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Combustion is a gas phase reaction. The heat of the flame vapourises the substrate and it's the vapour that reacts with the air. That's why heat is needed to get combustion started.
Anyhow, wood contains lots of relatively volatile compounds so it's not too hard to get combustion started. Once combustion has started the heat of the flame keeps the reaction ...
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$\alpha$-D-glucose and $\beta$-D-glucose are stereoisomers - they differ in the 3-dimensional configuration of atoms/groups at one or more positions.
$\alpha$-D-glucose
$\beta$-D-glucose
Note that the structures are almost identical, except that in the $\alpha$ form, the $\ce{OH}$ group on the far right is down, and, in the $\beta$ form, the $\ce{OH}...
24
Is there a chemical process as some people have suggested that is changing the 32g of powder into 33g of carbs once water is added?
No. Even if there was, it's not relevant in this case: the numbers on the label are an assay of the pure contents of the product before it's mixed with anything, or prepared or altered in any way. Nutrition labels refer to the ...
20
Aldehydes, including aldoses, are oxidized to their respective carboxylic acids in the presence of $\ce{Br2}$ in $\ce{H2O}$. The reason this reaction is often discussed with carbohydrates is that it is useful for differentiating aldoses from from ketoses, which cannot be further oxidized. A solution of $\ce{Br2}$ and $\ce{H2O}$ will lose its characteristic ...
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This is a nice well-defined question, and luckily there is excellent data for which we can provide a quantitative answer.
Richard Wolfenden's research group has sought for many years to characterize the spontaneous (i.e. not enzyme catalyzed) rate of many enzymatic reactions. In general this is so that the spontaneous rate can be compared to the enzyme-...
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The presence of the hydrogen atom in the aldehyde group gives reducing properties to the aldehyde molecule. Aldehydes are easily oxidized by mild oxidizing agents such as Tollens’ reagent ($\ce{Ag+}$ in aqueous ammonia), Fehling’s reagent ($\ce{Cu^2+}$ complexed with tartrate), or Benedict’s reagent ($\ce{Cu^2+}$ complexed with citrate). Ketones do not have ...
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No, the paper will not burn without oxygen being present.
Paper is made primarily of cellulose which is a polymer of glucose.
If you heat paper in a vacuum the cellulose simply decomposes to $\ce{H2O}$, $\ce{CO2}$, $\ce{CO}$ and carbon. As the paper decomposes it will "char" or turn brown to black as the cellulose polymer degrades. Here is a link to an ...
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Ah, a (fairly) common conundrum that assails us Chemistry students when we start Biochem. ;-)
At first glance, the fructose molecule in your first picture, and that in the second picture appear to have different conformations:
Believe me, they're the same. "How?" you ask? Stay with me here (it requires a little imagination):
Draw a vertical line/...
13
You're mostly right. Sucrose is a disaccharide composed of glucose and fructose.
It is a crystalline material. Heating sucrose results in a complex thermal process that involves both melting and decomposition. Depending how fast you heat it, you will observe different things. Basically, sucrose will start to decompose when it starts to melt. If there ...
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Starch is what plants primarily use as a glucose storage. As such, it is essential that they can break it back down into glucose otherwise it would defeat its own purpose and be removed by evolution very quickly. Think about it: Requiring a lot of energy to build up something that serves no further purpose — not exactly giving a plant an advantage in ...
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During pyrolysis, organic compounds are thermally decomposed in the absence of oxygen. The pyrolysis products are classified into categories based on their physical state of existence: char (solid), bio-oil (liquid) and non-condensable gases (gas). The relative proportions of these three product fractions significantly vary depending upon the process ...
12
The name "carbohydrate", which literally means“carbon hydrate,” arises from their chemical composition, which is roughly $\ce{(C.H2O)_n}$, where $n\ge 3$. The basic units of carbohydrates are monosaccharides.
Monosaccharides or simple sugars are aldehyde or ketone derivatives of straight-chain polyhydroxy alcohols containing at least three carbon atoms. ...
11
Potassium chlorate is a source of oxygen. After heating, it decomposes to $\ce{O2}$ and $\ce{KCl}$:
$$\ce{4 KClO3 → KCl + 3 KClO4}$$
$$\ce{KClO4 → KCl + 2O2}$$
The gummy bear is mainly composed of sugar and other carbohydrates. Those carbohydrates will react with oxygen, combustion occurs. For example, glucose will react in this manner:
$$\ce{6O2 + C6H12O6 -...
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With hydrocarbons a certain amount of oxygen (n) and a certain amount of heat energy (Q) are required for complete combustion. In complete combustion the byproducts are carbon dioxide and water in the form of steam.
$$\ce{C_xH_yO_z + nO_2 + Q -> xCO2 + \frac{y}{2}H2O}$$
Hypothetically, if wood and sugar were assumed to have an identical chemical ...
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This is based in the underlying redox rection. If we take e.g. mannose and attempt to oxidise that, the (unbalanced) half-reaction we need is the following:
$$\ce{C6H12O6 -> C6H10O6 + 2e-}\tag{Ox1}$$
To balance it, we either need to add protons to the right-hand side (acidic medium) or negative charges to the left-hand side. Since we are asked about ...
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Many sugars are drawn in the cyclic, closed form where the carbonyl group has been converted to a hemiacetal.
Once you realize that a hemiacetal can equilibrate with a carbonyl (e.g. it is a carbonyl in disguise), identification of reducing sugars becomes easier.
Here is an example from Wikipedia, it is Maltose, the same as your third compound
See the ...
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This is the concept called stereochemistry.
Each carbon with 4 bonds has an approximately tetrahedral geometry. We assign a configuration to the situation by numbering the substituents in terms of priority. Hydrogen is lowest. Oxygen is highest. The top carbon is second, and the other carbon is third. We then trace a route from 1 to 2 to 3. This ...
8
Your structures don't clearly show the different configurations between α- and β- linked glycopyranosides. Cellulose is a non-branching (poly) β-glycopyranoside. Amylose (a component of starch) is a non-branching (poly) α-glycopyranoside. The enzymes in your mouth and digestive system are good at breaking apart the α-glycopyranoside linkage, but not the β-...
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Lighthart gave the theoretical explanation of what enantiomers (‘right foot — left foot’) and diastereomers (‘right foot and right shoe — right foot and left shoe’) can be thought of. Here is the practical explanation for glucose.
You drew glucose in a chain structure. This is not wrong, but only $0.25~\%$ of all glucose molecules in any aquaeous solution ...
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I guess, your actual problem (in understanding why the orientation of the $\ce{-OH}$ groups in glucose is relevant) is caused by the Fischer projection, which is used in your picture.
The Fischer projection may be misleading since the carbon chain is actually not a flat, straight line. The four bonding partners of each carbon atom are arranged spatially in ...
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Does cooling a potato change the nature of its carbohydrates?
Yes, retrogradation is a reaction that takes place when the amylose and amylopectin chains in cooked, gelatinized starch realign themselves as the cooked starch cools.
Resistant Starch:
There is weak evidence that resistant starch can improve fasting glucose, fasting insulin, insulin ...
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What you are doing when you heat a sugar like sucrose is you are dehydrating it. The crystalline structure of sucrose breaks down and the molecules decompose into glucose and fructose and then lose water and then become isomers and polymerize to form caramel, a red-orange colored solid at room temperature. However there are several other products other than ...
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According to IUPAC definition:
‘carbohydrate’ includes monosaccharides, oligosaccharides and polysaccharides as well as substances derived from monosaccharides by reduction of the carbonyl group (alditols), by oxidation of one or more terminal groups to carboxylic acids, or by replacement of one or more hydroxy group(s) by a hydrogen atom, an amino group, ...
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The metabolic pathway you are talking about is how fructose is converted into energy and how its concentration in the blood is regulated. It is indeed true that blood fructose level does not affect blood insulin level, and this is why it is more "fat-inducing", as it cannot be effectively regulated like glucose.
It is also true that in glycolysis, glucose ...
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The Haworth projection depicts the pyranose (or furanose, as the case may be) as a planar ring. The Fischer projection depicts the open chain carbohydrate with the carbon backbone in a single plane. Importantly, the groups on the horizontal bonds are coming out of the plane and the vertical bonds are going into the plane. The result of this is that the ...
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Chemists actually still use D- and L- prefixes today, especially for carbohydrates where using it saves a lot of characters: Compare D-ribose to (2R,3R,4R)-2,3,4,5-tetrahydroxypentanal. The nomenclature is also in common use for amino acids, since all proteinogenic amino acids (except for glycine) in animals are L-configured (Bacteria, smart as they are, ...
7
The difference between the two molecules is highlighted in red.
The functional group in 3-phosphoglycerate is a carboxylic acid. That in glyceraldehyde-3-phosphate is an aldehyde.
Sugars have the general formula $\ce{C_$n$H_{$2n$}O_$n$}$. To accomplish this specific ratio of carbon to hydrogen to oxygen, the requirement is that one of the carbons must have ...
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The distal hydroxy groups are in a perfect distance to form pyranose (six-membered oxane) or furanose (five-membered oxolane) rings. Any equilibrium reaction that leaves the carbonyl group liberated for extended periods is prone to lose free aldehydes to the (unreactive) hemiacetal forms. Any reaction whose equilibrium is strongly product-sided or which ...
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A carbohydrate is a biological molecule consisting of carbon (C), hydrogen (H) and oxygen (O) atoms, usually with a hydrogen–oxygen atom ratio of 2:1 (as in water); in other words, with the empirical formula Cm(H2O)n (where m could be different from n).This formula holds true for monosaccharides. Some exceptions exist; for example, deoxyribose, a sugar ...
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For a given configuration there is a corresponding and fixed rotatory effect. However, the actual rotatory effect - that seen by a polarimeter - can either be + or −, independently of the D- or L-configuration of the stereogenic centre that shows up in the compound name (in the case of sugars, it refers to the carbon furthest away from the =O group).
In ...
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