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Pressure Cooker at High Altitudes and Low Altitudes
I just have a doubt if I'm right or wrong about this. Please explain more if possible.
Also, I have some questions regarding boiling point and vapour pressure.
 High Altitude Cooking
At high altitudes, the atmospheric pressure is lower than that at sea level, so the boiling point at high altitudes is quite low, which means water boils very fast and at low temperatures. The food inside it does not get enough heat to get cooked and thus food is difficult to cook at high altitudes.
Using a pressure cooker at such conditions helps increase the boiling time as the pressure inside the pressure cooker increases due to the vapour produced inside it. The boiling point increases and thus the time taken to reach the temperature increase and the cooking is better because the food is getting heat for a long period of time.
At Low Altitudes
The boiling point of liquid at low altitudes is more than that at high altitudes, so we do not have to worry much about less time being available for cooking the food. In fact, we try to decrease the cooking time in order to get food cooked faster. Even less amount of water can produce a good vapour pressure and get equal to the pressure produced inside the cooker. But less amount of water takes a lesser time to reach a particular temperature than more amount of water ( as $Q=mc{\delta}t$ ) and thus lesser the mass, lesser the time  to get the required heat, hence faster the cooking.
Questions
1. Have I understood this perfectly or is there something more to it?
2. Does the vapour pressure of the water inside the cooker have to become       equal to the vapour pressure produced due to itself in order to boil?
3. How does pressure cooker increase time for cooking at high altitudes and decrease time for cooking at low altitudes being basically the same thing?
4. The Critical temperature is the temperature at which gas just becomes liquid. The boiling point is the temperature at which all the bulk of the liquid starts turning into gas. How are Critical and Boiling Points different? How are they similar?Is there any mathematical relation between them?

Edit - As 1e9dB's answer points out I made a huge mistake in this answer. The inside pressure of the pressure cooker is ambient pressure plus the extra pressure of the valve. The pressure cooker does not cook at an absolute pressure independent of altitude.

You have it all backwards. The pressure cooker allows the pressure inside the cooker to be above the sea-level atmospheric pressure which also increases temperature inside the cooker and thus decreases cooking time. So the critical factor is the increase of the boiling temperature of water as a function of the pressure inside the cooker.
The pressure cooker raises the pressure inside the cooker to the same absolute pressure, regardless of the altitude at which the cooking is being done. A typical value for the internal pressure would be about 2 atmospheres absolute which makes the internal cooking temperature about 120 °C. (The steam is pushing against the pressure regulator (weight) to escape, not the outside atmosphere. It is like the safety valve on an old steam locomotive that keeps the pressure at a certain level so that the pot doesn't blow up.)
The inside temperature of a pressure cooker when steaming (120 °C) is way below the critical temperature of water (about 374 °C) so that factor doesn't enter into the cooking time at all.

Increase the pressure (of the surrounding) above water  increases its boiling point.  Water will boil at a higher temperature when the pressure above it is higher. If we consider the vapour pressure (of the water), if it is high, boiling temperature will be less since it will easily be equal to the pressure of the surroundings. This is what I meant. Though I never considered the weight of the regulator, which I'm really glad you made me aware about. How about 'increase of cooking time' for high altitudes?

Altitude doesn't matter!! The pressure regulator sets an absolute pressure, not a pressure relative to current atmospheric pressure. So cooking time at sea-level and 10,000 feet would be the same.

https://en.wikipedia.org/wiki/High-altitude_cooking  This means time at high altitudes with cooker is increased relative to cooking time at high altitudes without the cooker?

You have it backwards. Let's say that I'm trying to boil some carrots. By putting them in the pressure cooker the absolute internal pressure goes to about 2 atmospheres and the temperature inside the pot goes to 120 °C - regardless of altitude! Water in an open pot at sea-level boils at 100 °C. At 10,000 feet water in an open pot boils at about 90 °C. So a pressure cooker always cooks faster. https://en.wikipedia.org/wiki/High-altitude_cooking#Boiling_point_of_pure_water_at_elevated_altitudes

Then why the f is it said that they increase cooking time at high altitudes? ;-;

MaxW has it right except for one thing - the pressure inside the cooker does in fact does depend on altitude. For example, a cooker that will reach 15 psi at sea level will only reach about 12.5 psi in Denver.
Although it is true that the pressure is regulated by a weight which closes off a small hole, and the weight is constant regardless of altitude, remember that the ambient pressure outside the cooker is also imparting a downward force on the weight that is in opposition to the internal pressure that is imparting an upward force. At a higher altitude the internal pressure can't rise quite as much before it overcomes the lower external pressure in order to raise the weight. The result is that the internal pressure is lower by exactly the same amount as the ambient pressure difference due to altitude. That means that the boiling point inside the cooker is somewhat lower at higher altitude, although still higher than it would be outside of a pressure cooker.
Thus at higher altitude you need to increase the cooking time:
Pressure Cooker PSI FAQ

You're right, I blew it. Thanks for the correction.

Or you have to adjust the valve. I've seen pressure cookers (with a spring-loaded) valve where you can adjust the valve. Of course, the cookers has to be able to withstand a higher pressure difference if the outside pressure is lower.

This thread is very confusing. I studied Physics decades ago so I admit I need to really consider the variables. But I have to comment even when unprepared.

The weight of the regulator DOES change with altitude.  I do not know how significant is this variable.
If the volume of air in the pressure cooker is thinner because of altitude,  the pressure cooker will take longer to reach maximum pressure.
If the regulator is so heavy that the internal pressure never exceeds the weight of the regulator, then the regulator is irrelevant to the internal pressure achieved.
The lower ambient pressure at high altitudes does affect the regulator since less atmospheric pressure is exerted against the internal pressure.  Again, if no steam escapes then this is irrelevant just as the variable weight is irrelevant.
The pressure inside the electric pressure cookers is regulated by an internal pressure gauge. If the gauge can only detect relative pressure increases then the pressure cooker will not reach the same internal pressure at a higher altitude.
These are my very old laws of physics thoughts.

The pressure in a pressure cooker is controlled by the weight of the valve and weight is the same regardless of altitude as you know you weigh the same ingredients whatever height you live in the world.

There would be a slight difference in the inside pressures based on the altitude as the weight of the valve adds on to the ambient pressure. See the other answers.

Most of the responses here are overthinking a very simple device.
Pressure cookers are considered essential devices in places such as Nepal because at high altitude (substantially lower atmospheric pressure) water boils at a lower temperature making it slow or even impossible to adequately cook food.
Atmospheric pressure at sea level is around 14 pounds per square inch (psi). (This 14 psi figure is sometimes referred to as 1 Atmosphere of pressure) now forget all about "atmospheres" as a measurement, it just confuses things.
At the surrounding pressure of 14psi (normal at sea level) water boils at 100 degrees Celsius.
Put the water in a pressure cooker, seal it except for a very small hole with a weight on top. (Forming a pressure-control valve) The weight is calibrated to need let's say 28 psi before it lifts up and releases any excess pressure.
That has only to do with the amount of weight on the very small hole which is the valve.
At 28psi the water will not boil until it reaches about 120 deg C. as the greater pressure inside the cooker prevents the water turning to steam until it reaches a higher temperature than at 14psi. ("Why" is a whole different subject that isn't  relevant here)
That weight on the valve is the same at 10,000 feet altitude as it is at sea level.
Therefore 28psi inside the cooker is required to lift the valve at either altitude.
Therefore the temperature is the same (around 120 deg C)and the food cooks the same.
Same time, same temperature.
That's it. Why people in Nepal all love pressure cookers. End of story.
But, to deal with some minor matters raised previously.
The weight does weigh essentially the same at 10,000 feet as it does at sea level.
Sure it's a little bit further away from the major mass of the Earth so the effect of gravity is a tiny, tiny, tiny bit less but it's extremely negligible and not relevant. I lose weight when I shave in the morning but it's equally not relevant, I still weigh too much!
Now the notion that there is less atmospheric pressure pushing down on the weight at high altitude: Well that's true but totally forgetting that air pressure is applied equally from all directions.
There's also less atmospheric pressure pushing UP on the weight as well and from all sides for that matter. It makes no difference on the downward force exerted by the weight on the valve.
More pressure pushing on the weight from all directions at sea level is true also, and also makes no difference at all as it's coming from ALL directions.
If you got this far through my musings you did well! Thanks for sticking it out :-)  PS. I own 4 pressure cookers of different sizes!

Some qualifications and clarifications:

Altitude affects average air pressure, it's quite possible for pressure to be higher than 1 bar (101 kPa) at 1 km above sea level.
You should be careful to clearly state what the heat flow is.

In general "heat" is a vague term which is often used to describe energy or the flow of energy. The energy required to increase the temperature of a substance depends on the mass of the substance. The rate of temperature increase depends on the heat flow (as well as the mass and other things (such as heat loss)). Assuming the same heat flow, on average a given mass of water will boil sooner (and at lower temperature) at higher altitude.
Most food cooks faster at higher temperature. So, low altitude cooking can be faster if the limiting variable is the boiling of water. Likewise, high altitude cooking will be slower if the boiling of the water is the limiting variable. This will have little to do with frying an egg or roasting meat.
Your question mixes (confuses?) time and temperature. In an open container, a given amount of water will boil quicker at high altitude (on average, with a fixed rate of energy input). The temperature rise will be the same as at low altitude, until the boiling point is reached. At the boiling point (which will be lower at higher altitude (on average)) the temperature of the liquid becomes almost constant until (almost) all of the water has boiled away. IF this temperature is the temperature the food is cooked at, then higher altitude means slower cooking.
=========
The critical point of a substance Tc, is the temperature and pressure (Pc) at the "end" of the vapor-liquid curve. It is the temperature above which no liquid phase can exist, regardless of pressure. (It is the highest pressure at which both liquid and gas can co-exist.) I think of it as the temperature at which (and above which) the liquid and gas are indistinguishable.
The Wikipedia article has a diagram showing a 'typical' phase diagram. [see Critical point (thermodynamics)]. That diagram shows many different regions. If you count carefully, and include the dashed lines, you can count at least 11 different areas. The major problem with this is that the dashed lines don't represent real phases, and also the names given in that diagram are arbitrary. For instance, there's no technical difference between vapor and gas. Similarly, liquid phase and compressible liquid aren't inherently different.
A gas above Tc is indeed "supercritical", by definition, however. As temperature is increased towards Tc, the properties of the liquid phase converge towards the equivalent properties of the gas (vapor). Keep in mind that when speaking of these phase diagrams, we are speaking of systems at equilibrium. Each of the two solid lines represent the points of P and T at which two phases can co-exist at equilibrium. The triple point, when it exists for a substance, is the point at which all three phases can co-exist (and shouldn't be confused with Tc).
Of course, in the real world, I'm sitting here typing with a glass of liquid water, ice, which no doubt has water vapor over it. However, if I were to wait long enough, the ice would melt, and I'd be left with liquid and vapor (gas), and if I were to wait a really really long time, the liquid would all evaporate (unless I sealed the glass). So, the real world only works "near" equilibrium, rarely exactly at it.
The phase diagram works by describing which phases exist at a given T and P. It also enables you to predict how much increase (or decrease) in P or T is necessary to move the system to a different region in the phase space. I believe it is most useful to think of Tc as the point at which there is 'no difference' between gas and liquid, rather than the point at which liquid "just becomes" gas. It is incorrect, imho, to think of a supercritical gas as a liquid, but certainly it has some properties (depending on pressure) which are similar to ones we associate with liquids.

This answer would be greatly improved by paragraphs. At the moment I really don't want to read it because it is just a wall of text.

Interesting comment about the weather. So making pasting will take longer when you are in the eye of the storm...


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(stereochemistry)epoxide from alkene

picture above describes two ways of forming epoxides from alkene.
(a) peroxy acid
(b) Br2 and H2O / NaOH
it says R groups stay in the same position.
and I want to ask some about (b).
is it possible R groups that were on the same side before reaction occurs can be opposite side of where the double bond used to be after product is formed?
I've drawn this in the picture below.

Draw all three rotomers using Newman projections.  Determine the lowest energy conformer. I don't have a table of gauche energy costs handy.

The problem with NK Yu's diagram is that the bond rotation is unnecessary to obtain the correct conformation for epoxide formation. The epoxidation and the bromination/base give the same product and both processes are stereospecific. (Z)-Butene (1) gives the cis-oxirane 2 while (E)-butene affords the trans-oxirane (not shown). Peracid epoxidation adds oxygen to either face of the π-system of (Z)-butene to provide meso-oxirane 2. Bromination of 1 affords an intermediate meso-bromonium ion 3 by the same type of addition as was seen in the peroxidation. When hydroxide effects an SN2 (blue arrow) inversion of stereochemistry occurs (R $\rightarrow$ S), the S,S-bromohydrin 3 is formed. With equal probability the path of the red arrow leads to the R,R-enantiomer 4. Deprotonation of the racemic bromohydrin leads to a second inversion of stereochemistry to form meso-oxirane 2, i.e., R,R $\rightarrow$ R,S and S,S $\rightarrow$ R,S. The commonality here is that direct, peracid epoxidation involves no inversions of stereochemistry while the bromination/base route involves two inversions of stereochemistry. The bottom line is that an even number of inversions, zero being considered even, produces the same stereochemistry.

Good answer. The only minor suggestion for improvement I have is that, for beginners, if would be clearer if you formed the bromonium ion on the bottom face of the alkene so that the product epoxides have the exact same orientation as the one formed via the peracid.

@Zhe: Good suggestion that I had debated. Will do.

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Stereochemistry in Sn2 reaction
I'm confused why the absolute configuration always has to change after an $S_N2$ reaction:

For example, in primary alkyl halide above there are 4 planes Nu can attack; and absolute configuration is changed after an $S_N2$ reaction because the nucleophile (Nu) only attacks the 'black plane'.
I can understand that the nucleophile wouldn't prefer red or green plane because of steric hindrance, but here's my question:
Why doesn't the nucleophile attack the 'blue plane'?

The attacking nucleophile does not just shove its electron pair to the carbon from some random direction. No, it must donate that pair to the antibonding orbital ($\sigma^*$) of the $\ce{C-X}$ bond, thus weakening and eventually breaking that bond. And it just so happens that the antibonding orbital is mostly located on the side opposite to X.
Upd. Thought I'd better look up a nice picture. Here's one:

(from this page)

Ug.  F as a leaving group.  Ug.

Yeah, yeah, I know, but it would suffice to illustrate the basic idea.

The reason behind which 'plane' the nucleophile attacks has to do with Molecular Orbital Theory.
But here's a short explanation for this: the C-X bond appears because two atomic orbitals interact to form two molecular orbitals: a filled bonding orbital ($\sigma$) and an unfilled anti-bonding orbital ($\sigma^*$).
For the C-X bond to break, the nucleophile has to attack $\sigma^*$, which means that the nucleophile has to get a good overlapping of its filled orbital with the unfilled anti-bonding orbital of C-X. Because of this attack, the C-X bond breaks and the C-Nu bond forms.
To give you an image of where the anti-bonding orbital is located, think of the C-X $\sigma$-bond; behind that bond on the carbon side, there lies the biggest portion of $\sigma^*$ (see Ivan's image). If we consider that, this orbital intersects the 'black plane' that you were talking about.
Therefore, the nucleophile can't attack the other 'planes' besides the 'black plane' because that doesn't assure a good overlapping between orbitals.

As for the change in configuration, in primary alkyl halides (the example you gave) we don't have a change of configuration because the carbon atom that's bonded to the halogen is not chiral.
But with a chiral compound, (i.e. (R)-2-chlorobutane), we get a change in configuration because the nucleophile attacks 'from the back', reversing it (i.e. (S)-butan-2-ol.)

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Why do some periodic tables have 18 groups while others have 8?
In school I have been taught about the periodic table with 8 groups, representing the number of electrons in the outer shell (with exceptions). But online almost all I can find is periodic tables with 18 groups. Why do both exist? (Is the one with 18 more modern?) What is the purpose of 18 groups?

The only official periodic table is one with 18 groups.

See this answer as well: https://chemistry.stackexchange.com/a/133169/72973

Unfortunately, life sometimes is more complicated than school :)
Using 8 groups works well to teach the basic principle, but higher shells have space for more electrons. (And lower shells for fewer - ever wondered why there are only two elements in the first period?)
Hence, you'll get 18 electrons in the next shell (the "transition metals"), then 32 electrons ("Lanthanides/Actinides") - the general rule is $2n^2$, where $n$ is the number of the period. This is unfortunately made even more complex, as the energies of the shells are not quite what one would expect and they are not filled "in order." One probably should stop speaking of "shells" at this point and move towards "orbitals."
Anyway, the community has pretty much settled on using 18 groups nowadays. The idea behind the periodic system is to find regularities (hence the name) between elements.
It turns out, that individual groups and trends within them are similar in the 8 "main" groups (as you have learned). In the transition metals, there are still a fair few regularities, that justify this arrangement - but they are already a lot more similar to each other than say alkali-metals and halides. The Lantanides/Actinides finally are so similar to one another, that they are "grouped" (although they in fact are part of a period) together - they in fact so similar, that they usually only occur in mixtures in nature, and separating them is quite hard.
Finally: no heavier atoms have been discovered so far that would have to go into an even bigger "shell"/orbital. Hence, one could justify up to 32 groups, but for practical reasons, 18 are conventionally used.


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Why is the hypoiodite anion unstable in solution, whereas hypochlorite is relatively stable?
$\ce{Cl2}$ and $\ce{Br2}$ can disproportionate in a cold, aqueous, basic environment to form $\ce{ClO-}$ and $\ce{BrO-}$ respectively, yet $\ce{I2}$ directly forms $\ce{IO3^-}$. Why isn't $\ce{IO-}$ isolatable in solution?

At any temperature, IO- itself are rapidly redox reaction I- and IO3-.

@user6006786 Sure, but why? $\ce{ClO-}$ doesn't showcase this kind of behaviour at lower temperatures, yet $\ce {IO-}$ does.

I think this is because the cl-o bond is stronger than i-o, but I could not find the data

When iodine is added to water, the following reaction results:
$$\ce{I2(l) + H2O(l) -> OI^{-}(aq) + 2H+(aq) + I^{-}(aq)}$$
$\ce{I2}$ molecules and water molecules react to form hypoiodite ($\ce{OI-}$). The reaction can move both ways of the equilibrium, depending on the pH of the solution. Iodine may also occur as $\ce{I3-}$(aq), $\ce{HIO}$(aq), $\ce{HIO3}$(aq) . The compounds that are form behave differently when they come in contact with water. When iodine ends up in surface waters, it may escape as iodine gas.
$\ce{IO-}$ rapidly disproportionates to form $\ce{IO3-}$ and $\ce{I2}$
$$\ce{IO- -> IO3- + I2}$$.
See here for more details.
For the disproportionation reaction, consider the Latimer diagram for iodine:-

A species has a tendency to disproportionate into its two neighbor species if the potential on the right of species in a Latimer diagram is more positive than that on the left. In this case, right hand potential($\ce{+0.42}$) is greater than the left hand potential ($\ce{+0.15}$). So , it disproportionates.

Yet, my question still remains: why $\ce{IO-}$ rapidly disproportionates, whereas $\ce{ClO-}$ can be isolated?

Well, in your edit, I see that $\ce{IO-}$ is a radical oxide. That doesn't sound right since all electrons are found in pairs. $\ce{IO}$ does have an unpaired electron, but not $\ce{IO-}$.

Your answer does prove that $\ce{IO-}$ is unstable according to the standard reduction potentials of $\ce{IO^{3-}}$ and $\ce{IO-}$ - but that's not what I'm looking for. I'm trying to find the reason behind the numbers, not the numbers themselves.

@L3ul the reason behind the standard reduction potentials is the lower electronegativity of iodine compared to chlorine. See my answer below.

Chlorine has a higher electronegativity than iodine. Therefore, the Cl-O bonds are less stable than the I-O bonds (generally, oxygen reacts better with elements of lower electronegativity). Thus $\ce{ClO3-}$ is comparatively less stable than $\ce{IO3}$ which means that in case of chlorine the equilibrium is closer to $\ce{ClO-}$ while - in comparison - for iodine the equilibrium is closer to $\ce{IO3-}$.
The reason for the lower electronegativity of iodine compared to chlorine is that iodine has two more electron shells than chlorine, meaning that the outer electrons of iodine are further away from the nucleus.

From the comments, one apparent reason is that the $I-O$ bonds, being weaker than $Cl-O$ bonds, are more labile.
There may also be a steric effect.  The +5 oxidation state ions are pyramidal, and we have to "squeeze" multiple oxygen atoms around "one side" of a (positively charged, thus somewhat shrunken) halogen atom.


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Reduction of chlorate with phosphorus
Given
$$\ce{10KClO3 + 3P4 -> + 3P4O10 + 10KCl}$$
If 52.9 grams of $\ce{10KClO3}$ reacts with excess $\ce{P4}$, what mass of $\ce{3P4O10}$ could be produced? How many moles of $\ce{KCl}$ could be produced?

I am not sure if I have to calculate the excess mass of $\ce{P4}$ and then find the mass of $\ce{3P4O10}$ or if it is just telling me that $\ce{P4}$ is excess reactant. If it is calculating the excess mass, how would I do it if I do not know the original mass of  $\ce{P4}$?

When a question says Excess, it pretty much means "you don't need to worry about this chemical". The question is saying KClO3 is the Limiting reagent.

The number of moles produced for each product will be the number of moles times of the limiting reagent times the reaction stoichiometry (the coefficients).
So to solve this you determine the number of moles of $\ce{KClO3}$ and use that as the base for determining the amount of $\ce{P4O10}$ and the moles of $\ce{KCl}$ produced.  Then multiple by the molecular weight per usual to get your answer.


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Why don't labile protons such as -OH and -NH have a characteristic chemical shift?
Is it simply because the H instantly disassociates so it is a proton with zero shielding?

Many $\ce{-OH}$ and $\ce{-NH2}$ protons do have characteristic shifts.  However, their characteristic ranges tend to be much broader than those of $\ce{C-H}$ protons.
For example alcohol chemical shifts ($\delta$) range from roughly 1-5 ppm  while pheonlics range from 4-8ppm and carboxylic acid protons are from 10-13 ppm or so.  The ranges are 3-5 ppm in width, while similar ranges for $\ce{C-H}$ are 1-2ppm in width.  Amines and amides are somewhat in between these two in terms of width of the ranges.  So, saying they are not characteristic is an unhelpful description of the actual situation.
The next (implied) question is why would the ranges be broader for these groups?
Chemical shift is a measure of electron density near the proton being measured.  The magnetic moment of electrons 'shields' the proton from the external magnet, changing the amount of energy require for the quantum mechanical transition of nucleus aligned with field versus opposed.  Normally, we consider electron density to be a reflection of the bonding arrangements.  However, for  $\ce{-OH}$ and $\ce{-NH2}$ protons, hydrogen bonding is common.  As such, often times nearby functional groups or NMR solvent are also contributing electron density, increasing the variability of the observe chemical shift from case to case.
Finally, in some cases the hydrogen bonding is dynamically changing slowly compared to the NMR time scale.  In this cases you will observe very broad peaks.  For example, water in $\ce{DMSO-d6}$ (a high-viscosity solvent) occurs around 3.3ppm.  In very old bottles of the solvent (with lots of atmospherically absorbed water), I have observed this peak as much as 1.5 ppm wide baseline to baseline, always with a peak at 3.3ppm.  However, in $\ce{CDCl3}$ water tends to have a pretty sharp peak at 1.5ppm (presumably because of limited water solubility).

Labile protons, such as alcohols and amines have protons which exchange with other labile protons (typically water), and what is observed in the NMR spectrum is an average chemical shift for these species. The actual position will depend on the relative concentrations of the exchanging species, and the exchange rates.
If you consider two protons in very slow (or no) exchange, you will observe two independent peaks. Two protons that exchange so fast that you cannot tell whether a proton is attached to one molecule or the other will appear as a single averaged peak, with a chemical shift at a population-distributed average of those two exchanging peaks. For protons that exchange at an intermediate rate, peaks will become broadened, and have their chemical shift move towards the peak that they are exchanging with.
Hence, alcohols, amines, amides in many common solvents will have peaks that appear broad or have coalesced with the residual water peak. So, for example, the normal residual water peak in chloroform comes at 1.6. As the amount of labile proton signal increases in the sample, either from more water or alcohols etc, the shift of residual water slowly moves to higher chemical shift. Initially, the water and other labile peaks will appear as separate, but broadened peaks. As the concentration of these increase, the peaks broaden more and move together. At infinite concentration of labile peaks exchanging with water, the peak will appear at 4.7ppm.
If you exclude all water from your sample, and other pathways of exchange, the these peaks will have very characteristic chemical shifts, just as other protons do.


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Why do NMR spectrums have the higher chem shift values closer to x-axis?
Im not understanding what the intuition is behind why the graph is represented so that the higher chem shift value is closer to x-axis (shouldnt it be the other way around?)
I understand that downfield = deshielded and will have higher ppm value which is closer to x-axis, but i dont understand why this is reversed to how an x/y graph is usually represented.? Thanks for any help

Just convention. There is no good reason for it as far as I know.

I heard about reversed scale with 10=0; 0=10 IIRC

@Mithoron that was the old tau scale.  The delta scale is an atrocity on the English language, because we talk about higher shifts as more deshielded, leading to implicit double negatives when comparing alkanes to alkenes for example.  As a side-note we don't usually call that the x-axis when referring to NMR.

The chemical shift axis is the $x$-axis. I don’t understand your question.

This convention stems from historical reasons only. In the early days of NMR, there were a number of different ways to report NMR data; values for 'shielding constants' of nuclei was of most interest, rather than what we commonly refer to as chemical shifts. The two most common scales were the tau and delta scales . The tau scale had the chemical shifts running in the opposite direction, with increasing shielding constants running from left to right.
A third scale, the sigma scale was very commonly used, especially in heteronuclear analyses, and again had the shielding constant increasing to the right. This scale was favoured often by theorists, as it was usually a direct measurement of the shielding constant.
A mixture of tau and delta values was regularly seen in the literature up until IUPAC made their initial recommendations, in 1970. This was to adopt the delta scale, and report chemical shift relative to TMS at delta 0ppm, as values of chemical shift, which are dimensionless and, hence, field independent.


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Naming of bicyclo[2.2.1]hept-5-ene-1,4,5,6-tetramethyl-3-bromo-2-ethyl carboxylate

I've synthesized a molecule (depicted above) using a Diels-Alder reaction, and I can't find it in any database (I used scifinder for my search), so I need advice about the name.
The name I came up with, based on similar structures is bicyclo[2.2.1]hept-5-ene-1,4,5,6-tetramethyl-3-bromo-2-ethyl carboxylate. Is that correct?

ChemDraw gives me: ethyl 3-bromo-1,4,5,6-tetramethylbicyclo[2.2.1]hept-5-ene-2-carboxylate, so you're close at least. You may well be right, depending on the method of naming you're using.

The individual parts of the proposed name are correct; however, their order of citation in the name is wrong.
The name of the parent hydride is ‘bicyclo[2.2.1]hept-2-ene’. Note that the ending ‘ene’ receives the lowest locant possible (here: ‘2’ since the bicyclic ring system is numbered starting with one of the bridgeheads).

The principal characteristic group is expressed by means of a suffix, which yields the name ‘bicyclo[2.2.1]hept-5-ene-2-carboxylate’. Note that now the characteristic group receives the lowest locant possible (here: ‘2’), which
is cited immediately in front of the suffix. Thus, the locant for the ending ‘ene’ in the functionalized parent hydride is changed to ‘5’.

The other substituents are cited as prefixes in substitutive nomenclature: ‘3-bromo-1,4,5,6-tetramethylbicyclo[2.2.1]hept-5-ene-2-carboxylate’

Finally, the name of the ester is formed according to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), by placing the ‘hydroxylic’ component in front of the name.

P-65.6.3.3.1 Monoesters
Monoesters formed from a monobasic acid and a ‘monohydroxylic’ component are named systematically by placing the ‘hydroxylic’ component denoted by an organyl group (alkyl, aryl, etc.) in front of the name of the acid component expressed as an anion derived from the appropriate acid (…).

Therefore, the preferred IUPAC name (PIN) of the compound given in the question is
‘ethyl 3-bromo-1,4,5,6-tetramethylbicyclo[2.2.1]hept-5-ene-2-carboxylate’.


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Lithium ion battery - measure C rate
How does one determine the C rate of an aged cell?
I would like to find the discharge rate just before any cliff edge of capacity loss. Can the C rate of a cell be estimated by the cells rated heat dissipation and measured internal resistance? Where I could find the current that will use up all the cells heat dissipation capacity.
I have a number of recycled laptop battery 18650 cells.
Are there any negative effects of the cells being below room temperature?

see: http://batteryuniversity.com/learn/article/what_is_the_c_rate

To determine the C-rate, you should charge and discharge two cycles very slowly say at 0.1 mA, between say 1.5 V to 4.3 V for your Li-ion (unless it's a very old laptop battery it should be LiCoO2), and measure how long it'll take. The C-rate would be your capacity divided by one hour.
Don't know of a way to measure it using heat or internal resistance.
Batteries generally have an adverse effect when heating up, so unless it is much below freezing, they should be ok.

1.5-4.3v is pretty extreme isn't it, very little capacity is past 3-4.2v from what I have seen? I already have measured the capacity of my cells, but that doesn't help me choose a C rate, it only lets me know what current I would get for the C rate I do choose.

A fresh LiMnO2 battery will have max around 3.5 V. A new LiCoO2 battery around 4.2 V, so it'll depend on what type of Li-ion battery you have. Choose the voltage range to determine the capacity.
Not sure what you mean, as the C-rate is just another way to express your discharge rate... If you know what current you want to choose, and you know your capacity, you can just divide to work out your C-rate.


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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?

You could use IR, but IR is more limited in specificity. With IR n-pentane looks very similar to n-hexane for example. IR would be cheaper and easier to maintain, so there are pluses.

@MaxW Using IR to distinguish n-pentane and n-hexane would indeed be a bad idea. Perfect for distinguishing isomers of ethyltoluene though. IR as a technique has may advantages over MS such as determining aromatic substitution, ring isomers, cis/trans isomers, ring junctions etc. Perfect for fatty acid analysis where only subtle differences in double bond position exist. Like all methods, the key is to pick the right tool for the job.

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.

Wow, Bruker does more than NMR?


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Calculating equilibrium constant given Ka value?

$\ce{NH_3(aq) + H_3O^+(aq)  <-> NH_4^+(aq) + H_2O(l)}$
$\ce{K_a}$ for $\ce{NH_4^+}$ is $5.6\times10^{-10}$.

I am completely lost and I have no idea how to approach this question.

Totally lost leads to totally lost.  You have to give us something to work with here.  What do you know?

I know how to find the equilibrium constant with concentrations, but I don't know how I could get the concentrations of the substances with the Ka value

try writing out the Ka expression for NH4+, then compare it to the equilibrium expression.

Never mind, I figured it out. All I had to do is reciprocate the value since Ka is the reciprocal of the equilibrium constant for that reaction.

Good job on figuring it out.


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Titration curve, how to find equivalence point
If you have a graphed titration curve and have points near the equivalence point like
6.85 pH at 54.60mL titrant added and
8.01 pH at 54.73mL titrant added
is there any good way to find a more exact volume at which the equivalence point was reached other then taking a guess between those two volumes?

You can take the derivative of the graph to find exactly where the slope is the steepest.

how do you do that? I know what a derivative is, but derivative of the graph?

So essentially the slope of your slope. I've only ever done it using some wacky program in my school. But you can use excel to solve for the slope at each point, then graph that vs volume. This graph will tell you which of your volume points is closer to the actual eq. point, but unfortunately won't give you any new volume points.

Hmmm. Alright then.

With the limited information given assume the equivalence point is pH 7.00, and do a linear interpolation between  the two known points.


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How to calculate the electrochemical potential of Iron(II/III) and saturated calomel electrode?

Set up an electrochemical cell where the cathode is the system in question B, $\ce{Fe^2+ -> Fe^3+}\ E = 0.68~\textrm{V}$, whereas the anode is a saturated calomel electrode ($0.244~\mathrm{V}$ relative to standard hydrogen electrode). Determine potential of this electrochemical cell.

\begin{align}
\ce{Fe^2+ +e^- &-> Fe^3+} & E &=0.68~\mathrm{V}\\
\ce{2Hg &-> Hg^2+ +2e^- } & E &=0.244~\mathrm{V}
\end{align}
Usually, I just do $\text{Cathode (Fe) - Anode (Hg)}$ and get the correct answer. However, the correct answer is listed as $1.16~\mathrm{V}$. There are only two numbers to manipulate. I found that
$$E(\ce{Fe}) + 2 E(\ce{Hg}) = 0.68~\mathrm{V} + 2\cdot 0.244~\mathrm{V}  = 1.168~\mathrm{V}$$
Why is this? The amount of substance is not supposed to effect the electro-potential.

Please check your equations. Calomel is $\ce{Hg2Cl2}$! The equation should be $\ce{2Hg <=>Hg2^{2+} +2e-}$.

First of all, you have to use reduction potentials as IUPAC considers the reduction potentials to be called the standard potentials.
It is not clear in your question whether you have used reduction or oxidation potentials.
Secondly, your equations don't make sense. The first equation shows Fe in +2 state getting reduced to Fe in +3 state.
Your second equation shows oxidation of mercury from pure state to +1 state.
And you are right that stoichiometry should not affect the EMF of cell.
The correct cell reactions will be iron(II) getting oxidised to iron(III) and Mercury(I) being reduced to Mercury(0).
At Cathode, Reduction of mercury takes place.
At Anode , Oxidation of Iron takes place. But we always use only reduction potential to calculate EMF. Don’t use oxidation potential of one with reduction potential of the other.
And yes EMF= E(Cathode) - E(Anode) provided both are reduction potentials.
But with the given electrode potentials I see no way to get 1.16V as an answer for the EMF of cell.
I would suggest that you check the question and electrode potentials again. And check the answer key again. Maybe there is an error there. It would be best to check about that with your course instructor.

See, it's calomel electrode. There is no mercury (II).

I see. I didn't notice that for some reason. I only saw the equation with Hg(II)  and assumed that was correct. It must be Hg(I) not Hg(II).


text:
acid-catalyzed ring opening of epoxide

I have some questions about the mechanism above:

why does ethanol attack the tertiary C before O-C bond breaks?  Shouldn't the mechanism proceed via an $S_{N}1$ mechanism?
Why is ethanol, as a weak nucleophile, able to attack tertiary carbon?
In a lecture it's said that a partial tertiary carbocation is involved, but left secondary carbon is also losing its electrons to oxygen, therefore becoming a partial carbocation; and if two carbon are both partial carbocation, then the tertiary carbon would be more stable and ethanol should attack the secondary carbon.

The reaction proceeds via a transition state, where the more stable one is the one to go for.
As you said, the tertiary carbon would form a more stable carbocation than the secondary carbon; therefore, if ethanol were to attack the secondary carbon, the transition state form would be relatively unstable compared to an attack at the tertiary carbon.

Thank you! I'm not sure if I fully understood it, but mentioning the transition state really helped.

Ring-opening reactions can proceed by either SN2 or SN1 mechanisms, depending on the nature of the epoxide and on the reaction conditions.
If the epoxide is asymmetric, the structure of the product will vary according to which mechanism dominates.
When an asymmetric epoxide undergoes solvolysis in basic methanol, ring-opening occurs by an SN2 mechanism, and the less substituted carbon is the site of nucleophilic attack.
In the case of acid catalysed ring opening, the mechanism is not exactly SN1 or SN2. It is a hybrid of both.
In the first step, the oxygen gets protonated.
Second step the C-O bond starts to break ( the C with more alkyl substituents is preferred for cleavage of C-O bond. In your case the tertiary carbon atom is preferred to the secondary C atom)
Third step the nucleophile ( or alcohol in this case) attacks the electrophilic carbon (the C with which C-O bond is still breaking). Unlike in SN1 mechanism the reaction does not wait for the carbocation to be formed completely. The incoming nucleophile attacks the carbon from a direction opposite to that of direction of C-O bond because C and O are still bonded to some extent.
So hence the mechanism proceeds similar to what you have shown.
In acid catalysed ring opening the nucleophile attacks the more substituted carbon because this carbon has more fractional positive charge then the secondary one.


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Theoretical yields and determining the number of moles in a reaction
I'm working on a question which asks me to do two things. I've balanced the related equation as:
$$\ce{6H2O + 6CO2 -> C6H12O6 + 6O2}$$
Now, if I have $44$ grams of $\ce{CO2}$ and $36$ grams of $\ce{H2O}$, how many moles of glucose will I have? The molecular mass of $\ce{CO2}$ is $44\mathrm{~g/mole}$, and that of $\ce{H2O}$ is $18\mathrm{~g/mole}$. Do I need to determine how many moles of each element are in the certain amount of grams for each compound, and then use that to determine how many moles are in the glucose?
Secondly, for the same amounts of $\ce{CO2}$ and $\ce{H2O}$, I need to figure the theoretical yield. Am I correct in saying since for ever 6 molecules of $\ce{H2O}$ and $\ce{CO2}$ I get 1 molecule of glucose, I would do
$$36\text{ grams of } \ce{H2O} \times \left( \frac{1~\mathrm{mol}~\ce{H2O}}{18\mathrm{g}~\ce{H2O}} \right) \times \left(\frac{1~\mathrm{mol}~\ce{C6H12O6}}{6~\mathrm{mol}~\ce{H2O}}\right)$$
Then take the same for $\ce{CO2}$? I'm unsure of what to do with the values after that.

You are correct in your thinking.
First, knowing the mass of the compounds, use the molar mass relationship for each (g/mol) to convert them all to moles. This is because when it comes to reactions, compounds react on a mole or number basis.
This lets you know the "starting amounts" of $\ce{H_2O}$ and $\ce{CO_2}$. Now, our forum has a policy against providing direct answers to homework, so I'll give examples. If you hypothetically had, say, 12 moles of both water and carbon dioxide, then using the numbers in the balanced chemical equation, they can fully react to make 2 moles of glucose. Thus, 2 moles is your theoretical yield. However, in another case, if you had 9 moles of water and 14 moles of carbon dioxide, the $\ce{H_2O}$ limits the amount of glucose you can produce, and it is the limiting reactant. (14 moles of $\ce{CO_2}$ would require 14 moles of $\ce{H_2O}$. There will be some $\ce{CO_2}$ left over because you are limited by only 9 moles of $\ce{H_2O}$). In this hypothetical case, the theoretical yield would be 1.5 moles of glucose. (Your problem may or may not involve limiting/excess reactants).
And if you're wondering, it's called "theoretical" yield because if you were carrying out this reaction in real life, some inefficiencies in combustion would yield less than the calculated amount. Hope this helps.

This is immensely helpful, thank you! So since 44g of CO2 is equal to one mole, and 36g H2O is equal to 2 moles, and I need 6 moles of each for a proper reaction without any left over, then the theoretical yield would be 1/6 a mole of glucose correct?

It's not that you need a minimum of 6 moles of each to make a complete reaction. What the equation tells you is that carbon dioxide and water react in a 1:1 ratio to form 1/6 the amount of glucose. Since you have 2 moles of water and only 1 mole of carbon dioxide, your amount of carbon dioxide determines the maximum amount of glucose you can make. And like you said, 1 mole of carbon dioxide is stoichiometrically equal to 1/6 mole of glucose.


text:
Inter-classification of organic chemistry functional groups (carboxyl vs alcohol & ether vs ester)
I have a few questions about whether organic functional groups can be classified as another.
First, can carboxylic acids be also be considered alcohols (they have $\ce{-OH}$)? If not, can they at least behave like alcohols?
Similarly, can esters ($\ce{RCOO-R'}$) be considered ethers because the $\ce{R'}$ group is linked to the other half of the ester via an $\ce{-O -}$ linkage? In whichever case, what are some chemical characteristics that distinguish ethers and esters?

See also Glossary of class names of organic compounds and reactivity intermediates based on structure (IUPAC Recommendations 1995)

related http://chemistry.stackexchange.com/questions/10713/are-all-amides-also-amines

Carboxylic acids are not also alcohols.
In general carboxylic acids are just that acids. They are prone to ionizing in water. Alcohols in general don't ionize in aqueous solution, and they certainly don't ionize in water to give acidic solutions.
Esters are not also ethers.
Esters react with water to form a carboxylic acid and an alcohol. Ethers are relatively stable in water.
PS - It isn't a bad thought! Everyone who has taken organic chemistry has had to ponder the points raised by your question.

"Alcohols in general don't ionize." Not especially true -- they just require a considerably higher $\mathrm{pH}$ compared to carboxylic acids. Phenol's $\mathrm pK_\mathrm{a}$ is actually quite modest, only $\sim 10$.

@Brian, ok I qualified it a bit more...

Would it make sense to classify a pistol as a type of pipe? The barrel of a pistol is a cylindrical metal tube, after all.
Clearly, no. The other key defining components of a pistol (stock, trigger, etc.) combine with the cylindrical metal tube to radically change its function and behavior.
Similarly, even though a carboxyl group does contain an $\ce{-OH}$ group bonded to a carbon atom, its behavior is sufficiently different from that of an alcohol that it makes no sense to group them together. A parallel argument holds for ethers and esters as well, as ably noted by MaxW.


text:
What is the balanced equation for H2O2 + KI + dish soap
I am not sure how to balance $\ce{H_2O_2 + KI + dish soap}$.   Is the dish soap involved in the equation or just to show the produced gas?  Is it a double replacement or does the $\ce{KI}$ just act as a catalyst and $\ce{H_2O_2}$ is deconstructed to $\ce{2H_2O + O_2}$? It has been a while since I had chemistry.

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

Ah!! The elephant toothpaste experiment.
First thing first, this reaction is actually a decomposition experiment as the following equation.
$$\ce{2H2O2 -> 2H2O + O_2}$$
Second, $\ce{KI}$ acts as a catalyst in this experiment. You also can use $\ce{MnO4}$ too as the catalyst for this experiment.
Third, the dish soap makes the product of the experiment more soapy and bubbly. So, you can see it will create look like a toothpaste. If you didn't put the dish soap, it will only look like gas bubbling out if the solution.

I haven't seen this reaction, but I'd expect a solution that was white and somewhat opaque like milk, i.e milky. I wouldn't think that the solution would get thick like a toothpaste.

It's not viscous like toothpaste. You should search one vudeo and see how is it.

Here is one such video. https://sciencebob.com/fantastic-foamy-fountain/ Really looks more like canned shaving cream and is expelled from the container.


text:
Determine moles of limiting reagent
In the reaction involving 0.1 g of calcium nitrate and 1.5 g of sodium phosphate, what is the number of moles of the limiting reagent?
Here is what I have so far:
Balanced equation:
$$\ce{3Ca(NO3)2 + 2Na3(PO4) -> Ca3(PO4)2 + 6NaNO3}$$
$\ce{Ca(NO3)2}$ has MW = 164, mass = 0.1 g and determined moles using n=mass/MW = 0.000609
$\ce{Na3PO4}$ has MW = 164, mass = 1.5 g and determined moles using n=mass/MW = 0.0091496
I was able to identify limiting reagent by:
g $\ce{Ca(NO3)2}$ --> n $\ce{Ca(NO3)2}$ -->n $\ce{Na3(PO4)}$--> g $\ce{Na3(PO4)}$
    0.1 g --s1-> 0.000609 --s2--> 0.000406 -s3--> 0.06658 g
S1 = n = 0.1/164 = 0.000609
S2 = want/have x $\ce{Ca(NO3)2}$ = 2/3 x 0.000609 = 0.000406
S3 = m = nxMW = 0.000406 x 164 = 0.06658g
Therefore as all $\ce{Ca(NO3)2}$ did not use all the $\ce{Na3(PO4)2}$ that means the $\ce{Ca(NO3)2}$ is the limiting reagent.
Now to answer the question. What is the number of moles of the limiting reagent. I am not sure if it is 0.000609 OR 0.001828 (which is moles multiplied by the stoichiometric coefficient, which is 3).

Welcome to Chemistry! This seems like a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

Sorry this would be my first post on a forum. I have edited what i have so far under the question. Hope you can clarify my query!

Ok, please excuse my ignorance as mentioned above this is my first thread on a forum. I believe i have fulfilled the requirements of the question type to allow it to be on-topic. Can anyone assist me in my confusion?

You can already see from your first calculation that Ca(No3)2 is the limiting reagent, because you had more moles of Na3PO4 than Ca3(No3)2 and the equation tells us that 3 moles of Ca3(No3)2 react with 2 moles of Na3PO4. How much did exactly react? 0.000609 moles Ca3(NO3)2, 2/3 x 0.000609 moles Na3PO4 and the amount of moles NaPO4 left after the reaction: 0.0091496 - (2/3 x 0.000609).

user21398 I understand exactly what you have said BUT I am still confused!
Would the solution to this problem of "What is the number of moles of the limiting reagent?" be 0.000609 OR 0.001828

It's just adding up to the existing answer since I don't have enough rep to comment. 0.000609 is the no. of moles of limiting reagent. You don't multiply it by it's stoichiometric co-efficient because it is the no. of moles you were given. Stoichiometric equations just signify the proportions in which compounds react.


text:
Kinetic isotope effect in nitration of benzene
What is the effect of changing all the hydrogen atoms in benzene to deuterium? For example, which of the following reactions occurs at a faster rate?

If we were to still compare, the $Ar H$ molecule would take the $NO_2 +$ ion easiest right?

Page 513 of Organic Reaction Mechanisms by Bansal references the work of Melander in Arkiv för Kemi, 2, 211 (1950). This states that the rates of nitration for benzene vs benzene-d6 are about the same, implying that the addition of the nitronium ion is rate controlling and not the weakening of the C-H / C-D bond. However in case of sulphonation H containing benzene reacts faster than D containing benzene.

This would imply an early transition state for the nitronium-benzene complex.

This is an example of an inverse kinetic isotope effect.
When a transition state increases in coordination then differences gained from vibrational bending energy for the $\ce{C-D}$ vs $\ce{C-H}$ bond favor $\ce{C-D}$, lowering the transition state energy relative to starting materials, and increasing reaction rate.
In this case, the intermediate is the protonated benzene changing a reaction center from $\mathrm{sp^2}$ to $\mathrm{sp^3}$, increasing coordination.
The perdeuterated benzene will react faster. The effect will probably be on the order of half again as faster.

"order of half again as faster.", can you explain it a bit

I would guess the rate constant to be 50% higher. However, Beerhunter's data says my guess is incorrect.

I would refer to the book and see what happens

Please provide your source.


text:
Why is this a Lewis structure of ethyl acetate?
I read somewhere that the structure in the following picture is a Lewis structure of ethyl acetate.

However, I don't understand why:

The $\ce{C}$ has a positive charge, but only three bonds
The $\ce{C}$ does not satisfy the octet rule.

The gist is that no bond is purely ionic or purely covalent. Those are two extremes like pure black and pure white. You drawing shows how the C=O double bond can have some ionic character.

The two rules you cite are guidelines for the 'best' Lewis structure.  However, in the example provided, all electrons are accounted for and all atoms are connected making it a 'valid' Lewis structure.
In the context of resonance forms, this is likely a minor contributor.
However the electrophilic center in ethyl acetate is the carbon you marked as positive, this form has some descriptive  and predictive power in discussing reactions of ethyl acetate.  In general, for $\ce{R2C=O}$ molecules (esters, aldehydes, ketones, etc), The carbocation (sextet) carbon has some relevance.


text:
How to get 3D molecular structures for 270 compounds?
I need to calculate 10 properties for 270 molecules using Gaussian 09. Could you please suggest me what is the best and shortest way to have the structure of 270 molecules (in the format .com or .xyz)?
Is there any website which collected all the molecular structures?

I'm confused. You want 270 random molecules? Or you have a list of 270 molecules that you'd like to run? (In which case, what format do you have the list?)

What keeps you from drawing or writing the initial geometries by yourself?

@Geoff Hutchison: I have the list of 270 molecules. I found the .xyz for some of them but for most of them I did not find .xyz

@Klaus Warzecha: As you know drawing 270 molecule is time consuming procedure and unfortunately I do not have enough time for that. I first want  to find a faster way

There are a few resources that can get you what you want, but you'll probably want to consider some level of scripting automation. A search for 270 molecules isn't enormous, but doing it by hand would still be tedious.

PubChem: Run by the NIH, it offers many types of searching, and 3D coordinates for most compounds.
ChemSpider: Run by the RSC, it allows searching by name and has 3D structures for many (if not most) compounds. I'm less sure on quality, since they don't clarify how they generate their 3D coordinates.
NIH Chemical Resolver: Run by the NIH as well, it converts between different representations, including name to 3D structure.

Each offers automation tools, including Python interfaces called PubChemPy, ChemSpiPy, and CIRpy.
My choice would probably be the chemical resolver, partly because the access is very simple to automate:
https://cactus.nci.nih.gov/chemical/structure/"structure identifier"/"representation"
So you'd set up a URL like: https://cactus.nci.nih.gov/chemical/structure/aspirin/sdf?get3d=true
The result will be in SD format, but it's easy to convert that to XYZ or Gaussian files with Open Babel.

Dear Geoff  Hutchison. Thank you very much for your nice suggestions. I have another question: I found that the structure e.g. for Ethanol is different when I used the .xyz from two different resources. How we can find that the initial structure is correct? (I am new Gaussian user)

Most likely both are correct. BTW, are you going to run geometry optimization on all 270 molecules?

I need to calculate HOMO, LUMO, Ionization Potential, Heat of formation, Dipole Moment, Electronegativity, Total energy at 0 K, Entropy at 298 K for 270 molecules using DFT/B3LYP/6-31G(d,p) and then use these properties as descriptors in QSPR study. Which type of calculation is proper for that?

Don't calculate the values you can get experimentally.


text:
How can I tell which parts of this example drug are necessary parts of its pharmacophore?
$\hspace{10mm}$
$\hspace{10mm}$(Click image for larger version)
This picture identifies a generic drug-enzyme complex taken from an exam script. There is no name given.  One part of the exam question states:

Propose a structure based pharmacophore for the family of drugs to which this one belongs.

I understand that a pharmacophore is the minimum structure required to exert a certain biological activity. Therefore, my answer would go something like the following:

We would need a drug with similar steric conformities to fill the entire active site (competitive inhibition) and that has some form of interaction at either end which keeps it in place, e.g., $\pi-\pi$ stacking with Phe groups or the H-bond and electrostatic interactions with Arg.

But would that paragraph above be a viable answer? Would this drug work just as well if it didn't have the $\pi-\pi$ stacking with the Phe groups, for instance? Are the interactions between the drug and the Lys and Glu residues dispensable - maybe that region of the drug is only there for its length?

From this? You can't. You need more information. Examples:

Do you know other drugs of this family? Which chemical features do they have in common? The interactions mediated by these features are likely important then.
Do you know how the drug is supposed to function (you mention competitive inhibition)? What does the natural substrate look like, and what are the similarities and dissimilarities to your example drug?

If you know only the figure shown above, you're really asked to identify electronic and steric features of the drug, that can possibly interact with the amino acid residues in the active site shown. In this case, all amino acid residues are capable of favourable interaction with a part of the example drug, so I would mention which interactions are present.
There are many good ionic interactions/H-bond interactions, which are qualitatively the strongest.
One possibility is that it is sterically unfavourable to jam something into the hydrophobic pocket, but the payoff is, that this is what stabilises a certain (active or inactive) conformation of the protein, and the ionic/H-bond interactions simply serve to increase binding affinity / to keep the drug in place.
Another possibility is that the interaction with the hydrophobic pocket only increases binding affinity. The drug then might share the same ionic interactions as the real substrate, but placed differently in space, so they don't induce the same conformation change as the natural substrate.
The point is, without knowing what the action of the drug is, and some knowledge about other drugs of the same family or the natural ligand/substrate, anything beyond a listing of the interactions you can spot between the amino acid residues shown, and the chemical features of the drug will really just be guess work :-).


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Why p,d and f orbitals have non spherical shapes
I know what orbitals are. The orbitals are solutions (also called wave functions) to the (time-independent) Schrödinger equation[ref].
What I don't understand is why p,d,f orbitals have non spherical and s has a spherical shape. Our universe is made up of many circular or spherically (not perfect)  shaped objects like stars, planets, bubbles, droplets, wheels etc. I always thought (or guessed) circular and spherical shapes are the most efficient shapes in our universe. Then why p,d and f have non-spherical shapes or non efficient? Is there any other logic other than solution to Schrodinger equation which explains these unique shapes?

http://chemistry.stackexchange.com/questions/14247/why-do-atomic-orbitals-have-their-unique-shapes http://chemistry.stackexchange.com/questions/39987/please-explain-the-shapes-of-the-orbitals

A separate answer would be that the Laplace functions are a complete basis set, and a set of spherically-symmetric functions can not be a complete basis set.

There is no such thing as efficient shape.


text:
What is the difference between reversible and irreversible work done in adiabatic process?
The work done in an adiabatic reversible process is given by:
$$W = R\frac{\Delta T}{1-\gamma}$$
The work done in an irreversible process is given by:
$$W = P_{ext}\Delta V$$
By the 1st law of thermodynamics, the change in internal energy must be equal to the work done. The change in internal energy is given by:
$$\Delta U = nC_v\Delta T$$
where $C_v$ is the molar heat capacity at constant volume.
Can't this equation be applied to both the processes? Why are there different formulae for the two types of adiabatic processes?

Even the formula for reversible expansion, is pooison's ratio is expanded, we get the same result as heat capacity*change in temperature.

Related: Calculate Work Done for Reversible and Irreversible Adiabatic process, Reversible and Irreversible adiabatic expansion, and Why does the energy difference of a reversible process not equal that of an irreversible process in an adiabatic expansion?

Forget about the book.  Show us how YOU would solve the entire adiabatic expansion problem of an ideal gas (a) for the reversible case and (b) for the irreversible case in which the external pressure is held constant.  Let's see what you can do.

Lets say 1 mol of gas in a piston expanded from v1 litre to v2 litre. Now we want to calculate the work done. a) work done= -p(external)*dv but p(ext)=p(gas) for reversible process. PV^gamma(poisson's ratio)=k(some constant). P=k/v^gamma. Putting in work equation and integration. Rest answer in next comment.

Work=-(k/1-gamma)|v/v^gamma|. Using the PV^gamma=k relation. Work=-(k/1-gamma)|pv/k| = -|pv|/1-gamma. Now using pv=rt and putting limits work= R(T1-T2)/1-gamma. Its the same as -molar heat capacity at constant volume*change in temperature. B) work =-P(ext)dv. Integrating from v1 to v2. Work=-P ext (v2-v1).

Now comes the problem. According to 1st law of thermodynamics q= Change in internal energy(du)+ work done. As it is adiabatic process q=0. Work=-du. We now that change in internal energy = heat capacity at constant volume* change in temp. So work becomes -heat capacity at constant volume*change in temp. This the same as for irreversible process. What am I doing wrong?

The different conditions is0thermal, adiabatic, reversible & non-reversible are given in this answer http://chemistry.stackexchange.com/questions/70596/how-to-calculate-the-final-temperature-of-a-gas-when-it-undergoes-adiabatic-expa/71038#71038

Your problem lies in your analysis of the irreversible expansion.
In the case of both a reversible expansion and an irreversible expansion, the gas pressure at the interface with the piston matches the externally applied pressure during the deformation. However, in the irreversible expansion, the gas within the cylinder is not uniform with respect to pressure or with respect to temperature.
So the interface pressure (which determines the amount of work done) does not match the average pressure of the gas within the cylinder. In addition, the pressure at the interface depends not only on the gas volume but also on the rate at which the gas volume is changing. So for an irreversible process, we have very little control over what is happening inside the cylinder.
The only thing we can control is the pressure at the interface $p_{ext}$ where the work is being done. We do this by whatever means necessary, including feedback control systems. In the typical scenario, we drop the pressure at the interface to a constant value, less than the initial pressure in the cylinder, and then hold it at that value until the gas volume has increased to the point where the gas in the cylinder is again at equilibrium. At this point, the gas pressure throughout the cylinder is described by the ideal gas law again.
If we follow this game plan for the irreversible process, even if the final pressure matches the value be obtain in the reversible process, the final temperatures and the final volumes will not match.
If you would like me to provide a detailed analysis of the irreversible case, illustrating in detail what transpires, I will be glad to do so. But right now, I'll give you a chance to ask some questions.
Irreversible and Reversible Expansions
Initial conditions:  $T_i, P_i, V_i$
Irreversible expansion:
$$\Delta U=-\int_{V_i}^{V_f}{P_{ext}dV}$$
In the irreversible expansion we are considering, $P_{ext}$ is controlled to be constant at the final pressure $P_f$ with $P_f<P_i$. The expansion is allowed to continue until the gas re-equilibrates at the final pressure and volume. Therefore, we have:
$$nC_v(T_f-T_i)=-P_f(V_f-V_i)\tag{1}$$
As noted in my discussion above, the ideal gas law cannot be applied to the intermediate states during the irreversible expansion, but it can be applied to the two equilibrium end states. Therefore, $$V_f=\frac{nRT_f}{P_f} \tag{2}$$
and
$$V_i=\frac{nRT_i}{P_i}\tag{3}$$If we combine equation $(1)$, $(2)$ and $(3)$, we obtain:$$C_v(T_f-T_i)=-R\left(T_f-\frac{P_f}{P_i}T_i\right)\tag{4}$$
We can solve this equation for $T_f/T_i$ and a function of $P_f/P_i$ to obtain:
$$\frac{T_f}{T_i}=\frac{1+(\gamma-1)(P_f/P_i)}{\gamma}\tag{5}$$
Note that this equation differs from the relationship one obtains if the expansion is carried out reversibly.
Reversible expansion:
For a reversible expansion, we have $$dU=nC_vdT=-PdV=-\frac{nRT}{V}dV$$ or $$C_vd\ln T=-Rd\ln V=-R(d\ln T-d\ln P)$$This integrates to:
$$\frac{T_f}{T_i}=\left(\frac{P_f}{P_i}\right)^{\frac{(\gamma-1)}{\gamma}}\tag{6}$$
Note from equations 5 and 6, that if the temperature ratios are the same in both cases (equal work), the pressure ratios are not, and if the pressure ratios are the same in both cases, then the temperature ratios are not (unequal work).

In both cases change in internal energy would equal the magnitude of work? Or not. If yes then change in internal energy= specific heat at constant V*dT. Both would have the same work done if I knew the limits for the integral.

Yes.  In both cases, the change in internal energy would equal the magnitude of work.  The two amounts of work could turn out to be different and the two changes in internal energy could turn out to be different in the reversible and irreversible cases.  Do you want me to do a reversible problem and an irreversible problem so that you can compare?  If somehow you could force the work and internal energy changes in the two cases to be the same, the final pressures and the final volumes will not be the same.  It is much easier just to set up two focus problems for comparison.

If you could do separate problems that would be really awesome.

I'll be glad to.  It shouldn't take much.  I'll be back tomorrow morning.  Bed time now.

See edited version of my answer.


text:
Logical way to compare dipole moment
I have to find the decreasing order of dipole moment of $\ce{CHCl3}$, $\ce{CH3Cl}$, $\ce{CH2Cl2}$, and $\ce{CCl4}$. Now I know that in $\ce{CCl4}$ there will be no dipole moment. Now to compare the other three compounds I don't have logical reasoning to compare the dipole moments but I have the data to compare but in the examination we won't be having these data so I was looking for any logical reasoning to compare the dipole moments of the remaining three compounds. Thank you

http://chemistry.stackexchange.com/questions/8977/difference-of-dipole-moments-of-dichloromethane-and-trichloromethane

Look at dipole moment as a vector with the head of vector pointing towards the more electronegative atom. Ccl4 has 4 vectors pointing outwards but these vectors cancel, hence 0 dipole. Chcl3, 1 vector from H towards c, 3 vectors from c to each cl atom. Note they have tetrahedral geometry. The net vector would be along one of cl atoms but as other cl atoms have components of their vectors opposite to the victorious cl atom, dipole would be reduced. Ch2cl2, net vector would be somewhere between the 2 c==>cl vectors but there are not much opposite components of vectors as compared to chcl3, it has greater dipole than the former. Ch3cl, net dipole is along the cl atom and further each H==>C dipole is increasing the magnitude of the vector. These problems are based on visualization and vectors help you visualize better.

The electronegativity difference between carbon and hydrogen is roughly the same as the difference between carbon and chlorine. So why is the magnitude of the dipole moment between methyl chloride and chloroform so different?

Correct me if I'm wrong but as all the atoms around c are not the same, the structure won't be perfect tetrahedral. Cl atoms are much larger than the H atoms and would face greater repulsion from each other than 2 H atoms. The shape would be altered in space and cl atoms would be spaced away from each other.

Now the explanation. In trichloromethane the cl atoms would be spaced far and their bond angle would be increased, thus decreasing the net dipole. While in chloromethane H atoms won't face much repulsion from each other rather they would feel more repulsion from cl, hence they will be closer and increase the net dipole.

The bond angles in chloroform and methyl chloride are all very close to the ideal tetrahedral angle.  Focus on the different bond lengths.  No need to respond to me, just incorporate your thinking into your answer

So c-cl bonds would be longer and have greater dipole. Then in trichloromethane you have 3 relatively bigger vectors contributing to the net vector and one small h-c dipole contributing. In chloromethane we would have only 1 bigger c-cl dipole and 3 h-c dipoles contributing towards it. The common vector arrangement is there are 3 vectors pointing upwards but away from each other and one pointing upwards. In one case upward pointing vector is the larger c-cl vector and smaller h-cl in the other. But we just switched their dipoles. Where am I wrong?

If I understand your comment then I think you are on the right track.  In both molecules the individual bond dipoles are aligned in the same direction.  The $\ce{C-Cl}$ dipole is larger due to the larger bond length.  Then just vectorially add the individual bond dipoles

Let us continue this discussion in chat.

Sorry, I don't have time now.


text:
3000 mg/L NH4-N solution, pH 7.8
Can anyone suggest a protocol for preparing 1 Litre ammoniacal solution (pH 7.8, NH4-N, 3000 mg/L) from NH4Cl and NaOH?
I have worked out that I need 10.7g NH4CL and 0.45 g NaOH -is this correct? I am assuming that 1st step is to prepare 0.45 g NaoH in 0.5 L distilled water and then to that add 10.7 g NH4CL solution in 0.5 L of  distilled water.Total volume is 1 L?
Or are there better methods?

You say you want "(pH 7.8, NH4-N, 3000 mg/L) from NH4Cl and NaOH" what does the "3000 mg/L" mean?!? Is that the total mass of NH4Cl and NaOH?


text:
Thermodynamic control or kinetic control?

In my notes, it is written that the case 1 product will be major one. The explanation given is as follows:-

However, since the pi clouds of alkenes are perpendicular to each other, the resonance factor will be reduced. Till the time resonance occurs, water would have already attacked the positive charge centre and hence the thermodynamic-controlled product will be favoured.

I don't get this thing!

RE: "Thermodynamic control or kinetic control?" That is just a bad way to look at reactions. All reactions are controlled both by thermodynamics and kinetics.

This appears weird to me since the thermodynamic control vs kinetic control  pathways are used to describe as to what will be the major product. Let me elaborate. The kinetic controlled reaction is one which "happens faster but leads to less stable products" and the thermodynamic controlled reaction is the one which "happens slower but leads to the more stable products." But in this example, the kinetic controlled reaction appears to be the thermodynamic controlled reaction as well.

The comparison of kinetic control and thermodynamic control is supposed to show you why reactions do not give you the most stable products. It should also introduce you to activation barriers, and what not. I am not sure if your class notes are correct. But take it with a grain of salt, it has been a while since I took organic chemistry

Well I can safely say that my class notes are correct. I just don't get the explanation.


text:
What is the amount of limiting reagent in this chemical reaction?

In the reaction involving $\pu{124 g}$ of aluminium and $\pu{601g}$ iron (III) oxide, what is the amount of substance in moles of the limiting reagent?

Here is what I have so far:
Balanced equation:
$$\ce{2Al + Fe2O3 -> Al2O3 + 2Fe}$$
$\ce{Al}$ has MW = $27$, mass = $\pu{124 g}$ and determined its amount using $n=\frac{\text{mass}}{MW} = 4.59$
$\ce{Fe2O3}$ has MW = $160$, mass = $\pu{601 g}$ and determined its amount using n=mass/MW = 3.756
I was able to identify limiting reagent by:
g $\ce{Al}$ --> n $\ce{Al}$ -->n $\ce{Fe2O3}$--> g $\ce{Fe2O3}$
$124 g --s1-> 4.59 --s2--> 2.295 -s3--> 367 g$
$S1 = n = 124/27 = 4.59$
$S2 = \frac{\text{want}}{\text{have}} \times \ce{Al} = 1/2 \times 4.59 = 2.295$
$S3 = m = n\times MW = 2.259 x 160 = 367 g$
Therefore as all $\ce{Al}$ did not use all the $\ce{Fe2O3}$ that means the $\ce{Al}$ is the limiting reagent.
The amount of substance in moles of limiting reagent $\ce{Al}$ (answer to the question) is $4.59$ OR $9.18$ (which is moles multiplied by the stoichiometric coefficient, which is $2$).

If you only start with 4.59 moles of Al, how are you going to react 9.18 moles of Al with the iron (III) oxide?!?

I think you have written the wrong number for moles of rust.  Regardless of the typo, the number of moles you have is the number of moles you have.  It is not doubled by coefficients.  The coefficient means you need twice as many moles of aluminum as rust for a complete reaction.  So... ?

I am confused as 1 mole of Al = 4.59 but we start with 2 molecules of Al, so don't we start with 9.18 mole of Al? This is my confusion about this. *Thanks for the typo I changed it.

$4.59$ is not "something" (you haven't specified the unit) of an $\ce{Al}$ mole, but your amount of $124\mathrm{~g}$ of $\ce{Al}$ is exactly $4.59$ moles of $\ce{Al}$, it's just a unit change. In reaction you wrote, every two moles of $\ce{Al}$ reacts with one mole of $\ce{Fe2O3}$. This may help you to find the solution.

Just calculate which reactant gives the lower amount of Al2O3.
From Al you have  $n_\text{Al$_2$O$_3$} = \displaystyle \frac{124 \text{ g Al}}{26.9815386 \text{ g/mol}} \cdot \frac{1\text{ mol Al$_2$O$_3$}}{2 \text{ mol Al} }= 2.30 \text{ mol}$,
From Fe$_2$O$_3$ you have  $n_\text{Al$_2$O$_3$} = \displaystyle \frac{601 \text{ g Fe$_2$O$_3$}}{159.6882 \text{ g/mol}} \cdot \frac{1\text{mol Al$_2$O$_3$}}{1 \text{ mol Fe$_2$O$_3$} }= 3.76 \text{ mol}$.
Clearly Al is the limiting reagent with $n_\text{Al} = \displaystyle \frac{124 \text{ g Al}}{26.9815386 \text{ g/mol}} = 4.60$ mol.
Check out this site for tutorials on limiting reagents:  http://www.grandinetti.org/limiting-reagents

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Shape of Weak-Strong Acid-Base Titration
I can understand why in any titration, the pH changes really quickly near the equivalent point.
However, in a titration of either weak base-strong acid or weak acid-strong base, the weak acid/base first has a steep rate of change in pH in the beginning, then slows down at the midpoint, then becomes steep again closer to the equivalence point. Why is this the case? (See picture below). Also, it seems that the weaker the acid/base, the more pronounced the initial quick rate of change.
Also, why is the "midpoint" (not the equivalence point) so special anyways?
You can refer to this picture. The steep drop portion is from 0-10 mL.

see: http://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html

There are also several question on this site about that.

Suppose $\ce{NH3(aq)}$ is being titrated with $\ce{HCl(aq)}$. The products are $\ce{NH4+ (aq)}$ and $\ce{Cl- (aq)}$. As the titration proceeds [$\ce{NH3}$] decreases and [$\ce{NH4+}$] increases. The presence of $\ce{NH3}$ and $\ce{NH4+}$ together constitutes a buffer system, in this case a weak base and its conjugate acid. When molarities of both species are relatively large, the buffering capacity is significant.
Halfway to the equivalence point, the molarities of both species are equal, which allows for the greatest possible buffering action. In the very early part of the titration, the molarity of $\ce{NH4+}$ is not great enough to allow much of a buffering effect, so the pH falls rapidly with the addition of $\ce{HCl}$. But more toward the point where [$\ce{NH3}$] becomes equal to the [$\ce{NH4+}$], the buffering action is increasingly able to "resist" changes in pH so the slope of the titration curve becomes more shallow. Past this point, the [$\ce{NH3}$] steadily diminishes to the point that buffering is no longer possible and the pH dives through the equivalence point as we expect.
Past the equivalence point, only $\ce{NH4+}$ is present so no buffering can occur and there are no inflections like the one that occurs at the start of the titration. The importance of the "midpoint" as it is labeled in the illustration is that pH vs. titrant volume analyses are sometimes done specifically to determine the K$_a$ or K$_b$ of a weak acid or weak base, respectively. Continuing with the $\ce{NH3}$ and $\ce{HCl}$ example, the K$_b$ for $\ce{NH3}$ is $[\ce{NH4+}][\ce{OH-}]/[\ce{NH3}]$. Halfway to the equivalence point, [$\ce{NH4+}$]=[$\ce{NH3}$] and we have K$_b = [\ce{OH-}$] or pK$_b$ = pOH.


text:
Can diastereomers form a racemic mixture?
In my book, it says racemic mixture is the mixture of + and - optical isomers and it is optically inactive. This means enantiomers (optical isomers which are mirror images to each other) can form racemic mixture. But can diastereomers (optical isomers which are not mirror image to each other) form a racemic mixture?

The short answer is no.
A racemic mixture is defined as an equimolar mixture of enantiomers. Since each enantiomer rotates plane polarized light equally, but in opposite directions, the overall optical activity is zero.
Describing diastereomers as optical isomers is incorrect. Diastereomers are stereoisomers (that is, compounds that have the same connectivity of atoms but differ by the orientation of atoms in space) but are not enantiomers. In many cases, diastereomers are not even chiral (have no optical activity). If there's a mixture of diastereomers that are chiral, this is not described as racemic. Most likely that mixture would be optically active, but even if it was optically inactive, that would just be coincidental and not considered racemic.
The situation where a mixture is made up of two enantiomers in equal ratios is so common (and important) that it gets the special name of racemic.

It's not like “enantiomers (…) can form racemic mixtures” and “Who else can form racemic mixtures?”; it's by definition. Also note that the term “racemic mixture” is discouraged.

Racemic Mixture (usage strongly discouraged)
The term is confusing since it has been used as a synonym for both racemate and racemic conglomerate.
Racemate
An equimolar mixture of a pair of enantiomers. (…)
Racemic Conglomerate
An equimolar mechanical mixture of crystals each one of which contains only one of the two enantiomers present in a racemate. (…)

(as in Basic Terminology of Stereochemistry, IUPAC Recommendations 1996).


text:
Making benzoyl chloride - yellow liquid and a precipitate?
I am attempting to produce benzoyl chloride from benzoic acid (I need to convert an isotopically labeled version, so I can't just buy the acyl chloride). I am reacting benzoic acid with excess oxalyl chloride in DCM, and the reaction looks like its proceeding quantitatively, but when I try to rotovap off everything except the product my solution turns pale yellow and I get a white solid.
Both the yellow liquid and white solid appear identical on TLC, and the NMR is consistent with benzoyl chloride, but the physical appearance doesn't seem to match.
Can anyone suggest what might be going on?

This isn't really an answer to your question, but supposed that you're planning to synthesize an ester or amide: Did you consider to activate the labelled benzoic acid in situ using tBOC or DCC?

Have you prepared benzoic anhydride?

Water is pretty ubiquitous. I wonder if TLC is the way to test for benzoyl chloride. Wouldn't benzoyl chloride readily hydrolyze back to benzoic acid?


text:
Is there any possibility of tautomerisation to en-ol form?

In the picture, you can see the bridge whose end points are in para relation (for making the image clear).
Now, what my friend said is that there is no possibility of tautomerisation as there is no H to the right of =O and to the left of =O, on removing H, the bridge will be loaded(?!?).
But according to me, positive charge on the bridge end-points may be unstable as the bridge is out of the plane of the planar molecule and making it sp2 hybridized isn't a kind thing to do. But on removing H from the bridge end point (for tautomerisation), there will be negative charge on the bridge end point, which is perfectly normal.
So will it tautomerise or not?

Have you taken non-classical ions into account? Your molecule is well suited for it: https://en.wikipedia.org/wiki/2-Norbornyl_cation

Is there any kind of anomaly related to Norbonyl derivatives? Please answer if you know. It will be a great help.

If you're asking specifically about the tautomerisation, then I made this comparative computational study using B3LYP theory at the 6-31g* level, and the results are:

So as is evident, the bridge causes the alkene to be dramatically strained. You can compute the Boltzmann ratio and find that the tautomerisation in the first structure is completely negligible at room temperature.
If you're talking about the formation of an enolate ion, the results shouldn't differ much, because the alpha carbon still needs to become sp2 hybridised, in order to delocalise the carbanion into the carbonyl group. And this, as we know, is the reason for the increase in acidity of protons adjacent to carbonyl groups.
All this is of course to the best of my knowledge.

Excellent answer, completely got what I was looking for! So finally, the intermediate step of formation of carbanion shall not be much affected due to bridging, as the carbanion(-ed) carbon will be still sp3 hybridised, right?

@ReeshabhRanjan Just to be clear, the difference in these two compounds has everything to do with bridging.

Note that the increased strain of an alkene at a bridgehead is well-known enough to have been named: Bredt's rule

Well as per your answer, giving enough heat will make the bridge carbon sp2 hybridised. I think the bridge will break easily later?

I can confirm Brian's findings, who beat me to it. On the DF-BP86/def2-SVP level of theory, the energy difference is $\Delta G = 238.8~\mathrm{kJ\, mol^{-1}}$. Which means, that the enol tautomer does virtually not exist.
The straining can be quite easily explained by two pictures.
The newly created $\pi$ bond will certainly prefer a planar arrangement. Due to the rigidity of the whole structure, this is not possible. In one case the angle deviates by $35.5^\circ$ from the ideal conformation. In the other case it deviates by $58.3^\circ$. And this is just looking at the most distorted angles.
Since the geometrical distortion is so big, there cannot be a good enough overlap between the p-orbitals of the new $\ce{C=C}$ bond and the oxygen, hence delocalisation and therefore stabilisation is not possible.

Under sufficiently vigorous conditions (strong base, high temperatures}, 3,3-dimethyl[2.2.2]bicycloctan-2-one will probably undergo homoenolisation and a ring opening. The product most likely is a bicyclo[3.2.1]octanone.

But under normal conditions, the strain should not be enough for the ring to open. Please consider normal condition only for tautomerisation.


text:
Is silicon dioxide a polar molecule or not?
Is $\ce{SiO2}$ is a polar molecule or not?
The difference in electronegativity between the two atoms is 1.7.
I read that $\ce{SiO2}$ doesn't exist as a single molecule in which the silicon atom is connected to four oxygen atoms. Can I consider it polar?

$\ce{SiO2}$ doesn't exist as a molecule rather exists as network of $\ce{Si}$ and $\ce{O}$ bonds with $\ce{Si}$ bonded to four $\ce{O}$ atoms. $\ce{SiO2}$ simply denotes their ratio in the complete structure. $\ce{Si-O}$ bonds are very polar but there are four $\ce{Si-O}$ bonds which cancel the dipole, making the structure as a whole, non-polar.

thanks a lot it's very now, one question if i may ask can attract or interaction happen with this molecule and other moleculue although SiO2 nonpolar.

Yes interactions can happen. Glass is made up of siO2 and we know glass and water are adhesive, as they form a concave meniscus. This is because of very weak bonds between the O atoms of sio2 and the H atoms of water.

thanks for the info. i was searching for this kind of explanation, so can i consider a some kind of  attraction even small one between SiO2 bonds and c=o bond ?

As c=o bond is polar, yes.

so it can be interaction between c=o and sio2, thank u so much for the clear info.

The meniscus is related to Si-OH groups on the surface of the SiO2 network covalent structure, see https://chemistry.stackexchange.com/questions/30791/why-is-sand-glass-polar-how-does-lattice-structure-change-polarity/30798#30798. The process of silanizing derivatizes Si-OH groups into Si-OCH3 groups, making the sample's surface non-polar.


text:
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.

Yes, both graphite and borazine are aromatic in nature as each ring of a plane have six π-electrons (similar to benzene). This aromaticity explains why graphite and borazine are unusually thermodynamically more stable than diamond.
That's why graphite has higher melting point than diamond, and on pyrolysis  diamond (unstable) turns to graphite (more stable). Graphite is only a few $\pu{eV}$ more stable than diamond. Diamond does not contain any delocalised electrons. Graphite contains one delocalised electron per carbon. These cause greater attraction between carbon atoms hence giving stronger bonds, and more stability to the structure. Graphite has greater van der Waals forces because of the oscillating delocalised electrons which induce temporary dipoles in graphite increasing attraction and stability. Diamond's all delocalised electrons are used in covalent bonding.
The standard enthalpy of formation of diamond ($H^\circ_\mathrm{f} = \pu{2.425 kJ/mol}$) is slightly larger than the enthalpy of formation of graphite, which is the most stable form of carbon at $\pu{25 °C}$ and $\pu{1 atm}$ pressure. At very high temperatures and pressures, diamond becomes more stable than graphite (Source).
The decay of diamond is thermodynamically favorable $(ΔG = \pu{−2.99 kJ/mol})$ under normal conditions, it would take an extremely long time (possibly more than the age of the Universe) for diamond to decay into graphite.
Borazine, with a standard enthalpy change of formation $ΔH^\circ_\mathrm{f}$ of $\pu{−531 kJ/mol},$ is thermally very stable.

In the above pictures, each curved arrow represents the flow of two electrons. The structure on the far right has the electrons in the "double bonds"; all localized in the nitrogen p orbitals.
Borazine is made up of boron (electropositive: Lewis acid) and nitrogen (electronegative: Lewis base). Therefore, its pi electron cloud is "lumpy", with the electrons spending more time near the nitrogens than near the borons. Note that although the electrons in the cloud spend more time near the nitrogens, the nitrogens have a positive formal charge!
This should reinforce your understanding of formal charge as a book-keeping system only. It does give us valuable insight; nitrogen does get less electron density than it would like for having to share. Still, though, its natural electronegativity assures that it will get the lion's share of the π-cloud.
We can see this from the resonance possibilities: one of the resonance structures has the electrons fully localized on nitrogen. Since benzene’s electron cloud is more fully delocalized, and since electron delocalization is what gives aromatic molecules their special stability, benzene would be more stable than borazine.

what about boron nitride.Is it too aromatic?

Maybe you can answer this yourself. What is the structure of boron nitride?

Both compounds are aromatic.
To determine aromaticity, count all of the electrons that can contribute to a resonance structure. This includes electrons within a Pi-bond, and lone-pair electrons.
If they follow Huckel's rule of 4n+2, and if it is a planar molecules, then it is aromatic.

Counting electrons does not determine everything.  10-annulene follows the electron count rule (10) but can't assume a planar structure unless modified; disulfur dinitride with six conjugated electrons is messed up by interactions between formally nonadjacent atoms; I could say more but am running low on comment space.


text:
Pressure of carbon dioxide due to two simultaneous equilibriums

$1~\mathrm{mol}$ each of barium carbonate and calcium carbonate are put in a closed vessel which is maintained at $300~\mathrm{K}$ temperature. Two equilibria are established, with the respective equilibrium constants:
  $$\begin{align}
&\ce{BaCO3(s) <=> BaO(s) +CO2} & K_p&=3\mathrm{~atm} \\
&\ce{CaCO3(s) <=> CaO(s) +CO2} & K_p&=5\mathrm{~atm}
\end{align}$$
  Find the final pressure exerted by carbon dioxide.

Let $n_1$ moles of $\ce{CO2}$ form from barium carbonate and $n_2$ moles of $\ce{CO2}$ form from calcium carbonate.
Then $$K_{p1}=(n_1+n_2)RT$$
$$K_{p_2}=(n_1+n_2)RT$$
But the above two equations are not meaningful because $K_p$ values are different. (what mistake am I making here?)
Let barium carbonate be put into the vessel first. Then there will be $5\mathrm{~atm}$ of carbon dioxide. Then If I put calcium carbonate, will it affect the equilibrium mixture? The answer he gave was $5\mathrm{~atm}$.

That's the problem with solid-to-solid reactions. Well, this means one of them will be shifted all the way to the left, i.e., will not be in an equilibrium.

This is a fictional reaction which I detest. Part of learning chemistry is to get a feel for how real chemicals behave. Real barium carbonate and calcium carbonate do not react like this.

It is a conceptual subtlety. It is needed all the species in the hypothetical reaction to require that the equilibrium constant equation be satisfied. This is normally non understood in most chemistry textbooks. The sentence Two equilibria are established is unhappy.
Briefly you can not use this two equations system till you be sure that all the species exist in equilibrium state.
Initially, both solids will react to form the products, once a pressure of 3 atm is reached the only the second reaction will progress. But, as it will increase pressure above of 3 atm the reaction 1 will go in reverse. This situation will continue while CaCO$_3$ is present. Notice that until now there is no requirement that equilibrium constant equation be satisfied due to the states in which the system was were non equilibrium state.
This situation will continue till all BaO disappears. So, the pressure will increase to 5 atm and the first equilibrium won't take place.
Edit: "what happens if CaCO$_3$ is exhausted? then wont CaO react back to reduce the pressure of CO$_2$?"
If CaCO$_3$ is exhausted before reach 5 atm (after that it is not possible), let say at 4 atm, CaO could react back, but as a soon as a it happens (that a little piece of CaCO$_3$ enough big to have the intensive properties of a macroscopic crystal is formed), there will be more tendency to go forward again.
In such case the final pressure will be 4 atm, different from both equilibrium pressure. It will be another equilibrium state. Nothing will change for BaCO$_3$ in this analysis.
Above I exposed two possibilities: The first one, before the edition, that the vessel is small enough to allow the CO$_2$ pressure reach 5 atm. This is the supposed "right answer".
The later where final pressure of CO$_2$ greater than 3 atm and smaller than 5 atm.
If the vessel is very big, it can happen that reactives are exhausted before reach 3 atm, so other equilibrium state will take place.

what happens if $\ce{CaCO3}$ is exhausted? then wont $\ce{CaO}$ react back to reduce the pressure of $\ce{CO2}$?

@AdityaDev Does the edit answer your question?


text:
Confusion in finding electrophilic centre
In the compound CH3-CH2-CN the electrophilic centre according to my NCERT TEXTBOOK(class 11) is the carbon of cyanide but I think it should be the the carbon adjacent to cyanide since, during bond cleavage between cyaninde and ethyl the electron pair will be taken up by the carbon of cyanide so in this way it is a neucleophilic centre . Hence I think that the carbon of ethyl should be electrophilic centre. Am I right or iam missing something or my textbook is wrong?
Any help is greatly appreciated.

In all of the compounds illustrated, the starred carbon is connected to a more electronegative element.
Let's examine this one by one- I shall begin with methyl iodide. Here, the starred carbon is connected to a Iodine which is more electronegative, thus naturally, we can guess that the electron density shifts towards the halogen. Thus, the carbon in question is slightly electron deficient (slightly positive or electrophilic)
Now, let's look at the examples with the multiple bonds.
Consider ethanal: the carbon is again connected to a more electronegative atom i.e oxygen. So, if you were to cleave the pi bond, the electron density would shift to the oxygen, and again you are left with an electrophilic carbon right next to the oxygen.
The same can be said for the acetonitrile: the electron density shifts to the more electronegative element i.e nitrogen and the carbon right next to is left "slightly positive" or electrophilic.
The electron density would always be slightly greater in the vicinity of a more electronegative element across a bond between two atoms differing in electronegativity (essentially, what we mean when we say a bond is polar)