyl-1-pentene. What is the name of (CH3)2CHCH2C≡CH?
3. Do alkynes show cis-trans isomerism? Explain.
4. Draw the bond-line structure & write the condensed structural formula for each compound except (a). For part (a) write the condensed formula and full Lewis (Kekule) structure.
a) acetylene
b) 3-octyne
c) 3-methyl-1-hexyne
d) 4,4-dimethyl-2-pentyne
e) trans-3-hepten-1-yne
5. Give the IUPAC (Systematic) name for each compound.
a) CH3CH2CH2C≡CH
b) CH3CH2CH2C≡CCH3
Answer
1. Alkenes have double bonds; alkynes have triple bonds. Both bonds are rigid and do not undergo rotation, however, the pi bonds allow both alkenes and alkynes to undergo addition reactions.
2. 4-methyl-1-pentyne
3. No; a triply bonded carbon atom can form only one other bond and has linear electron geometry so there are no "sides". Allkenes have two groups attached to each inyl carbon with a trigonal planar electron geometry that creates the possibility of cis-trans isomerism.
4.
5. a) 1-pentyne or pent-1-yne
b) 2-hexyne or hex-2-yne
c) 2-methylnon-4-yne or 2-methyl-4-nonyne
d) (4E)-hex-4-en-1-yne or (E)-hex-4-en-1-yne (alkynes have higher priority over alkenes if they occur sooner in the parent chain)
e) pent-1-en-4-yne (alkenes have higher priority over alkynes when they have the same position in the parent chain)
f) pent-4yn-2-ol (alcohols have higher priority than alkynes)
Reference
1. Vollhardt, Peter, and Neil E. Schore. Organic Chemistry: Structure and Function. 5th Edition. New York: W. H. Freeman & Company, 2007.
Contributors and Attributions
• A. Sheth and S. Sujit (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.07%3A_Alkynes.txt |
learning objectives
• classify alcohols as primary, secondary, or tertiary
• name alkanes using IUPAC (systematic) and selected common name nomenclature
• draw the structure of alkanes from IUPAC (systematic) and selected common namesAlcohol Classification
Alcohol Classification
Alcohols may can be classified as primary, , secondary, & tertiary, , in the same manner as alkyl halides. This terminology refers to alkyl substitution of the carbon atom bearing the hydroxyl group (colored blue in the illustration). This classification system is based on the neutral bonding pattern for oxygen. The structure of alcohols requires one of the oxygen bonds to form with hydrogen and the other oxygen bond to form with carbon. All oxygen atoms of alcohols look the same. To distinguish between alcohol classifications, we must look at the carbon atom bonded to the hydroxyl group. The bonding pattern of this carbon atom determines the classification of the alcohol.
Primary alcohols
In a primary (1°) alcohol, the carbon which carries the -OH group is only attached to one alkyl group. Some examples of primary alcohols include:
Notice that it doesn't matter how complicated the attached alkyl group is. In each case there is only one linkage to an alkyl group from the CH2 group holding the -OH group. There is an exception to this. Methanol, CH3OH, is counted as a primary alcohol even though there are no alkyl groups attached to the carbon with the -OH group on it.
Secondary alcohols
In a secondary (2°) alcohol, the carbon with the -OH group attached is joined directly to two alkyl groups, which may be the same or different. Examples:
Tertiary alcohols
In a tertiary (3°) alcohol, the carbon atom holding the -OH group is attached directly to three alkyl groups, which may be any combination of same or different. Examples:
Nomenclature - IUPAC Introduction & Common Names
Alcohols are designated by an ol suffix if they are the highest priority functinal group. For example, ethanol contains a hydroxyl group to form an alcohol as shown in the follwowing condensed formula: CH3CH2OH. Note that a locator number is not needed on a two-carbon chain, but on longer chains the location of the hydroxyl group determines chain numbering and must be specified in the name. For example: CH3CH(OH)CH2CH2CH3 is 2-pentanol or pentan-2-ol.
Other examples of IUPAC nomenclature are shown below, together with the common names often used for some of the simpler compounds. For the mono-functional alcohols, this common system consists of naming the alkyl group followed by the word alcohol.
Many functional groups have a characteristic suffix designator, and only one such suffix (other than "ene" and "yne") may be used in a name. When the hydroxyl functional group is present together with a function of higher nomenclature priority, it must be cited and located by the prefix hydroxy and an appropriate number. For example, lactic acid has the IUPAC name 2-hydroxypropanoic acid.
The terms "vicinal" and "geminal" can be applied to any two functional groups that are part of the same compound. Typically, these terms are first encountered with alcohols. Vicinal is used to describe the structure of a compound in which the two groups are bonded to neighboring carbons. Geminal is used when both functional groups are bonded to the same carbon. In Latin, "gemini" means twins. In the same way that twins are connected to the same mother, geminal groups are bonded to the same carbon. Similarly, the vicinal groups are in vicinity of each other.
Naming Alcohols
1. Find the longest chain containing the hydroxy group (OH). If there is a chain with more carbons than the one containing the OH group it will be named as a subsitutent.
2. Place the OH on the lowest possible number for the chain. With the exception of carbonyl groups such as ketones and aldehydes, the alcohol or hydroxy groups have first priority for naming.
3. When naming a cyclic structure, the -OH is assumed to be on the first carbon unless the carbonyl group is present, in which case the later will get priority at the first carbon.
4. When multiple -OH groups are on the cyclic structure, number the carbons on which the -OH groups reside.
5. Remove the final e from the parent alkane chain and add -ol. When multiple alcohols are present use di, tri, et.c before the ol, after the parent name. ex. 2,3-hexandiol. If a carbonyl group is present, the -OH group is named with the prefix "hydroxy," with the carbonyl group attached to the parent chain name so that it ends with -al or -one.
Examples
Ethane: CH3CH3 ----->Ethanol: (the alcohol found in beer, wine and other consumed sprits)
Secondary alcohol: 2-propanol
Other functional groups on an alcohol: 3-bromo-2-pentanol
Cyclic alcohol (two -OH groups): cyclohexan-1,4-diol
Other functional group on the cyclic structure: 3-hexeneol (the alkene is in bold and indicated by numbering the carbon closest to the alcohol)
A complex alcohol:4-ethyl-3hexanol (the parent chain is in red and the substituent is in blue)
Exercise
1. Give the IUPAC (Systematic) name for each compound. For parts (a)-(c), classify the alcohols as primary, secondary, or tertiary.
Part (d) is a challenge question and sneak preview of coming attractions.
2. Draw the bond line structures, condensed strucutre, and name all the alcohols with the molecular formula C3H9O.
3. Oleic acid, a commonly occurring fatty acid in vegetable oils has the structure below.
a) Describe the stereochemistry of oleic acid - cis or trans?
b) Write the condensed formula for oleic acid.
4. Give the IUPAC name for each compound. For parts (a), (b), (d), and (e) classify the alcohol as primary, secondary, tertiary, or allylic. Part (c) is also a challenge question.
Answer
1. a) 2-butanol or butan-2-ol; secondary
b) 2-methylcyclohexan-1-ol or 2-methyl-1-cyclohexanol; secondary
c) cyclopentylmethanol (alcohol is higher priority over carbon chain or ring size); primary
d) 2-(butan2-yl)-6-tert-butyl-4-(propan-2-yl)phenol or 2-sec-butyl-6-t-butyl-4-isopropylphenol (some common names are recognized by IUPAC)
2.
3. a) cis alkene
b) CH3(CH2)6CHCH(CH2)6CO2H
4. a) 2-methyl-1-cyclohexanol or 2-methyl-cyclohexan-1-ol (no stereochemistry communicated even though it is possible); secondary
b) 1-methyl-1-cyclohexanol or 1-methyl-cyclohexan-1-ol (no steereochemicstry because both groups are bonded to the same carbon); tertiary
c) 4-nitrophenol
d) 2-propen-1-ol (alcohols have higher priority than alkenes so determine numbering and suffix); allylic
e) (1S, 2S)-2-methylcyclopentan-1-ol (For students who have learned about chirality); secondary
There are two equivalent answers for students who have not yet learned about chirlatiy:
trans-1-methyl-2-cyclpentanol or trans-1-methylcyclopentan-1-ol
Contributors
• William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.08%3A_3.8_Alcohols_-__Classification_and_Nomenclature.txt |
learning objectives
• name ethers, epoxides, and sulfides using IUPAC (systematic) and selected common name nomenclature
• draw the structure of ethers, epoxides, and sulfides from IUPAC (systematic) and selected common names
Note: Heterocyclic oxygen compounds are included for the sake of completion. Their nomenclature may or may not be required by the professor are requires additional instruction. Make sure to ask.
Ethers
Ethers are compounds having two alkyl or aryl groups bonded to an oxygen atom, as in the formula R1–O–R2. The ether functional group does not have a characteristic IUPAC nomenclature suffix, so it is necessary to designate it as a substituent. To do so the common alkoxy substituents are given names derived from their alkyl component (below):
Alkyl Group Name Alkoxy Group Name
CH3 Methyl CH3O– Methoxy
CH3CH2 Ethyl CH3CH2O– Ethoxy
(CH3)2CH– Isopropyl (CH3)2CHO– Isopropoxy
(CH3)3C– tert-Butyl (CH3)3CO– tert-Butoxy
C6H5 Phenyl C6H5O– Phenoxy
The smaller, shorter alkyl group becomes the alkoxy substituent. The larger, longer alkyl group side becomes the alkane base name. Each alkyl group on each side of the oxygen is numbered separately. The numbering priority is given to the carbon closest to the oxgen. The alkoxy side (shorter side) has an "-oxy" ending with its corresponding alkyl group. For example, CH3CH2CH2CH2CH2-O-CH2CH2CH3 is 1-propoxypentane. If there is cis or trans stereochemistry, the same rule still applies.
Example
Examples are: CH3CH2OCH2CH3, diethyl ether (sometimes referred to as ether), and CH3OCH2CH2OCH3, ethylene glycol dimethyl ether (glyme).
Common names
Simple ethers are given common names in which the alkyl groups bonded to the oxygen are named in alphabetical order followed by the word "ether". The top left example shows the common name in blue under the IUPAC name. Many simple ethers are symmetrical, in that the two alkyl substituents are the same. These are named as "dialkyl ethers". If we read the word "ether", the author is most likely communicating the compound CH3CH2OCH2CH3, ethoxyethane (diethyl ether), but we do not know with certainty - another example of the importance of accurate nomenclature.
Epoxides
An epoxide is a cyclic ether with three ring atoms. These rings approximately define an equilateral triangle, which makes it highly strained. The strained ring makes epoxides more reactive than other ethers. Simple epoxides are named from the parent compound ethylene oxide or oxirane, such as in chloromethyloxirane. As a functional group, epoxides feature the epoxy prefix, such as in the compound 1,2-epoxycycloheptane, which can also be called cycloheptene epoxide, or simply cycloheptene oxide.
A generic epoxide.
The chemical structure of the epoxide glycidol, a common chemical intermediate
A polymer formed by reacting epoxide units is called a polyepoxide or an epoxy. Epoxy resins are used as adhesives and structural materials. Polymerization of an epoxide gives a polyether, for example ethylene oxide polymerizes to give polyethylene glycol, also known as polyethylene oxide.
Sulfides (Thioethers)
A thioether is a functional group in organosulfur chemistry with the connectivity C-S-C as shown below. Like many other sulfur-containing compounds, volatile thioethers have foul odors.[1] A thioether is similar to an ether except that it contains a sulfur atom in place of the oxygen. The grouping of oxygen and sulfur in the periodic table suggests that the chemical properties of ethers and thioethers are somewhat similar.
General structure of a thioether with the blue marked functional group.
Nomenclature
Thioethers are sometimes called sulfides, especially in the older literature and this term remains in use for the names of specific thioethers. The two organic substituents are indicated by the prefixes. (CH3)2S is called dimethylsulfide. Some thioethers are named by modifying the common name for the corresponding ether. For example, C6H5SCH3 is methyl phenyl sulfide, but is more commonly called thioanisole, since its structure is related to that for anisole, C6H5OCH3.
Structure and properties
Thioether is an angular functional group, the C-S-C angle approaching 90°. The C-S bonds are about 180 pm.
Thioethers are characterized by their strong odors, which are similar to thiol odor. This odor limits the applications of volatile thioethers. In terms of their physical properties they resemble ethers but are less volatile, higher melting, and less hydrophilic. These properties follow from the polarizability of the divalent sulfur center, which is greater than that for oxygen in ethers.
Heterocycles with Oxygen
In cyclic ethers (heterocycles), one or more carbons are replaced with oxygen. Often, it's called heteroatoms, when carbon is replaced by an oxygen or any atom other than carbon or hydrogen. In this case, the stem is called the oxacycloalkane, where the prefix "oxa-" is an indicator of the replacement of the carbon by an oxygen in the ring. These compounds are numbered starting at the oxygen and continues around the ring. For example,
If a substituent is an alcohol, the alcohol has higher priority. However, if a substituent is a halide, ether has higher priority. If there is both an alcohol group and a halide, alcohol has higher priority. The numbering begins with the end that is closest to the higher priority substituent. There are ethers that are contain multiple ether groups that are called cyclic polyethers or crown ethers. These are also named using the IUPAC system.
Thiophenes
Thiophenes are a special class of thioether-containing heterocyclic compounds. Because of their aromatic character, they are non-nucleophilic. The nonbonding electrons on sulfur are delocalized into the π-system. As a consequence, thiophene exhibits few properties expected for a thioether - thiophene is non-nucleophilic at sulfur and, in fact, is sweet-smelling. Upon hydrogenation, thiophene gives tetrahydrothiophene, C4H8S, which indeed does behave as a typical thioether.
Example
Examples of ethers include CH3CH2OCH2CH3, diethyl ether (sometimes referred to as ether), and CH3OCH2CH2OCH3, ethylene glycol dimethyl ether (glyme).
Common names
Simple ethers are given common names in which the alkyl groups bonded to the oxygen are named in alphabetical order followed by the word "ether". The top left example shows the common name in blue under the IUPAC name. Many simple ethers are symmetrical, in that the two alkyl substituents are the same. These are named as "dialkyl ethers".
• anisole (try naming anisole by the other two conventions.
• oxirane or 1,2-epoxyethane, ethylene oxide, dimethylene oxide, oxacyclopropane,
• furan (this compound is aromatic)
tetrahydrofuran
oxacyclopentane, 1,4-epoxybutane, tetramethylene oxide,
• dioxane or 1,4-dioxacyclohexane
Exercise
Give the IUPAC and common name (if possible) for each compound respectively.
Answer
1. ethoxyethane; diethyl ether
2. 2-ethoxy-2-methyl-propane; etheyl t-butyl ether (ethyl tert-butyl ether)
3. cis-1-ethoxy-2-methoxycyclopentane; no common name possible
4. 1-ethoxy-1-methylcyclohexane; no common name possible
5. 1,2-epoxyethane; ethylene oxide or dimethylene oxide or oxacyclopropane or oxirane
6. 2,2-dimethyloxirane
Contributors and Attributions
• Wikipedia (used with permission) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.09%3A_Ethers_Epoxides_and_Sulfides.txt |
learning objectives
• name benzene and its derivatives using IUPAC (systematic) and selected common name nomenclature
• draw the structures of benzene and its derivatives from IUPAC (systematic) and selected common names
Benzene, as shown on the left, is an organic aromatic compound with many interesting properties. Unlike aliphatic (straight chain carbons) or other cyclic organic compounds, the structure of benzene (3 conjugated π bonds) allows benzene and its derived products to be useful in fields such as health, laboratory synthesis, and other applications such as rubber synthesis. Benzene is unique because we can write a condensed formula for its ringed structure, C6H6.
Introduction
Benzene derived products are well known to be pleasantly fragrant. For this reason, organic compounds containing benzene rings were classified as being "aromatic" (sweet smelling) amongst scientists in the early 19th century when a relation was established between benzene derived compounds and sweet/spicy fragrances. There is a misconception amongst the scientific community, however, that all aromatics are sweet smelling and that all sweet smelling compounds would have a benzene ring in its structure. This is false, since non-aromatic compounds, such as camphor, extracted from the camphor laurel tree, release a strong, minty aroma, yet it lacks the benzene ring in its structure (See figure 1). On the other hand, benzene itself gives off a rather strong and unpleasant smell that would otherwise invalidate the definition of an aromatic (sweet-smelling) compound. Despite this inconsistency, however, the term aromatic continues to be used today in order to designate molecules with benzene-like rings in their structures. For a modern, chemical definition of aromaticity, refer to sections Aromaticity and Hückel's Rule.
Figure 1. Top-view of camphor, along with its monoterpene unit. Notice how camphor lacks the benzene ring to be "aromatic".
Many aromatic compounds are however, sweet/pleasant smelling. Eugenol, for example, is extracted from essential oils of cloves and it releases a spicy, clove-like aroma used in perfumes. In addition, it is also used in dentistry as an analgesic.
Figure 2. Eugenol, an aromatic compound extracted from clove essential oils. Used in perfumes and as an analgesic.
The benzene ring is labeled in red in the eugenol molecule.
Is it cyclohexane or is it benzene?
Due to the similarity between benzene and cyclohexane, the two is often confused with each other in beginning organic chemistry students.
If you were to count the number of carbons and hydrogens in cyclohexane, you will notice that its molecular formula is C6H12. Since the carbons in the cyclohexane ring is fully saturated with hydrogens (carbon is bound to 2 hydrogens and 2 adjacent carbons), no double bonds are formed in the cyclic ring. In contrast, benzene is only saturated with one hydrogen per carbon, leading to its molecular formula of C6H6. In order to stabilize this structure, 3 conjugated π (double) bonds are formed in the benzene ring in order for carbon to have four adjacent bonds.
In other words, cyclohexane is not the same as benzene! These two compounds have different molecular formulas and their chemical and physical properties are not the same. The hydrogenation technique can be used by chemists to convert from benzene to cyclohexane by saturating the benzene ring with missing hydrogens.
IMPORTANT NOTE: A special catalyst is required to hydrogenate benzene rings due to its unusual stability and configuration. Normal catalytic hydrogenation techniques will not hydrogenate benzene and yield any meaningful products.
What about Resonance?
Benzene can be drawn a number of different ways. This is because benzene's conjugated pi electrons freely resonate within the cyclic ring, thus resulting in its two resonance forms.
Figure 4. The figure to the left shows the two resonance forms of benzene. The delocalized electrons are moved from one carbon to the next, thus providing stabilization energy. Ring structures stabilized by the movement of delocalized electrons are sometimes referred to as arenes.
As the electrons in the benzene ring can resonate within the ring at a fairly high rate, a simplified notation is often used to designate the two different resonance forms. This notation is shown above, with the initial three pi bonds (#1, #2) replaced with an inner ring circle (#3). Alternatively, the circle within the benzene ring can also be dashed to show the same resonance forms (#4).
The Phenyl Group (The Foundation of Benzene Derivatives)
The phenyl group is formed by removing one hydrogen from benzene to create the fragment is C6H5. NOTE: Although the molecular formula of the phenyl group is C6H5, the phenyl group would always have something attached to where the hydrogen was removed. Thus, the formula is often written as Ph-R, where Ph refers to the phenyl group and R refers to the R group attached where the hydrogen was removed.
Figure 5. Figure demonstrating the removal of hydrogen to form the phenyl group.
Different R groups on the phenyl group allows different benzene derivatives to be formed. Phenol, Ph-OH, or C6H5OH, for example, is formed when an alcohol (-OH) group displaces a hydrogen atom on the benzene ring. Benzene, for this very same reason, can be formed from the phenyl group by reattaching the hydrogen back its place of removal. Thus benzene, similar to phenol, can be abbreviated Ph-H, or C6H6.
As you can see above, these are only some of the many possibilities of the benzene derived products that have special uses in human health and other industrial fields.
Nomenclature of Benzene Derived Compounds
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 7. Example of simple benzene naming with chlorine and NO2 as substituents.
Figure 8. More complicated simple benzene naming examples - Note that standard nomenclature priority rules are applied here, causing the numbering of carbons to switch.
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)
Continuing with the example above in figure 8 (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:
Derivatives as Parent Names
In addition to simple benzene naming and OMP nomenclature, benzene derivtives are also sometimes used as the "parent" in the name of a larger compound. name.
For example, phenol (C6H5OH) is the parent name of the compound below because hydroxyl groups have higher nomenclature priority than halides. The chlorine atom is considered a branch at the ortho- position to the hydroxyl group. Accordingly, the compound is named 2-chlorophenol or o-chlorophenol.
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.
In order 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 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:
Exercise
1. True or False? The compound below contains a benzene ring and thus is aromatic.
2. Benzene unusual stability is caused by a combination of the _________________ conjugated pi bonds in its cyclic ring.
specify the number
3. 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?
4. Predict the product of the reaction below or explain why no reaction occurs.
5. At normal conditions, benzene has ___ resonance structures.
6. 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
7. Convert 1,4-dimethylbenzene into its common name.
8. TNT's common name is: ______________________________
9. Name the following compound using OMP nomenclature:
10. Draw the structure of 2,4-dinitrotoluene.
11. Give the IUPAC name the following compound:
12. 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
13. a) True or False? Benzyl chloride can be abbreviated Bz-Cl.
b) What is the condensed formula for benzyl chloride?
c) What is the condensed formula for phenyl chloride?
14. Benzoic Acid has what R group attached to its phenyl functional group?
15. True or False? A single aromatic compound can have multiple names indicating its structure.
16. List the corresponding positions for the OMP system (o-, m-, p-).
17. 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
18. 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
1. False, this compound does not contain a benzene ring in its structure.
2. 3
3. No, a substance that is fragrant does not imply a benzene ring is in its structure. See camphor example (figure 1)
4. No reaction, benzene requires a special catalyst to be hydrogenated due to its unusual stability given by its three conjugated pi bonds.
5. 2
6. b, d
7. p-xylene
8. 2,4,6-trinitrotoluene
9. p-chloronitrobenzene
10.
11. 4-phenylheptane
12. a) 3,4-difluorobenzyl bromide
13. a) False, the correct abbreviation for the benzyl group is Bn, not Bz. The correct abbreviation for Benzyl chloride is Bn-Cl.
b) C6H5CH2Cl.
c) C6H5Cl.
14. COOH or CO2H
15. True. TNT, for example, has the common name 2,4,6-trinitrotoluene and its systematic name is 2-methyl-1,3,5-trinitrobenzene.
16. Ortho - 1,2 ; Meta - 1,3 ; Para - 1,4
17. 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. The chemical formula would allow a determination.
18. d
• David Lam | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.10%3A_Benzene_and_its_Derivatives.txt |
learning objectives
• name aldehydes and ketones using IUPAC (systematic) and selected common name nomenclature
• draw the structure of aldehydes and ketones from IUPAC (systematic) and selected common names
Aldehydes and ketones contain the carbonyl 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.
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. When writing the condensed formulas for aldehydes and ketones, it is important to note that the carbonyl bond is not drawn. It must be recognized. The generic condensed formula for aldehydes is RCHO (CHO is our aldehyde CHUM) and RCOR' for ketones (no cute memorization aid - if you have one please share it.)
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.
Aldehyde Common Names to Memorize
Aldehydes often called the formyl groups. There are some common names that are still used and need to be memorized. Recognizing the patterns can be helpful.
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.
Ketone Common Names to Memorize
There are some common names that are still used and need to be memorized. Recognizing the patterns can be helpful.
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
Exercise
1. Give the IUPAC name for each compound and write the condensed formulas for parts (a), (b), (c), (d), (e), (f), (h), (i), and (l).
2) Draw the bond-line structure and write the condensed formula {except for (b), (d) and (h)} corresponding to each name:
A) butanal
B) 2-hydroxycyclopentanone
C) 2,3-pentanedione
D) 1,3-cyclohexanedione
E) 4-hydoxy-3-methyl-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
Answer
Solutions
1.
a) 3,4-dimethylhexanal; CH3CH2CH(CH3)CH(CH3)CH2CHO
b) 5-bromo-2-pentanone; CH2BrCH2CH2COCH3
c) 2,4-hexanedione; CH3COCH2COCH2CH3
d) cis-3-Penenal; cis-CH3CHHCH2CHO
e) 6-methyl-5-hepten-3-one; CH3C(CH3)CHCH2COCH2CH3 or (CH3)2CCHCH2COCH2CH3
f) 3-hydroxy-2,4-pentanedione; CH3OCH(OH)COCH3
g) 1,2-cyclobutanedione
h) 2-methyl-propanedial; CHOCH(CH3)CHO
i) 3-methyl-5-oxo-hexanal; CH3OCH2CH(CH3)CH2CHO
j) cis-2,3-dihydroxycyclohexanone
k) 3-Bromo-2-methylcyclopentanecarboaldehyde
l) 3-bromo-2-methylpropanal; CHOCH(CH3)CH2Br
2. condensed formulas below and bond-line structures to the right
a) CH3CH2CH2CHO
c) CH3COCOCH2CH3
e) CH2(OH)CH(CH3)COCH3
f) CH3CHC(CH3)COCH2CH2CH3
g) CH3COCH2CHO
i) CHOCH2CH2CHO
j) CH3CH2CHCHCH(CH3)CHO | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.11%3A_Aldehydes_and_Ketones.txt |
learning objectives
• classify amines as primary, secondary, tertiary, quaternary, or heterocyclic
• name amines using IUPAC (systematic) and selected common name nomenclature
• draw the structure of amines from IUPAC (systematic) and selected common names
Amine bases are classified according to the number of alkyl or aryl groups attached to nitrogen. Amines are classified differently from alkyl halides and alcohols because nitrogen has a neutral bonding pattern of three bonds with a single lone pair. To classify amines, we look at the nitrogen atom of the amine and count the number of alkyl groups bonded to it. This number is the classification of the amine.
There are two additional classifications of amines. When the nitrogen is double bonded to carbon, then it is call am imine. When nitrogen is part of a ring that includes double bonds, then it is classified as heterocyclic, as seen in the aromatic nitrogen bases shown below.
Nomenclature
Amines are derivatives of ammonia in which one or more of the hydrogens has been replaced by an alkyl or aryl group. Amino compounds can be named either as derivatives of ammonia or as amino-substituted compounds:
The nomenclature of amines is further complicated by the fact that several different nomenclature systems exist, and there is no clear preference for one over the others. 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. The bottom row shows the structures for some common amines that need to be memorized.
• 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).
To be consistent and logical in naming amines as substituted ammonias, they strictly should be called alkanamines and arenamines, according to the nature of the hydrocarbon grouping. Unfortunately, the term alkylamine is used very commonly in place of alkanamine, while a host of trivial names are used for arenamines. We shall try to indicate both the trivial and the systematic names where possible. Some typical amines, their names, and their physical properties are listed in he Table below. The completely systematic names give in the Table illustrate the difficulty one gets into by using completely systematic names, and why simpler but less systematic names continue to be used for common compounds. A good example is \(\ce{N}\),\(\ce{N}\)-dibutylbutamine versus tributylamine. The special ways of naming heterocyclic amines can be referenced in the appendix of this chapter.
Common Amines and Their Properties
Alkaloids
Many biologically important compounds are amines. Alkaloids are amines synthesized by plants to protect them from being eaten. Humans primarily use alkaloids medicinally as pain killers. All alkaloids are toxic and addictive. The Greeks killed Socrates with (S)-coniine. Mild cases of alkaloid poisoning produce psychological effects resembling peacefulness, euphoria or hallucinations.
Ammonium Salts
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. Salts of amines with inorganic or organic acids are named as substituted ammonium salts, except when the nitrogen is part of a ring system. Examples are
\(^1\)Note the use of azonia to denote the cationic nitrogen in the ring, whereas aza is used for neutral nitrogen.
Perhaps the most noteworthy aspect of ammonium salts is that they have low odor and are water soluble. These qualities are explored more fully in the amine chapter
Heterocyclic Amine Nomenclature
Heterocyclic amines are amines in which the nitrogen is part of a ring that contains at least one double bond. 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. For further details, refer to the appendix of this chapter for the full IUPAC rules of organic compound nomenclature.
Exercise
1. Draw the bond-line structure for each compound and write the condensed structural formula for parts (a) & (d) - (g).
a) 3-bromo-pentan-2-amine
b) cyclopentanamine
c) trans-3-ethylcyclohexanamine
d) sec-butyl tert-butyl amine
e) N,N-dimethyl-3-pentanamine
f) 4-methyl-2-hexanamine
g) 6-bromo-4-amino-2-heptanol
2. Give the IUPAC name and condensed structural formula for each compound.
Answer
1.
2.
Contributors and Attributions
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_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.12%3A_Amines_-_Classification_and_Nomenclature.txt |
learning objectives
• name carboxylic acids using IUPAC (systematic) and selected common name nomenclature
• draw the structure of carboxylic acids from IUPAC (systematic) and selected common names
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 that were chosen by chemists to usually describe the origin of the compound. 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
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.
Common Names for selected dicarboxylic acids
The following common names for these selected dicarboxylic acids are important to memorize; they are prevalent in biochemistry or industrial applications.
The saying, "Oh my, such good apple pie!", can help us remember these common names by correlating the first letters of each word with the common names:
oxalic acid, malonic acid, succinic acid, glutaric acid, adipic acid, and phthalic.
Exercise
1. Draw the bond-line structure and write the condensed structural formula for each compound.
a) octanoic acid
b) 4-hydroxypentanoic acid
c) cis-4-hexenoioc acid or cis-hex-4-enoic acid
d) (E)-5-bromo-3-heptenoic acid or (E)-5-bromohept-3-enoic acid.
e) 2-aminopropanoic acid
2. Give the IUPAC name and condensed structural formula for each compound.
Answer
1.
2. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.13%3A_Carboxylic_Acids.txt |
learning objectives
• name acid halides, anhydrides, esters, amides, nitriles, and dicarboxylic acids using IUPAC (systematic) and selected common name nomenclature
• draw the structure of acid halides, anhydrides, esters, amides, and nitriles from IUPAC (systematic) and selected common names
Note: Nomenclature of thioesters and phosphoesters is also discussed. Ask the professor if this information is required for your course.
Introduction
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. The carboxylic acid derivatives can all be hydrolyzed to carboxylic acids. The specific reaction conditions are discussed in the corresponding chapter later in this text, however, the shared pattern of chemical reactivity is summarized in the diagram below.
IUPAC Nomenclature - One Pattern, so Many Variations
Apply the IUPAC nomenclature format to carboxylic acid derivatives as summarized below using the suffix or substituent names listed in the table. Some students find esters challenging to name. Carboxylates can be described as independent ions, but require a contain to form compounds. It can be helpful to think of esters as "alkylated carboxylates": identify and name the carboxylate, this name is preceded by the alkyl group branch on the carboxyl oxygen.
Functional Group Structure Suffix Name Substituent Name
carboxylic acid -oic acid carboxy-
carboxylate -oate see above
ester -oate alkoxycarbonyl-
dicarboxylic acid -dioic acid not applicable
acyl halide -oyl halide not applicable
anhydride
-anhydride
not applicable
amide -amide amido-
nitrile -nitrile cyano-
Common names
Most common names were derived from older systems of nomenclature that some may argue were "not systematic at all". However, it is helpful to note that the older systems of nomenclature were often based on shared structural features and/or chemical reactivity. Understanding the older nomenclature systems can offer insights into chemical reactivity and structural patterns. There are some common names that are so prevalent, they need to be memorized. Common names frequently exist when the group bonded to the carbonyl carbon is a methyl group (indicated with "acet" or "acetyl" in the common name) or a hydrogen atom (indicated with "formyl" or "form"). For example CH3C≡N is ethanenitrile (or acetonitrile) and HCONH2 is methanamide (or formamide).
The common names for compounds with carbonyl groups often use Greek letters to specify the carbon position relative to the carbonyl carbon.
Example
In this text, we will learn about the patterns of reactivity for compounds with for beta-hydroxy carbonyl structures like the one shown below.
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
Nomenclature of Anhydrides
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.
Symmetrical anhydrides
A symmetrical anhydride is a carboxylic acid anhydride that has the following general structural formula.
where R1=R2= hydrogen atoms, alkyl groups, aryl groups
Unsymmetrical Anhydrides
A mixed or unsymmetrical anhydride is a carboxylic acid anhydride that has the following general structural formula.
where R1≠R2, but are hydrogen atoms, alkyl groups, aryl groups. When naming unsymmetrical acid anhydrides, name both using alkanoic general method and then put the two names alphabetically. Hence, first name each component and alphabetically arranged them followed by spaces and then the word anhydride.
propanoic anhydride
ethanoic propanoic anhydride
Exercises
1. Draw the bond-line structure for benzoic anhydride.
Solution:
2. What is the common name for the compound below?
Solution: acetic benzoic anhydride
Common anhydride names to 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
Exercises
3. Draw the bond-line structure for phenyl hexanoate.
Solution
4. What is the IUPAC name for the compound below?
Solution
methyl benzoate
Nomenclature of Lactones (Cyclic Esters)
Cyclic esters are called lactones. A Greek letter identifies the location of the alkyl oxygen relative to the carboxyl carbonyl group.
Nomenclature of Amides
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)
Exercises
5. Draw the bond-line structure for 3-chloro-N-ethylbenzamide.
Solution
6. What is the IUPAC name the compound below?
Solution
pentanamide
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
Exercises
7. Draw the bond-line structure for N,N-dimethylformamide.
Solution
8. What are the IUPAC and common names the compound below?
Solution
N-phenylethanamide and N-phenylacetamide, respectively
Cyclic amides are called lactams. A Greek letter identifies the location of the nitrogen on the alkyl chain relative to the carboxyl carbonyl group.
Nomenclature of Nitriles
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. A nitrile substituent, e.g. on a ring, is named carbonitrile.
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
(3-methylbutanenitrile (or isovaleronitrile) cyclopentanecarbonitrile
Nomenclature of Thioesters
Thiosters are made from a carboxylic acid and an thiol.
Thioesters 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 from the carboxylic acid part of the thioester named as an alkane with the ending –thiooate.
Example
Nomenclature of Phosphates
Phosphoryl groups are derivatives of phosphoric acid, a strong acid that is commonly used in the laboratory. The fully deprotonated conjugate base of phosphoric acid is called a phosphate ion, or inorganic phosphate (often abbreviated 'Pi'). When two phosphate groups are linked to each other, the linkage is referred to as a 'phosphate anhydride', and the ion is called 'inorganic pyrophosphate' (abbreviation PPi).
When a phosphate ion is attached to a carbon atom on an organic molecule, the chemical linkage is referred to as a phosphate ester, and the whole species is called an organic monophosphate. Glucose-6-phosphate is an example.
If an organic molecule is linked to two or three phosphate groups, the resulting species are called organic diphosphates and organic triphosphates.
Isopententyl diphosphate and adenosine triphosphate (ATP) are good examples:
Oxygen atoms in phosphate groups are referred to either 'bridging' and 'non-bridging', depending on their position. An organic diphosphate has two bridging and five non-bridging oxygens.
When a single phosphate is linked to two organic groups, the term 'phosphate diester' is used. The backbone of DNA is composed of phosphate diesters.
The term 'phosphoryl group' is a general way to refer to all of the phosphate-based groups mentioned in the paragraphs above.
Recall that phosphate groups on organic structures are sometimes abbreviated simply as 'P', a convention that we will use throughout this text. For example, glucose-6-phosphate and isopentenyl diphosphate are often depicted as shown below. Notice that the 'P' abbreviation includes the oxygen atoms and negative charges associated with the phosphate groups.
Exercise
1. Name the following compounds using IUPAC conventions
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
Answer
1.
a) 3-methylpentanoyl chloride
b) 2-cyclopentylacetamide
c) propyl2-methylpropanoate
d) cyclohexylbutanoate
e) tert-butylcyclopentanecarboxylate
f) 1-methylbutylcyclopentane carboxylate
g) N-methyl-3-butenamide
h) (S)-2-hydroxypropanoyl phosphate
i) propyl 2,3-dimethyl-2-butenethioate | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.14%3A__The_Carboxylic_Acid_Derivatives.txt |
Alkane Nomenclature
3-1 Draw the structure for each of the following compounds
(a) sec-butylcyclohexane
(b) 2,2,7-trimethyloctane
(c) 3-ethyl-2,4,5-trimethylhexane
(d) 1-cyclopropylbutane
(e) 3-ethyl-2,5-dimethyl-4-propylheptane
(f) 1-tert-butyl-4-iodocyclohexane
(g) trans-1,4-diisopropylcyclohexane
(h) 3-bromo-1-cyclopropyl-2-methylbutane
(i) Z-1,2-dichloro-1-methylcyclopentane
(j) 1-chloro-3-ethyl-3-methylhexane
(k) Norbornane (Bicyclo[2.2.1]heptane)
(l) 1-[2-(bromomethyl)cyclopropyl]-4-tert-butylcyclohexane
3-2 Give the IUPAC names of the following alkanes.
(a) (CH3)2CHCH2CH2CH(CH2CH3)CH(CH3)CH2CH3
(b)
(c)
(d)
(e)
(f)
(g)
(h)
3-3 The following names are incorrect or incomplete, but they represent the real structures. Draw each structures and name it correctly.
(a) 2-ethyl-3-methylhexane
(b) 2-isopropylpentane
(c) 2-bromo-5-isopropylheptane
(d) 4-ethyl-3-isopropylheptane
(e) 2,4-trimethylpentane
(f) 1-chloro-2-ethyl-5-methylcyclohexane
Alkyl Halide Nomenclature
3-4 Draw the structures of these following compounds
a) 2-chlorohexane
b) 1-bromo-2-chlorocyclopentane
c) Isobutyl chloride
d) 5-chloro-2,3-dimethylhexane
e) (2S) 2-bromohexane
3-5 Give IUPAC names to the following compounds
a)
b)
c)
d)
3-6 Classify each of the following compounds as primary, secondary, or tertiary alkyl halides
a)
b)
c)
d)
Alkene Nomenclature
3-7 Give the IUPAC name for the following alkene structures.
3-8 Give the structures of the following compounds.
a) [1,1'-bi(cyclohexan)]-1-ene
b) 2,3-dimethylbut-2-ene
c) (4E)-4-ethylidene-2-methylhept-1-ene
d) [(1E,3E)-hexa-1,3-dien-1-yl]cyclopentane
3-9 State whether the following compounds are Z, E, or neither.
3-10 Give the IUPAC name for the following structures.
Alkyne Nomenclature
3-11 Draw the bond-line for these following compounds:
a) 3-hexyne
b) 2-bromo-5-octyne
c) 3,3-dimethyl-6-decyne
d) Cyclopentylacetylene
e) 2-chloronon-4-en-6-yne
f) 2,4-heptadiyne
3-12 Give IUPAC names for the following compounds
a) b) c) d)
Alcohol and Phenol Nomenclature
3-13 Classify each alcohol as primary, secondary, or tertiary.
3-14 Give a systematic (IUPAC) name for each alcohol.
3-15 Draw the structures of the following compounds.
a) 2-methylhexan-1-ol
b) 2-methylpentan-2-ol
c) 2-chlorophenol
d) 4-(chloromethyl)hexan-1-ol
e) cyclopent-2-en-1-ol
f) 3-bromo-2-(2-bromoethyl)pentan-1-ol
3-16 Give a systematic (IUPAC) name for each phenol.
Ethers
3-17 Give the IUPAC name for the following chemical structures.
(g) (h) (i)
3-18 Draw structures of the following.
(a) 3-isopropoxypentane (b) 1-(4-chlorophenoxy)-3-methylbenzene (c) 2-(tert-butoxy)-2-methylpropane
3-19 Name the following ethers and sulfides.
(a) (b) (c) (d) (e)
Benzene and its Derivatives
3-20 State wither the following is para, meta, or ortho substituted.
3-21 Name the following compounds.
3-22 Draw the following structures
1. p-chloroiodobenzene
2. m-bromotoluene
3. p-chloroaniline
4. 1,3,5-trimethylbenzene
3-23 Give the IUPAC name for the following benzene derivatives.
3-24 Draw the following molecules given by their IUPAC nomenclature.
1. 5-formyl-2-hydroxybenzoic acid
2. 1,2-diethyl-3-fluorobenzene
3. 4-amino-3-ethyl-5-methylphenol
3-25 Rank the functional groups on the following molecule in order of priority (highest to lowest).
Aldehydes and Ketones
3-26 For the following molecules, give the IUPAC nomenclature.
Amines
3-27 Draw the structures given by the following IUPAC names.
1. 4-methylpentan-2-amine
2. N-ethyl-N-methylpropan-1-amine
3. 6-aminocyclohex-1-ene-1-carboxylic acid
4. (dimethylamino)acetonitrile
Carboxylic Acids
3-28 Give the correct IUPAC nomenclature for the following compounds.
Carboxylic Acid Derivatives
3-29 Name the following compounds using IUPAC conventions
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
3.16: Solutions to Additional Exercises
Alkane Nomenclature
3-1 The structure that corresponds with each name is drawn below.
(a) sec-butylcyclohexane
(b) 2,2,7-trimethyloctane
(c) 3-ethyl-2,4,5-trimethylhexane
(d) 1-cyclopropylbutane
(e) 3-ethyl-2,5-dimethyl-4-propylheptane
(f) 1-tert-butyl-4-iodocyclohexane
(g) trans-1,4-diisopropylcyclohexane
(h) 3-bromo-1-cyclopropyl-2-methylbutane
(i) Z-1,2-dichloro-1-methylcyclopentane
(j) 1-chloro-3-ethyl-3-methylhexane
(k) Norbornane (Bicyclo[2.2.1]heptane)
(l) 1-[2-(bromomethyl)cyclopropyl]-4-tert-butylcyclohexane
3-2 Give the IUPAC names of the following alkanes.
(a) 5-ethyl-2,6-dimethyloctane
(b) 3,4,7-trimethyl-6-(1-methylethyl)nonane
(c) 4-(1-methylpropyl)-2,3-dimethyloctane
(d) 1,3-dimethyl-2-(1-methylethyl)cyclobutane
(e) 1-tert-butyl-4-ethyl-2-methylcyclopentane
(f) trans-1,3-diethylcyclohexane
(g) cis-1,4-di(1-methylethyl)cyclononane
(h) 3-methylhexane
3-3 The following names are incorrect or incomplete, but they represent the real structures. Draw each structures and name it correctly.
(a)
(b)
(c)
(d)
(e)
(f)
Alkyl Halide Nomenclature
3-4
a)
b)
c)
d)
e)
3-5
a) 3-chloropentane
b) 1-chloro-3-methylcyclohexane
c) 5-chloro-2,2-dimethylhexane
d) (1R,2R)-1-bromo-2-methylcyclohexane
3-6
a) Secondary
b) Primary
c) Tertiary
d) Tertiary
3-7
3-8
3-9
a) Z
b) Neither
c) E
3-10
Alkyne Nomenclature
3-11
a)
b)
c)
d)
e)
f)
3-12
a) 1-chloropent-3-yne
b) 2,2-dimethyl-6-methyl-hex-4-yne
c) cyclohexylhept-1-yne
d) 4-bromo-2-chloro-3-methylnon-7-yne
Alcohol and Phenol Nomenclature
3-13
a) Primary
b) Secondary
c) Tertiary
d) Secondary
3-14
3-15
3-16
Ethers
3-17
(g) oxydicyclopentane
(h) 2-phenyloxirane
(i) 1-cyclohexylethane-1-thiol
3-18
(a) (b) (c)
3-19
(a) diisopropylsulfide (b) 1,3-dimethoxybenzene (c) 2-Methyltetrahydro-2H-pyran (d) methyl 3-sulfanylbenzoate (e) methyl(phenyl)sulfide
Benzene and its Derivatives
3-20
A – meta; B – para; C – ortho
3-21
1. 1,3-Dibromobenzene
2. 1-phenyl-4-methylhexane
3. 1,4-Dichloro-2,5-dimethylbenzene
4. 2-methyl-1,3,5-trinitrobenzene. (Also known as trinitrotoluene, or TNT)
13-22
3-23
3-24
3-25 In order of priority: Benzoyl bromide > Amino > Nitro
3-26
3-27
3-28
Carboxylic Acid Derivatives
3-29
1. 3-methylpentanoyl chloride
2. 2-cyclopentylacetamide
3. propyl 2-methylpropanoate
4. cyclohexylbutanoate
5. tert-butyl cyclopentanecarboxylate
6. 1-methylbutylcyclopentane carboxylate
7. N-methyl-3-butenamide
8. (S)-2-hydroxypropanoyl phosphate
9. propyl 2,3-dimethyl-2-butenethioate
3.17: Appendix - IUPAC Nomenclature Rules
Wikipedia Summary
Full Text of IUPAC Rules
Nomenclature 101 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.15%3A__Additional_Exercises.txt |
Learning Objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• distinguish between the different hydrocarbon functional groups - refer to section 4.1
• explain & predict the physical properties of alkanes including relative bp and solubility in a mixture - refer to section 4.2
• interpret and draw the rotation about a carbon-carbon single bond using Newman projections and sawhorse structures - refer to section 4.3 - 4.5
• correlate energies of conformations with rotational energy diagrams and predict the most stable conformations for butane, higher alkanes, cyclohexane, mono-substituted cyclohexanes, and disubstituted cyclohexanes - refer to sections 4.3, 4.3, 4.4, 4.5, and 4.7, 4.8, and 4.10 respectively
• explain the partial rotation of carbon-carbon single bonds in rings - refer to section 4.6
• explain ring strain and its relationship to cycloalkane stability - refer to section 4.6
• draw cyclohexane conformations (chair & boat) - refer to section 4.7
• draw mono-substituted cyclohexane conformers (chair only) - refer to section 4.8
• identify & draw the geometric (cis/trans) isomers of cycloalkanes - refer to section 4.9
• draw di-substituted cyclohexane conformers (chair only) - refer to section 4.10
• recognize and draw the three ways to join two rings - refer to section 4.11
• describe the uses and sources of alkanes - refer to section 4.12
• recognize and distinguish between the two major reactions of alkanes (combustion and halogenation) - refer to section 4.13
• 4.1: Hydrocarbon Functional Groups
Hydrocarbons are organic compounds that consist entirely of carbon and hydrogen atoms. Hydrocarbons can form different functional groups based upon their carbon bonding patterns as alkanes, alkenes, alkynes, or arenes.
• 4.2: Physical Properties of Alkanes
Alkanes are not very reactive and have little biological activity; alkanes are colorless, odorless non-polar compounds.
• 4.3: Structure and Conformations of Alkanes
The carbon-carbon single bonds of alkanes rotate freely. Conformers are the same molecule shown with different sigma bond rotations. Newman projections are one way to communicate bond rotation.
• 4.4: Conformations of Butane
The conformations of butane are studied to introduce the language and energetic considerations of single bond rotation when alkyl group interactions can occur.
• 4.5: Conformations of Higher Alkanes
Pentane and higher alkanes have conformational preferences similar to ethane and butane. Each dihedral angle tries to adopt a staggered conformation and each internal C-C bond attempts to take on an anti conformation to minimize the potential energy of the molecule.
• 4.6: Cycloalkanes and Ring Strain
For cyclic alkanes, only partial rotation of carbon-carbon single bonds can occur. The actual shape of the carbon ring distorts from the traditional geometric shapes to reduce steric hindrance and ring strain to lower the overall potential energy of the molecule.
• 4.7: Cyclohexane Conformations
Cyclohexane rings are notably stable. Understanding the conformations of cyclohexane and their relative energies is helpful when studying the chemistry of simple carbohydrates (monosaccharides).
• 4.8: Conformations of Monosubstituted Cyclohexanes
For monosubstituted cyclohexanes, the axial or equatorial orientation of the substituent influences the overall potential energy of the conformation.
• 4.9: Cis-trans Isomerism in Cycloalkanes
Stereoisomerism is possible for cycloalkanes with two different substituent groups (not counting other ring atoms). Cis and trans isomers are unique compounds.
• 4.10: Conformations of Disubstituted Cyclohexanes
Because six-membered rings are so common among natural and synthetic compounds and its conformational features are rather well understood, we shall focus on the six-membered cyclohexane ring to study the energetic relationship of conformation and overall potential energy.
• 4.11: Joined Rings
Two or more rings can be fused together into a bicyclic or spirocyclic system. The steroid fused ring system is a notable example.
• 4.12: Uses and Sources of Alkanes
The primary sources for alkanes are oil and natural gas. Alkanes are important raw materials for the chemical industry and are used as fuels for motors.
• 4.13: Reactions of Alkanes - a Brief Overview
Alkanes (the most basic of all organic compounds) undergo very few reactions. The two reactions of more importaces is combustion and halogenation, (i.e., substitution of a single hydrogen on the alkane for a single halogen) to form a haloalkane. The halogen reaction is very important in organic chemistry because it opens a gateway to further chemical reactions.
• 4.14: Additional Exercises
This section has additional exercises for the key learning objectives of this chapter.
• 4.15: Solutions to Additional Exercises
This section has the solutions to the additional exercises from the previous section.
04: Structure and Stereochemistry of Alkanes
Learning Objective
• distinguish between the different hydrocarbon functional groups
Saturated vs. Unsaturated Molecules
Hydrocarbons are organic compounds that contain only carbon and hydrogen. The broadest distinction between hydrocarbons is whether they are saturated and unsaturated. Saturated hydrocarbons only contain carbon-carbon single bonds with the maximum number of hydrogens relative to the number of carbon atoms. It can be said that the carbon atoms are "saturated" with hydrogen atoms in the same way a saturated solution has dissolved the maximum amount of solute. Hydrocarbons that contain pi bonds as carbon-carbon double or triple bonds are classified as unsaturated hydrocarbons. Unsaturation indicates that some of the carbon-hydrogen bonds were lost to from pi bonds between carbon atoms. There are less than the maximum number os hydrogens relative to the number of carbon atoms.
1. Saturated hydrocarbons (alkanes) are the simplest of the hydrocarbon species. They are composed entirely of single bonds and are saturated with hydrogen. Saturated hydrocarbons are the basis of petroleum fuels and are found as either linear or branched species.The simplest alkanes have their C atoms bonded in a straight chain; these are called normal alkanes. They are named according to the number of C atoms in the chain. The smallest alkane is methane:
2. Unsaturated hydrocarbons have one or more double or triple bonds between carbon atoms. Those with double bond are called alkenes and those with one double bond have the formula $C_nH_{2n}$ (assuming non-cyclic structures). Those containing triple bonds are called alkynes, with general formula CnH2n-2.
The smallest alkene—ethene—has two C atoms and is also known by its common name ethylene:
The smallest alkyne is ethyne, which is also known as acetylene:
3. For now, we will focus on benzene as the representative aromatic hydrocarbon. Aromatic compounds were first noted for their strong aromas and low chemical reactivity compared to other saturated hydrocarbons. Aromatic compounds will be discussed in greater detail in the second semester to organic chemistry.
Hydrocarbon Functional Groups
The four distinct hydrocarbon functional groups are: alkanes, alkenes, alkynes and arenes. Aromatic compounds derive their names from the fact that many of these compounds in the early days of discovery were grouped because they were oils with fragrant odors.
Alkanes are organic compounds that consist entirely of single-bonded carbon and hydrogen atoms and lack any other functional groups. Alkanes have the general formula CnH2n+2. Alkanes can be subdivided into the following three groups: the linear straight-chain alkanes, branched alkanes, and cycloalkanes. Alkanes are also saturated hydrocarbons. Alkanes are the simplest and least reactive hydrocarbon species containing only carbons and hydrogens. The distinguishing feature of an alkane, making it distinct from other compounds that also exclusively contain carbon and hydrogen, is its lack of unsaturation. That is to say, it contains no double or triple bonds, which are highly reactive in organic chemistry. Though not totally devoid of reactivity, their lack of reactivity under most laboratory conditions makes them a relatively uninteresting, though very important component of organic chemistry. As you will learn about later, the energy confined within the carbon-carbon bond and the carbon-hydrogen bond is quite high and their rapid oxidation produces a large amount of heat, typically in the form of fire.
The general formula for saturated hydrocarbons is CnH2n+2(assuming non-cyclic structures) as shown in hexane (C6H14) below.
Cycloalkanes are hydrocarbons containing one or more carbon rings to which hydrogen atoms are attached. The general formula for a cyclic hydrocarbon containing one ring is CnH2n as shown in cyclohexane (C6H12) below.
Akenes contain at least one carbon-carbon double bond and alkynes contain at least one carbon-carbon triple bond. Alkenes have the general formula CnH2n. Alkynes have the general formula CnH2n-2. The ratio of carbon to hydrogen increased because hydrogen atoms are replaced with pi bonds as shown in trans-2-butene (C4H8) and 2-butyne (C4H6) below. Since both double and triple bonds include pi bonds, Alkenes and alkynes share similar chemical reactivity.
Aromatic hydrocarbons, also known as arenes, are hydrocarbons that have at least one aromatic ring. Aromatic compounds contain the benzene unit. Benzene itself is composed of six C atoms in a ring, with alternating single and double C–C bonds:
Calculating Degrees of Unsaturation (DU)
There are many ways one can go about determining the structure of an unknown organic molecule. Although, nuclear magnetic resonance (NMR) and infrared radiation (IR) are the primary ways of determining molecular structures, calculating the degrees of unsaturation (DU) is useful information. Knowing the degrees of unsaturation tells us the combined number of pi bonds and rings within a compound which makes it easier to figure out the molecular structure.
Degree of Unsaturation (DU) can be calculated with the equation below and the molecular formula
DU= (2C+2+N-X-H)/2
where: C is the number of carbons; N is the number of nitrogens; X is the number of halogens (F, Cl, Br, I); and H is the number of hydrogens from the molecular formula.
As stated before, a saturated molecule contains only single bonds and no rings. Another way of interpreting this is that a saturated molecule has the maximum number of hydrogen atoms possible to be an acyclic alkane. Thus, the number of hydrogens can be represented by 2C+2, which is the general molecular representation of an alkane. As an example, for the molecular formula C3H4 the number of actual hydrogens needed for the compound to be saturated is 8 [2C+2=(2x3)+2=8]. The compound needs 4 more hydrogens in order to be fully saturated (expected number of hydrogens-observed number of hydrogens=8-4=4). Degrees of unsaturation is equal to 2, or half the number of hydrogens the molecule needs to be classified as saturated. Hence, the DoB formula divides by 2. The formula subtracts the number of X's because a halogen (X) replaces a hydrogen in a compound. For instance, in chloroethane, C2H5Cl, there is one less hydrogen compared to ethane, C2H6.
For a compound to be saturated, there is one more hydrogen in a molecule when nitrogen is present. Therefore, we add the number of nitrogens (N). This can be seen with C3H9N compared to C3H8. Oxygen and sulfur are not included in the formula because saturation is unaffected by these elements. As seen in alcohols, the same number of hydrogens in ethanol, C2H5OH, matches the number of hydrogens in ethane, C2H6.
The following chart illustrates the possible combinations of the number of double bond(s), triple bond(s), and/or ring(s) for a given degree of unsaturation. Each row corresponds to a different combination.
• One degree of unsaturation is equivalent to 1 ring or 1 double bond (1 $\pi$ bond).
• Two degrees of unsaturation is equivalent to 2 double bonds, 1 ring and 1 double bond, 2 rings, or 1 triple bond (2 $\pi$ bonds).
DU
Possible combinations of rings/ bonds
# of rings
# of double bonds
# of triple bonds
1
1
0
0
0
1
0
2
2
0
0
0
2
0
0
0
1
1
1
0
3 3 0 0
2 1 0
1 2 0
0 1 1
0 3 0
1 0 1
Remember, the degrees of unsaturation only gives the sum of double bonds, triple bonds and/or rings. For instance, a degree of unsaturation of 3 can contain 3 rings, 2 rings+1 double bond, 1 ring+2 double bonds, 1 ring+1 triple bond, 1 double bond+1 triple bond, or 3 double bonds.
Example
What is the Degree of Unsaturation for benzene?
Solution - "Thinking it through"
The molecular formula for benzene is C6H6. Thus,
DU= 4, where C=6, N=0,X=0, and H=6. 1 DU can equal 1 ring or 1 double bond. This corresponds to benzene containing 1 ring and 3 double bonds.
Even though there are other possible structures with a DU = 4, like the ones shown below. We will learn the benzene rings have unusual stability and occur frequently in the world of organic chemistry. When the DU for a compound is > 4, we can assume the presence of at least one benzene ring.
Exercise
1. Are the following molecules saturated or unsaturated:
1. (b.) (c.) (d.) C10H6N4
2. Using the molecules from 1., give the degrees of unsaturation for each.
3. Calculate the degrees of unsaturation for the following molecular formulas:
1. (a.) C9H20 (b.) C7H8 (c.) C5H7Cl (d.) C9H9NO4
4. Using the molecular formulas from 3, are the molecules unsaturated or saturated.
5. Using the molecular formulas from 3, if the molecules are unsaturated, how many rings/double bonds/triple bonds are predicted?
Answer
1.
(a.) unsaturated (Even though rings only contain single bonds, rings are considered unsaturated.)
(b.) unsaturated
(c.) saturated
(d.) unsaturated
2. If the molecular structure is given, the easiest way to solve is to count the number of double bonds, triple bonds and/or rings. However, you can also determine the molecular formula and solve for the degrees of unsaturation by using the formula.
(a.) 2
(b.) 2 (one double bond and the double bond from the carbonyl)
(c.) 0
(d.) 10
3. Use the formula to solve
(a.) 0
(b.) 4
(c.) 2
(d.) 6
4.
(a.) saturated
(b.) unsaturated
(c.) unsaturated
(d.) unsaturated
5.
(a.) 0 (Remember-a saturated molecule only contains single bonds)
(b.) The molecule can contain any of these combinations (i) 4 double bonds (ii) 4 rings (iii) 2 double bonds+2 rings (iv) 1 double bond+3 rings (v) 3 double bonds+1 ring (vi) 1 triple bond+2 rings (vii) 2 triple bonds (viii) 1 triple bond+1 double bond+1 ring (ix) 1 triple bond+2 double bonds
(c.) (i) 1 triple bond (ii) 1 ring+1 double bond (iii) 2 rings (iv) 2 double bonds
(d.) (i) 3 triple bonds (ii) 2 triple bonds+2 double bonds (iii) 2 triple bonds+1 double bond+1 ring (iv)... (As you can see, the degrees of unsaturation only gives the sum of double bonds, triple bonds and/or ring. Thus, the formula may give numerous possible structures for a given molecular formula.)
• Kim | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.01%3A_Hydrocarbon_Functional_Groups.txt |
Learning Objective
• explain & predict the physical properties of alkanes including relative bp and solubility in a mixture
Overview
Alkanes are not very reactive and have little biological activity; all alkanes are colorless and odorless non-polar compounds. The relative weak London dispersion forces of alkanes result in gaseous substances for short carbon chains, volatile liquids with densities around 0.7 g/mL for moderate carbon chains, and solids for long carbon chains. The differences in the physical states occurs because there is a direct relationship between the size and shape of molecules and the strength of the intermolecular forces (IMFs).
Because alkanes have relatively predictable physical properties and undergo relatively few chemical reactions other than combustion, they serve as a basis of comparison for the properties of many other organic compound families. Let’s consider their physical properties first.
Boiling Points
Table \(1\) describes some of the properties of some of the first 10 straight-chain alkanes. Because alkane molecules are nonpolar, they are insoluble in water, which is a polar solvent, but are soluble in nonpolar and slightly polar solvents. Consequently, alkanes themselves are commonly used as solvents for organic substances of low polarity, such as fats, oils, and waxes. Nearly all alkanes have densities less than 1.0 g/mL and are therefore less dense than water (the density of H2O is 1.00 g/mL at 20°C). These properties explain why oil and grease do not mix with water but rather float on its surface.
Table \(1\): Physical Properties of Some Alkanes
Molecular Name Formula Melting Point (°C) Boiling Point (°C) Density (20°C)* Physical State (at 20°C)
methane CH4 –182 –164 0.668 g/L gas
ethane C2H6 –183 –89 1.265 g/L gas
propane C3H8 –190 –42 1.867 g/L gas
butane C4H10 –138 –1 2.493 g/L gas
pentane C5H12 –130 36 0.626 g/mL liquid
hexane C6H14 –95 69 0.659 g/mL liquid
octane C8H18 –57 125 0.703 g/mL liquid
decane C10H22 –30 174 0.730 g mL liquid
*Note the change in units going from gases (grams per liter) to liquids (grams per milliliter). Gas densities are at 1 atm pressure.
The boiling points for the "straight chain" isomers and isoalkanes isomers are shown to demonstrate that branching decreases the surfaces area, weakens the IMFs, and lowers the boiling point.
This next diagrams summarizes the physical states of the first six alkanes. The first four alkanes are gases at room temperature, and solids do not begin to appear until about \(C_{17}H_{36}\), but this is imprecise because different isomers typically have different melting and boiling points. By the time you get 17 carbons into an alkane, there are unbelievable numbers of isomers!
Cycloalkanes have boiling points that are approximately 20 K higher than the corresponding straight chain alkane.
There is not a significant electronegativity difference between carbon and hydrogen, thus, there is not any significant bond polarity. The molecules themselves also have very little polarity. A totally symmetrical molecule like methane is completely non-polar, meaning that the only attractions between one molecule and its neighbors will be Van der Waals dispersion forces. These forces will be very small for a molecule like methane but will increase as the molecules get bigger. Therefore, the boiling points of the alkanes increase with molecular size.
Where you have isomers, the more branched the chain, the lower the boiling point tends to be. Van der Waals dispersion forces are smaller for shorter molecules and only operate over very short distances between one molecule and its neighbors. It is more difficult for short, fat molecules (with lots of branching) to lie as close together as long, thin molecules.
Example: Structure dependent Boiling Points
For example, the boiling points of the three isomers of \(C_5H_{12}\) are:
• pentane: 309.2 K
• 2-methylbutane: 301.0 K
• 2,2-dimethylpropane: 282.6 K
The slightly higher boiling points for the cycloalkanes are presumably because the molecules can get closer together because the ring structure makes them tidier and less "wriggly"!
Solubility
Alkanes (both alkanes and cycloalkanes) are virtually insoluble in water, but dissolve in organic solvents. However, liquid alkanes are good solvents for many other non-ionic organic compounds.
Solubility in Water
When a molecular substance dissolves in water, the following must occur:
• break the intermolecular forces within the substance. In the case of the alkanes, these are the Van der Waals dispersion forces.
• break the intermolecular forces in the water so that the substance can fit between the water molecules. In water, the primary intermolecular attractions are hydrogen bonds.
Breaking either of these attractions requires energy, although the amount of energy to break the Van der Waals dispersion forces in something like methane is relatively negligible; this is not true of the hydrogen bonds in water.
As something of a simplification, a substance will dissolve if there is enough energy released when new bonds are made between the substance and the water to compensate for what is used in breaking the original attractions. The only new attractions between the alkane and the water molecules are Van der Waals forces. These forces do not release a sufficient amount of energy to compensate for the energy required to break the hydrogen bonds in water.; the alkane does not dissolve.
The energy only description of solvation is an oversimplification because entropic effects are also important when things dissolve.
The lack of water solubility can lead to environmental concerns when oils are spilled into natural bodies of water as shown below.
Oil Spills. Crude oil coats the water’s surface in the Gulf of Mexico after the Deepwater Horizon oil rig sank following an explosion. The leak was a mile below the surface, making it difficult to estimate the size of the spill. One liter of oil can create a slick 2.5 hectares (6.3 acres) in size. This and similar spills provide a reminder that hydrocarbons and water don’t mix. Source: Photo courtesy of NASA Goddard / MODIS Rapid Response Team, http://www.nasa.gov/topics/earth/features/oilspill/oil-20100519a.html.
Solubility in organic solvents
In most organic solvents, the primary forces of attraction between the solvent molecules are Van der Waals - either dispersion forces or dipole-dipole attractions. Therefore, when an alkane dissolves in an organic solvent, the Van der Waals forces are broken and are replaced by new Van der Waals forces. The two processes more or less cancel each other out energetically; thus, there is no barrier to solubility.
Looking Closer: Gas Densities and Fire Hazards
Table \(1\) indicates that the first four members of the alkane series are gases at ordinary temperatures. Natural gas is composed chiefly of methane, which has a density of about 0.67 g/L. The density of air is about 1.29 g/L. Because natural gas is less dense than air, it rises. When a natural-gas leak is detected and shut off in a room, the gas can be removed by opening an upper window. On the other hand, bottled gas can be either propane (density 1.88 g/L) or butanes (a mixture of butane and isobutane; density about 2.5 g/L). Both are much heavier than air (density 1.2 g/L). If bottled gas escapes into a building, it collects near the floor. This presents a much more serious fire hazard than a natural-gas leak because it is more difficult to rid the room of the heavier gas.
Also shown in Table \(1\) are the boiling points of the straight-chain alkanes increase with increasing molar mass. This general rule holds true for the straight-chain homologs of all organic compound families. Larger molecules have greater surface areas and consequently interact more strongly; more energy is therefore required to separate them. For a given molar mass, the boiling points of alkanes are relatively low because these nonpolar molecules have only weak dispersion forces to hold them together in the liquid state.
Looking Closer: An Alkane Basis for Properties of Other Compounds
An understanding of the physical properties of the alkanes is important in that petroleum and natural gas and the many products derived from them—gasoline, bottled gas, solvents, plastics, and more—are composed primarily of alkanes. This understanding is also vital because it is the basis for describing the properties of other organic and biological compound families. For example, large portions of the structures of lipids consist of nonpolar alkyl groups. Lipids include the dietary fats and fatlike compounds called phospholipids and sphingolipids that serve as structural components of living tissues. These compounds have both polar and nonpolar groups, enabling them to bridge the gap between water-soluble and water-insoluble phases. This characteristic is essential for the selective permeability of cell membranes.
Tripalmitin (a), a typical fat molecule, has long hydrocarbon chains typical of most lipids. Compare these chains to hexadecane (b), an alkane with 16 carbon atoms.
Exercise
1. Without referring to a table, predict which has a higher boiling point—hexane or octane. Explain.
2. If 25 mL of hexane were added to 100 mL of water in a beaker, which of the following would you expect to happen? Explain.
1. Hexane would dissolve in water.
2. Hexane would not dissolve in water and would float on top.
3. Hexane would not dissolve in water and would sink to the bottom of the container.
3. Without referring to a table or other reference, predict which member of each pair has the higher boiling point.
1. pentane or butane
2. heptane or nonane
4. For which member of each pair is hexane a good solvent?
1. pentane or water
2. sodium chloride or soybean oil
Answer
1. octane because of its greater molar mass
2. b; Hexane is insoluble in water and is less dense than water so it floats on top.
3. a) pentane
b) nonane
4. a) pentane
b) soybean oil | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.02%3A_Physical_Properties_of_Alkanes.txt |
Learning Objective
• interpret and draw the rotation about a carbon-carbon single bond using Newman projections and sawhorse structures
• correlate energies of conformations with rotational energy diagrams
Single Bond Rotation and Conformational Isomerism
Conformational isomerism involves rotation about sigma bonds, and does not involve any differences in the connectivity or geometry of bonding. Two or more structures that are categorized as conformational isomers, or conformers, are really just two of the exact same molecule that differ only in terms of the angle about one or more sigma bonds. The carbon-carbon single bonds of alkanes rotate freely. Conformers are the same molecule shown with different sigma bond rotations. Newman projections are one way to communicate bond rotation.
Ethane Conformations
Although there are seven sigma bonds in the ethane molecule, rotation about the six carbon-hydrogen bonds does not result in any change in the shape of the molecule because the hydrogen atoms are essentially spherical. Rotation about the carbon-carbon bond, however, results in many different possible molecular conformations.
In order to better visualize these different conformations, it is convenient to use a drawing convention called the Newman projection. In a Newman projection, we look lengthwise down a specific bond of interest – in this case, the carbon-carbon bond in ethane. When there are multiple carbons,then we specify the bond of interest using the carbon numbers from the IUPAC name. We depict the ‘front’ atom as a dot, and the ‘back’ atom as a larger circle.
The six carbon-hydrogen bonds are shown as solid lines protruding from the two carbons at 120°angles, which is what the actual tetrahedral geometry looks like when viewed from this perspective and flattened into two dimensions.
The lowest energy conformation of ethane, shown in the figure above, is called the ‘staggered’ or 'anti' conformation, in which all of the C-H bonds on the front carbon are positioned at dihedral angles of 60°relative to the C-H bonds on the back carbon. In this conformation, the distance between the bonds (and the electrons in them) is maximized.
If we now rotate the front CH3 group 60°clockwise, the molecule is in the highest energy ‘eclipsed' conformation, where the hydrogens on the front carbon are as close as possible to the hydrogens on the back carbon.
This is the highest energy conformation because of unfavorable interactions between the electrons in the front and back C-H bonds. The energy of the eclipsed conformation is approximately 3 kcal/mol higher than that of the staggered conformation.
Another 60°rotation returns the molecule to a second eclipsed conformation. This process can be continued all around the 360°circle, with three possible eclipsed conformations and three staggered conformations, in addition to an infinite number of variations in between.
The carbon-carbon bond is not completely free to rotate – there is indeed a small, 3 kcal/mol barrier to rotation that must be overcome for the bond to rotate from one staggered conformation to another. This rotational barrier is not high enough to prevent constant rotation except at extremely cold temperatures. However, at any given moment the molecule is more likely to be in a staggered conformation - one of the rotational ‘energy valleys’ - than in any other state.
Free Rotations Do Not Exist in Ethane
The carbon-carbon bond is not completely free to rotate – there is indeed a small, 3 kcal/mol barrier to rotation that must be overcome for the bond to rotate from one staggered conformation to another. This rotational barrier is not high enough to prevent constant rotation except at extremely cold temperatures. However, at any given moment the molecule is more likely to be in a staggered conformation - one of the rotational ‘energy valleys’ - than in any other state. The potential energy associated with the various conformations of ethane varies with the dihedral angle of the bonds, as shown below.
The potential energy associated with the various conformations of ethane varies with the dihedral angle of the bonds.
Although the conformers of ethane are in rapid equilibrium with each other, the 3 kcal/mol energy difference leads to a substantial preponderance of staggered conformers (> 99.9%) at any given time. The animation below illustrates the relationship between ethane's potential energy and its dihedral angle
Exercise
1. Draw the Newman projections for the staggered and eclipsed conformers of propane along the C1-C2 axis.
Answer
1. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.03%3A_Structure_and_Conformations_of_Alkanes.txt |
Learning Objective
• interpret and draw the rotation about a carbon-carbon single bond using Newman projections and sawhorse structures
• correlate energies of conformations with rotational energy diagrams and predict the most stable conformations for butane
Butane Conformations
Now let's consider butane, with its four-carbon chain. There are now three rotating carbon-carbon bonds to consider, but we will focus on the middle bond between C2 and C3. Below are two representations of butane in a conformation which puts the two CH3 groups (C1 and C4) in the eclipsed position, with the two C-C bonds at a 0o dihedral angle.
If we rotate the front, (blue) carbon by 60° clockwise, the butane molecule is now in a staggered conformation.
This is more specifically referred to as the gauche conformation of butane. Notice that although they are staggered, the two methyl groups are not as far apart as they could possibly be.
A further rotation of 60° gives us a second eclipsed conformation (B) in which both methyl groups are lined up with hydrogen atoms.
One more 60 rotation produces another staggered conformation called the anti conformation, where the two methyl groups are positioned opposite each other (a dihedral angle of 180o).
As with ethane, the staggered conformations of butane are energy 'valleys', and the eclipsed conformations are energy 'peaks'. However, in the case of butane there are two different valleys, and two different peaks. The gauche conformation is a higher energy valley than the anti conformation due to steric strain, which is the repulsive interaction caused by the two bulky methyl groups being forced too close together. Clearly, steric strain is lower in the anti conformation. In the same way, steric strain causes the eclipsed A conformation - where the two methyl groups are as close together as they can possibly be - to be higher in energy than the two eclipsed B conformations.
The diagram below summarizes the relative energies for the various eclipsed, staggered, and gauche conformations.
The following diagram illustrates the change in potential energy that occurs with rotation about the C2–C3 bond at smaller rotational increments.
Potential curve vs dihedral angle of the C2-C3 bond of butane.
Because the anti conformation (staggered) is lowest in energy (and also simply for ease of drawing), it is conventional to draw open-chain alkanes in a 'zigzag' form, which implies anti conformation at all carbon-carbon bonds. The figure below shows, as an example, a Newman projection looking down the C2-C3 bond of octane.
Exercise
1: Using free rotation around C-C single bonds, show that (R,S) and (S,R)-tartaric acid are identical molecules.
2: Draw a Newman projection, looking down the C2-C3 bond, of 1-butene in the conformation shown below (C2 should be your front carbon).
Solutions to exercises
Online lectures from Kahn Academy
Newman projections part I
Newman projections part II
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
4.05: Conformations of Higher Alkanes
Learning Objective
• interpret and draw the rotation about a carbon-carbon single bond using Newman projections and sawhorse structures
• correlate energies of conformations with rotational energy diagrams and predict the most stable conformations for higher alkanes
Pentane and Higher Alkanes
Pentane and higher alkanes have conformational preferences similar to ethane and butane. Each dihedral angle tries to adopt a staggered conformation and each internal C-C bond attempts to take on an anti conformation to minimize the potential energy of the molecule. The most stable conformation of any unbranched alkane follows these rules to take on zigzag shapes:
Let's analyze the staggered conformations of pentane in more detail, considering conformations about the \$C_2 –C_3 \$ and \$C_3 –C_4 \$ bonds. shows a few possible permutations. The most stable conformation is anti at both bonds, whereas less stable conformations contain gauche interactions. One gauche-gauche conformer is particularly unfavorable because methyl groups are aligned with parallel bonds in close proximity. This conformation is called syn. This type of steric hindrance across five atoms is called a syn-pentane interaction. Syn-pentane interactions have an energetic cost of about 3.6 kcal/mol relative to the anti-anti conformation and are therefore disfavored.
Exercises
1. Draw Newman projections of the eclipsed and staggered conformations of propane.
2. Draw a Newman projection, looking down the C2-C3 bond, of 1-butene in the conformation shown below.
3. Draw the energy diagram for the rotation of the bond highlighted in pentane.
Answer
1.
.
2.
3. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.04%3A_Conformations_of_Butane.txt |
Learning Objective
• explain the partial rotation of carbon-carbon single bonds in rings
• explain ring strain and its relationship to cycloalkane stability
Cycloalkanes (aka Rings)
Cycloalkanes have one or more rings of carbon atoms. The simplest examples of this class consist of a single, unsubstituted carbon ring, and these form a homologous series similar to the unbranched alkanes. The IUPAC names of the first five members of this series are given in the following table. The last column gives the general formula for a cycloalkane of any size. If a simple unbranched alkane is converted to a cycloalkane two hydrogen atoms, one from each end of the chain, must be lost. Hence the general formula for a cycloalkane composed of n carbons is CnH2n. Although a cycloalkane has two fewer hydrogens than the equivalent alkane, each carbon is bonded to four other atoms so such compounds are still considered to be saturated with hydrogen.
Table: Examples of Simple Cycloalkanes
Name Cyclopropane Cyclobutane Cyclopentane Cyclohexane Cycloheptane Cycloalkane
Molecular
Formula
C3H6 C4H8 C5H10 C6H12 C7H14 CnH2n
Structural
Formula
(CH2)n
Line
Formula
Although the customary line drawings of simple cycloalkanes are geometrical polygons, the actual shape of these compounds in most cases is very different. Cyclic systems are a little different from open-chain systems. In an open chain, any bond can be rotated 360 degrees, going through many different conformations. Complete rotation isn't possible in a cyclic system, because the parts that you would be trying to twist away from each other would still be connected together. Cyclic systems have fewer "degrees of freedom" than aliphatic systems; they have "restricted rotation". Because of the restricted rotation of cyclic systems, most of them have much more well-defined shapes than their aliphatic counterparts. Let's take a look at the basic shapes of some common rings. Many biologically important compounds are built around structures containing rings, so it's important that we become familiar with them. In nature, three- to six-membered rings are frequently encountered, so we'll focus on those.
The Baeyer Theory and the Experimental Evidence of Ring Strain
Many of the properties of cyclopropane and its derivatives are similar to the properties of alkenes. In 1890, the famous German organic chemist, A. Baeyer, suggested that cyclopropane and cyclobutane derivatives are different from cyclopentane and cyclohexane, because their C—C—C angles cannot have the tetrahedral value of 109.5°. At the same time, Baeyer hypothesized that the difficulties encountered in synthesizing cycloalkane rings from C7 upward was the result of the angle strain that would be expected if the large rings were regular planar polygons (see Table 12-3). Baeyer also believed that cyclohexane had a planar structure like that shown in Figure 12-2, which would mean that the bond angles would have to deviate 10.5° from the tetrahedral value. However, in 1895, the then unknown chemist H. Sachse suggested that cyclohexane exists in the strain-free chair and boat forms discussed in Section 12-3. This suggestion was not accepted at the time because it led to the prediction of several possible isomers for compounds such as chlorocyclohexane (cf. Exercise 12-4). The idea that such isomers might act as a single substance, as the result of rapid equilibration, seemed like a needless complication, and it was not until 1918 that E. Mohr proposed a definitive way to distinguish between the Baeyer and Sachse cyclohexanes. As will be discussed in Section 12-9, the result, now known as the Sachse-Mohr theory, was complete confirmation of the idea of nonplanar large rings.
Table: Strain in Cycloalkane Rings and Heats of Combustion of Cycloalkanes
Compound n Angle Strain at each CH2 Heat of Combustion ΔHo (kcal/mol) Heat of Combustion ΔHo per CH2/N (kcal/mol) Total Strain (kcal/mol)
ethene 2 109.5 337.2 168.6 22.4
cyclopropane 3 49.5 499.9 166.6 27.7
cyclobutane 4 19.5 655.9 164.0 26.3
cyclopentane 5 1.5 793.4 158.7 6.5
cyclohexane 6 10.5 944.8 157.5 0.4
cycloheptane 7 19.1 1108.1 158.4 6.3
cyclooctane 8 25.5 1268.9 158.6 9.7
cyclononane 9 30.5 1429.5 158.8 12.9
cyclodecane 10 34.5 1586.1 158.6 12.1
cyclopentadecane 15 46.5 2362.5 157.5 1.5
open chain alkane 157.4 -
Ring Strain in Cycloalkanes
Ring Strain occurs because the carbons in cycloalkanes are sp3 hybridized, which means that they do not have the expected ideal bond angle of 109.5o ; this causes an increase in the potential energy because of the desire for the carbons to be at an ideal 109.5o. An example of ring strain can be seen in the diagram of cyclopropane below in which the bond angle is 60o between the carbons.
The reason for ring strain can be seen through the tetrahedral carbon model. The C-C-C bond angles in cyclopropane (diagram above) (60o) and cyclobutane (90o) are much different than the ideal bond angle of 109.5o. This bond angle causes cyclopropane and cyclobutane to have a high ring strain. However, molecules, such as cyclohexane and cyclopentane, would have a much lower ring strain because the bond angle between the carbons is much closer to 109.5o.
Below are some examples of cycloalkanes. Ring strain can be seen more prevalently in the cyclopropane and cyclobutane models
Below is a chart of cycloalkanes and their respective heats of combustion ( ΔHcomb). The ΔHcomb value increases as the number of carbons in the cycloalkane increases (higher membered ring), and the ΔHcomb/CH2 ratio decreases. The increase in ΔHcomb can be attributed to the greater amount of London Dispersion forces. However, the decrease in ΔHcomb/CH2can be attributed to a decrease in the ring strain.
Certain cycloalkanes, such as cyclohexane, deal with ring strain by forming conformers. A conformer is a stereoisomer in which molecules of the same connectivity and formula exist as different isomers, in this case, to reduce ring strain. The ring strain is reduced in conformers due to the rotations around the sigma bonds.
Other Types of Strain
There are many different types of strain that occur with cycloalkanes. In addition to ring strain, there is also transannular strain, eclipsing, or torsional strain and bond angle strain.Transannular strain exists when there is steric repulsion between atoms. Eclipsing (torsional) strain exists when a cycloalkane is unable to adopt a staggered conformation around a C-C bond, and bond angle strain is the energy needed to distort the tetrahedral carbons enough to close the ring. The presence of angle strain in a molecule indicates that there are bond angles in that particular molecule that deviate from the ideal bond angles required (i.e., that molecule has conformers).
Cyclopropane
A three membered ring has no rotational freedom whatsoever, so the three carbon atoms in cyclopropane are all constrained to lie in the same plane at the corners of an equilateral triangle. The 60º bond angles are much smaller than the optimum 109.5º angles of a normal tetrahedral carbon atom, and the resulting angle strain dramatically influences the chemical behavior of this cycloalkane. Cyclopropane also suffers substantial eclipsing strain, since all the carbon-carbon bonds are fully eclipsed.
Furthermore, if you look at a model you will find that the neighboring C-H bonds (C-C bonds, too) are all held in eclipsed conformations.
Cyclopropane is always at maximum torsional strain. This strain can be illustrated in a line drawing of cyclopropane as shown from the side. In this oblique view, the dark lines mean that those sides of the ring are closer to you.
However, the ring isn't big enough to introduce any steric strain, which does not become a factor until we reach six membered rings. Until that point, rings are not flexible enough for two atoms to reach around and bump into each other.
The really big problem with cyclopropane is that the C-C-C bond angles are all too small.
• All the carbon atoms in cyclopropane appear to be tetrahedral.
• These bond angles ought to be 109 degrees.
• The angles in an equilateral triangle are actually 60 degrees, about half as large as the optimum angle.
• This factor introduces a huge amount of strain in the molecule, called ring strain.
Cyclobutane
Cyclobutane is a four membered ring. In two dimensions, it is a square, with 90 degree angles at each corner. Cyclobutane reduces some bond-eclipsing strain by folding (the out-of-plane dihedral angle is about 25º), but the total eclipsing and angle strain remains high. Cyclopentane has very little angle strain (the angles of a pentagon are 108º), but its eclipsing strain would be large (about 10 kcal/mol) if it remained planar. Consequently, the five-membered ring adopts non-planar puckered conformations whenever possible.
However, in three dimensions, cyclobutane is flexible enough to buckle into a "butterfly" shape, relieving torsional strain a little bit. When it does that, the bond angles get a little worse, going from 90 degrees to 88 degrees.
In a line drawing, this butterfly shape is usually shown from the side, with the near edges drawn using darker lines.
• With bond angles of 88 rather than 109 degrees, cyclobutane has a lot of ring strain, but less than in cyclopropane.
• Torsional strain is still present, but the neighbouring bonds are not exactly eclipsed in the butterfly.
• Cyclobutane is still not large enough that the molecule can reach around to cause crowding. Steric strain is very low.
• Cyclobutanes are a little more stable than cyclopropanes and are also a little more common in nature.
Cyclopentane
Cyclopentanes are even more stable than cyclobutanes, and they are the second-most common paraffinic ring in nature, after cyclohexanes. In two dimensions, a cyclopentane appears to be a regular pentagon.
In three dimensions, there is enough freedom of rotation to allow a slight twist out of this planar shape. In a line drawing, this three-dimensional shape is drawn from an oblique view, just like cyclobutane.
• The ideal angle in a regular pentagon is about 107 degrees, very close to a tetrahedral bond angle.
• Cyclopentane distorts only very slightly into an "envelope" shape in which one corner of the pentagon is lifted up above the plane of the other four, and as a result, ring strain is entirely removed.
• The envelope removes torsional strain along the sides and flap of the envelope. However, the neighbouring carbons are eclipsed along the "bottom" of the envelope, away from the flap. There is still some torsional strain in cyclopentane.
• Again, there is no steric strain in this system.
Rings larger than cyclopentane would have angle strain if they were planar. However, this strain, together with the eclipsing strain inherent in a planar structure, can be relieved by puckering the ring. Cyclohexane is a good example of a carbocyclic system that virtually eliminates eclipsing and angle strain by adopting non-planar conformations. Cycloheptane and cyclooctane have greater strain than cyclohexane, in large part due to transannular crowding (steric hindrance by groups on opposite sides of the ring).
Exercise \(1\)
1. If cyclobutane were to be planar how many H-H eclipsing interactions would there be, and assuming 4 kJ/mol per H-H eclipsing interaction what is the strain on this “planar” molecule?
2. In the two conformations of cis-cyclopentane one is more stable than the other. Explain why this is.
Answer
1. There are 8 eclipsing interactions (two per C-C bond). The extra strain on this molecule would be 32 kJ/mol (4 kJ/mol x 8).
2. The first conformation is more stable. Even though the methyl groups are cis in the model on the left, they are eclipsing due the conformation, therefore increasing the strain within the molecule. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.06%3A_Cycloalkanes_and_Ring_Strain.txt |
Learning Objective
• draw cyclohexane conformations (chair & boat)
• correlate energies of conformations with rotational energy diagrams and predict the most stable conformations for cyclohexane
Introduction
Rings larger than cyclopentane would have angle strain if they were planar. However, this strain, together with the eclipsing strain inherent in a planar structure, can be relieved by puckering the ring. Cyclohexane is a good example of a carbocyclic system that virtually eliminates eclipsing and angle strain by adopting non-planar conformations. Cycloheptane and cyclooctane have greater strain than cyclohexane, in large part due to transannular crowding (steric hindrance by groups on opposite sides of the ring).
Cyclohexane Conformations (aka Chair Flips)
Cyclohexane is rapidly rotating between the two most stable conformations known as the chair conformations in what is called the "Chair Flip" shown below.
Several other notable cyclohexane conformations occur during the transition from one chair conformer to the other - the boat, the twist, and the half-chair. The relative energies of the conformations is a direct reflection of their relative stabilities. These structural and energetic relationships are summarized in the conformational energy diagram for cyclohexane below.
The Chair Conformation - a closer look
Since the chair conformation has the lowest potential energy, it is the most relevant to the conformation of cyclohexane. On careful examination of a chair conformation of cyclohexane, we find that the twelve hydrogens are not structurally equivalent. Six of them are located about the periphery of the carbon ring, and are termed equatorial. The other six are oriented above and below the approximate plane of the ring (three in each location), and are termed axial because they are aligned parallel to the symmetry axis of the ring.
In the figure above, the equatorial hydrogens are colored blue, and the axial hydrogens are in bold. Since there are two equivalent chair conformations of cyclohexane in rapid equilibrium, all twelve hydrogens have 50% equatorial and 50% axial character. The figure below illustrates how to convert a molecular model of cyclohexane between two different chair conformations - this is something that you should practice with models. Notice that a 'ring flip' causes equatorial hydrogens to become axial, and vice-versa.
How to draw stereo bonds ("up" and "down" bonds)
There are various ways to show these orientations. The solid (dark) "up wedge" I used is certainly common. Some people use an analogous "down wedge", which is light, to indicate a down bond; unfortunately, there is no agreement as to which way the wedge should point, and you are left relying on the lightness of the wedge to know it is "down". The "down bond" avoids this wedge ambiguity, and just uses some kind of light line. The down bond I used (e.g., in Figure 5B) is a dashed line; IUPAC encourages a series of parallel lines, something like What I did is a variation of what is recommended by IUPAC:
• In ISIS/Draw, the "up wedge" and "down bond" that I used, along with other variations, are available from a tool button that may be labeled with any of them, depending on most recent use. It is located directly below the tool button for ordinary C-C bonds.
• In Symyx Draw, the "up wedge" and "down bond", along with other variations, are available from a tool button that may be labeled with any of them, depending on most recent use. It is located directly below the "Chain" tool button.
• ChemSketch provides up and down wedges, but not the simple up and down bonds discussed above. The wedges are available from the second toolbar across the top. For an expanded discussion of using these wedges, see the section of my ChemSketch Guide on Stereochemistry: Wedge bonds.
As always, the information provided on these pages in intended to help you get started. Each program has more options for drawing bonds than discussed here. When you feel the need, look around!
How to Draw chairs
Most of the structures shown on this page were drawn with the free program ISIS/Draw. I have posted a guide to help you get started with ISIS/Draw. ISIS/Draw provides a simple cyclohexane (6-ring) hexagon template on the toolbar across the top. It provides templates for various 6-ring chair structures from the Templates menu; choose Rings. There are templates for simple chairs, without substituents (e.g., Fig 1B), and for chairs showing all the substituents (e.g., Fig 2B). In either case, you can add, delete, or change things as you wish. Various kinds of stereo bonds (wedges and bars) are available by clicking the left-side tool button that is just below the regular C-C single bond button. It may have a wedge shown on it, but this will vary depending on how it has been used. To choose a type of stereo bond, click on the button and hold the mouse click; a new menu will appear to the right of the button.
The free drawing program Symyx Draw, the successor to ISIS/Draw, provides similar templates and tools. A basic chair structure is provided on the default template bar that is shown. More options are available by choosing the Rings template. See my page Symyx Draw for a general guide for getting started with this program.
The free drawing program ChemSketch provides similar templates and tools. To find the special templates for chairs, go to the Templates menu, choose Template Window, and then choose "Rings" from the drop-down menu near upper left. See my page ChemSketch for a general guide for getting started with this program.
If you want to draw chair structures by hand (and if you are going on in organic chemistry, you should)... Be careful. The precise zigs and zags, and the angles of substituents are all important. Your textbook may offer you some hints for how to draw chairs. A short item in the Journal of Chemical Education offers a nice trick, showing how the chair can be thought of as consisting of an M and a W. The article is V Dragojlovic, A method for drawing the cyclohexane ring and its substituents. J Chem Educ 78:923, 7/01. (I thank M Farooq Wahab, Chemistry, Univ Karachi, for suggesting that this article be noted here.)
Aside from drawing the basic chair, the key points in adding substituents are:
• Axial groups alternate up and down, and are shown "vertical".
• Equatorial groups are approximately horizontal, but actually somewhat distorted from that, so that the angle from the axial group is a bit more than a right angle -- reflecting the common 109 degree bond angle.
• As cautioned before, it is usually easier to draw and see what is happening at the four corners of the chair than at the two middle positions. Try to use the corners as much as possible.
Because axial bonds are parallel to each other, substituents larger than hydrogen generally suffer greater steric crowding when they are oriented axial rather than equatorial. Consequently, substituted cyclohexanes will preferentially adopt conformations in which the larger substituents assume equatorial orientation.
When the methyl group in the structure above occupies an axial position it suffers steric crowding by the two axial hydrogens located on the same side of the ring.
The conformation in which the methyl group is equatorial is more stable, and thus the equilibrium lies in this direction
Exercise
Questions
1. Consider the conformations of cyclohexane, chair, boat, twist boat. Order them in increasing strain in the molecule.
2. Draw two conformations of cyclohexyl amine (C6H11NH2). Indicate axial and equatorial positions.
3. Draw the two isomers of 1,4-dihydroxylcyclohexane, identify which are equatorial and axial.
4. In the following molecule, label which are equatorial and which are axial, then draw the chair flip (showing labels 1,2,3).
Answer
1. Chair < Twist Boat < Boat (most strain)
2.
3.
4. Original conformation: 1 = axial, 2 = equatorial, 3 = axial
Flipped chair now looks like this.
Contributors and Attributions
Dr. Dietmar Kennepohl FCIC (Professor of Chemistry, Athabasca University)
Prof. Steven Farmer (Sonoma State University)
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
>Robert Bruner (http://bbruner.org) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.07%3A_Cyclohexane_Conformations.txt |
Learning Objective
• draw mono-substituted cyclohexane conformers (chair only)
• correlate energies of conformations with rotational energy diagrams and predict the most stable conformations for butane, higher alkanes, cyclohexane, mono-substituted cyclohexanes, and disubstituted cyclohexanes
Introduction
Because axial bonds are parallel to each other, substituents larger than hydrogen generally suffer greater steric crowding when they are oriented axial rather than equatorial. Consequently, substituted cyclohexanes will preferentially adopt conformations in which the larger substituents assume equatorial orientation.
When the methyl group in the structure above occupies an axial position it suffers steric crowding by the two axial hydrogens located on the same side of the ring. The conformation in which the methyl group is equatorial is more stable, and thus the equilibrium lies in this direction.
In examining possible structures for monosubstituted cyclohexanes, it is useful to follow two principles:
1. Chair conformations are generally more stable than other possibilities.
2. Substituents on chair conformers prefer to occupy equatorial positions due to the increased steric hindrance of axial locations.
Experimental Measurements of Steric Hindrance
The relative steric hindrance experienced by different substituent groups oriented in an axial versus equatorial location on cyclohexane may be determined by the conformational equilibrium of the compound. The corresponding equilibrium constant is related to the energy difference between the conformers and collecting such data allows us to evaluate the relative tendency of substituents to exist in an equatorial or axial location.
Looking at the energy values the table, it is clear that the apparent "size" of a substituent (in terms of its preference for equatorial over axial orientation) is influenced by its width and bond length to cyclohexane, as evidenced by the fact that an axial vinyl group is less hindered than ethyl, and iodine slightly less than chlorine.
A Selection of AG° Values for the Change from Axial to Equatorial Orientation of Substituents for Monosubstituted Cyclohexanes
Substituent \(-\Delta{G}^o\) kcal/mol Substituent \(-\Delta{G}^o\) kcal/mol
\(\ce{CH_3\bond{-}}\) 1.7 \(\ce{O_2N\bond{-}}\) 1.1
\(\ce{CH_2H_5\bond{-}}\) 1.8 \(\ce{N#C\bond{-}}\) 0.2
\(\ce{(CH_3)_2CH\bond{-}}\) 2.2 \(\ce{CH_3O\bond{-}}\) 0.5
\(\ce{(CH_3)_3C\bond{-}}\) \(\geq 5.0\) (CH3)3C- 0.7
\(\ce{F\bond{-}}\) 0.3 F- 1.3
\(\ce{Cl\bond{-}}\) 0.5 \(\ce{C_6H_5\bond{-}}\) 3.0
\(\ce{Br\bond{-}}\) 0.5
\(\ce{I\bond{-}}\) 0.5
Exercise
1. In the molecule, cyclohexyl ethyne there is little steric strain, why?
Answer
1.
The ethyne group is linear and therefore does not affect the hydrogens in the 1,3 positions to say to the extent as a bulkier or a bent group (e.g. ethene group) would. This leads to less of a strain on the molecule.
Contributors and Attributions
>Robert Bruner (http://bbruner.org)
4.09: Cis-trans Isomerism in Cycloalkanes
Learning Objective
• identify & draw the geometric (cis/trans) isomers of cycloalkanes
Geometric Isomerism of Cycloalkanes
The carbon ring of cycloalkanes forms a pseudo-plane that can be used to assign the relative orientation of atoms or substituents bonded to the ring (stereochemistry). One side of the ring is called "up" while the other side is called "down". By agreement, chemists use heavy, wedge-shaped bonds to indicate a substituent located above the average plane of the ring (up), and a hatched line for bonds to atoms or groups located below the ring (down).
Disubstituted cycloalkane stereoisomers may be designated by nomenclature prefixes such as cis and trans. Cis and trans isomers are also called "geometric isomers".
For the cis isomer, both substituents are aobe or below the carbon ring. For the trans isomer, one substituent is above the ring while the other substituent is below the ring.
The cis and trans isomers for 1,2-dibromocyclopentane are shown as an example below.
While the carbon-carbon single bonds of the rings can rotate partially, there is NO way to inter-convert between the cis and trans isomers. Cis and trans isomers are unique compounds with their own unique melting points, boiling points, densities, etc.
Further explanation:
In general, if any two sp3 carbons in a ring have two different substituent groups (not counting other ring atoms) stereoisomerism is possible. This is similar to the substitution pattern that gives rise to stereoisomers in alkenes; indeed, one might view a double bond as a two-membered ring. Four other examples of this kind of stereoisomerism in cyclic compounds are shown below.
If more than two ring carbons have different substituents (not counting other ring atoms) the stereochemical notation distinguishing the various isomers becomes more complex. However, we can always state the relationship of any two substituents using cis or trans. For example, in the trisubstitutued cyclohexane below, we can say that the methyl group is cis to the ethyl group, and trans to the chlorine. We can also say that the ethyl group is trans to the chlorine. We cannot, however, designate the entire molecule as a cis or trans isomer.
Exercise
1. Draw the following molecules:
trans-1,3-dimethylcyclohexane
trans-1,2-dibromocyclopentane
cis-1,3-dichlorocyclobutane
Answer
1. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.08%3A_Conformations_of_Monosubstituted_Cyclohexanes.txt |
Learning Objective
• draw di-substituted cyclohexane conformers (chair only)
• correlate energies of conformations with rotational energy diagrams and predict the most stable conformations for disubstituted cyclohexanes
Conformational Structures of Disubstituted Cyclohexanes
In a sample of cyclohexane, the two identical chair conformers are present in equal concentration, and the hydrogens are all equivalent (50% equatorial & 50% axial) due to rapid interconversion of the conformers. When the cyclohexane ring bears a substituent, the two chair conformers are not the same. In one conformer the substituent is axial, in the other it is equatorial. Due to steric hindrance in the axial location, substituent groups prefer to be equatorial and that chair conformer predominates in the equilibrium.
When cycloalkanes have two substituents on different ring carbon atoms, then a pair of configurational stereoisomers exist. Now we must examine the way in which favorable ring conformations influence the properties of the configurational isomers. Remember, configurational stereoisomers are stable, unique chemical compounds, whereas, conformational isomers are different rotations of the same compound. In examining possible structures for disubstituted cyclohexane, it is useful to follow two principles:
1. Substituents on chair conformers prefer to occupy equatorial positions due to the increased steric hindrance of axial locations.
The following equations and formulas illustrate how the presence of two or more substituent on a cyclohexane ring perturbs the interconversion of the two chair conformers in ways that can be predicted. When there is a potential energy difference between the conformers, then the lower energy conformation is favored as indicated by the equilibrium reaction arrows.
1,1-dimethylcyclohexane
1-t-butyl-1-methylcyclohexane
cis-1,2-dimethylcyclohexane
trans-1,2-dimethylcyclohexane
cis-1,3-dimethylcyclohexane
trans-1,3-dimethylcyclohexane
cis-1,4-dimethylcyclohexane
trans-1,4-dimethylcyclohexane
In the case of 1,1-disubstituted cyclohexanes, one of the substituents must necessarily be axial and the other equatorial, regardless of which chair conformer is considered. Since the substituents are the same in 1,1-dimethylcyclohexane, the two conformers are identical and present in equal concentration. In 1-t-butyl-1-methylcyclohexane the t-butyl group is much larger than the methyl, and that chair conformer in which the larger group is equatorial will be favored in the equilibrium( > 99%). Consequently, the methyl group in this compound is almost exclusively axial in its orientation.
In the cases of 1,2-, 1,3- and 1,4-disubstituted compounds the analysis is a bit more complex. It is always possible to have both groups equatorial, but whether this requires a cis-relationship or a trans-relationship depends on the relative location of the substituents. As we count around the ring from carbon #1 to #6, the uppermost bond on each carbon changes its orientation from equatorial (or axial) to axial (or equatorial) and back. It is important to remember that the bonds on a given side of a chair ring-conformation always alternate in this fashion. Therefore, it should be clear that for cis-1,2-disubstitution, one of the substituents must be equatorial and the other axial; in the trans-isomer both may be equatorial. Because of the alternating nature of equatorial and axial bonds, the opposite relationship is true for 1,3-disubstitution (cis is all equatorial, trans is equatorial/axial). Finally, 1,4-disubstitution reverts to the 1,2-pattern.
The conformations of some substituted cyclohexanes may be examined as interactive models by .
It can be helpful to add the hydrogen atoms at the axial positions to help recognize the equatorial position.
Exercise
1. Draw the two chair conformations for cis-1-ethyl-2-methylcyclohexane using bond-line structures and indicate the more energetically favored conformation.
2. Draw the most stable conformation for trans-1-ethyl-3-methylcyclohexane using bond-line structures.
3. Draw the most stable conformation for trans-1-t-butyl-4-methylcyclohexane using bond-line structures.
Answer | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.10%3A_Conformations_of_Disubstituted_Cyclohexanes.txt |
Learning Objective
• recognize, classify, and draw the three ways to join two rings
Bicyclic Ring Systems
There are three ways to join two rings. If two rings share two or more atoms, then the structure is called a bicyclic compound. If the two rings share a single atom, then the structure is called a spirocyclic compound. Examples of each way to join rings is shown below.
Bicyclic Compounds
Fused rings share two adjacent carbon atoms. Decalin is a fused bicyclic compound. Its IUPAC name is bibyclo[4.4.0]decane to communicate the bonding arrangement.
Bridged rings share two non-adjacent carbon atoms and one or more carbon atoms between them. Bicyclo[2.2.1]heptane shows the difference between the bridgehead carbons and the bridge carbons.
Spirocyclic Compounds
A spirobicycloalkane is a molecule in which only one carbon atom is shared by the two rings in the molecule. The carbon atom shared by the two rings is called the spirocarbon. A chain of bonds originating and ending at the spirocarbon is called a bridge. The compound below is named spiro[4.5]decane to communicate the number of carbons in each bridge with the spirocarbon.
Steroids
Steroids include such well known compounds as cholesterol, sex hormones, birth control pills, cortisone, and anabolic steroids.
The best known and most abundant steroid in the body is cholesterol. Cholesterol is formed in brain tissue, nerve tissue, and the blood stream. It is the major compound found in gallstones and bile salts. Cholesterol also contributes to the formation of deposits on the inner walls of blood vessels. These deposits harden and obstruct the flow of blood. This condition, known as atherosclerosis, results in various heart diseases, strokes, and high blood pressure.
Much research is currently underway to determine if a correlation exists between cholesterol levels in the blood and diet. Not only does cholesterol come from the diet, but cholesterol is synthesized in the body from carbohydrates and proteins as well as fat. Therefore, the elimination of cholesterol rich foods from the diet does not necessarily lower blood cholesterol levels. Some studies have found that if certain unsaturated fats and oils are substituted for saturated fats, the blood cholesterol level decreases. The research is incomplete on this problem.
Structures of Sex Hormones
Sex hormones are also steroids. The primary male hormone, testosterone, is responsible for the development of secondary sex characteristics. Two female sex hormones, progesterone and estrogen or estradiol control the ovulation cycle. Notice that the male and female hormones have only slight differences in structures, but yet have very different physiological effects.
Testosterone promotes the normal development of male genital organs ans is synthesized from cholesterol in the testes. It also promotes secondary male sexual characteristics such as deep voice, facial and body hair.
Estrogen, along with progesterone regulates changes occurring in the uterus and ovaries known as the menstrual cycle. For more details see Birth Control. Estrogen is synthesized from testosterone by making the first ring aromatic which results in mole double bonds, the loss of a methyl group and formation of an alcohol group.
Exercise
1. Someone stated that trans-decalin is more stable than cis-decalin. Explain why this is incorrect.
Answer
1. Cis-decalin has fewer steric interactions than trans-decalinbecause each ring can assume the chair form in both conformations. Working with models can be helpful.
Contributors and Attributions
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_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.11%3A_Joined_Rings.txt |
Learning Objective
• describe the uses and sources of alkanes
Occurrence
The most important sources for alkanes are oil and natural gas. Oil is a mixture of liquid alkanes and other hydrocarbons. Higher alkanes (which are solid) occur as residues from oil distillation ("tar"). One of the largest natural deposits of solid alkanes is in an asphalt lake known as the Pitch Lake in Trinidad and Tobago. Natural gas contains primarily methane (70-90%) with some ethane, propane and butane; some gas sources deliver up to 8% CO2. Traces of methane (about 0.00017% or 1.7 ppm) occur in the Earth's atmosphere, the content in the oceans is negligible due to the low solubility of methane in water.(1)
Use of Alkanes
Alkanes are important raw materials of the chemical industry and the principal constituent of gasoline and lubricating oils. Natural gas mainly contains methane and ethane and is used for heating and cooking purposes and for power utilities (gas turbines). For transportation purposes, natural gas may be liquefied by applying pressure and cooling it (LNG = liqid natural gas). The Sultanate of Oman, for example, exports most of its natural gas as LNG - see the LNG plant at Qalhat which has been designed to liquefy 6.6 million tons natural gas per year. Crude oil is separated into its components by fractional distillation at oil refineries. The different "fractions" of crude oil have different boiling points and consist mostly of alkanes of similar chain lengths (the higher the boiling point the more carbon atoms the components of a particular fraction contain - see the list of alkanes for details about the boiling points).
The following table provides a short survey of the different fractions of crude oil:
C3..C4 Propane and butane can be liquefied at fairly low pressures, and are used, for example, in the propane gas burner, or as propellants in aerosol sprays. Butane in used in cigarette lighters (where the pressure at room temperature is about 2 bar).
C5..C8 The alkanes from pentane to octane are highly volatile liquids and good solvents for nonpolar substances. They are used as fuels in internal combustion engines.
C9..C16 Alkanes from nonane to hexadecane are liquids of higher viscosity, being used in diesel and aviation fuel (kerosene). The higher melting points of these alkanes can cause problems at low temperatures and in polar regions, where the fuel becomes too viscous.
C17..C35 Alkanes with 17 to 35 carbon atoms form the major components of lubricating oil. They also act as anti-corrosive agents, as their hydrophobic nature protects the metal surface from contact with water. Solid alkanes also find use as paraffin wax in candles(2).
>C35 Alkanes with a chain length above 35 carbon atoms are found in bitumen (as it is used in road surfacing). These higher alkanes have little chemical and commercial value and are usually split into lower alkanes by cracking.
Notes:
(1) Methane can co-crystallize with water at high pressures and low temperatures, forming a solid methane hydrate. The energy content of the known submarine methane hydrate fields exceeds that of all known natural gas and oil deposits put together.
(2) Paraffin wax should not be mixed up with true animal or plant wax, which consist of esters of various carboxylic acids and alcohols. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.12%3A_Uses_and_Sources_of_Alkanes.txt |
Learning Objective
• recognize and distinguish between the two major reactions of alkanes - combustion and halogenation
Combustion
Complete combustion (given sufficient oxygen) of any hydrocarbon produces carbon dioxide and water. It is quite important that you can write properly balanced equations for these reactions, because they often come up as a part of thermochemistry calculations. Some are easier than others. For example, with alkanes, the ones with an even number of carbon atoms are marginally harder than those with an odd number!
Example: Propane Combustion
For example, with propane (C3H8), you can balance the carbons and hydrogens as you write the equation down. Your first draft would be:
$C_3H_8 + O_2 \rightarrow 3CO_2 + 4H_2O$
Counting the oxygens leads directly to the final version:
$C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$
Example: Butane Combustion
With butane (C4H10), you can again balance the carbons and hydrogens as you write the equation down.
$C_4H_{10} + O_2 \rightarrow 4CO_2 + 5H_2O$
Counting the oxygens leads to a slight problem - with 13 on the right-hand side. The simple trick is to allow yourself to have "six-and-a-half" O2 molecules on the left.
$C_4H_{10} + 6\dfrac{1}{2}\, O_2 \rightarrow 4CO_2 + 5H_2O$
If that offends you, double everything:
$2C_4H_{10} + 13 O_2 \rightarrow 8CO_2 + 10 H_2O$
The hydrocarbons become harder to ignite as the molecules get bigger. This is because the bigger molecules don't vaporize so easily - the reaction is much better if the oxygen and the hydrocarbon are well mixed as gases. If the liquid is not very volatile, only those molecules on the surface can react with the oxygen. Bigger molecules have greater Van der Waals attractions which makes it more difficult for them to break away from their neighbors and turn to a gas.
Provided the combustion is complete, all the hydrocarbons will burn with a blue flame. However, combustion tends to be less complete as the number of carbon atoms in the molecules rises. That means that the bigger the hydrocarbon, the more likely you are to get a yellow, smoky flame. Incomplete combustion (where there is not enough oxygen present) can lead to the formation of carbon or carbon monoxide. As a simple way of thinking about it, the hydrogen in the hydrocarbon gets the first chance at the oxygen, and the carbon gets whatever is left over! The presence of glowing carbon particles in a flame turns it yellow, and black carbon is often visible in the smoke. Carbon monoxide is produced as a colorless poisonous gas.
Note: Why carbon monoxide is poisonous
Oxygen is carried around the blood by hemoglobin, which unfortunately binds to exactly the same site on the hemoglobin that oxygen does. The difference is that carbon monoxide binds irreversibly (or very strongly) - making that particular molecule of hemoglobin useless for carrying oxygen. If you breath in enough carbon monoxide you will die from a sort of internal suffocation.
Halogenation of Alkanes
Halogenation is the replacement of one or more hydrogen atoms in an organic compound by a halogen (fluorine, chlorine, bromine or iodine). Unlike the complex transformations of combustion, the halogenation of an alkane appears to be a simple substitution reaction in which a C-H bond is broken and a new C-X bond is formed. The chlorination of methane, shown below, provides a simple example of this reaction.
CH4 + Cl2 + energy → CH3Cl + HCl
Since only two covalent bonds are broken (C-H & Cl-Cl) and two covalent bonds are formed (C-Cl & H-Cl), this reaction seems to be an ideal case for mechanistic investigation and speculation. However, one complication is that all the hydrogen atoms of an alkane may undergo substitution, resulting in a mixture of products, as shown in the following unbalanced equation. The relative amounts of the various products depend on the proportion of the two reactants used. In the case of methane, a large excess of the hydrocarbon favors formation of methyl chloride as the chief product; whereas, an excess of chlorine favors formation of chloroform and carbon tetrachloride.
CH4 + Cl2 + energy → CH3Cl + CH2Cl2 + CHCl3 + CCl4 + HCl
In the presence of a flame, the reactions are rather like the fluorine one - producing a mixture of carbon and the hydrogen halide. The violence of the reaction drops considerably as you go from fluorine to chlorine to bromine. The interesting reactions happen in the presence of ultra-violet light (sunlight will do). These are photochemical reactions that happen at room temperature. We'll look at the reactions with chlorine, although the reactions with bromine are similar, but evolve more slowly.
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. One would not 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.
Larger alkanes and chlorine
You would again get a mixture of substitution products, but it is worth just looking briefly at what happens if only one of the hydrogen atoms gets substituted (monosubstitution) - just to show that things aren't always as straightforward as they seem! For example, with propane, you could get one of two isomers:
If chance was the only factor, you would expect to get three times as much of the isomer with the chlorine on the end. There are 6 hydrogens that could get replaced on the end carbon atoms compared with only 2 in the middle. In fact, you get about the same amount of each of the two isomers. If you use bromine instead of chlorine, the great majority of the product is where the bromine is attached to the center carbon atom.
Cycloalkanes
The reactions of the cycloalkanes are generally just the same as the alkanes, with the exception of the very small ones - particularly cyclopropane. In the presence of UV light, cyclopropane will undergo substitution reactions with chlorine or bromine just like a non-cyclic alkane. However, it also has the ability to react in the dark. In the absence of UV light, cyclopropane can undergo addition reactions in which the ring is broken. For example, with bromine, cyclopropane gives 1,3-dibromopropane.
This can still happen in the presence of light - but you will get substitution reactions as well. The ring is broken because cyclopropane suffers badly from ring strain. The bond angles in the ring are 60° rather than the normal value of about 109.5° when the carbon makes four single bonds. The overlap between the atomic orbitals in forming the carbon-carbon bonds is less good than it is normally, and there is considerable repulsion between the bonding pairs. The system becomes more stable if the ring is broken.
Exercise
1. Classify the following reactions as combustion or halogenation.
Answer
1. a) halogentation
b) combustion | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.13%3A_Reactions_of_Alkanes_-_a_Brief_Overview.txt |
Structural and Geometric Isomerism
4-1
a) There are five alkane isomers of hexane C6H14. Draw and name all of them.
b) The heat of combustion of hexane is 4163.2 kJ/mol. Heat of combustion of neohexane is 4159.5 kJ/mol. Predict the relative stability of these two compounds?
c) Draw and name all cycloalkane isomers of C5H10, including all possible geometric (cis-trans) stereoisomers.
4-2 Which of the following structures represent the same compound? Name the structures given in part (a), (d), (e), (f), (g)
a)
b)
c)
d)
e)
f)
g)
4-3 Each of the following descriptions applies to more than one alkane. In each case, draw and name two structures that match the description.
(a) a sec-butylheptane
(b) a trans-dimethylcyclobutane
(c) a cis-di-tert-butylcyclohexane
(d) an isopropyloctane
(e) a (1,2-dimethylpropyl)cycloalkane
(f) a bicycloheptane
4-4 Write structures for a homologous series of alcohols (R-OH) having from one to five carbons.
4-5 In each pair of compound, which compound has the higher boiling point? Explain your reasoning.
(a) Nonane or 3-ethylhexane
(b) Pentane or 2-methylbutane
(c) Octane or 2,2,4-trimethylpentane
4-6 There are four isomeric four-carbon alkyl groups. Draw them, give their systematic names and label the degree of substitution (primary, secondary, or tertiary) of the head carbon atom which is bonded to the main chain.
4-7 Draw Newman projection of the most stable conformation of the following compounds as viewed from the indicated bond.
(a) 3-methylhexane viewed at C3-C4 bond
(b) 2,2-dimethylbutane viewed at C2-C3 bond
4-8
(a) Draw two chair conformations of trans-1,2-dimethylcyclohexane and label all position as (a) for axial or (e) for equatorial.
(b) Determine the higher-energy and the lower-energy conformations
(c) Calculate the energy difference in these two conformations
4-9 Draw the two chair conformations of each compound and label the substituents as axial or equatorial. In each case, determine which conformation is more stable.
(a) cis-1-ethyl-4-methylcyclohexane
(b) trans-1-ethyl-4-methylcyclohexane
(c) cis-1-bromo-3-methylcyclohexane
(d) trans-1-bromo-3-methylcyclohexane
(e) cis-1-methyl-2-isopropylcychlohexane
(f) trans-1-methyl-2-isopropylchyclohexane
4-10 Glucose with molecular formula C6H12O6 is by far the most abundant sugar in nature. Glucose can take form as an open chain or as can be closed into a ring form. Below are chair conformations of α and β D-glucose. Using what you know about the conformational energy of substituted cyclohexane, predict which of the two isomers predominates in equilibrium. Explain your reasoning.
4-11 Provide a line drawing corresponding to each of the following Newman projections and name them using IUPAC rules.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
4-12 Draw Newman projections along the C3-C4 bond to show the most stable and the least stable conformation of 2,3,5-trimethylhexane.
4-13 In β-D-glucose, the hydroxyl group in C1 position is cis to the CH2OH group in C5 position, as shown in the figure below.
There are two chair conformations of β-D-glucose. Draw them and identify which conformation is more stable.
4.15: Solutions to Additional Exercises
Structural and Geometric Isomerism
4-1 a)
b) The isomer that releases the least energy is the most stable, so neohexane is more stable than hexane.
c)
4-2
a) The following structures all represent hexane.
The following structure represents 3-methylpentane:
b) Both structures represent cis-2-pentene:
and
Both structures represent trans-2-pentene:
and
Both structures represent 1-pentene:
and
c) Both structures represent 3-ethyl-3-heptene:
and
These three structures all represent 3-ethyl-2-heptene:
d) Both structures represent (3R)-3-methylhenxane:
All four structures represent (3S)-3-methylhenxane:
e) Both structures represent trans-1,2-dichlorocyclobutane:
All four structures represent cis-1,2-dichlorocyclobutane:
f) All three structures represent trans-1,2-cyclohexanediol:
All three structures represent cis-1,2-cyclohexanediol:
Both structures represent trans-1,4-cyclohexanediol Cis-1,4-cyclohexanediol:
g) Both structures represent 3,4-dimethylpentan-2-ol:
All three structures represent 2,4-dimethylhexan-3-ol:
4-3
(a) a dimethylnonane
(b) a trans-dimethylcyclobutane
(c) a cis-di-tert-butylcyclohexane
(d) an isopropyloctane
(e) a (1,2-dimethylpropyl)cycloalkane
(f) a bicycloheptane
Bicyclo[2.2.1]heptane Bicyclo[3.1.1]heptane
4-4
4-5
(a) Nonane has the higher boiling point because it has the higher molecular weight. Recall that higher molecular weight compounds have more surface area, and therefore they have stronger London dispersion forces. As a result, higher molecular weight compounds have the higher boiling temperatures.
(b) Pentane has the higher boiling point. Pentane has a straight chain while 2-methylbutane is branched. Compared to a straight-chain isomer, a branched hydrocarbon has a lower boiling temperature because of its smaller surface area.
(c) Octane has the higher boiling point because 2,2,4-trimethylpentane is highly branched while octane is a straight-chain hydrocarbon.
4-6
4-7
(a)
(b)
4-8
(c) The first conformer has one gauche interaction between the –CH3 groups, so the strain energy of this conformer is 0.9 kcal/mol. The second conformer has four 1,3-diaxial interaction between H and –CH3 groups, so its strain energy is 4 x 0.9 = 3.6 kcal/mol. Therefore, the energy difference in these two conformations is 3.6 – 0.9 =2.7 kcal/mol.
4-9
(a)
(b)
(c)
(d)
(e)
(f)
4-10
In α-D-glucopyranose, the hydroxyl group at C1 occupies an axial position. In β-D-glucopyranose, the hydroxyl group at C1 of occupies an equatorial position, which is the more stable structure. So the β forms predominates in equilibrium.
4.11
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
4.12
4.13 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.14%3A__Additional_Exercises.txt |
Learning Objectives
After reading the chapter and completing ALL the exercises and homework, a student can be able to:
• recognize and distinguish between the four major types of organic reactions (additions, eliminations, substitutions, and rearrangements) - refer to section 5.1
• accurately and precisely use reaction mechanism notation and symbols including curved arrows to show the flow of electrons - refer to section 5.2
• identify nucleophiles and electrophiles in polar reactions - refer to section 5.3
• perform calculations using the equation $ΔGº = –RT \ln K = –2.303 RT \log_{10} K$ and explain the relationship between equilibrium and free energy - refer to section 5.4
• calculate reaction enthalpies from bond dissociation energies - refer to section 5.5
• draw Reaction Energy Diagrams from the thermodynamic and kinetic data/information - refer to se5.6
• use a Reaction Energy Diagram to discuss transition states, Ea, intermediates & rate determining step - refer to section 5.6
• draw the transition states & intermediates of a reaction - refer to section 5.6
• describe the structure & relative stabilities of carbocations, free radicals and carbanions - refer to sections 5.7 - 5.9 respectively
• Explain the mechanism & energetics of the free-radical halogenation of alkanes - refer to section 5.10
• Predict the products of chlorination & bromination reactions of alkanes based on relative reactivity and selectivity - refer to section 5.11
• describe the similarities and differences between reactions performed in the lab with biochemical reactions - refer to section 5.12
• 5.1: Types of Organic Reactions
The four main classes of organic reactions are additions, eliminations, substitutions, and rearrangements.
• 5.2: Reaction Mechanism Notation and Symbols
Arrows are used by chemists to communicate electron flow in mechanisms, reaction completion/equilibrium, and resonance relationships. It is important to use accuracy when selecting the type of arrow for reactions and precision in drawing the location of the arrow head and tail for the curved arrows of electron flow.
• 5.3: Polar Reactions- the Dance of the Nucleophile and Electrophile
Sterics and electronics are the underlying driving forces for polar organic reactions. The electron rich nucleophile (Nu:) reacts with the electron poor electrophile through a variety of pathways that can be limited and/or influenced by steric hindrance. We explore and learn the polar reaction pathways in subsequent chapters.
• 5.4: Describing a Reaction - Equilibrium and Free Energy Changes
The relationship between equilibrium and free energy is reviewed quantitatively and applied to organic reactions conceptually.
• 5.5: Homolytic Cleavage and Bond Dissociation Energies
The products of homolytic cleavage are radicals and the energy that is required to break the bond homolytically is called the Bond Dissociation Energy (BDE) and is a measure of the strength of the bond.
• 5.6: Reaction Energy Diagrams and Transition States
Reaction energy diagrams efficiently and effectively communicate the thermodynamics and kinetics of chemical reactions in a single diagram. They are a useful tool in learning organic chemistry.
• 5.7: Reactive Intermediates - Carbocations
A carbocation is a cation in which carbon has an empty p orbital and bears a positive charge creating a highly reactive intermediate. Comparing the relative stability of reaction intermediates helps elucidate reaction mechanisms and predict major and minor products.
• 5.8: Reactive Intermediates - Radicals
A radical (more precisely, a free radical) is an atom, molecule, or ion that has unpaired valence electron (half filled orbital) creating a highly reactive intermediate.
• 5.9: Reactive Intermediates- Carbanions and Carbon Acids
A carbanion is an anion in which carbon has an unshared pair of electrons and bears a negative charge creating a highly reactive intermediate.
• 5.10: The Free-Radical Halogenation of Alkanes
Free radical halogenation of alkanes is 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. We will apply the reaction concepts discussed in this chapter to this reaction to show how empirical data supports these theories.
• 5.11: Reactivity and Selectivity
In general, high reactivity correlates with low selectivity and vice versa. Depending on the structure of the substrate, reaction conditions can be optimized for high reactivity or high selectivity and occasionally for both.
• 5.12: A Comparison between Biological Reactions and Laboratory Reactions
Biochemical reactions occur within our body fluids at a typical pH of 7.4 and temperature of 98.6C. Our biochemistry relies on enzymes to catalyze physiological reactions within this narrow range of environmental conditions. Synthetic organic chemists can create extreme conditions within reactions flasks to catalyze and promote chemical reactions.
• 5.13: Additional Exercises
This section has additional exercises for the key learning objectives of this chapter.
• 5.14: Solutions to Additional Exercises
This section has the solutions to the additional exercises from the previous section.
05: An Introduction to Organic Reactions using Free Radical Halogenation of Alkanes
Learning Objective
• recognize and distinguish between the four major types of organic reactions (additions, eliminations, substitutions, and rearrangements)
Introduction
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.
Addition Reaction
Elimination Reaction
Substitution Reaction
Rearrangement Reaction
In an addition reaction the number of σ-bonds in the substrate molecule increases, usually at the expense of one or more π-bonds. The reverse is true of elimination reactions, i.e.the number of σ-bonds in the substrate decreases, and new π-bonds are often formed. Substitution reactions, as the name implies, are characterized by replacement of an atom or group (Y) by another atom or group (Z). Aside from these groups, the number of bonds does not change. A rearrangement reaction generates an isomer, and again the number of bonds normally does not change.
The examples illustrated above involve simple alkyl and alkene systems, but these reaction types are general for most functional groups, including those incorporating carbon-oxygen double bonds and carbon-nitrogen double and triple bonds. Some common reactions may actually be a combination of reaction types.
Example: substitution Reaction
The reaction of an ester with ammonia to give an amide, as shown below, appears to be a substitution reaction ( Y = CH3O & Z = NH2 ); however, it is actually two reactions, an addition followed by an elimination.
Example: Addition reaction
The addition of water to a nitrile does not seem to fit any of the above reaction types, but it is simply a slow addition reaction followed by a rapid rearrangement, as shown in the following equation. Rapid rearrangements of this kind are called tautomerizations.
Exercise
1. Classify each reaction as addition, elimination, substitution, or rearrangement.
Answer
1. A = Substitution; B = Elimination; C = Addition | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.01%3A_Types_of_Org.txt |
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