© Department of Chemistry, The University of Western Ontario Chem 2223b Intersession 2008: Lipids
• This section discusses the properties of triglycerides and fatty acids, the metabolism and biosynthesis of fatty acids and other lipids, soaps and detergents, and vitamins.
• Background material, from Chem 2213a or otherwise, that is important includes: o Reactions of carboxylic acids and acid derivatives o Reactions of carbonyl compounds o Reactions of the α-carbon o Dehydration reactions of alcohols o Substitution reactions o Reactions of alkenes and aromatic compounds Lipids 2 A. Classification of Lipids
• Lipids are a mixed collection of compounds that are water-insoluble, yet are soluble in less-polar solvents such as dichloromethane, acetone, and diethyl ether.
• Therefore, lipids encompass a large assortment of biomolecules, and they are divided into two general categories based on what they contain:
o A relatively large hydrophobic region and a smaller hydrophilic region • Fatty acids • Triglycerides (fats and oils) • Phospholipids • Prostaglandins • Fat-soluble vitamins (A, E, D, K)
o A common structural element: a fused tetracyclic ring system • Cholesterol, steroids, and other derivatives Lipids 3 B. Fats, Oils, and Fatty Acids
1. Triesters of Glycerol
• Fats are solids, while oils are liquids, at room temperature. Aside from this, both fats and oils are triesters of glycerol. Scientists and the medical community refer to these collectively as triglycerides (or triacylglycerides), and they can be hydrolyzed:
O
CH2 O C R CH2 OH O H+ or OH– CH O C R CH OH 3 RCOOH O
CH2 O C R CH2 OH
• The R groups are long alkyl chains (same or different) that may contain sites of unsaturation (C=C double bonds). These R groups are part of the fatty acids.
• In general, triglycerides containing one or more unsaturated fatty acids have melting points that are lower than those with fully saturated fatty acids.
• Fatty acids on their own (not esterified to glycerol) are not found in substantial amounts in the body, but they can be obtained from triglycerides by hydrolysis. Lipids 4
• Mechanism for the acid-catalyzed hydrolysis of an ester:
O O H+ RCOR' H ORCOH HOR' 2
• This is the reverse of a Fischer esterification, so it is an equilibrium reaction. Acid- catalyzed hydrolysis of esters does NOT go to completion. Lipids 5
• On the other hand, base hydrolysis (saponification) proceeds to completion. The reaction is stoichiometric, not catalytic, because base is not regenerated. The carboxylate and the alcohol cannot reform the ester.
• Homework: work out the mechanism of base hydrolysis (hint: remember that this is simply a nucleophilic acyl substitution, and compare it to amide hydrolysis in base).
O O – – RCOR' OH RCO HOR' Lipids 6 2. Properties of Triglycerides and Fatty Acids
• The presence of unsaturation (cis C=C bonds) in the fatty-acid portions lowers the melting point of the triglyceride.
• Saturated fatty acid has a regular shape, while unsaturated cis fatty acid does not. The introduction of cis fatty acid into the triglyceride makes it more difficult for the triglyceride to pack together.
• The cis C=C bond adds kinks to the structure. Thus, triglycerides with more cis unsaturation are even harder to pack, resulting in even lower melting points. (Analogous to the BP of straight and branched alkanes)
• Naturally occurring fatty acids are almost always cis and rarely trans. Lipids 7
• The presence of unsaturated fatty acids in triglycerides results in lower melting points, and the same trends apply to the fatty acids themselves.
O Stearic acid (18:0) MP = 70°C OH O OH Some trans (18:1) MP = 52°C O OH Oleic acid (18:1) MP = 16°C O Linoleic acid (18:2) MP = –5°C OH
A monounsaturated fatty acid has one C=C bond, while a polyunsaturated fatty acid has two or more C=C bonds. (See food labels)
• Due to the way they’re synthesized, fatty acids usually occur in integral numbers of two-carbon units, most commonly C16 and C18. Lipids 8
• Fatty acids have both a hydrophilic (polar) head and a hydrophobic (non-polar) tail.
O Tail Head OH
o In other words, we have a mixture of a highly water-soluble portion (head) with a portion that is highly water-insoluble (tail).
• Fatty acids are practically insoluble in water. This is because they are weak acids (pKa ~4.75), and the majority of the molecules will remain unionized. Only the charged, ionized form (RCOO−) is water-soluble.
o Although fatty acids are essentially carboxylic acids, we don’t call the short ones, such as acetic acid, fatty acids. Lipids 9
• Like most organic compounds, fatty acids are they are less dense than water, so they will form a layer at the surface if mixed with water.
o A portion of the molecule wants to be with water (the hydrophilic head), and a portion that wants to be away from water (the hydrophobic tail).
o The best arrangement is a monolayer, where the fatty-acid heads face the water and the tails are exposed to the air.
air hydrophobic interactions (van der Waals forces) between tails H-bonding and polar interactions between heads and with water
water surface
o Note that the monolayer is in equilibrium with the tiny amount of ionized fatty acid that is dissolved in the water. Lipids 10 B. In-the-lab Preparation and Biosynthesis of Fatty Acids
• In natural systems, fatty acids almost always have an even number of carbon atoms, usually in the range of 12-20 and most commonly 16 or 18. They are also biosynthesized not as acids, but as esters.
• This even-number phenomenon arises from an abundant carbon source used for biosynthesis: a derivative of acetic acid. Long fatty acids are generated by joining multiple acetyl units together, e.g. a C18 fatty acid has nine C2 units.
O # new C-C bonds? OH
• A crucial step in fatty-acid synthesis is the formation of a new C-C bond between two acetate-ester units. How many C-C bond forming reactions do you know of? o Cyanide as a nucleophile Å cyanide in a biological system? o Grignard reagent Å present biologically? o Aldol condensation Å works with aldehydes and ketones o Claisen condensation Å an α-carbon reaction that works with esters
• We’ll first determine how we could synthesize a fatty acid in the lab and then see how these in-the-lab concepts could also be extended to biosynthesis. Lipids 11 1. In-the-lab Synthesis of Fatty Acids
• The key step is the formation of a new C-C bond by joining two acetate esters together via a Claisen condensation. This is a nucleophilic acyl substitution, where the enolate of one ester acts as a carbon nucleophile, replacing the OR group.
O O O Na OR (strong base)
H3C OR H3C OR β-keto ester Lipids 12
• A series of chemical conversions takes us to a saturated chain.
O O multiple steps O
OR OR
repeat Claisen to yield C6 ester Lipids 13
• The four types of reactions that must occur biologically are exactly the same as those required in the lab synthesis. 1. Formation of C-C bond, resulting in a β-ketoester 2. Reduction of carbonyl group to produce a β-hydroxyester 3. Dehydration of alcohol to form an α,β-unsaturated ester 4. Reduction of C=C, which leads to the saturated alkyl chain
• The biosynthesis cannot be identical to the laboratory synthesis, because:
o Claisen reactions also have equilibrium constants that do not favour the products. One way to increase the products is to increase the amount of enolate, but this is difficult at physiological pH, as the α-H is not very acidic. Biological solution: modify it and perform a Claisen-type condensation
o Since we’re adding acetate units one at a time, we need a way to prevent the growing chain from diffusing away until the desired length is reached.
o NaBH4, conc H2SO4, and H2/Pt don’t exist biologically
Lipids 14 • Up to this point, some of the concepts we’ve covered include: o Properties of fats, oils, and fatty acids o Mechanisms of ester hydrolysis (nucleophilic acyl substitution) o Claisen Condensation o Chemical steps needed for fatty-acid synthesis in the lab
• Attempt:
o Practice problems in the lab manual: 1 – 7
o 2006 Midterm: 19
o 2007 Term Test #2: 13
• It is crucial at this stage that you thoroughly understand the Claisen condensation.
Lipids 15 2. Biosynthesis of Fatty Acids a. Modified Claisen condensation
• A modified Clasien condensation is used so that the reaction favours the products under physiological conditions. This is done by: o Using a thioester (RCOSR’) instead of a regular oxygen ester (RCOOR) o Using a α-carbanion equivalent instead of a true carbanion
1. Thioesters
• Thioesters (an acid derivative) are more reactive than their oxygen counterparts.
o The first reason is because RS− is a better leaving group than is RO−.
O O O + SR SR SR Lipids 16
o The second reason is because the reactivity of an acid derivative O depends on the amount of δ+ charge on the carbon of the C=O. How likely the C is going to be attacked by a Nu depends on its δ+. δ+ X For example, O esters are more reactive than amides, because nitrogen is a better resonance donor and can stabilize the δ+ charge. Similarly, S esters are more reactive than O O O esters, because S cannot donate very well by resonance: it cannot overlap a 3p orbital with 2p. OR SR + i.e. the δ charge is larger in thioesters. smaller δ+ larger δ+
A consequence of this the reduced pKa of the α-H on the thioester. The enolate formed on the thioester is more stable, because it is next to a larger δ+. (However, thioesters are still not acidic enough to have substantial amounts of enolate being made).
O O
H2C OR H2C SR
less stable anion more stable anion
• The most-common thol used in thioesters is Coenzyme A. Lipids 17
2. Carbanion equivalents
• A standard Claisen requires the use of a base to generate the α-carbanion (the enolate), which acts as the nucleophile. This is difficult at physiological pH.
• How can we make the enolate without a strong base? We rely on the decarboxylation of malonyl thioester. Decarboxylation also drives the not-so- favourable Claisen reaction to completion.
• Recall that decarboxylation reactions typically require a β-carbonyl group relative to the COO−. This is because the electrons from decarboxylation need to be accepted, and this is performed by the carbonyl group.
O O O O –CO2 O SR SR SR
malonyl enolates of acetyl thioester thioester Lipids 18
• Although we have made the enolate without the use of a base, it is it is generally accepted that the loss of CO2 is concomitant with nucleophilic attack. In other words, we avoid the enolate completely. i.e. malonyl thioester is the carbanion equivalent.
O acetyl C2 O O O O –CO2 SR SR O O SR SR SR O SR C4 β-keto thioester malonyl C3
• Malonyl thioester is made from acetyl thioester using ATP and biotin, a CO2 carrier. This enzymatic reaction uses CO2 dissolved as bicarbonate.
O O O – Biotin + HCO3 SR O SR ATP ADP, Pi
• i.e. The nucleophile is malonyl thioester, while acetyl thioester (or the growing chain) is the electrophile. The same malonyl C that is lost is the one that came from CO2. Lipids 19 • Mechanism of biotin-dependent carboxylation:
− o ATP first phosphorylates HCO3 to form carboxyphosphate (should know mechanism... same as the phosphorylation reactions already seen).
o Carboxyphosphate is used to make carboxybiotin
O O O O P O HO O P O O O
O O O
H N NH N NH HO
R R R R
Lipids 20 o Carboxyl group is subsequently transferred to acetyl CoA
O O O
N NH HN NH HO
R R R R
O O O
SR HO SR Lipids 21
• The overall cycle that we have for growing a long fatty acid is:
O O O
SR 1 O SR acetyl thioester 2 malonyl thioester (made from acetyl thioester)
Once at the butyryl thioester, use it instead of acetyl thioester O O for the next cycle. (And so forth)
O SR2 β-ketothioester SR2 butyryl thioester
O OH O
SR2 SR2
α,β-unsaturated thioester β-hydroxythioester Lipids 22 b. The diffusion problem
• Fatty acids are typically synthesized as a C16 molecule. (Additional modifications are possible, such as extension to C18 and other reactions). Since the fatty acid grows two carbons at a time, it must not be allowed to diffuse away until it reaches C16.
• The reactions are done on a multi-enzyme complex, fatty acid synthtase (FAS). The growing fatty acid thioester (CoA) also is replaced by acyl carrier protein (ACP). ACP is very strongly bound to FAS, so the growing chain does not escape and float away until it is deliberately detached from ACP.
O O H3C CH3 O C O HS O O CH N N P H H NH
OH O polypeptide chain
from pantothenic acid (Vitamin B5)
• The CoA thioesters can also be replaced by a cysteine, an amino acid in FAS that has an SH side chain, to form cysteine thioesters.
• Diagrammatically, we have... Lipids 23 O O O C H3C SCoA The acetyl group is loaded onto - HS the fatty acid synthase O CH2 SCoA complex HS FAS ACP The malonyl-ACP component is bound to an ACP-binding site
-CoA on FAS.
O Both reactions are catalyzed by C enzymes with transacylase O O H3C S activity - FAS O CH2 S ACP
The long arm of ACP permits movement for the Claisen-
O type condensation reaction and allows the product to O- H3C C S access other units of FAS for O CH subsequent transformations 2 FAS
O S ACP
Lipids 24
OO HS An enzyme catalyzes the H3CCH2 S FAS Claisen-type condensation β-ketothioester ACP
Reduction: NADPH + H+
The reduction is OH O HS enzymatically performed
H3CCH2 S FAS β-hydroxythioester ACP
Dehydration
Dehydration is also O HS enzymatic
S H3C FAS α,β-unsaturated thioester ACP
Lipids 25
O HS Enzymatic reduction S FAS of C=C bond to give H3C
ACP a fully reduced tail
swap positions O
S H3C growing chain moves HS FAS to spot of original electrophile for next ACP cycle
new electrophile add new malonyl O thioester Nu S H3C Ready for next cycle O O FAS - O CH2 S
Lipids 26 • Note that the growing fatty acid is always attached to either ACP, which is in turn attached to FAS, or directly to one of the enzymes in the FAS complex.
o The electrophile is always the growing chain, and it grows at the thioester end.
o The nucleophile is always malonyl thioester, regardless of the length of the growing chain. Remember that the use of malonyl thioester, a C3 compound, only adds an acetate unit (C2), because one carbon is lost as CO2.
o There are two reductions and one dehydration per cycle.
• After seven cycles, we will have made seven new C-C bonds and a C16 fatty acid.
7 new C-C bonds O C C C C C C C C C C C C C C C C SR acetyl malonyl-derived C2 units derived
• Since the malonyl-derived units are made from acetyl thioester + CO2, and the CO2 is later lost, we can also say the C16 compound originates from eight acetyl units. Lipids 27 c. Reduction and dehydration reagents
• In the laboratory synthesis of fatty acid, we had the:
o Reduction of the β-ketoester to the β-hydroxyester with NaBH4
o Dehydration of the β-hydroxyester to the α,β-unsaturated ester with conc H2SO4 o Reduction of the α,β-unsaturated ester by catalytic hydrogenation
• These reagents are nonexistent in biology, so we need a different approach.
1. Reduction of ketone to alcohol
O O OH O
SR SR β-ketothioester β-hydroxythioester
• NADPH is used as the reducing agent (same mechanism as NADH).
Lipids 28
2. Dehydration of alcohol to alkene
OH O O
SR SR β-hydroxythioester α,β-unsaturated thioester
• Compare this to the formation of phosphoenolpyruvate, as seen in Carbs.
3. Reduction of alkene to alkane
O O
SR SR α,β-unsaturated thioester saturated tail
• Chemically, we would require H2/metal, because hydrides do not attack the non- polar C=C bonds. However, biologically, the C=C alkene is reduced to an alkane by NADPH, which is a hydride source! Lipids 29 • Why does this work? It is because the alkene is conjugated to the C=O. Resonance structures can be drawn such that we can place a positive charge on the β carbon.
o i.e. the β carbon is electrophilic and can be attacked by hydride
O O
SR SR
• Reaction with NADPH would generate the enol form, which tautomerizes to the product. The addition of a nucleophile to the β carbon of an α,β-unsaturated carbonyl is named a Michael Addition.
• This product, attached to ACP, is formed after one cycle/round of biosynthesis. For the subsequent cycle, it becomes the new ELECTROPHILE. Lipids 30 d. Energetics of fatty-acid biosynthesis
• To elongate an existing fatty acid thioester by 2 C, (one acetyl unit), we require:
o 1 acetyl CoA
o 1 ATP to convert the above acetyl CoA to malonyl CoA
o 2 NADPH for the respective reductions e. Other fatty acids
• The normal product of fatty-acid biosynthesis in animals is palmitic acid (C16). Separate elongase and desaturase enzymes are responsible for elongating this fatty acid and producing monounsaturated fatty acids, respectively. Desaturase enzymes are remarkable in that they introduce a cis alkene using atmospheric oxygen as the oxidizing agent. (We won’t look at these in detail).
Palmitic acid (C16) O elongase OH Stearic acid (C ) 9 18 Oleic acid C18Δ desaturase Lipids 31 • There are some polyunsaturated fatty acids (PUFA’s) that humans cannot synthesize and must obtain in their diet. These two are the “parent” PUFA’s, as they can be converted to other necessary PUFA’s by the liver when consumed.
O OH
9,12 Linoleic acid C18Δ
An omega-6 fatty acid, which has a double bond six carbons from the omega end (methyl)
O OH
9,12,15 Alpha linolenic acid (ALA) C18Δ
An omega-3 fatty acid, which has a double bond three carbons from the omega end (methyl)
Lipids 32 • Up to this point, some of the concepts we’ve covered include: o Mechanism of biotin o Reactivity of thioesters o Biological purpose of malonyl CoA o The diffusion problem o Acetyl CoA (or the growing chain) as the electrophile o Energy requirements
• Attempt:
o Practice problems in the lab manual: 7 – 17
o 2006 Midterm: 20 – 23
o 2006 Midterm Intersession: 27 – 32
o 2007 Term Test #2: 14 – 18
Lipids 33 C. Catabolism of Triacylglycerides and Fatty Acids
1. Hydrolysis of Triacylglycerides
• Triacylglycerides are poorly absorbed from the digestive tract, but the hydrolysis products can be absorbed. Hydrolysis is performed lipase enzymes.
O
CH2 O C R CH2 OH O lipases CH O C R CH OH 3RCOOH O
CH2 O C R CH2 OH
• Like chymotrypsin, lipases use the same catalytic triad (Asp-His-Ser) and are mechanistically identical, except the substrate is an ester instead of an amide.
• Lipases are also the target of some weight-loss drugs. Orlistat (sold as Xenical and Ally) inhibits the enzymes, and the triacylglycerides pass through the digestive tract.
Lipids 34 2. Phosphorylation and Oxidation of Glycerol
• Glycerol, a three-carbon compound, is readily converted into a glycolysis intermediate via a phosphorylation followed by an oxidation.
CH OH CH OH CH OH 2 ATP 2 NAD+ 2 H OH H OH O 2- 2- CH2OH CH2OPO3 CH2OPO3 dihydroxyacetone phosphate • For practice, work out the mechanisms of these two reactions (refer to the structures of ATP and NAD+ if needed).
Lipids 35 3. β-Oxidation of Fatty Acids
• Fatty acids are degraded into acetyl CoA via a pathway known as β-oxidation, where the β carbon oxidized. The pathway is quite similar to the reverse of fatty-acid biosynthesis, where a β-keto was involved, with some differences.
• First, the fatty acid is enzymatically converted to the CoA thioester.
O
O
ATP O
2- OPO3
HS-CoA If we start with a C18 O fatty acid, we will obtain 9 acetyl CoA. Where are the 9 units, and which SCoA carbons need to be oxidized? Lipids 36
• FAD oxidizes the saturated tail, resulting in an α,β-unsaturated thioester. (Note: in biosynthesis, NADPH was used to reduce the C=C).
O
SCoA O N NH
N N O R O
SCoA H O N NH
N N O R H Lipids 37 • Conjugate addition (Michael Addition) of water forms a β-hydroxy thioester. Subsequent oxidation by NAD+ forms a β-keto thioester. These two steps are mechanistically opposite to those seen in biosynthesis.
HOH O
SCoA
OH O
SCoA
NAD+ O O
SCoA
Lipids 38 • The last step is a unique C-C bond-breaking reaction that uses cysteine residue to expel acetyl CoA as a leaving group. The enolate (α-carbanion) of acetyl CoA is protonated as it departs (compare to retro-aldol reaction).
O O
SCoA Enz S H O O
S-Enz SCoA
HS-CoA
O and repeat.... oxidation, hydration SCoA oxidation, cleavage
Lipids 39 D. Soaps and Detergents
• Both soaps and detergents perform the same task, but soaps are made from naturally occurring materials, while detergents are considered to be synthetic.
1. Soaps
• Soaps are metal salts of fatty acids, prepared by the base hydrolysis of triglycerides.
O
CH2 O C R CH2 OH O NaOH O CH O C R CH OH 3 R C O Na O
CH2 O C R CH2 OH
• The sodium salts are soluble at low concentrations. However, at very high [Na+], the equilibrium below shifts to the left and the salt precipitates (this is useful for isolating and purifying soap… Lab 4).
O O insoluble soluble R C O Na RCO Na Lipids 40
• Since the water-soluble soap molecule has a very hydrophillic head (ionic) and a hydrophobic (lipophilic) tail, it will form spherical structures known as micelles.
• The exterior s ionic, so the micelle is water-soluble. (The entire exterior is surrounded by water, whereas in fatty acid monolayers, one side was air).
• The inside is hydrophobic and can act as a solvent for oil, grease, and other hydrophobic dirt.
Lipids 41 2. Detergents
• Soaps are easily made from natural sources, but there is one major problem with them. Their Na+ and K+ carboxylate salts are soluble, but the carboxylates precipitate in the presence of polyvalent ions: Ca2+, Mg2+, Fe3+, etc.
• These polyvalent ions that are found in hard water. Soaps precipitate out and perform poorly in hard water, and the precipitates can also dull clothing.
O 2+ RCO M
• The solution is to use synthetic detergents that replace the carboxylate with alternatives that won’t precipitate out in hard water. Usually, there are two different choices: sulfonates or monoesters of sulfate. e.g. sodium dodecylsulfate (SDS)
OH OSO3H H2SO4 NaOH
OSO3 Na Lipids 42
• Note that sulfates and sulfonates are different functional groups. With the latter, the S is linked directly to a C, with no O in between them.
• The first inexpensive sulfonate detergents were developed in the 1940’s, and these were the alkylbenzene sulfonates. Similar to a natural soap, they featured an ionic head and a hydrophobic alkyl-aromatic tail.
O O Na S O
• The starting materials for such detergents were cheap, as they were easily obtained from the petrochemical industry. Their synthesis was also easy: three steps! Lipids 43
1. Formation of the alkyl tail
• The first step is an acid-catalyzed, self-polymerization of propylene (electrophilic addition to alkenes).
• Propylene is protonated to form a carbocation, which acts as an electrophile to attack another propylene (review electrophilic addition to alkenes). The process repeats and is done under controlled conditions to give the desired chain length.
H
-H
• For these detergents, an alkyl tail of about C12 was ideal. The ones that were too long were too insoluble to act as a detergent, and the ones that were too short didn’t form micelles. Lipids 44 2. Attachment of tail to benzene
• The tail, presently an alkene, is attached to benzene by an electrophilic aromatic substitution (EAS). Acid protonates the alkene to make a carbocation.
H and
3. Sulfonation
• Sulfonation is another electrophilic aromatic substitution. This gives the sulfonic acid, which can be treated with a base to yield the sodium sulfonate salt.
SO3Na
1. H2SO4 2. NaOH
• Sulfonation occurs at the para position – why? Lipids 45
• This is because the alkyl group, which is an electron-donating group, can stabilize carbocations resulting electrophilic addition (E+) to the para position.
EDG EDG EDG
+ charge adjacent a group that donates electrons (inductively or by resonance) E H H E H E
• Addition to the ortho position also results in a positive charge next to an EDG, but steric hindrance makes this addition less likely than para addition.
E EDG EDG E H H
Lipids 46
• However, if we have an electron-withdrawing group, we want to keep the positive charge away from it. Addition to the meta position accomplishes this. (This is ideal because the carbon next to the EWG already has a δ+ due to the EWG on it).
EWG EWG
not possible to draw a resonance E structure that places the + charge immediately next to the EWG H E H
• Thus, EDG are ortho/para directors, and EWG are meta directors.
• Although these alkylbenzene sulfonate detergents were cheap and worked well, they had a serious problem. The non-natural branched alkyl chain was not biodegradable by bacteria, so they accumulated in the environment.
• Today, they have been replaced by linear tails, which are more difficult and expensive to synthesize (we won’t look into the details). Lipids 47 3. Additives
• Synthetic detergents did not precipitate in the presence of hard-water ions, and thus they overcame the problem of natural soaps. However, the hard-water ions were still undesirable, because they combined with clays, mud, and other contaminants to leave laundry grey. O O • Phosphate-based additives solved this problem by chelating P the positively charged ions. O O O O • However, phosphates are also fertilizers and caused algal P P blooms in lakes and rivers, thereby reducing oxygen levels O and suffocating other organisms (fish, etc.). O O
M2+ O O • The modern solution nowadays is to use silicates, which are functionally similar, except Si with more negative charges. That also makes the O O water more basic, which cleans better! O O Si Si O O O Lipids 48 E. Waxes
• Waxes are different from fats and oils in that they are monoesters of long-chain alcohols and fatty acids. They are substantially different from triglycerides.
• In general, waxes are solids at RT. They are secreted to form protective, waterproof coatings for feathers, fur, exoskeletons (insects), and leaves and fruit (plants).
alcohol portion acid portion O
Bee's wax O 29 O 14
Carnauba wax O 33 30
• Apples are coated in FDA-approved carnauba wax after harvest and washing to keep them fresher, and to make them look shiny… a marketing incentive. Lipids 49 F. Phospholipids
• Phospholipids, also called phosphoacylglycerols, are the second-most abundant group of naturally occurring lipids. They are found almost exclusively in plant and animal cell membranes.
• They consist of glycerol esterified to two fatty acids and one phosphate. This skeletal structure, which is a monoester of phosphoric acid, is a phosphatidic acid. (Compare this structure, below, to a fat or an oil: one tail replaced by phosphate).
Hydrophillic head Hydrophobic tails O
CH2 O C O CH O C O
O P O CH2 O Monoester of phosphoric acid
• The fatty acids are usually a mix a saturated and monounsaturated ones, which forms a semi-fluid, liquid-like combination. This allows for membrane fluidity. Lipids 50
• Phospholipids contain the phosphatidic acid backbone esterified to a second alcohol at the phosphate, forming a phosphodiester (two esters at the phosphate).
• Various alcohols may be used, all of which result in a very hydrophillic head.
NH3
COO Some common alcohols: O O P O O NH3 ethanolamine HO CH2 CH CH2 O O O O NMe3 choline C C HO
OH OH HO OH inositol HO OH
Phosphatidylserine
Lipids 51
• Like soaps, phospholipids have a very hydrophilic head and a hydrophobic tail. However, instead of forming micelles, they form bilayers. These bilayers are the basis of biological membranes.
• Why form bilayers? Compared to soaps, phospholipids have an extra tail, which necessitates the rebalancing of two forces:
1. The driving force, caused by hydrophobic interactions, for the tails to be close to each other and away from contact with water.
2. The repulsive electrostatic forces between charged head groups.
• In other words, we want the tails close together, but the heads far apart. Lipids 52 • With soaps, the proper balance is found in a micelle structure, which is spherical. The heads are sufficiently far apart to minimize electrostatic repulsions between them, yet the tails are still close enough to each other.
• Due to this structure, there are gaps. Micelles are leaky, and molecules can enter and exit.
• The extra tail in phospholipids takes up additional room. This is beneficial, because it allows the tails to come together even more closely, while maintaining sufficient separation between the heads.
• Since the tails contact each other more than in micelles, there are no gaps; i.e. they are not leaky. This is an valuable, as the bilayers are used for cell membranes. Otherwise, the contents of the cell would escape.
• Note that both sides, which correspond to the interior and exterior of the cell, are aqueous.
Lipids 53
• In the fluid mosaic model of biological membranes, there are proteins, carbohydrates, and other lipids embedded within and on the surface of the phospholipid bilayer.
o The membrane is considered fluid in that the components are essentially floating in the membrane and can move around.
o The term mosaic indicates that the components exist as discrete units, so there is always a heterogeneous structure.
• Because of the non-leaky membrane, most hydrophilic and polar molecules do not diffuse across it. Such molecules need to be actively transported. Lipids 54 G. Prostaglandins
• Prostaglandins consist of a C20 prostanoic acid COOH skeleton characterized by a C7 carboxylic acid chain, a cyclopentane ring, and a C8 chain.
• Prostaglandins are not stored in the body, but they C20 prostanoic acid skeleton are synthesized from arachidonic acid only in presence of specific physiological triggers: injury or infection. These resulting prostaglandins can cause various effects, including inflammation.
• Prostaglandins originate from the compound formed from the oxidation of arachidonic acid by O2. The breakthrough discovery of the first selective inhibitor of PG biosynthesis, Vioxx, by Merck in Montreal is a Canadian success story.
Arachidonic acid Prostaglandin G2 9 O 8 6 5 1 COOH O2 COOH 10 7 13
11 12 14 15 O OOH Enzyme: Inhibitors: Lipids 55
• Some of the many other prostaglandin derivatives are shown here (don’t memorize).
Lipids 56 H. Leukotrienes
• Like prostaglandins, leukotrienes are also synthesized from arachidonic acid. However, they are made by leukocytes in response to allergens. (Histamines are also produced, but they are not classified as lipids).
• The first step in the biosynthetic pathway involves the epoxidation of arachidonic acid by the enzyme 5-lipoxygenase.
O 6 5 1 COOH O2 COOH
5-lipoxygenase Arachidonic acid Used to make other leukotrienes
• In asthmatics, leukotrienes can trigger asthma symptoms. New treatments:
o 5-lipoxygenase inhibitor: Zyflo (not available in Canada)
o Leukotriene receptor antagonists (LTRA): Accolate (by AstraZeneca) and Singulair (the first LTRA and the first new class of anti-inflammatory in 20 years for the treatment of asthma, also by Merck). Lipids 57
• Up to this point, some of the concepts we’ve covered include: o β-Oxidation o Soaps and detergents, EAS, and and o/m/p directors o Phospholipids o Micelles vs. bilayers o Prostaglandins and Leukotrienes
• Attempt:
o Practice problems in the lab manual: 18 – 34
o 2006 Midterm: 24 – 27
o 2006 Midterm Intersession: 33 – 35
o 2007 Term Test #2: 19 – 22
Lipids 58 I. Terpenes
• Many naturally occurring compounds, including fragrances, have carbon skeletons that contain n × 5 carbons, where n = integer. These compounds, terpenes, are comprised of a common structural unit known as an isoprene.
H2C
CH CH2
H2C CH2
• Realize however, that the actual compound isoprene is not present in the terpene; it only forms the characteristic structure of the carbon skeleton. This is because once the isoprene units are linked together, then can undergo a variety of chemical transformations, including reduction, hydration, cyclization, etc.
• What is important is the way the units are linked together. There are isoprene rules that specify the proper and necessary linkages.
Lipids 59 1. Isoprene Rules
• Although terpenes consist of C5 units, they are for historical reasons named and classified according to the number of C10 units.
# Carbons Classification # Isoprenes Must have… 10 Monoterpene 2 1 head-to-tail linkage
15 Sesquiterpene 3 2 head-to-tail linkages
20 Diterpene 4 3 head-to-tail linkages
30 Triterpene Derived from two sesquiterpenes (2 × C15)
40 Tetraterpene Derived from two diterpenes (2 × C20)
• How are the two sesquiterpenes and diterpenes joined in the C30 and C40 compounds, respectively?
• Important: other linkages may be present, but the above rules must be obeyed in order for a compound to be classified as a terpene. Lipids 60 2. Examples of Terpenes
• Monoterpenes (C10 = 2 isoprenes) must have one HÆT linkage (shown in blue).
• Other linkages maybe be present (shown in red).
CHO OH
Myrcene Limonene Menthol Citral
(bayberry) (lemon/orange) (peppermint) (lemon grass)
• Cyclizations, oxidations, etc. can be done, but the isoprene rules are obeyed. Lipids 61
• Sesquiterpenes have three isoprene units and two HÆT linkages
O
Bisabolene Turmeron Caryophyllene (termite pheromone) (tumeric) (misc essential oils)
• Diterpenes have four isoprene units and three HÆT linkages
OH OH
Steviol (from the sweet-tasting Stevia shrub)
COOH COOH Lipids 62
• Triterpenes are two sesquiterpenes linked TÆT (with 4 total HÆT)
OH
HO Dammarenediol (ginseng)
• Tetraterpenes are two diterpenes linked TÆT (with 6 total HÆT)
Lycopene (red colour of tomatoes)
Lipids 63 3. Biosynthesis of Terpenes
• The precursors for terpene biosynthesis are C5 units: dimethylallyl pryrophosphate and isopentenyl pyrophosphate. While the reaction occurs somewhere between SN1 and SN2, it is helpful to analyze it as though it proceeded entirely by SN1.
OPP DMA-OPP Both are resonance-stabilized, but we need to (electrophile) add a Nu to the primary in order to link the tail
unstable × E OPP IPP-OPP E OPP (nucleophile) Ionization forms unstable CC, but the addition of E+ forms
a much more stable tertiary, so IPP-OPP acts as a Nu instead
• DMA-OPP easily loses OPP to make a stable carbocation; hence, DMA-OPP will behave as an electrophile. IPP-OPP, on the other hand, will not ionize, because it does not give a stable carbocation. Thus, IPP-OPP acts as a nucleophile. Lipids 64 • The reaction between the DMA-OPP and IPP-OPP is better shown as follows:
H + OPP δ OPP T
OPP OPP H
new double bond is trans
OPP Geranyl-OPP (C10 with one H-T link)
• The C10 Geranyl-OPP (GPP) can be used to make monoterpenes, or, because the pink portion resembles DMA-OPP, it can act as an electrophile and be elongated. Lipids 65
• If geranyl pyrophosphate is diverted to the biosynthesis of monoterpenes, various reactions can occur. These are just some of the many possibilities:
H
OPP
resonance OPP and rotation =
Geranyl-OPP = H2O H
OH
limonene geraniol myrcene Lipids 66
• If the elongation route is taken, then a maximum of C20 is achieved. The C15 and C20 products can be used to make sesqui- and diterpenes.
Farnesyl-OPP Geranyl-OPP (C15 with two H-T)
OPP
OPP OPP
OPP
Geranylgeranyl-OPP OPP (C20 with three H-T) Lipids 67
• If elongation stops at C20, where do the C30 and C40 terpenes come from? They arise from the reductive dimerization of two C15 or C20 units. The reaction uses NADPH, but its mechanism still remains a mystery.
O O O P O P O Farnesyl pyrophosphate O O
O O O P O P O O O
NADPH
Squalene
• Squalene is subsequently used for the biosynthesis of other triterpenes and steroids. Lipids 68 4. Origin of DMA-OPP and IPP-OPP
• The branched C5 compounds used for terpenes are made from O O acetoacetyl thioester, an intermediate in fatty-acid biosynthesis. SR o Recall that this β-keto thiolester is formed from the Claisen condensation between malonyl thioester and acetyl thioester.
• A critical reaction that initiates the biosynthesis of the C5 compounds is an aldol reaction between acetoacetyl thioester and acetyl thioester.
o In the aldol, acetyl thioester is used, and not malonyl thioester, because the equilibrium constant for an aldol condensation is much more favourable (unlike the Claisen condensation).
o This reaction forms a branched C6 compound, a β-hydroxy dithioester, where the beginnings of an isoprene skeleton are recognizable.
O O O OH O + H H CH C CH C SR CH3 C SR CH C CH2 C SR 3 2 3 N O
CH2 C SR B T Lipids 69
• If we compare the aldol condensation product to IPP-OPP, we can predict the reactions that need to occur.
OH O H H IPP-OPP CH C CH C SR CH C CH 3 N 2 3 2 O
CH2 C SR CH2 CH2 OPP B T
o The extra C attached to one of the heads needs to be lost.
o The tail needs to be reduced to an alcohol, and then phosphorylated.
o The OH on the neck needs to be removed.
o A double bond between the neck and a head needs to be formed.
• The intermediate relating the aldol condensation product and IPP-OPP is a compound called mevalonic acid. Lipids 70
• First, one of the thioesters is hydrolyzed by an enzyme to a carboxylic acid. The starting material is achiral, but the product is chiral. However, the enzyme ensures that only one stereoisomer is formed.
• A couple of enzyme-catalyzed reductions then take us to mevalonic acid.
OH O OH CH C CH C SR CH C CH COO 3 2 H O 3 2 O 2 * O
CH2 C SR CH2 C SR
NADH
OH OH NADH CH3 C CH2 COO CH3 C CH2 COO
CH2 CH2OH CH2 CHO Mevalonic acid Lipids 71
• The primary alcohol of mevalonic acid is then phosphorylated twice, to give a pyrophosphate. The alcohol on the neck is also phosphorylated. • Why phosphorylate the neck OH? Instead of a β-keto group, a β-phosphate leaving group accepts the electrons generated during the decarboxylation reaction. OH O P O OH O O 3 ATP 3 ADP CH3 C CH2 COO CH3 C CH2 C O
CH2 CH2OH CH2 CH2 OPP
CH C CH OPP 3 2 Isopentenyl pyrophosphate CH2 CH2 OPP Lipids 72 • The biosynthesized IPP-OPP can then be converted to DMA-OPP by an isomerization reaction.
H H
OPP OPP OPP Isopentenyl Dimethylallyl pyrophosphate pyrophosphate
• As we have seen, in the biosynthesis of terpenes,
IPP-OPP Æ
DMA-OPP Æ
And the reaction joins them together in a head-to-tail manner.
• Try and see if you can determine how many ATP and NADH are required to synthesize IPP-OPP or DMA-OPP using acetyl CoA as the starting material.
Lipids 73
• Up to this point, some of the concepts we’ve covered include: o Fundamental structure of terpenes o Head-to-tail and tail-to-tail connections o Biosynthesis using IPP and DMA pyrophosphates o Aldol reaction o Mevalonic acid
• Attempt:
o Practice problems in the lab manual: 35 – 47
o 2007 Term Test #2: 23 – 24
Lipids 74
J. Steroids
• These compounds are plant and animal lipids that consist of a characteristic tetracyclic fused-ring system: three 6-membered rings and one five-membered ring.
17 11 13 CH3 CD H 1 14 CH3 9 15 10 8 A B 5 H 3 H H 6
• Usually, the rings are fused in a trans manner. The H and CH3 groups are also trans to each other, and they are also in the axial positions.
• Such an arrangement results in a planar, rigid structure, and this flatness is important as cell-membrane components – flat molecules take up less space.
• Steroids are synthesized from terpenes (and not fatty acids), and they usually also have axial methyl groups at C-10 and C13. Lipids 75 1. Biosynthesis of Cholesterol
• Cholesterol is biosynthesized from the triterpene (C30) squalene. If we redraw squalene, we can notice where the cyclization reactions need to occur so that new carbon-carbon bonds are formed.
Squalene
The green dashes are to-be-made bonds that will form the rings.
• The necessary C-C bonds are formed by the reaction of a carbocation with an alkene, similar to the acid-catalyzed self-polymerization reaction observed in the synthesis of alkylbenzene sulfonate detergents. Lipids 76
• First, squalene is enzymatically oxidized to an epoxide. Because of its strained nature, epoxides open very easily in the presence of an acid (review the acid- catalyzed opening of epoxides). The Nu is the electron-rich alkene.
H
O H O
HO HO HO
Lipids 77 H
H H H H H
= HO HO HO H A rearragement caused by a series of 1,2-hydride and 1,2-alkyl shifts occurs. These involve the movement of H H atoms and CH3 groups, along with their bonding electrons.
HO Lanosterol
• These reactions lead to lanosterol, an intermediate in cholesterol biosynthesis.
• Note that there are 7 chiral centres in lanosterol, and none in squalene. Although there are 27 = 128 possible stereoisomers, only one is formed. This is because the enzymes that perform these cyclization and rearrangement reactions place the starting materials in the proper orientation. Lipids 78
This is the "side view" of the same reaction, and it shows HO the ring conformations H
H
ChairBoat Chair
Lanosterol H HO
HO H
H H H
HO H H Lipids 79 • The product formed from these cyclization and rearrangement reactions is lanosterol, which is a principle component of lanolin (lana = wool, oleum = grease).
• Various (complicated) transformations to lanolin result in cholesterol.
H H
Lanosterol C30 Cholesterol C27 H H
HO HO
Lipids 80
K. Fat-Soluble Vitamins
• Vitamins are divided into two broad classes, those that are water-soluble and those that are fat-soluble, which are considered to be lipids (A, D, E, K).
1. Vitamin A (Retinol)
• Vitamin A is formed only in animals, but it arises from the metabolism of a provitamin found in plants: β-carotene.
cleave at T-T
β−carotene
CH2OH
Retinol (stored in liver as ester)
• Because vitamin A is fat-soluble, it can also be accumulated in the body just like other fat-soluble compounds. These include PCB’s, dioxins, DDT, and methyl-Hg. Lipids 81
• Humans require about 800 μg a day, while Species μg per g liver toxicity is observed starting at 5,000 ug/day. Cow and Man 200 Weddel Seal 150 • Some animals have adapted to tolerate high Cod 200 (oil) Southern Elephant Seal 400 levels of vitamin A. Yet, if humans consume Antarctic Husky 3500 just small amounts of polar bear liver, death Arctic Bearded Seal 4500 can result. Halibut 10000 (oil) Polar Bear 10000
• The best understood role of vitamin A is its participation in the visual cycle (rod cells). Retinol is oxidized to retinal (vitamin A aldehyde) in the body, which forms an imine with an NH2 group in the protein opsin. This retinal-opsin complex is responsible for the vision process.
Lipids 82
enzyme
H +
Light absorption causes double-bond isomerization and release of 11-trans- retinal from the enzyme imine formation
Lipids 83 2. Vitamin E
• Vitamin E is a terpenoid (a O compound derived from a terpene). HO terpenoid (isoprenoid) • It possesses a sterically hindered phenolic OH that is responsible for the antioxidant action of vitamin E. It loses H atoms (proton and an electron) to give a resonance-stabilized free radical.
• The Hy quenches other free radicals that may damage biomolecules. The vitamin E radical is stable and hindered, so it does not cause damage.
OH
• Synthetic antioxidants such as butylated hydroxy toluene (BHT) are used as food preservatives. Lipids 84 3. Vitamin K O • The name of vitamin K comes from the Vitamin K1 German word koagulation, which signifies its importance in the blood-clotting process. Vitamin K is synthesized by bacteria living in O 2 the intestine, but newborn babies are not O capable of doing this and are given menadione Menadione (synthetic vitamin K) after birth. (drug)
O
4. Vitamin D
• Vitamin D is prepared from 7-dehydrocholesterol in a multistep process. The first step is photochemical and requires UV light. Why won’t visible light work?
1 1
vitamin D 7 7 HO 5 5 HO Lipids 85
• Up to this point, some of the concepts we’ve covered include: o Structures of steroids o Cyclization reactions o Alkyl and hydride shifts o Retinal and the visual cycle o Fat-soluble vitamins
• Attempt:
o Practice problems in the lab manual: 48 – 58, and lab questions 59 - 61