Need for enthalpy of vaporization in calculating bond dissociation enthalpy In our chemical bonding class we were given the following problem:

I am confused by how the question says that you need to consider the heat of vaporization/sublimation for HBr and HI.

Bond dissociation enthalpy is defined as "the energy needed to break one mole of the bond to give separated atoms - everything being in the gas state."

So if you can find the heat of formation of that molecule in the gas phase, why would you need anything else?
My attempt:

The above answer obtained is not the one provided by the professor so I assume I'm missing something here. The answer provided is-

364 kJ/mol

Well, I think that you have to look up the actual way bond dissociation enthalpy is defined in your textbook/lecture notes, since it is a not quite standard term. IUPAC defines only bond-dissociation energy but not enthalpy.

bond-dissociation energy, $D$
The enthalpy (per mole) required to break a given bond of some
specific molecular entity by homolysis, e.g. for $\ce{CH4 -> .CH3 + H.}$, symbolized as $D(\ce{CH3−H})$ (cf. heterolytic bond dissociation energy).

Many resources on the Internet claim that the IUPAC definition of the bond-dissociation energy refers to $\pu{0 K}$ (see e.g. Wikipedia, UC Davis ChemWiki) and that it differs from bond dissociation enthalpy, which refers to the enthalpy change at room temperature ($\ce{298K}$).
Wikipedia

The bond-dissociation energy is sometime also called the
bond-dissociation enthalpy (or bond enthalpy), but these terms are not
strictly equivalent, as they refer to the above reaction enthalpy at
standard conditions, and differ from $D_0$ by about $\pu{1.5 kcal mol-1} \pu{(6
kJ mol-1)}$ in the case of a bond to hydrogen in a large organic molecule.

UC Davis ChemWiki

Officially, the IUPAC definition of bond dissociation energy refers to
the energy change that occurs at $\pu{0 K}$, and the symbol is $D_0$. However,
it is commonly referred to as BDE, the bond dissociation energy, and
it is generally used, albeit imprecisely, interchangeably with the
bond dissociation enthalpy, which generally refers to the enthalpy
change at room temperature ($\pu{298K}$). Although there are technically
differences between BDEs at $\pu{0 K}$ and $\pu{298 K}$, those difference are not
large and generally do not affect interpretations of chemical
processes.

If we follow this definitions, then yeah, at standard conditions $\ce{HBr}$ is liquid so we have to add up $\pu{12 kJ mol-1}$ for its enthalpy of vaporization, and (almost) get the desired $\pu{364 kJ mol-1}$. The small discrepancy is probably due to slightly wrong value of enthalpy of vaporization which I just quickly found somewhere.


Mechanism for the Cleavage of Diborane? The cleavage of diborane was presented in a recent lecture, and it was said that the borane could be cleaved unsymmetrically if sterics allowed, or symmetrically if they did not, as shown in the image below:

However, no mechanism for this cleavage was provided, nor can it be found in Housecroft and Sharpe's "Inorganic Chemistry", our course textbook, nor can I find much about it online.

I have postulated such a mechanism:

Is it known how this cleavage occurs, and if it does indeed follow this mechanism? Is my attempt a reasonable proposition?

Is it known how this cleavage occurs?

The mechanism proposed for the cleavage reaction involves an initial attack by the donor on one boron atom in diborane, leading to cleavage of one $\ce{B-H-B}$ bridge.

This is followed by the attack of a second donor molecule, cleaving the remaining $\ce{B-H-B}$ bridge. If the boron atom in the pendant $\ce{BH3}$ group is attacked, two moles of adduct result, whereas attack on the atom already carrying a donor produces cation and borohydride:

Support for this view comes from the fact that the intermediate ($\text{I}$) is detectable by tensimetric titration of amine-borane with diborane and by $\ce{^{11}_{}B}$ nmr.

Source: Boron Hydride Chemistry, Earl Muetterties


Why does water weaken ion ion attractions? My lecturer has told me that water, having a high dielectric value, will "shield" ions and reduce ion-ion attractions.

I really don't see why. Having water molecules surround these two ions doesn't mean that Coloumb's force doesn't exist anymore - nothing in $F=\frac{kQq}{r^2}$ has changed just because water molecules are surrounding them. I recognise that the water molecules will arrange to face the ions in a polarised way (oxygen to the $+$ ion and hydrogens to the $-$ ion). However, I don't see how having a shell of water surrounding you will reduce the attraction. I guess it would prevent them from coming closer due to the sheer physical size of the water shell, but wouldn't there still be attraction?
Can someone explain to me why the ion-ion attraction decreases when water is there?

Electrostatic field caused by ions or any other external source causes partial orientation of molecular electric dipoles along the field vector. Their orientation causes displacements of center of positive and negative charges. This displacement causes induced electrostatic field, that decreases the primary field intensity by the value of relative permittivity.
By other words, you replace the vacuum permittivity in the Coulomb law by the water permittivity.
$$F = \frac {q_1 \cdot q_2}{4 \cdot \pi \cdot \epsilon  \cdot r^2}$$


Alkene reactions with sodium tert-butoxide and CHBr₃ In a problem they gave me some reagents and they want me to figure out the end product.

The reagents given are:

2-butene reacting with sodium tert-butoxide and $\ce{CHBr3}$.

Now, what I was thinking was that the oxygen would pull the hydrogen from $\ce{CHBr3}$ making a positive charge on the carbon. (I'm not quite sure if the positive charge would be strong since the net dipole moment on $\ce{CHBr3}$ would be zero). However, from there, I immediately thought of a carbene reaction, yet, it has 3 bonds with $\ce{Br}$.

So, am I on the right track, and what are the next steps from here? Sorry if I didn't give you much to work with.

Strong bases react with chloroform $(\ce{CHCl3})$ and bromoform $(\ce{CHBr3})$ to form dihalocarbenes by alpha elimination.

The net reaction is as follows:

$$\ce{RO- + CHCl3 -> ROH + Cl- + CCl2}$$

The base removes the hydrogen atom from chloroform or bromoform to form an anion. Remember, when you deprotonate something, the electrons stay behind.

$$\ce{RO- + Cl3CH <<=> ROH + Cl3C:^{-}}$$

The trihalomethyl anion is unstable and fragments to lose a halide ion and generate the carbene.

$$\ce{Cl3C:^{-} <=>> Cl- + Cl2C:}$$

The mechanism using arrow-pushing formalism can be found at this website from the University of Liverpool.


Can Fe(CO)5 adopt both square pyramidal and trigonal bipyramidal geometries? I know the hybridization of $\ce{Fe(CO)5}$ is $\mathrm{dsp^3}$. According to my book, coordination compounds with coordination number 5 can interchange between square pyramidal and trigonal bipyramidal geometries. Is it true for this compound? Doesn't seem so according to the picture:

Was my understanding wrong somewhere?

Your book was correct that a five coordinate metal complex is able to adopt both square pyramidal and trigonal bipyramidal geometries, and in both cases the sp3d hybridisation scheme applies (if you believe in hybridisation...).

Which geometry is adopted depends upon a combination of steric and electronic factors, and isn't necessarily trivial to predict, though MO theory and crystal field theory can help understand which might be more favourable.

The structure of iron pentacarbonyl is widely agreed to be trigonal bipyramidal, with three CO ligands occupying equatorial positions, and two CO ligands occupying axial positions.

  Trigonal bipyramidal structure of iron pentacarbonyl

Whilst both geometries are possible (and indeed observed, IF5 is square pyramidal), I suspect that your confusion arises from the fact that iron pentacarbonyl is able to undergo a phenomenon known as Berry psuedorotation.

  Mechanism of Berry psuedorotation

Berry psuedoroation is the process in which two of the equatorial ligands are switched out for the two axial ligands via a square planar intermediate. This process is sufficiently rapid that if observed by NMR, only one environment for the carbon of the CO ligand is observed (i.e. the psuedorotation is faster than the NMR timescale).


Is cis- or trans-cycloundecene more stable? While reading about cis/trans isomerism in cycloalkenes, I came across the following statement:

When the number of carbon atoms in a cycloalkane is less than 11, the cis form is more stable than the trans form. When it is greater than 11, the trans form is more stable than the cis form.

An obvious question came to my mind: What if the number of carbon atoms in the ring is "equal" to 11? Which form will be more stable - cis or trans? I think both will be of the same stability, but would like to have some clarification in this regard.

I searched my books and didn't find any specifications on the stability of cis or trans forms, for a ring containing exactly 11 carbon atoms. I read the answers to the question Stability of cis vs trans isomers? but they discuss only the cases "less than 11" and "greater than 11" but not "equal to 11".

There is experimental data for the methoxy derivative: https://webbook.nist.gov/cgi/cbook.cgi?ID=C78289131&Units=SI&Mask=8

I think that means the (Z) form is favored. I would think the situation is similar in the absence of the methoxy group.


How to prepare iron(II) carbonate from iron(III) chloride? Now, this is an awkward reaction first of all. I know how to prepare $\ce{Fe2(CO3)3}$ from $\ce{FeCl3}$:

$$\ce{2FeCl3 + 3Na2CO3 -> 6NaCl + Fe2(CO3)3}$$

But the question is asking for $\ce{FeCO3}$. It seems to be an impossible task but since this was asked in a public examination, does anybody know how to prepare $\ce{FeCO3}$ from $\ce{FeCl3}$ (in one, two or three reactions)?


How long can polyethylene molecules be? In this video, Steve Spangler says the substance he is using (a polyethylene oxide): http://www.youtube.com/watch?v=m-88M75_PCI

He mentions that one molecule would go to the moon and back twice! I have a hard time believing this, but I guess it must be true.. Is it true that polyethylene molecules can be that long?

Short answer? Possibly, but these ones aren't.

So, the other thing mentioned in the video is that the specific material he's using has a molecular weight of approximately four million. If we assume it's ordinary poly-ethylene oxide, we can get the rough number of $\ce{(CH2)2O}$ subunits by just dividing by the molecular weight of the subunit:

$(12 \times 2) + (1 \times 2) + 16 = 42$
$\frac{4000000}{42} \approx 95000$

So, if we assume that each $\ce{C-O-C}$ back-bone unit is stretched out like a di-methyl ether molecule, a quick check on a fairly standard Ghemical force field gives me a subunit length of approximately 3.7 Å.

$3.7 \times 10^{-10} \times 95000 = 0.00003515 \text{m}$

So, these molecules should be about 0.04 millimetres long. Not exactly astronomical distances, but impressively long for a single molecule.

In theory, I don't think there's anything stopping you making a molecule as long as they claim, but these ones certainly aren't. I think also that the longest (synthetic) polymers in common use are of similar molecular weights and thus similar lengths: ultra-high molecular weight polyethene seems to top out around 4.5 million for commercial applications, from a quick search.

Biological molecules frequently get longer than this, though, especially DNA: the average human chromosome apparently contains a single DNA molecule approximately 5 centimetres long.

The man explaining in the video may have erroneously given the figure for the total length of all the molecules in that beaker.


A "Bible" for molecular orbital theory? Linus Pauling wrote the famous "Nature of the Chemical Bond" which is a Bible of sorts for valence bond theory.

However, VB theory was combined with crystal field theory to form molecular orbital theory, which eventually was condensed into the Hartree–Fock method.

Are there any "Bible" type reference books for these other methods?

I think an excellent book for understanding molecular orbital theory from a qualitative basis is Orbital Interactions in Chemistry by Albright, Burdett, and Whangbo. It covers a wide range of topics from molecules to solids.

The book doesn't go into particular methods, but instead illustrates how various MO treatments can understand a great deal of chemistry.


Finding molarity of hyposulfite solution used in titration $\pu{0.1 g}$ of $\ce{KIO3}$ is treated with excess $\ce{KI}$ solution. The liberated iodine required $\pu{44 ml}$ of $x$ molar hypo solution. Calculate the value of $x$.

I have used the equivalent concept and got the answer as $\pu{0.026 M}$.
But the correct answer to this question is $\pu{0.063 M}$.
Can you tell me where I have gone wrong in my solution?

$$
\begin{array}{cc}
\ce{&KIO3 &+ KI -> &I2}\\
&\frac{0.5}{214}~\mathrm{meq} & &\frac{0.5}{214}~\mathrm{meq}
\end{array}
$$

$$
\begin{array}{cc}
\ce{&I2 &+ &2 Na2S2O3 &-> 2 NaI + Na2S4O6}\\
&\frac{0.5}{214}~\mathrm{meq} & &\frac{0.5}{214}~\mathrm{meq}
\end{array}
$$

$$∴\frac{0.5}{214} = 44\times10^{-3}\times x\times 2 \to n\text{-factor of hypo}$$

$$∴x = \pu{0.026 M}$$

Your method in using the equivalent concept is correct, but the miscorrect in your solution is in determining the n-factor.

Refer to the above equations and the calculation N-factorPoutnik
N-factor Is the change of atom oxidation state multiplied by the number of atoms changing this state.
Because the iodine is just an intermediate product to be titrated by $\ce{Na2S2O3}$, the oxidation number change is by  $6$.Therefore, n-factor of  $\ce{KIO3}$ is $6$.So, the equivalent mass is M/6.And , n-factor of  $\ce{Na2S2O3}$ is $1$.So, the equivalent mass of $\ce{Na2S2O3}$ is equal to its molar mass.
$$
\begin{array}{cc}
\ce{&KIO3 &+ 5KI + 3H2SO4 -> &3I2 +&3K2SO4 + 3H2O}\\
&\frac{0.6}{214}~\mathrm{meq} & &\frac{0.6}{214}~\mathrm{meq}
\end{array}
$$
$$
\begin{array}{cc}
\ce{&I2 &+ &2 Na2S2O3 &-> 2 NaI + Na2S4O6}\\
&\frac{0.6}{214}~\mathrm{meq} & &\frac{0.6}{214}~\mathrm{meq}
\end{array}
$$
$$∴\frac{0.6}{214} = 44\times10^{-3}\times x\times 1 \to n\text{-factor of hypo}$$
$$∴x \approx{\pu{0.064 M}}$$


How to calculate the molecular weight for a volatile substance introduced into a Dumas bulb? In a Dumas bulb a volatile substance is introduced. After a few minutes when the liquid has evaporated, the bulb is sealed. It is known that the initial weight of the bulb with air is $12.0468~\mathrm{g}$, $12.4528~\mathrm{g}$ with the volatile substance and $350.6264~\mathrm{g}$ with water. Calculate the molar mass of the substance if all these measurements were done at $25~^\circ\mathrm{C}$ and $1~\mathrm{atm}$ of pressure.

The correct answer is $29~\mathrm{g/mol}$.

What I've done is the following.

I've constructed a system of two equations to know the mass of the bulb and the volume it can contain.

$m_b+\rho_\mathrm{air}V = 12.0468~\mathrm{g}$
$m_b+\rho_\mathrm{water}V = 350.6264~\mathrm{g}$

The solutions are $m_b=11.6144~\mathrm{g}$ and $V=0.339012~\mathrm{m^3}$

With this I can find the molar mass of the volatile substance by knowing that its true mass was $m=0.8384~\mathrm{g}$ and the volume it occupied was $0.339~\mathrm{m^3}$ and using the ideal gas equation.

The ideal gas law, can be used to say

$M=\frac{m}{pV}RT=0.060~\mathrm{g/mol}$

Which is wrong. I've done all sorts of other things and can't get $29~\mathrm{g/mol}$. Maybe this is the craziest thing I've done until now.

There are a couple of things wrong.

First, the mass of the volatile gas is not $0.8384\ \mathrm g$. It's the difference between $12.4528\ \mathrm g$ and $12.0468\ \mathrm g$.

Second, the volume is incorrect. $11.6144\ \mathrm g + 999.97\ \mathrm{kg/m^3} \times 0.339012\ \mathrm{m^3} \ne 350.6264\ \mathrm g$


When it is said 'at constant pressure, at constant pressure ', is it about pressure & temp. of the system or surroundings? These are the following statements:

  [...] the thermal decomposition of $1.0$ mole $\ce {CaCO3}(\text{s})$ at $1$ bar results in an increase of volume of nearly $90~\text{dm}^3$ at $800^\circ \text{C}$ on account of the carbon dioxide produced.

Is the system at $1$ bar or the surrounding exerts pressure of $1$ bar? Is $800^\circ \text{C}$ the temperature of the system or the surroundings?

  The standard state of a pure substance is the pure substance exactly at $1$ bar & at $298.15~\text K$.

I think the pressure mentioned is of surroundings but the temperature? Is it of the system or he surroundings?

  We see that at constant temperature & pressure, the change in Gibbs energy of a system is proportional to the overall change in entropy of the system plus its surroundings.

Now is the system at constant temperature & pressure or the surroundings?

Could anyone please explain me to clear this ambiguity?

When we say

$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$

we are referring to $\Delta G^\circ$, $\Delta H^\circ$ and $\Delta S^\circ$ of the system. As such, the variable $T$ in the equation refers to the temperature of the system. The condition for constant pressure $p$ is implicit; the equation above does not directly show why $p$ has to be constant. For more information, refer to this question and answer.

The sentences that you have given are not the best examples to explain this, because they do not relate to Gibbs free energy. I will give you a different example:

  When hydrochloric acid is added to sodium carbonate at atmospheric pressure, the standard enthalpy change $\Delta H^\circ$ and the standard entropy change $\Delta S^\circ$ are measured to be $x~\mathrm{kJ~mol^{-1}}$ and $y~\mathrm{J~K^{-1}~mol^{-1}}$ respectively. Calculate the Gibbs free energy of the reaction and hence determine the equilibrium constant. (The temperature at which the reaction is carried out is $25~\mathrm{^\circ C}$.)

The wording of the question may seem to imply that it is talking about the temperature and pressure of the surroundings, and that is probably not a coincidence. For a simple chemical reaction like the one above, we would probably carry it out in a beaker. We weigh out a certain mass of $\ce{Na2CO3}$, and add a certain volume of $\ce{HCl}$, and pour the acid into the beaker and watch the bubbles.

Under such conditions, we can speak of the system being at mechanical and thermal equilibrium. Without going into too much detail about reversible and irreversible processes (I know there are loopholes in the explanation):

Mechanical equilibrium, in general, means that the net force acting on the system is zero. Correspondingly, this must mean that the pressure of the system and the pressure of the surroundings are equal at all points in time. If, at any point in time, this was not the case, e.g. if we had $p_\mathrm{syst} < p_\mathrm{surr}$, then the surroundings would "push" on the system and cause the system to contract until it reaches the point where $p_\mathrm{syst} = p_\mathrm{surr}$.
Thermal equilibrium means that there is no heat transfer into, or out of, the system. Analogously to above, this must mean that the temperature of the system and the temperature of the surroundings are equal at all points in time. We conventionally assume the surroundings to be a "reservoir" which maintains a constant temperature, and this is pretty true as far as our simple example is concerned. Therefore $T_\mathrm{syst} = T_\mathrm{surr}$.


Does using stonger soaps help me lose weight? Soaps are amphiphilic molecules that, practically, removes body oils from the surface of the skin. The body, on the other hand, always replenishes sebum through the sebaceous glands after a certain amount of time.

Applying Le Chatelier's principle to this complex chemical equilibrium, where the equilibrium shifts towards the right hand side, where oil is being extracted from the equation, essentially so:

(body · oil)  ⇌  body + oil       # RHS preferred

Because the process of sebum regeneration must be endothermic, this means the body is losing energy in this overall process:

body + (lipids + heat)  ⇌  body + oil  ⇌  (body · oil)

My questions are:

Are my assumptions correct?
Does removing oil from the body more readily, by means of a stronger detergent/soap, reduce body mass?
Does the temperature of the water bath used in the experiment affect the rate of sebum dissociation?

I'm afraid no: you are applying a chemical principle to a complex biological response.
The Le Chatelier principle state:

If a chemical system at equilibrium experiences a change in concentration, temperature, volume, or partial pressure, then the equilibrium shifts to counteract the imposed change and a new equilibrium is established.

We start with the definition of chemical system at equilibrium :

the temperature dependent condition in which chemical species are present at concentrations which have no further tendency to change with time.

So what you need are chemical species: molecules, ions. Although fatty acids could be a product "body" is not a compound or a molecule. So what you have is not a chemical system but a biological one. You could go a little more close if you think something like this:
$$\ce{Fatty acid <=> Chemical  \space Precursor_{Y} + Chemical \space Percursor_{X}}$$
However even this is a wrong oversimplification of many complex biochemistry reactions and equilibria.
Moreover you can't interact with these equilibria, perturbing them and so applying LC principle because sebum, which is a complex mixture of lipids, is stored inside sebaceous glands and is formed by the complete disintegration of the cell it self. So to from new sebum you have to from new sebaceous glands.
Soap simply remove sebum not seabaceous glands, so using stronger soap by the way don't affect sebum production. At this point is clear that is no more chemistry but physiology. In this case an equivalent of Le Chatelier's principle is the homeostasis. However from what I know the diet is one of the main factors that induce sebaceous glands to produce sebum so eat better seems to be the best way to lose weight and sebum!

References
Pub Med Sebaceous gland lipids


What defines the specific enthalpy of atomization? 1 mol of $\ce{F2}$ is turned into $\ce{2F}$ (atoms).

Is this the specific enthalpy of atomization, or two times that?

ChemGuide says that 'The standard atomisation enthalpy is the enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state.'

For example,
\begin{array}{rc}
\ce{1/2 Cl2 (g) -> Cl (g)} & \Delta H_{a}^{0} = +122~\mathrm{kJ/mol} \\
\ce{1/2 Br2 (l) -> Br (g)} & \Delta H_{a}^{0} = +112~\mathrm{kJ/mol} \\
\ce{Na (s) -> Na (g)} & \Delta H_{a}^{0} = +107~\mathrm{kJ/mol}
\end{array}


What is the number of lone pairs of central atom when it has an odd number of electrons? What does it mean in asking the number of lone pairs of central atom when it has an odd number of electrons?
As lone pairs means the no. of non bonding pairs of electrons does it exclude the one odd electron if present in the central atom?

This seems to be a tricky bit of chemical terminology that may not have a lot of practical relevance, so I wouldn't get too hung up on this, if I were you. Technically, I would say that you are correct in that lone pairs should be pairs of electrons, as this is how the IUPAC defines them.


Does the hydrogen ion actually exist? A chemistry test I did was an IGCSE test. One of the questions was

  Is Hydrogen Chloride ionic or covalent?

Now, on a sheet I have been given, it states the hydrogen ion is $\ce{H+}$. So, it forms a positive ion, like metals. I'm fine with that. I know that hydrogen is strange and I also know that the (I)GCSE chemistry curriculum simplifies/lies a lot. The bonding in Hydrogen Chloride is covalent. Now, since a 'Hydrogen ion' is just a proton, I wanted to know whether in reality it actually exists or not.

Yes free $\ce{H+}$ ions, protons, really exist.

Protons are constantly emanating from the sun and reaching Earth.

The proton flux is continuously monitored by satellite.

However, in a solution such as water, instead of bare $\ce{H+}$ ions, they are $\ce{H3O+}$ or larger ions such as $\ce{H5O2+}$ or $\ce{H9O4+}$.

When $\ce{HCl}$ dissolves, the proton is transferred to a water molecule to form $\ce{H3O+}$.

$\ce{HCl}$ only has 17% ionic character according to Pauling's The Nature of the Chemical Bond, so it is covalent, but reacts with water to form separate ions in solution.

For more information see Eigen or Zundel Ion: News from Calculated and
Experimental Photoelectron Spectroscopy:

  The appearance
  of the hydronium ion was and is
  currently debated as several possible
  structures, three of them being $\ce{H3O+}$
  (the Eigen cation), $\ce{H5O2+}$ (the Zundel
  cation) and $\ce{H9O4+}$ (the strongly solvated
  Eigen cation)

and

Both Zundel and Eigen Isomers Contribute to the IR Spectrum of the Gas-Phase $\ce{H9O4+}$ Cluster:

  The “Eigen cation”, $\ce{H3O+(H2O)3}$, is the most prevalent protonated water structure in the liquid phase


Can a first degree carbocation form if a methyl shift is possible? I was going through a reference book when I stumbled upon this question

Is that even possible?

The cation is produced when a methyl group migrate from C-2 to C-1 as the C-Br bond is broken. The simultaneous migration of alkyl group and departure of leaving group to form a tertiary cation  is therefore faster and more favourable than formation of primary cation due to simple loss of leaving group.

The products observed are those that result from further reaction of the so formed tertiary cation. The mechanism mentioned in the pic you've attached is incorrect, 1° carbocation intermediate is not formed at all.

Several comments correctly mention that formation of primary carbocation is not favourable. So, it is important to note that methyl shift and departure of leaving group in this case are simultaneous processes - primary carbocation is not formed at all!

P.S.
Also, I'd like to quote some evidence I found for a very similar reaction of 1-bromo 2,2-dimethyl propane:

Above is the solvolysis reaction of 1-bromo 2,2-dimethyl propane.

Source: Reactive Intermediates in Organic Chemistry: Structure, Mechanism, and Reactions - By Maya Shankar Singh (Wiley Publishers)


Torquoselectivity: Disrotatory ring opening In organic chemistry lecture we discussed the beginning of the Woodward Hoffmann Rules with a key step of the vitamin B12 synthesis.

(The initial paper from their work can be found here:
http://pubs.acs.org/doi/pdf/10.1021/ja01080a054)

Let's first see the forward reaction (the electrocyclic ring closure) as Woodward and Hoffmann found (which goes by disrotatory cyclization; 6 electrons, thermal).

I'm interested in the reverse reaction, the ring opening (6 electrons, and therefore disrotatory:

I'm thinking now about the torquoselectivity of the ring-opening  (because in an exercise we had an example with 1-Ethyl-1-ethoxy(EDG)-cyclobut-2-ene).

Is there a rule how to see why the E (E: Electron withdrawing group) is in trans position when I start with the product?

EDIT:

I'm thinking if they are some stereoelectronic effects here which explain the obtained selectivity:

See also: http://pubs.acs.org/doi/abs/10.1021/jo00244a058
(Electronic control of the stereoselectivities of electrocyclic reactions of cyclobutenes against incredible steric odds)

Well here we have a conrotatory ring-opening. I don't know if they are still valid in our 6e case for disrotatory ring-opening of the reaction from W. and H..

No, there is no general orbital symmetry rule for predicting which of the two possible disrotatory motions will be favored.  However, by examining steric effects in the two possible transition states we can often make a good guess as to which product will predominate.  There are always a few "special cases" like the one you referenced in your post, where with certain substituents under certain conditions certain things will happen.  But to my knowledge there is no general rule for this situation.

In the following diagram there are 3 thermal reactions controlled by orbital symmetry (6 pi electrons, therefore disrotatory motion).

The top line shows the symmetry allowed disrotatory closure of trans, trans-1,6-dimethylhexatriene.  There are two possible disrotatory motions - either the tops or the bottoms of the p-orbitals on carbons 1 and 6 can move towards each other in a disrotatory fashion.  One motion produces the dimethylcyclohexadiene with the two methyl groups above the plane of the screen, the other motion produces the mirror image molecule with the methyl groups below the plane of the screen.

Similarly, cis, cis-1,6-dimethylhexatriene can undergo two disrotatory motions to produce the same two products (Note: the cis, cis isomer will react much slower and probably in lower yield; steric interactions between the methyl groups will disfavor the s-trans conformation about the two single bonds that is required for the ring closure to be able to occur).

The bottom line shows the reverse reaction, thermally converting a cyclohexadiene to a hexatriene.  As shown, two possible isomers can be formed.  The ratio of these two isomers will depend on steric interactions that develop during the two possible disrotatory motions.  In this case, we will primarily obtain the trans, trans isomer; the cis, cis isomer has steric problems in its transition state from the two methyl groups bumping into each other (build a model).

In this last reaction, if we replaced the methyl groups with deuterium, we would obtain a roughly equal mixture of the two possible products, cis, cis- and trans, trans-1,6-dideuteriohexatriene.  So again, there is no orbital symmetry "rule" for predicting whether a group would wind up cis or trans on the double bond.  But if we examine the steric effects in the two possible disrotatory transition states, we can often make a good guess about which isomer should predominate.


Isopropyl alcohol still leaves evaporation residue after distillation I just finished distilling two liters of IPA because evaporating 5 ml on a watch glass revealed it to contain quite a bit of dissolved impurities. However, it still leaves some residue, even though I used a 200 mm vigreux column.

All glassware was thoroughly cleaned beforehand. I do not get such issues with distilled acetone, for example.

What gives?

I assume the problem is, like in the watch glass the solutes after IPA evaporates, i.e. higher boiling point solutes.

Making the distillation slower and reduce the amount of aerosol to escape to distillate flask will surely diminish the amount of impurities. More time, less temperature elevation, it should solve the problem.


Decomposition of LiNO3 I have a doubt about decomposition of alkali metals.
Why does $\ce{LiNO3}$ decompose to give $\ce{Li2O}$ and $\ce{NO2}$ where as other alkali metals decompose to give corresponding nitrite and oxygen?

Have you checked Wikipedia?

  Upon thermal decomposition, $\ce{LiNO3}$ gives lithium oxide ($\ce{Li2O}$),
  nitrogen dioxide, and oxygen:

  $$\ce{4 LiNO3 → 2 Li2O + 4 NO2 + O2}$$

  Other group I nitrates decompose differently, forming the nitrite salt and oxygen. Because of its relatively small size, the lithium
  cation is very polarizing, which favors the formation of the oxide.


Nitrogen molecule vs carbon-carbon triple bond stability Why does the triple bond in the nitrogen molecule convey stability, whereas a triple bond between carbon atoms conveys instability?

You might already know that elements like to follow the octet rule, which means that atoms tend to combine in a way that each atom can have eight valence electrons. Nitrogen can exist diatomically because through the triple bonds, each nitrogen will have one lone pair and three shared pairs. As seen in the picture, each nitrogen satisfies the octet rule.

Carbon, on the other hand, cannot satisfy the octet rule diatomically. Since carbon has one less valence electron than nitrogen, a triple bond would not work because the carbon does not have enough electrons to bond and satisfy the octet rule. It would leave each carbon atom with three bonded pairs of electrons and only one lone electron, giving each atom a configuration of only seven electrons per atom, not eight. Sorry I don't have a picture of a triple carbon bond. Hope this answers your question though!


Why is potassium monoxide (K2O) coloured? Potassium monoxide is rarely formed and is a pale yellow solid. I don't understand how is this compound coloured because in oxide anion electron cannot excite in the next orbital due to huge energy difference. Also there are no empty molecular orbitals like there are in peroxide and superoxide anion. The colour cannot come from $\ce{K+}$ cation because electron cannot be excited. So how is this compound coloured?

Probably as good an explanation as any can be found here.

https://uk.answers.yahoo.com/question/index?qid=20090729062525AA0gZlS

Basically, the electronic structure of an ionic compound like potassium oxide has "valence bands", which are based primarily on the anions (oxide ions), and empty "conduction bands" based on the cations  (potassium ions).  Typically, to excite electrons from the valence band to the conduction band you need to absorb ultraviolet photons and thus you get no color.  But potassium oxide has just a small enough energy gap between valence and conduction bands to allow some electrons to be excited with violet light.  That absorption gives the yellow color.

If we go further down the alkali metal oxides the band gap gets smaller, and more visible light with longer blue/green wavelengths may be absorbed.  Then the color becomes stronger and more reddish.


Why does having identical groups on either side of the pi bond lead to no geometrical isomerism? As far as I was taught, cis and trans isomers were categorised under stereoisomers and sub category of geometric isomers.  Qualifying properties like, rigid and restricted double bond rotation (due to pi bond electron cloud), no terminal alkenes (those with double bond at the end of carbon chain).
But, here was a quote from a LibreTexts article:

If one carbon of the double bond has two identical groups such as 2 H's or 2 Cl's or 2 CH3 etc. there cannot be any geometric isomers.

What is the basis of above claim? Why cis and trans with identical groups can't be qualified as a geometric isomer? If not what is this? Am I missing something here?

Hmm, you seem to be having a problem visualizing the structures in the case of 2 identical groups on one carbon. But never fear, 3D molecules are here.
Check out chloroethene:

It has two identical groups (hydrogens) on the left side. If you tried to make a geometrical isomer for this, you'd get:

But hey! That's just the previous image flipped upside down! That's not an isomer, it's the same thing!
Indeed it's the same molecule. These molecules can't have a geometrical isomer because it possess an plane of symmetry:

So there you go. Go ahead and visit this link if you want an interactive 3D model that you can spin around.

Added based on comments from @Zhe, there's a plane of symmetry in following angle,


What will happen to bottle of cola after 50 years? I'm going to make a time capsule for 25 or 50 years, and I don't know wether it's safe to put food, e.g. bottle of cola there. Will it explode and ruin the capsule? Will it be ok to drink it, if it won't explode?
Also, what about Snickers or a hamburger in a ziploc package with air sucked out of it with vacuum cleaner?

So: will Snickers, Hamburger in a ziploc, Bottle (or can) of Cola, all not opened, go crazy in 25 years? In 50 years? If so, can they go out of their packages and ruin the contents of the time capsule? If not, will it be safe to consume one of them?

I'm going to make a time capsule for 25 or 50 years, and I don't know whether it's safe to put food, e.g. bottle of cola there.

Depends on the food and the way it's packed, but food is not something I'd suggest you pack for such a long period..

  Will it explode and ruin the capsule ?

Food or drink are very unlikely to explode (maybe alcohol would in the wrong circumstances).

But food can be unsafe to eat in quite a short time, depending on the food, preparation and packing and how you store it.  Bacteria could be a problem.

Most food will go bad over a period as long as you're describing without very specific storage (like refrigeration, and even then I'd say most things won't work).

The safest ones would be :

popcorn (kernels)
honey (jar of)
sugar (most forms)
alcohol (spirits like vodka, whiskey)
dried beans, dried lentils

I would not be planning to eat any of these stored for 25 years myself.  And in general I'd suggest testing the items before trying them after 25 years or more (if you feel you must).

I would not expect cans or glass or plastic bottles of soda to be in good shape after anything like 25 years.  The plastic might not survive without degrading.  The can and plastic might react with the liquid over that timescale and the glass would survive but I'd be less optimistic about a sugar laced chemical soup like soda or cola not undergoing some changes.  Hard to say.

If you want more info on this try this website.

  Will it be ok to drink it, if it won't explode?

I would not try it.  At best it soda would be flat and possibly not taste the same (chemical changes over that timescale ?) and at worst it could actually be harmful.

Exploding seems very unlikely.

  Also, what about Snickers or a hamburger in a ziploc package with air sucked out of it with vacuum cleaner?

Air isn't the issue.  There are bacteria that will happily live (and increase in numbers) on what's in the food.  Well, it is food, after all.  There are bacteria that will survive refrigeration as well.  Over the timescale you're talking about I'd say it's all bets are off territory.

  So: will Snickers, Hamburger in a ziploc, Bottle (or can) of Cola, all not opened, go crazy in 25 years? In 50 years?

All of those could be dangerous over such a long time period, IMO.  At the very least they'd taste bad and at worst they'd kill you if you consumed them.

  If so, can they go out of their packages and ruin the contents of the time capsule? If not, will it be safe to consume one of them?

Depends on the packaging.  Glass would last indefinitely baring physical force or extreme of hot and cold (which might possibly cause fatigue cracking).  The other wrappers would last pretty well (structurally), but 25 years is way past their design intentions.  It would be a dice throw.


Wave and particle properties of protons and electrons Do both electrons and protons have wave and particle properties and if they do, do they show them at the same time?

Actually, everything has wave-particle properties and is given by the deBroglie equation: $${\mathrm p}\lambda = h$$ where $\mathrm{p}$ is momentum (mv), $\lambda$ is wavelength and $h$ is known as the Plank constant ($6.626\cdot 10^{-34}\;\mathrm{J\cdot s}$).

As you can see as the mass of an object increases, the wavelength decreases. So in macroscopic objects that we an see, such as ball, its mass (therefore momentum) is so big compared to $h$ that the wavelength is really, really small, so we can't observe it. For light waves, it is the opposite. Due to their relatively large wavelength, they have a really small mass. However we can do experiments that show the particle nature of light. Such as the photoelectric effect which can only be explained by thinking of light as a particle rather than a wave.

Electrons which have a mass of $9.011\cdot 10^{-31}\;\mathrm{kg}$ are in between the two examples that I have given you. It turns out that their mass is perfect for their wave and particle nature to be observed. Protons are roughly 1840 times heavier than electrons, however we are also able to observe their dual nature.

The answer your second question is that they do exhibit these wave and particle properties at the same time. An electron can't simply decide to be a particle for a certain instant than decide to become a wave in the next moment. All objects exhibit their wave-particle properties at the same time.


How to calculate the concentration of a monoprotic organic acid that was extracted into an ethyl ether layer? Compound A is a monoprotic organic acid. If $\pu{25.0 mL}$ of a $\pu{0.456 M}$ solution of A in water was extracted with $\pu{25.0 mL}$ of ethyl ether and it was determined experimentally that $\pu{5.45 mL}$ of $\pu{0.523 M}$ $\ce{NaOH}$ was required to neutralize the bottom layer from the extraction, what is the concentration of compound A in the ether layer?

My attempt:

$\pu{25.0 mL}$ of a $\pu{0.456M}$ solution of A in water

This gives the total millimoles of your unknown acid that is split between the two layers.

$\pu{5.45 mL}$ of $\pu{0.523 M}$ $\ce{NaOH}$ was required to neutralize the bottom layer

That gives the number of millimoles in the bottom layer.
So the number of millimoles in the ether layer must be the difference, and there are $\pu{25 mL}$ of ether.
I think I figured the concentration out: $0.523\times\frac{0.00545}{0.25} = \pu{0.114 M}$
Or do I need to determine the amount in the aqueous layer by subtracting this number? I.e. $0.456 - 0.114 = \pu{0.34 M}$
I am left with either the answer being $\pu{0.34 M}$ or $\pu{0.114 M}$.
I think the answer is $\pu{0.34 M}$, because the $\pu{0.114 M}$ applies to the bottom layer which is water and needs to be neutralized by the $\ce{NaOH}$ solution, therefore the top layer (ether) has the remaining $\pu{0.34 M}$ of compound A.

You arrive at the correct answer, but the way you get there is a bit messy. (And it only works because both phases have the same volume.) You should be a lot clearer in defining what properties you have been given in the question and which ones you need to calculate. Setting up the equation is a lot more important then calculating it in the end, because this does not only show that you know what you are doing, but it also is a way to condense a difficult sentence into a calculator understandable way.
\begin{align}
V(\mathrm{water})       &= \pu{25.0 mL}\\
c_\mathrm{tot}(\ce{AH}) &= \pu{0.456 mmol/mL}\\
V(\mathrm{ether})       &= \pu{25.0 mL}\\
V(\ce{NaOH})            &= \pu{5.45 mL}\\
c(\ce{NaOH})            &= \pu{0.523 mmol/mL}\\
\color{red}{c_\mathrm{ether}(\ce{AH})} &= ?
\end{align}
Then you must always carry the units with the quantity; they belong together. This way you will always have a way to tell if you have made a mistake setting up the equation. If the units don't work out, it cannot be correct.
Since it is a monoprotic acid the following will hold true:
\begin{align}
\ce{AH &<=>[H2O] A- + H+} &
&& n_\mathrm{tot}(\ce{AH}) &= c_\mathrm{tot}(\ce{AH}) \cdot V(\mathrm{water})
\end{align}
Now you extracted your solution and split the amount of substance between the water and the ether phase. You can calculate the remaining substance in the water phase from the neutralisation experiment.
\begin{align}
\ce{AH + NaOH &-> A- + Na+ +H2O} & n_\mathrm{water}(\ce{AH}) &= n(\ce{NaOH})\\
&& n_\mathrm{water}(\ce{AH}) &= c(\ce{NaOH}) \cdot V(\ce{NaOH})
\end{align}
The extracted amount of substance of the acid is simply the difference, from that you can easily calculate the concentration in the ether phase.
\begin{align}
c_\mathrm{ether}(\ce{AH}) &=
  \frac{n_\mathrm{ether}(\ce{AH})}{V(\mathrm{ether})}\\
c_\mathrm{ether}(\ce{AH}) &=
  \frac{n_\mathrm{tot}(\ce{AH}) - n_\mathrm{water}(\ce{AH})}
       {V(\mathrm{ether})}\\
c_\mathrm{ether}(\ce{AH}) &=
  \frac{c_\mathrm{tot}(\ce{AH}) \cdot V(\mathrm{water})
        - c(\ce{NaOH}) \cdot V(\ce{NaOH})}
       {V(\mathrm{ether})}\\
c_\mathrm{ether}(\ce{AH}) &=
  \frac{\pu{0.456 mmol/mL} \cdot \pu{25.0 mL}
        -  \pu{0.523 mmol/mL} \cdot \pu{5.45 mL}}
       {\pu{25.0 mL}}\\
c_\mathrm{ether}(\ce{AH}) &\approx \pu{0.342 mmol/mL}
\end{align}


How to find the moles of carbonic acid I need to add to make a solution of KOH acidic I wanted to know if it is possible to add enough carbonic acid to a $\ce{KOH}$ solution to make it an acidic solution.

The $\mathrm{pH}$ of my KOH solution was $13.2$. After adding $\pu{50 g}$ of carbon dioxide (and I do not know how much reacted), the $\mathrm{pH}$ decreased to $10.302$.

Is there a way to know how many moles of carbonic acid more to add to the $\ce{KOH}$ solution to make it acidic?

At first, I thought I could take the change in pH to calculate the additional moles of carbonic acid needed.

So I thought going from a $\mathrm{pH}$ of $10.302$ to $7$, I would calculate the moles of carbonic acid needed for a change in $\mathrm{pH}$ of $3.302$ $\mathrm{pH}$ ($10.302 - 7 = 3.302$). Thus, the calculation would be $10^{-3.302} = \pu{4.989 x 10^{-4} M}$ $\ce{H2CO3}$. There are 2 moles of $\ce{H+}$ in 1 mole of $\ce{H2CO3}$; thus, I need:

$4.989 \times 10^{-4}/2 = 2.49 \times 10^{-4}$ moles of $\ce{H2CO3}$

$\ce{H2CO3}$ is a $1$ to $1$ ratio with $\ce{CO2}$; thus, I need $2.49 \times 10^{-4}$ moles of carbonic acid in solution.

But I do not know if it is right to think it this way?

Can someone tell me if what I am doing is right, and if not, what I can do to theoretically calculate moles of carbonic acid needed to make the $\ce{KOH}$ solution as acidic as I could.

You have not stated the volume of your $\ce{KOH}$ solution. I therefore have to make some assumptions.

I will use $\mathrm{p}K_\mathrm{a1} (\ce{H2CO3}) = 6.37$. I further assume you have $\pu{1 L}$ of the $\ce{KOH}$ solution and that this volume will not change by the addition of the carbon dioxide.

For the following, we have also to assume that all the solid carbon dioxide will react, i.e. no evaporation of any gas -- tricky, but perhaps not impossible.

You start with a solution of KOH that has the $\mathrm{pH} = 13.2$. From this we calculate $\mathrm{pOH}$:

$$\mathrm{pOH} = 0.8 \quad \to \quad [\ce{OH-}] = \pu{0.16 M}~(\text{or}~\pu{0.1584 M})$$

First, we suppose that the reaction between the solid carbon dioxide
and the solution of KOH will result in only potassium hydrocarbonate:

$$\ce{CO2 + OH- -> HCO3-}$$

OK, how much carbon dioxide is needed to transform all hydroxide ions into potassium hydrocarbonate? You will need (per liter $\ce{KOH}$): $\pu{0.16 mol}$.

After addition of $\pu{0.16 mol}$ carbon dioxide we will face the same situation as if you had dissolved $\pu{0.16 mol}$ of potassium hydrogencarbonate in $\pu{1 L}$ of water. The $\mathrm{pH}$ of such a solution will still be alkaline $(\mathrm{pH} > 8)$.

You wanted the solution to be acidic, but you did not specify the pH. I suppose you will be happy with $\mathrm{pH} = 6.37$.

Accordingly, we have to add more carbon dioxide. The charge balance is:

$$\ce{[H3O+] + [K+] = [OH-] + [HCO3-] + 2[CO3^2-]}$$

At $\mathrm{pH} = 6.37$, both $[\ce{OH-}]$ and $[\ce{CO3^2-}] << [\ce{HCO3-}]$.
Therefore, we re-write the charge balance as:

$$\ce{[H3O+] + [K+] = [HCO3-]}$$

We know the values of $[\ce{H3O+}]$ and $[\ce{K+}]$, so we calculate:

$$[\ce{HCO3-}] = 10^{-6.37} + 0.16 \approx \pu{0.16 M}$$

We can now calculate $[\ce{H2CO3}]$:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} \quad \to \quad [\ce{HCO3-}] = [\ce{H2CO3}] = \pu{0.16 M} \quad \to \quad [\ce{HCO3-}] + [\ce{H2CO3}] = \pu{0.32 M}$$

Accordingly, in total we need $\pu{0.32 mol} \approx \pu{0.3 mol}$ of carbon dioxide to reach $\mathrm{pH} = 6.37$. Per liter KOH solution this will correspond to $\pu{0.3 g} \times \pu{44 g mol-1} = \pu{13 g}$ solid carbon dioxide.


Phenomenon where fumes can relight an extinguished flame I lit a mosquito coil and used it to melt down small down candle wax which had fallen on the floor. The coil then produced some dense fumes. I directed the fumes to the flame of a candle. As soon the fumes reached the candle, my coil caught fire again and started burning. I mean the flame actually traveled the path of the fumes and re lit my mosquito coil. I just wanted to know what this phenomenon is called? Why does this happen?

It looks like you accidentally created a combustion mixture.  I checked the Material Safety Data Sheet (MSDS) for several mosquito coils, and in addition to the insecticide, one of them also contained the oxidizer potassium nitrate $\ce{(KNO3)}$.  The burning candle likely had some vaporized paraffin (fuel) in the flame, so when you directed the $\ce{KNO3}$ vapor into the flame, you had heat + fuel + oxidizer = combustion!  You literally set that vapor on fire.  I'd be careful with that: although the MSDS indicated a small amount of $\ce{KNO3}$ in the coil, you could still be injured if you were not prepared.


Why does an increase in alkyl group substitution cause downfield shift H'NMR? Since alkyl groups are electron donating, wouldn't an increase in alkyl group subsitution mean the protons would feel a lower magnetic field? Because they feel a lower magnetic field, wouldn't this cause an upfield shift? Is it just that alkyl groups are less electron donating than a hydrogen bound to the carbon?

The short answer to this question is that C is more electronegative than H (H 2.2, C 2.5), and therefore one would expect a trend in increasing chemical shift with increasing carbon substitution.

The longer answer....

As you point out, alkyl groups are electron donating, and you are on the right track in surmising that alkyl groups are less electron donating than protons. Remember that all shielding is relative. Looking at shielding effects at the α-substituent;

a bare proton nucleus (H+) would come at δ35,
H-H comes at δ7.4, and
H-CH3 comes at δ0.23.

A clear upfield shift due to the electron-donating alkyl group. So, we need to look deeper at the effect of β-substituents:

H-CH3 (δ0.23),
H-CH2CH3 (δ0.86),
H-CH(CH3)2 (δ1.33), and
H-C(CH3)3 (δ1.56).

This longer answer involves looking at the magnetic anisotropy around the observed nucleus, and questioning some age-old depictions of the C-C bond. Historically, the C-C magnetic anisotropy was depicted as below, with shielding at the side, and deshielding along the ends. Have a look in your NMR textbook, and you'll probably find this. Taken in isolation, this model conforms with the observed trend of increasing shift with increasing substitution with C.

However, this model contradicts somewhat the notion that alkyl groups are electron donating, and indeed what is observed with α-substitution. And with good reason. Recent studies by Baranac-Stojanovic (DOI: 10.1002/chem.201204267) show that this anistropy model should be reversed, and that there are shielding zones at the end of the C-C bond. Why then do we observe the increase in chemical shift with increased C substitution? Again, it is because the shielding anisotropy of the contributing C-C bond component is considerably less than contributions from the geminal C-H anisotropy contributions.


Why isn't ammonium strong acid, since it's charged and ammonia is neutral? Why is ammonium such a weak acid $(\mathrm pK_\mathrm a = 9.24)$, if $\ce{NH4+}$ has a positive charge on nitrogen, while $\ce{NH3}$ is neutral. It means that conjugate base (ammonia) is really stable, which makes conjugate acid a strong acid.

Another argument why ammonium should be strong acid is comparison with hydronium ion $\ce{H3O+}$. Both nitrogen (EN = 3.04) and oxygen (EN = 3.44) are significantly electronegative atoms and "don't like" to be positively charged. That explains pretty good hydronium ion $\mathrm pK_\mathrm a = -1.74$, but why is then ammonium $\mathrm pK_\mathrm a$ is just $9.24$?

As chemists we are given experimental observations, such as the difference in measured pKa between ammonium and hydronium ions, and we invent concepts such as "electronegativity" to explain this result in conjunction with many other similar results. If this concept is insufficient in your mind to explain the experimental data, you may be motivated to seek a more satisfying and self-consistent explanation. For me electronegativity is sufficient in this case. The fact that ammonia is "stable" and can be put in a bottle does not mean that it is somehow unstable when protonated to give ammonium ion. Many ammonium salts can be put in a bottle, even salts of weak acids like carbonic. Having a positive charge on an atom is not in itself an unstable state-note how stable salts like sodium chloride (with a full positive charge on sodium) are, as well as species like tetramethylammonium ion-- you can make the hydroxide and put it in a bottle.


Usefulness of the outdated Bohr model? Whilst the Bohr model is incomplete and incorrect, it had limited usage in predicting spectral lines. In the same way, could it possibly with limited accuracy, be used to predict the outcome of chemical interactions to within "an order of magnitude" for which it was so famously known to do for many physical phenomena?

I think the importance of Bohr model comes from the fact that it's the first atomic model that uses the hypothesis of quantization. More precisely Energy Quantization. It introduces the first quantum number $n$. It allows to interpret the atomic emission spectrum of hydrogen.

So, it is an interesting pedagogic tool to teach students how the atomic models had evolved to arrive to the modern atomic theory (based on quantum physics).

It's also a beautiful example to teach students how empirical science, like chemistry, evolved by making observation (emission spectrum of hydrogen), formulate hypotheses (atomic model based on the model of solar system but with electrostatic forces), testing the model (it works for interpreting the discrete emission spectrum), but it fails to interpret the behaviour of the atom under magnetic field (so, scientists thought about rejecting the model and used another one, the modern model of atoms).


Need an explanation for how they got this compound name out of this structure My professor posted this question as practice in naming compounds and provided the answer. I have stared at this for 30 minutes and cannot seem to figure out how he got this name out of it... Can someone please explain.

The names that are proposed in the question are in accordance with obsolete IUPAC recommendations:

2-ethyl-5,5-dimethyl-1-hexene (1979 recommendations)
2-ethyl-5,5-dimethylhex-1-ene (1993 recommendations)

The corresponding rule in the 1979 recommendations is Rule A-3.4:

  3.4 – Unsaturated branched acyclic hydrocarbons are named as derivatives of the unbranched hydrocarbons which contain the maximum number of double and triple bonds. If there are two or more chains competing for selection as the chain with the maximum number of unsaturated bonds, then the choice goes to (1) that one with the greatest number of carbon atoms; (2) the number of carbon atoms being equal, that one containing the maximum number of double bonds. In other respects, the same principles apply as for naming saturated branched acyclic hydrocarbons. The chain is so numbered as to give the lowest possible numbers to double and triple bonds in accordance with Rule A-3.3.

Therefore, the parent structure is the unbranched chain that contains the maximum number of double and triple bonds (here: one double bond). Since there is still a choice, the parent structure is the longest chain that contains the double bond, i.e. hexene. Numbering starts at the end of the chain that gives the lowest numbers to the double bond, i.e. 1-hexene. The 1-hexene is substituted using the usual principles of substitutive nomenclature, which yields 2-ethyl-5,5-dimethyl-1-hexene.

According to the 1993 recommendations, locants are placed immediately before the part of the name to which they relate. Therefore, the name 2-ethyl-5,5-dimethyl-1-hexene is changed to 2-ethyl-5,5-dimethylhex-1-ene.

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) incorporates a major change from earlier recommendations given in 1979 and 1993.

  In acyclic parent structures the order of seniority between unsaturation and length of chain given in earlier recommendations is reversed. Thus, the first criterion to be considered in choosing a preferred parent acyclic chain is the length of the chain; unsaturation is now the second.

Therefore, the parent structure of the compound that is given in the question is now the longest chain, i.e. heptane. The heptane is substituted using the usual principles of substitutive nomenclature; i.e. the prefixes (here: ‘methyl’ and ‘methylidene’) are considered together in a series of increasing numerical order. This yields the preferred IUPAC name 2,2-dimethyl-5-methylideneheptane rather than 6,6-dimethyl-3-methyleneheptane since the locant set ‘2,2,5’ is lower than ‘3,6,6’.


In zero order why we don't need the coefficient of reactant in integration of rate law? For a reaction: $\ce{aA->product}$

The rate law is $\text{Rate}=-\dfrac{\mathrm d[A]}{a\,\mathrm dt} = k$

Where:

$k$ is the rate constant
$[\ce{A}]$ is the concentration of the reactant $\ce{A}$ in the reaction
$a$ is the stoichiometric coefficient of $\ce{A}$

If we integrate the relation, we get:

$$[\ce{A}]-[\ce{A}]_0=-akt $$

However, my instructor wrote the relation on board as below;

$$[\ce{A}]-[\ce{A}]_0=-kt$$

Why is $a$ not needed in the integrated rate law?

You are right, for consistency the stoichiometric coefficient should be retained.
Keeping track of the stoichiometric coefficients is essential to convert between rates for different species.

If you want to compare the rate of formation of reagent A and product, you write

$$\text{Rate}=-\frac{\mathrm d[A]}{a\,\mathrm dt} = \frac{\mathrm d[product]}{\,\mathrm dt}= k$$

and so:

$$\ce{[A]-[A]_0=-a*k*t} $$

or

$$\ce{\frac{[A]-[A]_0}{a}=-k*t}$$

and

$$\ce{[product]-[product]_0=k*t}$$

However you would say that the reagent A is consumed with an effective rate constant $\ce{k' = a*k}$.


Why are the C-C bonds in benzene closer in length to double bonds than single bonds? The $\ce{C-C}$ bonds in benzene are $\mathrm{140~pm}$ long which puts them in between single bonds and double bonds (given at $\mathrm{147-154~pm}$ and $\ce{134-135~pm}$ respectively, sourced from here and here) as expected. However, since the two resonance structures for benzene contribute equally, leading to six identical bonds, one might expect that the bond lengths would be exactly intermediate between a single and a double bond, but in fact the bond lengths seem to be closer to a double bond than a single bond. Why is this the case? My teacher asked me this a while ago when I was preparing for university interviews and I have never thought of a satisfactory answer.

one might expect that the bond lengths would be exactly intermediate between a single and a double bond

They actually are. The relevant reference structure, that can provide us lengths of single and double bond for $sp^2$ carbons with not conjugation involved is cyclooctatetraene. The article on the X-ray structure of cycloocta-1,3,5,7-tetrene gives lengths 1.47 and 1.33 angstrom, with their average (1.40) being the exactly value of C-C bond in benzene you provided.


Chemical plating of glass with copper? It is commonly known how to plate glass with silver using silver nitrate, ammonia and glucose.

With about the same effort, is there a way to plate glass with copper?

According to Chemical Demonstrations: A Handbook for Teachers of Chemistry (Vol 3, p226), a similar effect can be obtained for copper using copper(II) acetate, ammonia, and hydrazine.

  The copper metal forms when hydrazine reduced and ammonia complex of
  copper(II) in basic solution. Hydrazine slowly reduces the complex to
  copper metal. When the surface of the test tube is very clean and wet,
  the copper metal will adhere to it, forming a copper-colored mirror.

  The overall reaction that occurs in this demonstration can be
  represented by the equation $\ce{2[Cu(NH3)4]2+_{(aq)} + N2H4_{(aq)} +
> 4OH-_{(aq)} -> 2Cu_{(s)} + N2_{(g)} + 4H2O_{(l)} + 8NH3_{(aq)}}$

Whether or not this can be described as "about the same effort" depends entirely upon the access you have to hydrazine, and to appropriate handling/safety equipment for working with the reagents. While hydrazine in solution is -- if I recall correctly, but I am not an MSDS -- considerably more stable than pure hydrazine, it's still toxic and corrosive (among other things).


"Limit test" vs. "Titration" in a pharmaceutical guide From the "Technical guide for the elaboration of monographs" by EDQM, edition 7, 2015, page 35:

  Where the introduction of tests for foreign anions in organic substances is considered then a single one, either for chlorides, sulfates or − less commonly − nitrates, will usually suffice even when several could theoretically be present. The test is then to be carried out on the most abundant anion. When a test for chlorides is considered, up to 1000 ppm (0.10 %), a limit test should be used rather than titration.

I don't understand this. Isn't titration one of the possible ways to perform a limit test (to check whether the chloride content is within the specified limits)?

Maybe my understanding of "limit test" is flawed?

From what I know, rules for making a monograph are not about giving an exhaustive list of all the analyses possible. Guides describe international agreements about what is recommended and sufficient to identify a substance and its impurities.

A titration is considerably more tedious to perform and validate (see ICH Guideline). If you use titration in a monograph, people referring to it will also have to use this method. Unless you also included a limit test but, if you did so, why bothering carrying out titration? Overall, you would waste your time and, most importantly, others' time.

So the sentence "limit test should be used..." should be read "we agreed that a limit test gives enough information in this context and we prefer it over titration".

HTH


Can we use silver nitrate to distinguish between chloride and carbonate salts? Initially, we are given a solution containing two salts, they can be either carbonate or chloride salts (we do not know their composition initially).
Generally, $\ce{AgNO3}$ is used as a confirmatory test of $\ce{Cl-}$ (a white precipitate is obtained on adding silver nitrate to solution containing the chloride salt). Therefore, can we use this method to directly identify chloride in the solution or do we need to remove the possible carbonate ions to confirm the presence of chloride before performing the test as mentioned above?
$$\ce{NaCl + AgNO3 -> NaNO3 + \underset{white}{AgCl} \downarrow}$$
$$\ce{Na2CO3 + 2AgNO3 -> 2NaNO3 + \underset{brownish white}{Ag2CO3} \downarrow}$$
(A similar question may arise in the identification of bromide and iodide ions but I have ignored it for now as they are rarely asked in exams. Still I would appreciate it if someone could give an answer for this as well.)
The question has been asked mainly because silver nitrate test is generally taught to be performed without mentioning above precautions and it is also a very famous confirmative test. So answer is expected more (not only) from a laboratory technician than theorist.

The test for Chloride ion is silver nitrite in an Acidified solution [with nitric acid]. This removes carbonate (as carbonic acid), cyanide[careful] and sulfide (as hydrogen sulfide). Sulfate must be absent, if present it can be removed with barium nitrate (as barium sulfate).  The silver chloride formed is solubilized by addition of ammonia; insoluble precipitate remaining may be bromide or iodide. The exact procedure to follow is given in reputed indepth Qualitative Analysis lab books.


More than one way of balancing a chemical equation My textbook had this question on balancing chemical reactions

  Ozone reacts with nitric oxide to give nitrogen dioxide and oxygen gas

Here's how I balanced it: $$\ce{O3 + NO -> NO2 + O2}$$

However I realised that the equation is also balanced if I write it as: $$\ce{2O3 + 4NO -> 4NO2 + O2}$$

and when it is written as: $$\ce{3O3 + NO -> NO2 + 4O2}$$

This seems wrong as I am sure that there cannot be more than one way of balancing a reaction but I do not have a satisfactory explanation. What is going wrong here?

This happens when you have a sum of two (or more) independent reactions. You can balance each one, and then add them together in an arbitrary proportion. People usually run into this when trying to come up with an equation for combustion of gunpowder, where the oxidation of carbon and oxidation of sulfur are pretty much independent.

Now, in your case the reactions are:
$$\ce{2O3 -> 3O2}\tag A$$
and
$$\ce{2NO + O2 -> 2NO2}\tag B$$
Your first reaction is a combination of ${1\over2}A+{1\over2}B$, the second is $A+2B$, and the third is ${3\over2}A+{1\over2}B$. There are infinitely many possible combinations.

Ozone, being an extremely powerful oxidant, is known to participate in certain reactions where it spends one atom of oxygen and the rest falls off as the less reactive $\ce {O2}$, but this is not the case here. $\ce {O2}$ reacts with $\ce {NO}$ instantly, as well as $\ce {O3}$.


Nomenclature of saturated aliphatic hetrocyclic amines with an alkyl group on the nitrogen For example, how would I name the following amine?

Since two of the alkyl groups on nitrogen are linked to each other (in a ring), I can't figure out what they should be called, as they don't fit into the usual alkyl (methyl/ethyl/etc.) mould.

According to Subsection P-22.2.1 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred IUPAC name (PIN) of the functional parent compound is the retained name ‘pyrrolidine’, the nitrogen atom being numbered ‘1’.

Therefore, the name of the compound given in the question is ‘1-propylpyrrolidine’.


How is it that fructose has a different metabolic pathway than glucose but yet glucose is converted to fructose? Fructose is described to have a different metabolic pathway (a more fat-inducing one) than glucose (see: http://healthyeating.sfgate.com/difference-between-sucrose-glucose-fructose-8704.html) as it can only be metabolized by the liver and does not impact insulin.

What I don't understand, is that in glycolysis, a paramount step of metabolizing glucose, it is very quickly converted to fructose-6-phosphate. Why does the body want to convert glucose into fructose?

The metabolic pathway you are talking about is how fructose is converted into energy and how its concentration in the blood is regulated. It is indeed true that blood fructose level does not affect blood insulin level, and this is why it is more "fat-inducing", as it cannot be effectively regulated like glucose.

It is also true that in glycolysis, glucose is converted to fructose-6-phosphate. However, this should not be confused with fructose itself. The fructose phosphate that fructolysis produced is fructose-1-phosphate, not fructose-6-phosphate.


Safe chemical reactions to start a campfire I'm looking for a way to start a small fire electronically with widely available chemicals. It would be great if they are safe to use with metals (not necessarily). But the biggest requirement is that it should be SAFE at medium scales and should not make an explosion.
I have found out $\ce{H2SO4 + KMnO4}$ can start a fire. But is it possible to do it with diluted H2SO4 because concentrated is not safe to handle?
NB: I only want a small ignition. The rest can be continued with some fuel, I believe.
Also, I doesn't possibly need to be a chemical reaction. Anything that can trigger a safe fire is enough

Small magnesium metal bars are available commercially, packed with a little iron rod for starting fires. You smack the magnesium with the iron rod and get some sparks. If you have some light fluffy kindling, you are supposed to be able to get a fire going. I suppose a little steel wool would help.
If you soak some cotton balls with KNO$_3$, dry them completely and carry them in a sealed plastic bag, they would make excellent kindling.


Why don't molecular orbitals change when electrons are added or removed from a molecule? My lecturer says that molecular orbitals don't change when electrons are added or removed from a molecule, but I don't understand why. As electrons are described by wave functions, surely more or fewer electrons means that the electrons interact differently and consequently the atomic orbital wave functions interact i.e. interfere differently to form different MOs.

The molecular orbitals do change when electrons are removed or added. But what does not change is the relative order of these orbitals with respect to other orbitals. If a given orbital has a higher energy than another one, it will remain so if you add one electron or it you remove one electron, whatever the absolute values of these energies.


Core cutoff radius and energy cutoff radius in pseudo potential DFT From what I understand the core cutoff radius is the distance where the pseudo wavefunction and the full electron wavefunction start to resemble each other. In other words, the pseudo wavefunction and full electron wavefunction have the same energy and electron densities for distances larger than the core cutoff radius. The energy cutoff radius is the energy that determines the size of the plane wave basis set.

My question is how they are related. Most computational software control the energy cutoff radius, and the energy cutoff radius is lower for ultrasoft pseudopotentials. My question is why is this so? I understand conceptually that ultrasoft pseudopotentials are smoother. Hence the Fourier series is less complicated. Hence the basis set does not need to be large. However, I have trouble putting this vague (if correct) concept into concrete reasoning. What are relevant equations that directly relate the energy cutoff and core cutoff?

The reason pseudopotentials are use is because (at low pressures), core electrons do not participate in chemical bonding or reactions, and more over have high kinetic energies. If you expand and electronic wave function in a plane wave basis, high energy wave functions require high frequency basis functions, therefore in order to capture the wave function of a core electron accurately, you would need a high plane wave cutoff, which is computationally expensive.

To avoid this, the core electrons are replaced by a pseudopotential, which is by construction smooth. The core radius is to some extent a measure of the 'hardness' of the pseudopotential. A soft pseudopotential will generally have a larger core radius, and will be smooth. A hard pseudopotential will have a lower core radius, and in order to capture the rapidly varying (more jagged) potential near the core, will be less smooth, and therefore require higher frequencies in its Fourier transform to recover the correct electronic wave functions. Harder pseudopotential are necessary, for example, at high pressures, when the nuclei are squeezed together --- if the core parts of the pseudopotential overlap, then the calculation will fail catastrophically.


Chemical bonding based on hybridisation model A snippet from a textbook:

  Therefore, the hybridization model predicts that an $\mathrm{sp}$-hybridized carbon atom is more electronegative than an $\mathrm{sp}^3$-hybridized carbon atom. Evidence for this effect is that the positive end of the dipole moment in $\ce{N#C-Cl}$ is on the $\ce{Cl}$ atom. We conclude that the carbon atom in the cyanide group is more electronegative than a chlorine atom.

The paragraph suggests that the $\ce{N#C}$ bond has a different hybrid orbital compared to the $\ce{C-Cl}$ bond, doesn't it? I am a bit confused since there is only $1~\ce{C}$ atom but somehow it has $2$ hybrid orbitals of different state (namely $\mathrm{sp}^3$ and $\mathrm{sp}^2$). Is this due to different ligands with which the central atoms connect?

And point of interest is why the conclusion isn't

  the nitrogen atom in the cyanide group is more electronegative than a chlorine atom

instead of carbon? Because I would imagine that the electron density of both $\ce{N#C}$ and $\ce{C-Cl}$ bonds would lie dominantly on $\ce{N}$ and $\ce{Cl}$, respectively. Hence, it is more reasonable to me to compare the electronegativity of $\ce{N}$ and $\ce{Cl}$ instead that of $\ce{C}$ and $\ce{Cl}$.

The usual explanation for $sp$ orbitals being more electronegative than $sp^3$ is the fact that more $s$ character means a shorter average distance from the nucleus, thus experiencing a greater effective nuclear charge. This partially explains the decreasing magnitude of the dipole moment of $\ce{CH3-CH2-Cl}$ to $\ce{CH2=CH-Cl}$ to $\ce{HC#C-Cl}$ and the increasing acidity of ethane to ethylene to acetylene.

I agree with you that the example they give doesn't really demonstrate that, since the orientation of the dipole (as you pointed out) relies on the electronegativity of the nitrogen, not the carbon. Specifically, the stability of the left resonance structure below:

$[\ce{^-N=C=Cl+  <-> N#C-Cl}]$

which clearly shows the orientation of the dipole, even if this resonance structure is only a minor contributor.

As a counterexample, chloroacetylene has a dipole with the (-) end on the chlorine, even though the carbon is still $sp$ hybridized. The corresponding resonance structure with a negative charge on C is much higher in energy than in the nitrogen case and so contributes negligibly to the overall structure, so the inductive electron-withdrawing of Cl dominates over the resonance electron-donating.


Deriving the buffer formula Essentially I want to derive the buffer formula: $\ce{pH}$ = $\mathrm{p}K_\mathrm {a}$ + $\log$ $\left(\frac{\alpha}{1-\alpha} \right)$ to $\alpha$ = $\left(\frac{1}{10^{\mathrm{p}K_\mathrm{a}-\ce{pH}}+1} \right)$
I'm going to include everything I had done trying to solve this here (and also show where I got stuck):

$$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log(\alpha) - \log(1-\alpha)$$

$$\ce{pH} - \mathrm{p}K_\mathrm{a} + \log(1-\alpha) = \log(\alpha)$$

$$\alpha = 10^{\ce{pH}-\mathrm{p}K_\mathrm{a}+\log(1-\alpha)}$$

$$\alpha = 10^{\ce{pH}} \times 10^{-\mathrm{p}K_\mathrm{a}} \times 10^{\log(1-\alpha)}$$

$$\alpha = 10^{-\log\ce{[H+]}} \times 10^{\log(K_\mathrm{a})} \times 10^{\log(1-\alpha)}$$

$$\alpha = \ce{[H+]}^{-1} \times K_\mathrm{a} \times (1-\alpha)$$

$$K_\mathrm{a} = \left(\frac{\alpha \times \ce{[H+]}}{(1-\alpha)}\right)$$

So at this point I got stuck and have no clue how to get rid of an $\alpha$ to complete the derivation. Appreciate any help.

Edit: Managed to complete. For those who are interested in how I had done it in the end:
$$\ce{pH}-\ce{pK_\mathrm{a}} = \log\left(\frac{\alpha}{1-\alpha}\right)$$

$$10^{pH-pK_\mathrm{a}} = \left(\frac{\alpha}{1-\alpha}\right)$$

$$10^{pH-pK_\mathrm{a}} - (\alpha 10^{pH-pK_a}) = \alpha$$

$$10^{pH-pK_a} = \alpha + \alpha 10^{pH-pK_\mathrm{a}}$$

$$10^{pH-pK_a} = \alpha(1 + 10^{pH-pk_a})$$

$$\left(\frac{10^{pH-pK_a}}{1 + 10^{pH-pK_a}}\right) = \alpha$$

divide numerator by itself = 1 and thus also divide denominator by the numerator.

$$\left(\frac{10^{pH-pK_a}}{10^{pH-pK_a}}\right)/\left(\frac{1+10^{pH-pK_a}}{10^{pH-pK_a}}\right) =$$

$$\left(\frac{1}{1+10^{pH-pK_a}/10^{pH-pK_a}}\right)$$

working out the denominator:

$$\left(\frac{1}{10^{pH-pK_a}}\right) + \left(\frac{10^{pH-pK_a}}{10^{pH-pK_a}}\right) ->$$

$$\left(\frac{1}{10^{pH-pK_a}}\right) +1$$

now work out the last fraction:

$$\left(\frac{1}{10^{pH-pK_a}}\right) = \left(\frac{1}{10^{pH} \times 10^{-pK_a}}\right) = (10^{pH})^{-1} \times (10^{-pK_a})^{-1}$$
finally gives:

$$\alpha = \left(\frac{1}{10^{pK_a-pH}+1}\right)$$

$$\ce{pH} = \mathrm{p}K_\mathrm {a} + \log \left(\frac{\alpha}{1-\alpha} \right)$$
Multiply the equation by $-1$:

$$-\ce{pH} = -\mathrm{p}K_\mathrm {a} + \log \left(\frac{1-\alpha} {\alpha}\right)$$
Rearrange the equation
$$-\ce{pH} +\mathrm{p}K_\mathrm {a} = \log \left(\frac{1-\alpha} {\alpha}\right)$$

Apply the definition of $\log$

$$10^{-\ce{pH} +\mathrm{p}K_\mathrm {a}} = \frac{1-\alpha} {\alpha}=  \frac{1}{\alpha} -1$$
Rearrange the equation
$$1+ 10^{-\ce{pH} +\mathrm{p}K_\mathrm {a}} =  \frac{1}{\alpha} $$
So,
$$\alpha= \frac{1}{ 1+ 10^{-\ce{pH} +\mathrm{p}K_\mathrm {a}}  } $$


Reaction involved in Combustion of Nitrocellulose-Nitroglycerine(NCNG) solid propellant? I'm working on solid propellant based microthrusters and I am using NCNG as the solid propellant. I want to know the reaction taking place and the mass fraction of the products obtained on combustion of NCNG propellant.

Nitrocellulose is complex stuff. It is generally considered to be the fully nitrated version of cellulose, in which every available alcohol functional group is converted to a nitrate:

$$\ce{[C6H10O5]n +3HNO3 -> (C6H7O2(ONO2)3]n +3H2O}$$

In principal it probably contains some small amount of partially and/or non-nitrated units.

If we consider nitrocellulose to be fully nitrated, then the combustion reaction is:

$$\ce{ 2C6H7N3O11 + 9/2O2 -> 12CO2 + 3N2 + 7H2O}$$

For nitroglycerin $(\ce{C3H5N3O9})$:

$$\ce{2C3H5N3O9  -> 6CO2 + 3N2 + 5H2O + 1/2O2}$$

It will depend on the ratio of nitrocellulose to nitroglycerin what the overall $\ce{CO2}:
\ce{N2}:\ce{H2O}$ mole ratio is. I'll leave you to figure out the mass ratios using the molecular masses of the products.


Beryllium chloride - Highly hydrolysable? Beryllium chloride's geometric structure is linear, with a bond angle of 180 degrees. Drawing the Lewis dot structure for this covalent compound, one can see that the compound is electron deficient and requires two lone pairs to achieve stability. As a result it is a strong Lewis acid. Since water has two lone pairs and is a strong Lewis base, mixing beryllium chloride with water results in stability for beryllium chloride and the evolution of energy.
My question is, why don't two lone pairs from either chlorine atom or even just a single chlorine atom make a coordinate bond with beryllium, as can be seen in the case of carbon monoxide where one of oxygen's lone pairs is donated to carbon to achieve nearest noble gas configuration of neon. If such a compound does not naturally exist, why does it not? Is it a theoretical defect or is there some explanation?

The structures of $\ce{BeX_2}$ where $\ce{X = F}$ or $\ce{Cl}$ can be found in A.F. Wells, Structural Inorganic Chemistry 5th edition, 1984 pages 412 and 1047.  There you will see that while the predicted structures for $\ce{BeX_2}$ are linear (sometimes rationalized with Lewis dot structures displaying double bonds and ignoring formal charge considerations of the halogens).  However, the actual structure of $\ce{BeCl_2}$ in the vapor-phase is a 3-coordinate dimer that converts to the linear structure at higher temperatures.  This behavior supports your 'gut feeling' that the electron deficiency of $\ce{Be}$ should be compensated for by another $\ce{Cl}$ atom.  The dimer network formed in the solid state allows lone pairs from $\ce{Cl}$ to be donated to the $\ce{Be}$ in an attempt to fill the valence shell of beryllium.

The use of a double bond to describe the nature of the $\ce{Be-Cl}$ bond is supported experimental data which shows that the bond length is shorter than expected for a single bond.

The image below is taken from Inorganic Chemistry by Miessler, Fischer and Tarr, 5th edition.


Is benzenesulfonic acid or cyclohexanesulfonic acid more acidic? Which of the two sulfonic acids above is a stronger acid?

I think that cyclohexanesulfonic acid 2 should be more acidic as there is no competitive resonance as in 1. Is that correct?

Draw out the full line-bond or Lewis structures for each and you will see that there is no way for the benzene ring to contribute to resonance structures that involve the sulfonic acid group, or the sulfonate conjugate base. Instead, the important contribution is the inductive effect. Benzene rings are more electron-withdrawing than alkyl groups, so benzenesulfonic acid should be more acidic than cyclohexanesulfonic acid. A comparison of pKa values for $\ce{PhSO3H}$ ($-2.8$) vs $\ce{MeSO3H}$ ($-1.9$) supports this assertion (Can. J. Chem. 1978, 56 (17), 2342–2354).


What is the relationship between configuration & optical rotation? Is there a clear cut rule between the two? For example, is R always (+), or is S always (-), or is there no relationship? And if so, why? Wouldn't it make sense that plane polarized light would rotate in the direction ascribed to chiral molecules?

There is no general correlation between the the rectus (R) and sinister (S) prefixes and dextro/levorotation. In some cases the (R)-enantiomer is the dextrorotary enantiomer (as in the case of D-alanine), and in other cases the (R)-enantiomer is the levorotary enantiomer (as in the case of L-cysteine). The relationship can only be determined on a case-by-case basis with experimental measurements or detailed computer modeling.

For a really detailed explanation of exactly how chiral molecules rotate planes of polarized light, I recommend you check out this physics post, but to answer your other question, no it doesn't make sense for planes of polarized light to rotate according the Cahn–Ingold–Prelog Priority Rules. This system only takes into consideration the atomic number and number of bonds to atoms of a higher atomic number. Even though $\ce{-Cl}$ might rank higher than $\ce{-C(CH)3}$, wouldn't it make sense for the two groups to have different effects on rotation, in the cases of both direction and magnitude? This is why dextro/levorotation must be determined experimentally.


formation of a compound due to photo-isomerisation I was reading through a topic called photoisomerism, where I read that when a molecule is hit by a photon it changes its structure to a more stable one. Is this a one-way reaction, or is it reversible (as in is it an equilibrium)? If it was an equilibrium, what was stopping it from forming before the photon hit it?
update
since the question was a bit ambiguous, my question was after the light that was shone was removed, if the equilibrium still exists, then what was the hindrance that did not allow such equilibrium to exist before the light was shone. (i am not sure if such a reaction exists, my question is more of a hypothetical question, asking if such a reaction exists then what was hindrance from reaching equilibrium (that was achieved after the shone light was removed)  before the light was shone)(if no such reaction can exist please also explain why so)

Stilbene has been extensively studied with pump-probe absorption spectroscopy as well as time resolved fluorescence.

In the ground state (S0) there is a large activation energy to go from trans  to cis and vice versa so that both species are stable at room temperature. This is not the case in the first excited state (S1)and the trans converts to the cis in a few tens of picoseconds and it is even faster for the cis excited state to convert to the trans.
The potential energy scheme is shown below (Not to scale). On the left is the trans species with a large barrier to for the cis (on the right). This has a higher energy than the trans but also has a large barrier to form the trans.
Detailed analysis suggest that the reaction occurs by a lengthening of the double bond before twisting. Exciting the trans puts it into the first singlet excited state (S1) where there is a small barrier on the pathway to cis, but as it happens there is an intersection with the ground state (by 3ps label & twisted species) and from here either the ground state trans or ground state cis can be formed. If the cis ground state is excited then there is effectively no barrier to form the trans ground state and the excited cis molecule lasts for only a few ps.

For a review or organic photochemistry calculations see Bernardi et al. Chemical Society Reviews, p321, 1996, in particular fig 1. A lot of the original experimental work was done by the Hochstrasser group, see Annu. Rev. Phys. Chem. 2006. 57:37–64 for many references.


Electrolytes in electrochemical half cells In most of the examples in my book I have come across, the electrolyte for an electrochemical half cell has always been the aqueous version of the metal used. For example, a silver electrode is immersed in a silver solution and so on, but is why is this necessary? Surely any positive metal ion will do?

For example, a silver electrode is immersed in a silver solution and so on, but is why is this necessary? Surely any positive metal ion will do?

There is a fundamental reason, which is the Nernst equation. It requires that the metal be in contact with its own ions. What is the functional meaning of the electrode potential? It is the tendency of an ion to reduce itself. Yes one can make salt water battery by dipping an iron nail (or copper) in one half cell and a piece of aluminum in another half cell but you cannot predict the cell potential beforehand.


Why do acetylene and chlorine react to yield carbon and HCl, but acetylene and bromine react less intensely? In the famous "underwater fireworks" reaction we get

$$\ce{C2H2 + Cl2 -> 2C + 2HCl}$$

On the other hand, when we react acetylene with bromine, we get

$$\ce{CH#CH + Br2 -> CHBr=CHBr}$$
$$\ce{CHBr#CHBr + Br2 -> CHBr2-CHBr2}$$

Why don't we get carbon and HBr? Or do we get it?

Bromine is not as strong of an oxidizing agent as chlorine.  In essence the reaction of acetylene with chlorine is the same as burning, just with elemental halogen instead of elemental oxygen.

The higher electronegativity of chlorine explains why it is a good oxidant as compared to bromine.

Other notes:

Describing something as 'radically' (this phrasing removed  from original question edit) when you mean 'intensely' will
cause much confusion around here (Stack) since 'radically' refers to reactions involving single electron mechanisms.


When pH increases, does the effect of increased [OH-] or reduced [H+] control precipitation of CaC2O4 The image below shows how oxalic acid dissociates and forms calcium oxalate ($\ce{CaC2O4}$ or $\ce{Ca(OOC-COO) as written below}$).

When the pH increases, does this result in more or less $\ce{Ca(OOC-COO)}$ forming?

I can think of two factors:

As $\ce{pH}$ increases, $\ce{OH-}$ increases. Greater $\ce{OH-}$ means more of the $\ce{Ca^{2+}}$ will react with $\ce{OH-}$, forming $\ce{Ca(OH)2}$. So the reaction in the third line below will shift to the left, resuling in less $\ce{Ca(OOC-COO)}$.
As $\ce{pH}$ increases, $\ce{H+}$ decreases. Less $\ce{H+}$ results in the reactions (in line 1 and 2) shifting to the right. Hence, there will be more $\ce{(-OOC-COO-)}$ and so more $\ce{Ca(OOC-COO)}$.

These give opposing results. Which one outweighs the other, and why?

The original question from by textbook:

Drinking strong antacids to relieve indigestion produces kidney fluid that is more alkaline than usual.

Are kidney stones more likely to occur under these conditions?

Yes, as at a higher pH a higher concentration of oxalate ions will form.
Yes, as both calcium and oxalate ions are bivalent.
No, as the presence of hydrogen oxalate ion (HOOCCOO–) means that a buffer is present.
No, as the higher concentration of hydroxide ion (OH–) will remove calcium ion by forming calcium
hydroxide.

In the body we'll assume that:

There is some low level of $\ce{Ca^{2+}}$ present in the blood or urine.
The pH isn't nearly high enough to precipitate out $\ce{Ca(OH)2}$.
Under "normal" bodily conditions calcium oxlate doesn't precipitate out either so the overall oxalate concentration is low.

Let's let $x$ be the concentration of all the oxalate species. By knowing $x$, the pH and the two pka's we can calculate the exact concentrations of all three oxalate species.

$x = \ce{[HOOC-COOH] + [HOOC-COO^-] + [^-OOC-COO^-] }$

Now the pKa values for oxalic acid are far enough apart that at most two of the three oxlate species will have an appreciable concentration.
For normal bodily functions only $\ce{[HOOC-COO^-]}$ and $\ce{[^-OOC-COO^-]}$ will be the only significant species. Since only a couple of significant figures will be of interest, we can use the Henderson–Hasselbalch equation to calculate the relative concentration of the two species to reasonable accuracy. (The essence is that the third species has a negligibly small concentration.)

pH = $\text{pK}_{a2} + \text{log}(\dfrac{\ce{[^-OOC-COO^-]}}{\ce{[HOOC-COO^-]}})$

Thus as the pH increases relatively more $\ce{[^-OOC-COO^-]}$ is formed and less $\ce{Ca^{2+}}$ will be soluble.

But normal blood is alkaline already. Thus there would be a negligibly small amount of $\ce{[HOOC-COO^-]}$ in the blood to start with since the $\text{pK}_{a2} = 4.2$. So calcium oxalate will be unlikley to ppt in the blood.

However the kidneys make urine which has a pH from 4.5 to 8. So if the pH in the urine changes as much as from 4.5 to 8 then there could be a greater tendency to ppt calcium oxalate in the urine.

ANSWER

(4) is out since if body was that alkaline your tissue would dissolve.

(3) is out since you can see from the HH equation specified that the presence of another buffer would effect the absolute pH but not the relative oxalate species at that pH.

(2) is true, but the answer doesn't note any effect of pH. So the answer doesn't really address the problem.

Thus the answer must be (1).


Why are Tertiary Radicals more stable than Tertiary Carbocations? I think tertiary carbocation must be more stable due to Hyperconjugation and +I effect....

The magnitude of the effects needs to be considered. To form a radical from isobutane, you break a C-H bond. To form t-butyl radical requires about 93 kcal/mol (390 kJ/mol) https://en.wikipedia.org/wiki/Carbon–hydrogen_bond. To form a cation, you break the C-H bond AND ionize the radical, which costs you another 6.7 +/- eV (154 kcal/mol or 646 kJ/mol http://www.colby.edu/chemistry/PChem/Hartree.html) https://webbook.nist.gov/cgi/cbook.cgi?ID=C1605738&Mask=20

So, while the t-butyl carbocation is indeed stabilized by the indictive effect, and hyperconjugation, the energy level at which this is occurring is much higher than the energy level of the radical.

Hyperconjugation is a qualitative concept, with a possible energy of about 5 kcal/mol.  Like a fudge factor, which explains all possible effects if you need it and apply it correctly, but in this comparison, its effect is negligible compared to bond-breaking and ionization.  https://en.wikipedia.org/wiki/Hyperconjugation


Can chemical energy from food be stored in a battery? Suppose you built a machine that digested food in a similar way as humans do. Would it be possible, in principle, to extract the chemical energy from the digested food, turn it into electricity, and store it in a battery that powers the machine itself? Doesn't matter if it's practical or efficient, I just need to know if it's not something totally impossible. (I asked the same question on Electrical Engineering SE and someone suggested this SE would be more appropriate.)

Three ways to turn chemically stored energy into electricity:

a fuel cell: While that works fine for hydrogen or methanol, I'm not sure if sugars are possible, and general "food" won't work. Of course you can chemically convert food into hydrogen or alcohol, or into sth. very specially designed to power a biofuel cell.
dry, burn, run a turbine or steam engine with attached generator, very obvious and simple.
direct conversion into mechanical energy via an artificial muscle, which powers a generator. Horribly complicated (bio-)chemistry, theoretically possible, because our body does it, too.


How to reduce Tin Oxide We have a pilot atomization setup where we atomize molten tin to produce tin powder. Unfortunately a lot of tin is wasted and oxidized.

I have found this video on which looks promising. Is this the best way of reducing tin oxide? Or do you know of anything better / effective?

Thanks!

The video shows the "purification" of zinc metal (probably covered with an oxide layer) by heating/reacting it with powdered carbon.

This is a very common process to reduce metal oxides to the corresponding metals. In turn, carbon is oxidized to carbon monoxide and carbon dioxide. The most prominent example is the production of iron from iron oxide in a blast furnace.

However for some metals, this fails and metal carbides are formed instead.

For tin oxide, the reduction with carbon (or coke) is an established industrial process.

Alternatively, the reduction of tin oxide can can be achieved using hydrogen, see Materials Transactions, 2011, 52, 1814-1817 (PDF).