I’m impressed. Yes, you’ve got it. So a triglyceride (fat or oil) would be:
H
HC-c-O-CH2-CH2-CH2-(...)-CH2-CH3
|
HC-c-O-CH2-CH2-CH2-(...)-CH2-CH3
|
HC-c-O-CH2-CH=CH-(...)-CH2-CH3
H
Where c means C=O and the (…) means a changeable number of CH2 units. Here, the bottom fatty acid chain has a double bond, so it’s monounsaturated.
Someone will correct me if I’m wrong, but I think formic and acetic acids would technically be fatty acids, though you’d never see them referred to as such. Oh, yep. I just found an article here that supports that. Of course, they also say that “fatty acid” only refers to saturated chains. I don’t know if that holds true in everyday practice.
When there’s a double bond in a carbon chain, it can be cis or trans. Let’s see if I can diagram it here. This should be interesting:
H H H R
>C=C< >C=C<
R R R H
Cis Trans
Whew. Hope that comes through. R, here, means the rest of the carbon chain.
Cis means the H’s are on the same side of the double bond, while trans is the opposite. If you’re going to refer to a whole chain as cis or trans, there better only be one double bond in it, for obvious reasons. (If there were 2 cis db’s, it’d be better referred to as cis-cis.)
So I guess that means a saturated fat melocule would have all 3 long-carbon-chains consist of plain old run-of-the-mill fatty acids, while a monounsaturated fat molecule would have 2 of its long-carbon-chains be regular fatty acids and the 3rd long-carbon-chain be a transfatty acid or a cisfatty acid.
So then, why is the FDA thinking about including trans-isomers in the saturated fat content of a piece of food? Wouldn’t the non-trans-isomers be saturated and the trans-isomers be monounsaturated?
I can’t say for sure - I haven’t heard anything about this. My WAG: someone discovered that trans-fatty acids (monounsaturated, as you said) are as bad for you as saturated fatty acids, so for simplicity’s sake, the FDA wants to report them all together. Again, just a WAG.
Tracer asks such good questions. I like how he/she thinks
The problem with trans fatty acids are that they are more stable than cis-fatty acids.
The problem with the trans-double bond is that it is energeticly more stable (and thus harder to break) than the cis-double bond.
Unsaturated fats are better for you because the body can break them down easier. The double bond itself is a souce of localized negative charge, since it has more electrons, and so provides a place for chemical reactions to occur. This is a MAJOR oversimplification, but not so egregious as to be misleading. Now, one of the main factors influencing reactivity is someothing known as “sterics” which, in laymans terms, refers to the “easy of attack” for a chemical bond. Chemical bonds with lots of stuff in the way tends to block other things from reacting with them. Basicly, something has to get to a bond to break it, and if stuff gets in the way when something is attacking, it is said to be “stericly hindered”. Look again at the difference between cis and trans bonds. I hope this works:
H H H R
>C=C< >C=C<
R R R H
cis trans
in a cis bond, you have an entire “side” of the bond that is “open”, since hydrogen atoms are relatively small and not stericly hindering. Thus, your body can much easier break this down than it can a trans double bond, where there is no “easy way in.” No matter how you apprach the trans bond, there’s always a big, bulky “R” group in the way.
If i’m not mistaken, the trans double bond in a fatty acid is about the same reactivity as an ordinary single bond, and thus a trans-monounsaturated fatty acid is nutritionally equivalent to a saturated fatty acid. It is only the cis-monounsaturated fatty acids which provide the true benefit of unsaturation: easy of digestion.
Ah. And yet another layer of the Onion of Ignorance is peeled away from my eyes.
Does it also make a nutritional difference how far down the fatty-acid chain the double-bond occurs? I would think that a double bond way out near the tip of the fatty acid would hardly make a difference at all, but a double bond near the base, where it connects to the dehydrogenated carboxyl group (and thence to the de-OH’ed glycerol) would make a huge difference, since that way the molecule would break apart into one almost-complete fatty acid chain and two connected fatty acid chains with a tiny fatty acid stump.
What with all these chemists 'round here, I feel like I’m hijacking by posting this, but I didn’t see it until it got over to GQ.
I can think of a possible source for the hydrogenation-genetic engineering link. IIRC, Calgene or maybe someone else was working on a rapeseed that would produce a saturated oil, or maybe it was one that was more amenable to hydrogenation. Most canola oil margarine (canola oil = rapeseed oil) seems to be mixed, and I think they wanted to make it easier. Perhaps they had success, but I don’t know why they wanted to try.
BTW, I think genetic engineering is generally good, as long as we keep enough genetic variety to prevent disease susceptibility.
I just saw one of those “health conscious” brands of margarine in the supermarket. The label bragged about how they didn’t use the hydrogenation process, and that therefore, their product didn’t contain any trans-fatty acids. Of course, the Nutrition Facts section of the label still showed 1.5 grams of saturated fat per tablespoon, but, darn it, it didn’t contain any trans-fatty acids!
Sheesh. To read this label, you’d think the hydrogenation process caused trans-fatty acids to break free from their fat molecules and go on a rampage through downtown Tokyo or something.
(Um, the “evil” trans-fatty acids caused by the hydrogenation process do stay attached to the rest of the fat molecule, don’t they?)
Tracer, i hope you take some advanced chemistry classes some day, because you have a GREAT mind for this stuff. You ask all the right questions.
The location of the double bond has little effect for several completely independent and unrelated reasons:
The location of the double bond has little bearing on the sterics involved, merely the cis/trans nature is enough. The only real difference is if the double bond is VERY close to the terminal end (the one dangling out in space, as it were.) Try this on for size:
1 2 3 4
R H R H R H R H
>C=C< >C=C< >C=C< >C=C<
H *C-H H *C-H H *C-H H *C-CH3
/ \ / \ / \ / \
H H H C-H H C-H H CH3
/ \ / \
H H H CH3
Sterics is generally only a significant factor when dealing with atoms one or two “bonds” away from the double bond in question. Compare molecule 1, 2, 3, 4, paying attention to the carbon atom I have starred(*). When dealing with tetrahedral carbon atoms (carbon atoms involved ONLY in single bonds) we can call them methyl, primary, secondary, or tertiary depending on how many other tetrahedral carbon atoms they are attached to. Methyl means NO other tetrahedral carbons (and thus 3 hydrogens), primary means ONE other tetrahedral carbons (and thus 2 hydrogens) etcetera. Generally speaking, sterics is determined by looking at THIS carbon atom only. Therefore, the double bond in molecule 1 is less stericly hindered than the one in molecule 2, and the double bond in molecule 3 is less stericly hindered than the one in molecule 4. What about molecule 2 and 3? Well, since the starred carbon atom (the one we care about) in each case is a primary carbon atom (only 1 other tetrahedral carbon atom) the double bond in each of them experiences about the same level of steric hindrance. It doesn;t matter how far the chain goes past that, they will all have about the same reactivity. Thus, unless the double bond is RIGHT at the end, they are all roughly equivalent (I say roughly, because once sterics have been discounted, other lesser effects become noticible, those, however, we need not go into now). Nearly all unsaturated fatty acids have their double bonds towards the middle of the chain.
Double bonds (as well as other sources of electron density like oxygen atoms) particpate in something known as conjugation. Basically, when double bonds are located near eachother in EXACTLY the following manner:
R H
>C=C< H
H C=C<
/ R
H
they are said to be in conjugation. Basically, you have any time two pyrimidical carbons (carbons having exactly 1 double bond and 2 single bonds) next to each other, they participate in conjugation. What is actually going on here requires more chemistry than I have time to go into, but for simplicities sake, we can say that the whole 4 carbon segment involved in the conjugation, in effect sort of “share” some of the electrons involved in the double bond, what this does is to stabilize the double bonds, and make ithe whole mess less reactive. You can even have 6 or 8 or more carbon atoms involved in the conjugation, as well as some non-carbon atoms (such as nitrogen or oxygen) that can also take a “pyrimidical” structure. Lets look at the glycerol-fatty acid juncture(in essence, this gould also be generalized to any ester) in detail.
H O H H
| " | |
H-C-O-C-C-C-R
| * | |
R H H
" or = are double bonds
Notice the starred carbon atom. It’s pyrimidical (attached to three other atoms). This means that it CAN participate in conjugation. Check out the following.
H O H
| " | H
H-C-O-C-C=C<
| * # R
R
Now check out carbon atoms * and #… They are neigboring pyrimidical carbons… Whammo, conjugation. So, your original question IS somewhat confirmed: A double bond RIGHT next to the ester IS more stable than at another point in the molecule, however (just like point 1) almost no unsaturated fatty acids have the double bond here in real life.
The manner in which the body metabolizes fatty acids has a bearing on this as well. Basicly, the body “eats” down from the end of the fatty acid. The body’s enzymes “attack” a source of electron density, break the molecule there, and then “munch” segments off of the fatty acid (actually 3 carbon segments if I remember correctly. It’s kinda fuzzy, and I am NOT a biochemist, so perhaps any out there could make this more clear.) If there is a double bond in the molecule (preferably a non-conjugated * cis* double bond) the body now has TWO sites to begin metabolization (the ester end and the double bond site). Basicly, unless the double bond is RIGHT at either end, there is little difference in metabolic rates for digesting various monounsaturated fatty acids. Now, here’s the kicker. Generally, fats are (if I am correct. Again, I don’t remember this 100% accurately, and I am largely conjecturing here) CONSTRUCTED in three carbon segments, meaning that nature has established that any points of unsaturation that form will TEND to not be at either end, but rather in the middle of the molecule, where there is little matter exactly WHERE they occur.
So, to summarize my rather LONG chemistry lesson here: YES the location of the double bond DOES matter in certain specific instances, but not for the reasons you cited. The location matters only in terms of stability of the double bond, not in the length of the resulting segments that are broken off.
Every time I use the word “pyrimidical” replace it with the word “planar” My mind somehow confused the two terms, and they are NOT the same thing (Pyrimidical is a variant of the tetrahedral shape). I was trying to use descriptive terms to show the geometry of the carbon atoms involved rather then going into a lesson on atomic geometry nomenclature. If you are interested, chemists refer to tetrehedral atoms as “sp[sup]3[/sup]” atoms and planar atoms as “sp[sup]2[/sup]” atoms. Pyrimidical atoms are also “sp[sup]3[/sup]” atoms. Thus carbon atoms involved in 4 single bonds are sp[sup]3[/sup] shaped and carbon atoms involved in 1 double and 2 single bonds are sp[sup]2[/sup] shaped.