Determining absolute configuration when the lowest priority group is not facing backwards I have learnt that the absolute configuration of a chiral centre can be labelled as (R) or (S) according to the Cahn–Ingold–Prelog rules, when the hydrogen (or other lowest priority group) is facing backwards, i.e. going into into the page, and the remaining groups decrease in priority clockwise or anticlockwise respectively. On the other hand, if the hydrogen is coming out of the page (i.e. as a wedge) we swap our answer from (R) to (S) or vice versa (to get the hydrogen going into the page).
But what do we do when the hydrogen is flat and not a wedge or dash, as is the case here?

The priorities are

Chlorine atom
Carbon chain on left (propyl group)
Carbon chain on right (methyl group)

This gives an anticlockwise or (S) configuration. But the answer is given as clockwise or (R) configuration.
Why is this the case? The hydrogen isn't coming out of the page. Is there something else that has to be done when the hydrogen is flat or in the plane of the page?  If I visualise it rotating to bring the H to the back, I still end up with an (S) configuration.

You had the right idea when you said "if the hydrogen is coming out of the page (i.e. as a wedge) we swap our answer from R to S or vice versa (to get the hydrogen going into the page)".
The difficulty seems to have came with your manipulation of the drawing. The diagram you have implies that the hydrogen is coming out of the page towards us (i.e. anticlockwise = (R)).
You could also flip the entire molecule through by 180° to give you the chlorine pointing out of the page towards you.


Work in free expansion I have read that no work is done during free expansion and temperature of an ideal gas undergoing free expansion doesn't change.

Let us assume an ideal system of piston and gas which has vacuum around it. The gas will expand but we say there is no work done by the gas. Why so ? Gas applies a pressure(& hence a force) on the piston and the piston gets displaced. So acc to F.s = work , there is work done by the gas !

Little bit tricky question. As you say, the work is force acting along the path. But there is no force acting, as there is weightless piston with vacuum on the opposite side.

To visualize the process better, imagine an inflatable balloon in outer space, when you puncture it. All gas molecules continue to fly in the direction and speed they had in the moment of accident (therefore the temperature is the same) and they exert no work (no collision thereafter).


Why does pH increase as a weak acid becomes more dissociated? From Wade's Organic Chemistry [1, p. 71]:

  In most cases, the $\mathrm{p}K_\mathrm{a}$ of an acid corresponds to the $\mathrm{pH}$ where the acid is about half dissociated. At a lower (more acidic) $\mathrm{pH}$, the acid is mostly undissociated; at a higher (more basic) $\mathrm{pH}$, the acid is mostly dissociated.

I'm having a hard time intuiting this concept. Consider placing $\pu{1 mol}$ of $\ce{HF}$ in $\pu{1 L}$ of water. As the $\ce{HF}$ dissociation reaches equilibrium, I'm assuming the concentration of $\ce{H3O+}$ ions in solution increases, until the maximum concentration of $\ce{H3O+}$ ions are reached at equilibrium. If the concentration of $\ce{H3O+}$ ions in water increases as the dissociation of $\ce{HF}$ proceeds, then wouldn't the $\mathrm{pH}$ of the solution decrease (become more acidic) as the acid dissociates over time?

If any one has any thoughts on how to best interpret the quoted statement, that would be very much appreciated! Any and all new perspectives are welcomed. Thanks in advance for your generous help.

References

Wade, L. G.; Simek, J. W. Organic Chemistry, 9th ed.; Pearson: Glenview, IL, 2017. ISBN 978-0-321-97137-1.

A weak acid (HA) in water: $\ce{HA + H2O <=> H3O+ + A-}$, and thus,

$$K_\mathrm{a} = \frac{\ce{[H3O+][A-]}}{\ce{[HA]}}$$

By Le Chatelier's principle, if you add some $\ce{H3O+}$ from an external source (see Mithoron's comment above) to increase $\mathrm{pH}$, the equilibrium favors the backward reaction in order to reduce some $\ce{A-}$. Consequently, the amount of the dissociation of acid decreases and $K_\mathrm{a}$ will remain constant.

In similar manor, if you decrease $\mathrm{pH}$ of the solution by adding some $\ce{OH-}$, that will react with equal amount of $\ce{H3O+}$. Therefore, the equilibrium favors the forward reaction in order to produce some $\ce{H3O+}$ to keep $K_\mathrm{a}$ constant. Accordingly, $\ce{[A-]}$ in the solution increases while $\ce{[HA]}$ decreases, meaning the amount of the dissociation of acid increases. At one point, $\ce{[A-]} = \ce{[HA]}$, and hence, $K_\mathrm{a} = \ce{[H3O+]}$. At that point, therefore, $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$.

Applying this to your example ($\ce{HA} = \ce{HF}$), $\mathrm{p}K_\mathrm{a}$ of $\ce{HF}$ is $3.14$ ($K_\mathrm{a} = 7.2 \times 10^{-4}$).

When $\pu{1.0 mol}$ of $\ce{HF}$ is dissolve in $\pu{1.0 L}$ of deionized water (no added external acid to adjust $\mathrm{pH}$), $\ce{[HF]} = \pu{1.0 mol\cdot L^{-1}}$ and $\ce{[F-]} = \ce{[H3O+]} = \alpha \gg 1.0$. Thus,
$$\ce{[H3O+][A-]} = K_\mathrm{a} \cdot \ce{[HA]}$$
$$\alpha^2 = K_\mathrm{a} \cdot 1.0~~~~\text{or}~~~~\alpha = \sqrt{K_\mathrm{a}}= \sqrt{7.2 \times 10^{-4}}= 2.7 \times 10^{-2}$$

Therefore $\mathrm{pH}$ of the solution is as follows:

$$\mathrm{pH} = -\log \ce{[H3O+]} = -\log \alpha = -\log 2.7 \times 10^{-2}= 1.57$$

Thus, at $\mathrm{pH} = 1.57$, the ionization of the solution is: $\ce{[F-]} = \ce{[H3O+]}= \pu{2.7 \times 10^{-2} mol \cdot L^{-1}}$.

Now, if you added an external base to the solution to increase its $\mathrm{pH}$ until it becomes $\mathrm{pH} = 3.14$, then, at $\mathrm{pH} = 3.14$, $\ce{[F-]} = \ce{[HA]}= \pu{0.5 mol \cdot L^{-1}}$.


Test for chlorine and chloramine in city water supply I am looking to install a water filter system and I wanted to test the water for free and total chlorine.

I understand from the city water report that it uses chloramine and NO chlorine (test report)and that chloramine content averages around 4 ppm in the water supply. I purchased one of these to test it out. http://www.bulkreefsupply.com/insta-test-free-total-chlorine-strips-lamotte.html

The above product can measure free and total chlorine ( 0, 0.5, 1, 3, 5, 10 ) PPMs. I was surprised to find out after using these test strips that city supply has 0 PPM for both free and total chlorine.

Questions
1. Any idea how I can test if these test strips are indeed working?
2. Anything wrong with the way I have tested?

Test strips for chlorine should work for both chlorine- and chloramine-treated water. See Here-to-learn's comments to this answer for the chemical kinetic reason why chlorine test strips will also work for chloramine-treated water.

Notice that your city supplies the water with chloramine. Depending on how much chloramine is present in your tap water, leaving water to sit for days on the countertop would be necessary to remove a significant amount of chloramine.

Please see this research paper: http://hbd.org/ajdelange/Brewing_articles/BT_Chlorine.pdf

Page 17 estimates the half-lives of chlorine vs. chloramine.


Do Polar Bonds Affect Molecular Shape? Repulsion between shared electron pairs shapes the molecule in such a
  way that enables the pairs to be as far apart as possible.
  Furthermore, the repulsion between shared and unshared pairs forces
  bonded pairs closer to one another.  These principles of the VSEPR
  theory and orbital hybridization, in addition to Van der Waals
  forces, explain the overall shape of a molecule.

In my textbook,  however, there was little mention to how polar bonds and the partial charges they cause affect the structure of a molecule.  I would expect that poles of the same charge would repel each other and have a profound affect on molecular shape.  However, many compounds make me second guess this assumption, including $\ce{H2O}$.

Do polar bonds at all affect molecular shape? Why or why not?  If so, to what extent?

Yes, of course they affect the shape. But only slightly so. More precisely, bond angles can change, however this does not change the general shape predicted by VSEPR.

For example, if two atoms (one with a positive pole, and one with a negative pole)  are connected by another atom, then the two atoms would make an angle that is smaller than the angle predicted by VSEPR.

And also, polar bonds are capable of intramolecular bonding. Such as in protein-folding.  So they definitely contribute to molecular shape.


Does the slippery feeling of bases always appear at certain pH? My guess is that the saponification reaction is only dependent on the concentration of $OH^{-1}$ ions and of course the coposition and concentration of fats. I also guess that the fat part of the situation is rather constant. My guess means that the cation would have no or negligible impact.

This would suggest that the degree of slipperyness would be only pH dependant (and not dependent on the cation). Does this mean that there is a certain pH level where the solution starts to feel slippery? If that's the case about which pH level is it?

There probably is no one pH level where things get slippery. Capacity to react as a base has something to do with it.

You can make ammonia and sodium hydroxide solutions both with pH 11, but the sodium hydroxide solution is only 0.001 molar and thus cannot react with much oil on the skin.  Ammonia solution has a lot more base available to react with those fatty acids.  The weak base "hides" most of its reactivity in water alone but reveals when there are fatty acids to neutralize, so it's slip sliding away.

Probably anything that we consider basic and is sufficiently concentrated will slip in the fingers.


How does WD40 displace water? How does WD-40 displace water? Or does it even displace water? WD40.com explains what WD-40 stands for:

  WD-40® literally stands for Water Displacement, 40th formula. That's the name straight out of the lab book used by the chemist who developed the product back in 1953. The chemist, Norm Larsen, was attempting to concoct a formula to prevent corrosion—a task which is done by displacing water. Norm's persistence paid off when he perfected the formula on his 40th try.

If it displaces water then I always thought that that meant it was denser or heavier. I just tested to see: I sprayed a couple teaspoons into an empty glass then poured that into another glass half filled with water. The WD-40 floats! So what does it mean to say that it "displaces water"? Surely it doesn't mean that it's hydrophobic, all lubricants are hydrophobic.

Let's start with the fact that bouyancy of an object in open water is caused by displacement, but liquids that are less dense and may spread out do not displace water, they just lie on top of it. If I put a block of wood in a fish tank that is full, it displaces and spill over a mass of water equal to the mass of the block, but if I slowly pour (or rather spray) oil on a full tank of water, the oil will spill over, not the water. The definition for displacement is (per Google):

  the moving of something from its place or position.
  the occupation by a submerged body or part of a body of a volume that would otherwise be occupied by a fluid.

The first definition is what WD-40 is doing to water, the second one is what causes buoyancy. The WD-40 displaces the water in that it moves it from where it is, but how?

Looking at the ingredients from the SDS, we can get a feel for their purpose:

$$\begin{array}{|c|c|c|}
  \hline
\textbf{Material} & \textbf{Hazards} & \textbf{Purpose (not stated on SDS)}\\
\hline\text{Aliphatic Hydrocarbon} & \text{Flammable Liquid Category 3}  & \text{Thin oils and evaporate quickly}\\
 & \text{Aspiration Toxicity Category 1} & \text{}\\ \hline
\text{Petroleum Base Oil} & \text{Not Hazardous} & \text{Provide long term protection} \\ & & \text{from water and air}\\ \hline
\text{LVP Aliphatic Hydrocarbon} & \text{Aspiration Toxicity Category 1} & \text{thin oil and evaporate slowly}\\ \hline
 & \text{Simple Asphyxiant}\\
\text{Carbon Dioxide} & \text{Gas Under Pressure,} & \text{Propellant}\\
& \text{Compressed Gas}\\ \hline
\text{Non-Hazardous Ingredients }&\text{Not Hazardous} & \text{This does the real work}\\ \hline
\end{array}$$

The oils are nice because once the water is removed, the oil will remain, preventing it from returning. But it is the non-hazardous ingredients that do most of the work. Hard to say what exactly this 10% is made of but it has to get water to "unstick"/"unwet" the surface that is being treated. To do this WD-40 is likely using a non-ionic surfactant of some sort similar to what is used in shampoos to emulsify the water and allow the flow of the WD-40 spray to remove (displace) the emulsified water. $%The rest of the wD-40 components act as carrying agents to remove the water from whie$ This speculation about surfactants is further supported by the fact that WD-40 foams when sprayed which is not characteristic of pure hydrocarbons.

Additionally, there would likely be some material to adhere to the surface to prevent further interaction with water or oxygen. There is the petroleum base oil, but it needs a way to stick to the metal. Likely, the use an unsaturated fatty acid in some way whether it is pure oleic acid or boiled linseed oil or something similar to those to bond to most metals and other materials and prevent further corrosion.


Can a lone pair be involved in cis/trans isomerism? I was trying to find whether a molecule showed isomerism, and this is what I came up with.

I thought this was the case as I took the nitrogen to be sp2 hybridised, thus trigonal planar (with the lone pair not shown on the diagram). The molecule can't rotate about any single bonds because resonance makes the whole molecule planar. Am I right in thinking this shows cis trans isomerism? If not, what sort of isomerism might it show?

Yes, due to the lack of rotation about the C=N bond this will display the desired isomerism. For a much simpler example see $ \ce{N2F2} $

(taken from https://en.wikipedia.org/wiki/Dinitrogen_difluoride )


How do gelatin and aluminium sulfate strengthen paper? When wet, the hydrogen bonds between the cellulose chains in paper are disrupted due to the formation of hydrogen bonds with water. This leads to the paper becoming extremely weak. This is the case with most paper, however, additives such as gelatin and aluminium sulfate are added to the paper to strengthen it.

I can't see how the addition of either of these strengthens the paper. I would guess that it aids the bonding between cellulose chains, perhaps by creating a 'bridge' between them, but I'm not sure if this is correct or how it would happen. Another idea is that the aluminium could form a complex ion, and then perhaps form a chelate to strengthen the paper, though I have little confidence in this.

So, how do the gelatin and aluminium strengthen the paper - particularly filter paper, which is almost pure cellulose.

The process of adding gelatin and such to paper (cellulose) is called sizing. The idea behind using gelatin is that it has both a hydrophilic end and a hydrophobic end to the molecule. The aluminum sulfate (alum) is added as a hydrogen bonding linker between the cellulose and the hydrophilic end of gelatin (or some other similar molecule). Mostly this is just to prevent the paper from getting wet in the first place.

Your second paragraph is spot on. Typically other reagents are used to cross link ( the cellulose in order to retain strength upon getting wet (wet strength). Those reagents are a bit more aggressive. Like alkyl ketene dimer

or urea-formaldehyde polymer.


Sulphuric Acid in Esterification Reaction Why is it that adding the sulphuric acid slowly to a solution of alcohol and carboxylic acid (making an ester) will increase the yield of the product? It seems that not only does the rate at which it is dispensed affects the yield of ester, but really does change the colour of the solution when heated. For example, my team used it slowly and the colour was a more faint peach colour whilst everyone else, who added it quickly, produced pure black solutions!

What is happening? Why does adding the acid slowly increase yield and purity of the ester?

The esterification is not the only thing catalysed by sulphuric acid. Basically, it’s catalytic activity is protonation of whatever feels happy enough to be protonated. In your desired esterification mechanism, we want to protonate the acid:

$$\ce{R-C(=O)-OH + H2SO4 <=> R-C(=\overset{+}{O}H)-OH + HSO4-}\tag{1}$$

This will ease the nucleophilic attack of the alcohol onto the carboxy carbon. However, we also have other atoms that can be protonated.  For example, the hydrogen-carrying oxygen may also accept the proton:

$$\ce{R-C(=O)-OH + H2SO4 <=> R-C(=O)-\overset{+}{O}H2 + HSO4-}\tag{2.1}$$

This could then lead to the removal of water and the formation of a very reactive acyl cation:

$$\ce{R-C(=O)-\overset{+}{O}H2 -> R-C#\overset{+}{O} + H2O}\tag{2.2}$$

This in itself isn’t bad as it could also react with the alcohol in the desired fashion. However, the alcohol also can be protonated:

$$\ce{R-CH2-CH2-OH + H2SO4 <=> R-CH2-CH2-\overset{+}{O}H2 + HSO4-}\tag{3.1}$$

And depending on the type of alcohol you have, a number of side reactions may occur. For instance, you could simply eliminate $\ce{H3O+}$ from the protonated alcohol to give an alkene:

$$\ce{R-CH2-CH2-\overset{+}{O}H2 -> R-CH=CH2 + H2O + H+}\tag{3.2}$$

If you have a suitable neighbouring group that can stabilise positive charges, semi-pinacol-type rearrangements could occur. I won’t draw that with the simple primary alcohol I drew because it would be unlikely in e.g. propanol. It could be a problem in other cases, though.

Finally, the intermediates I drew up can react even further and unpredictably with sulphuric acid still present.

If you add sulphuric acid slowly, the local acid concentration will remain rather low and the reaction mixture is rapidly homogenised. This means that side reactions caused by sulphuric acid will be suppressed. If, however, you add it quickly, there will be areas of your solution with very high local sulphuric acid concentrations. In these, these side reactions are accelerated. Furthermore, concentrated sulphuric acid acts as a dehydrating agent which further speeds up the reactions we don’t want (they all include losing a water molecule).

These side reactions — especially the dehydrating ability of sulphuric acid — tend to end up in ‘black tar’ which would give a more or less uniform black solution. What exactly happened is hard to say except that some polymerisation probably took place.


Weight Percent of Solutions Given Without Units I have been performing haloform reactions using sodium hypochlorite. The most available source is common household bleach, which is a solution of $\ce{NaClO}$ in water. The bottles invariably give a concentration in unitless percentages: e.g., "$6\%$ Bleach."

I have assumed this to mean: (mass of solute) / (mass of solution) * 100.

Now I see that there are other (mathematically erroneous) uses of the percent symbol %, including a measure called WV (weight volume), which seems to be normalized by dividing gram weights of solute by 100 ml.

An easy question, impossible without the units:

How many grams of $\ce{NaClO}$ are in one litre of $6\%$ household bleach?

I have assumed this to mean: (mass of solute) / (mass of solution) * 100

Most likely you are correct in this assumption. Percentage compositions of ionic solutes in water are nearly always implicitly expressed in percent-by-weight ("% w/w").

$6\%$ bleach is relatively dilute and will have a density very near to $1\ \frac{\mathrm{g}}{\mathrm{mL}}$.  So, $1\ \mathrm{L} \approx 1\ \mathrm{kg\ solution}$, and thus one can expect to have approximately $60\ \mathrm{g}\ \ce{NaClO}$ per liter of such a solution.


Simplest way to convert glucose into water and carbon dioxide other than burning? What is the simplest pathway from glucose to $\ce{H2O}$ and $\ce{CO2}$, such that

all reactions happen in aqueous solution with pH between 5 and 8,
temperature is in range between 0 and 35 °C,
no (expensive) enzymes are required,
the reaction is completed in around 6 hours, and
no toxic or harmful intermediates are formed and no toxic or harmful catalysts are used?

I realize that one possible answer to this would be, in biochemistry,

glycolysis → pyruvate decarboxylation → citric acid cycle → oxidative phosphorylation;

this is a fairly complex reaction cascade, however. Is there a way chemistry could simplify this in comparable (more or less physiological) conditions?

Aside from a possible (retro-)synthetic approach, there might be a catalytic option. I've found a publication by Degering and Upson* that describes iron pyrophosphate as a catalyst, but the reaction times are far higher (around 3 weeks), the temperature is relatively high and the reaction appears to be relatively messy in terms of products.

* Edward F. Degering and Fred W. Upson, J. Biol. Chem. 1931, 94, 423-431. DOI: n/a; URL: http://www.jbc.org/content/94/2/423.short (pdf) (no https available)

I'd suggest you read into the Fenton process. Using an iron catalyst with hydrogen peroxide will break down (potentially) any organic matter into $\ce{CO2}$ and $\ce{H2O}$ within minutes or hours. The only drawback from your requirements is the pH, since I think this is only down in acidic solution, but you could do a little research to find out more.


Coffee and tap water reaction - results in gray tint I use filtered water to make my coffee.

If I mix a few drops of said coffee with our tap water, it turns a grayish color; it's much darker than the coffee would turn the water by itself (I compared by doing a side by side comparison and adding an equal amount of coffee to filtered water, resulting in a much lighter brown (not gray) color that is closer to what I'd expect from coffee). It's a very pronounced effect (i.e. easily visible when you rinse out a coffee mug that has just a few drops of coffee at the bottom). Because of the distinct color change and the fact that it's significantly darker than it should be from simple mixing, it's obviously a chemical reaction.

What common tap water contaminants would react with coffee in this manner? Iron perhaps?

The tap water might be slightly hard.  Hard water has a high content of minerals such as calcium or magnesium which makes the water more alkaline. Alkaline water will draw tannin out of the coffee much quicker than pure water.

Filtering the water with something like activated carbon filter then an ion exchange filter will lower the mineral content, thus lowering the pH. With a lower pH the dissolution of tannin will be slower.

Slowing the dissolution of tannin relative to more soluble portions like caffeine is part of the reason for steeping tea and coffee for only a limited amount of time.  More steep time, or more alkalinity in the steeping solution,  releases more tannin.  More tannin makes the coffee more anti-nutritional, more bitter, and it darkens the color.


Folin–Ciocalteu formula I am searching a long time now. I have a question about the Folin–Ciocalteu formula for quantification of polyphenol. How is the $62.5$ in the equation explained? Where does it come from?
This is the protocol [1, p. 38]:

$\pu{5.0 mL}$ of extract were placed in calibrated flask and filled up to $\pu{25 mL}$ by distilled water. $\pu{1.0 mL}$ of Folin-Ciocalteu reagent, $\pu{10.0 mL}$ of distilled water were added to $\pu{2.0 mL}$ of diluted extract (from $\pu{25 mL}$ calibrated flask) and filled up to $\pu{25 mL}$ by sodium carbonate solution ($\pu{290 g/L}$). An absorbance of prepared sample was measured by means of UV spectrophotometer at $\pu{760 nm}$ after $\pu{30 min}$ of incubation in darkness. The same liquid (with pure water instead of plant extract) was used as a blind test. All determinations were performed in triplicate. The content of polyphenols was calculated from the formula:
$$X = 62.5 \cdot \frac{A_1 \cdot m_2}{A_2 \cdot m_1}$$
where $A_1$ - absorbance of pyrogallol solution, $A_2$ - absorbance of investigated samples, $m_1$ - mass of investigated sample [$\pu{g}$], $m_2$ - mass of pyrogallol [$\pu{g}$].

Reference

Modnicki, D.; Balcerek, M. Herba Polonica 2009, 55 (1), 35–42.

Authors mention that pyrogallol (PG), among other standards, has been purchased from Sigma Aldrich US and used as is without further dilution. The amounts of polyphenols were calculated as PG equivalent by external standard method, and utilizing additivity of absorption coefficients the proportion has been established.

$62.5$ is just a dilution factor for the herb extract probe. From the protocol, first time the extract was diluted $5$ times (from $\pu{5 mL}$ to $\pu{25 mL}$), second time – $12.5$ times (from $\pu{2 mL}$ to $\pu{25 mL}$), resulting in total dilution factor of $5 \cdot 12.5 = 62.5$.


Using a H NMR spectrum to determine the structure of a protein supplement This question is about the H NMR spectrum of creatine.
Creatine is a supplement taken by athletes to increase body mass. Recently, supplements containing derivatives of creatine have been marketed. These are usually more lipophilic in an attempt to improve uptake.
The 1 H NMR spectrum of one of these supplements in D2O is shown, with some parts magnified:

This particular supplement exists in an ionised form at pH1 but does not exist in an ionised form at pH 12 (I am unsure about this and will come back to it in my question below!)
Using this information we are asked to suggest a structure for the supplement. I was almost right. The answer is:

The mark scheme says that three things can be extracted from the NMR spectrum and the information given:

My question(s):

I don't understand the relevance of the pH statement. How does that relate to the structure? How does that relate to the ester group? (this is mentioned in the mark scheme which is attached just above this)

How can we use the integrals to calculate that there are 10 C-H protons? I measured the heights of the traces and got, 2.7cm, 4.75cm and 6.9cm but I'm not sure how this leads to '10 C-H protons'...

Would really appreciate help on this question and a breakdown of the key points discussed. Many thanks!

Let me try to explain the point about the 10 atoms and the integral.
The area under the peaks and the ratios of these areas correspond to the ratios of the hydrogen atoms. The blue line adds all these areas up as it goes along. That means at the first peak on the left you simply get the area of that peak. But at the second peak you still have the amount from the first peak plus the new area, which is why you have to look at the difference.
If we denote the three peak areas as $A_1$, $A_2$, and $A_3$, the height of the blue line goes from $A_1$ to $A_1 + A_2$, then lastly $A_1 + A_2 + A_3$.
With your values we have,
$$\begin{aligned}
A_1 &= \pu{2.7 cm} \\
A_1+A_2 &= \pu{4.75 cm} \\
A_1+A_2+A_3 &= \pu{6.9 cm}
\end{aligned}$$
Which allows us to easily determine $A_2 = \pu{2.05 cm}$ and $A_3 = \pu{2.15 cm}$. (Alternatively you could have simply read off the difference of the blue line between two peaks directly when you measured it).
This gives us the ratios of $\pu{2.7 cm} : \pu{2.05 cm} : \pu{2.15 cm}$. If we assume that the signal at $\pu{1.25 ppm}$ (i.e. $A_3$) is a methyl group with $3$ hydrogen atoms, we can divide all values by $\pu{2.15 cm}$ and multiply by 3 to normalize the ratio. We get approximately hydrogen atom ratios of $3.76 : 2.86 : 3.00$. Rounding the result gives a ratio of $4:3:3$ which agrees with $10$ atoms in total.


Easy method of analysing iodine content in supplements? Most health-food stores carry supplements supposedly containing iodine for people with iodine deficiency. I recently bought some kelp tablets that are supposed to have iodine but as a consumer how can I be sure that there is any iodine in them at all, let alone the claimed minimum daily requirement.

How do I know that I'm not just taking green-colored sawdust? Is there a cheap way to test a kelp tablet for its iodine content? If not are there testing labs that I could send some kelp to for testing that isn't going to cost me thousands of dollars?

Take some potato starch dissolved in water (you can easily get it from the water when you boil potatoes), and put a tablet into the water. If the water becomes bluish black, there is iodine in the tablet.

Yes, there are testing labs that will determine the amount of iodine in the tablets which will cost somewhere around \$100 to \$200.


Is Inositol chiral or achiral? I thought Inositol was clearly chiral since it has 6 asymmetric C atoms. But friends of me argue that it is achiral and i've found websites say the same (http://bip.cnrs-mrs.fr/bip10/Inositol-EJB.pdf). So what am I missing? Is it chiral or achiral?

Having asymmetric carbon atoms does not in itself make a compound chiral.

A compound is chiral if and only if it is not super(im)posable on its mirror image.

If there's a plane of symmetry, then reflecting in that plane gives the same thing, by definition, and that meets the criterion for super(im)posability - the mirror image is superimposable on the original and therefore the molecule is not chiral.

Only D- and L- chiro-inositol are chiral - all the other stereoisomers have planes of symmetry and are therefore not chiral.


Chemical reaction between dichloromethane and solid carbon dioxide I studied heat transfer liquids with low melting point (coolants), and how fast their temperature decreases when they are contacting with solid carbon dioxide. Nearly all liquids (brines, glycols etc.) that I studied have similar behavior, except dichloromethane.

When I put solid carbon dioxide grains into dichloromethane, the temperature of the liquid dropped nearly immediately! At first, the temperature was $\pu{20^\circ C}$, and after just $\pu{5 s}$ the temperature of the liquid was $\pu{-70^\circ C}$! Do you know why is it possible? When I freezed $\ce{KCl}$-brine, the temperature had been decreasing from $\pu{20^\circ C}$ to $\pu{-40^\circ C}$ for more than $\pu{5 min}$!

This property of dichloromethane may be very useful. I think it is due to its low viscosity.

Can you help me and explain this phenomenon?

In brine cooling happens only on the surface of the solid carbon dioxide and causes the water to freeze there, forming an insulating layer (the same happens with ethylene glycol). In dichloromethane, which freezes below $-97$ degrees, no freezing occurs. Moreover, carbon dioxide is well soluble in dichloromethane, which makes the heat transfer more efficient. The latter property may cause you problems if you want to use your coolant in closed installations, because after warming the solution may release the excess of carbon dioxide.


Does caffeine isolated from coffee contain harmane? I'm not sure whether the caffeine contained in otc caffeine pills is made synthetically or isolated from coffee and other plants, but if the latter is the case, is it likely that it contains the chemical harmane?

It's claimed that isolated caffeine is 99% pure and I'm wondering whether the remaining 1% is likely to contain any appreciable amount of harmane.

Given that caffeine can be readily extracted from a variety of abundant and exceedingly cheap natural sources, I doubt it would be economical to synthesize. Various decaffeinated products are popular, and the caffeine removed during the decaffeination process is normally recovered, purified, and re-sold for use in various products, which makes the synthetic route even less economically attractive. As far as I'm aware, the vast majority of caffeine sold commercially is extracted from plant sources. You can do your own research to determine the exact statistics, but since your concern relates to caffeine extracted from natural sources that might also contain harmala alkaloids, I'll address that case specifically.

There are a number of different methods of caffeine extraction (see the Wikipedia page on decaffeination), but all use solvents and/or adsorbents at some stage in the separation and purification processes. From the perspective of chemistry, then, the question is how similar are the solubilities of caffeine and harmala alkaloids, and what percentage by mass do they constitute of the natural caffeine sources used? I was able to find one article that quantified the chemical constituents of coffee beans. If you examine the data table therein, you'll see that the main constituents of those beans are carbohydrates, lipids, amino acids, and chlorogenic acids, not one of which is a category that applies to the harmala alkaloids. Lipids, carbohydrates, and amino acids are generally the main constituents of virtually all plants, so I think it's probably safe to extend those statistics to other natural sources besides coffee beans. Purely as a matter of probability, it's reasonable to assume the impurities are likely to consist mostly of substances from those categories. With respect to solubility, the harmala alkaloids are mostly somewhat heavier nitrogenous bases which are less polar than the caffeine molecule (at least in their neutral, "free base" form). Given that, I'd expect that methods using water extraction, which appear to be popular, would separate caffeine from harmala alkaloids fairly efficiently (I found one paper indicating that harmala alkaloids are water insoluble, while caffeine is anywhere from moderately to highly soluble depending on temperature).

An impurity of 1% is already quite small, and that 1% comprises a variety of compounds, both organic and inorganic (the overwhelming bulk of which are likely not harmala alkaloids, given the natural composition of caffeine-containing plant sources I described above). Based on the above information, I'd suspect that only a tiny fraction of those impurities, if any, would be likely to consist of harmala alkaloids at a detectable level.


Which is the correct way to name pentan-3-one? I have seen this organic compound named in two different ways.

According to some sources it is written pentan-3-one.. Other sources call it 3-pentanone.

Which one is correct?

Whether you write pentan-3-one or 3-pentanone both are correct.

But according to IUPAC rule pentan-3-one this is correct format of writing the name of organic compound. So I prefer you to write pentan-3-one rather than 3-pentanone.

You should also read this.


How to draw wedged double bond in ChemDraw? How can I draw a wedged double bond in ChemDraw? When I try to draw BINAP, making the wedged double bond seems impossible. It is however possible to draw a bold double bond.

As Zhe pointed out in the comments, bolding both bonds looks 'weird', and if you look in textbooks or throughout the chemical literature, BINAP (and related systems) are commonly drawn with only the outer edge bolded (as shown on the left in the image below).

Although the feature you want isn't directly supported in ChemDraw,  one can get both lines bolded by simply selecting the bond in question, right-clicking, and changing the line width to match the bold width (this gives you the structure on the right hand side of the image below).

That said, just because you can do this, doesn't mean you should, it goes against the standard way of drawing these systems, and doesn't add anything in terms of clarity (the first diagram emphasises the stereochemistry perfectly well, the additional bold bond adds nothing).


Advantages of thioesters over esters in fatty acid catabolism In several metabolic pathways, Nature (live systems) chooses to make esterification using thioesters (CoA) and not esters. I would like to know what is the main aim of the Nature with this choice: faster reactions or more extensive reactions?

what is the main aim of the Nature with this choice: faster reactions
  or more extensive reactions?

Nature is frugal, its main aim is to spend as little energy as possible in a chemical reaction.  Nature will always evolve so as to use the lowest energy pathway.

If we compare the resonance structures for an ester and a thioester we see that we can draw the same three resonance structures to describe each structure.

However, the resonance structure on the right, the one with the double bond between the carbonyl carbon and the ester oxygen or the thioester sulfur, is less important in the sulfur case.  Why?  Because a double bond between second and third row atoms is not nearly as strong as a double bond between two second row elements (much better electronic overlap results between atoms of similar size).

As a result the thioester carbonyl is not as stabilized by resonance as the ester carbonyl.  Said differently, the thioester carbonyl is more reactive (lower activation energy to react) than the ester carbonyl.  The thioester motif provides nature with a more reactive carbonyl. This way nature doesn't have to spend as much energy to bring about reaction at this more reactive carbonyl.


What are the common products of the reaction: 2-bromobutane and NaOH I am self-studying organic chemistry to better my understanding, I cannot formally enroll at a university due to being active duty.

From my understanding there are 2 possible mechanisms when 2-bromobutane reacts with NaOH or any strong hydroxyl base: The E2 and SN2. The carbon attached to bromine is secondary. In the E2 reaction, the oxygen from the hydroxide attacks the "beta" hydrogen, deprotinating it and allowing the electron to travel and form a double bond. Bromide is a leaving group. We get 2-butene as well as water and the bromide ion.

But we have two products from this reaction: trans-2-butene and cis-2-butene.

We also have (2,S)-butanol.

My question is which of these are more favored and why? I know that cis-2-butene would not be favored due to sterics. If possible, I would also like to know when E2 reactions are favored over SN2.

A chemical reaction is the result of competition; it is a race that is won by the fastest runner. A collection of molecules tend to do, by and large, what is easiest for them.

An alkyl halide with β-hydrogen atoms when reacted with a base or a nucleophile has two competing routes: substitution ($\mathrm{S_N1}$ and $\mathrm{S_N2}$) and elimination.

Which route will be taken up depends upon the nature of alkyl halide, strength and size of base/nucleophile and reaction conditions.

Thus, a bulkier nucleophile will prefer to act as a base and abstracts a proton rather than approach a tetravalent carbon atom (steric reasons) and vice versa.
Similarly, a primary alkyl halide will prefer a $\mathrm{S_N2}$ reaction, a secondary halide – $\mathrm{S_N2}$ or elimination depending upon the strength of base/nucleophile and a tertiary halide – $\mathrm{S_N1}$ or elimination depending upon the stability of carbocation or the more substituted alkene.

So, I think preferably in aqueous NaOH or KOH there should be formation of alcohol but that too by $\mathrm{S_N1}$ if there's a tertiary; and, if primary halide, then it would follow $\mathrm{S_N2}$. It's bit tricky for secondary; again it depends on the reagents and solvents used.


Negative LUMO Energy During the (DFT-)computations that I have done over the last years, it happened quite often that I got negative energies for the LUMO and many times also for a bunch of virtual orbitals.

Is there a common explanation what negative energies for LUMO means?

I was told that it might have to do with the overestimation of correlation energy within DFT. But what would it mean if there is no error?

Negative compared to what?

Most chemical system have several unoccupied orbitals still in the negative energy range, meaning lower excited states are still bound systems (they don't fall just apart).


Approximation of electron density of a small molecule Starting point for Quantum Theory of Atoms in Molecules (AIM) is the electron density $\rho(r)$. It could be determined experimentally (X-ray diffractions). I want to know how can we estimate them ab initio. In AIM
it is often emphasized that the focus is on density rather than a wave function for the later is just a mathematical function while the former is experimentally observable. Are there ways to estimate electron density without first calculating wave function (which will obviously lead to density function).

My interest is to use AIM in quantitative structure-activity relationship. I will typically deal with molecules of around 50 atoms. Usually I will already have the graph (connectivity) of the molecule (eg .SDF file, smile notation etc), how can we incorporate this to speed up electron density calculation?

It would be helpful if brief comment on the computational cost is also included in the answer.

You are looking for density functional theory. The theory is somewhat involved and you may be interested in A Chemist's Guide to Density Functional Theory or similar text book. A full description of the variety of ways to compute the (ground state) electron density is beyond the scope of an answer and better suited to a textbook.

The general gist is to optimize the energy with respect to the electron density numerically.


Why does the liquid in an Oralmedic stick flow uphill after I turn it upside-down? I was experimenting with an Oralmedic stick, which is basically a cotton swab with a hollow stick that has liquid medicine inside, and observed something very unexpected.  I followed the usage instructions, which say to snap the cotton bud at one end. As expected, the liquid inside the stick began flowing down into the cotton bud at the other end.

Out of curiosity, I turned the stick upside-down, to see what would happen. I was really surprised to see that the liquid still kept flowing in the same direction, which was then uphill after I inverted the stick!

How is this possible?

The most likely explanation is a combination of two capillary effects, where a combination of surface tension and adhesion of the liquid to surrounding solid surfaces causes macroscopic flow to occur.

For the first, the internal diameter of the hollow stick is small enough that if one were to pull off the cotton buds at the top and bottom, capillary forces would likely cause the fluid to flow only very slowly through it. This provides some resistance to flow, sufficient to keep the liquid from gushing up to the 'snapped bud' end when you invert the stick. The surface tension associated with this capillary effect also results in the liquid behaving like a "plug" in the hollow tube: gas cannot "slide" around the liquid meniscus, which is also an important component to the physics.

The other role of capillary action is active in the fine threads of the lower cotton bud itself.  Once the top bud is snapped, the liquid starts to slowly flow down toward that unsnapped bud. Once the liquid reaches it, it wicks aggressively into the finely structured cotton.  Since no gas can get past the liquid in the tube, conservation of mass requires that the rest of the liquid continue to flow down toward the unsnapped bud.

The reason why flow continues toward the unsnapped bud when you invert the stick is a combination of all three of the above factors: the capillary action drawing the liquid into the unsnapped bud is essentially unaffected by the inversion of the stick, and since no gas can slip around the remaining liquid in the tube, that remaining liquid has to continue flowing up toward the unsnapped bud.

Another way to think about it is this: as each microscopic parcel of liquid wicks into the unsnapped bud, it leaves a region of slightly lower pressure behind it.  Since the end of the liquid column facing the snapped bud is open to atmosphere, that results in a pressure differential against the snapped-bud side of the liquid column that pushes the liquid up toward the unsnapped bud.


How to remove Sn from Sn60Cu40 alloy? I have practical question.

How can I chemically remove $\ce{Sn}$ from $\ce{Sn60Cu40}$ alloy?

I want to get something like copper sponge this way.

Something like nanoporous gold produced from $\ce{Au50Ag50}$ for heat exchanger, but cheaper.

You can remove tin from bronze (copper-tin alloys) and leave behind the copper. However, doing so may not leave you with copper in an easy to use form.

Tin is a less reactive metal than copper. Tin reacts with all acids, like $\ce{HCl}$, while copper only reacts with oxidizing acids, like $\ce{HNO3}$.

$$\ce{Sn(s) +2H+(aq) -> Sn^{2+}(aq) + H2 (g)^ }$$
$$\ce{Cu(s) + 2H+ (aq) -> No \ Reaction}$$
$$\ce{Cu(s) + 4H+(aq) + 2NO3- (aq) -> Cu^{2+} (aq) + 2H2O (l) + 2NO2 (g) ^}$$

Soaking your bronze in acid (even vinegar might work) will eventually dissolve the tin away from the copper. Heating, or using a stronger acid like the $\ce{HCl}$ in muriatic acid, will likely increase the rate. However, since the composition of the alloy is uniform throughout, the copper will likely come out of solution as a fine powder.


Isocyanides undergo addition reactions, but cyanides do not. Why? My book mentions addition reactions of isocyanides with halogens, ozone, sulphur, etc. but it doesn't mention any such reaction for cyanides. Instead, it bluntly states: "Cyanides do not undergo addition reactions". My question is: why is this so?
 I understand the structures of the ions, but am unable to relate the exact correct logic in this situation.

Here are the addition reactions of isocyanides I am talking about:

$$
\begin{align}
\ce{R-NC + O3 &->RNCO + O2}\text{(methyl isocyanate)}\\
\ce{R-NC + 2HgO &-> RNCO + Hg2O}\text{  (methyl isocyanate)}\\
\ce{R-NC + S &-> RNCS} \text{  (methyl isothiocyante)}\\
\ce{R-NC + Cl2 &-> RNC\cdot Cl2} \text{ (methyl imino carbonyl chloride)}\\
\end{align}
$$

Isocyanide can undergo addition reaction due to the fact that it is capable to attach both the electrophile and nucleophile in the same carbon atom while it is not achieved by cyanide. The negative charge on carbon atom accepts the electrophile to form an intermediate species in which the positive charge on nitrogen atom gets neutralized by the addition of the nucleophile to the carbon atoms and the nucleophile gets attached in the same carbon atom. Cyanide is incapable of undergoing such mechanism because of less nucleophilicity of nitrogen atom of cyanide and thus don't undergo addition reaction. The mechanism is as follows:


Iron(III) nitrate and hydrogen peroxide reaction When iron(III) nitrate is added to hydrogen peroxide as a catalyst, the temperature initially decreases and begins to increase after 30 seconds. Why is this the case? (is there any intermediate reactions of $\ce{H2O2}$ and $\ce{Fe(NO3)3}?)$

Nitrates tend to lower the temperature on dissolving, having a positive enthalpy change of solution. For example, $\ce{NH4NO3}$ has an enthalpy of solution in water of ~26 kJ/mol at ~300 K, and it is used in self-cooling cold-packs for that reason.

Sorry, I could not locate the enthalpy change of solution for $\ce{Fe(NO3)3}$, so this is not a definitive answer. You could measure that value, though, with some  $\ce{Fe(NO3)3}$, some $\ce{H2O}$ and a calorimeter.


Oxidation of styrene with permanganate My teacher told me that when we oxidise styrene with $\ce{KMnO4}$, it will stop at benzaldehyde. But I think that it will continue to go on till benzoic acid. Am I right?

In the heterogenous permanganate oxidation of styrene, that is when the reaction is performed in dichloromethane with potassium permanganate on a solid support (such as aluminium oxide), the reaction stops at the stage of the aldehyde.

The method was published in Synthesis 2001, 11, 1645-1648 and the authors report a yield of 90% for the oxidation of styrene to benzaldehyde.


Lowering energy in molecular orbital mixing When drawing molecular orbital diagrams for mixing two orbitals we always end up with one orbital which is lower in energy than the initial orbitals and another which is higher in energy. The principle extends to mixing of more than two orbitals with the total number of orbitals always being conserved.
Often we will use arguments involving mixing of molecular orbitals to form MOs of a slightly lower energy to explain phenomena such as hyperconjugation with diagrams such as this.

In this situation, when you have one filled and one empty orbital, the electrons always end up in a lower energy than they started.
What stops you from mixing these orbitals again to produce even lower energy orbitals, even if the energy difference is very small? I understand that symmetry might have something to do with it but I am not well versed in the more mathematical elements of MO theory such as symmetry groups. Or is my initial assumption that mixing two orbitals always results in a lower energy wrong?

The orbital $\psi$ is obtained as a mixture of the orbitals {$\phi_i$}

$$\psi= c_1 \phi_1 + c_2 \phi_2$$

choosing the {$c_i$} through energy minimization.

Once obtained $\psi$, if you try to mix it with the initial {$\phi_i$} to obtain a new function, lets say $\mu$, you'll have:

$$\mu = c_3 \psi + c_4 \phi_1 + c_5 \phi_2$$

(notice that the coefficients don't need to be the same)

or equally

$$\mu = c_3 (c_1 \phi_1 + c_2 \phi_2) + c_4 \phi_1 + c_5 \phi_2$$
$$\mu = c_3 c_1 \phi_1 + c_3c_2 \phi_2 + c_4 \phi_1 + c_5 \phi_2$$

$$\mu = (c_3 c_1 + c_4) \phi_1 + (c_3c_2+c_5) \phi_2$$

that has the same functional form that
$$\psi= c_1 \phi_1 + c_2 \phi_2$$

so as the criteria is the energy minimization it should be no surprising that we can not get better energy that making
$(c_3 c_1 + c_4) = c_1$ and $(c_3c_2+c_5)=c_2$.

The idea behind this is that you want to obtain the "better" function $\psi$, but you are restricted to a set of functions of the form: $c_1 \phi_1 + c_2 \phi_2$.  You impose this restriction because it only requires to obtain two coefficients, that is far simpler than obtaining an arbitrary function (defined with infinite images).
One restriction is that the equation $c_i \phi_i = c_k \phi_k$  can only be true if $c_i = c_k = 0$.

You can expand your search to a larger set of functions, for example: $c_1 \phi_1 + c_2 \phi_2 + c_3 \phi_3$, but it requires that the set still be linearly independent, that is, you can not get any of the functions with a linear combination of the others, (i.e. it is not possible to get $c_1 \phi_1  =  c_2 \phi_2 + c_3 \phi_3$ if at least one $c_i$ is different from 0). If not, as in the discussion above, the space of function will contain the same functions so the inclusion of the new function is in some sense redundant or useful.

You can add infinite linearly independent functions, but, it does not mean that the energy will be always get lower.


How to find the concentration of potassium permanganate in this experiment? I completed a titration experiment in a lab. The aim was to find the concentration of the titrant $\ce{KMnO4}$ solution by titrating it with a sodium oxalate solution of a known concentration.

I first made a standard solution of sodium oxalate solution, using $\pu{0.826 g}$ of sodium oxalate powder, then poured $\pu{50ml}$ of distilled water along with $15$-$\pu{20ml}$ of $\pu{3 M} \,\,  \ce{H2SO4}$. Then I pipetted out $\pu{25 ml}$ of the solution and poured it in to a volumetric flask. Then I filled the rest of the volumetric flask with distilled water, to form the standard solution. I then did the Mcbride experiment to find $75\%$ of the volume of $\ce{KMnO4}$ need to titrate the solution.

I then did the experiment three times to get an average volume of $\ce{KMnO4}$ to be $\pu{25.48 ml}$. And the volume of sodium oxalate solution is $\pu{25 ml}$.

I should find the concentration of $\ce{KMnO4}$ now.

I know the molar ratio is 2:5. How do I calculate this?

I know I need to find the moles of sodium oxalate then I multiply it by $\frac{2}{5}$ and then divide that value by the volume of $\ce{KMnO4}$ which is $\pu{25.48 ml}$. But how do I calculate the moles of sodium oxalate? What would the concentration be? I know the volume is $\ce{25 ml}$.

This is what I have done, but I am not sure if it is correct:

Moles of sodium oxalate = $\frac{\pu{0.8265 g}}{\pu{133.96 g mol-1}} = \pu{0.006169752 mol}$

$\frac{0.006169752}{\pu{0.025L}}$ = concentration of sodium oxalate is therefore $0.246790086$, $0.0248$ to 4 s.f.
($\pu{0.025 L}$ is the volume of the sodium oxalate)

I then multiplied that by $\frac25$.  $0.0248 \cdot \frac25$ gives me the concentration of $\ce{KMnO4}$ which is $0.09872$.

But I think this is wrong because I have not used the volume of $\ce{KMnO4}$ used, which was the whole point of the experiment.

I think your experiment has one setup mistake.:

You're mixing $\pu{0.8265 g}$ of oxalate with $\pu{50 mL}$ of water and $15$-$\pu{20 mL}$ of sulfuric acid. So your $\pu{25 mL}$ oxalate solution does not contain $\pu{0.8265 g}$  of oxalate (more or less $\frac{25 \cdot 0.8265}{ 65-\pu{70 mL}}$). So you can't know the exact oxalate concentration of this solution.

I recommend you to use a $\pu{100 mL}$ volumetric flask to prepare the oxalate solution. You can then calculate its concentration and the amount of oxalate you have in $\pu{25 mL}$.


Polyatomics -How Do They Work? In class, we recently learned about Ionic nomenclature. Specifically, polyatomics and monoatomics. Even though our teacher did a decent job (I guess) of explaining it to us, I'm still very confused.
I have multiply questions:

Question: How would you know what type a compound is when given only the formula?

For example: Our teacher started with $\text{AuCl}$ and wrote down Gold (I) Chloride as the nomenclature. How did he get from $\text{AuCl}$ to Gold (I)?
There are two possibilities: Gold (I) and Gold (III).
Another example: Another example would be $\text{Pb}_3\left(\text C_6\text H_5\text{O}_7\right)$. Which apparently corresponds with Lead (II) Citrate. What does the II inside the parenthesis mean?
And below, I have provided some more examples. Please leave some for me! :P

Examples: Write the formulas for the following compounds:

Copper (II) Chloride

Vandium (III) Selenide

Beryllium Oxide

Potassium Permanganate

Ammonium Nitrate

Tin (II) Sulfite

Chromium (IV) Cyanide

Aluminum Arsenide

I'm lost at this point. Any help and a few suggestions would help a lot!

Our teacher started with $\ce{AuCl}$ and wrote down Gold (I) Chloride as the nomenclature. How did he get from $\ce{AuCl}$ to Gold (I)?
There are two possibilities: Gold (I) and Gold (III).

True, but there is only one possibility for chloride ;)
Karl's comment under your question gives the crucial hint:
Start with the part that you know and that is unambiguous!
Let's have a look at some examples:

Chloride is (always) $\ce{Cl-}$; it has a charge of (-1).
If your teacher would have written $\ce{AuCl3}$, you would know that there are 3 chloride ions with a total charge of $3\times(-1)=-3$. Consequently, you would know that you're having $\ce{Au(III)}$ here. The copper (II) cloride example belongs to the same category and if you like, you can try iron(II) bromide, iron(III) chloride and copper(I) iodide.

Karl also mentioned oxygen. You can safely assume one or more $\ce{O^{2-}}$ units in your metal oxides, depending on the oxidation state of the metal. Can you assign the oxidation state in $\ce{Fe2O3}$?

Nitrite, nitrate, sulfite, sulfate are some anions that you should memorize sooner or later. You might know that the names that start with nitr~ refer to nitrogen, the sulf~ refers to sulfur. All these anions contain various number of oxygen atoms.


Alkali metal hydrides I know that stability of alkali metal hydrides decreases down the group,Hence the lowest member in the group easily breaks into its cation but does this easiness of breaking into cation has any relation with its reducing nature.I also wanted to know what is order of reducing character in alkali metal group?

Alkali metal hydrides can typically be viewed as labile hydride anions. The extent of lability can correlate with the electronegativity of the metal. Hence, less electronegative metals in alkali metal hydrides yield more labile hydride anions and more "reducing" nature.


Relation between R and k $R$ is the universal gas constant equaling $8.314\ \mathrm{J/(mol\ K)}$.
$k$ is the Boltzman constant equaling $1.38\times10^{-23}\ \mathrm{J/K}$.

$$R=nk$$

$n$ is no. of moles.
$n$ can change.
$R$ is a constant, $k$ is a constant,  but $n$ changes. Then how is $k$ a constant?
Is the equation correct?

The form of the equation is correct, but one of the variables is wrong. It should be:

$$\mathrm{R} = N_{\mathrm A}k_{\mathrm B}$$

The Boltzmann constant ($k_{\mathrm B}$) is for one molecule (or atom); the universal gas constant ($R$) is for one mole of molecules (or atoms).

Therefore, in this relationship - and which is clear from dimensional analysis ($R$ is per mole) - $N_{\mathrm A}$ is the Avogadro constant, and not number of moles: in your argument, $n$ is a dimensionless integer, and thus the units do not work out. You need to use $N_{\mathrm A}$ instead of $n$. Note that the Avogadro constant can be denoted by either $L$ or $N_{\mathrm A}$.

The proper relationship is:

$$R = (\pu{6.022*10^{23} mol-1}) \cdot (\pu{1.380648*10^{-23} J*K-1}) = \pu{8.314 J*K-1*mol-1}$$

Note that I have included more significant digits for the Boltzmann constant here to get your value for $R$.


Friedel–Crafts reaction of phenol Does phenol undergo Friedel–Crafts reactions or does it react with Lewis acids like aniline does?

Like aniline, phenol too reacts to a very less extent during Friedel-Crafts reaction.

The reason being that the oxygen atom of phenol has lone pair of electrons which coordinate with Lewis acid.

In fact most substituents with lone pair would give poor yield.

The two pathways involved in the reaction with phenol reduce the overall yeild:

Phenols are examples of bidentate nucleophiles, meaning that they can react at two positions:

on the aromatic ring giving an aryl ketone via C-acylation, a Friedel-Crafts reaction

or,

on the phenolic oxygen giving an ester via O-acylation, an esterification


How to verify that ethanol is infinitely soluble in water? My textbook [Chemistry Live, Declan Kennedy] states that ''Ethanol is said to be infinitely soluble in water''. It then says ''verify this for yourself by mixing some ethanol with water in a test tube''.
Both water and ethanol are colourless. How can I observe that they are infinitely soluble ?

The better term is miscible in all ratios. The idea behind the term is that you can have a mixture from 99.9999… % water and 0.0000…1 % ethanol to a mixture of 0.0000…1 % water and 99.9999…% ethanol. No infinity necessary.
The obvious opposite to miscible is immiscible (in some or all ratios). A common pair of household liquids that are immiscible are fatty oils and water. In the chemistry lab, hexane and water might be a better example.
To verify the miscibility of water and ethanol, set up a parallel experiment with water and hexane/a fatty oil such as sunflower oil. You will notice that a fully immiscible pair will never mix, there will always be two different bits of liquid visible. You can try a little oil/hexane and lots of water and vice-versa. On the other hand, a miscible pair will mix and you get a solution where you cannot tell one from the other.
For completeness, partially miscible: if you repeat the experiment with ether and water you will observe that the first couple of drops of water will dissolve in sufficient volumes of ether (and vice-versa) but if more is added you will again get two phases.


Why is incubation in dark place necessary? While estimating the  total protein through Lowry method one needs to incubate the test tubes in dark once after adding alkaline solution and once after Folin-ciocalteu solution. My teacher said it is necessary for proper colour development. But I am curious to know what is the chemistry behind it?

Studies have not yet defined the exact mechanism of Lowry method. However, incubation in dark is suppose to promote the stability of copper tartrate complex with Folin-ciocalteu solution and thus their reaction/interaction with protein which ultimately leads to the oxidation of aromatic residues. You can read chemistry behind Lowry method for more details.


How is phosphoryl chloride attaching an aldehyde to benzene? This reaction was part of an organic scheme I was doing:

I thought that $\ce{POCl3}$ just converts the OH group into Cl, but looking at the chemical formula for A, I noticed that A has an extra carbon compared to the starting material.

It turns out that A is

I'm very confused as to how that aldehyde group was added to the benzene ring. Where is the additional C coming from?

It is a Vilsmeier–Haack reaction.

You are actually neglecting the effect of DMF (Dimethylformamide).

The Mechanism:

1. Formation of Iminium Cation:

The reaction of the DMC i.e. Dimethyl formamide (which is basically an amide) with $\ce{POCl3}$ produces an iminium cation.

2. Aromatic Electrophilic Substitution Reaction:

The formed iminium cation now act as an electrophile and does electrophilic aromatic substitution as a result of which we get an aryl ketone or an aryl aldehyde. So that extra aldehyde group which you are speaking about came from the electrophilic substitution reaction of your initial structure with iminium ion.

Here I have drawn the whole mechanism for you.


Which is the cheapest plastic that retains its structure at 85 °C or higher temperature? In an academic project, I need to protect some electronics components from heat (around 85 °C). I am searching for both a cheap plastic film and a cheap plastic to be used in a injection molding process.

As matt_black referred, the main criteria of the selection is "the point where structural integrity is reduced by repeated use at that temperature or the point where the polymer starts to creep".

By now I am using cork, but I would like to use a more robust material.

Those materials should be both thermal and electrical insulators and should not absorb water. They should also be resistant to the external environment: rain, sun (UV radiation), dust, etc.

Which materials are the best option in this case? From what I have searched, these are the main options in terms of thermal requirements:

PPO/PPE based
Polypropylene
Polyurethane
Vinyls
Polybutylene
Acetals
ABS and SAN
Polystyrene
ABS/Polycarbonate alloy
Acrylics
Cellulosics
Polyethylene and copolymers

The melting point is not going to give you the answer you want.  To properly select a polymer you need to look at the glass transition temperature $(T_\mathrm{g})$.  This is the temperature at which a material loses crystallinity and starts to more easily flow.  This parameter varies widely with thermal history and internal stress of a polymer material and can be raised by annealing.

Based on UV resistance, I would exclude any olefin or vinyl polymers as well as ABS.

For water resistance, I would exclude cellulose polymers and acetals.

Based on the temperature requirement I would say polystyrene ($T_\mathrm{g} = \pu{100^\circ C}$), acrylics $(T_\mathrm{g} = 85 - \pu{165^\circ C})$, polycarbonate ($T_\mathrm{g} = \pu{135^\circ C}$), or PPE/PPO ($T_\mathrm{g} = \pu{215^\circ C}$) have sufficiently high $T_\mathrm{g}$'s to remain as acceptable for the mentioned specifications.


How to find oxidation state of Co and Cr in [Co(NH₃)₄(NO₂)₂][Cr(NH₃)₃(NO₂)₃]? Which of the following options represents the oxidation state of $\ce{Co}$ and $\ce{Cr}$ in the given complex?
$$\ce{[Co(NH3)4(NO2)2][Cr(NH3)3(NO2)3]}$$
(A) $2, 3$
(B) $3, 2$
(C) $3, 3$
(D) $2, 2$

I am unable to judge the net charge on complex when it breaks into ions and therefore I am having trouble with this question. I assumed that some unknown charge with opposite signs must be on the complex ions.

Conventionally, first complex ion is positively charged and the latter is negatively charged. Suppose thess charges are $p+$ and $q-$, respectively. Since two ion ratio is 1:1, $|p+|= |q-|$.

Suppose oxidation number of $\ce{Co}$ is $n$ and that of $\ce{Cr}$ is $m$. Hence, for cation complex (Note that the net charges of $\ce{NH3}$ and $\ce{NO2-}$ is $0$ and $-1$, respectively):
  $$n+0\times4+(-1)\times2=+p = n-2$$
and for anion complex:
 $$m+0\times3+(-1)\times3=-q=m-3$$
Since, $|p+|= |q-|$,
$$m-3= +p \; \text{and  }  n-2=-p$$
Or, $m+n=5$.
The only answer agree with this is (A) or (B):

If (A) is correct, then, $\ce{Co^2+}$ and $\ce{Cr^3+}$. Substitute these values in original two equation: $+2+0\times4+(-1)\times2=+0 $ and $+3+0\times4+(-1)\times3=+0 $. This can't be.

If (B) is correct, then, $\ce{Co^3+}$ and $\ce{Cr^2+}$. Substitute these values in original two equation: $+3+0\times4+(-1)\times2=+1 $ and $+2+0\times4+(-1)\times3=-1 $. This is correct.

Therefore, your answer is (B), and the oxidation numbers of $\ce{Co}$ and $\ce{Cr}$ ared $+3$ and $+2$, respectively.


Protactinium-234 half-life This is going to seem like a rather daft question, but what is the definitive half-live for protactinium-234 (234Pa)?  I'm seeing lots of different answers in various academic papers etc.

For instance, in Bonotto & Andrews (2000) the half-life of protactinium-234 (234Pa) is given as 1.18 minutes, but in Edwards et al. (2003) they report the 234Pa half-life as 6.67 hours.  So which one is it?

Bonotto, D.M. & Andrews, J.N. The transfer of uranium isotopes 234U and 238U to the waters interacting with carbonates from Mendip Hills area (England). Applied Radiation and Isotopes. 52, 965-983. 2000.  DOI: http://dx.doi.org/10.1016/S0969-8043(99)00151-7

Edwards, R.L., Gallup, C.D. & C Uranium-series Dating of Marine and Lacustrine Carbonates.  Reviews in Mineralogy & Geochemistry. 52, 363-406 . 2003. URL: http://www.minsocam.org/msa/rim/Rim52.html

So which one is it?

Both! Proactinium-234 exhibits nuclear isomerism. The metastable form $\ce{^{234m}Pa}$ has a half-life time in the range of 1 min, whereas $\ce{^{234}Pa}$ shows a half-life time around 6 hrs.


Which complexes are stable: ammine complexes or carbonyl complexes? According to Wikipedia,

A spectrochemical series is a list of ligands ordered on ligand
strength and a list of metal ions based on oxidation number, group and
its identity.
$$\ce{I- < Br- < SCN- < Cl- < S^2- < F- < OH- < C2O4^2- < H2O < NCS- < edta^4- < NH3 < en < CN- < CO}$$
Ligands arranged on the left end of this spectrochemical series are generally regarded as weaker ligands and cannot cause forcible pairing of electrons within 3d level and thus form outer orbital octahedral complexes. On the other hand ligands lying at the right end are stronger ligands and form inner orbital octahedral complexes after forcible pairing of electrons within 3d level.

According to above passage $\ce{CO}$ is a stronger ligand than $\ce{NH3}$. So, carbonyl complexes are more stable than ammine complexes.
Now, there are many other factors which determine the stability of complex:

Nature of central metal atom
Nature of ligand
Crystal field effects

Nature of ligand has further classification:

Basic character of ligands
Charge on ligands
Chelate effects

Classification source: see here and here
Basic character of ligands states that the more basic a ligand is, the greater the ease of ligation with the central metal atom is and the more stable the complex is. Now, $\ce{NH3}$ is more or less a basic compound and $\ce{CO}$ is a neutral molecule.
So, according to above assumption, $\ce{NH3}$ is a stronger ligand than $\ce{CO}$. So, ammine complexes are more stable than complex complexes.
So, which one is the stronger ligand: $\ce{CO}$ or $\ce{NH3}$? or which complexes are stable: ammine complexes or carbonyl complexes?

That quotation from Wikipedia is bad and should be reworded (I can’t think of good ways to edit it, though). Ligands in a spectrochemical series are not ordered by bonding strength or anything, they are ordered solely by the field split they induce.

Both $\ce{CO}$ and $\ce{NH3}$ are great at forming complexes — as is iodide. The main difference is that they won’t form similarly strong complexes with the same cores. Just to give a few examples:

Carbon monoxide is neutral and a π-acid: It has an empty orbital with π symmetry which can take part in metal-to-ligand back-bonding. As such, it stabilises low oxidation states of metals. The neutral carbonyl complexes such as $\ce{[Fe(CO)5]}$ are well known, but also complexes such as $\ce{[Fe(CO)4]^2-}$ (yes, that’s iron($\mathrm{-II}$)!) are stable.
Ammonia is also neutral and a very weak π-base. However, it is a much stronger Brønsted base than carbonyl is. Therefore, it is much better at stabilising medium to high oxidation states such as in $\ce{[Cu(NH3)4(H2O)2]^2+}$.
The halides, finally, are anionic and good π bases and thus able to stabilise high oxidation states. They are classed as weak-field ligands because they generally generate a very low field split. An example of a halide complex would be the bright yellow $\ce{[FeCl6]^3-}$.

You cannot take ligand $\ce{X}$ and call it strong or weak and thus expect it to make stable or unstable complexes. You cannot even take the same ligand and expect it to make inert or labile complexes all the time. One ligand-metal combination may be stable, another may be very unstable — $\ce{[Fe(CO)6]^2+}$ can only be synthesised under a $\ce{CO}$ atmosphere in $\ce{SbF5}$ from $\ce{FeF2}$ where the antimony pentafluoride abstracts any free fluoride ions from the solution leaving a ‘naked’ iron ion which will grab the only ligands it can reach. Iron pentacarbonyl is stable. The same goes for ammonia: Some will readily accept it as a ligand, others won’t. There is no black and no white, only shades of grey.

(Although since coordination compounds are generally colourful let me rephrase that: There is no black and no white, only a wheel of colours.)


What is the difference between ''1 C atom and 1 mole of C atoms?'' Does 1 C atom mean 1 single C atom out of Avogadro's number of C atoms = 1 single C atom out of 6.023 X 10^23 numbers of C atoms?
Does 1 mole of C atoms mean Avogadro's number of C atoms = 6.023 X 10^23 numbers of C atoms?

I am confused. Could you clarify it, please?

What is the difference between one egg and one dozen eggs?  A dozen is simply a certain count, in this case 12, that is rendered as one unit for our convenience.  Avogadro's number s just the same thing as a dozen, except we made the unit count larger than 12 in order to match it up with our conventionally used units.

Your assumptions about a "mole" are thus correct.


Is molten potassium nitrate possible? Is it possible to melt potassium nitrate without it decomposing into potassium nitrite and oxygen? If so when it solidifies does it keep its oxidizing properties?

https://en.wikipedia.org/wiki/Potassium_nitrate gives a melting point of $\pu{334°C}%edit$ without decomposition, but then decomposition sets in at $\pu{400°C}$ so you have a narrow liquid range.

If you avoid decomposition the oxidizing property still works.


Concentrating Isopropyl alchohol using non-iodized salt I am using 99% isopropyl alchohol to clean soldering flux residues for my electronics projects. Due to a certain viruse running amok in the world today, I am not able to get from my usual supplier. I am running low on supplies. So i got this idea that i would concentrate the more common (but still not so easy to get in my area) 40% or 70% ispropyl alchohol.

Searching the internet on guides to do it, it would seem that there is an easy way to do it. According to this instructables; add non-iodized salt to the solution until its past its saturation point. By this time the heavier water should sink and the lighter isopropyl should float. Extracting it would now be easy with a syringe.

For my question. Water and especially salt is an enemy to any metal. I would like to ask if this process mixes tiny amount of salt to the isopropyl? if so by how much %? Is NaCl even soluble to isopropyl or are these just tiny particles floating around and if i leave the solution to settle for a longer period of time those particles would sink down. up until what percent can i bring the concentration to?

As for the Salt to use what other ingredients should i look out other than iodine( or other close sounding).

P.S. I am no chemist any way, the only chemistry background i have is during my high-school days. It was a time that i was still not able to appreciate class lessons. So go easy on me

Common table salt can have additives, so I would recommend working with high purity Epsom Salts (magnesium sulfate hydrate). Heat to remove water of crystallization at 100 degrees Celsius for two hours.

Add your isopropyl alcohol with the MgSO4 in excess and wait until there is a separation. I would decant and allow for a safety margin above the visible separation line.

Here is a reference.


Why does orbital optimization remove spin contamination? I've come across numerous articles asserting that orbital optimized methods, where minimization of the energy is achieved through a series of unitary rotations (achieved by exponentiated single-electron excitations from the ground state reference) of the reference determinant, are exceptionally good at removing spin contamination. I've seen the numbers they report, and I believe them! I get the gist that orbital optimization allows us to overcome the error introduced by the UHF orbitals, but nowhere can I find the actual explanation of why these rotations effectively cure spin contamination.

Both references and explanations warmly welcomed.

Upon further thought, I've summarized it as follows. If anyone comes along with a better explanation, they'll get the cake!

Spin contamination is a fictitious effect arising from not achieving the full CI solution to the Schrödinger equation. For, had we found the exact solution, then the solution we found would be an exact eigenfunction of the ${\hat{S}}{}^2$ operator, as $[\hat H, {\hat{S}}{}^2]=0$.

In a general sense, orbital optimization can be viewed as simply allowing for more correlation to be recovered on top of whatever other improvements you've applied to your initial Hartree-Fock reference. So, by recovering more correlation, you've brought your solution closer to FCI. In doing so, you've brought your solution closer to a solution that contains no spin contamination.

However, there seems to be a stronger statement that could be made. This type of correlation seems especially potent in removing spin contamination.


Combustion of non-hydrocarbon organic compounds When we learned about combustion, we learned about the combustion of hydrocarbons and compounds with only hydrogen, oxygen, and carbon. However, how would combustion work with compounds that contained nitrogen or other compounds? For instance, what would the combustion of an amino acid look like in a formula?

Simultaneous elemental analysis of organic carbon, hydrogen, and nitrogen is classically accomplished by combusting the material under a stream of $\ce{O2}$, usually in the presence of a catalyst to ensure complete combustion, at a temperature of 950-1050$^o$C.  Carbon, hydrogen, and nitrogen are quantitatively converted to $\ce{CO2}$, $\ce{H2O}$ and $\ce{N2}$, respectively.  This is a common method in particular for determining nitrogen concentration in amino acids.


What are the chemical reactions behind fire? I've always wondered what the chemistry behind fire is. What are the basic chemical reactions behind a simple wood fire, and how do they manifest into this phenomenon?

Even though fire is one of the Greek classical elements, it is the only one that is not matter in our current understanding. What we experience as fire is the energy (in the form of light and heat) given off by the exothermic combustion of (usually) organic materials, like wood. The large amounts of thermal energy released by combustion often cause the gases in the fire to incandesce, that is, some of the kinetic energy produced by the fire is converted into electromagnetic radiation that we see (visible) and feel (infrared).

The chemical reaction for combustion is pretty simple. Take, for example, the combustion of n-octane $\ce{C8H18}$, which produces more than 5 MJ of energy per mole during combustion (from webbook.nist.gov).

$$\ce{2C8H18 +25O2->16CO2 +18H2O} \space\space \Delta_cH^o=-5430 \text{ kJ/mol} $$

The combustion of wood is more complex. The majority of the organic mass of dry dead wood is lignin and cellulose. Lignin is a highly-cross-linked copolymer of p-coumaryl alcohol, $\ce{C9H10O2}$, coniferyl alcohol, $\ce{C10H12O3}$, and sinapyl alcohol, $\ce{C11H14O4}$. Cellulose is a linear polymer of glucose, with formula $\ce{(C6H10O5)}_n$.

Different species will have different ratios of lignin to cellulose, different cross-link densities in the lignin, and different ratios of coumaryl to coniferyl to sinapyl alcohols. The formula for lignin could be expressed then as $\ce{(C9H10O2)}_x \cdot \ce{(C10H12O3)}_y \cdot \ce{(C11H14O4)}_z$.

The equation for the combustion reaction for cellulose is:
$$\ce{(C6H10O5)}_n +6n\ce{O2->}+6n\ce{CO2}+5n\ce{H2O} $$

Since lignin is more complex, its combustion equation is more complex:
$$2[\ce{(C9H10O2)}_x \cdot \ce{(C10H12O3)}_y \cdot \ce{(C11H14O4)}_z]+(21x+23y+25z)\ce{O2}\\ \ce{->}(18x+20y+22z)\ce{CO2}+(10x+12y+14z)\ce{H2O}$$

And assuming the variable composition of wood as a ratio (A:B) of cellulose to lignin, the overall combustion reaction becomes the following monstrosity:

$$A\ce{(C6H10O5)}_n +2B[\ce{(C9H10O2)}_x \cdot \ce{(C10H12O3)}_y \cdot \ce{(C11H14O4)}_z]+[6An+B(21x+23y+25z)]\ce{O2}\\ \ce{->}[6An+B (18x+20y+22z)]\ce{CO2}+[5An+B(10x+12y+14z)]\ce{H2O}$$

Given the number of variables ($A, B, n, x, y, z$), the $\Delta_cH^o$ for this reaction is difficult to determine (but not impossible, assuming we know some thermochemical reference data). Whatever its value, it is exothermic enough to induce incandescence.


How to find the number of hydrogen atoms in 0.5 mol of hydrogen gas? How do I calculate the number of hydrogen atoms in $\pu{0.5 mol}$ of hydrogen gas?
I presume the answer would probably somehow use the equation
$$n = \frac{m}{M},$$
where $n$ is the amount of substance, $m$ is the mass, and $M$ is the molar mass. However,  I'm not sure how to calculate it.

There is no need to use
$$n = \frac{m}{M},$$
since you already have been given the amount of substance. You would need to use the equation, if you were given the mass $m$, e.g. as $\pu{0.5 mg}$.

If you want to calculate the number of hydrogen atoms in $\pu{0.5 mol}$ hydrogen gas, then you should consider that they are diatomic molecules, $\ce{H2}$.
The unit mole is represented by the Avogadro constant and is
$$N_\mathrm{A} = \pu{6.02214076×10^23 mol−1}
\approx \pu{6.022×10^23 mol−1}$$
Each hydrogen molecule has two hydrogen atoms, therefore
\begin{align}
n(\ce{H}) &= 2 \cdot n(\ce{H2}),\\
N(\ce{H}) &= n(\ce{H}) \cdot N_\mathrm{A},\\
N(\ce{H}) &= 2 \cdot n(\ce{H2}) \cdot N_\mathrm{A},\\
N(\ce{H}) &= 2 \times \pu{0.5 mol} \times \pu{6.022×10^23 mol−1},\\
N(\ce{H}) &= \pu{6.022×10^23}.
\end{align}


What is the chemical equation for wine fermentation? I am researching about the impact of sugar on wine fermentation, so I am looking for the equation of that reaction. However, there are a few different versions from different sources, so I was wondering which one is the right one to use and what does each equation represent?

\begin{multline}
\ce{Glucose + 2 NAD+ + 2 Pi + 2 ADP ->}\\
 \ce{2 pyruvate + 2 NADH + 2 ATP + 2 H+ + 2 H2O + heat}\tag{1}
\end{multline}
\begin{gather}
\ce{C6H12O6 -> 2CH3CH2OH + 2CO2 + ATP}\tag2\\
\ce{C6H12O6 -> 2C2H5OH + 2CO2}\tag3
\end{gather}

Generally there are two types of fermentation in biochemical processes:

homolactic fermentation
alcoholic fermentation

Both these processes occur under anaerobic conditions to replenish NAD+
by the reduction of pyruvate in an extension of the glycolytic pathway.

The overall reaction in muscles is:
$$\ce{Glucose + 2ADP + 2Pi -> 2 lactate + 2ATP + 2H2O + 2H^+}$$
For alcoholic fermentation:

Glucose is first converted to pyruvate by glycolysis, and the pyruvate is converted to ethanol and $\ce{CO2}$ in a two step process:

Note: For the second one we just started with pyruvate (to show mechanism and enzymes involved)Otherwise, for alcoholic fermentation, the net reaction is:

$$\ce{Glucose + 2ADP +2Pi + 2H+ -> 2CO2 + 2C2H5OH +2ATP +2H2O}$$

Both homolactic and alcoholic fermentation have the same function: the anaerobic regeneration of NAD+ for continued glycolysis. Their main difference is in their metabolic products

Having said that lets go back to experimental wine fermentation, it is essentially alcoholic fermentation (2).

Enzymes in yeast (zymase a mixture of enzymes) and  cozymase  a mixture of cofactors-coenzymes such as NAD+ , ATP, and ADP, as well as metal ions) help carry out the fermentation reactions.


pH of a mixture of base and an acid I have the following question: Assume that you have two solutions. One of  $50\ \mathrm{cm^3}$ $\ce{HCl}$ $\mathrm{pH} = 1.3$ and one of $150\ \mathrm{cm^3}$ $\ce{Ba(OH)2}$ $\mathrm{pH} = 12.3$. If we mix these two solutions what will the $\mathrm{pH}$ of the mixture be?

I have done the following:
$\mathrm{pH} = 1.3 \Rightarrow [\ce{H+}] = 10^{-1.3} = 0.05\ldots\ \mathrm M  \Rightarrow n = 0.05\ldots \times 0.05 = 0.0025\ldots$
$\mathrm{pH} = 12.3 \Rightarrow \mathrm{pOH} = 1.7 \Rightarrow [\ce{OH-}] = 10^{-1.7} = 0.019\ldots \Rightarrow n = 0.019\ldots \times 0.15 = 0.0029\ldots$

My problem is I do not know how to proceed

The OH- and the H+ will react, neutralize each other to form H2O, H+ + OH- = H2O. Hence the ions remaining is: 0,0029... - 0,0025... = 0,00048... which will be OH- ions. I believe you can proceed on your own from here. Just calculate the concentration and then the pH.


Can the molecule CH2(ONa)2 be stable? While studying about Cannizzaro reaction, it came to my mind as a wild thought what if this HCH(ONa)(ONa) molecule could pop out in some step of this reaction (not giving much details about how I thought it as it would be irrelevant). Then I thought this molecule might have some problem with stability. The $\mathrm{sp^3}$ hybridized carbon might have some have problem with spacing two -ONa groups close together.

So, is this molecule stable or not? If not why?

The species you are proposing would be a methanediol dianion (CH2O22–) with two sodium cations “attached” to it in some way, depending on the nature of the solvent.

While methanediol can indeed readily form an anion, with a pKa of 13.3, I find no mention of the dianion or any measure (or prediction) of the second pKa of methanediol.


Does H3O+ donate a H+ to Water? Does the hydronium ion donate a proton to water? If the collision of 2 molecules of water can cause a proton to be transferred, does hydronium 'pass' a proton to water?

$\ce{H3O^+}$ is really just a "shorthand notation" for what really occurs in solution.  Typically higher hydrates of the proton, particularly $\ce{H5O2+}$ the "Zundel" cation and $\ce{H9O4+}$ the "Eigen" cation are thought to be the main players in the proton transfer process.  As permeakra has pointed out, Grotthuss was one of the early pioneers studying this process and his name is still used to describe the mechanism.  The mechanism is still an active area of research with considerable controversy remaining over the details of the mechanism.  This reference provides some diagrams of the general process.

Basically, just as electrons are said to use a "hopping" transport mechanism, the same has been said of proton transport.  The making and breaking of hydrogen bonds, accompanied by the twisting and turning of protons completes the process.  This proton hopping mechanism between these larger water aggregates is thought to be a key step at room temperature.  However proton tunneling plays a role as well, particularly as the temperature is decreased.


Nomenclature of Coordination compounds containing nitrogen As Per rules , amine when present as a ligand is written as "ammine" with double m's , so a compound having a complex ligand such as ethane-1,2-diamine , will be written with 1 'm' amine or double 'm' ammine?

As per rules, double m is used when the ligand is just $\ce{NH3}$, ammonia:

Cite from IUPAC Nomenclature of Inorganic Chemistry, Recommendations 2005:

  IR-7.1.3 Representing ligands in additive names

  Additive names are constructed by placing (sometimes modified) ligand names as prefixes
  to the name of the central atom. For anionic ligands, the anion endings ‘ide’, ‘ate’ and ‘ite’
  (see Section IR-5.3.3) are changed to ‘ido’, ‘ato’ and ‘ito’, respectively,
  when generating
  these prefixes. Names of neutral and cationic ligands are used unchanged, except in a few
  special cases, most notably water (prefix ‘aqua’), ammonia (prefix ‘ammine’), carbon
  monoxide bound through carbon (prefix ‘carbonyl’), and nitrogen monoxide bound through
  nitrogen (prefix ‘nitrosyl’) (cf. Section IR-9.2.4.1).

E.g. in $\ce{[CoCl(NH3)5]^{2+}}$ pentaamminechloridocobalt(2+).

For amines, the ligand name is unchanged, e.g.
$\ce{[Co(en)3]^{3+}}$ tris(ethane-1,2-diamine)cobalt(3+).


How does the human body process Bromelain in pineapple? (A meat tenderising/dissolving enzyme) It turns out pineapple juice can dissolve a jello set, and turn a beef roast to paste.

It turns out this is caused by Bromelain, an enzyme sometimes used in meat tenderising.

I'm trying to work out how you can eat pineapple without dissolving on the spot.

My question is: How does the human body process Bromelain in pineapple? (A meat tenderising/dissolving enzyme)

The enzyme is slow acting, so it does not dissolve your mouth, just slightly tenderises it, which is the sandpaper feel that you get on your tongue after eating it. When the pineapple reaches your stomach the enzyme gets dissolved by your stomach acids, so cannot harm you any longer.


Examples of molecules with D4 and C9 point groups Trying to come up with examples of (hypothetical) molecules/shapes that have $D_4$ and $C_{9h}$ point groups.

For $D_4$, I thought a cyclobutane structure, but with 4 substituents coming off of the ring all pointing in the same direction. However, I'm not sure if this is $D_4$ because in a book I am reading it says that all the $D_n$ point groups show chirality, which this molecule does not.

For $C_{9h}$, I really can't think of any shape/structure that doesn't just end up being $D_{9h}$. $C_9$ would be a 9-bladed propellor shape, although I'm not really sure how to draw this, and I don't see how I could alter that to get $C_{9h}$.

Thanks

For nearly any point group which is actually feasible (including $D_4$), look up the Symmetry gallery. Your all-cis tetra-substituted cyclobutane would be $C_{4v}$, BTW.

As for $C_{9h}$... well, I don't quite believe this is possible, but probably we may come up with something that would look almost realistic.

A substituted nanotube fragment will do, I think.

Surely it is hypothetical, so what? Cyclo-$\ce{C18}$ is hypothetical as well, and in my opinion, has lower chances to be stable. Also, its symmetry is kinda dubious.


Why do some solutions eventually stop boiling and start "popping" as they get more concentrated? A very long time ago I performed an experiment in my kitchen, part of which was to remove all water from a solution. Unfortunately I don't remember what was in it, but the point is that as the solution got more concentrated, it gradually stopped boiling, and would instead remain completely still for seconds, until a huge bubble would form instantly at the bottom of the test tube, making a pop sound and shooting some of the solution out of the tube. The effect got stronger and stronger until it became too strong for me to feel safe to continue.

I was unable to get an explanation of this effect out of my school teachers at the time. What I expected was that as the solution got more concentrated, the dissolved compound would start to precipitate, until eventually no water is left. Why did the popping occur instead?

What you witnessed was probably something similar to superheating/bumping. Essentially, when a solution is heated too rapidly, large pockets of vapor can form beneath the surface of the liquid that have an internal temperature higher than the expected boiling temperature of the liquid. In those situations, even small physical disturbances and/or further addition of heat can provoke the bubbles to burst explosively as the vapor is released. Usually, boiling chips (made of inert, inorganic materials like aluminum oxide, silicates, carbonates, etc.) are added to the reaction vessel in order to prevent that occurrence. The boiling chips provide an uneven and porous surface around which smaller and more numerous vapor nuclei can form, rather than the large bubbles that can form in smooth containers which have relatively few nucleation sites. Furthermore, the chips themselves provide additional surface area for the radiation of heat, allowing more uniform heat distribution. The likelihood of bumping can also be reduced by using wider reaction vessels, which allow superheated bubbles to escape without disturbing all of the surrounding liquid above them.


Water and olive oil intereactions This is really basic knowledge, but I am a little confused. I understand that if you put a tablespoon of water and a tablespoon of olive oil in the same container the water descends at the bottom of the container due to its density. In other words, since we put the same volume, water weighs more and thus it will move below the olive oil.

My question is why is the volume irrelevant with this experiment? Had we put a much bigger volume of olive oil in the container and just a small amount of water, shouldn't the water move above oil, since its smaller amount weighs less?

Any help would be appreciated.

In this experiment, the volumes are totally irrelevant. This is because all that matters (when regarding immiscible liquids) is density.
Density is the mass of a selected volume of matter (in this case olive oil and water). What is important, is that the water is denser, so it sinks to the bottom.
You can think of it, as an interaction between one volume oil and one volume water. Since the water is denser, the same volume of the stuff must weight more. From here, you can use the same logic as before, the heavier one, the water sinks. If you imagine it this way, on the surface of the two fluids there will always be some of both of them, so they'll arrange themselves in the correct order.

Even if the volumes are not the same, we can use Archimedes' principle, which

  indicates that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to theweight of the fluid that the body displaces and it acts in the upward direction at the centre of mass of the displaced fluid. Wikipedia

This ask sounds very technical, but what is means, is that the weight of anything submerged in a liquid is decreased by the weight of the liquid it displaced. To put it in other words, it's weight will decrease by the weight of the displaced liquid. Since these two have the same volume, we can compare only their densities.

Think of the experiment as oil droplets submerged in water. When imagining is this way, each droplet will displace the same amount of water (which is denser), so its weight decrease by more, than it ever was, resulting in a negative weight. Since weight is the force an object exerts on other ones "downwards" to keep it in place. We know that the system will at some point reach equilibrium. When this happens, if there was any oil in the water, it would mean, that it has negative wight, so it has to exert a force upwards to keep it in place. Since this is obviously not true, we know, that at equilibrium state all the water will be below all the oil.


Most basic nitrogen in Adenine I need to find the most basic site of Adenine:

The $\ce{NH2}$ group and $\ce{NH}$ cant be strong base as electrons are delocalised. This leaves us with $1,7,3$. I thought that electron density on $1,3$ nitrogens is enhanced due to $+R$ effect of $\ce{NH2}$.

But is this correct, I still cannot decide which is the strongest basic site.

To add: Both $1,3$ nitrogen have two nitrogen in neighbourhood which destablise their conjugate acids by $-I$ effect. So still I can't decide.

According to Ionization constants and thermal stabilities of uracil and adenine under hydrothermal conditions as measured by in situ UV–visible spectroscopy Geochimica et Cosmochimica Acta Volume 93, 15 September 2012, Pages 182-204

  Adenine undergoes two ionization reactions. Density function calculations in the gas phase carried out by Russo et al. (1998) indicate that protonation occurs preferentially on the N1 nitrogen of the 6-member ring, in good agreement with the assignment given by Christensen et al. (1970b). The anionic species with a proton-deficient 5-member ring is the most stable.

Here, "Russo" is Protonation of thymine, cytosine, adenine and guanine DNA nucleic acid bases: theoretical investigation into the framework of density functional theory J. Comput. Chem., 19 (1998), pp. 989-1000 which says (emphasis added) :

  HOMO of adenine is formed essentially by the lone pairs of all nitrogens. The orbital of the amine group contributes with the higher coefficient, and that of N1 with the lower one. On the basis of this information, N1 protonation might be the less favorable process. On the contrary, the molecular electrostatic potential map (Fig. 5) indicates that N1, N3, and N7 are the better candidates for proton attachment. In particular, the N1 and N3 regions appear to be favored over N7. Experimentally, the preferred protonation sites of adenine are N1 and N3 with the former slightly favored.[reference 18] Our computation favors protonation at N1 over those at N3 and N7 by 1.12 and 7.38 kcal/mol, respectively. This trend is in agreement with previous theoretical computations.[reference 17] From bond orders analysis (see Fig. 6), and due to the small energy difference it is possible to view the N1 and N3 protonated forms in N1 the double bonds occur in C4—C5 and in the C2—N3 positions, whereas, in.the second, in the C4—C5 and N1—C2 positions as two resonance hybrids of a unique structure with a consistent charge delocalization. This observation allows better insight into the reason forw hich the protonation on the five-membered ring is unfavorable. In fact, as shown in Figure 6, the positive charge appears to be distributed only on the C8 and N9 atoms, because it is evident that no interactions exist between the two rings upon protonation.

"Christensen" of the first reference is the same as reference 18 of the second reference and is Thermodynamic pK, ΔH°, and ΔS° and ΔCp° values for proton dissociation from several purines and their nucleosides in aqueous solution Biochemistry, 9 (1970), pp. 4907-4913


Would trans hydride ligands on a metal complex be beneficial for a catalyst? Would trans hydride ligands on a metal complex be beneficial for a catalyst?
Such as alkene metathesis or hydrodefluorination

In alkene metathesis you need large ligands that direct the stereoselectivity of the reaction, and a hydride ligand is very small.

Also for hydrodefluorination I don't see an advantage of a trans hydride ligand on the catalyst. cis hydride ligands could be beneficial, as they then take the place of the fluorine atom on the substrate.


Why do toothbrushes have to be dried? I carry around a toiletries bag. I put my toothbrush there right after brushing in a restaurant or some public area.

These strangely similar articles(1, 2) suggest toothbrushes have to be aired. Why exactly? What harm does the wetness on the bristles cause?

The good people at the American Academy of Pediatric Dentistry have a number of useful resources available to shed light on questions such as yours.

I direct you to a manuscript entitled Microbial contamination of toothbrushes and their decontamination in which it is written:

  Storage conditions of toothbrushes are an important factor
  for bacterial survival. Dayoub et al. and Meier et al. reported
  that the number of microorganisms in the toothbrushes kept
  in aerated conditions was lower than in toothbrushes stored in plastic bags. Several authors have reported that bacterial contamination can be reduced by washing toothbrushes after use, and drying in aerated conditions. Caudry et al. reported that a wet environment increases bacterial growth and cross contamination. Therefore, as time increases between one toothbrushing and another, more microorganism development can occur in the toothbrushes a wet/moisture environment.

The references cited in the above passage are detailed in the linked PDF. Happy brushing!


Causes of reversible reaction It is generally said that reactants react so that they can achieve a lower energy state. Then why does a reversible reaction occur in the first place? Since turning reactants into products increases their energy - an undesirable effect from the point of view of reacting molecules.

First, let's tackle what an irreversible reaction is. Baking a cake, for example, is commonly considered "irreversible." The eggs, flour, milk, etc. which you started with are no longer present in their original forms after the cake has been baked.

Now, can you reconstitute the eggs, flour, milk, etc.? Of course you could take the molecules of the cake and piece them together to make eggs, flour, and milk, but doing so would be a very formidable task. A lot of energy will have to be expended to reconstitute the egg, for example, because eggs are made of protein and with heat (baking) proteins are denatured (lose their original 3D conformation and possibly their linear makeup or amino acid sequence as well).

Another example of an irreversible reaction is combustion of hydrocarbons - i.e. burning gasoline in your car. This reaction is definitely reversible in the long run - it just takes a few million years, a lot of heat, and a lot of pressure for the hydrocarbons to be formed again. But for semantic's sake we say that combustion is irreversible because it's not reversible on a really relevant time scale.

So in general, the difference between irreversible and reversible isn't exactly well defined - i.e. it might be hard to reverse a reaction (turn a cake back into eggs, etc.) or it might take a long time to reverse a reaction, but what's hard? What's long? Tellinging, the IUPAC (International Union of Pure and Applied Chemistry) does not try to define irreversible vs. reversible reaction.

A deeper treatment would involve discussion of equilibrium constants (symbolized by capital K) but I'll let another poster get into this if necessary.

  It is generally said that reactants react so that they can achieve a
  lower energy state. Then why does a reversible reaction occur in the
  first place?

Good question. Remember that we can always add energy to make an unfavorable reaction proceed. For example, the sodium ion, which is isoelectronic with neon, is stable with a full octet of electrons. However, we can still take away more electrons. It just takes a rather sizable application of energy.