text:
Ring contraction in a carbocation due to ring strain and back bonding

Today our teacher told us that the following carbocation rearrangement occurs due to back bonding. I could not really follow what he meant. Can someone please explain what is actually happening during the following carbocation rearrangement, and why it happens?

The rearrangement has little to do with ring strain and a lot to do with resonance stabilization.
The ring strain in a cyclobutane ring is ~26.3 kcal/mol, while the ring strain in a cyclopropane ring is ~27.5 kcal.mol (reference).  This isn't much of a difference, certainly not enough to drive a reaction one way or the other.
The reason that cyclobutyl carbocations generally rearrange to cyclopropylcarbinyl carbocations is due to resonance stabilization.  Here are some drawings that may help, if you build a model and look at the model that will help even more. The $\ce{C-C}$ bonds in a cyclopropane ring are approximately $\ce{sp^{3.74}}$ hybridized (see here), said differently, they have a lot of p-character in the bond.  In fact, the p-character is so high that a cyclopropane compound will absorb bromine much like an olefin. If that cyclopropane high p-character bond (orbital) can align itself with the p-orbital on the carbocation center, it will stabilize the carbocation, just like a p-orbital on an adjacent double bond (the allyl system) stabilizes a carbocation.

here's another representation

Typically we discuss two conformations of the cyclopropane ring with the carbocation p-orbital, a bisected and a perpendicular conformation.  In the bisected conformation the cyclopropane $\ce{C-C}$ bond lies in the same plane as the carbocation p-orbital.  The carbocation p-orbital and the $\ce{C-C}$ orbital which resembles a p-orbital (because it is high in p-character) overlap!  In the perpendicular conformation, the cyclopropane $\ce{C-H}$ bond is in the same plane as the carbocation p-orbital, the $\ce{C-C}$ bond is not - there is no carbocation stabilization by the cyclopropane ring in this conformation.  Molecules where the bisected conformation can be attained show stabilization due to the stabilizing overlap between the carbocation p-orbital and the p-like $\ce{C-C}$ cyclopropane ring bond.  Sometimes we use resonance structures such as the following to denote this stabilization.


text:
What would happen if I mix salt and sugar in a cup of water?
If I mix water in a cup with sugar at some stage the solution will become saturated and no more sugar will dissolve into the water. If I now add salt to the solution will the salt dissolve into the water or is the solution already saturated?
I have not tried an experiment as I cannot figure out how to measure the dissolved component of the 2, intuitively I think that some salt will dissolve, but not as much as in a pure salt water mixture.

Unfortunately, sugar is very soluble in water and will make a thick syrup (i.e. karo syrup) before you reach saturation.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
How can an aromatic ketone be reduced in the presence of a nitro group?
Is it possible to reduce an acyl group to an alkane, but still keep a nitro group intact without its reduction to an amino group? A Clemmensen reduction would reduce the nitro group, and it seems like other reducing agents will also reduce the nitro group to an amino group. Would some other mechanism be better suited for this reaction?
For instance, how can I selectively reduce 1-(3-nitrophenyl)propan-1-one to 1-nitro-3-propylbenzene?

You will most likely be successful if you convert the ketone to the corresponding tosylhydrazone and reduce that with $\ce{NaCNBH3}$ (DOI).
A previous study (Bull. Chem. Soc. Jpn., 1974, 47, 2323-2324) has shown that the tosylhydrazone of cyclohexanone can be reduced to cyclohexane with $\ce{NaBH4}$ in polar aprotic solvents.
$\ce{NaCNBH3}$ is an even milder hydride transfer agent, that has been widely used in the reduction of imines and iminium salts. A review by C.F.Lane, Sodium Cyanoborohydride - A Highly Selective Reducing Agent for Organic Functional Groups in Synthesis, 1975, 3, 135-146 gives some background information on the versatility of this reagent.
The nitro group is definitely safe with $\ce{NaCNBH3}$. However, considering the high toxicity and the price for this reagent, you might want to run a small scale reduction of your tosylhydrazone with the much cheaper $\ce{NaBH4}$ first.

A discussion about alternative reduction methods was here Clemmensen reduction vs. Wolff-Kishner reduction.  I've not seen nitro groups reduced by the acid-triethylsilane or tetramethyldisiloxane treatment. One used on plant that I've seen (and used)  is TFA - triethylsilane. The carbonyl is protonated and the first hydride delivery gives the alcohol. A second protonation and hydride addition gives the methylene. As a side note, any aromatics that are not electron deficient e.g. the solvent toluene, can undergo an electrophilic addition during the second reduction step when water is lost.

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

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
What are the spectroscopic signatures that differentiate triple state water from water at 1 atm?
According to the Wikipedia article Triple point of water, the triple point of water occurs at

The single combination of pressure and temperature at which liquid water, solid ice, and water vapor can coexist in a stable equilibrium occurs at exactly 273.16 K (0.01 °C; 32.02 °F) and a partial vapor pressure of 611.657 pascals (6.11657 mbar; 0.00603659 atm).

Shown in the illustration below (from the Wikipedia article):

Crucially, to put this into a context, the 611.73 pascals quoted in the article corresponds to the

minimum pressure at which liquid water can exist. At pressures below the triple point (as in outer space), solid ice when heated at constant pressure is converted directly into water vapor in a process known as sublimation.

Potentially useful information to determine if an exoplanet can possibly have water (if all other conditions are suitable).
What are the spectroscopic signatures that differentiate triple state water from water at 1 atm?

The triple point of water is point of dynamic equilibrium between three phases of water. Any small change in Pressure or Temperature will lead to one phase regions or two phase lines on the phase diagram.
The triple point is not a state of matter rather a combination/mixture of three phases . So i very much doubt we can get any unique spectroscopic signatures . We would get a distorted result as this point has three different phases.
Also Emission and Absorption spectra depends on  temperature and pressure.

Great answer (+1), a question though - would the distorted result be 'characteristic' enough to be differentiated from water at 1 atm?

water will have specific peaks at 1 atm and room(or experimental) temperature. But when we have three phases together like this , we will be  seeing random peaks (i think). So we can get the spectra of the individual phases by spatial and spectral seperation of components.


text:
Why would hexacyanidocobaltate(II) be uncommon while pentacyanidoaquacobaltate(II) is common?
In Huheey's Inorganic Chemistry it is written that hexacyanidocobaltate(II) is uncommon but pentacyanidoaquacobaltate(II) is common. Something is mentioned about Jahn-Teller distortion, but I don't understand how it applies here.

We use it in the lab all the time, I don't think it's uncommon.

that is artificial. i think huheey said not common in nature

In Inorganic Syntheses (1946), a synthesis of $\ce{K3[Co(CN)6]}$ is described, which boils down to the following three equations:[1]

$$\begin{align}\ce{CoCl2 (aq) + 2 KCN (aq) &-> Co(CN)2 v + 2 KCl (aq)}\tag{1}\\
\ce{Co(CN)2 + 4 KCN (aq) &-> K4[Co(CN)6] (aq)}\tag{2}\\
\ce{2 K4[Co(CN)6] (aq) + H2O &-> 2 K3[Co(CN)6] (aq) + 2 KOH (aq) + H2}\tag{3}\end{align}$$

The procedure requires precipitation of the cobalt(II) cyanide by carefully adding a $\ce{KCN}$ solution to a cobalt(II) solution. This precipitate is filtrated and washed, then redissolved in excess cyanide solution giving a deep-red solution. This in turn is heated to reflux for some 15 minutes, whereupon the colour changes to yellow indicating the formation of the hexacyanidocobaltate(III).[1]
I personally doubt equation $(3)$. Instead, I would assume that the oxidation of hexacyanidocobaltate(II) to the corresponding cobaltate(III) proceeds unter oxygen consumption, not hydrogen liberation. That would give equation $(\text{3'})$:
$$\ce{4 K4[Co(CN)6] + O2 + 2 H2O -> 4 K3[Co(CN)6] + 4 KOH}\tag{3'}$$
In any case, however, the presence of $\ce{[Co(CN)6]^4-}$ or hexacyanidocobaltate(II) as a stable intermediate in solution suggests to me that your premise is false and thus the book you cite from incorrect.

Reference:
J. H. Bigelow and J. C. Bailar (1946) Potassium Hexacyanocobaltate(III), in: Inorganic Syntheses, Volume 2 (ed W. C. Fernelius), John Wiley & Sons, Inc., Hoboken, NJ, USA. DOI: 10.1002/9780470132333.ch72.


text:
How is it possible, that the hydrolysis of ethers proceed via a SN2 mechanism?
Ethers on hydrolysis with dilute $\ce{H2SO4}$ under pressure produce alcohols.
This is basically a type of acidic cleavage of ethers by the SN2 mechanism.

(source)
There is inversion of configuration.
Now if $\ce{Nu}$ is $\ce{OH-}$ its called hydrolysis of ether.
Is it right, that for SN2 the leaving group must be more stable than the approaching group? In case of ether hydrolysis the leaving alkoxy group is less stable than the entering $\ce{OH-}$ group.
So how does the reaction succeed? Can someone please explain?

The leaving group is not $\ce{RO^{-}}$, but rather $\ce{ROH}$.  The reaction is run in the presence of acid, so the first step is protonation of the ether oxygen.  Neutral alcohol is a much better leaving group than alkoxide.  BTW, depending on the ether substituents the reaction can proceed by SN1 or SN2.

While the mechanism may look simple in the way you described it there, it doesn’t work exactly like that.
First, you are in the presence of (diluted) $\ce{H2SO4}$ so aside from extremely short-lived accidental self-deprotonation of water, $\ce{OH-}$ ions will not be around. Therefore, $\ce{OH-}$ cannot act as an attacking group, whatever the mechanism.
Second, an alcoholate is one of the worst leaving groups in $\mathrm{S_N}$ reactions one can imagine. The only case where you will frequently see leaving alcoholates is for epoxide openings. But that’s not a problem here, we have $\ce{H2SO4}$. We can protonate the ether oxygen so that the leaving group will now be an alcohol.
Third, since $\ce{OH-}$ cannot be the attacking species, we need to choose from the nucleophiles present. Either $\ce{HSO4-/SO4^2-}$ or $\ce{H2O}$ can attack the $\alpha$ carbon (in $\mathrm{S_N2}$ or $\mathrm{S_N1}$ manner). In the former case, a later attack of $\ce{H2O}$ to liberate the sulphate needs to be assumed. Both attacks work, but neither nucleophile is strong, so the driving force must be the liberated alcohol.
Fourth, the attacking $\ce{H2O}$ will create a protonated alcohol. This needs to deprotonate to give the free alcohol. Thereby the proton we used in second (step 1 of the mechanism, if you wish) is reliberated and therefore the reaction is acid-catalysed.
Finally, one needs to consider the possibility of equilibrium. The reaction can proceed in both directions. Usually by the choice of solvent one can direct it into one or the other. Hydrolysis is usually achieved with a large surplus of water, driving the reaction towards the alcohols by Le Chatelier’s principle. The reverse reaction would be triggered by removal of water or high concentrations of either alcohol.


text:
What is the mechanism of trialkylborane oxidation with hydroperoxide?

This is the mechanism of trialkylborane. In this mechanism I can't understand why should the -OH group leave and the -R group enter in its place to form borate ester. Can someone please explain me that step?

The carbon migrates because peroxide bonds are generally weak and dissociate quite easily.

Also, the mechanism is drawn poorly. In the species labeled "Unstable intermediate," the left arrow should not point towards the boron or the bond; the arrow head should terminate at the left oxygen.

@Zhe Yes. In the upper mechanism, curved arrows are drawn in a wrong way. The arrowhead should be placed pointing towards boron in two of the curved arrows.

One aspect is the release of the stable anion $\ce{OH-}$.
Another aspect is that $\ce{B-O}$ bonds are much stronger than $\ce{B-C}$ bonds because the empty p-orbital of boron overlaps with one p-orbital with a lone electron pair of oxygen.
$\ce{B-C}$ $D_{298}^ \circ  = 448 \pm 29\ \mathrm{kJ\ mol^{-1}}$
$\ce{B-O}$ $D_{298}^ \circ  = 809\ \mathrm{kJ\ mol^{-1}}$
values taken from “Bond Dissociation Energies”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.


text:
Acid catalyzed hydrolysis of ester

During acid catalyzed hydrolysis of ester why does the carbonyl oxygen attack the $\ce{H^{+}}$ ? Why doesn't the alkoxy oxygen which is more electron rich and less electronegative not attack the $\ce{H^{+}}$ with its lone pair? Any ideas?

Why do you think that it is more electron rich?

Because one is sp2 hybrid while other is sp3 hybrid.

The difference between the carbonyl oxygen (red in colour) and alkoxy oxygen (blue in colour) is that the highly distorted pi-electron density (green shaded) towards to the carbonyl oxygen, due to more electro-negativity of oxygen that is the feature which is more on this oxygen than alkoxy oxygen . That feature provide the cavity for the H+ ion from acid, and the distortion of sigma- electron density at alkoxy oxygen is comparatively less than the distortion of  pi-electron density at carbonyl oxygen.


text:
Between ethene and propene which is more acidic and why?
Between ethene and propene, which is more acidic and why? If they are they are the same then why so?
 Their $\mathrm{p}K_{\mathrm{a}}$ is the same $\mathrm{p}K_{\mathrm{a}}$ as ~44.

related http://chemistry.stackexchange.com/questions/18805/abstraction-of-a-proton-in-cycloheptatriene

I dont think that quite answers my question.

The difference between there acidity arises only due to the difference in inductive effect[or the electron donating capacity of the alkyl group] of the alkyl group attached to the $\ce{C=C}$ in both the cases.
By this site there exists a difference of total one unit
For more info on Inductive Effect
Relative acidity of a compound is checked by removing the acidic hydrogen and then comparing the stability of the anions.The more stable the anion is the faster it is formed and hence the acid is stronger.
The anions formed are $\ce{H3C-CH=CH-}$ and  $\ce{H2C=CH-}$ . In the former one the anion or the negative charge on the anion is destabilized by the electron donating ability or the inductive effect of $\ce{CH3-}$, hence it is less stable and less rapidly formed and hence less acidic.

Why $\ce{H3C-}$ has electron donating effect?

In an alkyl group, for example, in methyl group carbon is linked to three hydrogen atoms. Carbon is more electronegative than hydrogen. So carbon pulls bonded electrons nearer to carbon and carbon acquires a slight extra negative charge (delta-) and correspondingly hydrogen lose hold over bonded electrons and acquires a slight positive charge (indicated by delta +). Thus carbon get a little more electronic charge which the carbon will push towards the group to which it is linked.When this electron rich carbon centre (methyl group) attached with other compounds it behaves as electron donor.
And thus methyl group acts as electron donating group.
Be in mind, the electronegativity difference between carbon and hydrogen is very less.
$\ce{H3C-CH=CH-}$ is less stable due to presence of above factor.
You can also use conjugate base method or charge holding capacity method to compare the acidity of compound to give the same result as above.


text:
Smoke alarm ion neutralisation
Smoke alarms are a necessary part of many modern day buildings, especially in laboratories and of course, the family home.
According to the World Nuclear Association page Smoke Detectors and Americium (2014), a major part of the operation of a smoke detector is when the Americium-241 emits alpha particles collide with atmospheric oxygen and nitrogen, ionising them - which form part of an electric current between two electrodes.
When there is smoke (from the article) - bolding mine:

When smoke enters the space between the electrodes, the smoke particles attach to the charged ions, neutralizing them. This causes the number of ions present – and therefore the electric current – to fall, which sets off an alarm.

What is the chemical process that causes smoke to neutralise the charged oxygen and nitrogen ions in smoke alarms?

From this paper, found the mechanism for neutralisation,

Smoke particle charge interaction
When smoke particles enter in the ionization chamber, they are
  subjected to the electric field generated by electrodes but also to the
  one generated by the charges. Close to the charges, the electric field is
  highly divergent resulting in dielectrophoretic force acting on smoke
  particles. The force tends to bind the smoke particles to the charges.
  When smoke particles are considered as isotropic neutral dielectric
  bodies linearly and homogeneously polarisable, dielectrophoretic
  force F can be expressed as
$ F = \frac{1}{2} \alpha v \overrightarrow{grad} |E^{2} |$
where $\alpha$ depends on smoke particle polarizability, $v$ is the smoke
  particle volume and $E$ is the electric field produced by the charges
  [11]. Coefficient $\alpha$ is typically of the order of $\epsilon_{0}/3$ [12] where $\epsilon_{0}$ is the
  dielectric constant of air. Volume $v$ is approximately $4 \mu m^{3} $ for a 1-$\mu m $
  diameter smoke particle.
  Amplitude of electric field $E$ produced by charges is given by
  coulomb’s law:
$ E = \frac{1}{4\pi \epsilon_{0}} \frac{q}{r^{2}}$
in which $r$ is the distance between the charge and the smoke particle.
  Due to the divergence of the field, the dielectrophoretic force
  decrease as $\frac{1}{r^{5}}$. If a smoke particle is close to a charge then they can bind. If the smoke particle is relatively far from the charge, the time
  of interaction may be relatively long and it is necessary that the
  charge drifts sufficiently slowly in the chamber to optimize the
  interaction. Indeed if the charge passes too rapidly there will be not
  enough time for the smoke particle to bind and thus nothing would be
  detected. In fact the capture radius reduces with the electric field and
  is of the order of 10 $\mu m $ for $E= 1  V/mm$.
When the smoke particles attach to the free charges due to
  electrophoresis force, the charges become heavy bodies then the
  charge mobility $\mu$ decreases resulting in a drop of the electric current
  measured between the electrodes of the ionization chamber.
  Even if ionic detectors are very efficient, according to new
  standards they are no longer authorised in almost all of the developed
  countries because of the radioactive sources they include for
  generating the ions. Such radioactive sources are indeed difficult to
  collect and to recycle.


text:
Why can't helium be solidified at 'ordinary' pressures?
According to the UC Davis ChemWiki Chemistry of Helium, helium has a comparatively unusual property, specifically:

Helium is the only element that cannot be solidified by lowering the temperature at ordinary pressures.

'Ordinary' referring to standard air pressure (1 atmosphere).  In order to solidify, there needs to be a corresponding pressure increase, with a projected density of:

$0.187 \pm 0.009~\mathrm{g~mL^{-1}}$ at $0~\mathrm{K}$ and $25~\mathrm{bar}$.

So why can't helium be solidified at 'ordinary' pressures?

https://en.wikipedia.org/wiki/Zero-point_energy

The short answer is that even the London dispersion forces between helium atoms (molecules) are extremely weak. So helium boils at a lower temperature than any other element as well.

When the temperature of helium gas is decreased to about 5.2 K, a phase transition to ordinary liquid helium ($\ce{He}$ I) occurs. The behavior of this liquid phase is normal and identical to any other liquid phase.
As temperature is decreased more (at moderate pressures), helium does not solidify. In fact, it undergoes a phase change to a second liquid phase known as $\ce{He}$ II at a temperature of approximately 2.17 K. As shown in the phase diagram below, the $\ce{He}$ II phase persists until absolute zero. Due to quantum effects, $\ce{He}$ II has remarkably unique properties. It is considered a superfluid.

For more details, please click here. Pressure is required to produce solid helium (25 atmospheres or more), because of the small mass and extremely weak forces between the helium atoms.

Great answer, the properties of helium are indeed remarkably unique - are there any other elements/compounds having 2 liquid phases?

@santiago helium-3 is even stranger than helium-4: https://en.wikipedia.org/wiki/Helium-3#Cryogenics

It should be noted that helium is the only substance whose isotopes have radically different phase diagrams.

@santiago Liquid crystalline materials often have even three liquid phases.

There are a numer of liquid sulphur modifications with varying viscosity.


text:
Difference in ozonolysis reagents
What is the difference when following up ozonolysis with DMS (dimethyl sulfide), Zn/water, or just water? Or is there really no substantial difference at all?
In my textbook, the alkenes section uses DMS or Zn/water but then in the alkynes section it uses water with no explanation as to why that happens. My professor also uses them interchangeably, which gets confusing.

Dimethylsulfide (DMS) is commonly added to an ozonolysis reaction after it is complete (e.g. it is the first step in the reaction work-up).  I can see how zinc could serve the same purpose.  Both of these reagents can reduce oxidized materials.  When the ozonolysis is complete the reaction contains the raw ozonide along with numerous undesirable side-products such as peroxides, hydroperoxides and tetroxanes - all of these materials are potentially explosive.  The DMS or zinc will reduce them to carbonyls and alcohols and remove the explosion hazard.  I suppose that tap water is somewhat acidic or basic depending upon where you live, and it might, therefore, also serve to reduce these materials.
It is one thing to say or write "add water during workup" in a lecture or text, it's another thing to actually deal with explosive materials in the lab.  Personally, I wouldn't chance it. I'd use a real reducing agent such as DMS or zinc, instead of a water work-up.


text:
Basicity Alteration Due To Ortho Effect In Methoxy Aniline System
I wanted to ask something more in reference to this question.

Suppose instead of toluidine, 2-methoxyaniline and 3-methoxyaniline are present. Now, why is 2-methoxyaniline more basic than 3-methoxyaniline (even though 2-methoxyaniline has reduction of basicity due to ortho-effect but 3-methoxyaniline does not)?
Also what would be the correct order of basicity if nitro group was present in place of methoxy group? Would ortho-effect still take place?

Why would ortho effect reduce basicity? Shouldn't it increase?

Take home message: Just because you have an ortho substituent does not mean you will have an ortho effect.
To have an ortho effect two conditions must be met:

you need an ortho substituent
and, importantly, it must be of sufficient steric bulk to disturb the adjacent substituent.

One measure of steric bulk is the cyclohexane $A$-Value, a table of which can be found here.  From the table we see that methoxy $(A=0.6)$ and nitro $(A=1.1)$ groups have a smaller steric size than that of a methyl $(A=1.7)$ group.  Methoxy and nitro groups generally do not produce an ortho effect.
Therefore, the basicities of the various methoxy anilines (the anisidines) are controlled by a combination of resonance and inductive effects.
\begin{array}{l|r} \hline
\text{Substituent} & \mathrm{p}K_\mathrm{b} \\ \hline
\ce{-H} & 9.13 \\
\ce{o-methoxy} & 9.5 \\
\ce{m-methoxy} & 9.8  \\
\ce{p-methoxy} & 8.7  \\ \hline
\end{array}
(data source)
The basicity of p-methoxy aniline is controlled primarily by resonance effects (too far away for inductive effects to be significant).  Drawing resonance structures we see that the p-methoxy group is pushing electron density into the ring, this tends to keep the nitrogen lone pair more localized on nitrogen (if we also delocalize the nitrogen lone pair into the ring like charges repel leading to destabilization). This results in p-methoxy aniline being more basic than aniline.
The basicity of m-methoxy aniline is controlled purely by inductive effects (resonance is not possible from this position).  The inductively electron withdrawing methoxy group removes electron density from nitrogen, making it less basic than aniline.
The basicity of o-methoxy aniline is controlled by a mix of resonance and inductive effects operating in opposite directions.  Apparently the inductive effect is stronger and wins out (this shouldn't be surprising, note that the inductive effect from the meta position had a 0.7 unit effect on the basicity, and now in the ortho position it will be even stronger, while the resonance effect in the para isomer only changed the basicity by 0.4 units) making o-methoxy aniline less basic than aniline.

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


text:
Why was the prefix 'bi' used in compounds, such as for bicarb of soda?
According to the Wikipedia article IUPAC nomenclature of inorganic chemistry,

he prefix bi- is a deprecated way of indicating the presence of a single hydrogen ion

A very common example is the commonplace 'bicarb of soda', or sodium bicarbonate (or using its correct chemical name sodium hydrogen carbonate). On the Wikipedia page for sodium bicarbonate there is a note that states that:

The prefix "bi" in "bicarbonate" comes from an outdated naming system and is based on the observation that there is two times as much carbonate ($\ce{CO3}$) in sodium bicarbonate ($\ce{NaHCO3}$) and other bicarbonates as in sodium carbonate ($\ce{Na2CO3}$) and other carbonates.

However, the statement above is not particularly clear, especially as carbonates are not the only compounds that used the 'bi' prefix, such as:
Bisulphite $\ce{HSO3-}$
Bisulphate $\ce{HSO4-}$
Why was the prefix 'bi' used to indicate the presence of a single hydrogen ion in compounds?

According to the Online Etymology Dictionary, the word bicarbonate is derived as follows:

1814, bi-carbonate of potash, apparently coined by English chemist William Hyde Wollaston (1766-1828), from bi- + carbonate.

The question then becomes, what is the meaning of the prefix bi- here?  We know it means two, but it also can mean double or twice.  If that is the case, then sodium bicarbonate is sodium double carbonate.
Consider a given quantity of acid, let's say $\ce{HCl}$, that we are going to neutralize with sodium carbonate, $\ce{Na2CO3}$, and an equal quantity of acid that we are going to neutralize with sodium bicarbonate, $\ce{NaHCO3}$.  We add the sodium carbonate to the acid, and as the neutralization reaction goes forward, it produces some quantity of carbon dioxide.  The same process is repeated for sodium bicarbonate, and the carbon dioxide is again collected.
$\ce{Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O}$
$\ce{NaHCO3 + HCl -> NaCl + CO2 + H2O}$
In the first reaction, one unit of sodium carbonate is able to neutralize two units of acid.  In the second reaction, one unit of sodium bicarbonate can only neutralize one unit of acid.  Another way of looking at this is that double the quantity of carbon dioxide is produced from the sodium bicarbonate neutralization.  That is, the compound is a double carbonate.
Similar explanations make sense for bisulfate and bisulfite.  The acid-base reactions that a sulfate or sulfite ion might undergo are already partially completed in the bi- versions of the compound, so it takes twice as much sulfate or sulfite to do the same job as the non-hydrogen form.
The hydrogen phosphates don't fit this pattern as nicely.
Additionally, if the cation in these salts is a relatively light weight one, then the required mass of the bicarbonate or bisulfate is substantially more than that of the non-hydrogen form of the compound, though it's not quite double.  53g of $\ce{Na2CO3}$ could neutralize one mole of $\ce{HCl}$, but it takes 84g of $\ce{NaHCO3}$ to do the same job.  This is closer to accurate for the more massive sulfur and phosphorus based oxyanions.

The use of "bi" for 'two" denotes that there are two different cations (e.g. $\ce{Na+, H+}$ in the examples), but it is confusing, because it could imply that there are two of a single cation, so its use is deprecated. Similarly, in English, biweekly can mean either twice a week or every two weeks.
The use of "di" for "two" is more consistent, i.e. it always means two of a single species, as in carbon dioxide, $\ce{CO2}$. [Perhaps "always" is a bit strong... anyone know of contrary examples?]

Diphosphate, as in adenosine diphosphate (ADP) is the most common contrary example. "Diphosphate" is a side group composed of two esterified phosphates, $\mathrm{H_3P_2O_7—}$, whereas a bisphosphate compound such as fructose 1,6-bisphosphate bears two $\mathrm{H_2PO_4—}$ side groups, which may or may not be bound to the same atom. (The protonation state of all of these will of course depend on pH.)

I'm curious, do you have a source for this answer?  Everything I can find indicates that the answer suggested by santiago in the question is correct.


text:
When, how and why does sodium metal react with cycloalkenes, cycloalkadienes, cycloalkatrienes?
I've been searching for the conditions necessary for the reaction of sodium (Na) metal with cycloalkenes, cyclo-di-alkenes, cyclo-tri-alkenes all over the internet but I cannot find a proper answer. Can someone explain to me the necessary conditions?
For example why does sodium metal react with cyclopentadiene but not with cycloheptatriene?

related http://chemistry.stackexchange.com/questions/29428/acidity-of-unsaturated-five-membered-ring-relative-to-seven-membered-analogue

http://chemistry.stackexchange.com/questions/18805/abstraction-of-a-proton-in-cycloheptatriene

The negatively charged particles are conjugate bases of respective compounds, which (reactants) acts due to presence of active methylene group. Depends on the stabitily of conjugate base the reaction goes forward.


text:
How can I choose an indicator for determining the concentration of a base using its pKb values?
I have a base $\ce{M(OH)_{3}}$ having 3 $\mathrm{p}K_{\mathrm{b}}$ values as,
$$
\mathrm{p}K_{\mathrm{b}_{1}} = 0.5\\
\mathrm{p}K_{\mathrm{b}_{2}} = 3.7\\
\mathrm{p}K_{\mathrm{b}_{3}} = 7.7\\
$$
If a strong acid is given as the titrant and phenolphthalein and methyl orange are given as possible indicators, how can I choose an indicator using above $\mathrm{p}K_{\mathrm{b}}$ values ?

Since:
\begin{align}
K_\mathrm{w} &=K_\mathrm{a} \cdot K_\mathrm{b}\\
K_\mathrm{a} &=\frac{K_\mathrm{w}}{K_\mathrm{b}}\\
\log{K_\mathrm{a}} &=\log K_\mathrm{w}-\log K_\mathrm{b}\\
K_\mathrm{w} &=1 \times 10^{-14}\\
\log{K_\mathrm{a}} &= -14-\log K_\mathrm{b}\\
-\log{K_\mathrm{a}} &=14+\log K_\mathrm{b}\\
-\log{K_\mathrm{a}} &=\mathrm{p}K_\mathrm{a}\\
-\log{K_\mathrm{b}} &=\mathrm{p}K_\mathrm{b}\\
\mathrm{p}K_\mathrm{a} &=14-\mathrm{p}K_\mathrm{b}
\end{align}
Using the above formula, the three $\mathrm{p}K_\mathrm{a}$ for the conjugate acid of the conjugate base are: $13.5~(\mathrm{p}K_\mathrm{a_1})$, $10.3~(\mathrm{p}K_\mathrm{a_2})$, and $6.3~(\mathrm{p}K_\mathrm{a_3})$, where $\mathrm{p}K_\mathrm{a_1}=14-\mathrm{p}K_\mathrm{b_1}$, etc.
An important equation to note is that the $\mathrm{pH}$ at the equivalence point is given by the equation: $$\mathrm{pH_\mathrm{eq}}=\frac{\mathrm{p}K_\mathrm{a1}+\mathrm{p}K_\mathrm{a2}}{2}$$
This means that two of the equivalence points exist at $$\mathrm{pH_\mathrm{eq_1}}=\frac{13.5+10.3}{2}=11.9$$ and $$\mathrm{pH_\mathrm{eq_2}}=\frac{10.3+6.3}{2}=8.3$$
The $\mathrm{pH}$ range for methyl orange is $3.0~-~4.4$ and the $\mathrm{pH}$ range for phenolphthalein is $8.2~-~10.0$. Phenolphthalein would be useful for detecting the second equivalence point $(\mathrm{pH_\mathrm{eq_2}})$, since this shift occurs within phenolphthalein's range.
The final equivalence point cannot be calculated from the information given, but given that methyl orange's color transition occurs in an acidic medium, and also the relatively basic $\mathrm{p}K_\mathrm{a_1}$, methyl orange can be eliminated as a possible indicator.

@LDC3 the only $\mathrm{p}K_\mathrm{a}$ close to being within the useful range of one of the indicators is $\mathrm{p}K_\mathrm{a_2}$ to phenolphthalein's range, but it is still not actually within the range. Unless I am misunderstanding how an indicators work (which I guess could be possible) then I don't see how either indicator is useful.

@ringo You want the indicator to change color after all the base has been converted (distant from a $\ce {pK_{b}}$), not when it is being converted (near a $\ce {pK_{b}}$). So the indicators you would choose would not be near any of the $\ce {pK_{b}}$ of the base.

I'll add that the reason to convert to pKa values is that tables for acid-base indicators are usually give in terms of pH not pOH. https://en.wikipedia.org/wiki/PH_indicator

The first step is to determine the $\mathrm{p}K_{\mathrm{a}}$ values for the indicators and convert them to $\mathrm{p}K_{\mathrm{b}}$ values. Then determine how you would titrate the solution using the indicator(s).

Do you mean that the $pK_{b}$ values of $M(OH)_{3}$ have no part to play in choosing the indicator ?

Yes it does, but you need to know when the indicators change colors.

OK. Say I have methyl orange with pH range of 3.0 - 4.4. How do I proceed ?


text:
What is the meaning of cmmol in chemistry?
I came across a unit called $\text{cmmol dm}^{-3}$ in buffer solutions. What does this unit mean?

I guess it simply lack space between C and m - it's concentration in mmol per dm3

Can you give us more context? Copy and paste here the paragraph where you found it.

@NicolauSakerNeto: I'm sorry can't paste it because it was a handwritten one. But I've found some other references online. Such as http://www.thenakedscientists.com/forum/index.php?topic=35027.0  and  http://www.labfert.agr.br/ferramentas.php

That first reference doesn't seem clear to me, and the second one uses $\rm{cmol}$ rather than $\rm{Cmmol}$, in which case it simply means centimol, i.e. a hundredth of a mole ($\rm{1\ cmol} = \rm{10\  mmol}$).

@NicolauSakerNeto I agree that the references are not relevant. The original text I referred mentions $cmmol,dm^{-3}$ and not $cmol,dm^{-3}$. And I have changed 'C' into 'c'.

The unit $cmmol\,dm^{-3}$ is the same as $mmol_{c}\,dm^{-3}$ which denotes the millimoles of charge per litre. I think the former notation is not well used anymore. On the other hand, latter is used often and a google search would return a number of references. For example,
http://micromaintain.ucanr.edu/Prediction/Source/Groundwater/Potential_for_clogging/CP/Water_analysis_for_hazard/Levels_of_concern/
http://www.slidefinder.net/c/converting_various_units_mmol_mmolc/32208520
http://www.scielo.br/scielo.php?pid=S0103-90162011000400012&script=sci_arttext

Upon researching this, I found that this a measurement of charge:
$$\mathrm{cmol=10~\mathrm{meq}}$$
and
$$1~\mathrm{eq}= \mathrm{F}$$
where $\mathrm{F}$ is Faraday's constant and is:
$$\mathrm{F}=96 485.3365~\mathrm{C}$$
So:
$$\mathrm{cmmol}=\frac{\mathrm{cmol}}{1000}=\frac{\mathrm{eq}}{100000}=.96 4853365~\mathrm{C}$$
Sources:
http://en.wikipedia.org/wiki/Equivalent_%28chemistry%29
https://www.agronomy.org/files/publications/jnrlse/pdfs/jnr014/014-02-0084.pdf

Do you mean that $C,mmol,dm^{-3}$ refers to milli moles of charge per litre ?

@chemkatku actually, since Faraday's number is 1 mole of charge, which is equal to 1 equivalent, $1~\mathrm{cmmol}~\mathrm{dm^{-3}}$ is 10 micro moles of charge (per liter).

Do you have any reference to back this up? One mole is similar but not the same as one equivalent, and I've never heard anything relating $\mathrm{eq}$ with $\mathrm{F}$. I agree with @Mithoron and I think this is simply a typo. It was supposed to be either $\mathrm{cmol}$ or $\mathrm{mmol}$ (or the $\mathrm{C}$ is from concentration and a space is missing) per $\mathrm{dm^{-3}}$, which is a litter, and this sounds a lot like a buffer concentration, which in turn can be related to it's "capacity" as the full question mentions on the link provided above.

Also, the reference given uses lower case $\mathrm{c}$ and not $\mathrm{C}$, which is another point against the Coulomb thing.

@Molx sources cited

OK I changed Cmmol to cmmol. My bad, the original text mentions $cmmol,dm^{-3}$ with a simple 'c'

@ringo the agronomy paper talks about $\mathrm{cmol}$ which is centimole. They really like using that in agronomy, as well as equivalents, which is related to Normality. I still think it's a typo but I don't see how one can post an answer saying the question is "wrong".


text:
Why sodium sulphate has water absorbing capability?
At first I thought maybe sodium sulphate in contact with water produces sulphuric acid which absorbs water but I do not think it is actually a valid reason.

Sodium sulphate reacts readily with water at room temperature to form hydrates up to sodium sulphate decahydrate, $\ce{NaSO4\cdot10H2O}$. $$\ce{NaSO4 + 10H2O -> NaSO4\cdot10H2O}$$ This means that $\ce{NaSO4}$ can absorb up to 10 mol of water for every 1 mol of salt that is used, making it one of the most effective drying agents in terms of sheer capacity.
Using data from this source we can calculate the Gibbs energy change for the above reaction at different temperatures. $$\mathrm{\Delta G_{20^\circ C} = -1.33~kJ~mol^{-1}}$$ $$\mathrm{\Delta G_{30^\circ C} = 1.28~kJ~mol^{-1}}$$
Since the entropy change for the reaction is negative, we can see that the cooler the solution you are trying to dry is, the more effective $\ce{NaSO4}$ will be as a drying agent.


text:
Transformation of acyl group to carboxyl group in biphenyl system with two acyl groups at para position
How do I perform this reaction?
I have prepared 4,4'-acylbiphenyl and want to make one acyl group into $\ce{-OH}$ and other into $\ce{-COOH}$ group.

Is it correct that you are aiming to synthesize 4-(4-hydroxyphenyl)benzoic acid [58574-03-1]?

The target compound is available in one step from biphenyl using potassium peroxydisulfate $(\ce{K2S2O8})$ as an oxidant, Palladium(II)-trifluoroacetate, $\ce{Pd(OCOCF3)2}$, as a catalyst, and formic acid as the carbon source for the $\ce{-COOH}$ group. The synthesis was published in
Org. Lett., 2004, 6, 2437-2439 (DOI).

EDIT
Limited resources and experience will probably rule out Stille-type coupling reactions too.
Alternatives pretty much depend on which 4,4'-disubstituted biphenyl are available.
A methoxyarene can be converted to the corresponding phenol by $\ce{BBr3}$
or the cheaper hydrobromic acid. The $\ce{-OCH3}$ group resists oxidation, is a +M substituent and is not acidic. All this might be helpful (see below).
In order to introduce the $\ce{-COOH}$ substituent, two approaches seem to be possible in an undergrade lab:

oxidation of a methyl group by permanganate
Grignard reaction of a bromoarene with carbon dioxide

Thank you sir. But I am just a UG student and I don't think these reagents will be available for me sadly.


text:
Synthesis of bromobutane
What will be the cost efficient and high yielding method of synthesis for 1-bromobutane?
I can react butanol with $\ce{HBr}$ but I am worried butanol is not present at the moment in lab. But I need to hurry and make bromobutane.

Well, clearly what you do have in the lab will the rate determining step. What potential candidates have you got in the inventory?

Current prices are about 34 € (500 g) for 1-bromobutane and 32 € (500 mL) for 1-butanol in decent lab quality.
The most cost efficient method is ordering 1-bromobutane from a chemical supplier of your choice!
On a technical scale, one would probably react 1-butene with anhydrous hydrogen bromide under radical conditions, using peroxides or UV light to initiate homolytic cleavage of the $\ce{H-Br}$ bond.

Bromine radical is selective. You cannot expect to obtain a high yield of 1-bromobutane from such a reaction. Besides, such reactions are a mess to purify and this will be very tedious and difficult to achieve.

Start with butanal or butanoic acid, reduce them with LiAlH4 and aqueous work-up and you have 1-butanol.
If you have 1-butene, you do hydroboration/oxidation then phosphorus tribromide.
Are you trying to make a Grignard? or a substitution reaction?


text:
What are some books, websites, etc. where I can find the detailed protocol for standardization of sodium thiosulfate with dichromate?
I am looking for a book or website such as "Vogel's Textbook of Quantitative Chemical Analysis" where I can find methods for analyzing/standardizing a particular species easily.
I am not looking for an exact sub area such as water analysis or drug analysis; rather I want a resource like Vogel's which indexes generic analytical methods, both classical and modern.
Edit:
Actually I am looking for the method of standardizing a sodium thiosulfate solution using a potassium dichromate solution. I can't find the method in "Vogel's Textbook of Quantitative Chemical Analysis 6th Edition".

Finally I found the book I was looking for. The method is listed in "Vogel's Textbook of Chemical Analysis 5th Edition" (page 392). They have removed it from the 6th edition.

My go-to for this is typically Kolthoff & Sandell's Textbook of Quantitative Inorganic Analysis.  They unfortunately do not report a detailed protocol for dichromate standardization of thiosulfate solutions therein, but instead cite Volumetric Analysis, Vol III by Kolthoff and Stenger (Amazon and Google Books).  I don't have access to this volume; WorldCat shows numerous copies in university libraries in my area of the U.S.; more than likely there's a library where you are that has a copy.

@chemkatku Sure! I was rather surprised that K&S didn't have it in detail. It must be that authors are either considering it an obsolete method, or so commonplace that it's not needed to include any more.

Book:
I can find one book which should have analytical methods for most of the species.
Encyclopedia of Separation Science
It is ten volume set.
websites:
There are too many websites on internet but I will choose following two:

Chemistry Stack Exchange
Wikipedia

Thanks. Encyclopedia of Separation Science is great. But I want some resource that indexes analytical methods.


text:
Why does the gauche conformation of ethane-1,2-diol show hydrogen bonding?
Why does the gauche conformation of ethane-1,2-diol show hydrogen bonding even though there is no 6-membered ring system? If we don't consider one of the $\ce{H}$ (which is what is usually done), then there is only a 5-membered ring formed. How then does hydrogen bonding take place?

(Also, I feel that there must be some reason apart from 6-member ring for hydrogen bonding which I'm not aware of)

What's the deeper reason for you to go looking for a six-membered and only a six-membered ring system?

Ok thats what i've learnt from my teacher till now.If you know of any other reason please let me know!

How is it that your reputation never increases?

Cyclopentane has a relatively low strain energy, something on the order of 6.5 kcal/mol more than cyclohexane (reference).  In the case of ethane-1,2-diol, the strain energy in the 5-membered ring formed by hydrogen bonding will be even less because some of the $\ce{H-C-C-H}$ and $\ce{C-C-C-C}$ interactions that produce the strain in cyclopentane are no longer present; the oxygens don't have the hydrogens attached like a methylene group does and the hydrogen used in the hydrogen bond is also much smaller than a methylene group.
A typical $\ce{O-H..H}$ hydrogen bond can stabilize a system by ~7 kcal/mol (reference).  So to form a 5-membered ring in ethane-1,2-diol and create a hydrogen bond it will cost something less than 6.5 kcal/mol to overcome the strain energy in the ring, but we get 7 kcal/mol back due to hydrogen bond formation - a net stabilization.
$%edit$

Updated your first reference.


text:
What is torsional strain?
I am having difficulty in understanding the concept of torsional strain. Can someone explain it to me in a more elaborate manner?
I mean why should bond restrict movement in eclipsed conformation? Doesn't that make it unstable! I tried this Wikipedia article but I'm still having difficulty in understanding the concept.

For the case of a single bond, consider a simple case like ethane, which can take an eclipsed conformation, a staggered conformation, or anything in between depending on the torsion angle of the C-C bond.  Each carbon has a region of especially high electron density where it's bonded to a hydrogen.  When the torsion angle is changed from staggered to eclipsed, these regions approach eachother and repel, and thus the eclipsed conformation is higher in energy than the staggered.
In the case of a double bond, as in ethylene, the energy associated with torsional strain is much higher because the P orbitals involved in the pi bond have to be oriented the same way.  Thus, applying torsional strain involves breaking this pi bond, and has a high energy.

Well, you don't have to break the $\pi$ bond to apply torsional strain. Consider biphenyl, where the lowest-energy dihedral angle is definitely not flat. I'd consider some torsional strain. There are even "torsional balance" molecules designed to measure this.

How energy is increased if nothing adds energy to the molecule? Also when in chemistry we speak about energy what is this energy? Potential or kinetic or both?


text:
Why do Grignard reagents or Organolithium compounds not displace halides as easily as acetylide ions do?
This is a snapshot from LG Wade Organic Chemistry. I have a doubt with a particular statement in the book. It says that Grignard reagents do not displace halides as easily as acetylide ions do.
But shouldn’t it be the other way round? Acetylide ions, having an sp-hybridised carbon, are much more electronegative compared to the sp³-hybridised carbon in Grignard reagents. So acetylide ions should have less tendency to displace halide ions by attacking with their lone pair compared to the alkyl group in Grignard reagent. Shouldn’t it?
Please help me sort this out!

According to you, the sp carbanion is more stable and less reactive ?

Then your statement is contradicting the book's one.I request you to elaborate. @user97554

An acetylide ion proceeds in a nucleophilic substitution pathway (SN2 in the picture due to primary halide). The substitution involves the attack of a highly nucleophilic sp-carbon on a highly electrophilic primary carbon. The enthalpy of the C-C bond formation is relatively low due the pKa of the nucleophile being ~ 25. The ion is treated as a "soft" nucleophile that prefers to substitute alkyl halides instead of acting as a base and conducting an elimination reaction.
A grignard reagent is essentially a highly polarized bond between Carbon and the Bromide(Halide)/Magnesium complex. The electronegative difference between Carbon and Magnesium is 1.3 (http://www.thecatalyst.org/electabl.html <-- Source of values). The Carbon atom in a Gringnard is treated as a carbonanion. This carbonanion characterization is highly unstable in nature and treated as a "hard" nucleophile with pKa of ~ 50. The highly basic character of a Grignard reagent often results in an elimination reaction or no reaction at all. The transition state to substitute the alkyl halide is less stable than the Magnesium/Bromide(Halide) complex. This is due to a ligation formation between the solvent and the Magnesium atom. The solvent (ex. diethyl ether) shares non-bonding electrons with the cation and creates a stable electronic atmosphere with a lower energy than a possible C-C bond with the alkyl halide.
Substitution isn't preferred with a hard nucleophile and therefore, the acetylide ion is more likely to preform a nucleophilic substitution reaction with an alkyl halide. Note that the electronegative difference between Carbon and Magnesium supersedes the hybridization of the sp or "triple" bond complex. The bond is so high in energy that the nucleophilic character of the grignard far surpasses the electronegative character of the sp-hybridization of the acetylide ion.


text:
Mixing solvents and other compounds - safety issues
I would like to know if it is safe to mix organic solvents together or organic solvents with some compounds listed below.
In a mixture with random ratios (greater than zero - all chemicals present) we have:

Ethanol
Methyl ethyl ketone (as denaturant for ethanol)
Bitrex (aka Denatonium - as denaturant for ethanol)
Isopropanol
n-butanol (as denaturant for isopropanol)
n-propanol (as denaturant for isopropanol)
Ethyl acetate
Acetic acid (in case that ethyl acetate hydrolyzes)
Water (from ethanol and isopropanol)
Acetone
Methanol
Naphtha (Hydrogenated)

In general, we have alcohols, ketones, denatonium, ester, organic acid, water, and a liquid hydrocarbon mixture (Naphtha).Ratios can be taken as equal for all chemicals except those used for denaturation.(I think ratios for denaturants  wouldn't matter in case they would react.)
Could some of these chemicals react together and produce any unstable or hazardous compound?
What conditions should be avoided? (Light,heat,etc.)
Any reaction (slow) that could happen with atmospheric oxygen? (opening and closing a bottle containing all these chemicals)
I don't know too much about organic chemistry, just asking to be sure this is safe.I want to test something that possibly involves mixing all the above chemicals and I don't know if they could probably react together.
Note: Question unrelated to waste disposal. I'm experimenting with a DIY electronics cleaner.

First, I would suggest you read the material safety data sheets (MSDS) for each of those products to be sure of any incompatibility that may arise. They are available throughout the web or on distributor web sites like Sigma or Fisher.
For your list of waste disposal (as it seems to be) most of your products are alcohols or water so they can be mixed without issues.
The ethyl acetate will react with water to create ethanol and acetic acid and acids may be an issue if you have great quantities. So I would suggest to react and neutralise with $\ce{NaOH}$ prior disposal.
Same thing for acetic acid: you should neutralise prior disposal.
You can also request an evaluation of price from your chemical waste service contractor, which will give you the price of each disposal category with the incompatibility management included.

I read all MSDSs but I want to be sure about this.I have very basic knowledge of organic chemistry,I only know some reactions like hydrolysis.I don't know if something could react violently (even in case of hydrolysis) or there is a buildup of something dangerous over time.
The mixture is about to be used as a cleaner for electronics - I should have mentioned it. I also edited the original post (check the list).
"Acids may be an issue if you have great quantities" is this about reactivity or disposal?

In presesnce of acetic acid could the following esters form?
Isopropyl acetate,Butyl acetate,Propyl acetate,Methyl acetate ,(Ethyl acetate already in the list)
After searching MSDSs for esters above I didn't found anything related to corrosivity and reactivity except for Methyl acetate:"Reactive with oxidizing agents, acids, alkalis, moisture."
Does it has to do with acetic acid or just inorganic acids?
Regarding safety of each one,they seem to be ok but further from this I don't know if something could react with these new chemicals in the list - since they will probably be formed.


text:
How to calculate the number of geometrical isomers in allenes?
Can someone explain how to calculate the number of geometrical isomers in allenes like $\ce{CH3-(CH=CH)4-CH3}$ or $\ce{CH3-(CH=CH)5-CH3}$

In the linear cases you drew, the answer is always two. They’re just sometimes called differently. (E/Z vs. R/S)

But suppose one terminal group is in a different plane compared to the other,will still there be 2 geometrical isomers? I don't think so.

Other than searching the net, what have you tried to figure this out? Have you drawn any compounds?

Ah, sorry, I misunderstood ‘geometrical isomer’ But what is the reason to consider axial R and S differently from E and Z? Is ther eany?

Yes I have drawn.But i''m not sure whether i'm right.For the first one i'm getting 0 and for the second one 2.

I wanted to ensure geometrical isomerism concept is still valid when the terminal groups are in different planes. @jerepierre

@SanchayanDutta Put that information in the question. Show some of the compounds you have drawn. You will get better answers, and your question won't be seen as homework.

an important point is that allenes with even no. of carbons can not exhibit geometrical isomerism
but odd no.can

If we have one double bond in a hydrocarbon compound we have an olefin or alkene.  Ethylene is the simplest example of this class of compounds.  The carbons in the double bond and the 4 atoms attached to them lie in the same plane.  One pair of cis-trans isomers is possible in compounds with a single double bond.

If we add a nother double bond directly on to the end of the double bond in ethylene such that an sp hybridized carbon is created in the process, then we have formed an allene.  The 4 substituents at the end of the double bonds in allene lie in planes that are oriented 90° to one another.  Look at the allene (bottom line) in the following drawing.  One set of substituents (R3, R4) are located in the plane of the screen; the other set of substituents (R1, R2) are located in a plane perpendicular to the screen.  Allenes cannot generate cis-trans isomers, but they can generate enantiomers.

Now let's add one more double bond to allene such that we now have 2 sp hybridized carbons (top line in the above drawing).  We see that, like in ethylene, the 4 atoms connected to the double bond lie in the same plane as the double bond and one pair of cis-trans isomers is possible.
Any compound with 3 or more cumulative double bonds is a member of the cumulene family. Any cumulene with an odd number of double bonds is geometrically structured like ethylene (the 4 atoms connected to the double bond lie in the same plane as the double bond) and is capable of having one pair of cis-trans isomers. Cumulenes with an even number of double bonds are structured like allene (the 4 atoms connected to the double bonds lie in perpendicular planes) and cannot display cis-trans isomerization, but can have enantiomers.

As a general rule, an even number of cumulative double bonds with different groups on terminal carbon atoms will be optically active and odd number of cumulative double bonds with two different groups on terminal carbon will be optically inactive.
The reason behind this is : For a even number of cumulative double bond compound(generally: Allene)  the plane of π bond of that Allene are perpendicular to each other. This geometry of this π bond causes the groups attached to terminal C atom to lie in perpendicular planes. Such Allene with different substitutents on end C are chiral  .Such Allene do  not show cis trans isomerism, but show optical activity.  (Due to the presence of stereo centres not chiral centres.)
Penta12diene do not show optical activity ,since terminal carbon has two same groups. (H atoms)

Please consider using some punctuation.


text:
Nomenclature of an organic compound
Which substituent should be placed first (in case of alphabetisation)?

isopropyl-
methyl-

I have come across two compounds: 1-methyl-3-methoxy-4-isopropylbenzene and 1-isopropyl-4-methylcyclohexane
So, is “iso” taken care of in case of alphabetisation? And, are there any special criteria for this operation?

The name “1-methyl-3-methoxy-4-isopropylbenzene” is not in accordance with IUPAC nomenclature. The systematic name is 1-isopropyl-2-methoxy-4-methylbenzene.

(Note that the prefix ‘isopropyl’ is retained for use in general nomenclature but the preferred IUPAC name is ‘propan-2-yl’.)
Simple prefixes (simple substituent groups consisting of just one part that describes an atom, or group of atoms as a unit, for example methyl, methoxy, and isopropyl) are arranged alphabetically disregarding any multiplicative prefixes. Any multiplicative prefixes are inserted later and do not alter the alphabetical order.
For example, ‘1,2-dibromo-’ is considered to begin with ‘b’; ‘isopropyl’ is considered to begin with ‘i’.
On this matter, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.5 ALPHANUMERICAL ORDER
Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved
Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.
(…)
P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Therefore, the correct alphanumerical order in your first example corresponds to the name x-isopropyl-y-methoxy-z-methylbenzene.
In your second example, the correct alphanumerical order corresponds to the name x-isopropyl-y-methylcyclohexane.
Furthermore,

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Note that Rule f takes precedence over Rule g.
Therefore, your first example is named as 1-isopropyl-2-methoxy-4-methylbenzene rather than 4-isopropyl-3-methoxy-1-methylbenzene since the locant set ‘1,2,4’ is lower than ‘1,3,4’.

Your second example is named as 1-isopropyl-4-methylcyclohexane rather than 4-isopropyl-1-methylcyclohexane since isopropyl is cited first as a prefix in the name.


text:
Preparation of 6-bromo-1-hexanol
I am refluxing $\ce{HBr}$ with 1,6-hexanediol for 16 hours at $80\ ^\circ\mathrm{C}$ in toluene. How can I be sure that more than $90\,\%$ of the product is brominated only at one terminal? If the amount of double brominated product is in more than $10\,\%$ of the total yield, then the product mixture will be useless for me.

Would having unreacted 1,6-hexanediol be acceptable? If so, use it in excess.

@ringo the ° is not typeset well in math equations in mathjax, it's probably better to use ${}^\circ$.

jerepierre what I am concerned about is the by product 1,6-dibromohexane which I can't allow to be there.

In an article by Katharina C. Kress, Martin Kaller, Kirill V. Axenov, Stefan Tussetschläger, and Sabine Laschat, published in
Beilstein J. Org. Chem., 2012, 8, 371-378 (DOI), this procedure is described. From the Supporting Information, I read that the authors obtained the product in 72% yield after

column chromatography on silica gel (hexanes/EtOAc, 1:1)

I wouldn't be astonished if that's as good as it can get.


text:
Purification of Ethanol
In Vogel's organic book, they have mentioned that in the Lund and Bjerrum method of purification of ethanol, magnesium is added, which forms ethanolate. But they said to activate the magnesium first by iodine.
Why is the activation needed and how exactly does this activation process work in the activation of magnesium?

The refluxing of methanol or ethanol over magnesium is indeed the method of choice to dry these solvents.
As far as the iodine is concerned, this is a classical trick, which is also used to initiate Grignard rections. Is is assumed that magnesium iodide is formed, which is soluble in the alkanols. As a result, a fresh, activated metal surface is left behind.
This might even work, when the magnesium metal is covered with a very thin oxide layer.

Top answer here supports this. It reacts at the defects of the passivated surface and is soluble., thus increasing the surface. On another note, reading papers like this (German) make me very thankful that I was born almost a century later... mercury chloride vapour O.o


text:
Why are there more fluoride compounds formed with Xenon?
Based on WebElements, of the Noble Gases, $\ce{He}$ and $\ce{Ne}$ do not react with any of the halogens; however:

$\ce{HArF}$ has been detected at low temperatures (thank you to @bon and @Martin in the comments)
$\ce{KrF2}$ can be synthesised at low temperatures.
Xenon can react with fluorine to produce $\ce{XeF2}$, $\ce{XeF4}$ and $\ce{XeF6}$
Radon can react with fluorine to produce $\ce{RnF2}$

I understand that fluorine is the only halogen that seems to react with the Noble Gases (partly due to the related question Why Aren't Chlorides Of The Noble Gases As Prevalent As Their Fluorides?.  My question is
Why are there more fluoride compounds formed with Xenon compared to the other Noble Gases?

Do you know the three-center four-elcetron bound ?

Also $\ce{HArF}$ has been detected at very low temperatures  and confirmed to have proper bonds between the atoms.

Calculations for HArF show that there is a covalent HAr and an ionic ArF interaction: http://www.nature.com/nature/journal/v406/n6798/full/406836a0.html

Thank you @bon and Martin, I have added this compound into the question

The reason that we discovered more fluorides of xeon than helium, argon, and krypton is quite obvious and you might already know it. As lighter noble gases have more stable shell configuration, it is more difficult to make any kind of compound, including fluorides.
Also, the higher amount of fluorine atoms in a molecule of the compounds means drawing more electron from the noble gas, which is thermodynamically not preferred. Thus, $\ce{KrF4}$ and $\ce{KrF6}$ are far less stable than $\ce{KrF2}$. The quantum chemical calculation confirmed this (Dixon 2007). The same reference said that they are stable, but have not been synthesized yet, and the synthesis would be very difficult.

For radon it is a different problem. Briefly, there might be more fluoride compounds of radon than xenon, but the synthesis is difficult to be confirmed.
Even though radon is less inert than xenon and its compounds were predicted and calculated to be more chemically stable than xenon compounds (Chernick 1962, Thayer 2005), it is difficult to study. Radon is is rare and radioactive (half-life of 3.8 days). Studying chemical compounds in trace amount is a difficult task. There are claims for discoveries of $\ce{RnF4}$ and $\ce{RnF6}$ (also in the above references), but confirming that you have synthesized new compounds in trace amount is extremely difficult because the evidence is not obvious. $\ce{RnF8}$ was calculated to be unstable, but should be more possible than $\ce{XeF8}$.

References:
1 David A. Dixon, Tsang-Hsiu Wang, Daniel J. Grant, Kirk A. Peterson, Karl O. Christe, Gary J. Schrobilgen, "Heats of formation of krypton fluorides and stability predictions for $\ce{KrF4}$ and KrF6$\ce{KrF46}$ from high level electronic structure calculations," Inorganic Chemistry 2007, 46(23), 10016-10021 (https://doi.org/10.1021/ic701313h).
2 C. L. Chernick, H. H. Claassen, P. R. Fields, H. H. Hyman, J. G. Malm, W. M. Manning, M. S. Matheson, L. A. Quarterman, F. Schreiner, H. H. Selig, I. Sheft, S. Siegel, E. N. Sloth, L. Stein, M. H. Studier, J. L. Weeks, M. H. Zirin, "Fluorine Compounds of Xenon and Radon," Science 1962, 138(3537), 136-138 (DOI: 10.1126/science.138.3537.136).
3 John S. Thayer, "Relativistic Effects and the Chemistry of the Heaviest Main-Group Elements," Journal of Chemical Education 2005, 82(11), 1721-1727 (https://doi.org/10.1021/ed082p1721).

Another reason why the fluorides of the lighter noble gasses are less stable is, that the elements are smaller, which leads to a smaller fluorine-fluorine distances and higher repulsion between the non-bonding electrons (Pauli repulsion).

Most of the noble gases compounds involve Xenon, Fluorine and Oxygen. The only known binary compounds of noble gases are fluorides and oxides of Xenon and to a lesse extent of Kryton. For example, $\ce{XeF2, KrF2,  XeF4, XeF6,  XeO3,  XeOF2,}$ etc. but not $\ce{NeF4}$.
Noble gases form compounds with Fluorine and Oxygen only because Fluorine and Oxygen are the most electronegative elements. Heavier noble gases (Xenon and Krypton) are able to form compounds with other elements because  Xenon and Krypton, both have low ionisation enthalpies as compared to Helium and Neon. The noble gas elements do not react with other elements as they are less electronegative and are not able to attract the lone pair of electrons from noble elements. Fluorine and Oxygen are more electronegative elements. They are strong oxidizing agents. They can easily attract the lone pair of noble gas elements.
See also Why are inert gas (especially Xenon) compounds powerful oxidizing agents? and Why Aren't Chlorides Of The Noble Gases As Prevalent As Their Fluorides?


text:
What does it mean when it is said that an sp³ orbital has 25% s character?
My textbook frequently mentions:

$\mathrm{sp^3}$ hybrid orbital has 25% $\mathrm{s}$-character and 75% $\mathrm{p}$-character.

What are these "characters"? And how do these characters influence the bond and the hybrid orbital?

Hybridisation is a purely mathematical concept, which makes it possible to explain experimentally found structures. The most prominent example for this is methane, where you can consider the central carbon atom to be $\mathrm{sp^3}$ hybridised. Formally, the $\mathrm{s}$ orbital and the three $\mathrm{p}$ orbitals can be linearly combined to form four equivalent orbitals. Hence $\mathrm{sp^3}$ is a contraction of $\mathrm{s^{\frac{1}{4}}p^{\frac{3}{4}}}$. Therefore you can see, that the orbital has $25\,\%$ $\mathrm{s}$ character and $75\,\%$ $\mathrm{p}$ character.
Generally, the concept of hybridisation has to be treated as a concept only, it is nothing which can be physically observed. It is a mathematical model to understand geometries better. Please also note, that hybridisation is a function of the geometry, not the other way around. Having said that, it is fairly obvious, that I am not a big fan of the concept. I would not go as far as Alexander Grushow* and preach to abandon the whole thing, but I advise caution in the use of this concept to not obtain wrong conclusions.

Despite significant experimental evidence and theoretical advances to indicate that hybrid atomic orbitals do not exist and do not appropriately describe molecular bonding, their description still permeates chemical education at many levels, and the model still finds its way into modern chemical literature.

* “Is It Time To Retire the Hybrid Atomic Orbital?” Alexander Grushow, J. Chem. Educ., 2011, 88 (7), 860–862.

Wouldn't it be better to say "hybridisation is a simplification of the actual physical orbitals" rather than saying it's not physically observable outright? The orbitals do behave as hybrid orbitals, after all - that's the whole point of why they were postulated in the first place. Unless you're just saying this to avoid confusion, of course - a lot of chemistry education takes the path of "we already know this is completely false, but it mostly works and it's a lot less work, so learn to use it" :D

@Luaan I am not sure I understand what you have written. Do you have any reference for the statement, that orbitals behave like hybrid orbitals? We might be able to look at energy differences of molecular orbitals and in that way we might be able to conclude that they are observable, but hybrid orbitals are no molecular orbitals, they are just linear combinations of atomic orbitals, which are one of the more fundamental approximations, we already include. I also always thought, that they were a by-product VB theory and the obsolete VSEPR model.

Well, I meant just the fact that the resulting orbitals are all equivallent (just with "different orientation"), not the "25%s + 75%p" part; explaining the molecular structure of methane and friends. But you're right that they don't really make any sense outside of molecules, and even there they're the result of trying to map the individual parts of the molecule rather than modelling the whole molecular orbitals. Perhaps you might want to add a paragraph on just the basics of MO? Although you're right that it wouldn't really help to answer the OP's question. My bad.

The hybridisation concept allows for thinking about individual atoms (and their orbitals) within a molecule. In a molecule, it is the molecular orbitals which are important, but within reasonable approximation, you can dissect them back to atomic orbitals, but they come out mixed (hybrydised) in a proper way.

@ssavec Please see my edit in response to your comment. I really believe that thinking in localised orbitals does not help understanding structure and reactivity of compounds. I also do not see why the hybridisation concept would be necessary to think about the individual atoms in a molecule. After all, the quantum theory of atoms in molecules does not require orbitals at all. And I do not understand, what you mean with 'mixed in a proper way'.

@Martin-マーチン: AIM theory operates only on density, that's fine. But it will never tell you "energy of the bond". On the other hand, Natural Bond Orbitals (NBO) analysis gives you exactly these hybridised atomic orbitals we are talking about. So if you want to think in bonds, it is this method which shows them to you. Both AIM and NBO are just interpretation tools for the (possibly many-electron) wavefunction. AIM is atom-centric, NBO is bond-centric. I agree that hybridisation is not an important pedagogic concept. But if you wish, you can rigorously obtain the hybridised AOs.

@ssavec NBO is a localisation technique that will also never tell you the energy of a bond. It is in itself just a mathematical transformation and you lose all of the information about the eigenvalues of the orbitals. The only theory which would make sense in that framework would be valence bond theory. AIM on the other hand can be performed on a physical observable, the electron density.

@Martin-マーチン I think the discussion got too technical for the OP. The analysis of wavefunction is obviously hot topic and better be cooled with beer.

I think a good way to understand the importance of hybridization is to start with an illustration.
You have two structures: $\ce{SnCl2}$ and $\ce{SnCl4}$, and you are asked to rationalize why $\ce{SnCl2}$ has a stronger bond than $\ce{SnCl4}$ (this information is given). How would you do that?
Now there are two ways to approach this:
(You can skip this if you are not interested in the argument, but it is fairly straightforward)

The oxidation state of Sn is +4 in $\ce{SnCl4}$, while it is +2 in $\ce{SnCl2}$. The higher the oxidation state of the atom, the valence orbitals will be more contracted and lower in energy as the effective nuclear charge increases without a compensating increase in electron-electron repulsion (no electrons are added obviously). Therefore, as the coordination number is higher for $\ce{SnCl4}$ and the orbitals are more compact, this will weaken the $\ce{Sn-Cl}$ bond in $\ce{SnCl4}$.
A simpler way though, would be to say that according to the $\ce{SnCl4}$ geometry, it has to be $sp^3$ hybridized ($\ce{Sn}$ is group IV like carbon), and in $\ce{SnCl2}$, it will be $sp^2$ hybridized (2 $sp^2$-$p$ $\sigma$ bonds, 1 lone pair of electrons and 1 empty orbital). As the $sp^3$ bond has 25% $s$ character but the $sp^2$ bond has 33% $s$ character, and as the $s$ orbital is lower in energy than the $p$ orbital due to penetration effects, this means that the $\ce{Sn-Cl}$ bond in $\ce{SnCl2}$ will be stronger than that of $\ce{SnCl4}$.

So that's one use of the hybridization argument. Hope this helps in some way.

$\ce{SnCl2}$ is not $\ce{sp^2}$ hybridized.  The $\ce{Cl-Sn-Cl}$ bond angle is close to 90° (reference) implying that the central tin atom is essentially unhybridized, it uses its p-orbitals to bond to chlorine. But your thinking is correct.  Just compare bond strengths or lengths in C(sp3)-H vs. C(sp2)-H vs C(sp)-H


text:
Chemical equilibrium and Le Chatelier's principle
$$\ce{energy + 2KClO3 (s) <=> 2KCl(s) + 3O2 (g)}$$
The reaction will be more active in the reverse direction if:

$\ce{O_2}$ is added
more $\ce{KCl}$ salt is added.
More $\ce{KClO3}$ is added
Increasing the temperature.

My thoughts:
I think (1) and (2) are both right. But the book states that only (1) works and I can't find reason for that searching google was useless.
$K_c = [\ce{O2}]^3$, so if we increased $\ce{KCl}$ or decreased $\ce{KClO3}$ it won't affect the equilibrium?

$$\ce{Fe(OH)(s) <=> Fe^{2+}  + 2OH^-}$$ The reaction will be more active in the reverse direction if : ... is added

$\ce{KOH}$
$\ce{Na_2S}   $
$\ce{Fe(OH)_2}  $
$\ce{Fe(NO3)2}$

Won't it be (1) and (4)? The book says it's (4) only.

Why do you think (1) and (2) are correct?

Can you write out the formula for the equilibrium constant?

Because if I added KCL le chattele ( never written his name right)  says the reaction will go in the direction which makes it get back to equilibrium and so will happen if we added o2.  maybe it's misunderstanding from me.

Ok ,    Kc = [0_2]^3

You're on the right track there, please consider adding your thoughts presented in your comments into your question.

Now I think I understand. When I add o2 I will increase the number of moles  and as a result the concentration will increase but the KCl is out of the Kc so it won't affect the equilibrium , right ? The problem is that my teacher told me that is wrong ...

I think your reasoning is sound,  (have upvoted as well)

Why do you think  KCl is out of the Kc? You're wrong...

Because it's solid (long explanation). H2O liquid is not counted too

@Mithoron perhaps expand on that into an answer?

Help , now he confused me more.

Bro, I think I am right check this : http://www.chemguide.co.uk/physical/equilibria/kc.html       (Writing an expression for Kc for a heterogeneous equilibrium)  You will see "The important difference this time is that you don't include any term for a solid in the equilibrium expression." this too http://bouman.chem.georgetown.edu/S02/lect8/lect8.htm

You are correct @HoytVolker - and the answer below provides a good explanation emphasising what you found.

But I don't understand what he does mean by oxygen pressure I think we are not playing with pressure , are we ?

@mamun's comment below explains what is happening

@HoytVolker I would suggest taking the 2nd question you edited in and writing that as a separate question, linked to this one.

Ok , thanks for the suggestion , I will. Ops , seems that I can't I have reached the limited 2 questions.

1) Increasing Oxygen pressure will increase the reverse rate because to reestablish equilibrium some of the oxygen should be consumed.
2) Out of system due to precipitation.
3) Out of system due to precipitation.
4) This is endothermic reaction. So increasing temperature will increase the forward rate to consume the added heat.

"Oxygen pressure" ??

Adding gas phase oxygen means increasing the partial pressure of oxygen.

Because we are increasing the number of atoms in Unit of volume ?  p(ro) = m/Vol

Yeah. Think about it from mathematical point of view: You have a constant K which has numerical value of 100. K is also equal to ${O_2^3}$. So that means ${O_2}$ is 4.64 bar roughly. Now you increase it to 5 bar and the equilibrium wants it to stay at 4.64 bar. So the equilibrium will consume 0.35 bar worth of oxygen molecule to reestablish it to 100.

A reference that helps explain what is happening with the oxygen is at http://chemwiki.ucdavis.edu/Physical_Chemistry/Equilibria/Chemical_Equilibria/The_Equilibrium_Constant/Effect_of_Pressure_on_Gas-Phase_Equilibria

Thanks I am checking it atm @santiago. Thanks too mamun maths explanation is better , but please check the new example it confused me.

authors: เพชรลัดดา บุดดี


text:
Clarifications about the mechanism of the Wittig reaction
Ok so this is the proposed mechanism for the Wittig Reaction as stated by my textbook, LG Wade. I have some doubts regarding this mechanism:

1) Why is only triphenyl phosphine used here? Why not some stronger
  nucleophile like say trimethyl phosphine?

  2) Why would the lone pair on P attack the C of the alkyl halide in the
  first place? Isn't the lone pair in resonance with the benzene ring?
3) Why is butyllithium used particularly.Can't any other reagent be
  used to abstract H ?
4) Why is the four membered ring formed in step 2?
5) Why does the ring collapse in step 3?

Note: I'm just a beginner in organic chemistry (Org 2) so it would be helpful if you can keep adjust your answer to suit my level of understanding.

It is preferred if you can post separate questions instead of combining your questions into one. That way, it helps the people answering your question and also others hunting for at least one of your questions. Thanks!

That would be making 5 posts on the same topic ! @Loong

related: http://chemistry.stackexchange.com/questions/5313/why-is-the-wittig-reaction-syn-stereoselective

Yes, but splitting the questions helps keep the answer chains clean.  It's perfectly all right to hyperlink among all the various questions to make it clear that they're part of the same broader search for understanding.

@SanchayanDutta - Multiple question posts is preferred. Think about it this way: What if I know the answer to questions 1) and 2) from your post, but  I do not know how to answer questions 3), 4), and 5)? I might not feel that I can post anything since I cannot answer all five parts. If you post them separately, then I can answer the ones I know how to answer and leave the others for people who know more than I do.

@SanchayanDutta - You can also think of it this way - More posts means more chances for everyone to earn rep, badges, privileges, etc.

Ok I understand your point @BenNorris.But think from the asker's perspective also.Chem SE allows one post per 30 mins.So it will take 30*5 mins for asking this question part by part.That means almost 2 and a half hours!!Thats really a taxing job.But,well as you are saying I'll try that from my next post.

1) Why is only triphenyl phosphine used here. Why not some stronger
  nucleophile like say trimethyl phosphine?

Trialkyl phosphines can particpate in the Wittig reaction just fine, but they create some other problems.  Let's say you were going to create the phosphonium salt from ethyl iodide and trimethyl phosphine.  It would form in high yield, but in the next step where you deprotonate it and form the ylide you would have a problem.  You could form the ylide from deprotonation of the desired ethyl group, or you could form an ylide by deprotonating any of the 3 methyl groups.  This would give rise to a mixture of ylides and consequently a mixture of products.  On the other hand, there are no protons on triphenyl phosphine that can be removed once the phosphonium salt is formed.
The following additional reasons also favor usage of triphenyl phosphine:

it is cheap
it is very cheap
it is a solid and easy to handle and weigh out, many of the smaller trialkyl phosphines are liquids
triphenyl phosphine is less toxic and doesn't smell as bad as many of the smaller trialkyl phosphines

2) Why would the lone pair on P attack the C of the alkyl halide in the
  first place? Isn't the lone pair in resonance with the benzene ring?

Triphenyl phosphine has the aromatic rings arranged in a non-planar, propeller-like geometry which makes delocalization less likely.  Also $\ce{2p-3p}$ overlap would be involved which is not as effective as $\ce{2p-2p}$ overlap.

3) Why is butyllithium used particularly. Can't any other reagent be
  used to abstract H?

In the phosphonium salt, the protons alpha to phosphorous are not very acidic (pKa ~ 20-35), a strong base is required to remove them.  Other strong bases could be used, but BuLi is commercially available in high purity at a reasonable cost.  When BuLi abstracts the proton butane is formed and escapes as a gas - one less side product to worry about separating out.

4) Why is the four membered ring formed in step 2?

The 4-membered ring may not be as strained as you imagine.  Remember, phosphorous is in the third row of the Periodic Table.  It has longer bond lengths and can tolerate smaller bond angles.

5) Why does the ring collapse in step 3?

The bond between phosphorous and oxygen is an extremely strong bond, in other words its formation is very exothermic.  Therefore, this last step is being driven by thermodynamics.  It is the exothermicity of this step that drives the reaction to completion.

Actually, $\ce{KO}t\ce{Bu}$ is often used for Wittig reactions in my group (phosphane usually being $\ce{PBu3}$) and my supervisor even said that it be way to strong — prompting me to try diazabicyclo[5.4.1]undecene (DBU) which worked like a charm on the test reaction. (Not on my product because my product is ugly, though.)

This is indeed a very good answer! Have you been gone a while? I don't remember reading something from you for at least two weeks... For your enjoyment, I found a quite comprehensive article about the Wittig reaction: Bruce E. Maryanoff , Allen B. Reitz, Chem. Rev., 1989, 89 (4), pp 863–927.. Especially the part about betaines and oxophosphetanes is interesting.


text:
Clemmensen reduction vs. Wolff-Kishner reduction
In which cases is the Clemmensen reduction is preferred over the Wolff-Kishner reduction?
One thing is that the Clemmensen reduction is done under acidic conditions whereas the Wolff-Kishner reaction is done under basic condition, and that I think decides somewhat which to choose depending of the nature and other groups present in the reactant.
But are there any other points to keep in mind while choosing the best suitable reaction?

hydrazine is known to produce heterocycles with 1,3-diketones.

Clemmensen reduction is particularly effective at reducing aryl ketones, such as those formed from Friedel-Crafts acylations. However, acid-base is the primary consideration. Also remember the Mozingo reduction as a mild alternative.

A key consideration for the Wolff-Kishner reduction is thermal stability of the substrate. High temperature is required for the classical reaction using hydrazine and KOH, sometimes up to 200 °C. This can rule out plenty of compounds.
The Clemmensen reduction may be fine for the laboratory,  but fundamentally many industries wouldn't use it due to scaling up the use of mercury.
Scalable alternatives are molecular hydrogen with a catalyst under acidic conditions or, the trifluoroacetic acid-silane combination, such as triethylsilane or tetramethyldisiloxane.

The basic difference is the medium as Clemenson is in acidic medium so if $\ce{OH}$ is present so it will be attacked too, but in Wolf-Kishner it is in basic medium so $\ce{OH}$ group can be saved which is sensitive, but if there is halogen, ester, or amide group it will not tolerate the reaction.


text:
Selective alkylation of carboxylic acid versus phenol

Starting from the ketone shown, I plan to use a haloform reaction to transform the acetyl group into a carboxylic acid.
Following that, there are two alkylations to be carried out: one esterification on the carboxylic acid (using propargyl bromide) and one etherification on the alcohol (using 6-bromohexan-1-ol). In what order should these be performed?

Using the haloform reaction to convert the methyl ketone 1 to the corresponding carboxylic acid 2 is a neat idea. According to an old review by R. C. Fuson and B. A. Bull, published in Chem. Rev., 1934, 15, 275-309 (DOI), this seems to work for 4-hydroxyacetophenone. Your starting material 1 is a phenyl homolog of that.

In 2, you have two acidic protons ($\ce{-COOH}$ and $\ce{-OH}$). Based on our knowledge and by comparison with 4-hydroxybiphenyl ($\mathrm pK_\mathrm a = 9.55$), benzoic acid ($\mathrm pK_\mathrm a = 4.2$) and 2-phenylbenzoic acid ($\mathrm pK_\mathrm a = 3.46$) (all data are taken from the Rubber Book), you know that

the carboxylic acid is more acidic than the phenol
the difference is acidity is quite large

Potassium carbonate ($\ce{K2CO3}$) in a polar aprotic solvent, such as N,N-dimethylformamide (DMF) should be sufficient to deprotonate both centres.
Back to your question on the order of events:
I would add 1 equivalent $\ce{K2CO3}$ of to a solution of 2, stir at room temperature for a while and then add the propargyl bromide, assuming that the large differences in acidities will make sure that the carboxylic acid is deprotonated, while the phenol is not.
Monitoring the reaction by TLC will show whether the esterification is quantitative.
If this is the case, I would dare to run a one-pot reaction, add another equivalent of $\ce{K2CO3}$, stir again and then add the bromohexanol to form the ether.
It might be a good idea to try this first with very small amounts, just enough to check via TLC.


text:
Why doesn't EDTA complex with alkali metal ions?
EDTA complexes with all the other metal ions in the periodic table except those from the group 1. Why is this so ? What is the coordination chemistry behind this ?

That's funny, the label on my bottle says Ethylenediaminetetraacetic Acid, Disodium Salt.

@LDC3: That is $\ce{Na2EDTA.H2O}$. In that compound, $\ce{Na+}$ ions are not complexed to EDTA.

@chemkatku Can you explain how you differentiate between "complexed?" I guess your argument against Na is that it's ionic and not a coordination bond?

It does. I’m willing to bet my life it does. Only that the energy gained from complexation is so little and the process so easily reversible that there is no inert/favoured complex form that could be observed easily.

@GeoffHutchison Yes that is what I think. I don't know about any way of  complex formation without formation of coordination bonds.

Alkali metal cations are very "hard", in the HSAB theory of acid-base interactions (roughly corresponding to 'poorly polarizable'), and thus only coordinate in the conventional sense of the term to "hard" ligands such as ethers.  Crown ethers, in particular, are known for their alkali metal ligating/chelating capability.
The amine nitrogens and carboxyl oxygens of EDTA are sufficiently 'soft' that, as Jan noted in a comment to the question, there is sufficiently little energetic benefit derived from EDTA chelation of a 'very hard' alkali cation such that the complexes are not practically observed.  Instead, the chemistry is dominated by electrostatic interactions of the positively-charged alkali cations with the negatively-charged carboxylate groups (when deprotonated, anyways).

If you're going to be picky that a "complex" doesn't form without covalent coordination bonds, then that's your answer. Alkali metals will tend to form cationic salts with EDTA.
Personally, I agree with the other comments. EDTA does form complexes with alkali metals.
I'd classify the disodium salt of EDTA as a complex. It has a metal and an organic ligand, and there's no direct metal-carbon bond (which would make it organometallic). As discussed in another question, it's a pretty arbitrary distinction between an ionic bond and a covalent one:

What happens if the electronegativity difference is exactly 2.1?

This distinction is particularly cloudy in the case of metal complexes, since most metals readily gain positive charge.

I think crown ethers and cryptands provide unquestionable examples of alkali metal ion coordination? If those do, then that establishes a scale of coordination strength which never truly hits zero.

@NicolauSakerNeto This is my opinion as well. My feeling is that there's a metal, a ligand, and they're close enough for some measure of covalent bonding. All bets are off if the ions are far apart, but in EDTA, it's at least partially covalent.

@NicolauSakerNeto After looking closely at some $\ce{Cu(II)}$ aquo complexes, I have to agree with you, Geoff. Copper indisputably forms complexes, but its interactions with water ligands appear almost purely ionic. In my computation of $\ce{[Cu(H2O)5]^{2+}}$, the line critical point joining the axial $\ce O$ with the $\ce{Cu}$ has $\rho=0.05$ and $\nabla^2\rho=+0.26$, and all of the equatorial LCPs have $\rho\approx 0.07-0.08$ and $\nabla^2\rho \approx +0.4$. Thus, ionic EDTA-$\ce{Na+}$ interactions could also be called coordination.

$\ce{Na2EDTA.H2O}$:
First of all the EDTA sodium salt is an ionic compound with complexed water. A sodium ion forms an ionic salt with the carboxylate group in two of the four acetic acid molecules bound to the amine groups. The salt has one water of crystallisation.
Solution Chemistry:
 I'd say the high ionic radius and hardness of group I metal ions against the soft carboxylate make chelation less favourable with respect to softer metal ions, such as group III metals and transition metals.
Affinity constants:
\begin{array}{lr}\hline
\ce{Al (III)} &  16.13\\
\ce{Ca (II)}  &  10.96\\
\ce{Mg (II)}  &   8.69\\
\ce{Pb (II)}  &  18.04\\\hline
\ce{Li+}      &   2.79\\
\ce{Na+}      &   1.66\\
\ce{Cs+}      &   0.15\\\hline
\ce{Fe (II)}  &  14.33\\
\ce{Fe (III)} &  25.10\\
\ce{Mn (II)}  &  14.04\\
\ce{Mn (III)} &  24.80\\\hline
\end{array}


text:
How can an inexperienced chemist determine the chemical structure of a molecule?
I'm a fairly inexperienced, but nonetheless interested chemist, currently planning a small project for the coming summer: I want to isolate an unknown chemical from a plant and determine it's chemical structure.
I have ample access to Carex grayi, a sedge plant, and have chosen this plant for a few reasons.

Of the plants growing near where I live, I can definitively identify Carex grayi. I am relatively inexperienced in taxonomy and plant identification, and I want to know which species I am working with.
As far as I can tell, the plant is relatively unresearched, and may or may not contain alkaloids.

This brings me to my next point. I know it may seem immature or disheartening, but my interest in chemistry stemmed from my interest in neuroscience and neurotransmitters. From countless hours of reading, I have learned much about the pharmacological (and sometimes psychoactive) tendencies of alkaloids. I realize that ingesting unknown alkaloids is extremely dangerous, and I do not intend to ingest anything. However, due to their numerous applications in medicine and recreational use, there is a massive amount of information on the internet regarding their extraction and isolation. I am entirely open to the idea of isolating a chemical from another group, but as I am inexperienced and my interest stems from neuroscience, I only really know about alkaloids. I am only somewhat familiar with terpenoids, and know little about any other group.
So, assuming I am able to extract an unknown chemical of some specified group from a plant, is there any way for me to determine the chemical's structure? The Chemical structure Wikipedia page lists the following methods for determing chemical structure:

X-ray diffraction
Proton NMR
Carbon-13 NMR
Mass spectrometry
Infrared spectroscopy

I don't really have access to these tools as student over the summer. I am willing to spend a small amount of money on this project, but no more than about one hundred dollars. Is it too much or too difficult for a student to determine a chemical's structure over the summer? Should I do anything differently (as in, is there a certain group of chemicals that would make this project possible or easier)?
Also, is there a "classical" chemical extraction similar to this that most fledgling chemistry students would conduct? This is obviously a ridiculous made-up example: extract dandylionine from a dandy lion flower.

A nice and quite easy exercise for practising is the extraction of limonene from orange peel. Maybe you would like to try that first to get a hold on things...

So, assuming I am able to extract an unknown chemical of some specified group from a plant, is there any way for me to determine the chemical's structure?
Yes.
I don't really have access to these tools as student over the summer. I am willing to spend a small amount of money on this project, but no more than about one hundred dollars.
Then probably no (at least not with the same accuracy you would if there weren't these constraints).
Mass spectrometry and nuclear magnetic resonance are the two most powerful analytical tools to identify completely unknown substances. This doesn't mean it's as trivial as putting your purified substance in the equipment and finding its structure. For reliable results you need a purified substance, an optimized experimental routine and of course the instrument properly set up.
In fact, before trying to determine the structure of the substance, you must purify it, and that alone is a challenge. As I'm sure you are aware, the plant contains thousands, millions of different molecules. How are you going to extract something you haven't defined? The first step would have to be defining what is it that you're looking for.
Let's consider you're looking for whatever makes plants green:
You could macerate the plant, add some water, filter the mixture, check the colors of the filtrate and of the residue, and you would notice most of the green wasn't extracted. You would find different results with different solvents though, and a better option would be ethanol. Even then, it wouldn't mean you have a pure chlorophyll solution, so several other steps would probably be required to purify the substance*. Generally speaking, you would use some kind of chromatography.
For your project, I suggest you try to first precisely define what is it you're looking for. The substance that gives the plant its color? Its smell? Its pharmacological effects? You must use this property to analyze your experimental results. You can visually inspect the color, you can smell extracts, and perhaps test physiological effects (this is obviously dangerous and shouldn't be done unless you really know what you are dealing with!). Use the target substance properties' to decide what's the best way of extracting and purifying it.
Then, considering you did purify something, without the high-tech analytical tools mentioned, there are some experiments which can indicate what you have, but they require a reference substance and usually can't be used to say that a substance is the same as another, but rather that a substance isn't the same as the reference. For example, you can compare boiling/melting points and/or thin layer chromatography retardation factors, both which can be done at home home (to some extent), and also UV/Visible light absorption/emission, which might not be trivial to do at home but require instruments that are a lot easier to find in chemistry teaching labs.
For precise tips on the extraction and identifications, it would be necessary to know what you're looking for.
It's also important to note that there's a lot you can do besides determining a chemical structure. There are several properties of a substance you can study in order to characterize it. Some examples are solubility (in different solvents, at different conditions (temperature, pH, salinity)), density, reactivity (in face of pretty much anything you find at home) and physical properties such as texture, color, smell, etc.
Finally, do not think I'm saying you should give up, your wish to discover what something is made of is what made chemistry a science, but you need to have a clearer problem to be solved. Decide precisely what it is you're looking for (in terms of macroscopic properties, chemical group or something like that), and then research (and ask!) what could you do about that without a full lab.
*Chlorophyl is actually a group of substances which have a critical role in photosynthesis, and are, of course, green.
In summary:

You already have your sample. Now choose your analyte, the molecule(s) you want to study.
Do research. Look for information on how people usually study this (group of) substances. You'll have better luck looking in education-related sources, which usually have simpler experiments.
Once you've found something that fits your interests and available material, do experiments. Observe the results. Change parameters, do them again, try to understand what's happening.
If you can't explain the results, do research, check your references, and of course, ask here. We'll be glad to help.

Nicely done! But "your wish to discover what something is made of  is what made chemistry a science." In fact, it was what made physics a science. Chemistry used to be a branch of physics, and it was till the eighteenth century. But you might go on, I'm being overly pedantic here! :P

@M.A.Ramezani I actually thought I wasn't being very precise when I added that, but I just wanted to point how curiosity is important to science as a whole!

Indeed, the usual thing to do with a plant that has interesting properties is isolate the substance with interesting properties (colour, smell, psychoactivity, whatever) confirm that you have indeed isolate it (pretty easy with colour or smell, less so with psychoactivity) and only then invest the considerable effort required to elucidate its structure. Finally, spend several months synthesizing it from scratch to prove (hopefully!) thatyou got the structure right.


text:
How to perform oxidation of keto group in presence of hydroxyl group?
How to perform oxidation of keto group in presence of hydroxyl group. I fear there may be halogenation at $\ce{OH}$ group. Is there a way to protect the $\ce{OH}$ group ?

Protect the hydroxyl group, e.g. as an ether. Have a look here.