Effects Guiding Electrophilic Aromatic Substitution This is a question which I come across in an examination. I know the answer and also know the solution. Well when I first saw this problem I didn't know how to approach this problem and I think I will still not be able to predict the correct answer of a question which will be based on this question's type . I am not asking that this is "this" reaction and it must proceed like" this " ,but what I want is how will you approach this problem . What will you think on first look of the problem. What all things you will observe from question and also do write what you think the answer must be . Also BEWARE THIS HAS MULTIPLE ANSWERS CORRECT. I am not telling you the answer because it would be easier to explain that answer . Also, please those who have seen this problem earlier. Please resist yourself from posting answers.

The first thing I notice is the hydroxyl group, a very powerful activating group and an ortho-para director.  The hydroxyl group is such a strong activator that if you take a solution of phenol and drip bromine into it, the phenol will react with the bromine so rapidly that the bromine color will be immediately removed from the solution as 2,4,6-tribromophenol is formed.

The next thing I notice is the very bulky t-butyl group, also an activator, but much milder than hydroxyl and also a mild ortho-para director.

Next I notice that the two substituents are arranged meta to one another.  There is one position in between the two groups that will be sterically blocked (double-ortho effect); the position on the other side of the t-butyl group will also be sterically blocked, but not as bad.  Also the directing effect of the hydroxyl group will overwhelm the directing effect of the t-butyl group.

Finally, in the halogen series size increases from chlorine to iodine.

In the reaction with iodine, the position ortho to the hydroxyl group and para to the t-butyl group is the only position that is electronically activated and sterically accessible to the bulky iodine.

In the reaction with smaller bromine, the position ortho to the hydroxyl and para to the t-butyl and the position para to the hydroxyl and ortho to the t-butyl are electronically activated by both groups and sterically accessible.  The "double-ortho" position is still sterically blocked.

With the smaller chlorine atom, all 3 positions - even the "double ortho" - position are sterically accessible to the smaller chlorine.

Based on this reasoning, I would select answers A, B, and C.  I would probably also include D because the results are (weakly) reinforced by the electronic effect of the t-butyl group.


Can a saturated hydrocarbon have side chains? Saturated hydrocarbons: Each carbon atom is bonded to as many hydrogen atoms as possible. In other words, the carbon atoms are saturated with hydrogen (Source).

or

  Saturated means that the hydrocarbon has only single bonds and that the hydrocarbon contains the maximum number of hydrogen atoms for each carbon atom Latech.edu.

But if a carbon chain has side chains, the carbon from the side chain takes the place of a hydrogen. So it doesn't have as many hydrogens as it could have right?

But then I guess I could say that anything except methane has less than the max (4) number of hydrogens.

So what exactly is meant by maximum number of hydrogens?

Standard organic chemists’ understanding is that saturated/unsaturated is a word pair entirely independent of branched/linear/cyclic. Thus, all six combinations are possible:

branched/saturated (e.g. neopentane or 2,2-dimethylpropane)
branched/unsaturated (e.g. 2,3-dimethylbut-2-ene)
linear/saturated (e.g. n-pentane)
linear/unsaturated (e.g. hex-3-ene)
cyclic/saturated (e.g. cyclohexane)
cyclic/unsaturated (e.g. cyclohexene)

I chose these examples for a reason. The central carbon atoms in both neopentane and 2,3-dimethylbut-2-ene are only bonded to other carbon atoms while all other carbons in the branched examples are methyl groups ($\ce{CH3}$). Yet, no organic chemist will classify neopentane as ‘unsaturated’ — so clearly there must be an additional criterion other than ‘maximise the number of bonds to hydrogen atoms’.

On the other hand, hex-3-ene has almost maxed out its number of hydrogen atoms: two carbons with three, two carbons with two and only two carbons which are lacking a hydrogen each. But that does not make it any less unsaturated. Note that cyclohexane and hex-3-ene have the same formula $\ce{C6H12}$ and thus the same carbon/hydrogen ratio. Yet, the cyclic compound is considered saturated and the linear isn’t.

Thus, forget about the maximum hydrogen definition. A saturated molecule is one that is made up of single bonds only.

Note: Of course, cyclic compounds can also be branched or ‘substituted’. But that doesn’t add too much.


How exactly are ions in a Mass Spectrometer accelerated? After a molecule is ionised, the ion is accelerated by an electric plate up to a certain kinetic energy by the attraction of the ion to the electric plate.
However, after the ion has passed the electric plate, won't the plate still exert an attractive force on the ion?  If so, by the attraction between the ion and the electric plate, the kinetic energy of the ion were attenuated, perhaps even to the point where there would not be a net gain in kinetic energy of the ion.

The acceleration of the freshly generated ions is stepwise (similar to a linear particle accelerator), i.e. the beam of ions is guided an array of electrodes with a small opening in their centre, as schematically depicted below:

(image source)

Hence when ions approach the first accelerator electrode from the left hand side, their interaction with the subsequent accelerating electrode already phases in.  Ions passing the slit of the first accelerating electrode will experience an even stronger accelerating force by the subsequent accelerating electrode.  Eventually, their momentum will increase further; and on the same time, the ions preferentially will keep their direction of propagation towards the mass dispersive element (here: the magnet), regardless the attractive force between them and the last accelerating electrode.


Bomb calorimeter and heat capacity of calorimeter Can anyone describe the process through which we determine the heat capacity of calorimeter, for me?

The formula is
$q_{calorimeter} = C_{calorimeter} \times \Delta T$

The heat of bomb calorimeter is shown to be  the same as  negative heat of reaction. But without mass, their unit would be different, no?

The heat capacity of a calorimeter, $C_{calorimeter}$ is given in units of kJ/°C, with no units of mass included.  This is because the calorimeter's mass is a constant and each calorimeter's heat capacity (and mass) is different.

The general procedure that is used to determine the heat capacity of a calorimeter is as follows:

Place a carefully measured quantity of a compound whose combustion energy is well known and that is available in a pure form in the bomb, usually benzoic acid.
Seal the bomb and add oxygen so that the sample will burn completely.
Submerge the bomb and ignite the sample.
Measure the temperature change of the water bath in which the bomb is submerged.

Since you know the energy of the reaction and you know the heat energy in the water, you can find the heat energy of the calorimeter.

$\Delta E = q_{water} + q_{calorimeter}$

The heat of the calorimeter is the difference between the energy we know we put into the system through the combustion of our standard and the heat that was measured in the water bath.


Shape of urea molecule around N atom I was doing a question where I was supposed to predict the shape of the urea molecule $[\ce{H2N(CO)NH2}]$ using Lewis and VSEPR. I identified that the N atoms will have a lone pair on them and so I assigned a trigonal pyramidal shape around N, having the ammonia structure in mind. However when I looked up some 3D structures of the urea molecule it had a trigonal planar shape around N making the whole molecule lie in the same plane. I did some further research and read on wikipedia that

"The urea molecule is planar in the crystal structure, but the geometry around the nitrogens is pyramidal in the gas-phase minimum-energy structure."

What causes this change in geometry when the molecule crystallizes? Does the same thing go for other molecules? Would there have been a way of predicting this as I was solving the question?

I was supposed to predict the shape of the urea molecule $[\ce{H2N(CO)NH2}]$ using Lewis and VSEPR.

Don't forget that there are assumptions in VSPER to simplify the model. From Wikipedia:

  Valence shell electron pair repulsion (VSEPR) theory is a model used in chemistry to predict the geometry of individual molecules...

Essentially VSPER is for molecules in the gas phase where the only electrostatic interactions are within the molecule. Once the molecule is in a condensed phase, the assumption that it does not interact with external forces breaks down.

  What causes this change in geometry when the molecule crystallizes?

For one let's not forget that urea is a resonant structure, and does not have fixed bonds which also goes against assumptions of VSPER.

Further you have neighboring molecules with the hydrogens of those molecule taking some electron density from the resonance structure which is shown in the unit cell representation shown below:

In summary you have resonance which will flatten out the molecule and other molecules taking some electron density that combined will contribute further causing the molecule to flatten out.

  Does the same thing go for other molecules?

Yes, other molecules can condense and form structures that deviate from VSPER. Water in ice for example has a $109.5 ^\circ$ bond angle (tetrahedral angle) where as VSPER would predict $104.5^\circ$ (bent angle).

  Would there have been a way of predicting this as I was solving the question?

No, you would have to know how urea crystallizes and many other factors, then used a computer simulation to calculate (estimate) such.


Are graphite and hexagonal boron nitride aromatic Are graphite and hexagonal boron nitride aromatic?

Graphite has a planar network of 6-membered rings with each carbon connected to three other carbons. Since the valency of carbon is not satisfied, there will be some $\pi$-bonds. Can graphite act like an aromatic system? Can I explain this using Hückel's rules?

Hexagonal boron nitride has a similar structure. So if graphite exhibits aromatic character, then so will boron nitride (like inorganic benzene and benzene).

Graphite is definitely aromatic and boron nitride is at least partially aromatic. For instance, in this paper, the authors calculate the percent resonance energy (%RE) of graphite as a comparison to other known aromatic systems. The %RE is a measure of the total resonance stabilization per carbon atom relative to some reference system.

They find the %RE of graphite to be 3.061 and that of benzene to be 3.528. This value for graphite is larger than that calculated for naphthalene (2.924), thus for all practical purposes, graphite is essentially as aromatic as benzene.

In the case of hexagonal boron nitride(h-BN), I haven't been able to find any reference which says explicitly that the system is aromatic, but there certainly is some delocalization of $\pi$ electrons as I've seen a few papers which note that aromatic systems adsorb to BN nanotubes (which ought to be electronically similar to h-BN) quite strongly and a large portion of this binding interaction comes from dispersion (i.e. $\pi$ interactions). See here for instance.

You mention "inorganic benzene" which is the molecule borazine. Borazine actually only exhibits about half the total resonance stabilization per atom as benzene. This is attributed to the polarity of $\ce{B-N}$ bonds which inhibits delocalization of electrons. See this paper on the aromaticity of borazine for instance.

However, the first reference I gave found that graphite is nearly as aromatic as benzene, so I would be willing to speculate that h-BN is about as aromatic as borazine.

  Can I explain this using Huckel's rules?

I would be hesitant to use Huckel's rules on a $\ce{B-N}$ system as Huckel's rules come from Huckel theory which is really only meant to describe fully conjugated hydrocarbons. Nonetheless, people will point out when other systems follow these rules just in case it turns out to be significant.

In contrast to what I just said, however, there are boron clusters which have been found to follow Huckel's rule, but this doesn't mean you can extend Huckel's rules to heteroatomic systems such as h-BN at will.


How can I measure amount of ozone byproduct coming out of hydroxyl generator A Hydroxyl generator releases hydroxyl radicals into the air to neutralize odors and kill mold.

Most hydroxyl generators also release some amount of ozone as a byproduct. How much ozone is released varies from one hydroxyl generator to the next.

Hydroxyl is supposed to be safe to use in occupied premises. Large amounts of ozone are not.

How can I measure the amount of ozone released by my hydroxyl generator?

Ozone is very reactive, so unless you have an industrial scale hydroxyl generator, the ozone will readily react with organic matter in the air and surrounding surfaces before it builds up to dangerous levels. There are commercial portable instruments that are sold to monitor ozone concentration, they will be in the \$500 to $1500 price range for the instrument. You can also look at this simple experimental apparatus described in the Journal of Chemical Education for a cheaper alternative.


Boron trihalides acidity A boron trihalide ($\ce {BX3}$) is a Lewis acid, because it can accept an electron pair, as per Gilbert N. Lewis' definition.

Now my question is-

  Arrange $\ce {BF3}$, $\ce {BCl3}$, $\ce {BBr3}$ and $\ce {BI3}$ from the weakest to the strongest acid properties.

The following order is experimentally found:

$$\ce{BF3}<\ce{BCl3}<\ce{BBr3}<\ce{BI3}$$

This stands in contrast to what is expected when the electronegativity of the halides is considered. A more electronegative halide should be able to stabilize the negative charge at $\ce{B}$ in the Lewis acid-base complex better, and this would suggest that $\ce{BF3}$ would be the strongest Lewis acid.

However, the nature of the $\ce{B-X}$ bond needs to be taken into account. The change of hybridization from $\ce{sp^2}$ in $\ce{BX3}$ to $\ce{sp^3}$ in $\ce{[BX4]-}$ involves the loss of partial $\pi$ bonding which results from the overlap of the $\ce{p_z}$ orbitals of $\ce{B}$ and $\ce{X}$. The $\ce{p_z}$ orbitals of $\ce{B}$ and $\ce{F}$ are of similar size, because both elements are members of the same period. Therefore, they have the best overlap and form the strongest $\pi$ bond. Because the reaction with a Lewis base would require some energy to break this bond, $\ce{BF3}$ is a comparably weak Lewis acid. The larger $\ce{p_z}$ orbitals of the heavier halides have increasingly less overlap with the smaller boron $\ce{p_z}$ orbital. Therefore, the $\pi$ bonding is weaker and formation of the tetrahedral Lewis acid-base adduct via rehybridization becomes energetically more favorable.


Why is excluded volume in van der Waals equation divided by 2? Source: Physics Stack Exchange

The excluded volume calculated based on the diagram is 8 times the volume of a molecule. The volume is then divided by 2  to give the so-called excluded volume per molecule. Why divided by 2 only?

Obviously more than two molecules can come close together, and their assigned excluded volumes can overlap with each other, giving a smaller $b$. Is $b$ just an upper limit of excluded volume?

Collision is predominantly a bimolecular act. The probability of three gas molecules simultaneously colliding is incredibly low, therefore it quite accurate to say that the excluded molar volume $b$ is overcounted by a factor of 2.


Is it dangerous to inhale the steam coming from electrolysis of salt-water solution Is chlorine in there that accumulates (at the anode) and emerge out as chlorine gas? I was worried since I added a pinch of salt to the steamer in order to hasten the heating of tap water when I caught a cold (I should've just boiled a pot of water on the stove). I'm worried since chlorine gas is bad for the respiratory system. Also I'm curious to know what is this brown precipitate under the solution - hydroxides or cupric-oxides from the electrode?

So if I understand the question correctly, you heated up water with some salt by performing electrolysis.

If so, you would also generate chlorine gas. However, the heat capacity of water is really large, so I doubt that you would even be capable of bringing water to a boil by just jolting current through it.

To answer your question: Yes, when performing electrolyzation of any solution you should apply the usual precautionary measures, such as working in an open space or under a fume hood. Under no circumstances should it be possible for you to inhale the gases produced.


Reaction with carbon Usually maximum time we see then after cooking by gas there is no black spot on the pot. But continually burning can affect the pot surface and makes black powder.

My question is:

  What is the most acceptable reason that black powder(carbon) take places on the pot surface. Is it related to the burning temperature? or fuel reacts with the pot surface?

For combustion to occur you need 3 things at the same place: fuel, oxygen and heat.

If the air intake is dirty or not big enough, you get a flame that has an orange color, and on a cooler surface deposits C before it has a chance to burn.

The reaction with the pot is only a physical one of heat transfer and maybe C deposition.


Is there Any Advantage to Using Multiple Steel Plates in an Electrolyzer? If I have an electrolyzer (For water) is there any advantage to having additional steel plates in between the electrodes in solution? For some reason this appears to be a popular design and I'm not sure why. I'd think that at worst the voltage would have to be slightly higher to conduct through the extra solution in a 2 electrode system.

As the reactions in electrolysis take place on the electrodes surface, it can therefore be concluded, that the more electrode surface (contact surface) exists, the more water molecules can be electrolyzed in one instance of time. Therefore, more steelplates result in more surface, which increases the productivity of the electrolyzer.


The desirable property of plastics My teacher once asked me that "one desirable property of plastics is that they can be made into very complicated shapes, Explain how this is done as a reaction"
I never know how to answer this question, can anyone help

The property of "moldability" can be based on several properties of plastics
There are broadly two types of polymer commonly used. Polymers that set and thermoplastics that can be molded when hot. Both of these types can have the desirable property of being liquids that can fill a complex mold before becoming rigid (though by far the commonest is the thermoplastic that merely requires heat to make it flow).
Thermoplastics can be supplied to manufacturers as beads of the already polymerised compound that can be melted in the feed to the moulding machine. Examples include polystyrene, polypropylene, polyethylene and many others. These form flowable liquids when hot enough so that a mold can be filled easily during manufacturing and the polymer will set into a rigid shape when cooled. This makes for very easy manufacturing.
This is how a very wide variety of complex shaped can be made. PET bottles, Lego bricks, toy soldiers, plastic shells for computers and printers, car bumpers and many, many other complex components are made this way.
But other complex components can be made from plastics that set from chemical reactions and cannot be thermally molded. These work as long as the setting reaction is slow enough and the components are liquid. There are a variety of such materials which usually involve mixing a (fairly) liquid monomer or oligomer (examples include methyl methacrylate, some silicones, epoxies and some polyesters. Usually the process involves mixing a controlled amount of the monomer (eg methyl methacrylate which was often the basis for home modelling kits) with another ingredient that triggers polymerisation (often a radical initiator). With the right mix of the two ingredients, the polymerisation speed can be controlled so the speed of polymerisation is slow enough that the liquid can fill a complex mold before starting to set. This process is more often used in small batches as it isn't always well adapted to speed because the curing process can take minutes to hours.
The thing that matters in both thermoplastics and thermosets is that the components are liquids that can fill complex molds before setting rigid. In some cases the setting mechanism is based on cooling; in others it is based on chemical reactions.


What is the purpose of froth in washing clothes? I stay at a rented place and there is no washing machine. I, now, wash clothes in a bucket. When I add detergent, sometimes there is froth, sometimes there is not. I wonder if this had any impact on the washing process. Though I am aware of micelles and all that, but I wonder if froth formation is nearly redundant.

The froth has little or no effect on the detergent action. In fact detergent manufacturers have to add anti-foaming agents to stop excessive foam generation in automatic washing machines.

Froth/foam is generated because surfactants in the detergent adsorb at the air water interface and stop the water film that makes up the walls of the bubbles from collapsing. Note that the surfactant does not cause bubbles to form - bubble formation is always energetically disfavoured and the foam is only kinetically stabilised. The foam is caused by agitation entraining air into the surfactant solution so the amount of foam will depend on the intensity of the agitation as well as the type and concentration of the surfactant.

It is conceivable that in some cases the foam may assist with dispersal of mineral soils. This is the basis of the industrial froth flotation process. However this plays at most a very minor role in domestic washing machines. Not that it's directly relevant, but I mention it out of interest, foam is thought to play a role in cleaning teeth i.e. in toothpaste. It entrains particles of food so they can be spat out along with foam.


Changing the spectroscopy type to infrared in GC-MS I know you can use GC-MS for the analysis of gaseous molecules, but would you be able to instead use an infrared spectroscopy method to analyse gaseous molecules, or does it need to be mass spectroscopy?

GC-MS is just one common method of what is known as 'hyphenated' analysis. GC-IR instruments do exist (see examples), although usually involve some pretty specialised cells to allow for gas flow through the sensor. Hyphenated instruments are limited pretty well by your imagination (and budget) only. LC-NMR-MS is a great example of extended hyphenation, and is a powerful emerging tool for automated analysis of complex sample mixtures.


What's the source of "2.303" in Van't Hoff equation? I've seen a "2.303" term multiplied to $R$ in Van't Hoff equation several times.

Here is one example:

http://en.wikipedia.org/wiki/Van_%27t_Hoff_equation#Error_propagation

  $$\Delta H^\circ=2.303R\frac{\log K_1-\log K_2}{\frac1{T_2}-\frac1{T_1}}$$

What is this number and where does it come from?

It is the multiplier used when converting the equation to use a base 10 log instead of a natural log.

It comes out to be the natural log of 10.


Calculating heat of combustion in a bomb calorimeter In a constant volume bomb calorimeter, we measure the change in temperature of the calorimeter and find out it's change in  internal energy. We then convert it to change in enthalpy.

To determine $\Delta H_c$ we write the balanced chemical equation to find $\Delta n_g$ and using the value of $\Delta U$ obtained from calorimeter, we apply $$\Delta H = \Delta U + \Delta n_g RT$$

The above equation is only valid for constant pressure and temperature. How can we apply this equation on calorimeter, where the temperature change is observed?

In a constant volume bomb-calorimeter, we can make chemical reactions take place. If we let chemical reactions freely take place in it, then the total internal energy change is zero. This is due to the fact there is no expansion work since the system is a constant volume by definition and the heat transferred is zero since the bomb calorimeter is well insulated. The heat released in a reaction is automatically absorbed by the bomb calorimeter device. Hence the total internal energy change is zero.
Now, consider another path of the reaction. We start with reactants and turn them into products under constant volume and constant temperature conditions (*) and then these products we raise the temperature as to reach the final temperature we got in the previous path.
The energy associated with moving from reactants to products isothermally is $\Delta U_{rcxtn}$ and the energy change for raising up the products to the temperature of first path is $ \Delta U_{heat}$
Now, we can write:
$$ \Delta U_{heat} = C_{v}^{cal} \Delta T$$
And,
$$ \Delta U_{rcxtn} + \Delta  U_{heat} = 0$$
Or,
$$ \Delta U_{rcxtn} = - C_{v}^{cal} \Delta T$$
Now, consider the enthalpy change of reaction:
$$ \Delta H_{rcxtn} = \Delta U_{rcxtn} + \Delta (PV)$$
Now, if we were to assume our reactants as ideal gases a, then $PV=nRT$ and also considering constant temperature, we get:
$$ \Delta (PV) = \Delta (nRT) = RT \Delta n$$
Using the above result and expression for internal energy, we finally arrive at the following equation:
$$ \Delta H_{rcxtn} =  - C_{v}^{cal} \Delta T + RT \Delta n$$

Refernces:
Around 42:49 of this video
Page-2 of these notes


Bonding and Antibonding So I understand, electrons exist in orbitals, mainly s, p, d, f and that when they bond with one another it will form sigma and pi bonds and that whether it bonds in a certain orientation a molecular bonding orbital or an anti bonding orbital will form.

What I DON'T understand is that when an element bonds it's energy levels will ALWAYS form a series of energy levels that contain bonding and anti bonding orbitals so when they come together, the electrons are always orientated such that these orbitals will form? I also don't really understand what happens when these two atoms meet (the elements that are bonding) how are the orbitals interacting with one another? And what are the orbitals involved?

For those who would like to answer the question feel free to loosely address these questions and refer to fundamental concepts or explain bonding through orbitals briefly.

Orbitals are defined by a set of wavefunctions. Hence, when two orbitals overlap they can either interfere constructively (bonding) or destructively(anti bonding).

Lets take two pz orbitals orbitals overlapping with each other. You must understand that the p orbitals look like a dumbbell. One side of the dumbbell is + and the other side is -.

If - meets - we have constructive overlap. If - meets + we have destructive overlap.

The former gives a bonding orbital ($\sigma$) and the latter an antibonding orbital ($\sigma*$).

Depending on the molecule it is often energetically favourable to be in a bonding state hence bonding orbitals are lower in energy to their anti-bonding counterparts.


Why is EDTA most soluble at a pH 8? (addition of NaOH) Although Wikipedia says that EDTA is a very soluble salt, preparing a solution (.5M) is a pain in the neck. It takes a lot of time and still doesn't dissolve. Although addition of a few drops of NaOH makes the process a bit easier (still takes a lot of time)

Now, it did not make sense to me of why NaOH made EDTA more soluble. The salt I'm using is a disodium salt of EDTA, so wouldn't Na+ turn out to be the common ion only to decrease the solubility? Then, it might be the effect of pH(?) But, being a salt of a weak acid and a strong base, theoretically shouldn't such a salt be more soluble in lower pH? The EDTA divalent ion thus formed would be a stron conjugate base and accept H+ increasing the solubility?

I know both my assumptions are wrong since adding NaOH actually helped. What might be the actual reason which is counterbalancing the above mentioned 2 effects (are they even valid in this case?)

A disodium salt of EDTA also has two remaining carboxylic-acid groups, which can be deprotonated by addition of base, like $\ce{NaOH}$. By the way, higher pH is generally better especially if you plan to use it for chelation, i.e. cation-capture. The labs I've taught in the past use buffers of pH ~10 with EDTA for complexometric titration experiments.

Deprotonating an organic acid in general improves solubility in water because the electrostatic interactions of $\ce{-O^-}$ and the solvent (the hydrogens on water) are stronger than interactions of $\ce{-OH^0}$ and water.

A similar example would be 4-t-butylphenol, which is nearly impossible to dissolve in water without addition of a base, e.g. $\ce{NaOH}$. In experiments I had to reach pH of about 12 to strip the compound of the alcohol-proton, yielding a negative charge which interacts more strongly with water.

Nonetheless you can still make a decently concentrated (~0.1M) solution of EDTA in water by stirring alone, if you allow a stir-plate to do the work for ~30 minutes.

In short, it's not the addition of $\ce{Na+}$ that matters, but rather the addition of $\ce{OH-}$, to increase solubility.


A reversible exothermic reaction between a molecule in an excited triplet state and a reducing agent Say I promote a molecule to a triplet state via repeated electronic excitation events.  Here, this is going to mean that an electron from the HOMO level, originally with a spin-UP and spin-DOWN electron, will be promoted to the LUMO level such that its spin is the same as the electron left-over in the HOMO level (we can then use the Pauli exclusion principle to partially understand that the triplet state will generally have a longer lifetime than the typically higher energy singlet state).  The classic example of a molecule in a triplet state is oxygen (though here, this is also the ground state, which is somewhat unusual):

Prior to returning to the ground state via phosphorescence or a reverse intersystem crossing event, let's say that the triplet $T_1$ state molecule reacts with a reducing agent via an exothermic bond formation process, and that the reaction is reversible.  Consider that there are numerous instances where triplet state reactivity is higher than ground state reactivity.

The following two things are unclear to me:

(1) When the reducing agent dissociates, is the molecule in the $T_1$ triplet state or the $G_0$ ground state?  If the molecule is still in the triplet state, can it enter the ground state while remaining bound to the reducing agent?

(2) Call the energy gap between the relevant $T_1$ triplet state and the $G_0$ ground state $E_1$.  Call the energy released into a surrounding thermal bath by the reducing agent reacting with the molecule in the $T_1$ triplet state $E_2$.  Must $||E_2|| > ||E_1||$?

My guess is that the pair of electrons donated by the reducing agent will necessarily set  $||E_2|| > ||E_1||$ and that the molecule will be left in the ground state upon dissociation of the reducing agent since the electrons will be reallocated to properly satisfy spin-pairing requirements.

We're looking at a case of PET (photoinduced electron transfer) with electron back transfer between an (electron) acceptor A and a donor D here.

For organic dyes, the ground state typically is a singlet state ($S_0)$, excitation of A yields an excited singlet state (typically $S_1$).

$\ce{A ->[h\nu]\  ^1A^{\ast}}$

Quinone based dyes (or aza-BODIPY derivatives) often have a high triplet quantum yield, the following ISC (intersystem crossing) to a triplet state is efficient then.

$\ce{^1A^{\ast} ->[ISC]\  ^3A^{\ast}}$

In the excited (triplet) state, electron transfer from a (ground state) donor may occur by filling the lower SOMO of the excited species:

$\ce{^3A^{\ast} + D -> A^{\bullet-} + D^{\bullet+}}$

Is the acceptor radical anion still in a triplet state? No, since there no longer an electron pair with opposite spins.

Q: So, what is this good for?

A: By electronic excitation, we managed to facilitate a redox reaction that would not have been possible in the ground state and generated a radical ion pair.

Radical ions are reactive intermediates.

For a spectroscopist, it can be favourable if the only reaction of the radical ion pair is  electron back transfer:

$\ce{A^{\bullet-} + D^{\bullet+} -> A + D}$

This regenerates acceptor and donor in their former spin states, which - as mentioned before - are typically singlet states in the case of organic molecules.

Practically, this means that no permanent bleaching occurs. You can excite the sample with laser pulses again and again to measure the transients in time-resolved uv-vis spectrosopy.
+
From a preparative perspective, this is the worst case: there is no product being formed.

We have examined more productive cases of PET in the context of PET-initiated radical cyclizations of isoprenoid polyalkenes a while ago and published about these, e.g. in J. Phys.Chem A, in J.Chem. Soc., Faraday Trans. and in Pure Appl. Chem.. (The last two should be freely accessible).

In the course of these reactions, an alkene is oxidized to the corresponding radical cation by the excited state of an electron acceptor. The radical cation
traps a nucleophile (water, alkanol), immediately deprotonates and thus form a radical, which undergoes 5-exo-trig cyclization to a stabilized dicyanomethyl radical. The latter is reduced to its corresponding anion by $A^{\bullet-}$ and eventually protonated to yield the product (as a pair of separatable diastereomers).


What does the adenosine do in ATP? This is an extremely basic question about biochemistry from someone who doesn't know much about organic chemistry. I'm aware of how organisms store energy ATP and release it by dephosphorylating it into ADP. My question is about the role that the adenosine part of the ATP molecule plays in this process.

Naively, it seems as if the triphosphate group could be attached to any old thing, and you'd get the same energy by dephosphorylating it. But as far as I'm aware, it's only ATP that's used for this purpose in biology, and not any other triphosphate. Why is this? Is the fact that the triphosphate is attached to an adenosine just a "frozen accident" like the genetic code, or does it play an important role in the phosphorylation/dephosphorylation reactions themselves?

The adenosine part is needed to make DNA and RNA.  It is logical for the triphospate to be attached to ADP, GDP, UDP or CDP since they are needed to make DNA and RNA.

In some instances, GTP is used instead of ATP.  In fact the Krebs cycle does involve phosphorylation of GDP.   GTP is used in the formation of microtubes by tubulin.

I've seen some give the rate of hydrolysis being greater for ATP than GTP as a reason for ATP being more utilized.

There is a related question here: https://biology.stackexchange.com/questions/11286/why-is-atp-the-preferred-choice-for-energy-carriers


How is more than one pinacolone is possible? This question came in one of my school tests. They had given us an unsymmetrical pinacol i.e 1,1-diphenyl-2-methylpropane-1,2-diol and asked us whether it will give only one pinacolone or not.

According to me, I had assumed that carbocation will be formed on the diphenyl-substituted carbon due to which a methyl shift will occur and we will get only one product. But my teacher marked it wrong and said that more than one pinacolone is possible here.

How is it possible?

The problem is what is a reaction which gives "only one product", the problem is that very few reactions give a 100 % yield of a substance. For almost all real life reactions there are a small fraction of molecules which react he wrong way.

I think that when more than 99 % of molecules react the "right" way that those which react the "wrong" way can be ignored normally. Unless the wrong product has some very harmful effect on the product when it is used for a purpose.

What I think will happen is that the carbocation formed by protonation of the hydroxyl bonded to the carbon bearing the two phenyl groups, followed by the loss of a water molecule will have a much lower energy than the other carbocation. Thus it will be more easy to form.

But there is a problem, for a rearrangement of a carbocation to occur normally we need to rearrange to a lower energy carbocation. By lowering the energy of the carbocation we can lose the thermodynamic drive for rearrangement. In the ideal world we would consider the rate of rearrangement of the two different carbocations and their formation. I think that the reactions forming the carbocation will be reversable while the pinacol rearrangement will be irreversable.

But in some cases such as many homogenious asymmetric rhodium hydrogenations of prochiral alkenes the lower energy alkene rhodium complex which is present at a far higher concentration than the higher energy isomer, the lower energy isomer has far too little reactivity. As a result while the majority of the complex might be in the R form the tiny trace of the complex in the S form is responsible for the vast majority of the hydrogenation reactivity.


Obtaining magnesium oxide from magnesium hydroxide What's the simplest and safest DIY way to obtain magnesium hydroxide to magnesium oxide?

Magnesium hydroxide decomposes at 332 °C (630 °F) :

$$\ce{Mg(OH)2_{(s)} -> MgO_{(s)} + H2O_{(g) }}$$

So heat the magnesium hydroxide in a covered crucible. The released water will cause the particles of magnesium hydroxide to "explode" so you want to do this covered.


Constructing an equation for the precipitation of a salt I was asked to write a balanced equation, as well as the net ionic equation, for the formation of $\ce{CaCO3}$ by precipitation. The equations should also include state symbols.

Unfortunately, I am not familiar with the concept of precipitation. Could anybody please show me a method for constructing these equations?

Precipitation is usually a single or double displacement reaction, and refers to the formation of an insoluble, solid salt from soluble ions. For instance, silver bromide, $\ce{AgBr}$, is insoluble in water. However, it is made up of $\ce{Ag+}$ and $\ce{Br-}$ ions, which we can introduce in the form of soluble salts:

$$\ce{LiBr(aq) + AgNO3(aq) -> LiNO3(aq) + AgBr(s)}$$

In an aqueous solution, lithium bromide and silver nitrate underwent a reaction, forming lithium nitrate and silver bromide. In the products, silver bromide is denoted with the state symbol (s), which stands for "solid". This reflects its insolubility in water and hence its tendency to "precipitate".

The net ionic equation is when we only write the ions that participate in the reaction. The ions that do not participate are called spectator ions. For example, in the reaction above, only $\ce{Ag+}$ and $\ce{Br-}$ participated in precipitation. Thus, the net ionic equation for the reaction above is:

$$\ce{Ag+(aq) + Br-(aq) -> AgBr(s)}$$

One way of deriving this net ionic equation is to realise that salts such as $\ce{LiBr}$ and $\ce{AgNO3}$ do not exist as molecules in water, but rather dissociate to give their constituent ions:

$$\begin{align}
\ce{LiBr(s) &->[dissolution] Li+(aq) + Br-(aq)} \\
\ce{AgNO3(s) &->[dissolution] Ag+(aq) + NO3-(aq)}
\end{align}$$

However, $\ce{AgBr}$ is insoluble and does not dissociate into ions. If we go back to the first equation and replace all the soluble salts with their constituent ions, we get

$$\ce{\underbrace{Li+(aq) + Br-(aq)}_{from LiBr} + \underbrace{Ag+(aq) + NO3-(aq)}_{from AgNO3} -> \underbrace{Li+(aq) + NO3-(aq)}_{from LiNO3} + \underbrace{AgBr(s)}_{does not dissociate}}$$

The spectator ions which do not take part in the reaction are $\ce{Li+}$ and $\ce{NO3-}$, as they are unchanged in this reaction. Subtracting them from both the left- and right-hand sides, we obtain the net ionic equation:

$$\ce{Ag+(aq) + Br-(aq) -> AgBr(s)}$$

If we take this idea and apply it back to $\ce{CaCO3}$, we first need to come up with soluble sources of the ions $\ce{Ca^2+}$ and $\ce{CO3^2-}$. A possible choice is $\ce{CaCl2}$ and $\ce{Na2CO3}$, although that is hardly the only option.

The overall equation may be written as

$$\ce{CaCl2(aq) + Na2CO3(aq) -> 2NaCl(aq) + CaCO3(s)}$$

and the net ionic equation as

$$\ce{Ca^2+(aq) + CO3^2-(aq) -> CaCO3 (s)}$$

where $\ce{Na+}$ and $\ce{Cl-}$ are the spectator ions.

Finally, here are some web pages you might find helpful:

http://www.occc.edu/KMBailey/Chem1115Tutorials/Net_Ionic_Eqns.htm

http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Reactions_in_Aqueous_Solutions/Precipitation_Reactions


Rate of reaction with HBr Rate of reaction is more when Carbon is more electrophile(less electron density)

So according to me the order should be D>B>C>A

As NO2 has -M -I effect , Cl has +M -I effect , while OCH3 has +M -I effect.

But the answer given as A>B>C>D

Stability of carbocation is directly proportional to +M effect .
So ,
 carbocation will more stable on A then on B and then C as NO2 has high -M. Effect it's last..


accurate measurement from inaccurate tools It is a very general question. Is there a good way to systematically increase the accuracy and precision of a measuring tool using only mathematical means?  For example, average 10 measurements can create a better one. Or using two independent tools.  I don't know if measuring theory or statistical quality control or other subject can help.

Generally, improvement of measurement process has much better positive impact on results, than sophisticated mathematical processing of low quality data. But, sometimes we are forced to data what we have.

There are 3 quantities:

resolution = the smallest difference of directly measured value, distinguishable by the tool,

precision = Level of uncertainty of repeated measurement of the same value, due random fluctuances. It is affected by the resolution as well, if comparable with precision.

accuracy = Level of agreement of the estimated and real value, due the tool/method bias.

There can be a very precise but inaccurate tool/method and vice versa,
even if one usually comes with the other.

Precision can be improved by measurement repetition. The arithmetic mean of $n^2$ measurements has $n$ times better precision than a single measurement.

The estimation of the standard deviation of normal Gauss distribution of measurements is

$$s=\sqrt{\frac{\Sigma ( x - \overset{-}x )^2}{n-1}}$$

The estimation of the standard deviation of normal Gauss distribution of real value estimation by arithmetic mean is:

$$s_{\overset{-}x}=\sqrt{\frac{\Sigma ( x - \overset{-}x )^2}{n(n-1)}}$$

For a method consisting of multiple measurement steps, each with  their own error, the error propagation rule has to be followed.

If 2 quantities, subjected to random errors, are added or subtracted, the squares of their standard deviation estimations are additive

$$(s_{A\pm B})^2=(s_{A})^2+(s_{B})^2$$

If quantities are multiplied or divided, their relative standard deviation estimations are propagated:

$$\left(\frac{s_{A\cdot B}}{A\cdot B}\right)^2 = \left(\frac{s_{A}}{A}\right)^2 + \left(\frac{s_{B}}{B}\right)^2$$
$$\left(\frac{s_{A/B}}{A/ B}\right)^2 = \left(\frac{s_{A}}{A}\right)^2 + \left(\frac{s_{B}}{B}\right)^2$$

If we e.g. determine the difference of the standard solution volumes $A - B$,
where $A=20.00 \pm 0.03$, $B=10.00 \pm 0.04$,

then $A-B=10.00 \pm 0.05$,

as $\sqrt{0.03^2 + 0.04^2}=0.05$.

Accuracy is harder to improve by mathematical means. It can be improved by measurement bias analysis, by calibration of tools or methods and by combination of more tools / methods / labs.

There is an additional decrease of the measurement method precision due statistical error of calibration via rules above.

$$(s_\mathrm{rel,method})^2 =  (s_\mathrm{rel, meas})^2+(s_\mathrm{rel, cal})^2$$

More tools/methods/labs lead to the  "mean of the means" and partially decreases the possible bias,
similarly as averaging of measurements improve precision.

But it is more costly.


Why lattice constant (unit cell) is always a parallelepiped? Wikipedia defines lattice constant as physical dimension of unit cells in a crystal lattice.

Unit cell is defined as:

  The unit cell is defined as the smallest repeating unit having the
  full symmetry of the crystal structure. The geometry of the unit
  cell is defined as a parallelepiped.

Why that elementary pattern that repeats itself in crystals is always a parallelepiped? In real life such elementary repeating unit/part/block is very often not a parallelepiped at all!

Is this some abstraction and a "bounding parallelepiped" is meant - the one into which definitely fits whatever real elementary repeating pattern is?

A unit cell is always a parallelepiped because this is the definition of a unit cell. In case when you don't have a repeating unit of this kind (which might indeed be the case in real life), you don't have a crystal, also by definition. You might be dealing with a quasicrystal or something.

3D periodicity is what defines a crystal. If you have it, this means you have three non-coplanar periodic vectors, and these inevitably define a parallelepiped.


Iterative method for calculating pH of a weak acid / base The problem asks to determine the $[\ce{H+}]$ in a $0.20~\mathrm{M}$ solution of $\ce{Na3PO4}$. The $K_\mathrm{a}$ of $\ce{HPO4-}$ was given as $4.5\times 10^{-13}$, which then allows one to calculate the corresponding $K_\mathrm{b}$ as $2.22\times10^{-2}$.

As any acid-base problem, I simply set up the expression $\frac{x^2}{0.20~\mathrm{M}}=2.22 \times 10^{-2}$ with $x$ being $[\ce{OH-}]$ and solved for $[\ce{H+}]$.

However, the solution involves an interesting iteration which I have never seen before for this type of problem:

Can anyone shed some light on this?

This method of repeating iterations is interesting, though I would suggest something perhaps a little more straightforward (and rigorous) than doing something until you are "acceptably close."

This method I learned is called a RICE diagram.

Reaction

$$\ce{PO4_{(aq)}^{3-} +H2O_{(l)}<=>HPO4_{(aq)}^{2-} +OH^{-}_{(aq)}}$$

Initial (concentrations)

$$[\ce{PO4^{3-}}]=0.20~~~~~[\ce{HPO4^{2-}}]=0~~~~~[\ce{OH-}]=0$$

Change (in concentrations)

$$[\ce{PO4^{3-}}]=-x~~~~~[\ce{HPO4^{2-}}]=+x~~~~~[\ce{OH-}]=+x$$

Equilibrium (concentrations)

$$[\ce{PO4^{3-}}]=0.20-x~~~~~[\ce{HPO4^{2-}}]=x~~~~~[\ce{OH-}]=x$$

We can write the reaction expression as:

$$K_\mathrm{b}=\frac{[\ce{HPO4^{2-}}]\cdot[\ce{OH-}]}{[\ce{PO4^{3-}}]}$$

We substitute in the equilibrium concentrations and the $K_\mathrm{b}$:

$$2.22\times10^{-2}=\frac{[x]\cdot[x]}{[0.20-x]}$$

And solve for $x$:

$$2.22\times10^{-2}=\frac{x^2}{0.20-x}$$

$$4.44\times10^{-3}-2.22\times10^{-2}\cdot x=x^2$$

$$0=x^2+2.22\times10^{-2}\cdot x-4.44\times10^{-3}$$

Use the quadratic formula:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

$$x=\frac{-(2.22\times10^{-2})\pm\sqrt{(2.22\times10^{-2})^2-4(1)(-4.44\times10^{-3})}}{2(1)}$$

$$x\approx-7.87\times10^{-2},~5.65\times10^{-2}$$

Obviously the negative solution is not applicable to our problem, so $x=5.65\times10^{-2}$.

Since $x=[\ce{OH-}]$, $[\ce{OH-}]=5.65\times10^{-2}$.

We know that:

$$[\ce{H+}]\cdot [\ce{OH-}]=1\times10^{-14}$$

so:

$$[\ce{H+}] \cdot [5.65\times10^{-2}]=1\times10^{-14}$$

$$[\ce{H+}] =\frac{1\times10^{-14}}{5.65\times10^{-2}}\approx1.77\times10^{-13}$$

If you have access to a calculator with a numerical solver function, I would recommend that instead of using the quadratic formula, but if not, this is how it is done.


Isotopes of Elements When we talk about the substance $\ce{H2O}$ with a molar mass of $\pu{18.0 g/mol}$, what hydrogen and oxygen isotopes are present in the $\ce{H2O}$ molecules? In other words, what isotopes do the symbols represent?

Most periodic tables quote the molar mass of an element as the average mass of the element given the natural abundance of the isotopes. Chlorine, for example, has a standard atomic weight of 35.45 which is a weighted average of ~75% 35Cl and ~25%  37Cl, the two common isotopes.

But isotopes of hydrogen and oxygen are rare. So the vast majority of water molecules will not contain them. So, at least to one decimal place, most water will have a molar mass of 18.0.

At it's natural abundance of about 0.25% 18O won't noticeably affect the molecular weight of water until you can measure it to at least 3 decimal places of precision. Slightly simplifying (assuming pure isotopic masses are exact integers, which they are not), water containing that isotope at its natural abundance will have an an average mass of 18.005 rather than 18.000.

When working with normal (unenriched) substances the best assumption is that they will contain the normal isotopic abundance of the component elements (which are well known and, for most elements, don't very much by source). But the effect on molecular mass will often be small and you will need a fair amount of precision to notice it. If you are not working to very precise masses, it is fair to assume natural isotopic abundance.


Modification to improve antibacterial activity I have difficulties with part b).

For C, it's clear to make a bulky dimethoxybenzene group in order to block the C=O group from reacting with b-lactamase.

However, in A, how does the structure of sulfonamide affect the bacterial activity in general?
Is there an alternative for C as well?

Essentially the basis of antibacterical activity of sulfonamides is the inhibition of  the enzyme dihydropteroate synthase, an
important enzyme needed for the biosynthesis of folic
acid derivatives and, ultimately, the thymidine required
for DNA.
Now they do this by competing at the active
site with p -aminobenzoic acid (PABA), a normal structural
component of folic acid derivatives.

PABA

So how does the structure of sulfonamide affect the bacterial
activity in general?

The basis of the structural resemblance of sulfonamides to PABA is clear. The
functional group that differs in the two molecules is the
carboxyl of PABA and the sulfonamide moiety of sulfanilamide.
Now there are requirements that are vital for activity:

Para-amino group:

unsubstituted (i.e $\ce{NH2}$).
very important for activity because any modification of it other than to
make prodrugs results in a loss of activity. For example, all
of the N4-acetylated metabolites of sulfonamide are inactive.

Sulfonamide:

$\ce{N}$ must be primary or secondary
$\ce{R2}$ can be substituted

Aromatic ring:

Para substituted

Modifications
The pKa of the carboxyl group of PABA is
approximately 4.9 thus an optimised sulfonamide has a pKa somewhere close to this value.

It was observed that replacement of one of the $\ce{NH2}$ hydrogens by
an electron-withdrawing heteroaromatic ring enhanced the acidity of
the remaining hydrogen and dramatically enhanced potency.
With suitable groups in place, the pKa is reduced to the same range as
that of PABA itself. Not only did this markedly increase the
antibacterial potency of the product, but it also dramatically
increased the water solubility under physiologic conditions.

Structural variation among the clinically useful sulfonamides
is restricted primarily to installation of various
heterocyclic aromatic substituents on the sulfonamide
nitrogen.
An example  is Sulfisoxazole (pKa 5.0) which can also be formulated as a prodrug, by inclusion of N-acetyl groups. Another advantage is the acetyl derivative
is tasteless and, therefore, suitable for oral administration, especially
in liquid preparations.

Is there an alternative for C as well?

Clavulanic acid is a naturally occurring, mechanism based
inhibitor , which causes potent and
progressive inactivation of β-lactamases.
There  are alternatives e.g sulbactam (A synthetic derivative of penicillins undergone the oxidation of the sulfur atom to a sulfone which greatly
enhances the potency) or tazobactam.
References

The 12th Edition of Wilson and Gisvold’s Textbook of Organic Medicinal and Pharmaceutical Chemistry
Foye’s Principles of Medicinal Chemistry 7th ed.


Nitrosation of benzene How does benzene react with $\ce{NaNO2/HCl}$? Will $\ce{PhNO}$ be obtained? I read that pyridine can be produced under special conditions, but what will happen in some less exotic conditions?

As you noted, the link you provided was to a reaction run under rather exotic conditions (low-temperature plasma). Under more everyday laboratory conditions, for example, in solution, activated aromatic rings can react with $\ce{NO^+}$ as shown below.

However if there is a ring-activating amino substituent present, then the $\ce{NO^+}$ will react with the nitrogen to produce the synthetically important diazonium ion.


Prevent molten glass from cracking when cooling down My mother works in a ceramics workshop. Besides giving lessons she creates various ceramic dishes, cups and plates. She often uses the effect of colored glass melting in the product and then leaving beautiful patterns at the bottom.

However it always cracks a lot as cooling down. Sometimes, it looks good when it's cracked, but other times she's trying to make it look like a gem - and cracked gem is not very gemmy.

So I was curious if there is something non toxic, insoluble in water and alcohol, that would melt at the similar temperature as glass and would prevent it from cracking in the cool down phase.

The glass doesn't have to be temperature resistant in the final products. We just want to avoid the cracks

The surface cracks due to a mis-match in the temperatures inside and outside the ceramic. Things contract as they cool. So, the cooler outer surface is contracting against the hotter inner surface, which is still not contracting. The outer surface wants to be closer together than the inner material, but the inner material forces the outer material to stay apart. When the ceramic on the outside transitions to a solid (at what is called its "glass temperature"), it can no longer flow like a liquid. The only way it can continue to stay apart due to the forces from the inside is to "fall apart" ... i.e. to crack.

Your mother may be able to coat the pottery with a different glaze that forms a glass at a lower temperature than the one she has now. Also, the suggestions to cool VERY SLOWLY are directly on target. The cooling should happen in a place that has no air flow, ideally in a closed box. Absent an annealing oven, a different option may be to put the pots in an conventional NON-CONVECTION oven that is held at a temperature just below the glass temperature of the glaze being used. When the entire pot reaches that temperature, it can be removed to cool further in a box at room temperature, again with no air flow.


Can every conductor be electroplated using copper and sulphuric acid? I really like electrolysis. Today I bought 30% sulphuric acid to electroplate something with pieces of copper.

I've poured half water and half acid in a glass. I connected some cuprextit-copper plate to the anode and outdated aluminium coin to the cathode.

After starting the process, I could see that nothing is appearing on the cathode. I scratched the aluminium of the oxides. After repeating the electrolysis, aluminium turned black wherever scratched. Then I found out I swapped the wires.

So, once I had the wires correct, this happened to my cathode:

The dust can be easily washed off.

According to the image from wikipedia, any metal should be "electroplateable". I actually thought that any conductor that doesn't react with the solution can be electroplated.

So what's the reality? Was I just impatient? Is electroplating only limited to some conductors? What about non metal ones? Can I electroplate carbon electrode?

While you can use basically any conductive material as an electrode to reduce the metal ions in solution, to actually plate the electrode with the metal, the metal has to be able to attach to the surface. If you have contaminants like surfactants, it can prevent deposition. The trouble with using aluminum as a cathode is that it readily forms an oxide layer that can be difficult to stick things to. I understand that it's possible to electroplate copper onto aluminum, but it's usually done with copper cyanide baths, so you may find it easier to plate copper onto steel or nickel. The other issue is that reduction of the copper ion is not the only possible reaction at the cathode. If your voltage is too negative, you may get significant hydrogen evolution and the bubbles will prevent a solid film from adhering properly. (which looks like it may be the problem in your photo) What you want is to deposit very slowly by keeping your voltage down so that you minimize gas evolution, though you may still have trouble with an aluminum cathode.

As to the sorts of things that can be electroplated, many metals have similar behaviour where it's possible to produce quite high quality films of a second metal over top. Other metals like mercury and bismuth tend to form alloys or intermetallic compounds. It is certainly possible to deposit thin metal films on materials like indium tin oxide, graphite or glassy carbon, thought adherence can be a problem.


Questions about Buffer System (Lab) I will be doing a lab on Hydrogen Phosphate buffer system tomorrow and I have a few questions about the buffer system. I am not sure if my assumptions are correct.

(1) What would happen to the rate of pH change if 5.0M HCl and NaOH were used instead of 0.50M?

My assumption is that, since the pH of the buffer does not depend on the actual concentration of the buffer, but on the ratio of the two parts, therefore, the rate of pH should not expect to change. Is this correct?

(2) Lets say I prepared a buffer at a different pH, how would the capacity of my buffer system change?

My assumption is that, the buffer capacity would not change upon the change of pH, is this correct?

Let's consider two buffer solutions of the same kind ($pK_a=4.75)$, but of different concentrations:

$\ce{CH3COOH}\,0.01\,\text{M}$ and $\ce{CH3COO-} 0.01\,\text{M}$
$\ce{CH3COOH}\,0.1\,\text{M}$ and $\ce{CH3COO-} 0.1\,\text{M}$

The starting buffer volume is the same for both and equal to $100\,\text{mL}$.

Now, add a strong base $\ce{NaOH}\,0.05\,\text{M}$, few $\text{mL}$ only: see below $pH$ as a function of $\ce{NaOH}$ added volume:

Have a look how "fast" the $pH$ changes and you will conclude that higher concentrations buffers are more effective. Differences can indeed be appreciated from the graph: they witness the evidence that larger concentrations better "buffer" the addition of the base.

I would suggest you to do some calculations (e.g. for $1\,\text{mL}$ of  $\ce{NaOH}\,0.05\,M$ added) and see whether you get the same values of the graph (hopefully I have done them right...). Remember that the addition of a strong base will provoke a decrease in concentration of the acid, $\ce{CH3COOH}$ and a consequent increase of its conjugated base, $\ce{CH3COO-}$.


Sample preparation for UV analysis of drug content in capsules/tablets I want to analyse the drug content of a sample of gelatin capsules filled with active drug (declared dose is known) and excipients such as lactose monohydrate, microcrystalline cellulose etc. with UV spectroscopy. However I have a question about the sample preparation of my capsules. So first what I want to do is dissolve the capsule content in an organic solvent (such as ethanol, chloroform etc.) because the solubility of the drug is bad in water, and then dilute the samples to a concentration which is suitable for UV analysis. The problem is that when I dissolve the content of the capsule, I'll likely get a suspension where the excipients won't dissolve in the organic solvent. This would interfere with my UV analysis. So I was wondering if it would be a good idea to centrifuge the samples, or just wait for it to precipitate on its own? And if centrifuge, would it be better to first centrifuge the stock solution and then dilute or first dilute and then centrifuge the acquired diluted solution? Thanks in advance.

I'd agree with Zhe that "this is a non trivial problem." Method development is tricky if you don't have a standard analysis procedure.

When you dissolve the capsule, if you do have suspended particles then yes they will interfere with the UV analysis. So do centrifuge.

But think of the removal of the liquid as an extraction. So add more solvent to the centrifuge tube, shake up the particles and centrifuge again. Assuming that you won't over dilute, I'd do the extraction three times, then dilute to the final volume.

The rub here in my mind is that UV is not very selective. I'd doubt that you're going to get a nice flat baseline with one nice peak. So you'll need a "pure" drug sample to compare to your extract samples to see if there is any other gunk causing interference.


Are metallic/ionic bonds weaker than covalent bonds? In mineralogy class, I was taught that metallic and ionic bonds are weaker than covalent bonds and that's why quartz and diamond have such a high hardness value. However, in organic chemistry class, I learned that covalent bonds are weaker than metallic and ionic bonds, thus organic substances have a much lower melting point than that of metals and ionic compounds.

What am I getting wrong? Are ionic and metallic bonds weaker than covalent bonds or not?

Quartz and diamond are stronger substances because their molecules form network covalent structures. These structures form a lattice-like structure, much the same as ionic compounds.

This molecular network is also the reason that diamond and quartz form a crystalline structures, just like you'd see in ionic substances such as NaCl. Some other structures you might want to look into are Graphite and Graphene, which are both allotropes of carbon (allotropes are, simply put, different molecular arrangements of an element).

The network structure combines to make the substance stronger than normal covalent bonded substances.

So to answer your question, substances with standard covalent bonds seem to be weaker than those with ionic bonds because the ionic bonds tend to form a lattice structure, that makes them much stronger. You can see this in the fact that the boiling points of ionic salts are much higher than that of a covalent substance like water. However, when covalent bonds form network covalent structures, atoms combine to form a singular macromolecule that is much stronger than singular covalent bonds.


Electrolysis involving K+ and Li+ ions Need clarification on Q31a) in the image.
Possibly a dumb question but the book's answer might be wrong. The book says lithium ions will be reduced instead of potassium ions. How is this true when K+ ions are stronger oxidants?

Their answer for part d even reiterates that's Li+ will react in preference to K+ so I think I may be wrong...not sure how though

Thanks

$\ce{Li+}$ 's low standard reduction potential value applies to aqueous solutions only.

$\ce{Li+}$ has very high polarising power due to its small size (and in turn a high hydration energy). This is why $\ce{Li+}$ has a very low reduction potential compared to other alkali metals in aqueous solutions.

When you deal with a solution of molten metals, the idea of hydration energy makes no sense anymore. Therefore, the electrochemical series order for aqueous solutions isn't applicable to the molten solution of alkali metals.

As you go down the alkali metal group, the ionisation energy decreases. The tendency of an atom to give its electron (or get oxidised) increases. Therefore, potassium oxidises more easily than lithium or in other words, potassium has a lesser tendency to get reduced compared to lithium.

As lithium has a higher tendency to get reduced than potassium, lithium gets deposited instead of potassium.


Reversibility of a reaction I was doing some questions on equilbrium. I got stuck in this particular question.

Identify which of the following reactions are reversible.
$$
\begin{align}
\ce{Fe + S &-> FeS }\label{rxn:1}\tag{1}\\
\ce{NaCl + AgNO3 &-> AgCl + NaNO3}\label{rxn:2}\tag{2}\\
\ce{H2SO4 + Ba(OH)2 &-> BaSO4 + 2 H2O}\label{rxn:3}\tag{3}\\
\ce{H2 +I2 &-> 2HI }\label{rxn:4}\tag{4}\
\end{align}
$$

I know that \eqref{rxn:2} and \eqref{rxn:3} cannot be reversible because one of the products is precipitated.
I could not reach a conclusion between \eqref{rxn:1} and \eqref{rxn:4} (although I know that \eqref{rxn:4} is reversible).
Is there a way to determine without actual experiments whether a reaction is reversible or not?

Strictly speaking, the answer depends on conditions and, as a consequence, states of aggregation.

Reaction (1) can be reversible if iron(II) sulfide is subsequently heated above $\pu{700 °C}$ in vacuum to prevent oxidation of the elements [1, p. 422]:

$$\ce{FeS(s) ->[\pu{700 °C}][vac] Fe(s) + S(g)}\tag{1a}$$

Reactions \eqref{rxn:2a} and \eqref{rxn:3a} are indeed irreversible if carried in aqueous solution — as you already mentioned, a poorly soluble salt is formed in both cases:

$$
\begin{align}
\ce{NaCl(aq) + AgNO3(aq) &-> AgCl(s) + NaNO3(aq)}\label{rxn:2a}\tag{2a}\\
\ce{H2SO4(aq) + Ba(OH)2(aq) &-> BaSO4(s) + 2H2O(l)}\label{rxn:3a}\tag{3a}
\end{align}
$$

Reaction \eqref{rxn:4a} is a canonical example of a reversible reaction used to study equilibrium in a gas phase:

$$\ce{H2(g) + I2(g) <=> 2 HI(g)}\label{rxn:4a}\tag{4a}$$

I'd say it's impossible to tell in advance what reaction is reversible or not without conducting an actual experiment or at least estimating free Gibbs energy.

References

R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)


What is a neutral atom? I was told that an atom's atomic number is defined as follows:

  The number of electrons or protons present in a neutral atom is called atomic number. It is represented by Z.

What does neutral mean here? Why isn't it just "..present in an atom..."?

Electrons and protons are charged particles. The electrons have negative charge, while protons have positive charge. A neutral atom is an atom where the charges of the electrons and the protons balance. Luckily, one electron has the same charge (with opposite sign) as a proton.

Example: Carbon has 6 protons. The neutral Carbon atom has 6 electrons. The atomic number is 6 since there are 6 protons.


Acidic Strength of Nitrobenzoic Acid The Correct acidic strength of the following :

But according to me pK value of 2-nitro benzoic acid should be more than 4-nitro benzoic acid because NO2 might have H-bonding  with COOH as NO2 have it in ortho-nitro phenol.

Correct me if i'm wrong and please do tell me how to identify the presence of H-Bonding.

Thanks !

When it comes to acidity, stability of the counter-ion is one of the main considerations. The nitro-group acts to stabilize the negative charge of the resulting nitrobenzoate by pulling the negative charge onto itself.

In the case of 2-nitrobenzoic acid, it is much nearer and therefore able to more effectively 'remove' the negative charge from the carboxylate group. In the case of 4-nitrobenzoic acid, it's still able to do so but less effectively as it is further away.

The hydrogen bonding effect that you describe does exist in 2-nitrobenzoic acid but this does not act to make it less acidic. While one might reason that the hydrogen bonding "holds the proton in place" and therefore makes it harder to leave, the presence of the hydrogen also means that both the nitro and the carboxylate group are highly unlikely to be simultaneously coplanar with the benzene ring (it's not obvious from the image because there's no bond between O and H). The energetics in this case are likely to favor the release of H+ in exchange for a coplanar configuration.


Amount of substance produced in chemical reaction If $15.0\ \mathrm{mol}$ of nitrogen are reacted with $30.0\ \mathrm{mol}$ of hydrogen, how much ammonia will be produced?

$$\ce{N2 + 3H2 -> 2NH3}$$

What I’ve tried is the following:

$$\dfrac{n(\ce{N2})}{n(\ce{H2})}=\dfrac{1}{3}=\dfrac{15.0\ \mathrm{mol}}{x}$$

$$\dfrac{n(\ce{N2})}{n(\ce{H2})}=\dfrac{1}{3}=\dfrac{x}{30\ \mathrm{mol}}$$

I’m not sure how to proceed from here. Any help would be appreciated.

In the reaction:

$$\ce{N_2 + 3H_2 \to 2NH_3}$$

If 15 moles of $\ce{N_2}$ is used along with 30 moles of $\ce{H_2}$, then $\ce{H_2}$ will be the limiting reagent (Reactant which is consumed completely). This is because 15 moles of $\ce{N_2}$ will need 45 moles of $\ce{H_2}$ ($15 \times 3$, by the balanced reaction) for reacting completely, but as there is only 30 moles available hence it will react in a fixed proportion, hence you will have to assume the reacting moles of $\ce{N_2}$ as $x$. By this the equation goes,

$$\frac{\ce{N_2}}{\ce{H_2}}=\frac{1}{3}=\frac{x}{30}$$
From this
$$x=\frac{30}{3}=10$$

Hence from 15 moles of $\ce{N_2}$ only 10 moles will react, hence the amount of $\ce{NH_3}$ formed will be 20 moles ($10 \times 2$, by the balanced reaction).


How to verify the accuracy of a pH meter One suggestion I received is that I prepare a solution of a known activity and probe the solution with the pH meter. The problem I see with this is that any error in solution preparation can lead us to conclude that the pH meter is faulty.

Is there any other way to test the accuracy of a pH meter? We need to make extremely accurate measurements (we are studying the activities of ionic solutes).

Also, what solution has an activity of exactly 1.0?

Your school should have a purchased buffer solutions that have been calibrated against a NIST standard.  If they don't, have them purchase several in order to span the entire pH range, here's one from Sigma-Aldrich.  For your calibration, pick a standard solution that is close to the range you plan to work in.  Here's a couple of links with some more "tips" on how best to actually perform the calibration.

tips #1
tips #2


How to determine which gas is more soluble I have a chemistry homework problem that asks whether $Br_2$ or $O_2$ is more soluble in water in room temperature. However, I can't seem to find a way to figure it out, short of looking it up.

Is there a guiding rule for such figures, such as relating to molar mass?

You could answer this question by looking at the possible interactions between a gas molecule and water molecules. The only I can see are van der Waals interactions (because these gas molecules cannot participate in hydrogen bonding).

$\ce{O_2}$ and $\ce{Br_2}$ have no permanent dipole so the only interactions are Debye forces (interaction between water permanent dipole and a gas molecule induced dipole) and London forces (between instantaneous dipoles).

The induced/instantaneous dipoles strength increases with the size of the atom because the bigger the electron cloud the easier it is to modify its shape.It happens because the electrons far away from the nucleus are less attracted by it and can move around a bit more easily than electrons very close to the nucleus.

This is the reason why van der waals forces are stronger for bigger atoms.
This would explain why $\ce{Br_2}$ has a better solubility than $\ce{O_2}$ in water.

As a side note:

London interactions are very important if you look at liquefaction temperature of gases. For instance, it explains why $\ce{Ar}$ becomes a liquid below 87K while $\ce{He}$ become a liquid only below 4K.


Type of structural isomerism What is the type of structural isomerism in which isotopes move around?

Examples

$\ce{CH2D-NH2}$ and $\ce{CH3-NHD}$
$\ce{CH2D-NH-CH3}$ and $\ce{CH3-ND-CH3}$
$\ce{CH2D-NH-CH2CH3}$ and $\ce{CH3-NH-CHD-CH3}$ and $\ce{CH3-NH-CH2-CH2D}$

here $\ce{D}$ refers to deutrium.

My progress: So I know that there are various types of isomerism, like

Tautomerism - Differ in the position of protons and electrons. Usually I see that this occurs with changes in bond orders. So I don't think it will be this.
Ring-chain isomerism - When there is a difference in the number of carbons in a ring
Functional isomerism - When the functional group changes. I don't know if $\ce{-NH2}$ is a different functional group from $\ce{-NHD}$.
Metamerism - This is when there is a change in carbon chain on both sides of a bridging group. I don't know if a change in isotopes would count as a change in carbon chain.
Chain isomerism - When the carbon skeleton changes. Here the skeleton carbon is the same.
Positional isomerism  - When a functional group attaches itself on different carbon of the same chain. Not applicable here.

This is not a homework question. This question was to fill the gap in my knowledge of isomerism with isotopes.

According to Wikipedia, this type of isomerism is known as Isotopomerism

  Isotopomers or isotopic isomers are isomers with isotopic atoms, having the same number of each isotope of each element but differing in their positions. The result is that the molecules are either constitutional isomers or stereoisomers solely based on isotopic location. For example, $\ce{CH3CHDCH3}$ and $\ce{CH3CH2CH2D}$ are a pair of constitutional isotopomers of propane

For a different source, you can also check out the IUPAC Gold book's definition of an isotopomer.


How to name this benzene alcohol? The following is the molecule I am confused about.

Earlier in my textbook it gives an example of how to name benzenes with 2 or more hydroxyl groups:

My textbook (Nelson 12 Chemistry) is infamous for its crappiness but my school is too lazy to change it, so I've been asking a few of these questions in recent weeks.

What is the proper, IUPAC name of the first molecule depicted in this question?
Are there any other commonly accepted names for this molecule and under what convention are they named?

Personally, I would name the first molecule benzene-1,2,4-triol, but the textbook gives the name 1,2,4-trihydroxybenzene, using the hydroxyl groups as substituents.

The compounds given in the question contain the characteristic group $(\ce{-OH})$. Since there is only one characteristic group, the seniority order of classes is not relevant in this case; thus, the $\ce{-OH}$ substituent corresponds to the principal characteristic group that is expressed as a suffix (‘ol’).

Therefore, the preferred IUPAC name (PIN) for compound (e) is benzene-1,2,4-triol.

Various traditional names exist for hydroxy compounds. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), only the name ‘phenol’ is retained as a PIN.

  P-63.1.1.1 Only one name is retained, phenol, for $\ce{C6H5-OH}$, both as a preferred name and for general nomenclature. The structure is substitutable at any position. Locants 2, 3, and 4 are recommended, not o, m, and p.

Therefore, the PIN for compound (c) is indeed phenol.

Furthermore, according to Subsection P-63.1.1.2, the names

pyrocatechol (benzene-1,2-diol),
resorcinol (benzene-1,3-diol), and
hydroquinone (benzene-1,4-diol)

are retained but only for general nomenclature and only when unsubstituted.

Therefore, the PIN for compound (d) is the systematic name benzene-1,2-diol (the traditional name ‘pyrocatechol’ may be used in general nomenclature).


Polymer Crystallinity See the following question:

I think the answer is A because of the hydrogen bond in the side group, thus the degree of secondary bonding between parallel chain segments is higher.  Is my thinking correct?

Question is, what is "better"? Faster, or higher melting enthalpy, or degree of crystallinity? The methyl group is bulkier and thus hard to get into the right backbone conformation, and the OH group is polar, as you've noted, leading to a melting point of PVA of 230°C (iPP ~175°C).

But then, it all depends on the actual tacticity, which is not specified. The isotactic species have a totally different crystal structure than the syndiotactic, and sample with low tacticity (PVA!) might still have another one. That leads to totally different crystallisation kinetics.

So while a lot speaks for "A", I say there can be no definite answer.


Electromagnetic energy to Chemical energy How can electromagnetic energy (photons) be efficiently converted to chemical energy? For instance, can water be efficiently split to hydrogen and oxygen using solar energy? As we are running out of non-renewable resources is there any way for us to use the solar energy so efficiently to generate power?

You can certainly turn electromagnetic energy into chemical energy, but the key is 'efficiently.'

A photo-voltaic cell can be used to split water into Hydrogen and Oxygen. The Hydrogen can be burned or used in a fuel cell.

But the key is how efficient is 'efficient.'

Each step will lose energy, but if solar is the input, there is a huge supply. If a solar powered system were continually operating, and we had many of these systems, a huge amount of Hydrogen gas can be generated. Even considering the costs of delivering to where it is needed (which can be paid with the product itself) there is a huge and realistically limitless supply.

That being said, there is much 'waste' which hampers a calculated efficiency, but once established a system that relies on water for a raw material is relatively efficient because there is very little required for raw material acquisition.


Can you synthesize phenol from salicyladehyde? Can you synthesize phenol from salicyladehyde? Well for starters I know that salicyladehyde can be obtained from phenol through Reimer–Tiemann reaction but I was wondering if the reverse was possible.

It is possible to convert salicylaldehyde into phenol in a two-step process. Firstly, salicylaldehyde is oxidized to salicylic acid with a standard reagent like $\ce{KMnO4}$ (see the answers to this question).

In the second step, salicylic acid is thermally decarboxylated to phenol in a simple distillation apparatus. Solid salicylic acid is strongly heated in a round-bottom flask equipped with a short path distillation head with thermometer. Heating is maintained to keep a reasonable distillation rate of the desired product (phenol passes over at a temperature range from 145 to 165 °C). Most by-products of the reaction, like phenyl salicylate, have higher boiling points and remain in the residue. The crude product can finally be purified by vacuum distillation, for example, at a pressure of 30 Torr and a temperature range of 85 - 90 °C (reference).


Why does H2 react with O2? We know $\ce{H2}$ is formed with covalent bonds (just as $\ce{O2}$ is formed via covalent bonding). Hence, these molecules gain some stability. In spite of this stability, why do they react with each other to form $\ce{H2O}$ or $\ce{H2O2}$?

Indeed, much of the chemistry that takes place in the universe involves the breaking of covalent bonds and the formation of new covalent bonds. The key is that the energy of the products needs to be in a lower state than that of the reactants. This is the thermodynamic requirement anyway. Lets consider one of your examples:

$$\ce{2H2(g) + O2(g) -> 2H2O(g) + \pu{572kJ}}$$

Although we had to break the covalent bonds of hydrogen and oxygen, the new bonds that were formed are even stronger, thus the reaction is thermodynamically favored.

The complicating point of the above example is the fact that there is a large kinetic barrier to the reaction. This basically means that there is an intermediate product in the steps of this reaction that is energetically very high (unfavorable), so some catalyst is required (say, a spark) for this reaction to get over that energetic barrier and proceed on any reasonable time scale.

Summary, TL;DR:
The breaking and creating of covalent bonds plays a major role in chemistry. The criteria for covalently bound compounds to participate in a chemical reaction is that the products have a lower energy state than than the reactants. If this is not the case, the reaction will not happen. Even if it is the case (that the products are lower energy / more stable than the reactants), there may be kinetic limitations requiring a catalyst in order for the reaction to proceed. But, no catalyst can make a reaction proceed if the above criteria is not met, that the products be more stable than the reactants.


Why does hybridization significantly matter for organic chemistry? I just recalled, and still see that almost every organic chemistry book starts with bonds, empahsizing hybridization. One thing is straight forward but the rest doesn't seem to connect well.

The thing that's clear is, energy level wise $\mathrm{sp < sp^2 < sp^3}$. When $\mathrm{p}$ character is higher, the energy is larger, the electrophilicity is higher and its affinity for reaction is high up as well. This is the opposite of what electrons constantly thriving to achieve : the lower energy state and be stable. However, hybridization allows molecules to have a shape minimizing the energy though. Through this bonding it also releases the energy (dissociation) by stabilizing itself - so bond formation is the tendency. Hybrids ($\mathrm{sp, sp^2, sp^3}$) form $\sigma$ bonds while pure-breeds (100% $\mathrm{s}$, 100% $\mathrm{p}$) form $\pi$ bond.

I had the impression hybridization regulates a standard bond among same species of atoms, e.g $\ce{C-H}$ bond in all alkanes are the same. But that's not the case when I read brief comments indicating that the $\ce{C-H}$ bond in $\ce{CH4}$ differ from $\ce{CH3-CH3}$.

What I am unable to reason is, why hybridization is so important for organic chemistry than to any other (general, inorganic..., physical)? Is it about bond strength (how fast one can bond and break) or is it about acidity or basity? Why should we (learners) or organic chemists care?

Please provide some application example. It's also great if one could point out how each $\ce{C-H}$ bond differ from alkane to alkane or any other functional group.

Orbital hybridization is mostly useful in the rationalisation of molecular geometry. In fact it was invented by Linus Pauling for this purpose: to rationalise structures of organic molecules using the language of atomic orbitals. Yes, hybridization is also used to rationalise bonding properties, as well as acidity or basicity, but its primary use is to rationalise molecular geometries.

  I had the impression hybridization regulates a standard bond among
  same species of atoms, e.g $\ce{C-H}$ bond in all alkanes are the same. But
  that's not the case when I read brief comments indicating that the $\ce{C-H}$ bond in $\ce{CH4}$ differ from $\ce{CH3-CH3}$.

Well, $\ce{C-H}$ bonds, of course, differ in $\ce{CH4}$ and $\ce{CH3-CH3}$, but not too much; they are approximately the same. Are you naive enough to hope that a model as simple as the orbital hybridization one would perfectly predict the molecular geometry of each and every organic molecule? Surely, it won't. It can just give a crude (but simple) approximation.

  What I am unable to reason is, why hybridization is so important for
  organic chemistry than to any other (general, inorganic..., physical)?

Where did you get that? Hybridization is a concept of general chemistry, it is used in both organic and inorganic one. Its relatively big importance for organic chemistry is due to the fact that it is the only simple model which can explain (approximately) molecular geometry of organic compounds. The original valence bond theory couldn't do it, so it was extended.

On explanation vs rationalisation

Now I would like to emphasize that hybridization provides a rationalisation of a molecular geometry (as well as some other properties), but not an explanation, so it can not be used to reliable predict the properties. What is the key difference between an explanation and a rationalisation in these context?

Well, an explanation of some phenomenon describes it in way that we can observe and test the consequences of our claims and premises. We can then tell from the observations are our claims and premises true or not, or, in other words, does our explanation work or not. The whole point of an explanation is to distinguish what is true from what is not true, in a sense, what corresponds to the actual reality from what does not do so.

In a rationalisation things work differently. We insist on some premises being true without testing them against the reality. We do observations, but we do not use them to tell are our premises true or not, rather on the basis of the desired premises and the observable evidence we start to make the observed outcome seem consistent with the premise.

In psychology rationalisation is also known as making excuses. It is a defence mechanism in which a controversial (from the society viewpoint) behaviour is justified "in a seemingly rational or logical manner to avoid the true explanation". With respect to scientific theories, it can also be thought of as a defence mechanism: when the real explanation is too complex for beginners we avoid it to defend them. Instead we first provide some simple rationalisation and later in the course of study the explanation is introduced.

The way rationalisation works for all these general chemistry models is as follows:

We do know geometry or some other property of a particular molecule, say, from an experiment, and we are seeking for some rationalisation of why the things are the way they are.
The rationalisation usually has to be as simple as possible, so that, for the case of molecular geometry, we usually start from VSEPR theory.
If the simplest possible theory can not rationalise the observed property, we choose a more sophisticated model. For molecular geometry, for instance, we could then try to use the original valence bond (VB) theory.
But even the VB theory (in its original formulation) won't work for organic compounds, so it has to be extended by two concepts: resonance and orbital hybridization. Note that this concepts are also required in inorganic chemistry as well, but not always: for some inorganic molecules the original VB theory is enough.

But however deep you go into this hierarchy of chemical models, all them are just rationalisation models: they provide simplified (and in many cases oversimplified) descriptions of the reality which is approximate (and in many cases rather crude). And besides all this process of seeking for an rationalisation is a bit like cheating:

If some model doesn't work, let us try another model.
If none of them works, let us introduce a new concept with some funky name in one of them, the purpose of which will be to rationalise this particular thing which couldn't be rationalised by the original model.

As a result, the predictive power of such models is quite small. They can be used to rationalise what is already known, but any predictions based on such models will be unreliable. And yeah, the concept of orbital hybridization is just one of these cheats mentioned above.


Triple Point of a compound and critical point Is it necessary for all compounds to have a triple point? Also, how is the triple point determined?
Suppose a substance does not have a triple point, so how do researchers agree on the impossibility of existence of a triple point in such a case? Also, can there be multiple triple points for a compound?
Also, since the vapour equilibrium curve ends at point E, what is the phase of water beyond that point?

The essential condition is the compound thermal stability.
If it decomposes below its melting point, it does not have a triple point.
If it decomposes before properties of gaseous and liquid phases converge to each other,  it has just estimated, extrapolated triple point.
Beyond CP we talk about supercritical fluid, it kind of shares many properties of both liquid and gaseous worlds.
It is usually determined experimentally, or as an extrapolation.
See the picture of liquid/gas phase system below, at and above the critical point on Wikipedia: Critical_point


Alternatives to Congo Red? I require Congo red indicator for an experiment I wish to perform involving dipping a cellulose sponge into an acid and a base.  I first want to dye the cellulose sponge with Congo red, and observe the color change of the sponge as I place it in acids and bases.

Congo red apparently binds to cellulose and thus is suitable for this experiment.  Unfortunately the laboratory doesn't have this indicator at the moment.  Are there any other indicators that could be used for this effect instead?

Yes, congo red bind the cellulose because is a substantive (or direct) dye these dyes don't need a mordant so it binds suddenly to the cellulose.
 There are others dyes for example safflower, cochineal with this property. In biology Congo Red is frequently substitute with Sirus Red F3B and Tioflavone S. At this link you can find some alternatives to Congo Red use for demonstrating amyloid, however I don't know if you can apply these dyes to your case, because I don't find any reference about their use as pH indicator. Although is reported that many substantive dyes are used as pH indicator I can't find any list.

Take in account that acid pigment are not substantive to cellulosic fiber and most of the pH indicator are acids.

I think that maybe you can try to use a mordant to obtain the same result with a pH indicator normally used, however I'm afraid that using a mordant you risk to change the behavior of your pH indicator.


Electric Attraction Vs Resonance Energy in Electrophilic Aromatic Substitution Reaction Image Credits: MHHE

In the electrophilic aromatic substitution attack of Benzene by the nucleophile, as shown in the figure, nucleophile attacks hydrogen of the intermediate, instead of attacking the positive charge distributed ring; leading to substitution reaction.

Bezene's closed shell of six pi electrons is said to give it an extra stability (resonance energy)$_1$, but it is not clear and precise in the explanation of how the electric attraction is overcome by this another tendency which is described. So, the question is, how is this electrostatic attraction towards positive ring overcome by other tendency of stability?

$_1$ Organic Chemistry 11e by Solomons, Fryhle, Snyder, pg.671.

The positive charge, instead of being localized, is actually delocalized throughout the ring because of resonance, making the charge shared between three carbons, lessening the ability for the nucleophile to attack the positive charge.


Synthesis of 1-methylcyclopentene from cyclopentene I have to put a methyl group on a cyclopentene. My teacher made this solution (the picture). But my first guess was just to use $\ce{CH3Cl}$ and $\ce{AlCl3}$. Why can't you do that?

In Friedel–Crafts alkylation you form a strong electrophile to break the aromaticity of a benzene ring, resulting in a species that is also electrophilic, but not on the carbon atom. It is much more energetically favorable to reform the aromatic ring by losing a proton to $\ce{AlCl4-}$, evolving $\ce{HCl}$ gas.

The reaction of the electrophile with an alkene forms another carbon centered electrophile, however, which can rearrange and will go on to react with other alkenes or with $\ce{AlCl4-}$. There is also nothing to stop the reformed $\ce{AlCl3}$ from continuing to react with the newly formed chloroalkanes, so overall it would probably be a big mess. I found a resource here that discusses the reaction of $\ce{AlCl3}$ with alkenes:

For your reaction, though, I would propose something a little simpler and probably higher yielding like this:


Is there any emerged master principle in chemistry, like the theory of evolution in biology? The question is about emergence and reductionism.

Except for a few exceptions, such as dark matter, most phenomena in Nature could be reproduced from the standard model and general relativity with great accuracy. Up to a Planck scale, there is no inconsistency between general relativity and quantum theory by means of effective theory. However, at different level of complexity, a new principle may emerge. One example is the theory of evolution in Biology.

Is there any similar emerged master theory in Chemistry that comparable with evolution in Biology?

Here, the emerged master theory means:

It is not just a brute-force ab initio computation (emergence). Here I have excluded explicitly correlated Gaussian.
The theory is universally valid. Not like 8- and/or 18-electron principle and/or VSEPR and/or periodic table, we know there are many exceptions.
The theory should generate experimentally testable predictions, hopefully quantitative. Here I have excluded Bader's QTAIM at its current form. Atom in molecule and types of chemical bonds are anyway not observables. One may see some correlation between Laplacian at bond critical point with other molecular properties, and do some fitting or hand-waving argument. Such fitted relation could be useful. However, it is a fitting, not a derivation from its principle. I do not regard it as a prediction.
The theory should not be discovered in condensed matter physics first. Here I have excluded spontaneous symmetry breaking.
I do not expect the statement like, there is no master principle in chemistry is the master principle in chemistry.

You may be asking for something quite a bit deeper, and it's rather hard to strictly adhere to the requirements in the question, but after Eric's interesting edit showing some outside-the-box thinking, I went along those lines and came up with some options. I'll let you decide which (if any) make the cut.

Two good candidates for master principles in chemistry may be acid-base theory and reduction-oxidation theory. I think ultimately they're just somewhat coarse measures of the relative positions between HOMOs and LUMOs for different species, but after a couple centuries of research and some tabulation, we've come up with general top-level rules for understanding the chemical behaviour of large classes of compounds, with significant predictive power, without necessarily resorting to complicated calculations.

The periodic table itself could also work; different elements possess certain chemical similarities which allows them to be grouped with one another. Mendeleev famously used this to predict a handful of new elements, such as germanium (though he did also predict some other elements which never came to be).  A related principle which is simple but of great importance is that elements are entirely characterized by the number of protons in their nuclei. Every single atom in the Universe with 79 protons in its nucleus is a gold atom, no matter how many neutrons and electrons are present.

Also, I wonder if you'd count materialism as a master principle which emerged in chemistry; every material in the Universe is made of atoms (not counting dark matter). If two materials are the same, then they must be made of the same types and proportions of atoms in the same configurations. If two materials are different, then either the atoms are different, or they are present in different proportions, or they are present in different configurations.

Finally, I myself would actually have included the octet rule as a master principle. It has many holes and limited applicability, but it is uncanny that such a simple concept works at all. I find it a surprising way to effectively skip tons of gritty quantum mechanical calculations, helping chemistry education to begin in school around ages 14-15.