Should compounds in a generic reaction be italicised (typography)? As of recent, I provided an answer to the question Law of mass action & collision theory. Therein I wrote a generic reaction

$$\ce{$r_1$R1 + $r_2$R2 + \ldots + $r_n$R_n <=> $p_1$P1 + $p_2$P2 + \ldots + $p_m$P_m}.\tag{1}$$

It seemed natural to write stoichiometric coefficents in Italic, and use upright Roman for the species themselves.

Italic would be in line with the fact that the coefficients are numbers, and vary with reactions (are different for different transitions).
Upright Roman is how we usually write chemical species, e.g., $\ce{NaCl}$.

Though if I were to overthink, the $\ce{R_$i$}$ would also seem to be variables as such. Since this is a generic reaction, $\ce{R_$i$} \in \{\mathrm{every\ imaginable \ chemical \ species}\}$. Using this logic, I ought to write $(1)$ instead as

$$r_1R_1 + r_2R_2 + \ldots + r_nR_n \ce{<=>} p_1P_1 + p_2P_2 + \ldots + p_mP_m.\tag{1'}$$

Is there an official example or guideline how to handle this situation?

Personally I wholeheartedly support option $(1)$ but is this what is recommended?

What about the subindices? Upright or Italic?

Note that this qualitatively differs from Proper typography of “$\ce{NO_x}$”. There the number of elements actually changes within a chemical species (marks the number of carbons in alkanes for example) whereas here it is more for labelling purposes.

Going by the basic rule of a variable (that can hold multiple values) being italic, I would go for the $r_nR_n$ style (4 times italics), but then I found the following in the IUPAC Green Book (printed page 53):

  $$0 = \sum_j \nu_j \mathrm{B}_j \quad \text{where }\mathrm{B}_j\text{ denotes a species in the reaction ...}$$

Here, the B is upright although it is definitely a placeholder. I don't understand, why.

For the subindices, I think (hope) there is no doubt that the $n$ and $m$ should be italic, because they clearly are variables standing for numbers (either any number as in the sum, or for an unknown maximum number $1, 2, 3, \ldots, n$ as in the generic reaction). Both form are well established in mathematics (and I hope chemists don't deviate).

The IUPAC Gold Book, R05156 uses the same convention.

  For the general chemical reaction :
  $$a\,\mathrm{A} + b\,\mathrm{B} \rightarrow p\,\mathrm{P} + q\,\mathrm{Q} + \dots% not using \ce to have proper formating regardless of mhchem version$$

And they do not mean Boron and Phosphorus.

(Spotted via this answer).


Mercury metal: Not toxic? Now I (and most of us here, I guess) have grown up around "Mercury's toxic! Avoid handling it! Do NOT screw around with this thing, if you value your life" and other equally heartening pieces of advice regarding mercury. Apparently mercury poisoning makes you, really lose it.

Now a while back I chanced upon a video on Youtube (I checked today, and now there're multiple videos on the same thing) where someone was messing around with elemental mercury … with his/her (I don't remember) bare hands.

Naturally I was taken aback. A quick check provided sufficient evidence to ascertain that he/she hadn't died as a result.

Now I posed this to a teacher of mine, and after some deliberation, he concluded that it isn't the mercury metal that's toxic. He told me that it's the various compounds of mercury that are formed by elemental mercury once it gets in your system (i.e- once you drink it) which are responsible for the toxicity. So apparently touching metallic mercury with your bare hands isn't an issue, since the metal isn't subjected to all those metabolic reactions that go on inside the body; heck you can even gargle it and live (provided you spit it out afterwards) since mercury's fairly inert.

So what I'd like to know is

Is it true that metallic mercury by itself, isn't toxic?

*Spin-off questions:

Is there any substance to my teacher's claim? How come those asses that performed this mundane task survived and now live (seemingly) normal lives without going insane? Were they already insane prior to touching the mercury? Should I just go and touch some mercury, partly to discover if it is indeed possible to touch metal mercury safely (and partly just to let out all the stress that built up while typing out this question)?

Mercury is toxic, but you need to carefully define what you mean by toxic or you draw incorrect conclusions

Toxic is a broad term. It means a lot of different things. The timescale matters. Some toxic things take years to exhibit their effects; others act instantly.

A binary distinction between toxic and not-toxic is pretty meaningless: you need to define the context and the timescale of the toxicity.

Mercury metal and mercury compounds are usually considered toxic. But their effects are varied in time and degree. Mercury metal is pernicious but only if you are exposed to it over a long time period. In fact you could probably drink it with few ill effects. The body just doesn't absorb it quickly. What is dangerous about mercury is not short term exposure to the metal but long term exposure to the vapour. This is why people don't suffer immediate ill effects when handling the metal even without skin protection.

Mercury vapour is readily absorbed in the body and will accumulate in tissue causing a variety of long term effects. This was discovered by mercury miners who often developed long term problems from their exposure. And it was documented for science by some chemists who started to suffer effects after working with the metal over long periods of time and managed to document their own decline (see Stock's work, for example). Mercury metal is often widely used in laboratories to provide a limited overpressure for gas distribution (you allow the gas to bubble through a mercury manometer).

Since the toxicity was recognised, chemists have been a lot more careful and always avoid vapour buildup by working in well ventilated spaces and making sure that manometers containing mercury are vented safely to the outside (via scrubbing filters) along with other potentially toxic vapours.

There is little immediate risk when working with metallic mercury as long as you don't spill it somewhere where it will collect and allow vapour to build up in the atmosphere.

Mercury compounds are a bigger risk. Some are readily absorbed into the body. The worst sort are mercury organometallic which are both volatile and penetrate the skin quickly. If you work with those you need to take extreme precautions. Even experienced chemists have been killed by accidents involving things like methyl mercury (see this tragic story).


Searching for a fireproof sponge? (I know this is a bit off topic, but I don't know where else to ask. If you know a better StackExchange site, please tell me)
I'm working on a project, and I want to basically create a "flaming sail". For this, I'm planning to drench a cloth in methanol and then lighting it, and for this, I need a fabric that both retains a lot of moisture and is flame proof, as this should be reusable.

Fiberfrax is the standard material for high temperatures in refinery, chem. plants, etc. There are also other brands , all readily available on the net. Ads do not give detailed composition other than "alumina and silica" .  Fiberglasses may not have  high melting /softening temperatures. There are higher temperature fibers like thoria, but are expensive and not easy to find.


What type of bonding is there among d-block metals? Metallic bond strength in $\mathrm{d}$-block elements increases up to the middle and then decreases. Why does pairing decrease the strength of metallic bond? What does pairing have to do with metallic bonding?

Some books [1, p. 1129] say there is covalent bonding, which is related to unpaired electrons, and that is why their inter-atomic interaction strength increases up to the middle:

  Hardness of these metals indicates the presence of covalent bonds. The unpaired $d$-electrons may overlap to form covalent bonds. The number of such covalent bonds would be regarded as a measure of overall bond strength and hardness of the metal. In a transition series tire number of unpaired $d$-electrons increases on moving from group IIIB $[(n-1)d^1,ns^2]$ to VIB $[(n-1)d^5,ns^1]$ after which pairing begins.
  ...
  The high melting and boiling points of transition elements show that they arc held by strong forces. This is due to the presence of (a) strong metallic bonding because of the overlapping of $(d-1)d$ orbitals, and (b) covalent bonding by the unpaired $d$-orbital electrons. As Zn, Cd and Hg are having completely filled $(d—1)d$-orbitals, their atoms arc not expected to form covalent bonding amongst themselves; hence they arc having relatively low m.p. than the other $d$-block elements.

Are unpaired electrons taking part in covalent bonding or metallic bonding? Why can't paired electrons take part in metallic bonding?

Gurdeep, R. Advanced Inorganic Chemistry, 3rd ed.; Krishna Prakashan Media, 2008; Vol. 1. ISBN 978-81-87224-03-7.

Unpaired electrons are not the only thing that are used for bonding in d-block elements. The outermost ns electrons are also present for bonding.
     and when you ask what type of bonding is present in d-block elements, it means you mean to ask the type of bonding present only amongst these metals, and not with some other element(like p-block elements, say F, Cl). If that is the case, then only metallic bonds are present in d-block elements. For example, copper atoms are held by metallic bonds in the metal crystal.

The factors affecting the strength of a metallic bond are-

No. Of valence electrons (more number means higher attractive forces, and hence stronger bond)
Distance of ionic core from the valence electrons (lesser distance means a smaller size of the metal ion, and hence higher value of the force)
Presence of vacant d-orbitals (the free valence electrons in metal travel through the metal crystal using these vacant d-orbitals. So, if there are more number of vacant d-orbitals, the free electrons are more easily "accessible" for attraction by the ionic cores)


Order of the reaction between hemoglobin and oxygen I have the reaction:

$$\ce{Hb + O2 <=>[$k$][$k'$] HbO2}$$

where $\ce{Hb}$ is hemoglobin, $k$ is the rate constant of the forward reaction, $k'$ is the rate constant of the inverse reaction. On my book there is the rate law:

$$W = kP(\ce{O2})[\ce{Hb}] - k'[\ce{HbO2}]$$

where $W$ is in $\pu{mol m-3 s-1}$, $P(\ce{O2})$ is a partial pressure in Pascal ($\pu{kg m m-2 s-2 = kg m-1 s-2}$).

The inverse reaction is a first order reaction because:

$$\pu{mol m-3 s-1} = [k'] (\pu{mol m-3})$$

thus $[k'] = \pu{s-1}$.

I am in trouble with the forward reaction because I got a strange unit of measurement:

$$\frac{\pu{mol}}{\pu{m3 s}} = [k] \frac{\pu{kg}}{\pu{m s2}} \frac{\pu{mol}}{\pu{m3}}$$

$$[k] = \frac{\pu{m s}}{\pu{kg}}$$

Which is the order of the forward reaction?

You seem to be overthinking this a great deal. The order of a reaction is not defined by the units of the rate constant, but rather by the sum of the exponents in the empirical rate equation. From the IUPAC Gold Book:

  If the macroscopic (observed, empirical or phenomenological) rate of reaction ($v$) for any reaction can be expressed by an empirical differential rate equation (or rate law) which contains a factor of the form $k[\ce{A}]^\alpha[\ce{B}]^\beta\cdots$ [...] then the reaction is said to be of [...] (total or overall) order $n = \alpha + \beta + \cdots$

Considering that the partial pressure of $\ce{O2}$ is directly proportional to its concentration, the forward reaction has a rate equation

$$k'[\ce{Hb}][\ce{O2}]$$

where the proportionality factor between $P(\ce{O2})$ and $[\ce{O2}]$ is absorbed into the new rate constant $k'$. The order is simply 2.


Migratory aptitude in pinacol-pinacolone rearrangement I am confused about the migratory aptitude of various groups, as there are many different orders for the same given in different places, especially about -Ph and -H. I would like to know if someone could reliably tell the order.

In the pinacol rearrangement, a 1,2-diol is treated with acid and rearranges to a carbonyl compound. Here is a reaction scheme showing a mechanism for the rearrangement.

In this case the molecule is symmetric and methyl migration is the only reaction pathway available.  Methyl migration can 1) help stabilize the developing carbocation center and 2) once fully migrated, produces a resonance stabilized hydroxyl-carbocation.

If one or more of the methyl groups is replaced with some other substituent (for example, alkyl, hydrogen, aryl), then our reaction possibilities become more complex; which hydroxyl group will protonate, which group will migrate?  Generally, both hydroxyls will protonate and lead to separate products.  In terms of which group will migrate, the only thing agreed upon is that a tertiary carbon will migrate in preference to a secondary carbon which will migrate in preference to a primary carbon.  Clearly, the ability of the migrating group to stabilize positive charge plays a role.

The reaction is reversible and product distribution can change as reaction conditions are changed (thermodynamic control vs. kinetic control).  All of these complexities make it difficult to sort things out in detail, thereby making it difficult to determine the true relative migratory preference of other groups like H and phenyl.  This is why the migratory aptitudes of these groups (H and phenyl) can change from textbook to textbook.

See section 2. Pinacol Rearrangement in this reference for a very nice, straightforward discussion of this reaction mechanism, and all its subtleties.


Gas laws, finding pressure What is the resultant pressure if 0.6 mol of
ideal gas at 273 K and 1.18 atm in a closed
container of constant volume is heated to
563 K?
Answer in units of atm

We have PV=nRT
(where symbols have there usual meaning.)

You see V is constant as the container doesn't change or deform ( as you mentioned)

so thinking that way PV/nRT=constant
as moles are constant and not leaking
R is universal gas constant .
so ;

P/T is constant
then $P_1/ P_2 = T_1/T_2$


Naming this iron complex: Fe2Cl6[H2O]5 This is the iron complex in question:

$\ce{Fe2Cl6[H2O]5}$

I'm pretty sure according to the nomenclature it should be: pentaaquahexachloroiron(III) but I wanted to check somewhere to be sure.

In coordination chemistry square brackets denote inner coordination sphere; $\ce{[H2O]5}$ makes little sense in this context.
It looks like the correct molecular formula for this compound appears to be $\ce{[FeCl2(H2O)4]+[FeCl4]- * H2O}$ according to the crystal structure determined by Szymański [1].

The IUPAC-compliant name proposed by the author is tetraaqua-cis-dichloroiron(III) tetrachloroferrate(III) monohydrate.

Figure 1. Asymmetric unit $\ce{[FeCl2(H2O)4]+[FeCl4]- * H2O}$ (ICSD-8198) with added hydrogen atoms.
Color code: $\color{#EEEEEE}{\Large\bullet}~\ce{H}$; $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#1FF01F}{\Large\bullet}~\ce{Cl}$; $\color{#E06633}{\Large\bullet}~\ce{Fe}$.

References

Szymański, J. T. The Crystal Structure of Tetraaqua-cis-Dichloroiron(III) Tetrachloroferrate(III) Monohydrate (Ferric Chloride $2\frac 1 2$-Hydrate). Acta Cryst. B 1979, B35 (9), 1958–1963. https://doi.org/10.1107/S0567740879008244.


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

\begin{align}
\ce{Fe^2+ +e^- &-> Fe^3+} & E &=0.68~\mathrm{V}\\
\ce{2Hg &-> Hg^2+ +2e^- } & E &=0.244~\mathrm{V}
\end{align}

Usually, I just do $\text{Cathode (Fe) - Anode (Hg)}$ and get the correct answer. However, the correct answer is listed as $1.16~\mathrm{V}$. There are only two numbers to manipulate. I found that
$$E(\ce{Fe}) + 2 E(\ce{Hg}) = 0.68~\mathrm{V} + 2\cdot 0.244~\mathrm{V}  = 1.168~\mathrm{V}$$

Why is this? The amount of substance is not supposed to effect the electro-potential.

First of all, you have to use reduction potentials as IUPAC considers the reduction potentials to be called the standard potentials.
It is not clear in your question whether you have used reduction or oxidation potentials.

Secondly, your equations don't make sense. The first equation shows Fe in +2 state getting reduced to Fe in +3 state.
Your second equation shows oxidation of mercury from pure state to +1 state.

And you are right that stoichiometry should not affect the EMF of cell.

The correct cell reactions will be iron(II) getting oxidised to iron(III) and Mercury(I) being reduced to Mercury(0).

At Cathode, Reduction of mercury takes place.

At Anode , Oxidation of Iron takes place. But we always use only reduction potential to calculate EMF. Don’t use oxidation potential of one with reduction potential of the other.

And yes EMF= E(Cathode) - E(Anode) provided both are reduction potentials.

But with the given electrode potentials I see no way to get 1.16V as an answer for the EMF of cell.

I would suggest that you check the question and electrode potentials again. And check the answer key again. Maybe there is an error there. It would be best to check about that with your course instructor.


pH of CH3CH2NH2 buffer solution For $\pu{550.0 mL}$ of a buffer solution that is $\pu{0.170 M}$ in $\ce{CH3CH2NH2}$ and $\pu{0.150 M}$  in $\ce{CH3CH2NH3Cl}$, calculate the initial pH and the final pH after adding $\pu{0.020 mol}$ of $\ce{HCl}$.

The following is my attempt on solving this problem:

Since Kb for ethylalamine = $5.6 \times 10^{-4}$.

pKb = $-\log( 5.6 \times 10^{-4}) = 3.25$

pOH = $3.25 + \log (0.150/0.170)$

pOH = $3.196$

pH = $14 - 3.196 = 10.80$ for Initial

Final pH:

$\pu{0.170 M} * \pu{0.55 L}= 0.0935 mol$                $\ce{CH3CH2NH2}$

$\pu{0.150 M} * \pu{0.55 L} = \pu{0.0825 mol} \ce{CH3CH2NH3Cl}$,

$0.0825 - 0.02 = \pu{0.0625 mol} \ce{CH3CH2NH3Cl}$,

$0.0935 + 0.02 = \pu{0.1135 mol} \ce{CH3CH2NH2}$

pOH = $3.25 + \log(0.0625/0.1135)$

pOH = $2.991$

pH = $14 - 2.991 = 11.01$ for final

However, these values turn out to be wrong. Please help, thank you!

Part A
-   Calculating  the initial $\pu{pH}$(before adding $\pu{0.02 mol}$ of $\ce{HCl})$

The equilibrium here is
$$\ce{CH3CH2NH2 + H2O <=> CH3CH2NH3+  + OH−}$$
For  $\ce{ CH3CH2NH2} , \mathrm{K_b} = 4.6\times{10^{−4} }$

$$\mathrm{K_b} = 4.6\times{10^{−4} }=\frac{[\ce{CH3CH2NH3+}][\ce{OH−}]}{ [\ce{CH3CH2NH2}]}$$
The nominal concentrations of the base and conjugate acid  are respectively $$\mathrm{C_b} =\pu{0.170 M}~\text{ and } \mathrm{ C_a } =\pu{0.150 M}$$.

The mass balance expressions are:
$$ [\ce{CH3CH2NH3+} ]+ [\ce{CH3CH2NH2}] = \mathrm{C_a} + \mathrm{C_b} = \pu{0.32M}~\text{ and}~ [\ce{Cl−}] = \mathrm{C_a}=\pu{0.150M}$$
and the charge balance is given by:
$$[\ce{Cl−}] + [\ce{OH−}] = [\ce{H+}] + [\ce{CH3CH2NH3+}] \approx{[\ce{ CH3CH2NH3+}]}$$
in which the approximation shown above is justified  that the solution will be weakly alkaline.

The equilibrium concentrations of the conjugate species are then:
$$[\ce{CH3CH2NH3+}] = 0.150 + [\ce{OH−}]$$

\begin{array}{  }
[\ce{CH3CH2NH2}] &= &(\mathrm{C_a} + \mathrm{C_b}) &-&([\ce{ CH3CH2NH3+} ])\\ &= &(0.32) &–&([\ce{ CH3CH2NH3+} ])\\ &= &0.320 &- &(0.150 + [\ce{OH−}])\\&=&0.170 &- &[\ce{OH−}]\end{array}
Because $\mathrm{C_a}$ and $\mathrm{C_b}$ are large compared to $[\ce{OH−}]$ (the solution is not expected to be strongly alkaline),
the $[\ce{OH−}]$ terms in the above expressions can be dropped and the equilibrium expression becomes
$$\mathrm{K_b}=5.6\times{10^{-4}} =\frac{ (0.150)[\ce{OH−}]}{(0.170)}$$
from which we find $[\ce{OH−}] = \pu{6.347\times{10^{-4}} M}$ , $\pu{pOH} = 3.2$ and $\pu{pH} = 14- 3.2 = 10.8$

Part B : Calculating  the final $ \pu{pH}$(after adding $\pu{0.020 mol}$ of $\ce{ HCl}$)

Calculate the initial amount of $\ce{ CH3CH2NH3+}$ and the initial amount of $\ce{CH3CH2NH2}$:
$$\begin{align}n_\ce{ CH3CH2NH3+} &=n_\ce{ CH3CH2NH3Cl}\\ &=V_\text{(solution)}\times{[\ce{ CH3CH2NH3Cl}]_\text{I}}\\ &=
\frac{\pu{550.0 ml}}{\pu{1000 ml/L}}\times{\pu{0.150 mol/L}}\\
&=\pu{0.083 mol}
 \end{align}$$
$$\begin{align}n_\ce{ CH3CH2NH2} &=V_\text{(solution)}\times{[\ce{ CH3CH2NH2}]_\mathrm{I}}\\ &=
\frac{\pu{550 ml}}{ \pu{1000 ml/L}}\times{\pu{0.170 mol/L}}\\
&=\pu{0.094 mol}
 \end{align}$$
Addition of the strong acid $\pu{ 0.020 mol}$ of $\ce{ HCl}$ , consumes this same amount of $\ce{ CH3CH2NH2}$ and produces $\pu{ 0.020 mol}$ of $\ce{ CH3CH2NH3+}$ ion as explained in the following table:

\begin{array}{c | c c c c c c c}
\text{rxn}&\ce{ CH3CH2NH2} &+ &\ce{HCl} &-> &\ce{CH3CH2NH3+} &+ &\ce{Cl-}\\\hline
 \text{I}&(0.094)&  &(0.020)&  &(0.083)& &(0.083)\\
\text{C} &(0.094-0.020)& &(0.020-0.020)&   &(0.083+ 0.020)&
   &(0.083+ 0.020)  \\
\text{F} &(\pu{0.074 mol})&  &(\pu{0.000 mol})&  &(\pu{0.103 mol})&  &(\pu{0.103 mol})
 \end{array}

We can continue using the approximations for the concentrations of the conjugate species
that we developed in the preceding part A.
The values of C$_\mathrm{b }$ and C$_\mathrm{a }$ are now

$$\mathrm{C}_\mathrm{b }= [\ce{CH3CH2NH2}]_\mathrm{I} =\frac{n_\ce{CH3CH2NH2}}{V}=\frac{\pu{0.074mol}}{\pu{0.550 L}}=\pu{0.133 M}\approx{[\ce{CH3CH2NH2}]_\mathrm{e}}$$
and
$$\mathrm{C}_\mathrm{a }= [\ce{CH3CH2NH3+}]_\text{I} =\frac{n_\ce{CH3CH2NH3+}}{V}=\frac{\pu{0.103 mol}}{\pu{0.55 L}}=\pu{0.187 M}\approx{[\ce{CH3CH2NH3+}]_\mathrm{e}}$$

Substituting these into the equilibrium constant expression yields
$$\begin{align}[\ce{OH−}] &=(\mathrm{K_b})\frac{[\ce{CH3CH2NH2}]} {[\ce{CH3CH2NH3+} ]}\\
&=(5.6\times{10^{-4}})\times{\frac{\pu{0.133 M}} {\pu{0.187M}}}\\&=\pu{3.983\times{10^{-4}} M }
\end{align}$$
the new $\pu{pOH}= 3.40$ ,and the new $\pu{pH}=14- 3.40=10.60$, so addition of the acid has changed the $\pu{pH}$, a decrease of only $0.20~~\pu{pH}$ unit.


Primitive cells and their variations Is there any intuitive way(without involving higher maths) to explain possible variations in seven primitive types of (unit)cells? I would like explanation in context of why end centred is possible only for orthorhombic and monoclinic but not in other unit cells.

I will not consider hexagonal cell in this answer.

Unit cell is generally parallelepiped, which has three sides $a$, $b$ and $c$ and three angles between them $\alpha$, $\beta$ and $\gamma$. The sides can be f.e. of equal length and angles can be same and equal to $90^\circ$, this way you will have cubic unit cell. You can try to vary sides and angles and you will discover, there is only those seven types of unit cells.

When I way learning this I found useful third chapter in "Foundations of material science and engineering". At least there is nice table and images to show those cells. You can find it by typing "Crystal structures and crystal geometry" into Google.

Well, now answer last question. Let's say we have cubic base-centered (more correct than end-centered, I think) cell. From top it would look like this:

I draw only two cubic base-centered cells (I can add more if it will be more clear) next to each other. However, as you can see, there is one smaller cell (red) inside of our base-centered lattice (green) and that is simple tetragonal cell (If you draw it in three dimension, you will see, that the height is $\sqrt{2}$ times greater). As it has lower volume, this is the primitive cell, not that base-centered.


Stabilisation of carbocations through hyperconjugation Our teachers told us that greater the number of alpha H, greater is the stability of carbocation.

But consider this:

$\ce{CH3CH2+}$      and.      $\ce{CH3CH2CH2+}$

The first one has 3 alpha H while the second one has 2. So first one is more stable, as our teachers say. But even C-C sigma bond can participate in hyperconjugation, isn't it?

Going with that, second one has equal no bond resonating structures and even higher inductive effect of ethyl group. So second one should be more stable (without overruling the dominance of hyperconjugation).

Hyperconjugation

Hyperconjugation, as you point out, is able to stabilise carbocation intermediates, IUPAC define this as:

  the interaction is ... between σ-bonds and an unfilled or partially filled π- or p-orbital

  Ref: IUPAC Gold Book

In a simple sense, we can visualise this interaction by drawing the carbocation and considering the adjacent σ bonds able to interact.

This can also be visualised in a more 'quantitative' way by considering a simple mixing of orbitals, in which we can see that the σC-H -> p+ lowers the energy relative to the vacant p-orbital.

Ref: Molecular orbitals and organic chemical reactions, Fleming

Stability of carbocations

If we now consider the various types of cation, we should be able to rationalise the order of stability.

The methyl group is completely unstabilised through hyperconjugation, due to the stereoelectronic requirement that the σ bond be located on the adjacant carbon (of which the methyl cation has none).

Primary carbocations have a maximum of 3 σ-bonds capable of hyperconjugation, secondary a maximum of 6, tertiary a maximum of 9– hence why the tertiary is the most stable cation out of the simply alkyl groups (the allyl and benzyl systems in the diagram above are stabilised through delocalisation of the charge onto the pi system).

Going back to your question of ethyl vs propyl carbocations, you can hopefully now see that they have the same number of σ-bonds in the correct position to participate in hyperconjugation.

An additional complication

Primary carbocations are sufficiently unstable that if they form, they tend to rearrange into more stable carbocations, such that the primary propyl cation will quickly undergo a 1,2-hydride shift to give the secondary propyl cation. A more complex example is shown below:


How does gefitinib affect the EGFR(ecptor) and why is there relapse Gefitinib has been shown to be an effective tyrosine kinase inhibitor in a fraction (~$10$%) of non-small cell lung cancer patients.

These patients are characterized as having a mutation (usually a deletion or missense) in the EGFR gene affecting the ATP-binding site.

I have two questions:

-- How does gefitinib impede the proliferation and apoptosis pathway? Does it in effect "ace out" the binding of EGF? If this is correct, how does it accomplish that? If not, I would appreciate a brief explanation as to what the successful mechanism is.

-- It seems that (similar to imatinib, another tyrosine kinase used to treat CML) successfully treated patients tend to relapse after a period of time. Why is this?

How does gefitinib impede the proliferation and apoptosis pathway? Does it in effect "ace out" the binding of EGF? If this is correct, how does it accomplish that? If not, I would appreciate a brief explanation as to what the successful mechanism is.

No, EGF binds on the extracellular side (to the ectodomain), and gefitinib binds on the intracellular side (to the tyrosine kinase, purple in the figure). The inhibitor interferes with ATP binding (see below, middle panel - the part boxed in green has structural similarities to adenine in ATP). The ATP would otherwise phosphorylate the tyrosines of other signalling proteins. So even if EGF binds to the ectodomain, there is no downstream effect because the tyrosin kinase domain is inhibited.

Source of images: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6523254/

I recommend section 3.2 to learn about how various inhibitors work.

  It seems that (similar to imatinib, another tyrosine kinase used to treat CML) successfully treated patients tend to relapse after a period of time. Why is this?

Additional mutations might directly prevent the drug from binding, or change signalling pathways in a different fashion to render the drug inefficient.


Is aromaticity of the 4 aromatic amino acids affected by pH? During a discussion of NMR, my lecturer told me to think about which amino acids contain protons in an aromatic ring, and whether any are likely to be affected by changes in pH.

The 4 aromatic amino acids are His, Phe, Tyr and Trp. I've marked Histidine as such in my notes, but can't remember why I thought this at the time. Contrary to this, Wikipedia states explicitly that “the imidazole ring of histidine is aromatic at all pH values”.

I also marked Histidine as "+ve" (as in, positive) but I'm stumped as to what I meant by it

I can't find any information on the others, but am guessing perhaps the fact that these have phenyl (rather than imidazole) rings means it's possible they respond differently to changing pH?

Any advice or hints would be appreciated, as I can't find any further information online and nothing in the structures jumps out at me to suggest an answer.

There are two nitrogen atoms in the imidazole ring in histidine, but the $sp^2$ nitrogen is significantly more basic than the $sp^3$ nitrogen, as the electron pair in the latter is delocalized over the ring. Protonating the $sp^3$ nitrogen would destroy the aromaticity, but that would require very strongly acidic conditions, possibly not attainable in aqueous solution. Protonating the $sp^2$ nitrogen does not affect aromaticity because the electron pair is approximately perpendicular to the plane of the ring, meaning it cannot overlap with other orbitals in the ring and therefore does not participate in mesomerism.

Tryptophan also has a $sp^3$ nitrogen atom in an aromatic ring, but like in the imidazole ring it is not very basic.


What is pre-concentration (context: Breath analysis)? What is the meaning of pre-concentration both in general and specific to this context of

a) "In addition to consistent sampling protocols,.. pre-concentration, and analysis of breath samples require standard methodology, calibration standards..."

b) "Offline breath analysis involves some form of pre-concentration of analytes followed by a separation step using high-resolution GC-MS based detection"

Pre-concentration just refers to the process of concentrating a sample before analysis, so that trace components won't be overlooked. If you're looking for something that's only present at a few parts-per-million, you could easily miss it if you used the raw sample, so ideally you want a way to separate off either large, common sources of noise/error, or to just extract the exact components you're looking for.

A quick look suggests, for example, that a key first component to remove in breath analysis is moisture, as some volatile components will dissolve readily in the water and make it harder to detect them.

Anyway, this paper on a few pre-concentration and detection techniques might be informative: Mochalski P, Wzorek B, Sliwka I, Amann A; Chromatogr B Analyt Technol Biomed Life Sci. 2009 Jul 1;877(20-21):1856-66.


Why does a tetrahedral molecule like methane have a dipole moment of zero? I was studying chemical bonding when I noticed something odd.

We say compounds like $\ce{CCl4}$ and $\ce{CH4}$ have a tetrahedral geometry (which is a 3D structure) but when we talk about their dipole moments, we say they have no dipole moment. We give the reason that as the H atoms are opposite each other (hence assuming it to be a 2D structure), they cancel out their bond moments.

But why? In the beginning we saw that they are 3D structures with one H being at the top and other 3 at the bottom, with the bond dipole moment directed from each H towards C. Due to this the component of the bond moment of the 3 downward H atoms along the line of the upper H atom would cause a net upwards bond moment which is not equal to zero.

But certainly I am wrong as these values have been calculated scientifically. So can someone please point out where i am going wrong in my understanding?

Karsten's answer is excellent, but here is a figure that shows the mathematics involved:

The central atom (green) is at the center of the cube, the four other atoms (purple) are at alternating vertices and the geometry should be clear.

Alternatively, if you orient the molecule so one peripheral (purple) atom is directly "above" the central (green) atom, then each of the other three atoms is just $1 - \theta$ ($\approx 70.52877940 ^\circ$) away from being directly "under" the central atom, so each contributes $\cos(1 - \theta)$ times the bond dipole. This is the downward component of the bond dipole from one of the lower atoms.

But $\cos(1 - \theta) = 1/3$, so this is simply (bond dipole)/3, and there are three of these lower atoms, so the three downward components exactly balance the one upward component.


Mixing Ca(OH)2 with Ca(NO3)2 Mixing $\ce{Ca(OH)_2}$ with $\pu{0.1 M}$ concentration of $\ce{Ca(NO3)2}$.

What will actually happen in the process? What I will see with my eyes? I know that $\ce{Ca^2+}$ cation with concentration of $\pu{0.1 M}$ will be the first to react with $\ce{OH-}$, am I right? Would there be any changes in color? What will be the timing to the end point (approximately) of the reaction?

Not much is going to happen. Due to low solubility of $\ce{Ca(OH)2}$, further suppressed by the presence of calcium cations from the dissolved $\ce{Ca(NO3)2}$, you get a suspension of calcium hydroxide, which unfiltered looks like white cloudy liquid and is also known as milk of lime.

Regarding the $\mathrm{pH}$ if the solution, $\ce{Ca(NO3)2}$ is prone to hydrolysis and alone would create slightly acidic medium:

Step 1:

$$
\begin{align}
\ce{Ca(NO3)2 + H2O &<=> CaOHNO3 + HNO3} \tag{R1.1}\\
\ce{Ca^2+ + 2 NO3- + H2O &<=> CaOH+ + 2 NO3- + \color{red}{H+}}\tag{R1.2}\\
\ce{Ca^2+ + H2O &<=> CaOH+ + \color{red}{H+}}\tag{R1.3}
\end{align}
$$

Step 2:

$$
\begin{align}
\ce{CaOHNO3 + H2O &<=> Ca(OH)2 + HNO3}\tag{R2.1}\\
\ce{CaOH+ + NO3- + H2O &<=> Ca(OH)2 + NO3- + \color{red}{H+}}\tag{R2.2}\\
\ce{CaOH+ + H2O &<=> Ca(OH)2 + \color{red}{H+}}\tag{R2.3}
\end{align}
$$

(Here .1, .2, .3 stand for molecular, complete ionic, and net ionic equation, correspondingly.)

But, when you add enough $\ce{Ca(OH)2}$ (saturated solution), which is a strong base:

$$\ce{Ca(OH)2 <=> Ca^2+ + 2 \color{blue}{OH-}}\, ,$$

I'd expect the resulting solution to be basic.


What is the product of magnesium reacting with carbon? $$\ce{Mg + C -> X}$$
  $$\ce{X + H2O -> Y}$$
  Determine unknowns X and Y.

Now the problem is I think the answer should be $\ce{X=MgC2}$ and $ \ce{Y=C2H2 + Mg(OH)2}$. But the given answer was $ \ce{X}= \ce{Mg2C3}$ and $\ce{Y}=\ce{C3H4} + \ce{Mg(OH)2}$

I tried to google the answer but where some website say $\ce{C3H4}$ is correct some say $\ce{C2H2}$ is.  Funny thing is in the exam just after that one we got the same question and this time the $\ce{C2H2}$ was the answer (and I wrote $\ce{C3H4}$)

This asnswer is intended to extend my comment. Most of the information is taken from this site.

Both types of magnesium carbide($\ce{MgC2, Mg2C3}$) is formed when acetylene gas is passed over magnesium. At different temperatures, one type of carbide predominates over the other. At 570°C-610° C, $\ce{MgC2}$ converts to $\ce{Mg2C3}$:

$$\ce{2MgC2 -> Mg2C3 + C }$$

Above 610° C, $\ce{Mg2C3}$ decomposes:

$$\ce{Mg2C3 -> 2Mg + C}$$

Both types of magnesium carbide can be hydrolysed:

$$\ce{MgC2 + 2H2O -> Mg(OH)2 + C2H2}$$
$$\ce{Mg2C3 + 4H2O -> 2Mg(OH)2 + C3H4}$$

For more information on the structure of magnesium carbide, see this old paper from 1943. Also, how magnesium carbide behaves at high pressure is discussed in this new paper(2015). Another type of magnesium carbide, $\ce{Mg2C}$ exist at high pressure and it along with $\ce{MgC2}$ behaves as semiconductor while $\ce{Mg2C3}$ behave as insulator-metal transition. (OP is requested to see both papers if interested)


Polyprotic acids and bases I just have a question on how you handle polyprotic bases.
Question:

Show how oxalate ion can be a polyprotic base

My answer:

My question is whether we stop doing the reaction until the oxalate ion has no charge or do we keep going on as below:

Another question I have is ;why hypochlorite ion would be a monoprotic base, rather than a polyprotic base?

If you have a molecule or ion with multiple sites that can be protonated (as you do here), and the sites can interact with each other (as they can here), the affinity for protons drops rapidly as more sites are protonated.

The proton affinity of the ion in your second reaction is a couple of orders of magnitude less than in your first reaction because of this effect.

Piling a third proton on there would be far less favorable than that---oxalic acid is a weak acid, and you're forcing it to act as a base by making it accept a proton1. You should stop with the two reactions you have.

It's really the same thing with hypochlorite ion; putting one proton on is fine (you get HClO, which is a weak acid). Putting a second proton on is tough, because you'd be making an acid act as a base.

1. This kind of thing does happen, though such protonated acids will quickly lose their extra proton to the surrounding water molecules. If you're using glacial acetic acid as a solvent, your acid ion is $\rm CH_3COOH_2^+$ instead of $\rm H_3O^+$,
and $\rm CH_3COO^-$ instead of $\rm OH^-$.


How to Find the Mass of an Individual Element in a Compound/Molecule What is the procedure/math for finding the mass of a specific element in a substance? For instance, how many grams of Oxygen are in 6.8 grams of Epinephrine (Adrenaline)?

Step 1: Find the Molar Mass of Adrenaline

To do this, you must need to know its chemical formula which is $\ce{C9H13NO3}$. Now to calculate its molar mass, we add up all of the molar masses of each atom: $$M =9\times12~\mathrm{g~mol^{-1}} + 1\times13~\mathrm{g~mol^{-1}} + 14~\mathrm{g~mol^{-1}} + 3\times16~\mathrm{g~mol^{-1}} = 183~\mathrm{g~mol^{-1}}$$

Step 2: Find the Mass of Oxygen in 1 mol of Adrenaline

In each molecule of adrenaline, there is 3 molecules of oxygen atoms. This means that in $1\ \mathrm{mol}$ of adrenaline, there are $3\ \mathrm{mol}$ of oxygen. Therefore the mass of oxygen in $1\ \mathrm{mol}$ of adrenaline is: $\mathrm{3~mol\times16~g~mol^{-1} = 48~g}$.

Step 3: Calculate Percentage Composition of Oxygen by Mass

We now know that in every mole of adrenaline, there are $48\ \mathrm{g}$ of oxygen. We also know that $1\ \mathrm{mol}$ of adrenaline weighs $183\ \mathrm{g}$. Therefore in $183\ \mathrm{g}$ of adrenaline, there are $48\ \mathrm{g}$ of oxygen. All have to do is simply divide by them by each other and multiply by $100$ to get the percentage composition of oxygen. Therefore the percentage composition is $26.23\,\%$.

Step 4: Calculate Mass of Oxygen in Sample

Now we know the percentage composition of oxygen which is $26.23\,\%$. This means that in every gram of adrenaline, there is $0.2623\ \mathrm{g}$ of oxygen. Therefore to find the mass of oxygen in the sample: $$m_\text{oxygen} = 0.2623 \times 6.8~\mathrm{g} = 1.78~\mathrm{g}$$


Does the weird 4s and 3d rule apply for much of the p-block from Ga-Rn? In the transition metals, $4s$ has a higher energy level than $3d$, because in ionization, the electrons from the $4s$ energy level are lost first.

I was thinking that this might be the same for much of the p-block from elements Ga-Rn because:

Ga has a electron configuration of: $1s^22s^22p^63s^22p^63s^23p^63d^{10}4s^24p^1$

Ga$^{3+}$ has an electron configuration of: $1s^22s^22p^63s^23p^63d^{10}$

Well the only exception in the orbitals is for $s$ and $d$ orbitals. Meaning you will first remove electrons from $(n+1)s$ and then onto $nd$ orbital. It is only because the $(n+1)s$ sublevel is only lower in energy if there are no electrons in the d sublevel. As per the Aufbau principle, the the s orbital is therefore filled before. Then as the $d$ becomes filled with electrons, then the energy starts to flunctuate relative to one another and the s orbital ends up higher in energy. That is why when electrons are removed, they are first removed from s and then $d$. Luckily this only applies to $s$ and $d$ orbitals. When are you working with $p$ orbitals, well then the $np$ orbital will always be higher than the $(n-1)d$ orbital. Therefore for making ions, you will have to first removed electrons from $p$ then $s$ and then $d$ and so on.


Elimination of bromine with sodium iodide and electrocyclization I've been trying to solve the following reaction scheme for a while now and I am having trouble especially with the products E and F.

For product D I got the following, but I'm unsure about compound E, which should be a reactive intermediate.

While trying to figure out the structure of compound E, I drew the following structure, but I'm not sure how the rearrangements will happen to get rid of the charges and get product E. I do expect that E won't be aromatic but that F will be.

All in all, I am wondering what products E and F will be and how to figure them out.

Pöytä: Your conclusion that D contains four bromine atoms symmetrically distributed is sound. The treatment of D with NaI serves to substitute one, if not two bromine atoms with iodine to form "J". Iodide acts as a reducing agent in a vinylogous fashion. (Compare this reaction with the reduction of a vicinal dibromide with iodide to form an alkene.) The stereochemistry of E is likely to be E,E which arises from a transition state for elimination that avoids steric interactions involving bromine. I am assuming that the "electrocyclization" is thermal since there is no indication that light was involved. Achiral E will undergo a conrotatory electrocyclization in two directions to give racemic F and, eventually, diiodide G via SN2 displacement. (The electrocyclization may be rationalized as involving the HOMO of a butadiene or octatetraene. Look here for a discussion.) Reductive elimination of iodine via Zn-Cu couple affords benzocyclobutadiene H. (The elimination is akin to the formation of ethylene from 1,2-dibromoethane with magnesium.) The strained, reactive double bond of H undergoes a Diels-Alder reaction with cyclopentadiene to form adduct I. The stereochemistry shown arises from a so-called "endo" transition state.I found this question somewhat amusing in that information was provided for all species save the final product. ;)


Why is H⁺ ion considered the source of acidity, and OH⁻ considered the source of basicity? As the title implies, what caused the association between $\ce{H+}$ and acidity, and between $\ce{OH-}$ and basicity?

Yours is a historical question. Note that in modern chemistry, it is not required that acids have H+ or bases have OH-. The words acidity or basicity referred to certain common behaviors such as acids will liberate hydrogen when come in contact with metals, or they will decompose carbonates to carbon dioxide. Similarly, bases had other common characters. Indicators behaved the same way with acids or bases. No one knows why bases were called bases. Acids we understand, that they tasted sour (=Latin). The term alkali (Group I bases) makes more sense (from Arabic for soda ash or potash).

Definitions are made by humans and I am glad you are not blindly accepting the definitions and at least thinking about them. Arrhenius was among the first ones to think why solutions conduct electricity. When Arrhenius proposed this idea in his PhD, but he knew that his mediocre committee members will not like it. How can anything ionize in water? It was a stupid idea to them. It turned out that most common acids and bases had very high conductivity better than anything else known, even better than salt solutions. Say 0.1 M HCl had way higher electrical conductivity than 0.1 M NaCl? Why? The "proton" in aqueous solution has the highest conductivity known today. What comes next? The hydroxide ion. Also in Arrhenius' time electrical decomposition was known. Faraday did a lot of electrochemical experiments then.

When you electrolyze an acid solution, what do you get at the electrode which has a negative electrostatic charge. Hydrogen gas! What appears at the positive electrode, chlorine gas. How can "H" and "Cl" reach the electrode, this is possible when they have a charge on them as well. So, "H" in acids must exist as "H+"; as a positively charged entity. I have not done justice to Arrhenius history or the history of acid bases.

If you are really (or better word genuinely) interested then search the Theory of Ionization+ Arrhenius History in Google Scholar. Partington's History of Chemistry will have it too but it is meant for serious scholars of chem. history (multiple volumes). It is available online from the Internet Archive.


Yellow stains after cleaning silver I've recently moved to a old house and found some stuff in the attic, along that I found some really beautiful silverware(??) that unfortunately was VERY tarnished, one of the pieces was totally black while some had small places where I could see, in the back of one of them there's a label stating Silver 90.

I googled and found that using a solution of hot water, baking soda, salt, vinegar and strips of aluminium foil would help clean them. I submerged them and after some time I took them out and started to scrub them and the gross black stuff was going out, showing some silver and some yellowish. I them took a sponge and kind got carried away.

The pieces now all have this yellowish stains.

The one that was completely black had no silver-color in any part of it after cleaning, except the bottom, while the rest is all yellow now.

Here's one of the pieces

(the back is a white paper, white-balance is crazy with this piece)

My question is: Is this yellow stain any kind of dirt that wasn't removed in the process, or is this the underlayer of the piece(meaning that the silver is gone in those parts)? And above all, is the method used the best way to remove dirt from silver?

I suggest to do the following:
Line the bottom of a pan with aluminum foil. Put the silver piece on top of the aluminum foil. The silver piece and aluminum foil must be in contact with each other.
To about 2L of hot boiling water add about a half cup of baking soda (Be careful!). Then add the mixture to the pan. Make sure to cover the whole silver piece. The tarnish will disappear within several minutes.
The chemistry underlying this treatment is the following:
The silver tarnish is the chemical product of the reaction of silver with sulfur-containing substances in the air, $\ce{Ag2S}$. Silver sulfide is black. In this process, we convert silver sulfide to silver according to the following reaction:
$$\ce{3Ag2S + 2Al->Al2S3 +6Ag}$$ In fact, aluminum has a greater affinity to sulfur than silver.  The reaction between silver sulfide and aluminum takes place when the two are in contact while they are immersed in a baking soda solution. And it becomes faster when the solution is hot.The silver piece should touch the aluminum because a small electric current flows between them during the reaction. The formed aluminum sulfide may adhere to the aluminum foil, or it may form tiny, pale yellow flakes in the solution.

The advantage of this method of cleansing is that we do not remove the tarnish with some silver. We simply transfer silver sulfide to silver.


Why is the product of sodium and water, sodium hydroxide and not sodium oxide? In the reaction $\ce{Na(s) + H2O}$ the products are listed as $\ce{NaOH(aq) + H2 (g)}$, but why would $\ce{Na2O}$ not be a possible product?

I searched and found that the reaction $\ce{Na2O + 2H2O -> 2NaOH}$ occurs as well. Does it have to do with the stability of oxides?

In the usual case every student learns, the reaction of sodium with water is the following — I made note to include the states of matter here as they matter (pun intended).

$$\ce{2 Na (s) + 2 H2O (l) -> 2 Na+ (aq) + 2 OH- (aq) + H2 ^ (g)}\tag{1}$$

In the typical setup, you have lots of water and little sodium. Therefore, whatever you get will be a solution or a precipitated salt. Alkali metals, however, often form very soluble salts and indeed all alkali hydroxides are well soluble. Thus, we will never get to see any solid by-products; all will happen in the solution phase.

Technically, it is possible for an oxide anion to be formed. However, this is when Brønsted and Lowry’s acid-base theory kicks in. Remember that water is amphoteric and can react as a base to form $\ce{H3O+}$ or as an acid to form $\ce{OH-}$. Well, equation $(2)$ expands that scheme.

$$\ce{H3O+ <=>[$K_\mathrm{a,1}$][H+] H2O <=>[$K_\mathrm{a,2}$][H+] OH- <=>[$K_\mathrm{a,3}$][H+] O^2-}\tag{2}$$

The oxide anion is connected to hydroxide and water by a simple acid-base reaction. From the definition of $\mathrm{p}K_\mathrm{a}$ values and the ionic product of water, we know that

$$\mathrm{p}K_\mathrm{a} (\text{acid}) + \mathrm{p}K_\mathrm{b} (\text{conj. base}) = 14\tag{3}$$

Since $K_\mathrm{w}$ doubles as the acidity constant of water, $\mathrm{p}K_\mathrm{a}(\ce{H2O}) = 14$. Likewise it can be shown, that $\mathrm{p}K_\mathrm{a} (\ce{H3O+}) = 0$. This is a difference of $14$ logarithmic units between two subsequent deprotonation steps. The acidity constants of sulfuric acid or phosphoric acid show that while this difference is large, it is not unusual for the conjugate base (if it is acidic) to have an acidity constand orders of magnitude lower than the original acid).

Condensing that into a conclusion, the acidity constant of hydroxide must be even lower still, to the point where we can say practically no oxide anions can be formed in equilibrium. Likewise, considering that there are many more water molecules than hydroxide anions, any oxide accidentally generated will get protonated immediately.

$$\ce{O^2- (s) + H2O (l) -> 2 OH- (aq)}\tag{4}$$

Therefore, it is not feasible for oxide anions to exist in water. This is known as the nivelling effect: no acid more acidic than $\ce{H3O+}$ can survive in aquaeous solution for extended time and no base more basic than $\ce{OH-}$.

Since the entire reaction produces products in solution, sodium oxide is not a possibility.


What is an active centre of a catalyst? I was studying catalytic poisoning and read that in temporary poisoning, the poisoners are held at active centres by weak forces.

What is an 'active centre' in this context?

The active centre is the site where the substrate binds to the catalyst in order for catalysis to take place. On a molecular level, most catalysts (or most substances really) are not just flat surfaces, they have complex three dimensional shapes. The substrate often has to bind to the catalyst in a very specific way in order for the catalysed reaction to take place - this is the active centre.


Single step transformation of a carboxylic acid into an aldehyde What are the different ways of transforming a carboxylic acid into an aldehyde in a single step ? (as opposed as for instance first transforming it to an ester then to an aldehyde)

The bottom of this page doesn't indicate any such way.

I am not looking for a "big-list", so the most common ways would be enough.

$\ce{Li, RNH2 (R=Me, Et)/NH4Cl/H3O+}$

Burgstahler, A. W.; Worden, L. R.; Lewis, T. B. Direct Reduction of Carboxylic Acids to Aldehydes by Lithium in Ethylamine. J. Org. Chem. 1963, 28 (10), 2918–2919.
Bedenbaugh, A. O.; Bedenbaugh, J. H.; Bergin, W. A.; Adkins, J. D. Synthesis of aldehydes and secondary amines from carboxylic acids via imines. J. Am. Chem. Soc. 1970, 92 (19), 5774–5775.

$\ce{BH3.SMe2/PCC}$

Brown, H. C.; Rao, C. G.; Kulkarni, S. U. A Convenient Conversion of Carboxylic Acids into Aldehydes. Synthesis 1979, 1979 (9), 704–705.

$\ce{(CH3)2CHC(CH3)2BH2}$

Brown, H. C.; Heim, P.; Yoon, N. M. Selective reductions. XVII. Reaction of thexylborane in tetrahydrofuran with selected organic compounds containing representative functional groups. Comparison of the reducing characteristics of diborane and its alkyl derivatives. J. Org. Chem. 1972, 37 (19), 2942–2950.

$\ce{(CH3)2CHC(CH3)2BHCl.SMe2}$

Brown, H. C.; Cha, J. S.; Nazer, B.; Yoon, N. M. Exceptionally facile reduction of acyclic and alicyclic carboxylic acids to aldehydes by thexylchloroborane-dimethyl sulfide. J. Am. Chem. Soc. 1984, 106 (25), 8001–8002.
Brown, H. C.; Nazer, B.; Cha, J. S.; Sikorski, J. A. Selective reductions. 38. Reaction of thexylchloroborane-methyl sulfide complex in methylene chloride with selected organic compounds containing representative functional groups. Comparison of the reducing characteristics of thexylchloroborane, thexylborane, and diborane. J. Org. Chem. 1986, 51 (26), 5264–5270.
Brown, H. C.; Cha, J. S.; Yoon, N. M.; Nazer, B. Selective reductions. 39. Partial reduction of carboxylic acids with thexylchloroborane-methyl sulfide. A direct and simple aldehyde synthesis. J. Org. Chem. 1987, 52 (24), 5400–5406.

$\ce{(CH3)2CHC(CH3)2BHBr.SMe2}$

Cha, J. S.; Kim, J. E.; Lee, K. W. Facile reduction of saturated and unsaturated carboxylic acids and their salts to aldehydes by thexylbromoborane-dimethyl sulfide. J. Org. Chem. 1987, 52 (22), 5030–5032.
Cha, J. S.; Kim, J. E.; Oh, S. Y.; Lee, J. C.; Lee, K. W. Direct conversion of saturated and unsaturated carboxylic acids into aldehydes by thexylbromoborane-dimethyl sulfide. Tetrahedron Letters 1987, 28 (21), 2389–2392.

9-BBN/Li 9-BBN

Cha, J. S.; Kim, J. E.; Oh, S. Y.; Kim, J. D. One-pot conversion of carboxylic acids to aldehydes through treatment of acyloxy-9-borabicyclo[3.3.1]nonanes with lithium 9-boratabicyclo[3.3.1]nonane. Tetrahedron Letters 1987, 28 (39), 4575–4578.

9-BBN/t-BuLi/9-BBN

Cha, J. S.; Kim, J. E.; Kim, Y. S. One-pot transformation of carboxylic acids into aldehydes through stepwise treatment of acycloxy-9-borabicyclo[3.3.1]nonanes with Full-size image (<1 K)-butyllithium and 9-borabicyclo [3.3.1]nonane. Tetrahedron Letters 1987, 28 (49), 6231–6234.

DIBAL-H

Zakharkin, L. I.; Khorlina, I. M. Zh. Obshch. Khim. 1964 34, 1029; J. Gen. Chem. USSR 1964 34, 1021.

bis(N-methylpiperazinyl)aluminum hydride

Muraki, M.; Mukaiyama, T. Aminoaluminum hydride as new reducing agents. I. Selective reduction of carboxylic acids to aldehydes. Chemistry Letters 1974, 3 (12), 1447–1450.
Hubert, T. D.; Eyman, D. P.; Wiemer, D. F. A convenient synthesis of bis(N-methylpiperazinyl)aluminum hydride: a reagent for the reduction of carboxylic acids to aldehydes. J. Org. Chem. 1984, 49 (12), 2279–2281.

activated silyl carboxylates

Corriu, R. J. P.; Lanneau G. F.; Perrot M. The one-pot conversion of carboxylic acids to aldehydes via activated silyl carboxylates. Tetrahedron Letters 1987, 28 (34), 3941–3944.

titanium-catalyzed Grignard

Sato, F.; Jinbo, T.; Sato, M. The Reduction of Carboxylic Acids to Aldehydes by Dichlorobis[π-cyclopentadienyl]titanium-Catalyzed Grignard Reactions. Synthesis 1981, 1981 (11), 871

N,N-dimethylchloromethyleniminium chloride and lithium tri-t-butoxyaluminum hydride

Fujisawa, T.; Mori, T.; Tsuge, S.; Sato, T. Direct and chemoselective conversion of carboxylic acids into aldehydes. Tetrahedron Letters 1983, 24 (14), 1543–1546.

$o\text{-}\ce{HSC6H4OH, POCl3, HClO4/LiAlH4/H2O, HgCl2}$

Costa, L.; Degani, I.; Fochi, R.; Tundo, P. Pentaatomie heteroaromatic cations. Note III. A Convenient Synthesis of Aldehydes from Carboxylic Acids via 2-Substituted 1,3-Benzoxathiolium Perchlorates. J. Heterocyclic Chem. 1974, 11, 943–948.

$o\text{-}\ce{(NH2)C6H4, PPA/NaOet/MeI/NaBH4}$ or $\ce{LiAlH4/H3O+}$

Craig, J. C.; Ekwurire, N. N.; Fu, C. C.; Walker, K. A. M. Conversion of Carboxylic Acids into Aldehydes and their C-1 or C-2 Deuteriated Derivatives. Synthesis 1981, 1981 (4), 303–305.


Organic reaction rate equations needed Where to find a list of common organic reaction rate laws?? Anybody has this?? I could not find one on internet

The Mayr group at LMU Munich (Germany) spent a great deal determining rate constants for organic reactions between nucleophiles and electophiles.  They established a public, searchable database (explained here), and grouped their findings on posters (C-, N-, P-, O-nucleophiles; entries relevant to organometallic chemistry), too.

(picture source)


Why is a double bond given priority over substituents during IUPAC nomenclature? My book names this compound as 6-ethyl-5-methylcyclohex-2-en-1-one.
I have learnt that after numbering the principal functional group which is ketone in this case, we have to treat all other groups as substituents and therefore I think the name should be 2-ethyl-3-methylcyclohex-5-en-1-one.
Why should the double bond be given priority?

The most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of 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);

  (…)

  (e) saturation/unsaturation:

    (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

    (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (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;

  (…)

You have correctly identified the principal characteristic group and assigned the lowest locant to the ketone in accordance with Rule (c).

Next, a low locant is assgined to the double bond according to Rule (e). Therefore, the name of the structure without further substituents is cyclohex-2-en-1-one rather than cyclohex-5-en-1-one since the locant 2 for the double bond is lower than 5.

Thus, the correct name is 6-ethyl-5-methylcyclohex-2-en-1-one.


Why is the para product major in the nitrosation of phenol? When phenol reacts with nitrous acid ($\ce{NaNO2 + \mathrm{conc}\ H2SO4}$), 4-nitrosophenol and not 2-nitrosophenol is formed. I cannot understand why the para-isomer should be preferred over the ortho-isomer.

I understand the reaction mechanism proceeds by electrophilic substitution by the $\ce{NO+}$ cation, and that the $\ce{OH}$ group activates the ring, which leads to ortho and para substitution.

I don't understand why the para-isomer should be preferred over ortho?

Ortho and para positions are activated due to the +R effect of the $\ce{-OH}$ but the preference of the para product can be explained by the mechanism.

The mechanism proceeds through a dienone. The ortho dienone intermediate will experience strain as  $\ce{=O}$ annd$\ce{=NOH}$ lie in the same plane. Para nitrosated intermediate on the other hand is free from strain.

Reference: González-Mancebo S., Lacadena J., García-Alonso Y., Hernández-Benito J., Calle E., Casado J. Monatsh. Chem. 2002, 133 (2), 157-166. DOI: 10.1007/s706-002-8245-0.


Is oil shinier than water? Why? The physics part: The part that talks about being shiny, I agree, is mainly physics. But proceed in my question, and you'll see how it's a chemistry one.

We know things are shiny if they either produce light, or deflect light well. Producing is not our topic in here, but deflecting is. Metals are considered shiny objects.

Mad scientist (the site, not our fellow friend!) has provided a nice explanation for why metals are shiny: In short, because free electrons (electrons are partially "free" in metals, due to the especial way of bonding) deflect more photons than usual and do not let the inner layers of atoms in the metal absorb them.!

(These three differ in shine because of impurity; not related)

Metals aren't the only materials that are exceptionally "shiny". We are aware of many crystals that are either well-cut or well-formed, which result in total internal refraction and that makes them very shiny too. There's a related question in physics.SE wherein the diamond's shiny properties are questioned.

The organic-chemistry part:

Saying the word "oil" talks about a vast variety of chemicals; therefore, in an attempt to narrow down the question, I only ask about liquid-at-RTP triacylglycerol (the famous triglycerides which exist usually in plants or fish). The chemical structure is as following:

My question is: Organic molecules like these aren't metals, are they?! So since a thorough research (maybe seemingly) didn't get me answers about why these are good light reflectors I'm asking for a way that will allow me to compare the reflectivity of these liquids, primarily from their structural formula. e.g.: Are these liquids "shinier" than water? Why? About water, I would say that for example, Hydrogen atoms in correspondence to the Oxygen atom, make water have hydrogen bonding, which is the main cause of surface tension. Surface tension causes the surface of water to be "smoother" and thus the reflectivity increases. On the other hand, it can't be only about Hydrogen bonding, can it?

(If the answer might be too broad, you may guide me to any resource helping me to find an answer, and that's why I add the tag: reference-request.)

Image sources: http://www.luxrender.net/wiki/images/2/2f/Shiny_yellow_specular_series.jpg ; http://library.med.utah.edu/NetBiochem/FattyAcids/7_2.html ; http://guweb2.gonzaga.edu/faculty/cronk/biochem/images/triglyceride.gif

The Fresnel equations, http://en.wikipedia.org/wiki/Fresnel_equations, are used to calculate reflectivity of transparent substances based on the index of refraction of the liquid vs. that of air or other medium. There are complications introduced by conductivity, and solids may display anisotropic reflectivity, see http://www.hindawi.com/journals/isrn/2012/732973/.

Dispersion staining is a technique for staining microscopic specimens depending on the change in the index of refraction with the color of light (dispersion), see http://en.wikipedia.org/wiki/Dispersion_staining. This uses the shininess of substances in a liquid, often an organic immersion oil, to help distinguish or identify them.

Note that metallic reflection is based on a "sea" of electrons bouncing back an inverted wave, and differs from refraction. Liquid metals such as mercury are shiny for that reason, but I do not know if there are sufficiently conductive organic liquids that would have metallic reflection.


Timescale of Protonation and Deprotonation across micelle layers I am curious about the timescales of protonation and deprotonation of solvent systems. As a followup, how is this affected when the proton source is separated by an organic phase? For instance, it is known that the pH inside a liposome can be maintained from the outside. What are the timescales of protonation when the protons need to be transported across various micelles thickness.

Permeation of protons, potassium ions, and small polar molecules through phospholipid bilayers as a function of membrane thickness

  Reported values for proton permeability coefficients vary over
  a
  wide
  range,
  from
  $10^{-2} \, \mathrm{cm/s}$ to $10^{-7} \, \mathrm{cm/s}$
  (Nichols
  and
  Deamer,
  1980;
  Nichols
  et
  al.,
  1980;
  Biegel
  and
  Gould,
  1981;
  Elamrani
  and
  Blume,
  1983;
  Grzesiek
  and
  Dencher,
  1986;
  Perkins
  and
  Cafiso,
  1986;
  Norris
  and
  Powell,
  1990).
  This
  variation
  relates
  to
  the
  use
  of
  different
  techniques,
  lipids,
  and
  bilayer
  systems
  (liposomes
  and
  planar
  bilayers)
  or
  different
  pH
  ranges
  (Perkins
  and
  Cafiso,
  1986).
  The
  proton
  data
  of
  this
  study
  agree
  with
  the
  higher
  reported
  values.

  As revealed by Fig. 4, proton permeability coefficients strongly
  depend on membrane thickness, starting at a maximum value of $1.3 \cdot 10^{-2} \, \mathrm{cm/s}$ for the shortest lipid and steadily decreasing to a value of $4.9 \cdot 10^{-5} \mathrm{cm/s}$ for the longest lipid. It follows that proton permeability decreases by a factor of approximately 250 as the thickness of the
  hydrophobic region is increased from $20 \, \mathring{\mathrm{A}}$ to $37 \, \mathring{\mathrm{A}}$.


Calculation of the pH of a mixture of a strong acid and weak acid The question is to find out the $\mathrm{pH}$ of a mixture of weak acid and strong acid. My book just states the formula as $$\mathrm{pH}=-\log
\frac{C_2+\sqrt{C_2^2+4K_\mathrm{a}C_1}}{2}$$

where $C_1$ is the concentration (in mole/litre) of the weak acid (ionisation constant $K_\mathrm{a}$), and $C_2$ is that of the strong acid.

I understand that the strong acid will dissociate completely and the weak acid will dissociate with an ionisation constant. So the $\ce{H+}$ ion concentration from strong acid will be simply $C_2$ and that from weak acid will be $\sqrt{K_\mathrm{a}C_1}$. But I could not derive the above formula for the $\mathrm{pH}$ using these facts. I need help here.

Chemical thermodynamics is easy in that it always has enough equations to derive all the variables involved, and guarantees to produce a unique solution. The derivation itself may be not that easy, but that's another story.

In short, we have some weak acid $\ce{HA}$ and the products of its dissociation. Mind you, $\ce{H+}$ comes also from the strong acid, which dissociates completely. Now, the equations:

$$
\ce{[HA] + [A-] = }C_1 \\
\ce{[H+]=[A-]}+C_2 \\
{\ce{[H+][A-]}\over\ce{[HA]}} = K_\mathrm{a}
$$

Is the rest clear?

   Use the second equation to exclude $\ce{[A-]}$, then use the first to exclude $\ce{[HA]}$, and end up with a quadratic equation on $\ce{H+}$.


What happens to iron electrodes (attached to a battery) in copper(II) sulphate solution? After a few hours, the following observations were made:

The solution turns green. My guess: $\ce{FeSO4}$ is formed.
There are reddish-brown and black powder deposits on both the electrodes.
The deposits fall down easily when the set-up is shaken a little.

What are the reactions that are occurring?

According to me, the copper is displaced from the $\ce{CuSO4}$ and is deposited on the iron electrode (the cathode), and $\ce{FeSO4}$ is formed as a green solution. What is getting deposited on the anode? Are there ferrous and ferric oxides/hydroxides being formed?

The reaction is spontaneous, you don't need put a power supply.

Let's see the reaction with the potentials (obtained from the table of standard reduction potentials):

$$
\begin{align}
\ce{Fe^0 &→ Fe^2+ + 2 e-} &\quad E^\circ &= \pu{0.44 V} \\
\ce{Cu^2+ + 2 e- &→ Cu^0} &\quad E^\circ &= \pu{0.337 V} \\
\hline
\ce{Fe^0 + Cu^2+ &→ Fe^2+ + Cu^0} &\quad E^\circ &= \pu{0.777 V}
\end{align}
$$

So, the solid formed in both electrodes is copper. The green solution is related to iron sulfate.


Product of the reaction between chalcone and sodium methoxide in methanol I'm a bit lost on how to do this, for Product C I can see two things happening:

1) The $\ce{-OCH3}$ attacks the alpha hydrogen, but that doesn't really make sense as then you would get two double bonds next to one another

2) The $\ce{-OCH3}$ attacks the electrophilic carbon, but then we know the negatively charged oxygen cannot collapse back because then the $\ce{-OCH3}$ would leave. So, would it just get printed by $\ce{HOCH3}?$ And if it does that how would it react with the second set of reactions?

The first step is called Michael addition. Have a look at this wikipedia page.

  The Michael reaction or Michael addition is the nucleophilic addition of a carbanion or another nucleophile to an α,β-unsaturated carbonyl compound. A newer definition, proposed by Kohler, is the 1,4-addition of a doubly stabilized carbon nucleophile to an α,β-unsaturated carbonyl compound.

Of course, in your case, the nucleophile is $\ce{OCH3-}$. Perform 1,4-addition by adding it to the double bond at the beta position (with respect to the carbonyl), and you'll expect an enolate ion, which looks like this -

Finally protonate the $\ce{O-}$ and tautomerise to get your product as @Zenix already stated.

For the second step, check out this link.

References: Michael reaction, wikipedia || Ketone halogenation, wikipedia || Enol, wikipedia


Redox reaction with tin and sulfuric acid solution I need to write the redox reaction from tin and a sulfuric acid solution. Here is the solution given:

  The first half-reaction: $\ce{Sn->Sn^2+\+2e^-}$
  The second half-reaction: $\ce{2H+\+2e^-->H2}$
  Adding both reactions will give you: $\ce{Sn\+2H+->Sn^2+\+H2}$

Why is the sulfuric acid solution replaced with $\ce{H^+}$?

The sulfuric acid dissociates in water: $$\ce{H2SO4 -> H+ + HSO4-}$$$$\ce{HSO4- -> H+ + SO4^{2-}}$$

When writing redox equations the spectator ions are left out because they do not participate in the redox reaction.

In this case it is the hydrogen ions which are oxidising the tin and the sulfate and hydrogensulfate ions play no part in the reaction so they are left out. Including them in the hydrogen half equation would give (with sulfate ions): $$\ce{2H+ + SO4^{2-} +2e- -> H2 + SO4^{2-}}$$

Clearly the sulfate ions are unnecessary and so they are omitted from the equation.


Sodium permanganate from manganese dioxide and sodium hypochlorite This page states that sodium permanganate can be formed by the reaction of manganese dioxide with sodium hypochlorite and sodium hydroxide:

$$\ce{2 MnO2 + 3 NaClO + 2 NaOH -> 2 NaMnO4 + 3 NaCl + H2O}$$

I assume sodium carbonate ($\ce{Na2CO3}$) will also function in this reaction as a substitute for $\ce{NaOH}$.

Does this reaction take place at standard conditions, or are increases in temperature/pressure etc. required? Approximately how long will it take (at STP)? Also, will sodium carbonate be sufficient, or is sodium hydroxide required?

I performed this reaction, and the answer is yes, concentrated sodium carbonate can be used in place of sodium hydroxide, though the latter works better. The reaction goes under normal pressure, but it requires heating. The rate of the reaction also depends on the quality of manganese dioxide - freshly prepared works best. You will notice the formation of permanganate as soon as you boil the reaction mixture, but complete conversion of dioxide may take several minutes to hours. If you want a complete conversion, better use the sodium hydroxide.


Pyrotechnic "star" without oxidizer In the design of a celebratory pyrotechnic (firework), there exists a gunpowder explosive charge at the center of the firework and pyrotechnic "stars" lining the circumference of the explosive. The "stars" are composed of an oxidizer and a fuel such as KNO3 oxidizer and charcoal as a fuel. If the oxidizer is removed from the "stars", is it reasonable, upon detonation, to assume that the now vapourized "star" will combust only with the detonation of the gunpowder explosive charge without the need for an oxidizer in the "star" composition?

See this page for a diagram and composition of materials: http://www.instructables.com/id/How-to-Make-Fireworks/

When the interior gunpowder charge explodes, it blows the stars in all directions and ignites them.  They then burn very rapidly (perhaps explosively) giving the star effect exactly because the stars are a mixture of fuel and oxidizer.  If the stars were all fuel (carbon), they would be essentially charcoal briquettes.  In that case the center would explode, throwing (possibly) burning charcoal chunks in all directions.  This could be a truly novel approach to barbecue...


Complexes formed by lanthanides and actinides A very simple test for the presence of alcohols in the lab involves adding ceric(IV) ammonium nitrate solution which gives a pinkish red colour in the presence of alcohol.

When I looked up the structure of this compound it turns out to be quite interesting. So a couple of questions about lanthanides came up in my mind (a Google search only comes up with very complicated research papers):

Wouldn't this compound be highly unstable since there 12 donor atoms (forming an icosahedral structure) surrounding it (but it is quite stable)?
Is the formation of complexes in lanthanides and actinides similar to that of transition metals? Do $\ce{f}$ orbitals undergo crystal field splitting just like $\ce{d}$ orbitals?
Are higher coordination numbers like 12 more common (than octahedral and tetrahedral structures) among lanthanides and actinides because they have more orbitals for bonding, or is this compound an exception?

After a lot of digging around, I finally found something about this in J.D. Lee's Concise Inorganic Chemistry. I thought I'd post whatever I found over here anyway since it's very interesting:

Answer to question 2

Complex formation by lanthanides is different from that of actinides. In lanthanides, the $\mathrm{4f}$ orbitals are well shielded by the larger $\mathrm{5d}$ and $\mathrm{6s}$ orbitals and are deep inside the atom. So $\mathrm{f}$ orbitals do not participate in any bonding and complex formation is similar to that of transition metals. However in the actinides, the $\mathrm{5f}$ orbitals extend outwards and participate in bonding much more easily making the interactions much more complex than in the transition metals.

Answer to question 3

Higher coordination numbers are apparently extremely common among f-block elements (strange right?).
Octahedral (6) and tetrahedral (4) structures are very rare except when bulky ligands are present. Most common coordination numbers among lanthanides are

7 (Capped trigonal prismatic) yttrium acetylacetonate hydrate (Yttrium seems to be grouped with the f-block elements due to similar properties)
8 (Square antiprismatic and Dodecahedral) cerium acetylacetonate and holmium tropolonate
9 (Tri-capped trigonal prismatic) nonaaquaneodymium(III) complex $\ce{[Nd(H2O)9]^3+}$

10 (very complex) and 12 (icosahedral) are seen only in the larger elements like cerium and thorium with small ligands like $\ce{NO3-}$ and $\ce{SO4^2-}$

The actinides also commonly form exotic structures like chained tricapped trigonal prismatic in $\ce{[ThF8]^4-}$ and $\ce{[PaF7]^3-}$ distorted cubic in $\ce{[PaF8]^3-}$

The text mentions that the nature of bonding in $\ce{[Ce(NO3)6]^2-}$ is still not understood because it would imply bond orders of less than 1 or participation of f orbitals.

Could anyone explain anything about the first question since there is nothing mentioned on the stability of these compounds.


Enclathrated - definition? When discussing the solvent uptake of a metal-organic framework would the correct term be "solvent molecules were enclathrated into the pores"? Despite searching for a definition online I have drawn a blank and its not a term I am familiar with. Having read many papers in polymer chemistry where this term is also used I wonder if I have misunderstood its usage.

The Oxford Dictionary does not have an entry for enclathrate, but does for clathrate:

  A compound in which molecules of one component are physically trapped within the crystal structure of another.

Using the word enclathrate should be redundant.


Does citric acid react with carbonated water? I've noticed that when adding lemon juice to carbonated water, large 'soapy' bubbles appear on the surface. Since I'm adding another acid to a solution of carbonic acid, I would maybe expect a faster release of $\ce{CO2}$, but the bubbles I'm seeing do not resemble the usual fizzy drink bubbles.

What's going on?

The chemistry you're describing the the one you expect: CO2 is formed, and released from the liquid. However, your question is more about kinetics than about the final product. We can assume that slow release of gas leads to larger bubbles, because nucleation rate for the bubble formation is lower. Now, in fizzy drink, there is a large quantity of carbon dioxide accommodated by the liquid due to pressurization. In your case, it will depend on the concentration of your lemon juice (natural lemon juice has pH ~ 2), and more importantly on that of the carbonated water: I just tried with a few grams of sodium bicarbonate dissolved in a glass of water in my kitchen, and the kinetics and bubble size really depends on initial carbonate concentration. (I haven't been able to make pictures or video, because I had to handle my kids who were very fast drawn to this impromptu experiment.)

Also, I second Richard's comment: there are many additional chemicals in lemon juice, some of which surely act as surfactants… which changes bubble formation and kinetics.


Vapor pressure and boiling point for solutions in open containers Question 1: If we have an open container which has water in it and is heated. If there is no lid over the container then how can this water ever boil as there will be no vapor pressure created over the water surface.

Question 2: What is the logic in the fact that boiling starts when vapor pressure becomes equal to atmospheric pressure.

Boiling is not about the upper surface at all. What happens there is but of little interest to us. Boiling means that bubbles are forming at the bottom and probably within the bulk of the liquid. Guess what is inside these bubbles? It is pure water vapor. That's where the equilibrium between vapor and water actually comes into play.

As for the second question, the bubbles must withstand the external pressure, otherwise they would not be able to form in the first place (or if they will, they will be instantly squeezed into nothing). That's why they actually appear only when vapor pressure becomes equal to atmospheric pressure.


Carbocation Stability A homework question I was given asked me to identify the most stable carbocation between these 4. I was able to narrow it down to either the trifluoromethylcyclohexane carbocation or the methoxycyclohexane carbocation, but I’m not sure which between the two of them.

My gut feeling is that the methoxycyclohexane is more stable because of its contributing resonance structures, but I’m not sure if that trumps the hyperconjugation that the fluorine atoms provide. Any insight would be appreciated!

The fact that the methoxycyclohexane carbocation is resonance stabilized greatly increases its stability over the the trifluoromethylcyclohexane carbocation.

The same power of oxygen’s ability to donate electrons can be seen in its effects on benzene rings. Both $\ce{-OH}$ and $\ce{-OR}$ groups are strongly activating groups, while $\ce{-CF3}$ is a ring deactivating group, meaning it actually pulls electrons from the benzene ring. This further shows that the methoxycyclohexane carbocation is the most stable of the 4.


The usage of 'limiting composition' I am reading wikipedia page on Molybdenum bronze and they mention this:

  Blue bronzes, with limiting composition $\ce{A_{0.3}MoO3}$ that is,
  $\ce{A_3Mo10O30}$.

I used to see these fractional numbers (e.g. 0.3) and always wondered whether technically speaking they are correct or not. I understand that, as the wikipedia page says, they are equivalent to the 'full' numbers (e.g. 3) but still I am not sure whether in literature one of them is preferred or more common than the other?

These are called non-stoichiometric compounds or berthollides. Because the ratio of the fractional component is not fixed, but can vary over a small but continuous range, it would be pointless to try to express the formula as least common multiples (LCM).

For example, iron(II) sulfide can have a ratio from about $\ce{Fe_{$0.8$}S}$ to $\ce{FeS}$. So in LCM terms, the formula varies continually from $\ce{Fe4S5}$ through $\ce{FeS}$, including $\ce{Fe9S10}$, $\ce{Fe19S20}$ and so forth, all with similar physical and chemical properties.


Converting Disolved Oxygen from micromole/L to millimole/kg I have dissolved oxygen data that is in units of µmol/liter.

I need to convert it to mmol/kilogram.

I have been told by someone who knows more about chemistry than I do (but only a little bit more) that I need to know the mass of my sample as well (which I don't know).

So only knowing my original DO data (in µmol/L) and the density of SW (assume 1.027 g/cm³), how does one convert the data to mmol/kg?

Oxygen is barely soluble in water, so honestly, treating the saltwater as though it were "pure" saltwater with no dissolved oxygen is roughly equivalent to treating it as heavily oxygenated.  The difference is in the fourth or fifth decimal place.  Note that this would start to fall apart if you are concerned about the solubility of high pressure oxygen in water, but for small quantities of oxygen (such as the micromoles and millimoles you're using) this is likely not a factor.

The conversion is relatively straightforward.  For a solution with concentration $X$:

$$(X\ \mathrm{\mu mol/L}) \times (1\ \mathrm{mmol}/1000\ \mathrm{\mu mol})\times(1\ \mathrm L/1.027\ \mathrm{kg}) = 0.0009737\ X\ \mathrm{mmol/kg}$$


What is the difference between infrared action spectroscopy and conventional IR spectroscopy? I'm familiar with infrared spectroscopy generally, which probes mainly the vibrational behavior of molecules under study.  However, I'm seeing references in various literature to 'infrared action spectroscopy' (e.g., here), which I'm having a hard time finding information about.

So, the title question: What is the difference between infrared action spectroscopy and conventional IR spectroscopy?

After 30 minutes of Googling, I'm thinking that we both (EDIT: nope, it was just me) misparsed the name.

When reading the name "infrared action spectroscopy," my first instinct was to mentally split it up like

$$\textrm{infrared action} \qquad\qquad \textrm{spectroscopy}$$

that is, spectroscopy of the "infrared action" flavor.

However, after running into more dead ends than a drunk rat in a maze with no exits, I realized that the correct way to split it might be

$$\textrm{infrared} \qquad\qquad \textrm{action spectroscopy}$$

or "action spectroscopy" that happens to use infrared light.

Lo and behold, the first hit on google was a wikipedia page on action spectroscopy. That link seems to have more of a biological tint to it (as do most results), but there are several physical science papers out there as well. (See Edit)

It seems that action spectroscopy in general refers to hitting a system with light and watching some kinetic property of the system (rate of photosynthesis, conversion rate of isomers, etc.) In the case of the paper you cited, they report the relative intensity, so my guess would be that they did something like a scan over frequencies and watched how the ions fragmented at each frequency, which is supported by a cursory glance over the text surrounding Equation (1). Unfortunately, I don't have time to read the paper in full, so let me know if I'm totally wrong.

The moral of the story is that English is hard.

=============

EDIT: It turns out I'm the dolt and everyone else processed this correctly. As Brian mentions in the comments, the action spectroscopy page on Wikipedia is pretty awful, so I'll try to expand on action spectroscopy below, given what little I know about the subject.

In your usual spectroscopic setup, you hit the sample with some form of EM radiation, then look at what's reflected or absorbed to learn something about the system (e.g. UV-vis, IR, NMR, rotational)

It seems that in action spectroscopy, your goal is to use the light to cause some sort of a change in the system, and measure that instead. In the example given on Wikipedia, you measure the rate of photosynthesis as you change wavelengths--this tells you at which wavelengths photosynthesis is most effective.

In the paper I linked above, they seem to mediate the conversion of cis-butene to trans-butene using light. Presumably, you could start off with low-energy photons, then gradually ramp the energy up until you see the isomerization happen. The wavelength at which the transformations first start happening would tell you the activation energy barrier for the cis-trans isomerization. (Note that I can't actually read that paper because of the paywall, so I may be wrong about this.)

In the paper Brian linked, they seemed to use IR to study the fragmentation of an ion sample. They mention "loss paths" as observed through an MS, so it seems like they hit a sample with IR light, then figure out what it breaks up into.

As Brian said in a comment, it seems that action spectroscopy is where light is used "to alter the system under study, not to just probe it passively." I'd like to add that, in general, it seems that the quantity of interest is not the amount of light that comes back out, but something else that happens in the system as a result of the exposure to light (e.g. isomerization, ion fragmentation, glucose production).


Finding final pH of the buffer solution without applying Henderson equation A is beaker with $\pu{0.1 M}$ $\pu{25 cm^3}$ solution of $\ce{NH3}.$ B is beaker with $\pu{0.1 M}$ $\pu{5 cm^3}$ $\ce{HCl}.$ Now the solution of A and B is mixed. What will be the final $\mathrm{pH}$ of the mixture? $K_\mathrm{b} = \pu{3.3E-5}.$

I believe the mixture will become a buffer solution and we should use Henderson equation to solve this.
$$n(\ce{NH3}) = \pu{0.1 mol L^-1} × (\pu{25E-3 L}) = \pu{2.5E-3 mol}\tag{1}$$
$$n(\ce{HCl}) = \pu{0.1 mol L^-1} × (\pu{5E-3 L}) = \pu{5E-4 mol}\tag{2}$$
$$\mathrm{pOH} = \mathrm{p}K_\mathrm{b} + \log\frac{n(\ce{NH4Cl})}{n(\ce{NH4OH})}\tag{3}$$
$$\ce{NH3 + HCl -> NH4Cl}$$
$$n(\ce{HCl}) = n(\ce{NH4Cl}) = \pu{5E-4 mol}\tag{4}$$
$$n(\ce{NH3}) = n(\ce{NH4OH}) = \pu{2.5E-3 mol} - \pu{5E-4 mol} = \pu{2E-3 mol}\tag{5}$$
$$
\begin{align}
\mathrm{pOH} &= \mathrm{p}K_\mathrm{b} + \log\frac{\pu{5E-4 mol}}{\pu{2E-3 mol}}\\
  &= 4.4819 - 0.602\\
  &\approx 3.88 \tag{6}
\end{align}$$
$$\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 3.88 = 10.12\tag{7}$$
However, the book from which I am solving this problem suggests that this problem should not be solved by Henderson equation without providing any reason. Can anyone tell me why this solution should not be considered a basic buffer? I would really appreciate some opinion on this.
P.S. The problem was given in Bangla. I tried my best to translate it. This is an admission question of KUET-2019 Bangladesh. My textbook has this question, but I'm not sure about the solution. Please note that this is an admission question of previous year (2019).

As the amounts of substance in the final solution are known to be $n(\ce{NH3}) = \pu{2 mmol},$ and $n(\ce{NH4^+}) = \pu{0.5 mmol},$ you may simply use the definition of the constant $K_\mathrm{b}:$
$$K_\mathrm{b} = \frac{n(\ce{NH4^+})[\ce{OH^-}]}{n(\ce{NH3})} = \frac{\pu{0.5 mmol}\times [\ce{OH-}]}{\pu{2 mmol}} = \pu{3.3E-5}$$
from where $[\ce{OH-}],$ $[\ce{H+}]$ and $\mathrm{pH}$ can be quickly obtained:
$$[\ce{OH-}] = \pu{1.32E-4 mol L^-1}$$
$$[\ce{H+}] = \frac{10^{-14}}{[\ce{OH-}]} = \pu{7.57E-11 mol L^-1}$$
$$\mathrm{pH} = -\log[\ce{H+}] = 10.12$$


activating and deactivating groups and directing effect In electrophilic aromatic substitution:
I have just learnt a 'rule' about benzene with more than 1 substituents attached (excluding the effect on steric hindrance):

if a strong activating group (EDG) and a strong deactivating group (EWG) are attached, activating group wins hence the electrophile adds on ortho & para position relative to the activating group.

if a weak activating group (EDG) and a strong deactivating group (EWG) are attached, deactivating group wins hence the electrophile adds on meta position relative to the deactivating group.

I understand rule 2 (strong must dominate the weak), but I do not understand what is the principle behind rule 1:

e.g. methoxide group wins over nitro group, although they are both strong EDG & EWG. I understand if the methoxide group is replaced by a methyl group, nitro group wins as $\ce{R-CH3}$ only has a much weaker inductive effect (compared to strong resonance effect). However, both methoxide and nitro groups use resonance to exert their respective directing effect, why activating groups always win?

I believe that if we ignore the inductive effects of the substituents for the time being, then the electron-donating group(which donates electrons via resonance effect) would manage to activate a few positions in one or more resonating structures, and the electron-withdrawing group can only take up this delocalized pair of $\pi$ electrons when it is present at exactly an ortho or para position where the $\ce{+M}$ effect operates, and the negative charge lies on the carbon alpha to the electron-withdrawing group.
It can be seen here that the negative charge in the resonance structures of $2,4$-dinitroaniline can be taken up by the nitro group only when the negative charge becomes alpha to it:

So the only way an electron-withdrawing group can "win" is if it placed against a very poor electron-donating group.
First, in the case of $\ce{-CH3}$ or halogens (who neither use $\pi$ conjugation for electron donation nor have an electron-donating nature), the electron density donated to the ring is unable to emerge in the intermediates without being delocalised by the EWG, or second when the EWG is present exactly at positions where the negative charge reaches by $\pi$ conjugation(in case of "strong" electron donators).
If the group donating via the $\pi$ bonds is even a relatively better donating group, then unless all the ortho and para positions are having electrons being withdrawn by the presence of electron withdrawing groups, there will always be resonance structures donated having some partial activation at ortho or para positions.
Hence, the electron-donating group "wins" in such situations unless there is no steric hindrance during the substitution taking place.


How can I find the heat that will be released in kJ per gram for the reaction of Al with Fe2O3? Find the heat that will be released in kJ per gram when aluminium ($\ce{Al}$) reacts with $\ce{Fe2O3}$ as follows:
  $$\ce{2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(l)}$$

The information given is:

$\Delta H = 12.40 \space \rm{kJ/mol}$ (for $\ce{Fe}$)
$\Delta H = -822.2 \space \rm{kJ/mol}$ (for $\ce{Fe2O3}$)
$\Delta H = -1669.8 \space \rm{kJ/mol}$ (for $\ce{Al2O3}$)

If it asked me to find the heat released in kJ/mol that would be easy:

$$
\Delta H^\circ = \sum \nu_{i}\Delta H^\circ_f (\rm{products}) - \sum \nu_{i}\Delta H^\circ_f (\rm{reactants})
$$

I assumed that there is one mole and I found the mass in grams by multiplying the molar mass of each compound by one mole.

Then I divided the $H^\circ_f$'s by those amounts to get the new values "expressed in kJ per gram"

Then I used the previous equation and I got an answer around 10 but the correct answer is 15.24 kJ/g.

The question is: is the way I used wrong, or is there something else I didn't pay attention to?

When you converted the heats to kJ/g, you inadvertently "broke" the Hess's law equation that you plugged them into.

Each of the enthalpies of formation in that equation is multiplied by the number of moles of the matching species, and so the units of the enthalpy of formation are kJ/mol.

If you assume one mole of product and convert that to grams, you then would have to convert all of the other species to grams (two moles of Al, two moles of Fe, etc).

It's easier to just do the whole calculation in moles, then divide the final answer by grams. However, you need to specify which species is the gram basis. In other words, you are asking for the heat per gram, but per gram of what?

Whatever the answer is, then find the total heat output in terms of kJ/mol of that substance, then divide it by the molar mass of the substance (in g/mol).

One way to avoid these kinds of problems is to always explicitly include the units in any calculation that you do. In this case, you would have quickly seen that the Hess's Law equation would not give you the correct units.


Is micelle formation in a soap solution a chemical reaction? I know that soap molecules have a hydrophilic end and a hydrophobic end which attract to water and oil respectively. However, does a chemical reaction take place?

Chemical 'reactions' are normally defined by changes in 'bonding'.  Related to soap micelles, for the micelle to exist and have integrity enough to be characterized from a situation where they did not previously exist, some 'bonding' must come into being.  That bonding is normally described as 'London Forces' or 'Vander Waals Bonds' or 'Hydrogen bonding' or 'electrostatic' interactions.  Since equivalent concentrations produce equivalent results, the micelle formation is also reproducible.  In the purest sense this does imply that soap micelle formation is a chemical reaction.

The question is more philosophical than anything.


Why the formation of a fog is observed when triethylamine is added? In the procedure for the synthesis of N-anisoyl-2-pyrrolidinone that was given to me, it is written that when triethylamine is added, a "fog" is observed inside the flask.

Triethylamine is added to neutralize the hydrochloric acid that is generated during the reaction, forming triethylammonium chloride.
I searched for similar procedures involving an acyl chloride and an amine, but the formation of a "fog" is never mentioned.
A quick search on the internet gave me no clue.
In my experience I have never witnessed the formation of a "fog" with this kind of reactions.
Since I cannot enter the lab due to the COVID-19 restrictions, I cannot perform the reaction now, can someone help me to understand if and possibly why in this case a "fog" is formed?

You are adding Et3N to a solution that contains HCl as a reaction product. That means there will be HCl vapour above the solution. The Et3N is fairly volatile so there will be Et3N vapour around the addition stream/droplets and this will react with the HCl vapour as it is added to give triethylammonium chloride. This gives you the fog.


Acidic strength of organic acids. Substituted PhCOOH acids Could someone please explain the order of acidity of ortho, para and meta substituted benzoic acids with the substituent as methyl group?
Methyl group operates +I and +R effect, I believe.

The ortho substituted benzoic acid is most acidic, followed by the meta substituted and finally para substituted benzoic acid.

The methyl group, when at ortho position, causes steric inhibition of resonance, which essentially means that the $\ce{-COOH}$ group goes out of the plane of the benzene ring (due to steric hindrance from the $\ce{-CH3}$ group). Due to this, the benzene's $\pi$ electrons can no longer be transferred to the central carbon of the $\ce{-COOH}$ group by resonance, hence increasing the positive charge on it. This makes the hydrogen of the $\ce{-COOH}$ group more acidic.

When the methyl group is on meta position, it operates by +I, thus reducing its acidity. When on para position, it operates by both +I and +R (actually +H, which means hyperconjugation effect), hence decreasing the acidic strength even more.


Which of these compounds does not liberate heat when water is added to it? Which of these compounds does not liberate heat when water is added to it: $\ce{KNO_3}$, $\ce{NaOH}$, $\ce{CaO}$, $\ce{H_2SO_4}$, $\ce{Na}$.

I know from experience that dropping sodium tablets into water has explosive results, but I am clueless about the method to solve this problem. I know the answer is $\ce{KNO_3}$, but I don't know why. If someone can provide the method, I think I can do the problem.

It is important to note that if a molecule or formula unit is capable of undergoing hydrogen bonding ($\ce{NaOH}$, $\ce{H2SO4}$), then the formation of this hydrogen bond will be extremely exothermic.

Also, if a substance can react to form a substance that will undergo hydrogen bonding ($\ce{CaO + H2O -> Ca(OH)2}$), this species will subsequently form these hydrogen bonds and liberate heat.

Lastly, as you have stated you know from personal experience, the reaction between a alkali metal ($\ce{Na}$) and water is extremely exothermic— so much so that many of these metals will melt from the heat liberated.

Otherwise, there is no fail-proof method for predicting the enthalpy of solution for any given substance. Generally, the enthalpies of solution are determined experimentally, so unless you are expected to perform all these reactions in a calorimeter yourself, I would recommend something like this: this.


Electron arrangement in Nickel I'm struggling to understand why the element Nickel can be written like so: [Ar]3d84s2, but cannot be written as [Ar]3d10.

I know that the d-orbital can 'house' 10 electrons, and that according to the Aufbau principle, electrons must be 'housed' in the vacant orbital which has the lowest energy, so the first way I presented seems wrong to me.

I'm sure that there's some confusion on my end here, or rather, I'm missing something that might seem obvious, but I really don't get it.

The available energy levels are arranged in the following order, going from the lowest level to the highest :
$1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p,$  etc. So the $4s$ level is filled before the $3d$.


Endothermic or Exothermic Is an endothermic reaction always endothermic at all conditions of pressure, temperature etc.? Similarly for exothermic... I mean is sign of ∆H always the same as sign of ∆H°?

This might seem to be a vague doubt but I just feel that there can be a reaction with trends of energies of formation of reactants and products such that endothermic reaction changes to exothermic or vice versa subject to change in concentrations of substances or external factors like temperature, pressure etc.

This is just a random thought of mine and even an example in this direction would be helpful.

Thanks!

Because $\Delta H$ is temperature-, pressure-, and concentration-dependent, it's certainly theoretically possible for $\Delta H$ for a chemical reaction to change sign in response to changes in any of those intensive properties.

I don't have an example of such a chemical reaction handy.  However, there are direct experimental measurements showing enthalpies of solution that change sign in response to both temperature and concentration changes:

For instance, the enthalpy of solution of polystyrene in decalin has been shown to change from ~ –10 J/g at ~20 C to ~ +6 J/g at ~100 C (the ~'s were needed because I'm eyeballing these from a plot).  Reference:

Tager, A. A., et al. "The temperature and concentration dependence of enthalpy, free energy and entropy of mixing polystyrene solutions with an upper and lower critical temperature of mixing." Polymer Science USSR 12.6 (1970): 1497-1511.

And the enthalpy of solution of $\ce{LiNO_3}$ has been shown to change from negative to positive as the concentration increases.  Reference:

Wolf, G., V. Vacek, and V. Pekarek. "Enthalpies of solution and of crystallization of lithium nitrate and of lithium nitrate trihydrate in water at 25° C." Journal of Solution Chemistry 19.10 (1990): 1029-1039.