Bromination of alkenes In regard to the mechanism of halogen addition, consider the situation when we add a halogen (say, bromine) to an unsymmetrical alkene about the double bond (say, propene) in the presence of $\ce{CCl4}$.

I knew the mechanism would be something of this sort:

  A cyclic bromonium ion would be formed with the two carbons of the double bond. The second bromide ion (acting as a nucleophile) attacks on the more substitued carbon from the side opposite to the bromonium ion to cause ring opening and formation of a vic-dibromide.

My teacher however tells me that other than ethene, no other alkene forms bridges like these, because of unstability of bromonium complex due to repulsion between substituents on carbon and the bridged element.

I have two questions in this regard:

Which of the two is the correct mechanism?
If the former is indeed the correct mechanism, then would the nucleophile  attack the least substituted carbon? (seeing as how the partial positive charge on the more substituted carbon is nuetralised to a greater extent due to $+I$ effect of attached methyl/ethyl groups).

The first mechanism is more likely unless specific conditions of the experiment dictate otherwise. In the presence of $\ce{CCl4}$, the addition of $\ce{Br2}$ takes place via the formation of a cyclic halonium ion as intermediate as follows:

Yah, there are two mechanisms for this, one is as you wrote and the other one is th stereospecific reaction where stereoisomeric starting substrate produce products which are stereoisomers of each other.

First $\ce{Br2}$ will dissociate into ions:-

$\ce{Br2-> Br+ (electrophile)  + Br-(nucleophile)}$
$\ce{H2C=CH2 + Br+ -> H2(Br)C-C+H2(carbocation)}$   slow
$\ce{+CH2-CH2Br  + Br- -> CH2Br-CH2Br}$

Symmetrical alkenes and Symmetrical reagents, addition reaction gives:-

$\ce{cis ~alkene + syn ~addition -> meso}$

$\ce{trans + syn~addition -> racemic}$

$\ce{cis + anti~ addition -> racemic}$

$\ce{trans +anti ~addition -> meso}$

You can watch this video (http://m.youtube.com/watch?v=5GQelnluHzE)

My teacher however tells me that other than ethene, no other alkene forms bridges like these, because of unstability of bromonium complex due to repulsion between substituents on carbon and the bridged element.

For this I have not heard any thing as such and also couldn't find it.

For the most common mechanism which i was taught was the first one with bromonium ion complex formation.

As for your doubt about nucleophilic attack,
negative bromide anion is attracted to the slight positive charge on the carbon atoms. It is blocked from nucleophilic attack on one side of the carbon chain by the first bromine atom and can only attack from the other side.

Personally how I solve it, is to check for the stability of the carbocation formed just before the nucleophilic attack practically because its $\ce{SN2}$ mechanism  carbocation is not formed but I just use it like this best way is to check for high electron density. Though when it comes to aryl group,  it is seen that it attack to the place with most substituted C-atom.

Here the stability was checked by hyperconjugation. Inductive effect is not the most effective way. You should know that the priority order of these effects  to check the stability,

$$\mathbf{aromatisation~\Rightarrow~resonance~\Rightarrow~hyperconjugation~\Rightarrow~inductive effect}$$


Why is the bond angle H-P-H smaller than H-N-H? $\ce{N}$ & $\ce{P}$ are in the same group. Both $\ce{NH3}$ and $\ce{PH3}$ have one lone pair and according to VSEPR theory, both the central atoms are predicted to be $\ce{sp^3}$ hybridized.

But in spite of that, the bond angle in the former is $107^\circ$ while that in the latter is $92^\circ$. What is the cause of such a difference?

Starting point: 2s orbitals are lower in energy than 2p orbitals.

The $\ce{H-N-H}$ bond angle in ammonia is around 107 degrees.  Therefore, the nitrogen atom in ammonia is roughly $\ce{sp^3}$ hybridized and the 4 orbitals emanating from nitrogen (the orbitals used for the 3 bonds to hydrogen and for the lone pair of electrons to reside in) point generally towards the corners of a tetrahedron.

In the analogous case for phosphorus (phosphine, $\ce{PH_3}$), the $\ce{H-P-H}$ bond angle is 93.5 degrees.  This angle indicates that the phosphorus atom is almost unhybridized (the bond angle would be 90 degrees if it were completely unhybridized). The 3 bonds from phosphorus to hydrogen roughly involve the three 3p orbitals on phosphorus and the phosphorus lone pair of electrons resides in the 3s orbital of phosphorus.

So the question becomes, why does the nitrogen atom in ammonia choose to hybridize, while the phosphorus atom in phosphine does not? Let's start by listing the factors that will stabilize or destabilize geometries in these compounds.

There are two choices for the central atom:

remain unhybridized: [$\ce{p}$ orbital - H1s] bonds will form and they will be arranged 90 degrees with respect to one another.  As a result,

substituents will be arranged closer together and destabilizing steric interactions will be increased
due to the absence of s-character in the $\ce{X-H}$ bonds emanating from $\ce{X}$, the electrons in these orbitals will be higher in energy
the lone pair electrons will be highly stabilized since they will reside in a low energy, pure s orbital

or

hybridize: [$\ce{sp^3}$ orbital - H1s] bonds will form and they will be arranged 109 degrees with respect to one another.  As a result,

steric interactions will be reduced because the tetrahedral orbital arrangement will space the attached substituents further apart
due to the presence of s-character in the $\ce{X-H}$ bonds emanating from $\ce{X}$, the electrons in these orbitals will be lower in energy
the lone pair electrons will be less stabilized since they will reside in a higher energy orbital that contains significant p-character

An example:

Let's now consider the example of $\ce{NH_3}$ and $\ce{NF_3}$.  Fluorine is much more electronegative than hydrogen, therefore we would expect electron density in the $\ce{N-F}$ bond to be shifted away from nitrogen towards fluorine.  Because of this electron redistribution, the $\ce{sp^3}$ orbital on nitrogen involved in this bond will contain less electron density.  Consequently it will rehybridize - if there is less electron density in the orbital, there is less of a need for lower energy, electron stabilizing s-character in this orbital.  The orbital will wind up with higher p-character and the s-character that has been "saved" can be used to stabilize other electrons (the lone pair!).  Our prediction would be that $\ce{NF_3}$ should have more p-character in its $\ce{N-F}$ bonds than $\ce{NH_3}$ has in its $\ce{N-H}$ bonds.  As a result we would expect the $\ce{F-N-F}$ bond angle in $\ce{NF_3}$ to be smaller than the $\ce{H-N-H}$ bond angle in $\ce{NH_3}$.  Indeed the bond angle in $\ce{NF_3}$ is 102 degrees compared to the 107 degrees observed in ammonia!

Back to our problem:

Nitrogen (3.04) is more electronegative than phosphorus (2.19), which is about the same as hydrogen(2.2).  In our $\ce{X-H}$ bonds, we would therefore expect more electron density around the central atom when $\ce{X~=~N}$ than when $\ce{X~=~P}$.  Using the same reasoning used in our example, we would then expect the $\ce{N-H}$ bonds in ammonia to have higher s-character (and a larger $\ce{H-N-H}$ angle) than the analogous bonds in phosphine, just as observed.  The fact that phosphorus, being a second row element, has longer $\ce{P-H}$ bonds (142 pm) than ammonia (102 pm) lessens steric problems in the unhybridized geometry and further lowers the energy of the unhybridized configuration for phosphine.

In the case of ammonia, the shorter $\ce{N-H}$ bond lengths (increased steric interactions) and the increased electron density in the $\ce{N-H}$ bonds makes the  hybridized case the lowest energy.  Whereas in the case of phosphine, steric interactions are of less consequence because of the longer bond lengths and the decreased electron density in the bonds around phosphorus make the energetics of the (nearly) unhybridized geometry more favorable.

Since Phosphorus has vacant 3d orbitals, the s-orbitals of hydrogen can directly overlap with the orbitals of Phosphorus. The procedure of bonding does not see much benefit in hybridisation as it is energetically expensive. Thus the hybridised character of the bonds is less dominant than in ammonia and the structure resembles the Lewis structure that we would expect instead of the tetreahedral (VSEPR) structure. This effect is called Drago's Rule and can also be observed in hydrogen Sulfide.


How do I take chemistry notes on Mac? I am learning Chemistry now, and I want to take some notes that involve huge amount of organic chemistry on Mac. How can I do that? Pages does not seem support scientific use.

If you know LaTeX and you want to see what you are writing you can use lyx, and some TeX package like mhchem and chemfig. So you can write all the things you want and add comments. This is what I have done with my Mac just now:

Here is the code, you can make these lines as a default template:

\documentclass[12pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}
\usepackage{chemfig}
\usepackage{tikz}
\usepackage[version=4]{mhchem}
\setlewis{4pt}{0.8em}{black}

From here you have to write! The date is automatically added!

\title{Introduction to chemistry notes}
\begin{document}
\maketitle
\section{mhchem}
You know that \ce{ CH4 + 2O2 -> CO2 + 2H2O}     this is a simple equation\ldots
Okay maybe you can do this with word, but if you are familiar with Chemistry S.E. you will be very fast with this notation.

\ce{Zn^2+
<=>[\ce{+ 2OH-}][\ce{+ 2H+}]
$\underset{\text{amphoteres Hydroxid}}{\ce{Zn(OH)2 v}}$
<=>C[+2OH-][{+ 2H+}]
$\underset{\text{Hydroxozikat}}{\ce{[Zn(OH)4]^2-}}$
}

\section{Chemfig}
\begin{figure}[h]
\setatomsep{1.8em}
\schemestart
\subscheme[90]{\chemfig{*6((-[@{a1}]H_2@{a2}\Lewis{26,N})=[@{r1}]-[@{r2}]=[@{b2}](-[@{b1}]N =[@{c1}] N -[@{c2}]*6(=[@{d1}]-[@{d2}] =[@{e1}](-[@{e2}] OH)-=-))-=-=)}\chemrel{<->}
\chemmove[-stealth,shorten <=1pt, shorten >=1pt]{%
    \draw(a2)..controls +(90:5mm) and +(135:5mm)..(a1);% first arrow
    \draw(r1)..controls +(225:12mm) and +(300:12mm)..(r2);% ring arrow
    \draw(b2)..controls +(10:3mm)..(b1);% ring arrow
    \draw(c1)..controls +(340:8mm)..(c2);% ring arrow
    \draw(d1)..controls +(225:12mm) and +(300:12mm)..(d2);% ring arrow
    \draw(e1)..controls +(20:4mm)..(e2);% ring arrow
    }
\chemfig{*6((=[@{a1}]H_2@{a2}\chemabove{N}{\oplus})-[@{r1}]=[@{r2}]-[@{b2}](=[@{b1}]N-[@{c1}] N =[@{c2}]*6(-[@{d1}]=[@{d2}] -[@{e1}](=[@{e2}] \chemabove{O}{\ominus}H)-=-))-=--)}
}
\schemestop
\caption{This is a coniguated system\ldots}
\label{fig:con}
\end{figure}
Okay the lesson will end and in one hour you'll draw this figure only, maybe you should use an external software\ldots %what the professor means whit this last phrase?!
\end{document}

Sincerely I think that is best to take handwrite notes because I think that writing and drawing chemical formula allow you to memorize better but of course this is only a thought.

Try Avogadro http://avogadro.cc/wiki/Get_Avogadro

It's an advanced molecule editor and visualizer designed for cross-platform use in computational chemistry, molecular modeling, bioinformatics, materials science, and related areas. It offers flexible high quality rendering and a powerful plugin architecture.


Claisen ester condensation Why is it a requirement that the ester should have 2 alpha hydrogens in this reaction although we actually need just one for the reaction to occur?

Great question!

The Claisen reaction happens under equilibrium conditions, and the addition of the ester enolate is highly reversible. Your mechanism omits the very important end of this reaction, which is that the β-keto ester product is highly acidic and easily deprotonated to yield the corresponding enolate.

The enolate formation drives the reaction forward according to Le Châtelier's principle. Very importantly, your question about the α-hydrogen. The deprotonation of the β-keto ester is not possible without two hydrogens on the original ester that will serve as the nucleophile.

Two hydrogen  are use , because if only one H is used then again the product form having acidic H which is abstracted by ethoxide ion then this ethoxide ion convert in Ethanol and final product is obtained.


The effect of NaCl on the ability of an acid to protonate benzoate I've been doing several experiments using soda to quantify the amounts of caffeine and benzoate. Because of our methods of analysis, we had to decarbonate our soda. The experiment offered two ways for decarbonating the soda: shaking it or heating it. Then, to make sure all the benzoate in the soda is converted to benzoic acid, we add a small amount of $\ce{HCl}$ (to protonate it.)

Several years ago, I did a similar experiment to quantify the amount of caffeine. To decarbonate the soda, I added salt ($\ce{NaCl}$.)

I'm wondering if adding salt would be inappropriate in the more recent experiment. None of the lab procedures mention adding salt and, to me, this is the easiest way to decarbonate soda. Therefore, I'm thinking there may be a specific reason the salt method was left out of the procedure.

My thought is there could be some interaction between $\ce{Na+}$ and $\ce{Cl-}$ that could "neutralize" the $\ce{HCl}$ so to speak, making the protonation of benzoate less effective.

Perhaps it would make sodium benzoate instead of benzoic acid? Perhaps they just didn't think to add salt. Any thoughts on this?

Some salts added obviously will alter the pH value of the soda ($\ce{NaHCO3}$, for example), and others not (like the example provided by you, $\ce{NaCl}$). Changing the ionic strength by addition of salts may render your mathematical model to describe the composition of the analyte more complex.

Hence, let me I suggest an alternative to outgassing the soda: immerse the open container into an ultrasound bath.  This works rapidly, however ensure the container is widely open and may accommodate the foam briefly generated.  Otherwise, it may surprise you as a sticky geyser.  Second hint: gently approach the active regions in the ultrasound bath to moderate the gas evolution.

(Degassing solvents with ultrasound is a common preparation for HPLC, too.)

Na+ + BzCOO- ◄► BzOONa is an equilibrium. Part of the answer is that increasing [Na] will shift the reaction towards the right. I guess heating the liquid makes sense, I've no idea how consistently you can remove the excess CO2 by shaking. I've also no idea how NaCl changes the solubility of CO2. Have you considered doing simple weight change experiments to compare the three techniques? You should. You should also consider that, in general, adding anything to an analyte increases the chances that there will be interference. Since the analytical techniques (as well as the work-up processes) weren't specified (to be honest, I wouldn't be able to help any more if they were, I'm almost certainly not familiar with the chemistry behind them), there's no way for me to say whether NaCl would interfere or not. I'm sure you understand there are other chemicals present in the "soda", so that your idea of adding NaCl would require pretty stringent verification of its "non effect" on the results. I sure wouldn't go there. I've been involved in the creation of analytical techniques and the process can be quite tedious as well as lengthy. (Co-incidentially, CO2 absorbed from the atmosphere turned out to influence not only the NaOH reagent solution (virtually no storage stability), but also the result of the titration (both time and mixing speed) itself. I sure wouldn't go there. But that's what literature searches are for.


How is (SO4)^2- Charged if S and O all are Neutral Themselves? As far as I understand $\ce{SO4^2-}$ structure,  sulfur makes two coordinate bonds with two oxygen atoms and two double bonds are formed with oxygen after sulfur expands it's valance shell to 6. At this point calculating formal charge of those two coordinate bonded oxygen gives total formal charge of −2 and the formal charge on $\ce S$ and the double bonded $\ce O$ is zero. Thus the minus 2 charge on $\ce{SO4^2-}$ can be shown but initially all the atoms were neutral. While making bonds electrons were shared only.

How did excess 2 negative charge pop up?

Please help, I am certainly making some mistakes

The ion is not one S atom and 4 O atoms together. The ion on the whole is having two negative charges, Thus it is somewhat like 1Sulphur + 4 Oxygen+2 electrons. It is not as though there was one sulphur and four neutral oxygen atoms, and they became negatively charged during bond formation. The bond must have formed in some reaction, and in that reaction the charge would have been conserved.

An S atom has 6 valence electrons and a O atom has 6 valence elections

SO4 => (6x1) + (6x4) = 30 valence electrons

We would need 32 valence electrons for 4 full outer shells(8x4) but only have 30. Thus the charge is 2- for the ion.


Resultant temperature of mixture **Question: ** There are 2 flasks of equal volume, connected by a narrow tube of negligible volume, filled with nitrogen gas. When both are immersed in boiling water, the gas pressure inside the system is 0.5 atm. Calculate the pressure of the system when one of the flasks is immersed in ice water mixture, keeping the other in boiling water.

Problem: There are two ways that I am going to describe that I can think of to solve the question, but they give different answers.

Way 1: Net Temperature of the system should be 50 degree Celsius because there is no heat loss and there is symmetry. As both the flasks are identical and at 100 and 0 so it should be 50.
Calculating the answer with this assumption and the formula
P1/V1=P2/V2
The answer comes out to be about 0.432 atm

Way 2: The sum of the moles of gas in both the flasks would remain same.
And using the ideal gas equation to get n = number of moles = PV/RT
After equating the values. The pressure come out to be about 0.422 atm

I was bothered with the 0.1 change. So I took the liberty to calculate the temperature from way 2, and it came out to be about 41.82 degree Celsius.

Note : calculations were done in Kelvin itself.

So we assumed that temperature would be 50 but way 2 contradicts that. I could not find a fault in either of the ways so could not understand what temperature would it really be.

Attempt to solve: is it something about Gibbs Paradox? I was Googleing and that came, but I couldn't understand...

Please help with proper explanation on why the other way is wrong.

In the first condition when both flasks are boiling, the temperature of the gases in both flasks are the same (100 °C).  Using the ideal gas law on the system that comprises both flasks, where $n$ is the number of moles initially in each flask and $V$ is the volume of one flask.

$$ P_{init} = \frac{(2 n) R\;T_{boil}}{2 V} = \mathrm{0.5~atm}$$

$$ 2n = \frac{P_{init}(2 V)}{R\;T_{boil}} $$

Now if Flask #2 is immersed in ice water, we should assume that the pressure in the system will be constant but that the flasks will have different temperature.  For the pressure to be constant at different temperatures, material will have to flow from the hot flask to the cold flask; now, $n_1$ and $n_2$, the moles of gas in flask #1 and flask #2, will not be the same, but since no gas leaves the system comprising both flasks, then we know $n_1 + n_2 = 2n$

$$ n_1 = \frac{P_{final}V}{R\;T_{boil}} $$
$$ n_2 = \frac{P_{final}V}{R\;T_{ice}} $$
$$ n_1 + n_2 = 2n$$

Substituting:

$$ \frac{P_{final}V}{R\;T_{boil}} + \frac{P_{final}V}{R\;T_{ice}} = \frac{P_{init}(2 V)}{R\;T_{boil}}$$

$$ P_{final}\frac{V}{R}\left(\frac{1}{T_{boil}} + \frac{1}{T_{ice}} \right)= 2 \frac{P_{init}V}{R\;T_{boil}}$$

$$ P_{final}\left(\frac{1}{T_{boil}} + \frac{1}{T_{ice}} \right)= 2 \frac{P_{init}}{T_{boil}}$$

$$ P_{final} = 2 P_{init}\frac{\frac{1}{T_{boil}}}{\left(\frac{1}{T_{boil}} + \frac{1}{T_{ice}} \right)} = 2 P_{init} \frac{\frac{1}{T_{boil}}}{\frac{T_{ice} + T_{boil}}{T_{ice}{T_{boil}}}}$$

$$ P_{final} = 2 P_{init} \frac{T_{ice}}{T_{ice} + T_{boil}}  $$

This answer makes sense in that (i) if the temperature $T_{ice}$ is actually the same as $T_{boil}$, then the final pressure would be the same as the initial pressure, and (ii) the answer doesn't depend on the volume or moles of gas in the flasks, just the temperatures and the initial pressure.

  Way 1: Net Temperature of the system should be 50 degree Celsius because there is no heat loss and there is symmetry.

This way is incorrect because a flask immersed in boiling water ($T=$100°C) is unlikely to have a temperature of 50 °C.  Similarly for a flask immersed in ice.

  Way 2: The sum of the moles of gas in both the flasks would remain same.

This is the correct method.  It does require assuming that the rate of heat transfer from the boiling / ice baths to the flasks is much much greater than the rate of heat transfer between flasks via diffusion through the tube.  However since the tube is small, this assumption is probably OK.

The gas law is applicable when both elements are at same temperature but it isnt the case here so your way 2 is right


Why isn't water an ionic compound? If two alkali metal atoms join with an oxygen atom, an ionic bond forms. Since hydrogen has the same number of valence electrons as alkali metals, why can't water be ionic?

This is what I'm thinking:

$$
\ce{(H^+)_2O^{2-}}
$$

Thank you

First of all, the difference between ionic and covalent bonds is not sharp. As electronegativity differences increase, you move away from covalent and towards ionic bonds. There are "in between" states like polar covalent, where one side of the bond is stronger but not fully ionic. And this I think is the main reason: hydrogen has fairly high Pauling electronegativity (2.20), rather close to oxygen (3.44), which seems polar covalent overall (and why we get hydrogen bonding with water). In contrast, the alkali metals all have electronegativity less than 1.00, a much bigger difference versus oxygen and thus a more ionic bond.

Simply put, H2O is not ionic because it is a nonmetal/nonmetal compound, meaning it's covalent. Water is an oddball in the chemistry world, but certain laws prohibit water being an ionic compound


What is lithium pyroborate? This component is mentioned in an article called "Novel geopolymer materials containing borate and phosphate structural units".
I've never heard about this before, and a google search didn't really give an answer it seems.

Lithium pyroborate ($\ce{Li2B4O7}$) is the salt of lithium ($\ce{Li}$) and pyroboric acid ($\ce{H2B4O7}$).

Nomenclature

The prefix ortho- designates an acid with the maximum number of hydroxyl ($\ce{OH}$) groups [reference]. For boron ($\ce{B}$), orthoboric acid is $\ce{H3BO3}$ (also written as $\ce{B(OH)3}$ to show that there are $3$ hydroxyl groups).

The prefix pyro- here seems to not match with the usual usage, so I will not be explaining the usual usage here. One common feature between this and the usual usage is that their molecular formulas are multiples of the ortho acid's molecular formulas, minus some water molecules:

$$\ce{4H3BO3 -> H2B4O7 + 5H2O}$$

Gallery

Some proposed structures of lithium pyroborate:

Some proposed structures of pyroboric acid:

The above are modified from two proposed structures of sodium pyroborate, with the first one referenced here and the second one referenced here.


Europium stability in +2 and +3 state I would like to ask a question about europium's stability in the $+2$ and $+3$ oxidation state.

The electronic configuration of europium in its neutral state is $\ce{[Xe] (4f)^7 (6s)^2}$.

Now, when in the $+2$ oxidation state, the electronic configuration is $\ce{[Xe] (4f)^7}$ and in the $+3$ oxidation state, it is $\ce{[Xe] (4f)^6}$.

Now, I thought the $+2$ oxidation state is more stable because it's a half-filled $\ce{f}$ sub-shell so there is less mutual repulsion between electrons in the same sub-shell.

However, an article from nature.com reads the following:

  Europium metal is now known to be highly reactive; the element's most stable oxidation state is +3, but the +2 state also occurs in solid-state compounds and water.

Now, I would like to ask, how is it possible for europium to be most stable in its $+3$ oxidation state, knowing what I have mentioned above?

This is going to be a rehash of Oscar Lanzi's answer, but this point needs to be driven home, so I make no apologies.

"Special" electron configurations - fully-filled or half-filled subshells - are only a relatively minor factor in determining the stability of oxidation states. I've written about this before in a slightly different context, but IMO it would be well worth reading this, as the exact same principles operate in this case: Cr(II) and Mn(III) - their oxidizing and reducing properties?

What determines the "most stable oxidation state"? We can consider what happens when we increase the oxidation state of a metal: firstly, we need to pay ionisation energies to remove electrons from the metal. On the other hand, though, metal ions in higher oxidation states can form stronger bonds (e.g. lattice energy in ionic compounds, solvation energy in solution, or more covalent bonds in molecular compounds). The balance between these two factors therefore leads to a "best" oxidation state.

For example, sodium could hypothetically go up to Na(II) and form $\ce{NaCl2}$. Theoretically, this compound would have a significantly larger lattice energy than $\ce{NaCl}$. However, sodium's second ionisation energy is prohibitively large, which means that it stays put in the +1 oxidation state. On the other hand, magnesium's second ionisation energy is not prohibitively large, and therefore Mg preferentially forms $\ce{MgCl2}$ over $\ce{MgCl}$.

For all lanthanides, the first three ionisation energies are all fairly comparable. There is a graph here which shows the variation in the ionisation energies of the lanthanides, which I reproduce below.

(source: Inorganic Chemistry 6ed, Weller et al., p 630)

For all lanthanides, these three ionisation energies are easily compensated for by the extra lattice energy / solvation energy derived from a more highly charged ion. The fact that Eu(II) has a $\mathrm{f^7}$ configuration only serves to make its third IE marginally larger than that of Gd. This difference is sufficient to make Eu(II) an accessible oxidation state (cf. different electronic behaviour of EuO and GdO), but not sufficient to make it the most stable oxidation state.

Note that "most stable oxidation state" depends on the conditions, too, but I assume we are talking about aqueous solution.

One reason is that in the f orbital, the +2 oxidation state makes it that one pair of electron in the orbital is not filled, and will at least try to fill it.
And the +3 orbitals filled those pairs, even though it didn't fill the whole orbital. Of course filling the whole orbital (like noble gases) will be great but it is easier for the metals to fill up it's pair.


Siiloxane polymer experiment I believe the first reaction would be a substitution of $\ce{SiMe3}$ for $\ce{I}$ and creating $\ce{SiMe3Cl}$.
The second process I think would involve the $\ce{Mg}$ removing the two $\ce{Cl}$ and forming diphenylsilane, which would likely be a repeating unit in a silicone polymer.
Could any explain the mechanism for these reactions?

A)SiMe3 Increases the inductivity and reduces Aromaticity,
so by free radical (uv termination) its likely to have p-iodideSi(Me)3Bencene.

B)Dimer crystal solution in Cl- ions.  (Ph)2Si(Cl)Mg+---ClSi(Ph)2
Or Solid crystals of (Ph)2Si(Cl)MgCl.


What is the numbering of substituted benzene derivatives according to IUPAC nomenclature? According to the lowest locant rule shouldn't the numbering of the compound 1 start from the carbon at the 4th position, thus giving the IUPAC name as 1,2-dimethyl-4-hydroxybenzene?

For compound 2 they could have prioritized the functional group, but they have prioritized the lowest locant rule instead.

There are several criteria for the numbering of locants. The most important simplified criteria for the numbering for substituted benzene compounds are:

lower locants for the principal characteristic groups, which are expressed as a suffix or implied by the name of the functional parent compound (If there is a choice, the principal characteristic groups are assigned in accordance with the order of seniority of classes of compounds.)
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording of the nomenclature rules (taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) reads as follows:

  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:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (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;

  (…)

The first compound given in the question has the functional parent compound phenol. The name ‘phenol’ implies the presence of the characteristic group $\ce{-OH}$. Since there are no other characteristic groups and low locants are assigned first to the principal characteristic group, the locant ‘1’ is assigned to the  substituent group $\ce{-OH}$. Therefore, the systematic name for this compound is 3,4-dimethylphenol.

The second compound given in the question does not have any principal characteristic group – neither expressed as a suffix nor implied by the name of a functional parent compound. Therefore, low locants are assigned to the prefixes ‘chloro’, ‘methyl’, and ‘nitro’, all considered together as a set in a series of increasing numerical order. Thus, the systematic name for this compound is 2-chloro-1-methyl-4-nitrobenzene and neither ‘1-chloro-2-methyl-5-nitrobenzene’ nor ‘3-chloro-4-methyl-1-nitrobenzene’ since the locant set ‘1,2,4’ is lower than ‘1,2,5’ or ‘1,3,4’.

Lowest set locant rule is considered after prioritising the functional group (i.e. giving it the first position).
so answer of the first question is 3,4-dimethylphenol.
In the second question the answer given in the book is wrong . It should have been 3-chloro-4-methylnitrobenzene


Why does the degree of dissociation change when we dilute a weak acid even though the equilibrium constant is constant? $K$ represents the ratio of concentrations of molecules in a solution at equilibrium, which means that $Q_\mathrm{r}$ (that ratio at any given point) looks to be identical to $K$. In other words, the molecules in that solution react accordingly so that they reach equilibrium and the ratio of their concentrations is equal to $K$.

If $K$ is large enough (bigger than $10^4$ in my curriculum), this means that the the concentration of the reactants is almost zero. In other words, the  equilibrium position of that solution looks very much like a reaction that went to completion.

The more we dilute an acidic/basic solution, the higher the degree of dissociation, even though $K$ stays the same. So, does that mean that the more we dilute a solution the harder it is for it to reach the point of equilibrium for that specific molecule/solution or what?

For instance, say you found $K$ of solution to be $10^{-5}$. This means that when the reaction happens there are lots of reactants left, and not much products produced, which means that the degree of dissociation is low. But the more we dilute a solution, the closer it gets to a "complete reaction" (if you pour a small amount of weak-acid molecules into a large tank of water, it's certain that all of the weak-acid molecules are going to react with the water, i.e. the degree of dissociation approaches $100\%$).

So, how come $K$ can be independent of the initial reactants concentrations, and tell if a reaction was complete or not, when the "completion" of a reaction (the degree of dissociation) depends on the initial concentrations of reactants?

See my comment above.

As a numerical example, take acetic acid ($\ce{AcOH}$), which has $K_\mathrm{a} = 1.8 \cdot 10^{-5}$.

This means that:

$$K_\mathrm{a} = \frac{[\ce{AcO-}][\ce{H+}]}{[\ce{AcOH}]}$$

And the total nominal concentration of acid is:

$$C_\mathrm{a} = [\ce{AcOH}] + [\ce{AcO-}]$$

Combining these two equations, you can see that the % of dissociated acid is:

$$\alpha = \frac{[\ce{AcO-}]}{C_\mathrm{a}} = \frac{K_\mathrm{a}}{K_\mathrm{a} + [\ce{H+}]}$$

i.e. not a constant, but a value that depends on the $\mathrm{pH}$. So no, there is no need for the 'degree of completion' of the reaction to be a constant.

You can even calculate the explicit concentrations of all the species.

Given that:

$$C_\mathrm{a} = \left(\frac{1}{[\ce{H+}]} + \frac{1}{K_\mathrm{a}}\right) \cdot ( [\ce{H+}]^2 - K_\mathrm{w}^2)$$

you can see that, for $\mathrm{pH} = 5$ you need:

$$C_\mathrm{a} = \left(\frac{1}{10^{-5}} + \frac{1}{1.8 \cdot 10^{-5}}\right) \cdot \left((10^{-5})^2 - (10^{-14})^2\right) = 1.5554 \cdot 10^{-5}$$

Then:

$$[\ce{AcOH}] = \frac{C_\mathrm{a} \cdot [\ce{H+}]}{K_\mathrm{a} + [\ce{H+}]} = 5.555 \cdot 10^{-6}$$

$$[\ce{AcO-}] = C_\mathrm{a} - [\ce{AcOH}] = 9.999 \cdot 10^{-6}$$

So the % of dissociated $\ce{AcOH}$ is:

$$\alpha = \frac{[\ce{AcO-}]}{C_\mathrm{a}} \approx 64 \%$$

Now repeat for $\mathrm{pH} = 3$:

$$C_\mathrm{a} = \left(\frac{1}{10^{-3}} + \frac{1}{1.8 \cdot 10^{-5}}\right) \cdot \left((10^{-3})^2 - (10^{-14})^2\right) \approx 0.05656$$

$$[\ce{AcOH}] = \frac{C_\mathrm{a} \cdot [\ce{H+}]}{K_\mathrm{a} + [\ce{H+}]} \approx 0.05556$$

$$[\ce{AcO-}] = C_\mathrm{a} - [\ce{AcOH}] \approx 0.001$$

$$\alpha = \frac{[\ce{AcO-}]}{C_\mathrm{a}} \approx 1.8 \%$$

So you see, in a more dilute mixture the $\mathrm{pH}$ is higher, and the dissociation reaction of acetic acid is 'more complete', whereas at higher concentration the pH is lower, and there is proportionally less acetate anion.

All of these systems are at equilibrium.

A really handwavy explanation would be that the system, attempting to maximize its entropy, will prefer dissociated states as the system dilutes since there will be an increasing number of microscopic configurations with atoms $A$ and $B$ far apart (i.e. molecule $AB$ dissociated) as opposed to close together (i.e. atoms $A$ and $B$ bound together into molecule $AB$).


Are lithium bicarbonate and magnesium bicarbonate both soluble in water? Are lithium bicarbonate and magnesium bicarbonate both soluble in water?

Source: JEE Mains 2017 (National Level Examination of India)

Please do cite a reputed source along with your answer. Currently I have found a line from JD Lee (Concise Inorganic Chemistry 5th ed):

According to this Wikipedia solubility table, lithium bicarbonate is soluble in water to $\pu{5.7 g/100mL}$ at $\pu{20^oC}$, while this Wikipedia page gives the solubility of magnesium bicarbonate as $\pu{0.077 g/100mL}$.

However, another source (found by Oscar Lanzi in his good answer), "The Solubility of Magnesium Carbonate in Aqueous Solutions of Certain electrolytes, The Journal of Physical Chemistry, Volume 7, pub. 1903, reports making two measurements of the solubility of magnesium bicarbonate and finding $\pu{3.1 g/100mL}$ and $\pu{2.0 g/100 mL}$.  Note that this source is from 1903, but also that the the Wikipedia link above does not give a source for their value at all. At least one reason for the dearth of information regarding magnesium bicarbonate is that it easily decomposes and it cannot readily be isolated outside of aqueous solutions. My feeling is that the Wikipedia value seems low, and that the values reported in the Journal of Physical Chemistry seem more reasonable.

Carbonates oxygenate the body and their alkalinity neutralise acidic deposits(blockages), especially in communication between cells, serotonin to dopamine to Nuerotransmitters !
Electrolysis enhances this process which is why lithium carbonate is used to treat mental illness, but with only some success, due to many side effects of lithium metal. Lithium is poisonous to the body, and we should be using magnesium bicarbonate! It's the carbonate that's helping mental illness, not the lithium content. Magnesium is great for the body and mind, not lithium !


Still struggling with Huckel rule and aromaticity I'm still learning huckel rule and aromaticity in different compound.
For the 1-methoxy-1,3,5,7-octatetraene, is it correct that it has 8 electrons inside the cycle, but since they will be 8, then it is antiaromatic?
And for the Imidazole, is it 6 electron inside because the N: doesn't contribute to the pi, or is it 8?

Advice
Take a closer look at the lone pairs and double bonds that participate only and only in the delocalization of electrons.
1-methoxy-1,3,5,7-octatetraene
As you can see, 1-methoxy-1,3,5,7-octatetraene has four double bonds in an octagonal cycle. The lone pairs of the adjacent oxygen atoms, even though they directly influence the delocalization by means of the resonance effect, aren't actually part of the cycle and cannot be included amongst the pi electrons of the double bonds. Therefore, your molecule is antiaromatic. Its structure is not planar, in fact it looks something like this (the picture below includes only the cyclic backbone):

For aromatic molecules, planarity is absolutely necessary. But Huckel's rule which I'm sure you're familiar with, provides the reasoning behind the planarity of the molecule.
The imidazole
As for the imidazole, your assumption is absolutely correct. Here we get to know a scenario similar to pyridine:

The lone pair does not participate in the aromatic cycle. This can be easily checked by taking a look at the hybridization of the nitrogen atom. If the lone pair were to be part of the aromatic system, it should have been perpendicular to the other orbitals of the cycle. However, this is not the case. The 2p orbitals of the nitrogen are actually involved in the aromatic cycle, while the aforementioned paired electrons get to be embedded in an sp2 orbital, placed perpendicular to the plane of the cycle. This actually has significant consequences on the basicity of pyridine.
On the other hand, in order not to interrupt the aromatic cycle, the lone pairs of the other nitrogen atos must come into play when counting pi electrons. Thus, the molecule is aromatic.


Reaction of orthocresol with butyrolactone in presence of AlCl3 and heat What will be the major product of the following reaction?

Please also help me with the mechanism. I'm not sure whether C- acylation or O- acylation will occur here. And if O- acylation does occur, I think the heat will cause a Fries' rearrangement. What will be the major product of this rearrangement?

Also, if simple Friedel Crafts acylation occurs, will AlCl3 affect the reaction due to formation of -O-AlCl3 (Lewis acid-base interaction), something like what happens in the case of Friedel Crafts reactions with aniline? (I know aniline cannot undergo FC reactions due to deactivating effect of -NH-AlCl3 interactions)

I'm unable to upload a pic of the question due to some technical problem, but what I mean is reaction of ortho cresol and butyrolactone in AlCl3, along with heat.

I can take a guess since no one else has provided an accepted answer.
First, I don't think lactones will take part in FC-acylation on an arene, but rather prefer alkylation as mentioned in this paper. In the case of o-cresol and heat, this could be a possible explanation for your question why the O-acylation takes place instead of the C-acylation.
Second, when the Fries rearrangement occurs, and the acylium ion is released in the intermediate, the phenolic oxygen is complexed with the aluminum chloride (second to last step here). If you imagine an ortho-methyl group, as well as the complexed aluminum chloride, it's plausible substitution to the remaining ortho-site is not sterically favorable. This would force it to substitute elsewhere on the ring.
As for why "elsewhere" prefers para to the methyl group and not the oxygen—going by what you wrote in your comment—I'm not sure. The para alumino-oxy substitution should produce the more stabilizing resonance structure for the arenium intermediate. At least, in one other reaction of electrophilic aromatic substitution with ortho-cresol, namely its nitration, the 3/5-nitro products are not found. Maybe it's a solvent matter. Wikipedia states a non-polar solvent prefers ortho in the Fries rearrangement; perhaps if the ortho site is blocked, it will settle for para to a methyl substituent.
Edit
In a summary of the comments, the m-hydroxy-ketone was called into question as the correct product. I mentioned none of the papers I found of similar reactions furnished anything but the para or ortho product. Here are the papers I looked through:
paper 1 - first paragraph: o-tolyl acetates yield predominantly p-hydroxy-ketones.
paper 2 - tables 2 and 3
paper 3 - table 1 (page 378)
paper 4 - 5th paragraph
A few others: book p 223, paper, paper

It will follow the same mechanism as friedel crafts acylation.Just break the C-O bond of that cyclic Ester and add it to para (with respect to OH group)  and the H removed will be added to the last O- and form OH group.


About pH of an aqueous solution of SO2 Probably we can have an aqueous solution of $\ce{SO2}$ by dissolving it in water, because we would have an equilibrium between $\ce{SO2(g)}$ and $\ce{SO2(aq)}$:

$$\ce{SO2(g) <=> SO2(aq)}$$

How can I compare pH of an aqueous solution of $\ce{SO2}$ with that of pure water qualitatively?
What happens to the pH of an aqueous solution of $\ce{SO2}$ if I aerate it by bubbling air through it?

Upon adding $\ce{SO2}$ gas to water, it shows its acidic oxide character by hydrolyzing to sulfurous acid:

$$\ce{SO2(g) + H2O(l) <=> H2SO3(aq)}$$

$\ce{H2SO3}$ is a weak diprotic acid in water, with the following acid dissociation equilibria:

$$\ce{H2SO3(aq) <=> H+(aq) + HSO3^{-}(aq)}\ \ \ \ \ \mathrm{K_{a1}=1.4\times 10^{-2}}$$
$$\ce{HSO3{}^{-}(aq) <=> H+(aq) + SO3{}^{2-}(aq)}\ \ \ \ \ \mathrm{K_{a2}=6.3\times 10^{-8}}$$

Thus, addition of $\ce{SO2}$ gas to water will produce an acidic solution.

What would happen if air were bubbled into the solution? Air is a mixture of mostly nitrogen and oxygen. It so happens that $\ce{SO2}$ reacts with oxygen to form $\ce{SO3}$:

$$\ce{2SO2(g) + O2(g) -> 2SO3(g)}$$

Here, the sulfur atom in $\ce{SO2}$ is oxidized (oxidation number goes from +4 to +6) while the oxygen atoms in $\ce{O2}$ are reduced (oxidation number goes from 0 to -2). Now, $\ce{SO3}$ also happens to be an acidic oxide, and it hydrolyzes in water to form sulphuric acid:

$$\ce{SO3(g) + H2O(l) <=> H2SO4(aq)}$$

This time, $\ce{H2SO4}$ is a strong diprotic acid:

$$\ce{H2SO4(aq) <=>> H+(aq) + HSO4^{-}(aq)}\ \ \ \ \ \mathrm{K_{a1} \approx 10^6}$$
$$\ce{HSO4{}^{-}(aq) <=> H+(aq) + SO4{}^{2-}(aq)}\ \ \ \ \ \mathrm{K_{a2}=1\times 10^{-2}}$$

Assuming that not much $\ce{SO2}$ is lost while bubbling air through the sulphurous acid solution, then you're effectively replacing a weak acid with a strong acid, which would cause the pH to decrease.

As a final remark, if the sulphurous acid solution were bubbled with pure nitrogen, then there would be no oxidation of $\ce{SO2}$ into $\ce{SO3}$, and after a while enough $\ce{SO2}$ will have been dragged away for the solution to become a less concentrated sulphurous acid, which would therefore have a higher pH.

When you dissolve $\ce{SO2}$ in water you get $\ce{H2SO3}$. $\ce{SO2(aq)}$ means $\ce{H2SO3}$ therefor,the pH of the solution will be slightly acidic when compared to that of pure water.

to your other question.when you aerate the solution $\ce{SO2}$ gas right above the solution goes away.The equilibrium is then disturbed.Therefore, $\ce{SO2}$ dissolved in the solution comes out.Because of that pH of the solution increases.


Why doesn't an sp3s hybridisation exist? Trying to prove that nitrogen doesn't split its lone pair to form 5 bonds, I thought of a situation that I couldn't rule out; the paired electron being excited to the 3s orbital, so that five bonds could be formed- in an "$\ce{sp3s}$" hybridisation.
I know hybridisation is just an observation-based explanation, but I'd like to know how to rule wacky situations like this out.
Can this - and all other nonstandard hybridisations you can come up with - be ruled out by reasoning that the electron excitations involved take up too  much energy?

The counter question that might help you rule this out is: "Which 3s orbital?" Note that atomic Nitrogen doesn't have a 3s orbital as N itself belongs to the 2nd period(n=2).

Also note that in hybridization, Atomic orbitals of same or almost similar energies intermix to form same number of new Hybrid orbitals of same shap and equivalent energy.
That being said, you cant expect 2s or 2p orbitals to intermix with a 3s orbital
(if it was a case. Since this case isn't possible with N)


How to convert glucose to fructose How to convert glucose to fructose? Moreover, can there be direct conversion(not necessarily in a single step), or we have to use indirect means, such as from sucrose, or other higher carbohydrates, and then separating fructose from the products?

After searching on Google about Isomerization of Glucose to Fructose, you can find so many sources that explain the process. I think that this one is easy to understand among them. It explains that

  The isomerization of glucose to fructose is part of the glycolysis cycle that converts glucose to pyruvate. The way this is done is to isomerize the aldehyde (hemiacetal) glucose to the ketone (as a hemiacetal) fructose,and make another phosphate ester. The isomerization takes advantage of the ease of breakage of a C-H bond which involves a carbon next to a carbonyl carbon.

This gives a clear Idea on the mechanism of Glucose Isomerage to fructose. :)

Also see this.

Below by that process we can convert from glucose to fructose. At first a mixture of glucose and excess phenyl hydrazene is heated, so glucose will form osazone. Then if we will give $\ce{Zn}$ and acetic acid with this, fructose will form.


Why is trimethylsilanol a stronger acid than tert-butanol? The Wikipedia article about trimethylsilanol says the following:

  Trimethylsilanol is a weak acid with a $\mathrm{p}K_\mathrm{a}$ value of 11. The acidity is comparable to that of orthosilicic acid, but much higher than the one of alcohols like tert-butanol ($\mathrm{p}K_\mathrm{a}$ 19).

Why is it so? A silicon atom is more electropositive than a carbon atom, therefore the conjugate base should be more destabilized by the +I effect of the $\ce{Si(CH3)3}$ group compared to the $\ce{t-Bu}$ group.

The involvement of d-orbitals in the stabilisation of silanol anions is minimal. The ions $\ce{H3CO-}$ and $\ce{H3SiO-}$ were investigated by Hopkinson and Lien,1 who concluded that:

  Silicon is a more electropositive element than carbon and consequently the silyl group might be expected to be an electron donor. As such the silyl group should stabilise cations and destabilise anions. However, ab initio molecular orbital calculations show both the α-silyl and β-silyl to strongly stabilise anions and to weakly destabilise cations. Inclusion of d orbitals on  silicon does not significantly alter these conclusions. This contradictory behaviour can be largely understood in terms of n $\to$ π* delocalisation from the lone pair to π* of the silyl group.

These π* orbitals are antibonding orbitals derived from linear combination of Si 3p and H 1s. A partial MO diagram for the $\ce{H3Si}$ fragment is provided here (from the same paper):

Delocalisation of the oxygen lone pair into the $\pi_x^*$ and $\pi_y^*$ orbitals therefore leads to stabilisation. The authors write that the π* orbitals of the $\ce{H3Si}$ fragment are lower in energy than the corresponding π* orbitals of the $\ce{H3C}$ fragment, and hence the stabilisation is greater for silicon than it is for carbon.

A good test for this is to examine how the Si–H bond lengths change upon deprotonation. If the delocalisation is into Si–H antibonding orbitals, then the bond lengths should increase upon deprotonation. In contrast, if the delocalisation is into Si 3d orbitals, then there should be no change. Upon deprotonation, the Si–H bond length increases from 1.473 Å to 1.528 Å, which indicates it is the former.

However, there isn't a thorough explanation of why the energy levels are so. I would surmise that there are two factors: one is the fact that Si–H overlap is poor, compared to C–H overlap. Therefore, Si–H antibonding orbitals are lower in energy than C–H antibonding orbitals, and can act as better acceptor orbitals. Another factor (and one commonly taught in textbooks) is that since Si is more electropositive, the π* orbitals have a larger coefficient on Si; the π* orbitals of the $\ce{H3C}$ fragment have a smaller coefficient on C.

There is also an issue of why the stabilisation is larger in the anion than it is in the neutral molecule: if both the anion and the neutral molecule were equally stabilised, then there would not be any difference in the acidity, which is essentially a measure of the relative stability of the two species. It may be easily explained - at least partly - by the fact that the lone pairs in the anion are higher in energy than those in the neutral molecule, and consequently act as better donor orbitals.

Now, this explanation can be easily extended to the methylated compounds $\ce{Me3CO-}$ and $\ce{Me3SiO-}$. Due to the poorer acceptor capability of Si–C antibonding orbitals, we expect the factors above to play a smaller role in the preferential stabilisation of the silicon anion. As such, the difference in $\mathrm pK_\mathrm a$ between $\ce{X3COH}$ and $\ce{X3SiOH}$ should be smaller when $\ce{X = Me}$ compared to when $\ce{X = H}$.

Is there a way to test this? A crude analysis is as follows. I will take the value of $\pu{27.6 kcal/mol}$ given by Hopkinson and Lien1 for the reaction

$$\ce{H3SiO- + H3COH -> H3SiOH + H3CO-}$$

which corresponds to a $\mathrm pK_\mathrm a$ difference of 20 units when $\ce{X = H}$. This is obviously not rigorous, but serves as a good preliminary benchmark. Now, if Wikipedia's sources are to be believed, then when $\ce{X = Me}$, the $\mathrm pK_\mathrm a$ difference is decreased to a mere 8 units - which is in line with our prediction.

The topic of negative charge stabilisation by silicon has been studied much more thoroughly in the context of carbanions, because they are synthetically useful species (e.g. Peterson olefination). In this context, Schleyer et al.2 also conclude that d-orbital participation is negligible. The myth of d-orbital participation in main group bonding has been thoroughly debunked decades ago,3 so it is regrettable that it is still being taught nowadays.

References

Hopkinson, A. C.; Lien, M. H. A theoretical study of the effects of silyl and methyl groups on acid-base reactions at oxygen, nitrogen and carbon. J. Mol. Struct.: THEOCHEM 1983, 92 (1–2), 153–166. DOI: 10.1016/0166-1280(83)80065-7.
Schleyer, P. v. R.; Clark, T.; Kos, A. J.; Spitznagel, G. W.; Rohde, C.; Arad, D.; Houk, K. N.; Rondan, N. G. Structures and stabilities of α-hetero-substituted organolithium and organosodium compounds. Energetic unimportance of d-orbital effects. J. Am. Chem. Soc. 1984, 106 (22), 6467–6475. DOI: 10.1021/ja00334a001.
Reed, A. E.; Schleyer, P. v. R. Chemical bonding in hypervalent molecules. The dominance of ionic bonding and negative hyperconjugation over d-orbital participation. J. Am. Chem. Soc. 1990, 112 (4), 1434–1445. DOI: 10.1021/ja00160a022.

You are correct in saying that Carbon is more electronegative than Silicon, but due to presence of empty 3d-orbitals in silicon, it can engage in back bonding with the lone pair of oxygen.

(Following is a crude example of back bonding)

$\hskip2.5in$

This leads to delocalisation of the negative charge lending it a stability comparable to that of resonance, resulting in Trimethylsilanol being a better acid than tert-butyl alcohol.


What is the chemical formula of KIT-6? I have seen KIT-6 used in literature as a catalyst support, but cannot find the chemical formula for it.

According to original paper (1), KIT-6 is a highly ordered cubic $Ia3d$ silica ($\ce{SiO2}$) templated by Pluronic P123 ($\ce{[EO20PO70EO20]}$) block-copolymer surfactant. As a source of $\ce{Si}$, tetraethoxysilane (TEOS) or sodium silicate are primarily used.

(1) Kleitz, F.; Choi, S. H.; Ryoo, R.; Chem. Commun. 2003, 0 (17), 2136–2137. DOI: 10.1039/B306504A

According to this data sheet (LINK) is SiO2.


Law of Mass action VS Rate law The Law of mass action has 2 parts - the equilibrium rate part & the reaction rate part.
The "reaction rate part" states that rate of a reaction is directly proportional to the product of active masses of the reactants with each reactant term raised to it's stoichiometric coefficient. The rate law states that rate of a reaction is directly proportional to the product of active masses of the reactants with each  reactant term raised to some power which may or may not be equal to stoichiometric coefficient of the reactant.

Why are the two laws different when in fact we are talking about the powers to which active masses of reactants should be raised in both the cases ? And which law is correct?

First of all, you need to understand the basic difference between the two.

Consider the reaction $$\ce{A + B -> C}$$ Law of mass action states that in a rate equation the powers that the active masses of components raised are always stoichiometric factors.

For a reaction they are always fixed. However, the rate of  reaction differs with the physical conditions. This is applicable to both equilibrium reactions and moderate reactions that are unidirectional.

This makes law of mass action kind of theoretical in nature. In the rate law, the rate equation consists of actual factors that changes with change in experimental conditions. These may or may not be equal to stoichiometric factors. These are generally responsible for determining the order of reaction.

Both the laws are correct because both have a wide area of application. Rate law is although quite experimental in nature and is generally considered to be more appropriate.

Your statement of the Law of Mass Action is wrong. From Wikipedia:"The law is a statement about equilibrium and gives an expression for the equilibrium constant, a quantity characterizing chemical equilibrium." You apparently confuse the dynamic equilibrium where concentrations of reactants and products do not change with the rate of chemical reactions not at equilibrium (generally at constant temperature, and always with changes in concentrations). Keq is the equilibrium rate constant, the Reaction Quotient is Q. Q changes as the reaction proceeds and reaches Keq as dynamic equilibrium is attained.
It is NOT true that the rate of a (forward) chemical reaction is proportional to the concentration of the active masses of the reactants raised to some empirical power. Some reactions proceed at constant rates. These are known as zeroth order reactions - catalyzed reactions can be examples of these. (While you can argue that if R → P and if d[R]/dt = -k then d[R]/dt = -k[R]° (since concentration taken to the zeroth power is 1) that the proportionality holds, but this is trivial and not useful.)
In the historical development of the Law of Mass Action, there have been many instances where the powers for the Reaction Quotient at equilibrium have matched (more or less) the empirically determined powers of the rate expression. This is quite different than CONFUSING a situation at equilibrium with the situation which must, by definition, be non-equilibrium (since concentrations are changing).
Which is correct?
Logically invoking a dynamic equilibrium requires concentrations to be unchanging. Say R → P. With [R] and [P] constant, then taking the ratio of [R]α/[P]ß for some arbitrary constants α & ß is also going to produce a constant. Choosing the stoichiometric coefficients for α & ß doesn't change that.
For the rate of a reaction, the claim is that the rate of (forward) reaction is proportional to [R1]α[R2]ß...[Rn]n. This often isn't even an acceptable approximation. (although over a small enough time interval, any continuous (non-pathological) data can be approximated by a straight line). The more difficult question to answer here is does the rate of reaction for simple single-step reactions follow this rule? I'd say in dilute gas phase reactions (without significant effects from products) (and of course at constant temperature and pressure) it generally will. The problem with this, is determination of which are and which are not "simple single step" reactions isn't trivial. You could start there and confirm the proportionality. You'd have to confirm it, which makes it mostly unnecessary to assume it to begin with, but you have to start somewhere.


Sensitive test for lead in water? Some of my friends are in Flint and I'm horrified by their situation. I would like to learn how to test for lead independently.

After all if you can't trust the state sending them water to test is not sufficient. The testing procedures I have read about give too many type II errors. A procedure with a bias toward type I errors would be better since then I could pay a professional lab to see if it was real.

For tests that only detect high levels of lead, could they be made more sensitive by evaporating the water then testing a larger sample?

I used to work in an environmental water testing laboratory.  The regulatory standard method for testing for lead in water uses an Atomic Absorption (AA) spectrophotometer.

It seems that the issue with Flint's water appears to be that the new source of water is corrosive, the water has not been treated for corrosion, and pipes in Flint are commonly made out of lead.  So if your pipes at home are made of lead and are corroding, you may have a problem that the lab at the water treatment plant was not able to test for.

If you are curious about checking your water at home, you can obtain a water testing kit from a local hardware store.  These kits come with test strips that allow you to check your water for things like nitrates, pH, alkalinity, hardness etc.  Make sure your kit allows you to test for lead.

A spectrophotometric method would be the cheapest available method, however, the lead would have to be dissolved using a complex ion formation that has a distinct color or absorption pattern that would be is difficult to figure out. There is a commercial lead test that does this, from the Hach Company, but the extraction method requires specific reactants and procedures that I am not versed on.


Why should 2,6-Dimethyl-4-nitrophenol be more acidic than 3,5-Dimethyl-4-nitrophenol I was trying to compare acidity of some organic compounds but can't understand some:

Why should 2,6-dimethyl-4-nitrophenol be more acidic than 3,5-dimethyl-4-nitrophenol?

I have been trying to figure it out using various factors like hyperconjugation and induction effect of Me group but its not working. Suppose i take H.C effect 2,6-dimethyl-4-nitrophenol be "less" acidic than 3,5-dimethyl-4-nitrophenol since it destabilizes the negative charge on conjugate base. But this logic is not working.

In the 3,5-dimethyl isomer, steric effects come into play and influence the acidity.

Resonance structures like B in the figure below, that stabilize the phenoxide anion explain why 4-nitrophenol ($\ce{R=H}$) is more acidic than phenol.  Resonance structure B requires that the nitro group be planar with the aromatic ring.  If methyl groups are placed adjacent to the nitro group ($\ce{R=CH3}$), the steric interactions between the two methyl groups and the nitro group becomes large and forces the nitro group to rotate out of the aromatic plane,  Hence resonance structure B can no longer contribute and the compound is less acidic.

In3,5 dimethyl 4 nitro phenol , the nitro group is between two ortho methyl groups . Therefore due to this steric inhibition of resonance occurs and this nitro group is pushed out of plane by two methyl groups. Such that nitro group does not take part in the resonance. And there have no extra stable structure. That is why it is less acidic than2,6 dimethyl 4 nitro phenol.


Why is 1-bromotriptycene inert to nucleophilic substitution? According to my textbook, 1-bromotriptycene is inert to any nucleophilic substitutions.

Due to steric hinderance we can say SN2 reactions cannot occur, but I think SN1 reactions should be favoured as formed carbocation will be resonance stabilised.

Can someone explain why it doesn’t undergo SN1 reactions?

Sn1 reaction proceed through carbocation intermediate which has planar structure. The structure of 1-bromo triptycene  is cage like structure so the bridgehead carbon cannot assume planarity , hence,the formation of a carbocation at the bridgehead position does not take place. Consequently 1-bromotriptycene is inert to sn1reaction.

Bredt's rule is a consequence of the fact that having a double bond on a bridgehead or carbocation  is not stable for small rings (fewer than eight atoms) due to a combination of ring strain, and angle strain (nonplanar alkene).Since SN1 reaction involves carbocation intermediate, and such intermediate is not possible here, hence SN1 IS RULED OUT.SN2, reaction involves a backside attack from the leaving group.Due to stearic hinderence to the backside attack, such a reaction is also not possible.Totally nucleophilic substitution is not possible.


Testing isopropyl alcohol purity I'm curious how to test purity of isopropyl alcohol? I simply got very curious today. I bought rubbing alcohol and it doesn't say what's the purity anywhere. It could be 70%, but it could also be 90%, or 99% or even 99.9%. I'm a curious man so I started googling but couldn't find anything.

If I were a chemist, how would I test the purity?

If the isopropanol mixture was bought commercially, the chances that it contains significant amounts of methanol or ethanol are fairly small, I think.  (I've never seen commercial isopropanol in the US that contains > 3% of these impurities, at least, but the situation may vary in other countries or markets.)
If you can assume that the only other main component is water, the easiest way to estimate purity is simply by measuring the density.  100% pure isopropanol has a density at 20 °C of 0.786 g per mL.  Pure water has a density of 1.00 g/mL.  At intermediate concentrations of isopropanol, the density is in between those values.  If you can accurately measure both (a) the volume and (b) the weight of a portion of isopropanol, you should be able to measure the density.  I know my kitchen scale and kitchen measuring cups would be precise enough for this.
If you can't assume that water is the only other impurity, other methods such as NMR, chromatography, or spectroscopy would probably be required.  I think NMR would probably be the most informative (but perhaps the least accessible to the home chemist.)

For very quick test, you can go for flame test. Pure IPA when poured on a cotton burns upto a certain flame length . when impurities are added , the length of flame decreases.
Do not forget to keep the size of cotton same while comparing


Why are some molecules unable to absorb infrared radiation? Methane $\ce{CH4}$ is symmetrical and non polar, and so is oxygen $\ce{O2}$. But $\ce{O2}$ doesn't absorb IR. Why is that? People say that if a molecule can change its polarity then it will absorb IR, but I don't understand how a molecule changes its polarity while absorbing IR.

Please try to keep your answer simple — I don't have a strong background in infrared spectroscopy.

In basic terms, for a molecule to absorb radiation there has to be an oscillating dipole being produced. This can occur by nuclear motion (vibrations, rotations) or electronic motion to produce electronically excited states. You ask about IR radiation, which does not have sufficient energy to produce electronic excitation, so this is ignored in the text below.

The restriction on producing a moving dipole exists because radiation consists of oscillating electric and magnetic fields and to allow energy to go from the radiation into the molecule, there has to be an interaction between radiation and the molecule and the origin of this is the oscillating molecule's dipole. (Of course the radiation also has to have the correct energy to bridge the gap between any two energy levels, but this is just another way of saying that the molecular dipole and the radiation oscillate at the same frequency, i.e. are in resonance.)

Thus in any homonuclear diatomic there is no IR absorption because these molecules have no  intrinsic dipole or one caused by vibrations. Also as they have no permanent dipole there is obviously no dipole generated when molecules rotate. Heteronuclear diatomics, HCl for example, all have a permanent dipole and one that is changed  by vibrational motion, thus there is a vibrational and rotational spectrum in all heteronuclear diatomics.

All homo-nuclear diatomics, $\ce{N2}$, $\ce{O2}$ for example, are transparent to IR radiation and this is of prime importance in global warming, since if these molecules obtain energy, say by collision with water or carbon dioxide that has been excited by IR radiation, they cannot radiate the energy away and so must heat the atmosphere by collisions with other molecules.

In a molecule such as $\ce{CO2}$ although it has no overall dipole it does have moving (oscillating) dipoles due to the way it vibrates. One mode is to stretch one bond at the same time as the other contacts (an asymmetric stretch), and so an oscillating dipole is formed and this interacts with radiation and $\ce{CO2}$ has an IR vibrational-rotational spectrum. The bending vibration is also IR active.

A similar effect occurs in methane, the way the molecule vibrates develops oscillating dipoles and these interact with the radiation and so absorb energy. Molecular group theory is usually used to sort out which of the $3N-6$ ways molecules can vibrate can have IR spectra.

There are complicated higher order effect even in homonuclear diatomics which may generate absorption but these effect are absolutely tiny and can be ignored for most every purpose. The relative magnitudes of electric dipole, quadrupole and magnetic dipole intensities are $I_\mathrm{dip}:I_\mathrm{quad}:I_\mathrm{mag} \approx 1:10^{-5}:10^{-5}.$

We can measure the vibrational frequency of homonuclear as well as heteronuclear diatomics by using Raman spectroscopy, which is a scattering effect (no photons are absorbed), and depends on the shape of the 'electron cloud' in the molecule. This is more properly called its polarisability.

Very simple answer: IR radiation (in very first approximation) interacts with the individual bond. While methane is unpolar, the bonds in it are polar.

Not terribly polar, but still.


Why is +M effect of OH greater than OR? In some books it is written that +M  (mesomeric effect)  effect of OH is less than that of OR . The reason they give is inductive effect of R (group) which sounds senseful . But at some places it is written +M of OR is less (which is correct). But I am not able to understand the reason which they give.

First of all, you must understand that the reactivity of OR changes as you change the alkyl group (-R).

For example, consider, the methoxy group (-OCH3), as per proton NMR spectra of anisole and phenol, the reactivity is almost similar in both the cases. In fact the reactivity is so similar that undergraduate texts just mention the fact that anisole has reactivity similar to phenol. (Source 1)

On the other hand, if you consider more bulky alkyl groups, steric hindrance might be a contributing factor to reactivity, competing with the mesomeric effect. But to a large extent, it is sufficient to say that both hydroxyl and (-OR) groups have similar reactivity.

In phenole it can form phenate ion which has lone pair in oxsigen which can donate to the benzene ring. but in C6H5OR we cannot observe that kind of charge separation.


Do more double bonds cause a bigger number of stereoisomers in cycloalkenes? If trans double bonds cause cycloalkenes to be chiral, does a second trans or maybe cis double bond induce another chiral plane and hence four stereoisomers are possible for the given cyclodiene?

The question is based upon these previous questions:
Are all higher cycloalkenes chiral?
Why is trans-cyclooctene chiral?

If trans double bonds cause cycloalkenes to be chiral

Keep in mind that a trans double bond in a cycloalkene does not necessarily produce a chiral cycloalkene.  Only when the chain connecting the ends of the double bond is short enough so that the trans double bond is twisted will a chiral conformation exist.

So in large trans-cycloalkenes the double bond is not twisted and cannot produce chirality.

Remember too, that even if the trans double bond is twisted and a chiral conformation results, the molecule may still be optically inactive if the chiral conformation rapidly interconverts with its enantiomeric conformation. This latter point was discussed in one of your earlier questions where it was pointed out that

  At room temperature, trans-cyclooctene is the only trans-cycloalkene
  where the two enantiomeric conformations are stable enough (e.g. the
  barrier to reach the planar conformation is high enough) to be
  isolated.  If we lower the temperature to 0°C, then we can isolate the
  enatiomers of trans-cyclononene.

Back to your question.

  does a second trans or maybe cis double bond induce another chiral
  plane

The key here is that, as mentioned above, if the second double bond is also twisted, then there will be the potential for an increase in the number of chiral stereoisomers (of course inserting a double bond will always increase the number of stereoisomers because we have additional cis and trans double bond possibilities, but I think your question is focused on chiral stereoisomers, not geometric stereoisomers).

Look at the above figure on the left, of a trans double bond in a small ring.  We are sighting down a trans double bond. To alleviate ring strain, the double bond if forced to twist.  It is this twisting out of plane that infers chirality upon the double bond.  Its insertion into the ring will produce an additional chiral plane and the R or S chirality of the new double bond will  need to be specified in the name of the compound.

Note the figure on the right of a cis double bond in a small ring. To alleviate strain here, the $\ce{C_{ring}-C=C}$ angle is pinched down; there is no out of plane twisting motion, the plane of the double bond still exists.  Therefore, in the cis double bond case, there is no chirality associated with the double bond; no additional chiral isomers will be produced.

  and hence four stereoisomer are possible for the given cyclodiene?

Not always.  Just as in more common compounds with 2 chiral centers, meso stereoisomers are possible.  As an example, if the known E,E-cycloocta-1,5-diene
adopted a conformation as pictured below, then it has a plane of symmetry and would be an achiral meso compound where one trans double bond has the R configuration and the other double bond the S configuration.

(image source)

The number of stereoisomers do not depend on number of double bonds. It depends on the number of chiral/stereogenic carbons present in that molecule.
A stereogenic centre refers to a carbon which is bonded to four different atoms or groups.
Based on the number of stereogenic centres in the molecule, the number of stereoisomers can be determined by the formula: 2 raised to the nth power, whre n= number of stereogenic centres.
Therefore if a molecule has 3 stereogenic centres, the maximum number of stereoisomers it can hav is 2 raised to 3 that is 8 stereoisomers.


Are there any solvents that dissolve talcum powder? I understand that talcum powder is insoluble to water, would alcohol or something be able to dissolve it?

Talcum powder is made from talc, a silicate mineral with formula unit: $\ce{Mg3Si4O10(OH)2}$. Silicates tend to be pretty much insoluble in water, but it is possible to dissolve them in strong acids. I don't know about talc specifically, but many silicates can be dissolved in concentrated nitric acid. Hydrofluoric acid is also commonly used for dissolving minerals for chemical analysis. Both are very very dangerous substances that really shouldn't be used without proper training and safety equipment. I don't think there's really anything accessible for a home experimenter, unfortunately.

I consider that there is no solvent that can dissolve talcum powder, indeed if you consider "solving" as having your material into solution but maintaining the crystalline structure, only some inorganic compounds (most of them transition metal complexes) are soluble even in water or in other organic solvents (like alcohol).

Quoting wikipedia, we find:

  Talc is not soluble in water, but it is slightly soluble in dilute mineral acids.

In any case, solving it will keep intact its structure. This is naturally true for the case of silicates, which are characteristics by their insolubility.


when Al2O3 melts which bonds are broken? Previously, I was trying to understand why $\ce{Al2O3}$ has a lower melting point than MgO. An answer on quora said

  Do not confuse the character of the bond between the metal and oxygen with the bonds between molecules in the compound. The polar character of the bonds between aluminum and oxygen is indeed less than between magnesium and oxygen but this describes the energy needed to separate the metal from oxygen. When a solid melts it doesn’t break down the bonds between oxygen and aluminum or magnesium. The intermolecular bonds (that is bonds between one $\ce{Al2O3}$ and another $\ce{Al2O3}$) are in fact weaker than the bonds between two magnesium oxide molecules. This results in the greater melting point of magnesium oxide in spite of the less covalent character of the bonds within the molecule.

But if $\ce{Al2O3}$ is ionic and forms a giant lattice, which bonds that are "between one $\ce{Al2O3}$ and another $\ce{Al2O3}$" are broken?

Right, the text you are mentioning is not clear enough. It is not correct  to say that the molecule $\ce{Al_2O_3}$ exists and is separated from the next $\ce{Al_2O_3}$. Such a molecule does not exist. A sample of $\ce{Al_2O_3}$ is made of a superposition of $\ce{Al^{3+}}$ ions and $\ce{O^{2-}}$ ions. It is not made of a superposition of molecules $\ce{Al_2O_3}$. Now the fact that MgO melts at a higher temperature than $\ce{Al_2O_3}$ is due to the stronger attraction between $\ce{Mg^{2+}}$ and $\ce{O^{2-}}$ ions. For some reason, the attraction between $\ce{Al^{3+}}$ ions and $\ce{O^{2-}}$ is weaker.

The quote you give  clearly summarises the situation. In a solid (or liquid) there has to be some significant interaction between individual molecules otherwise there would only be gasses. This inter-molecular interaction is not that forming chemical bonds, i.e. a solid is not a super big molecule, but is additional to these and is due to the nature of the molecules and has several different forms.

The general term is inter-molecular forces, sometimes also call van-der-Waals forces and these are generally much weaker than a chemical bond. These forces arise from charge-charge (Coulomb), charge-dipole, dipole-dipole, dipole--induced-dipole and induced-dipole--induced dipole (also called dispersion or London forces) interactions between molecules.

You can find a summary in many phys. chem. textbooks but for all the gory details see 'Intermolecular and Surface Forces' by J. Israelachvili, publ Academic Press.


Is there a type of detergent/surfactant that evaporates at STP? Are there any type of detergent/surfactant chemicals that would be good for removing dirt and grease from fabric that also would evaporate from the clothes within, say, 24 hours leaving no residue?

Googling "volatile surfactant" turned up fluorocarbons and something called Surfynol 61, which is really 3,5-Dimethylhex-1-in-3-ol as possible candidates. I don't know if these would be safe for laundry.

Volatile surfactants

There are many molecules that are both surfactants (in water) and volatile, but of the ones I can think of, none can be used safely in home laundry applications.

Hydrocarbon derivatives.  The examples you found by googling are in this class.  3,5-dimethyl-1-hexyn-3-ol is the principal component of Surfynol 61 (probably named Surfynol 61 because it is a surface-active (surf) alkyne alcohol (ynol) with six carbons (6) and the alkyne moiety at carbon one (1).  Other examples would be medium-chain fatty alcohols like hexanol, octanol, cyclohexanol, etc.  These molecules reduce surface tension of water and have OK-ish cleaning power (not as good as larger but non-volatile alternatives like SDS), but their volatility is also their downfall: their use in home environments creates a strong fire and explosion hazard when used in automatic heated dryers.
Fluorocarbon derivatives.  I found the same article you probably did about volatile fluorocarbon surfactants.  The most powerful at reducing the surface tension of water seems to be nonafluoro-tert-butyl alcohol, which reduces water surface tension by 78%!  However this experiment required saturating the water with the vapor of this compound.  And 1 gram costs >$80.  And toxicity is a concern, despite this compound apparently being researched as a component of or precursor to artificial blood mixtures in the 1970s.
Other halocarbon derivatives.  These include solvents like perchloroethylene, chlorobutanol, etc.  These are all highly toxic, and with some of them, there is a flammability/explosion risk as with hydrocarbon derivatives.

Non-surfactant use

You might consider just using organic solvents that are not (very) surface active as a way to degrease fabric.  However, these may leach dyes out of the fabric or otherwise damage them.  Isopropyl alcohol (IPA) is accessible to home users, and is relatively safe.  Be sure to do the soaking in a well-ventilated area, and do not use heated dryers to remove the IPA!  Instead, rinse out the IPA thoroughly with water.  Depending on local regulations, you will have to dispose of all of the dirty IPA and IPA-contaminated rinsing water as solvent waste.

The best option

Pay to have your clothes dry cleaned!

May the force of additives be of aid to you!

Beforehand, choose an easier type of detergent:

  "Surface-acting polyglucosides [which are] easily broken down by microbes, leaving no traces in the environment", is one (Quellen, 213).

What is slow to biodegrade may not be used as often-inform yourself, and go from there!

And please, be safe.

Source:

Why There's Antifreeze in Your Toothpaste: the Chemistry of Household Ingredients. P 213. Quellen, Simon. 2008.


Figuring out the theoretical mass of reactants I'm working on my chemistry lab report. The initial instructions were to test the hypothesis that the mass of products of a reaction will equal the mass of the reactants.
Basically we where given a equation, our job was to balance and figure out the theoretical mass of reactants, and the theoretical mass of products.
There we're a total of three questions. In the first one I got the following differential values, 0.2 and 0.1. This seemed to be the most logical for me since I knew the law of conservation of mass implied that mass cannot be created nor destroyed but could be transformed. I attributed this small differential value to this transformation theory.
However, the next two questions I attempted still gave a very large differential value (40-120), which rejects my initial hypothesis.
I believe I may have missed something in the second and third question, below are my attempted solutions.
Question 1:
Initial Equation: $\ce{CH4 + O2 -> CO2 + H2O}$
Balanced Equation: $\ce{CH4 + 2O2 -> CO2 + 2H2O}$
$$
\begin{array}{ccccccc}
\ce{CH4} & + & \ce{2O2} & \ce{->} & \ce{CO2} & + & \ce{2H2O}\\
1(12+4) & + & 2(32) & \ce{->} & 1(12+32) & + & 2(2+16)\\
1(16) & + & 2(32) & \ce{->} & 1(44) & + & 2(18)\\
16 & + & 64 & \ce{->} & 44 & + & 36\\
\end{array}
$$
Theoretical Calculation:
$$
\begin{array}{ccc}
\ce{CH4}=\pu{16g} & & \ce{H2O}=\pu{36g}\\
+ & \ce{->} & +\\
\ce{O2}=\pu{32g} & & \ce{CO2}=\pu{44g}\\
\end{array}
$$
Actual Calculation:
$
\ce{CO2}=\pu{43.8g}\\
\ce{H2O}=\pu{35.9g}\\
$
Differential Value(Theoretical Calculation - Actual Calculation):
$
\ce{CO2}=\pu{0.2g}\\
\ce{H2O}=\pu{0.1g}\\
$
Question 2:
Initial Equation: $\ce{C2H6 + O2 -> CO2 + H2O}$
Balanced Equation: $\ce{2C2H6 + 7O2 -> 4CO2 + 6H2O}$
$$
\begin{array}{ccccccc}
\ce{2C2H6} & + & \ce{7O2} & \ce{->} & \ce{4CO2} & + & \ce{6H2O}\\
2(24+6) & + & 7(32) & \ce{->} & 4(12+32) & + & 6(2+16)\\
2(30) & + & 7(32) & \ce{->} & 4(44) & + & 6(18)\\
60 & + & 224 & \ce{->} & 176 & + & 108\\
\end{array}
$$
Theoretical Calculation:
$$
\begin{array}{ccc}
\ce{C2H6}=\pu{60g} & & \ce{H2O}=\pu{108g}\\
+ & \ce{->} & +\\
\ce{O2}=\pu{224g} & & \ce{CO2}=\pu{176g}\\
\end{array}
$$
Actual Calculation:
$
\ce{CO2}=\pu{78.59g}\\
\ce{H2O}=\pu{48.2g}\\
$
Differential Value(Theoretical Calculation - Actual Calculation):
$
\ce{CO2}=\pu{97.41g}\\
\ce{H2O}=\pu{48.2g}\\
$
Question 3:
Initial Equation: $\ce{C3H8 + O2 -> CO2 + H2O}$
Balanced Equation: $\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$
$$
\begin{array}{ccccccc}
\ce{C3H8} & + & \ce{5O2} & \ce{->} & \ce{3CO2} & + & \ce{4H2O}\\
1(36+8) & + & 5(32) & \ce{->} & 3(12+32) & + & 4(2+16)\\
1(44) & + & 5(32) & \ce{->} & 3(44) & + & 4(18)\\
44 & + & 160 & \ce{->} & 132 & + & 72\\
\end{array}
$$
Theoretical Calculation:
$$
\begin{array}{ccc}
\ce{C3H8}=\pu{44g} & & \ce{H2O}=\pu{72g}\\
+ & \ce{->} & +\\
\ce{O2}=\pu{160g} & & \ce{CO2}=\pu{132g}\\
\end{array}
$$
Actual Calculation:
$
\ce{CO2}=\pu{82.5g}\\
\ce{H2O}=\pu{45.0g}\\
$
Differential Value(Theoretical Calculation - Actual Calculation):
$
\ce{CO2}=\pu{49.5g}\\
\ce{H2O}=\pu{27.0g}\\
$
Not sure if it matters, but we where given this program to enter our calculations to simulate the reactions.

Try to answer to these questions, don't cheat, check the answer passing over the yellow box.
I think this software is confusing you always start from how much reactants you have and then calculate how much products you will have, suppose you have $\pu{100g}$ of oxygen and $\pu{44g}$ of propane.
How many mole of propane and oxygen ($\ce{O2}$) do you have?

 Oxygen moles$=\frac{\pu{100g}}{\pu{32g/mol}}\sim \pu{3mol}$,  Propane moles$=\frac{\pu{44g}}{\pu{44.096g/mol}}\sim \pu{1 mol}$

If this is your balanced equation:
$$\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$$
How many oxygen molecules do you need to react with one molecule of propane?

 You need 5 oxygen molecules for every propane molecule. So you need 5 moles of oxygen for every mole of propane.

Which is the ratio?

 $\frac{1}{5}$ if you have $x$ moles of oxygen you need $\frac{x}{5}$ moles of propane. e.g. if you have 5 moles of oxygen you need? 1 mole of propane.

But you have only 3 moles of oxygen to get the products, oxygen is a limiting reagent! So how many moles of propane will react:

 $$\pu{\frac{3}{5}mol} =\pu{0.6 mol} $$

So how many moles of the products will you have?

 $3\times0.6$ moles of $\ce{CO2}$ and $4\times0.6$ moles of water

And so how many grams?

 for $\ce{CO2}$, $\pu{1.8mol}\times \pu{44g/mol}$ for water, $\pu{2.4mol}\times \pu{18g/mol}$

$\pu{0.1 gram}$ is ($\pu{E4 kg}$)($\pu{299792458 m/s2}$) = $\pu{8.988E12 J}$ or two kilotons nuclear yield equivalent.  Let's be conservative and say "experimental error."
Mass-energy is conserved because time is homogeneous throughout the universe.  Run this continuous symmetry through Noether's theorems to obtain the First Law of Thermodynamics.  However, time is not locally homogeneous!  We live at the bottom of a gravitational potential well.  GPS atomic clocks run slower in orbit for special relativity (orbital velocity), but faster still for general relativity (higher in the potential well).  Therefore, don't transform "missing mass" except in physics labs.  Their party, their mess.


What factors could lower the voltage of a galvanic cell Recently I conducted a lab in which the voltage of a galvanic cell between two metals was recorded. Both metals were placed in an solution of their own ions with nitrate and a salt bridge was made by dipping some paper in salt water. The only issue is the voltage recorded in the cell was lower than that calculated using standard reduction and oxidation potentials. Does anyone have any ideas as to why this might be? I known the metals were not super clean although I don't think that would have an affect on the voltage.

Although it seems simple to make a galvanic cell ( i.e. a battery) it is really fairly complex.

The theoretical voltage is given by the Nernst equation.
The actual measured voltage also depends on the current. The equation
$V=iR$ still applies. So various restance sources in the cell cause voltage drops.

    The current must be very low so a high impedance voltmeter must be    used. (Not really a problem these days with modern semiconductor
    voltmeters as opposed to galvanometers.)
    The salt bridge offers resistance, so it has a voltage drop.
    Impure metals for the anode or cathode can cause a voltage change.
    Surface coating can cause a voltage drop.
    The acid or base concentration can cause a voltage change.
    The ionic strength of the solution can change the voltage.

There are more factors to consider when determining the true voltage of the cell such as the development of overpotentials.

'In electrochemistry,  overpotential is the potential difference (voltage) between a half reactions thermodynamically determined reduction potential and the potential at which the redox event is experimentally observed.

In a galvanic cell overpotential means less energy is recovered than thermodynamics predicts. This extra/missing energy is lost as heat.

Overpotential is specific to each cell design and varies across cells and operational conditions, even for the same reaction.

Regarding the galvanic cell, two possible polarities of overpotential can occur:

the anode may be less negative, supplying less energy than thermodynamically possible.
the cathode is less positive, supplying less energy than thermodynamically possible.

The overpotential increases with growing 'current density', or rate as described by the 'Tafel equation'. Each step of the electrochemical reaction is associated with multiple forms of overpotential. The overall overpotential is the summation of many individual losses.

'Voltage efficiency' describes the fraction of energy lost through overpotential. For a galvanic cell it is the ratio of a cells experimental potential divided by the cells thermodynamic potential converted to a percentile.

Overpotentials can be grouped into three categories, activation, concentration and resistance but  can be divided into many different sub categories also that are not well defined, such as 'polarization overpotential'.'

Ref:overpotential/wikipedia

For more information on 'voltage efficency','Tafel equation','current density' I suggest to 'Google' them or to look at the page I have referenced.

Other factors such as loss of electrolyte and the temperature of the system can also cause loss of voltage.


How to calculate the percentage yield for a reaction with excess reagent? A chemist mixed $12~\mathrm{g}$ of phosphorus with $35.5\ \mathrm{g}$ of chlorine gas to synthesize phosphorus(III) chloride (phosphorus trichloride). The yield was $42.4~\mathrm{g}$ of $\ce{PCl3}$. The equation is
  $$\ce{2P + 3Cl2 -> 2PCl3}$$
  Calculate the percentage yield.

I think that the amount of phosphorus the chemist used is in excess. I figured that only 1/3 of the amount of phosphorus the chemist used reacted to produce phosphorus trichloride. But the question didn't mention anything about products in excess that is why I am not so sure of my answer, which is $92.51$.

First I found the amount of substance of phosphorus used which is $12/31$. Then I calculated the amount of substance of chlorine gas used (and it's $0.5$). But $12/31\ \mathrm{mol}$ of $\ce{P}$ needs $1.5 \times (12/31)\ \mathrm{mol}$ of chlorine gas, so $\ce{P}$ is in excess. The amount of substance of $\ce{P}$ that will react will be $$0.5/1.5 = 1/3$$ Now $1/3\ \mathrm{mol}$ of $\ce{P}$ produces $1/3\ \mathrm{mol}$ of $\ce{PCl3}$ and the molecular mass of $\ce{PCl3}$ is $137.5$ so the mass that is supposed to be produced is $137.5 \times 1/3$. Then I divided $42.4$ by $(137.5 \times 1/3)$ and multiplied by $100$.

The yield is typically calculated according to equation $(1)$.

$$\text{yield} = \frac{\text{actual amount of substance obtained}}{\text{theoretical amount of substance to be obtained}}\tag{1}$$

(Whether it is given in percent or as a fractional number is not important, but percentages are the usual value.) So for any reaction, you need to figure out:

the reaction equation
the amount of each reactant substance involved
the amount of desired product theoretically possible by stoichiometric coefficients
the actual amount of desired product.

The first is rather easy, especially since the equation $(2)$ has been given previously.

$$\ce{2 P + 3 Cl2 -> 2 PCl3}\tag{2}$$

Thus, for every mole of phosphorus, we expect one mole of product and for each mole of chlorine gas, we expect two-thirds of a mole of product. The amounts of the reactants are easily calculated using molar masses:

$$\begin{align}M &= \frac mn\tag{3}\\
n (\ce{P}) &= \frac{12~\mathrm{g}}{30.97~\mathrm{g/mol}}\\
&= 0.39~\mathrm{mol}\tag{4}\\[1em]
n(\ce{Cl2}) &= \frac{35.5~\mathrm{g}}{70.9~\mathrm{g/mol}}\\
&= 0.50~\mathrm{mol}\tag{5}\end{align}$$

For each of these two we would need to check the amount of product we would expect using the stoichiometric coefficients.

$$\begin{align}n_\text{theor}(\ce{PCl3}) &= \frac 23 n(\ce{Cl2})\\
&= 0.33~\mathrm{mol}\tag{6}\\[1em]
n_\text{theor}(\ce{PCl3}) &= n(\ce{P})\\
&= 0.39~\mathrm{mol}\tag{7}\end{align}$$

Since $0.33 < 0.39$, chlorine is the limiting reagent and our theoretically possible yield is calculated from the amount of chlorine added to the reaction. Therefore, we now know our denominator which is $0.33~\mathrm{mol}$. What is the numerator? Again, it is easily calculated:

$$\begin{align}n_\text{exp}(\ce{PCl3}) &= \frac{42.4~\mathrm{g}}{137.32~\mathrm{g/mol}}\\
&= 0.31~\mathrm{mol}\end{align}\tag{8}$$

And now all we need to do is calculate the yield:

$$\frac{0.31~\mathrm{mol}}{0.33~\mathrm{mol}} = 0.93 = 93~\%\tag{9}$$

This is consistent with your result.

I'd of done this using a different approach.

Frankly I'd first try a shortcut of sorts. Such chemistry problems usually consume all of one of the reactants.

Knowing that 42.4 g of $\ce{PCl3}$ were produced

g P = $\dfrac{30.97}{137.3} \cdot 42.4 = 9.56$ grams

the rest must be Cl so

g Cl = 42.4 - 9.6 = 32.8 grams

Since neither of these amounts is equal to the mass of the starting reagents I need to look at the moles of the reactants to determine which is the limiting reagent.

moles P = $\dfrac{12}{30.97} = 0.387 $

moles Cl atoms = $\dfrac{35.5}{35.5} = 1 $

But every mole of $\ce{PCl3}$ needs 3 moles of Cl atoms. So 35.5 grams of Cl can make at most 0.333 moles of $\ce{PCl3}$.

So theoretically the yield should have been
$0.333 \cdot 137.3 = 45.8$ grams

% Yield = $\dfrac{42.4}{45.8}\cdot 100\% = 92.6\%$


Na2CO3 + FeBr3 + H2O: why do we get Fe(OH)3? $$\ce{3Na2CO3 + 2FeBr3 + 3H2O -> 6NaBr + 2Fe(OH)3 + 3CO2}$$

Why $\ce{Fe(OH)3}$? Where do these $\ce{-OH}$ groups come from?

This is from a series of reactions in a Russian sample exam question.

P.S.

Following Ivan't hints, I've gone through the process of reaction, and I get some water in the right-hand side of the equation. I wonder why the answer has no water in the right-hand side.

You have certainly already recognized $\ce{Na2CO3}$ as the base in your equation.

Did you also consider the following:
When $\ce{FeBr3}$ is dissolved in water, the hexaaqua complex $\ce{[Fe(H2O)6]^3+}$ is formed. This complex is fairly acidic:
$$\ce{[Fe(H2O)6]^3+ + H2O<=> [Fe(H2O)5OH]^2+ + H3O+}$$

Ignoring ligand exchange processes, you might state that the $\ce{OH}$ groups come from the deprotonation of the water ligands of your initial complex.

Draw a lewis diagram of the molecules to give you visual clues. From a glance it looks like the halides that are connected to the iron take their electrons leave and then are captured in solution to the sodium molecules. (this wouldnt be such a great bond since its ionic and the sodium and bromine ions would really act more like spectators).

From thee the CO3 attacted to the Na would decompose to CO2 and the extra O would attach itself to the iron and and somewhere before or after that instant those oxygens would strip a hydrogen from the water.


The effect of pressure on boiling point? My textbook states the following:

Qualitatively, I understand why the boiling point of a substance increases when the pressure is increased. However, I learned that if the pressure is increased when the reaction system is at equilibrium, the volume of the system decreases and hence, the reaction will proceed in that direction in which the volume of the system increases. In the above scenario, however, it has been given that water in the gaseous form occupies more volume than its liquid form and so, shouldn't the given reaction proceed in the forward direction, hence reducing the boiling point?

Please note that I am convinced and I believe that it makes sense for the boiling point to increase when the pressure increases but can somebody please explain the flaw in my reasoning with regard to the given equilibrium?

Much thanks in advance :) Regards.

Your flaw in the logic lies in that you think that the volume of the system includes the liquid. Le Chatelier's principle on pressure equilibrium only applies to when the compound involved is in gaseous phase, in an ideal sense. Therefore if you increase the pressure (decreasing the volume), then the reaction will progress towards the direction where gas volume DECREASES, which is towards liquid phase. Overall, the entire apparatus increases the vapor pressure on the gas (gas pressure on the external pressure), which increases the boiling point.

It's because when we increase pressure, we are actually increasing external pressure so according to Le Chatelier's principle, to counteract the effect it should increase the internal pressure. This is the reason that the reaction shifts where internal volume decreases, so that internal pressure increases.


Measuring or calculating the pH of 4M NaOH? I prepared the solution in Milli Q water I prepared $4\ \mathrm{M}$ solution of $\ce{NaOH}$, and the glass electrode I use cannot measure it the $\mathrm{pH}$ correctly. What are the other options do I have to exactly know the solution's $\mathrm{pH}$?

For $\mathrm{pH}>12$ and $\ce{Li+}$ or $\ce{Na+}$ concentrations greater than $0.1\ \mathrm{M}$, glass electrodes experience alkali error. Because the concentration of $\ce{H3O+}$ ions in solution is so low, interference from alkali metal ions (like $\ce{Na+}$) becomes noticeable and causes the electrode to read lower than the true $\mathrm{pH}$ of the solution. $\ce{K+}$ typically cause less error than $\ce{Li+}$ or $\ce{Na+}$ due do its larger size.

Since sodium hydroxide is a strong base and dissociates completely into $\ce{Na+}$ and $\ce{OH-}$ ions, though, it's actually quite simple to just calculate the $\mathrm{pH}$.

$$K_\mathrm{w}=\ce{[H3O+][OH-]} \Rightarrow$$
$$ 1 \times 10^{-14} \ \mathrm{M^2}=\ce{[H3O+] \cdot 4\ \mathrm{M}} \Rightarrow $$
$$\ce{[H3O+]}=2.5 \times 10^{-15}\ \mathrm{M}$$
$$\mathrm{pH}=–\log(2.5 \times 10^{-15}\ \mathrm{M})=14.60$$

If you are worried about whether or not you truly have a $4\ \mathrm{M}$ solution of $\ce{NaOH}$, you should try titrating it with a known concentration of acid (in the past I always used potassium hydrogen phthalate). This process is called standardization in acid-base parlance.


What is the mechanism for the oxidative cleavage of diketones via permanganate under basic/hot conditions? The products of the reaction are carboxylates.
What exactly would cause the alkyne to cleave?

In J. Org. Chem. 1979, 44 (15), 2726–2730, the authors Lee and Chang have investigated the permanganate oxidation of alkynes to α-diketones. In addition, they investigated the oxidative cleavage of several α-diketones to carboxylic acids. In the case of hexadecane-8,9-dione, permanganate oxidation under at least three sets of conditions provided n-octanoic acid 8 (major) and n-heptanoic acid 7 (minor). They suggested that the oxidation of the dione 1 leads only to octanoic acid 8 via intermediate 4 while oxidation of enol 2 affords 2-oxononanoic acid 6 and ultimately octanoic acid 8 via oxidative decarboxylation. In addition, oxidation of heptanal 5 affords heptanoic acid 7. The three solvent systems employed were acetone/aqueous HOAc, aqueous H2SO4 and water. Starting material was recovered in each experiment. Tetradecane-7,8-dione showed similar behavior.
For the mechanism of permanganate oxidation of alkynes to α-diketones, see the accompanying post by @Raul Luciano.

Alkynes react just as alkenes in the oxidative cleavage with $\ce{KMnO_4-}$ (the mechanism is the same). The difference is that since there is a triple bond, $\ce{MnO_4-}$ doesn't break the molecule apart like it does with alkenes (only if you use high temperature and high base/$\ce{MnO_4-}$ concentration), instead it forms a diketone. What causes the alkyne to cleave is the oxidant ion $\ce{MnO_4-}$ that is reduced to $\ce{MnO_2}$.


Can these changes be reversed by changing temperature? An egg turning hard when boiled.
A spoon full of batter cooking on a hot griddle.

I said that they both cannot be reversed, because I think if you throw both the egg and batter into the freezer, it will freeze, but it will not be in its original form before you cooked both of them. Then again, I guess I could try this for real on my own but I don't know if that would be a good idea...

Both processes involved the denaturation of proteins and complex reactions so is impossible reverse them only with a change of temperature.

Boiling an Egg
Although in most case denaturation is not reversible there are some cases where it is.
However in the case of Egg white denaturation is irreversible (see here) and so the whole process is by the way irreversible.
Cooking batter
The same for the batter even if is unclear which flours are used, and the composition of the batter  you can be sure that after cooking batter on a hot griddle surely some irreversible reactions occur: denaturation, Maillard reaction, Amadori rearrangment.

When you boil an egg, you denature its proteins, changing their structure. It can't be reversed.
Once the butter is melted by heat, you can solidify back by freezing.