Mechanism of lithium-halogen exchange of primary alkyl iodide On Clayden's page 190 this reaction is given for preparation of organometallics:

What is the mechanism of this reaction? I expected the "almost carboanion" part of the $\ce{t-BuLi}$ to attack as a nucleophile and form a longer 7 membered chain? But instead, that almost anionic $\ce {t-Bu^-} $ has attached to iodine.

I realised that $\ce{t-Bu-}$ is too unstable because of electron releasing alkyls, but what is the exact mechanism of this reaction?

With primary alkyl iodides the (simplified) mechanism is likely similar to a $\mathrm{S_N2}$ process, but with attack of $\ce{t-Bu}$ at iodine instead of carbon.

Note that this is not necessarily the case for other substrates, e.g. aryl iodides, or aryl/alkyl bromides. In general lithium-halogen exchange can occur either via a polar mechanism (such as the one shown above) or via a radical mechanism (by two successive electron transfers from RLi to RX). There is also a solvent and temperature dependence. Why does the solvent matter? We know that ethereal solvents such as Et2O or THF are capable of acting as Lewis bases - or ligands for lithium - and possibly assist in deaggregation of the alkyllithium. This sort of thing can't be depicted easily with curly arrows, which is why I add the disclaimer that the mechanism above is (heavily) simplified.

In any case, this particular substrate (Tetrahedron Lett. 1986, 27 (17), 1861–1864) is fairly interesting because if the Li/I exchange were to occur via a radical pathway, the radical formed would cyclise onto the alkene via a 5-exo-trig pathway (even at −78 °C):

However, if you quench the reaction at −78 °C with methanol, no methylcyclopentane is observed, which rules out the intermediacy of a radical. [If you warm up the reaction before quenching then the organolithium can cyclise by nucleophilic attack onto the alkene.]

This particular case (and general mechanisms for Li/Hal exchange) are described quite thoroughly in Clayden's Organolithiums: Selectivity for Synthesis, Chapter 3.

There is also an excellent set of slides on Li/Hal exchange reactions, courtesy of the MacMillan group at Princeton; and also another set, courtesy of the Denmark group at the University of Illinois.


What is the reason for anomalous expansion of water? What is the reason for anomalous expansion of water? Why doesn’t it simply expand on heating or contract on cooling? Why it shows anomalous behavior at 4 degrees Celsius? Why not on 4.6 or 10 (or whatever) degrees Celsius?

So, the first question to ask would be: how "anomalous" is change in the thermal expansion coefficient?  It turns out that for most real materials the coefficient of thermal expansion is not a constant, and takes on both positive (expansion) and negative (contraction with increasing temperature) values.  For silicon as an example, see ioffe,  Above room temperature Si has a positive (but not constant) CTE, below about 100K silicon's CTE is negative.

The second question is why would this occur?  Taking a liquid or a solid (ignore gases for the moment), an individual atom (or molecule) sits in some kind of potential well which is what is causing the material to be bound together. At absolute zero, these atoms sit at the very bottom of the potential well.  As the temperature increases, heat is added and the atom starts rattling around in the potential well, and is able to explore more of the well.  There is no particularly good reason to believe that the shape of the well, across the entire temperature range of existence of the phase, will be such that a single linear coefficient will describe where the atom will most likely sit in the well at a given temperature. On top of that in a solid is the equilibrium concentration of point defects that is dependent on temperature.  As an example, Google for Simmons and Balluffi and their experiments combining x-ray measurements of the lattice parameter (where the atoms sit in the potential well) and thermal expansion (change in length of a heated bar) to determine the point defect concentrations vs temperature for various solids.


Calculating change in state function following different paths in thermodynamics, we learn that change in state function is independent of the path/process, and thus we can conveniently pick any arbitrary path that allows easier calculation. I'm TA'ing a class and found that it would be great to actually show students an example where that a simpler path can indeed be constructed to calculate the change of state function. However, I feel that there must be some blind spot that I myself am struggling with. And would like to get everyone's feedback.
Example and detail
We are asked to calculate $\Delta H$ for an isochoric process where a known amount of energy is transferred into the system (of a monatomic ideal gas) as heat. Assuming only P-V work is allowed in the system.
I can first calculate change of internal energy ($\Delta U$) using the first law, and subsequently $\Delta T$ with the known heat capacity (over constant volume) $c_V = \frac{3R}{2}$. Then to calcuate $\Delta H$, I'll follow an isobaric process instead (with $c_P = \frac{5R}{2}$), but it will give the same result. Here's the detail:

$w = -\int_i^f P_\text{ext} dV = 0$ (no work is done at constant volume)

$\Delta U = w + q = 0 + q = q$

$\Delta U (= q_V) = n c_V \Delta T \implies \Delta T = \frac{\Delta U}{n c_V} = \frac{q}{n c_V}$

$\Delta H = q_P = n c_P \Delta T = n c_P \frac{q}{n c_V} = \frac{c_P}{c_V} q$

Question
My question - under the context of this example - is how to make up a path connecting the same endpoints (the initial and final states) that is under constant pressure?
Without loss of generality, assume that $q > 0$, and thus the temperature at the end increases as the internal energy increases (no work is done due to the constant volume constraint). This suggests that the pressure also increases. But I just said that I'd like to construct an isobaric path that connects the initial and final state..., which now seems to be contradicting the fact.
My thought is - that instead of a single isobaric path (which connects the initial state and potentially an intermediate state), maybe I need another segment that connects the intermediate state and the final state? Such that overall the volume stays the same? If this is the approach, what would the strategy be for picking the intermediate state and the two segments making up the simpler path?
Would the following work?

1st segment: isobaric expansion to an intermediate state whose temperature is the same as the final state. The relevant heat is $q_P$ which will contribute to the overall change in enthalpy with its contribution $\Delta H_1 = q_P$.

2nd segment: isothermal compression to the final state (i.e., need to volume to go back to its initial/final state volume). For this part - there is going to be additional heat and work involved, but internal energy and enthalpy should remain unchanged (as same as the intermediate state), resulting in $\Delta H_2 = 0$.

Overall, $\Delta H = \Delta H_1 + \Delta H_2 = \Delta H_1 = q_P$.

I'm not very sure about this as it invokes that enthalpy change is zero for an isothermal process -- which I don't know if it will hold for non-ideal gas? And if not - this suggests my proposal doesn't work universally...
[Edits for fixing typo]

You wrote, in one of your comments, "My biggest question is how to demonstrate the proper use of $\Delta H = n C_p \Delta T$irrespective of the path...".
So:
In a closed system (no flow of matter across the boundaries) of constant composition (no chemical reactions),
$$dH = \left(\frac{\partial H}{\partial T}\right)_ p dT +  \left(\frac{\partial H}{\partial p}\right)_ T dp$$
$$= C_ p dT +  \left(\frac{\partial H}{\partial p}\right)_ T dp$$
$$= C_ p dT +  \left(V - \alpha T V \right) dp$$
$$= C_ p dT +  \left(V - T \left(\frac{\partial V}{\partial T}\right)_ p  \right) dp$$
Thus, at constant $p$,  $dp =0$, and
$$dH_p = C_ p dT \textit{, for all substances.*}$$
It is worth noting, as Chet Miller mentioned in the comments, that $C_p \equiv C_p(T,p)$ (i.e., $C_p \text{ is a function of both } T \text{and } p$ for real substances).
If there are no inter-particle interactions (e.g., an ideal gas), then $V = T \left(\frac{\partial V}{\partial T}\right)_ p $, and thus
$$\left(\frac{\partial H}{\partial p}\right)_ T = 0$$
You can confirm this yourself for an ideal gas by substituting $V = \frac{n R T}{p}$ for the second occurrence of $V$ in: $\left(V - T \left(\frac{\partial V}{\partial T}\right)_ p  \right)$.
Hence, for an ideal gas,
$$dH = C_ p dT \text{, always,} \textit{ even if the pressure isn't constant.}$$
In summary, if the substance is an ideal gas, $dH = C_ p dT$ regardless of whether there is a change in pressure.  If the substance is real, $dH = C_p dT$ only if the pressure is constant.

*Note that this relationship holds for all types of work.  If we add the constraint that the only work is $pV\text{-work}$, then we have the additional nice relationship that:
$$dH_p = C_ p dT = q_p \textit{, for all substances.}$$
If we keep the constant-$p$, closed system, and $pV\text{-work}$-only constraints, but remove the constant-composition constraint (allowing chemical reactions), then it will no longer be the case that $dH_p = C_ p dT$, but it will be the case that:
$$dH_p = q_p \textit{, for all substances.}$$


How to determine the angle between non-bonding electron pairs? Is there a way to determine the angle between non-bonding electron pairs that aren't obvious? Take $\ce{H2S}$ for example:

The two non-bonding electron pairs would be at the yellow end of this molecule, but I am curious as to what the angle is between them? Is there even a way to tell?

Yes, there are ways one could claim to calculate an angle between two non-bonding electron pairs.

BUT: As Mithoron points out, this Chem.SE question illustrates how quantum chemical calculations and photoelectron spectroscopy both demonstrate the non-equivalence of the lone pairs of $\ce{H2O}$, an analysis which presumably applies equally well to the analogous $\ce{H2S}$. Thus, the methods used for calculating such an angle will be controversial, and the results may or may not be of any particular practical value given that they're at odds with PES data. That being said, I'll show here one way that a putative 'non-bonding electron pair angle' can be calculated.

The calculation below is based on application of the quantum theory of atoms in molecules (QTAIM) to the quantity called the electron localization function (ELF), which is a scalar function in $\mathbb R^3$.  QTAIM is useful for identifying intrinsic features of various three-dimensional fields that arise in quantum chemical calculations, with a significant focus placed on the 'critical points,' where the field gradient is zero. It was originally developed for analysis of the electron density distribution, but has been extended to the ELF and other quantities. One of my favorite papers illustrating coupled QTAIM/ELF analysis is a 2009 review by Matito and Solà (Coord Chem Rev 253: 647); it also discusses the localization and delocalization indices (see here and here), which I won't go into in any detail here.

I'm going to start out by emphasizing the distinction between electron density and electron localization. The electron density at a point is fairly straightforward: it's a measure of, on average, how many electrons (or fraction thereof) are located at that point, per unit volume. Due to its waveform nature, each electron's position is a continuous distribution function, and the electron density of the system represents the collective distribution function of all the electrons present.

The electron localization at a point can be harder to get a grasp of: it's a measure of how "spread out" the "location distribution" is, of the electrons that contribute to the density at that point. Regardless of the magnitude of the electron density, the electron localization at a point is high when the electrons contributing to the density at that point overall contribute relatively little to the electron density in other parts of the system.  That is to say, these electrons associated with the point of interest don't "travel around very far" from that point, in their quantum-mechanical meanderings about the system.  Conversely, the electron localization is low when the electrons associated with the point do contribute significantly ("wander") to "distant" regions of the system.

My answer here provides a more concrete example of how the electron localization can differ significantly among systems whose electron density distributions are otherwise relatively similar.  It also anticipates the discussion below, noting features of the three-dimensional ELF field that suggest the locations of electron lone pairs on N, O and F atoms. It is this ELF field that I will be focusing on in the following analysis.

I performed DFT calculations of $\ce{H2O}$, $\ce{H2S}$, $\ce{H2Se}$, $\ce{H2Te}$ and $\ce{H2Po}$ for this analysis using ORCA v3.0.3, and I carried out the subsequent QTAIM/ELF analysis in Multiwfn v3.3.7.  The ORCA input for $\ce{H2S}$ was

! RKS PBE0 def2-TZVP def2-TZVP/J RIJCOSX
! OPT GRID4 GRIDX5 PRINTBASIS

* xyzfile 0 1 H2S.xyz

H2S.xyz was an initial geometry file generated by Avogadro, overwritten with the optimized geometry at the end of the ORCA run.  The inputs for $\ce{H2O}$ and $\ce{H2Se}$ were identical except for the identity of the central atom; I added relativistic effects for $\ce{H2Te}$ via the ! ZORA simple keyword, and for $\ce{H2Po}$ via the use of a 60-electron effective core potential by adding ! ECP{def2-TZVP,def2-TZVP/J}. (While there is an argument to be made that I perhaps should've included relativistic effects for $\ce{H2Se}$ also, I expect them to minimally affect the geometry and overall valence-electronic structure of that system, which is what is of interest here.)

I generated a MOLDEN wavefunction file for each computation, to serve as inputs to Multiwfn. Using $\ce{H2S}$ as an example, the shell command was:

$ orca_2mkl H2S -molden

I then renamed the resulting H2S.molden.input to H2S.molden, which is necessary for Multiwfn to correctly identify it as an ORCA-generated MOLDEN file.

After loading H2S.molden into Multiwfn, I used the following series of commands to locate the 3-D maxima ("attractors") of the ELF distribution:

Main menu:  [17] Basin analysis
[1] Generate basins
[9] ELF
[3] High quality grid

Multiwfn's search routine found six ELF attractors, one of which was identified as a cluster of 'degenerate attractors'. The figure below plots these attractors (via Multiwfn sub-command [0] and some MS Paint manipulation) atop two different views of the $\ce{H2S}$ molecule (click to enlarge):

As can be seen, there are two attractors located right where chemical intuition would expect lone pairs to sit, along with two more attractors located at the hydrogen atoms.  The final non-degenerate attractor is right in the center of the sulfur atom, and could be interpreted as representing the $n=1$ core electron shell. The degenerate attractor is distributed around the $n=1$ attractor, and could similarly be interpreted as the $n=2$ core electron shell. Interestingly, the two $n=3$ valence electrons of the sulfur that are involved in bonding to the $\ce{H}$ atoms do not exhibit independent attractors---each shares an attractor with the electron originating from its respective hydrogen atom.

The QTAIM method provides a way to subdivide the space occupied by a molecule based solely on the properties of the ELF distribution, and associate portions of the electron density to each of these attractors. Integrating these subdivisions of the electron density then provides the number of electrons associated with each attractor:

$$
\begin{array}{cccc}
\hline
\text{lp} & \text{H} & \text{S 1s} & \text{S 2sp} \\
2.114 & 1.855 & 2.143 & 7.865 \\
\hline
\end{array}
$$

All of these values seem pretty reasonable---the attractors corresponding to single orbitals $(\ce{lp}$, $\ce{H}$, $\ce{S 1s})$ have about $\ce{2 e-}$ associated with them, and the $\ce{S 2sp}$ attractor has about $\ce{8 e-}$. In my experience, the signs and magnitudes of these deviations from $2.0$ and $8.0$ are typical for ELF basins.

So, leaving the question of actual physical significance aside, I would argue that these ELF attractors provide a reasonable representation of the (non-)bonding structure of the molecule. Thus, to the main question: what is the structure?  Conveniently, angles and distances among the attractors and atoms can be calculated via Multiwfn sub-command [-2]:

$$
\begin{array}{cccc}
\hline
\angle\,\ce{H-S-H} & 92.3^\circ & r_\ce{S-H} & 2.535\,\mathrm{Bohr} \\
\angle\,\ce{lp-S-lp} & 127.8^\circ & r_\ce{S-lp} & 1.838\,\mathrm{Bohr} \\
\angle\,\ce{H-S-lp} & 108.5^\circ \\
\hline
\end{array}
$$

(In the above, the actual nuclear positions of the atoms were used where relevant, not the positions of the associated attractors. While each such attractor generally falls close to its associated nucleus, it's usually displaced by a tiny amount, $<0.1\,\mathrm{Bohr}$. I've always attributed this to numerical precision limits of the calculation methods, but I don't know for sure if that's what causes it.)

Therefore: By this QTAIM/ELF method, the angle between the two lone pairs of $\ce{H2S}$ is calculated to be $127.8^\circ$. This is considerably greater than the $\ce{H-S-H}$ angle, consistent with the intro-chem conception of the increased 'steric bulk' of a lone pair.

For comparison, I used the same procedure to generate results for $\ce{H2O}$, $\ce{H2Se}$, $\ce{H2Te}$ and $\ce{H2Po}$:

$$
\begin{array}{c|c|ccccc}
\text{Quantity} & \text{Units}  & \ce{H2O} & \ce{H2S} & \ce{H2Se} & \ce{H2Te} & \ce{H2Po} \\
\hline
\int_\text{lp}{\rho} & \ce{e-} & 2.265 & 2.114 & 2.206 & 2.265 & 2.390 \\
\int_{\ce{x}\ce{-H}}{\rho} & \ce{e-} & 1.667 & 1.855 & 1.861 & 1.807 & 1.806 \\
\int_\text{core}{\rho} & \ce{e-} & 2.129 & 10.008 & 27.273 & 44.567 & 77.609 \\
\hline
r_{\ce{x}\ce{-H}} & \mathrm{Bohr} & 1.813 & 2.535 & 2.772 & 3.129 & 3.289 \\
r_{\ce{x}\ce{-lp}} & \mathrm{Bohr} & 1.103 & 1.838 & 2.526 & 3.216 & 3.399 \\
\hline
\angle \ce{H-x}\ce{-H} & ^\circ & 105.1 & 92.3 & 91.2 & 90.9 & 89.6 \\
\angle \ce{lp-x}\ce{-lp} & ^\circ & 114.9 & 127.8 & 139.5 & 156.8 & 156.8 \\
\angle \ce{H-x}\ce{-lp} & ^\circ & 109.1 & 107.8 & 104.0 & 98.1 & 98.2 \\
\hline
\end{array}
$$

Due (presumably) to numerical precision limitations, the calculated values of the four different $\ce{H-x}\ce{-lp}$ angles for each system varied by $1^\circ$ or so; I've reported the means of the obtained values above.

Those at home who are doing their math carefully will note that there is some electron density missing in the systems with the heavier central atoms.  This is because at the 'high' grid quality used, the deep core electrons are poorly captured by the numerical integration involved.  (Indeed, the deep core electrons may not be accurately captured at just about any computationally reasonable grid size!) I think the $\int_\text{lp}{\rho}$ and $\int_{\ce{x}\ce{-H}}{\rho}$ values should be pretty reliable, though.

Naturally, $\ce{H2O}$ does not exhibit a degenerate attractor for the $n=2$ valence electron shell, though it does have a non-degenerate attractor for the $n=1$ core shell.  The two heaviest chalcogenides examined possess degenerate attractors representing their $n\geq 2$ core electrons, as with sulfur.

In general, the $\ce{x}\ce{-H}$ and "$\ce{x}\ce{-lp}$" bond lengths increase as the central atom is varied down the group, which is unsurprising ("bigger atoms are bigger"). Interestingly, while the $\ce{x}\ce{-lp}$ distance starts out considerably shorter than the $\ce{x}\ce{-H}$ bond length for $\ce{x}=\ce{O}$, by the time one reaches $\ce{x}=\ce{Te}$ the relative magnitudes are reversed.  I don't have a particularly good explanation for this trend; presumably it's due to some sort of counterbalance between the nucleus-nucleus repulsion and nucleus-electron attraction forces.

Where these results are particularly dramatic is in the clear trends in the bond angles: as the central atom is made heavier, the $\ce{lp-x}\ce{-lp}$ bond angle increases markedly while the $\ce{H-x}\ce{-H}$ and $\ce{lp-x}\ce{-H}$ angles steadily decrease. The lone pair 'steric bulk' apparently increases uniformly down the group for this series of analogous systems, at least as determined by this QTAIM/ELF approach.  The answers to this question provide more detail into the physical reasons for this decreasing trend in $\ce{H-x}\ce{-H}$ bond angles.


Can the Smirnov-Zamkow reaction be described as a pericyclic process? The Smirnov-Zamkow reaction involves the reaction of 2-butyne and chlorine in the presence of a Lewis acid catalyst to afford a dichlorocyclobutene:

At first glance, it appears to be an example of a thermal [2+2]-cycloaddition, which, by the Woodward Hoffman rules is forbidden.[*] This either means that (1) the reaction is stepwise, or (2) the intermediates are able to arrange themselves such that both components are not reacting suprafacially.

Very few mentions of the Smirnov-Zamkow reaction appear in the literature. The only real description is found in Fleming [1] where the reaction is proposed to involve initial formation of an allene. All other references on the internet simply link back to Fleming, with no mention made of this reaction in March, Carey, or any of the other usual suspects.

Clearly the reaction isn't synthetically that useful, and its initial reports occurred before pericyclic reactions had been widely studied/rules existed governing what was 'allowed',[**] but it is a curious example, to which I can't (yet) convince myself is an allowed pericyclic reaction on the grounds of orbitals.

  Can the Smirnov-Zamkow be satisfactorily described as a pericyclic process (i.e. can it be drawn in such a way as to obey the thermal Woodward Hoffman rules), or is it best described as a stepwise process?

References:

[*]: The reaction was so strange that allegedly R.B. Woodward was left 'open-mouthed, with nothing to say' when he first read the report of the reaction taking place, under such 'mild' conditions, if a set of second year undergraduate notes are to be believed.

[**]: The reaction is actually referenced in Conservation of Orbital Symmetry by Woodward and Hoffman, they actually give a slightly different explanation to Fleming/the lecture notes I first encountered this reaction in.

[1]: Fleming, I. Molecular Orbitals and Organic Chemical Reactions (Reference Ed.), Wiley: 2010

[2]: I. V. Smirnov-Zamkov, Doklady Akad. Nauk SSSR, 1952, 83, 869 (Chem. Abs., 1953, 47, 2711g); I. V. Smirnov-Zamkow and N. A. Kostromina, Ukrai. Khim. Zh., 1955, 21, 233 (Chem. Abs., 1956, 50, 9302e); R. Criegee and A. Moschel, Chem. Ber., 1959, 92, 2181.   (I also don't speak Russian to read some of these)

I was going to write a comment, but I think this is good enough for an answer.

The issue that symmetry rules forbid a [2+2] thermal cycloaddition if the addition is suprafacial-suprafacial.

However, you can achieve a suprafacial-antarafacial [2+2] thermal cycloaddition. This is symmetry allowed because the antarafacial symmetry is exactly opposite suprafacial symmetry.

I'm not familiar with your exact example, but I am familiar with [2+2] thermal cycloadditions of ketene. It is thought to proceed via suprafacial-antarafacial additions.

See for example: Burke, L. A. J. Org. Chem., 1985, 50, 3149.


Ostwald Law of dilution makes no sense According to wikipedia

the Ostwald's dilution law says that for a weak acid
 $K_{d}={\cfrac {[A^{+}][B^{-}]}{[AB]}}={\frac {\alpha ^{2}}{1-\alpha }}\cdot c_{0}$
when I read the derivation of it ,I knew something was wrong
specially this line :

  If α is the fraction of dissociated electrolyte, then αc0  is the concentration of each ionic species

here is how i see it :

$AB\rightleftharpoons A^{+}+B^{-}$
each Molecule of AB gets separated into A and B when put into water
so we can make a small table

|                     |     AB    |   A    |   B     |
| Before dissociation |   1 Mol   | 0 Mol  | 0 Mol   |
| Concentration Before| 1/V Mol/L = C acid Mol/L | 0 Mol/L| 0 Mol/L |

| After dissociation  | (1-X) Mol | X Mol | X Mol |
| Concentration After | (1-X)/V Mol/L | X/V Mol/L | X/V Mol/L |

V = Volume of the Acid AND Water,
After dissociation is an equilibrium between AB and (A,B), so we can apply the law of mass action
$$
K_{dilute}=\frac{\frac{\alpha}{v}\cdot \frac{\alpha}{v}}{\frac{\left(1-\alpha\right)}{v}}
$$
and then
$$
K_{dilute}=\frac{x^2}{v\cdot \left(1-x\right)}
$$
and as we referenced the AB Concentration Before dissociation with $c_{acid}$ = 1/V , then
$$
K_{dilute}=\frac{x^2\cdot c_{acid}}{\left(1-x\right)}
$$
and from here the non-sense starts,
the law says that X is a very tiny value so (1-x = 1) making x = 0 but forgetting that this way $x^2$ is even a tinier value and would also would also = 0,then $k_b$ = 0 , but in my opinion the law should be
$$
\alpha ^2+\frac{K_{dilute}}{C_{acid}}\cdot \alpha -\frac{K_{dilute}}{C_{acid}}=0
$$
to get $\alpha$, and to calculate the concentration:
$$
\left[A^+\right]=\left[B^-\right]=\frac{K_{dilute}\ \cdot \left(1-\alpha \right)}{\alpha }
$$
Any explanation would be appreciated

You made $K_b$ = 0. That is where you were wrong. Taking approximation of Neglecting $\alpha$ with respect to 1 only gives us advantage of not solving the quadratic equation.

From there you can write $\alpha = √(K_b/C)$ here you can see $\alpha$ comes inversely proportional to $√C_o$ or we can say directly proportional to $√V$. From here you can deduce if $V$ approaches infinity $\alpha$ approaches 1. And that's the law.


Difference between baking soda and baking powder I learnt that the chemical symbol for baking soda and baking powder is NaHCO3. However, when I do a simple experiment which is put vinegar into both substances. I can get different observations.

When I put vinegar into baking soda, I get a fast and fuzzy reaction. But when I put vinegar into baking powder, instead of a fast and fuzzy reaction, I get a slow and bubbly reaction.

So, my question is what properties do each of the both substances have to make different reactions with vinegar although their chemical symbols are the same?

As LDC3 says, Baking soda is simply sodium bicarbonate, $\ce{NaHCO3}$. When one adds an acid, the following reaction occurs: $\ce{HCO3- + HA -> H2CO3 + A-}$ and the carbonic acid releases carbon dioxide: $\ce{H2CO3 -> CO2 + H2O}$ (these are both equilibrium reactions, but as you can observe when you add vinegar, a lot of gas is produced and the gas is generally lost, pushing the equilibrium towards making carbon dioxide).

Baking powder is a mixture of sodium bicarbonate and one or more weak acids. When dry, no reactions occur, but when wet, the components dissolve and can react. The observed differences when mixing vinegar with baking powder probably result from the fact that there is less sodium bicarbonate in a given volume of baking powder than baking soda, and baking powder often contains starch, which can influence the properties of the added liquid, giving the bubbles a different appearance.


How to differentiate the notation of a compounds charge and an anions oxidations state? I have had the case a few times. Being given something of this form:

$$\ce{Cr2O7^2-}$$

How do I know whether the $2-$ is referring to the oxidation of the oxygen anion or the charge of the whole compound?

make sure to read up about that topic. The concept of charge and oxidation state are closely related but aren't the same.

Usually oxidation state is written with roman numbers (that's how I learned it) and are not explicitely written like in the example you showed.

Charges on the other hand are always written (if mentionned) on the upper right side of the anion.

I am far from being an expert, but thinks this might be an answer to your question.


Negative oxidation state Why doesn’t aluminum have negative oxidation states? In almost all compounds of aluminum  which I know of has positive  oxidation state.

It has valence EC [Ne] 3s2 3p1 , my teacher told it usually forms +1 and +3 oxidation state by completely sharing p orbital electron or both s and p orbital completely(and +2 o.s. is always average of aluminum)

So why can’t aluminum or any 13th group element for that matter accept electron and make its p orbital half filled i.e form compounds with negative oxidation state?

Aluminum does assume negative oxidation states, see the list of elemental oxidation states in Wikipedia. Aluminum assumes a negative oxidation state when forming compounds with more electropositive elements like magnesium.

The magnesium compounds referred to above include $\ce{Mg_{17}Al_{12}}$.  In this compound there is enough magnesium to half-fill the valence $p$ orbital on aluminum and thus attain a $-2$ oxidation state for the aluminum, which is the minimum in Wikipedia's list. (Part of the magnesium would then be below $+2$).  However, this is only a formal designation. Magnesium aluminides are far from ionic in character.


Why does the AraC arabinose sensing system only respond to the presence of L-arabinose? I understand that when arabinose is present it interacts with araC and changes it shape, promoting the binding of RNA polymerase with the promoter and GFP is produced. However, I don't understand why the sensing system in E.coli only responds to L-arabinose and no other pentose sugar nor their D-isomer.

I would truly appreciate it if someone could explain why this is.

Your question is really about the origin of specificity in molecular recognition.

The simple answer to this is that two molecules (like the AraC protein and arabinose) bind to each other because they make favorable interactions with each other. ("Favorable" here means that free energy is released in the "reaction" of going from separate molecules to the complex.) There's a number of different types of interactions, but the ones you typically talk about with non-covalent protein binding are van der Waals,  hydrogen bonding, and electrostatics.

The reason AraC binds to arabinose is that it has functional groups in the binding pockets which exactly match three dimensional organization of corresponding atoms in the arabinose molecule (see the experimental structure of the AraC/arabinose complex). You can see a schematic representation of these interactions here.

The thing about these interactions is that they only happen if the three dimensional geometry matches up -- we might write molecules in a 2D schematic form, but molecules are intrinsically 3D. So when you have the hydrogen bond between, for example, Tyr82 and the C4 hydroxyl of arabinose, it can only form that favorable interaction if the C4 hydroxyl ends up in the correct 3D position.

If instead of L-arabinose you have D-arabinose (or L-ribose), then you have a different spatial orientation of atoms. Your C4 hydroxyl no longer goes "down", but instead goes "up". Since Tyr82 is still in the same place, it can no longer form a favorable hydrogen bond to the C4 hydroxyl - the distance is too far. There's no way to put the 3D structure of D-arabinose into the binding pocket of AraC and keep all the interactions. If you lose enough of these interactions (and sometimes one is enough), the total energy of binding is no longer favorable, and there's no driving force for binding.

Keep in mind that it's not always an all-or-nothing thing. It all depends on which interactions are destroyed by changing the chemical structure of the ligand. For example, AraC has no difficulty binding D-fucose. The difference between L-arabinose and D-fucose is just an addition of a C6 methyl group. For AraC this addition doesn't disturb the existing interactions enough to significantly perturb them. Therefore, you still get a favorable energy of interaction when the AraC binds to fucose.


Acidity of naphthol vs hydroxynaphthoquinone I have the following two compounds whose acidity I need to compare. I tried drawing out the resonating structures but I seem to not be able to draw final conclusion.

I know that the conjugate base of Q has more resonance structures, but R is still more acidic according to the book, with its resonance looking like this:

Why such resonance makes it more acidic?

For naphtol both the acidic and the basic forms have many resonance structures along the two fused benzene cycles. However, in the case of 2-hydroxy-1,4-naphthoquinone, the basic conjugated form creates novel resonance structures that engage the double bond of the quinone cycle as seen in the written resonance structure in the question. The novel resonance structures of the basic forms will stabilize the basic form.
Therefore, the hydroxyl group in 2-hydroxy-1,4-naphthoquinone has higher tendency to lose the hydrogen ion than the hydroxyl group in naphtol. This explains the higher acidity of 2-hydroxy-1,4-naphthoquinone in comparison with naphtol.


Are imine or amine bonds similar to ethereal bonds, from the perspective of an enzyme? I'm interested in enzymes that can cleave (hydrolyse or however) an ethereal bond ($\ce{C-O-C}$). Enzymes can evolve to act on chemicals/bonds that are similar to what they previously acted on, so enzymes with similar functions are also interesting.

I've been told by a couple of chemists that carbon-nitrogen-carbon bonds are similar to ethereal, so I was wondering if, and in what way that is true. Are the bonds in an imine group ($\ce{C-N=C}$) or a amine group ($\ce{C-NH-C}$), more similar to those in an ethereal group compared to a $\ce{C-C}$ pair, for example?

I guess from the perspective of an enzyme, if you were approaching the imine from the direction of the lone pair, that would feel pretty similar to an oxygen maybe?

This does depend a lot on what the enzyme does and what it expects. Take the following three structures, one from an ether (e.g. diethyl ether), one from an amine (diethyl amine) and one from an imine (N-ethylethanimine).

Scheme 1: Ether linkage, amine linkage and imine linkage (from left to right) including hydrogen atoms on adjacent carbons.

When looking for a ether, an enzyme may well attempt to form a hydrogen bond to the ethereal oxygen. This might work in the case of an amine — there is a spare lone pair on nitrogen that can accept hydrogen bonds, but the additional hydrogen present can present ordering and space problems —, but would in the case of an imide.

However, note that the carbons next to ether and amine bonds have four atoms attached to them (two hydrogens and another carbon each), while the imine has only one hydrogen. The $\ce{C=N}$ bond is planar, meaning that the configuration of the neighbouring atoms is also relatively fixed. Thus, the imine may have a pretty different shape — and may not fit into the enzyme’s active pocket. This might actually also mean that the imine cannot accept the hydrogen bond offered by the enzyme because it cannot orient itself favourably.

As I said above, it depends on the enzyme whether it cares about hydrogen bonding more or whether the overall shape is more important. In general, I would consider an amine to be closer to an ether structurally, and therefore more likely to be accepted. Possibly, the amine hydrogen can point away, allowing for the amine to accept a hydrogen bond nonetheless. In that case, the amine would be the clear winner.

All these heteroatom bonds are clearly different from $\ce{C-C}$ bonds due to the bonds’ inherent polarity (electronegativity of $\ce{O}$ and $\ce{N}$).


Can precipitation reactions attain equilibrium? While studying about the characteristics of Chemical Equilibrium, I came across a point in my book which read

A chemical equilibrium can be established only if none of the products is allowed to escape out or separate out as a solid (precipitate)

I understand that if one of the products is a gas and escapes out the reaction won't be able to proceed in the backward direction since that gas is no more a part of the system.
But what I'm not able to get is why it's written that precipitation reactions can't attain equilibrium.
Is there no precipitation reaction which can attain equilibrium?
Or is it that the precipitate can't react back chemically with the solution to produce the reactants in reverse?
My teacher told me about about a reaction
$$\ce{Ca(HCO3)2(aq) + heat -> CaCO3 +CO2 +H2O}$$
and on cooling the solution and bubbling $CO_2(g)$ through it the $CaCO_3(ppt.)$ reacts back to form calcium bi-carbonate.
I'm still a bit confused as the reverse reaction doesn't happen spontaneously. Moreover if the reaction is carried out in a closed vessel and constant temperature (after heating), will the reaction reach equilibrium?
edit
in the answer recently provided, the solid-solution equilibrium between $\ce{AgCl}$ and $ \ce{Ag+}$ $\ce{Cl-}$ ions is discussed. But I want to know if in a reaction like $$\ce{AgNO3(aq) + NaCl(aq)\to AgCl(s,ppt) + NaNO3(aq)}$$
the precipitate $\ce{AgCl(s,ppt)}$ can react with $\ce{NaNO3(aq)}$ in the solution to give the reactants?
Can it reach chemical equilibrium?
I've seen this and this but couldn't understand.
Any of your help is of great value to me.

The statement is inaccurate. A better statement would be something like:

A chemical equilibrium can not be established if one of the products is continuously removed (i.e. its concentration always decreases).

If one of the products is a gas and can mix with earth's entire atmosphere, that would make its concentration (or partial pressure) drop to almost zero. In order to establish an equilibrium, however, all concentrations have to be non-zero. Otherwise, the equilibrium constant will not be equal to the reaction quotient, and there will continue to be a net reaction.
In case of the precipitation, the solute is not continuously removed. At some point, the solution will no longer be supersaturated, so then the concentration of solute will be constant.

My teacher told me about about a reaction
$\ce{Ca(HCO3)2(aq) + heat -> CaCO3 +CO2 +H2O}$
and on cooling the solution and bubbling CO2(g)
through it the CaCO3(ppt.) reacts back to form calcium bi-carbonate.

Once you change the temperature, you don't expect the reaction to stay at equilibrium. So this does not support either the original statement or the opposite statement.

But I want to know if precipitation reactions can attain equilibrium. Why does the book state otherwise?

Yes, it can. Here is an example:
$$\ce{Ag+(aq) + Cl-(aq) <=> AgCl(s)}$$
The equilibrium constant expression for this reaction is:
$$K = \frac{1}{[\ce{Ag+(aq)}][\ce{Cl-(aq)}]}$$
(I could have written the chemical equation in the other direction, and then the equilibrium constant would have been the solubility product.) So the product is a solid, and its concentration does not change. The silver and chloride ion concentrations will drop until the equilibrium has been reached.
Here is another set of reactions:
$$\ce{CoCl4^2-(aq) + 6 H2O(l) <=> Co(H2O)6^2+(aq) + 4 Cl-(aq)}\tag{1}$$
$$\ce{Ag+(aq) + Cl-(aq) <=> AgCl(s)}\tag{2}$$
Here, the equilibrium of the first reaction will be far on the right side because the second reaction removes a lot (but not all) chloride ions from the solution if the silver ions are in excess of the chloride ions. Still, the system will reach equilibrium.

Why does the book state otherwise?

I don't know. Maybe they consider an equilibrium where one of the species is at extremely low concentration not an equilibrium. Or maybe it is a typo.
It is great to have a critical mindset when reading books and hearing people talk about chemistry. Even if all the statements are true, it helps you learn.
edit

I want to know if in a reaction like
$$\ce{AgNO3(aq) + NaCl(aq)\to AgCl(s,ppt) + NaNO3(aq)}$$
the precipitate $\ce{AgCl(s,ppt)}$ can react with $\ce{NaNO3(aq)}$ in the solution to give the reactants?

Yes. I would rather write it like this, though:
$$\ce{Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) <=> AgCl(s) + Na+(aq) + NO3-(aq)}$$
and then leave out the spectator ions:
$$\ce{Ag+(aq) + Cl-(aq) <=> AgCl(s)}$$
The following "reaction" might serve as illustration why leaving the spectator ions in the equation is not a good idea in preparation for writing the equilibrium constant expression:
$$\ce{KCl(aq) + NaNO3(aq) <=> KNO3(aq) + NaCl(aq)}$$
For this "reaction", the concentration of species is undefined. Writing the equilibrium constant would be strange because how can I tell apart $\ce{KCl(aq) + NaNO3(aq)}$ from $\ce{KNO3(aq) + NaCl(aq)}$ if they fully ionize in solution? If I show the ions separately, it becomes clear that the "reaction" is not a reaction:
$$\ce{K+(aq) + Na+(aq) + NO3-(aq) + Cl-(aq) <=> \mathrm{the\ same???}}$$

I've seen this and this but couldn't understand.

Those two Q&A's are quite confusing, I agree.


Y-axis of the reaction co-ordinate graph In the graph of the reaction co-ordinate, is the $y$-axis that is used  $∆H,$ $∆U$ or $∆G$?

Single molecule reaction

Some diagrams refer to a single molecule reacting. If that is the case, several quantities are not defined or ill-defined:

pressure
entropy of mixing
temperature

In those diagrams, the best label would be potential energy, and the energy difference between reactant and transition state should be labeled activation energy.

Ensemble reaction

If the diagram refers to an ensemble reacting (reaction mixture with sufficient number of molecules to have macroscopic properties), the y-axis could correspond to Gibbs energy, enthalpy or internal energy.

If the y-axis represents Gibbs energy, the energy difference between reactant and transition state should be labeled $\Delta G^\ddagger$. Those diagrams make sense when discussing how equilibrium constants remain unchanged but reaction rates change when a catalyst is added.

If the y-axis represents enthalpy or internal energy, the energy difference between reactant and transition state should be labeled $E_a$ for activation energy. Those diagrams make sense when discussing reaction rates in the context of collision theory.

What is most popular?

Looking at about 50 images of reaction energy diagrams found in a google image search, I found a wide variety, and lots of mixing and matching (i.e. incompatible quantities in a single diagram). This tells you that the diagram is conceptual, and the concepts are not always clearly articulated in diagrams.


Difference in leaving group ability due to variation in nucleophiles Wikipedia writes:

  There can still exist contextual differences in leaving group ability in the purest form, that is when the actual group that leaves is not affected by the reaction conditions (by protonation or Lewis acid complexation) and the departure of the leaving group occurs in the rate determining step. In the situation where other variables are held constant (nature of the alkyl electrophile, solvent, etc.), a change in nucleophile can lead to a change in the order of reactivity for leaving groups. In the case below, tosylate is the best leaving group when ethoxide is the nucleophile, but iodide and even bromide become better leaving groups in the case of the thiolate nucleophile.

Why is it so?As far as I know, leaving group ability is basically how stabilized the leaving group is when it exists as an independent species. I can't see why it depends on the nucleophile

The data which you are using to explain a groups tendency to act as a good leaving group is via a nucleophilic substitution reaction.  Nucleophilic sub reactions are affected by 3 things mainly: the nucleophilicity of the nucleophile, the strength of the bond being broken and how easy is it for the leaving group to leave.

This means if you use a better nucleophile the substitution reaction will happen faster. so in the case of your data, you are comparing leaving groups using "relative rates". So the nucleophilicity of the nucleophile will end up impacting the rate. So by looking at two different nucleophiles you are looking at each nucleophiles nucleophilicity and the leaving group.


What will be the product of this reaction? We have an alpha beta unsaturated ether, along with a keto group in the same compound, which is being treated with a grignard reagent shown (assume the grignard reagent is in excess). Will it only attack the carbonyl group, or are things different here?

This reaction is one of a special kind. The one and onlyend product is 3-hydroxy,3- methylcyclohexan-1-one. As the Grignard reagent is in excess, and it is an $\alpha$- $\beta$ unsaturated ketone, there is a possibility of undergoing two different type of reaction mechanisms. But Surprisingly, both will give the same ultimate end product after two steps.   The major mechanism through which the reaction should undergo, is the direct nucleophilic attack of the grignard reagent at the electrophilic carbonyl carbon. As, Grignard reagent is a hard base , it's attack to hard acid centre of carbonyl carbon is more preffered. So, first it will undergo this addition after which it acidic hydrolysis is done, which cleaves the ether and forms an enol which tautomerises to give the final end product ketone. The reaction scheme is shown below,

Now, the minor mechanism is the process of conjugated addition of the methyl group to that $\alpha$-$\beta$ unsaturated ketone. But again after tautomerisation and  hydrolysis it actually forms the same compound, which is 3-hydroxy,3-methyl cyclohexan-1-one. The reaction scheme is shown below,

So, it is seen that however the presence of excess reagent may interfere with the reaction mechanism but the end product is $100 $$\%$ $3$-hydroxy,$3$-methylcyclohexan-$1$-one.


Can a molecule contain a linear series of conjugated pi bonds twisted through 180 degrees? Imagine a long chain of carbon atoms connected by single and double bonds in conjugated fashion. All of the p orbital pi bonds would normally be aligned to either side of the carbon atoms forcing the molecule to adopt a flat structure.
Would it be possible to take this structure and twist it slightly so as to form a long helical shape?

The bond overlap would not be as good but if the chain were long enough I would have thought the bond weakening would not be excessive.
Obviously both ends would need to be firmly bonded to some other inflexible linking components such as a large protein molecule or a long chain of aromatic rings.

Has anything like this ever been synthesised? Is it even possible?

Here is an example either end of this structure would be connected to a larger molecular segment that would hold it in place and prevent it from untwisting. I want to give this normaly flat structure a 180 degree twist.

The material you imagine has been a target for Natta, best known together with Ziegler for they work on isotactic polymers.

Natta was interested in what you describe - polyacetylene - as the extended conjugation would have likely led to unusual electronic properties.

Calculations and models were around theoretically, before and in  parallel to the actual synthesis.

The material prepared by Natta was however insoluble and basically no work up was possible.  This is not totally surprising if we look at the structure.

After decades, the fortuitous use of thousand time the amount of catalyst led to the formation of nice shiny films of polyacetylene. Still insoluble but so different from commons residues to attract sufficient attention to cause the gathering of brilliant scientists and finally the emergence of an entire field, that of conducting (or better, conjugated) polymers and so called plastic electronics.

https://www.nobelprize.org/nobel_prizes/chemistry/laureates/2000/

https://en.m.wikipedia.org/wiki/Polyacetylene

This is to answer the final part of the question.

Concernig the first part, if I understand correctly your picture, I can tentatively answer in an apparently convoluted way.

As far a chain of alternating single and double bonds is considered, and without accounting for delocalisation, we have basically free rotation around the single bound. As such, any statistical coil can be obtained.

Conversely, if one imagine a "metal like" structure with full electron delocalisation, the chain is than stiff.

The reality is somewhere in the middle. Real conjugated polymers should be regarded as delocalized pi systems (which at the end is the reason of the huge interest on them) inerhently not much flexible and not very soluble.

Still, a long chain is susceptible to adopt a variety of elical or coiled conformations as a result of collective little departures from planarity. Ideally, you can turn very rigid rod on themselves providing they are long enough.

One cannot apply Flory theory as in polyethylene, but adaptations exist to account for rigidity due to order bond > 1 in conjugated polymers.

Also let me note that independent of the extent of delocalisation (also known as effective conjugation) a linear series of sp2 hybridized C atoms can close to a ring - no need for examples here).

Additional notes:

1- Solubility in common solvents is a major challenge towards real  application of conjugated polymers. A common strategy is the side substitution with moderately long alkyl chains. This in generak results in improved solubility via an extra entropy term.

2- polyanilyne far preceded the interest on polyacetylene. This systems is more complicated and in a way interesting, but less as far organic electronics is in scope.

3*- a detailed and chemically exhaustive paper appeared well before the paper mentioned in the linked Nobel prize. It has been published in the Journal of the  Australian chemical society and it details the preparation and properties of polythiophene.

*(Making products and selling them are two different tasks. This is the lecture I got)


Steric effect of t-butyl group on SN1 on a cyclohexane ring This is a question from GRB Kota Question Bank Organic Chemistry, Chapter 3, Reasoning type, Q. 4:

Assertion: trans-1-t-butyl-4-chlorocyclohexane is less reactive than cis-1-t-butyl-4-chlorocyclohexane towards an SN1 reaction.
Reason: More the steric factor near the leaving group, the higher the leaving group tendency.

The answer given is:

 Both the assertion and reason is correct and the reason is the correct explanation for the assertion.

The answer given feels wrong.
My reasoning for saying so is because the rate determining step in an SN1 is the formation of a carbocation. Due to this, the steric effect seems irrelevant. However, when the intermediate is same for both compounds undergoing substitution, I assume that the compound having greater stability would be less reactive since the threshold energy required would be greater.
Using the above statement, we can say that trans-1-t-butyl-4-chlorocyclohexane would be less reactive since both the groups are in equatorial position making it more stable whereas in its cis-isomer the t-butyl group would be equatorial whereas the chloride group would be axial.
Due to this, I presume that the assertion is correct. However, the reason seems vague as it does not mention whether the reaction taking place is SN1 or SN2 since this could change with respect to the reaction involved.
Is there anything wrong with my above reasoning? Can the validity of the reason be proved/disproved beyond doubt?

I would argue that the reasoning makes sense.
There are some important things to discuss here.
The cis isomer ($C$) and the trans isomer ($T$) will both react via the same carbocation intermediate. This means the energies of the intermediate are the same. Chloride has an A-value of 0.43, so $C$ is higher in energy than $T$. Since the formation of cation is endothermic, we may invoke the Hammond Postulate: the first step has a late transition state that structurally looks like the intermediate and is similar to it in energy. Finally, we conclude that the transition state for the reaction from $C$ has a lower barrier and thus a faster reaction.

Now, there is no way to "prove" that this or any other rationale is correct. There is no way to prove any model is correct. That's not how science works. You can only have consistent models that are good enough for now. They may even be "right," but you will never know. On the other hand, you could a reject the model based on inconsistency with experimental evidence, but there's no contradictory experimental evidence here.


Quick and simple explanation of molar mass, molecular mass and atomic mass As a physics student, I hardly deal with such quantities and when I do, I blunder through.
I think it’s a good time to be given a good explanation for these since I’m in a class of Nuclear physics. I have tried googling but it seems that some sources are ‘misleading’ and most have more than 1 terminology for the same thing!
Horrifying but real.

Please tell me what is the difference between molar mass, molecular mass and atomic mass and what are their associated twin terminology.

The quantity molar mass (quantity symbol: $M$) for a pure sample is defined as
$$M = m/n$$
where $m$ is mass and $n$ is amount of substance.
The dimension of the molar mass is
$$\dim M = \mathsf{M}\;\mathsf{N}^{-1}$$
The coherent SI unit for molar mass is ‘kilogram per mole’ (unit symbol: $\mathrm{kg/mol}$).
(The usually used unit is gram per mole ($\mathrm{g/mol}$) rather than kilogram per mole ($\mathrm{kg/mol}$).)

The quantity relative molecular mass (quantity symbol: $M_\mathrm{r}$) is defined as the ratio of the average mass per molecule or specified entity of a substance to 1/12 of the mass of an atom of the nuclide $\ce{^12C}$; i.e. the relative molecular mass is the ratio of the mass of a molecule to the unified atomic mass constant:
$$M_\mathrm{r} = m_\mathrm{f}/m_\mathrm{u}$$
where $m_\mathrm{f}$ is mass of entity (molecule or formula unit) and $m_\mathrm{u}$ is the unified atomic mass constant.
The relative molecular mass is a quantity of dimension one (for historical reasons, a quantity of dimension one is often called dimensionless):
$$\dim M_\mathrm{r} = 1$$
The coherent SI unit for relative molecular mass is the unit one (symbol: $1$).
For historical reasons, the IUPAC accepts the use of the special name ‘molecular weight’ for the quantity relative molecular mass. The use of this traditional name is deprecated.

For atoms, the relative molecular mass $M_\mathrm{r}$ is the relative atomic mass, and the quantity symbol $A_\mathrm{r}$ may be used. The quantity relative atomic mass is defined as the ratio of the average mass per atom of an element to 1/12 of the mass of an atom of the nuclide $\ce{^12C}$:
$$A_\mathrm{r} = m_\mathrm{a}/m_\mathrm{u}$$
where $m_\mathrm{a}$ is the atomic mass (see below) and $m_\mathrm{u}$ is the unified atomic mass constant.
The relative atomic mass is a quantity of dimension one:
$$\dim A_\mathrm{r} = 1$$
The coherent SI unit for relative atomic mass is the unit one (symbol: $1$).
For historical reasons, the IUPAC accepts the use of the special name ‘atomic weight’ for the quantity relative atomic mass. The use of this traditional name is deprecated.

The quantity atomic mass (quantity symbol: $m_\mathrm{a}$) is defined as rest mass of a neutral atom in the ground state.
The dimension of the atomic mass is
$$\dim m_\mathrm{a} = \mathsf{M}$$
The coherent SI unit for atomic mass is ‘kilogram’ (unit symbol: $\mathrm{kg}$).


Failed synthesis of aspirin due to excess sulfuric acid Last week, I tried synthesizing acetylsalicylic acid - the reaction is shown below - using $\ce{H2SO4}$ as a catalyst. However, as the title suggests the synthesis failed as I used too much $\ce{H2SO4}$ - approximately four times more than the prescribed volume. Needless to say, I had to redo the synthesis.

But now I try to understand why the synthesis failed - I know that catalysts are used in small amounts, but I thought that much of it would only lead to equilibrium state being reached faster.

I think that maybe the acidity of the reaction environment could've somehow explained why the synthesis didn't take place. But I don't understand why - shouldn't more protons in the environment lead to faster protonation of the carbonyl oxygen of acetic anhydride? Could you please explain it to me?

But I don't understand why - shouldn't more protons in the environment lead to faster protonation of the carbonyl oxygen of acetic anhydride? Could you please explain it to me?

The short answer is "side reactions".  Without more information about your exact experimental conditions, or even better, data about the obtained yields of the desired product and of byproducts, it isn't possible to explain conclusively what went wrong, but here are some ideas.

Acetyl sulfate formation.  In the comments you you implied that you used anhydrous sulfuric acid.  Under those conditions, acetic anhydride can undergo solvolysis to yield acetyl sulfate and water.  $$\ce{CH3COOOCCH3 + 2H2SO4 <=> 2 CH3COOSO3H + H2O} $$  Acetyl sulfate might be a less active acetyl transfer reagent than acetic anhydride, or the water formed may hydrolyze some of the acetic anhydride or even the formed acetyl sulfate.
Dehydration/sulfation of the starting material instead of acetyl transfer.  If you really did use anhydrous sulfuric acid, it is a powerful dehydrating reagent.  It could be possible that you are sulfating the alcohol or even the acid group (or both?) of the substrate, forming either o-sulfoxybenzoic acid (sulfation at $\ce{-OH}$ group) or sulfosalicylic acid (sulfation at $\ce{-COOH}$ group).

The heat of mixing of concentrated sulfuric acid with water is about $-74~\mathrm{kJ~mol^{-1}}$, judging from Figure 34b of a NIST treatise on the subject.  The heat of reaction for hydrolysis of acetic anhydride is only $-41~\mathrm{kJ~mol^{-1}}$.  Thus under anhydrous conditions, I think sulfated products will be more thermodynamically favored than acetylated products.  Sulfated products will be far more soluble in water or sulfuric acid than would be acetylated products.

I would think that using more water in the reaction would help.  The reason is that the heat of mixing of sulfuric acid is a very strong function of water concentration, but the heat of hydrolysis of acetic anhydride is not.  So raising the concentration of water a little bit will disfavor sulfation significantly by lowering the heat of mixing of sulfuric acid, but will leave the heat of hydrolysis of acetic anhydride the same.  It's worth noting that this argument is purely thermodynamic, not kinetic.  But I don't think sulfation reactions are kinetically hindered under ambient conditions, so although it isn't always true, I think thermodynamic arguments are valid in this case.


entropy of vaporization vs entropy of fusion Is the entropy of vaporization greater than entropy of fusion, for elements? And in which case they can be same, if it so happens?
I have asked this here because I couldn't find any satisfactory answers anywhere on the Internet and most of them were very limited in their description.

Here I'm assuming you are interested in comparing the magnitude of $\Delta S^{\circ}_{fus}$ at $T_{fus}$ with that of $\Delta S^{\circ}_{vap}$ at $T_{vap}$, for the elements in their standard states (so, for instance, hydrogen would be $\ce{H_2}$ rather than $\ce{H}$).
Like you, I was unable to find a comparative tabulation of these values on the internet.  Fortunately, one can readily generate such a table using Wolfram Mathematica's chemical database.
For each element:
$$\Delta S^{\circ}_{fus} \text{ at } T_{fus} = \frac{\Delta H^{\circ}_{fus}}{T_{fus}}$$
$$\Delta S^{\circ}_{vap} \text{ at } T_{vap} = \frac{\Delta H^{\circ}_{vap}}{T_{vas}}$$
Wolfram has the above data for all of elements 1–93 (hydrogen through neptunium), except for helium (which can't be solidified at standard pressure, which is 1 bar), astatine, and francium.*
Here I've plotted $|\Delta S^{\circ}_{vap}|$ vs. $|\Delta S^{\circ}_{fus}|$* for these 90 elements, and added a y = x line. From the placement of the points relative to this line, you can see that all except one of the elements have $|\Delta S^{\circ}_{vap}| > |\Delta S^{\circ}_{fus}|$.
That single exception is hydrogen, for which:
$$|\Delta S^{\circ}_{fus}| \text{ at } T_{fus} = 39.8 \frac{J}{mol K}$$
$$|\Delta S^{\circ}_{vap}| \text{ at } T_{vap} = 22.3 \frac{J}{mol K}$$

*Note, however, this complication:  Most, but not all, of these measurements were done at standard pressure (1 bar). For instance:
"When heated at standard atmospheric pressure, arsenic changes directly from a solid to a gas, or sublimates, at a temperature of 887 K. In order to form liquid arsenic, the atmospheric pressure must be increased. At 28 times standard atmospheric pressure, arsenic melts at a temperature of 1090 K. If it were also measured at a pressure of 28 atmospheres, arsenic's boiling point would be higher than its melting point, as you would expect."
https://education.jlab.org/itselemental/ele033.html


Are these reaction mechanisms correct? Are these reaction mechanisms correct for Ammonia and Bromide anions acting as a base respectively?

Except for the missing equilibrium arrow (these reaction are acid-base equilibria), they are fine. Or do you have any objections :) (what makes you doubt) ?

Also take @jerepierre's comment in account. This would probably be considered wrong if at an exam.


Why does lithium forms no ethynide on reacting with ethyne? Sodium Metal when treated with ethyne forms mono and di sodiumethynide why doesn't  lithium ?

Sodium alkynide was formed by reacting sodium and acetylene but in case of lithium alkynide, it can be formed by reacting acetylene with organolithium reagents like BuLi, Lithium amide, LiHMDS. From wikipedia:

  Lithium amide, LiHMDS, or organolithium reagents, such
  as butyllithium, are frequently used form lithium acetylides:

  $\ce{H-C#C-H + BuLi ->[THF][-78℃] Li -#-H + BuH}$

The above process is well described in this paper which @Tyberius linked in the comment section.


Bonds within complex anions Many groups of bound atoms complexively bearing a (negative) charge are called complex anions. Examples are $\ce{NO3-}$, $\ce{CO3^2-}$, $\ce{BO3^2-}$, $\ce{CrO4^2-}$, $\ce{Cr2O7^2-}$, $\ce{MoO4^2-}$, $\ce{WO4^2-}$, $\ce{PO4^3-}$, $\ce{AsO4^3-}$, $\ce{VO4^3-}$ and the several, sometimes polymeric, complex anions occurring in silicate and borate minerals.

I apologise if I am asking something trivial, but I cannot find a source, except for silicates, for which I read that the $\ce{Si-O}$ bonds are basically covalent, explaining whether the bonds within complex anions are covalent in general.

Are they always covalent? If they are not covalent in general, are they in the examples I have written?

All complex anions you noted that include only non-metal atoms are always considered covalent. Except for a few extreme examples including fluorine and hydrogen, non-metals only form predominantly covalent bonds between themselves.

It gets a little more tricky when considering those anions whose central atom is a transition metal. In this case at first approximation, we can consider the anion a coordination compound and the coordinate bonds within coordination compounds are typically considered more covalent. Moving on to a second approximation clarifies the picture: the formal oxido-ligands are in fact very strongly attached to the metal and do not undergo ligand substitution. This means that they are even more covalent than other coordination complexes.

Therefore, it is typically safe to consider the bonds within complex anions to be predominantly covalent.


Why is it only electrons that take part in chemical reactions? I learnt that only electrons take part in chemical reactions and the protons do nothing (neutrons are electrically neutral, so they do nothing). Why is this so if the protons are positively charged and without them, electrons in the other atom won’t be attracted (and since electrons are like charges I do not think they would interact with each other)?

The nature of an electron is more complex than a single charged particle. Once you get down to the quantum level, our intuition about how matter interacts with other matter fails.

A simple answer to your question is that electrons occupy a much larger volume than nuclei. When two atoms get in close proximity to one another, their electrons will "bump into one another" and interact far before their nuclei will.

You are also correct that the nucleus of a neighboring atom is essential for the electrons to interact with one another. If there was no positive charge to attract another electron, a bond would not be formed. The point your instructor was trying to get across is that in most (excluding nuclear reactions) chemical reactions, electrons are the species moving around and forming/breaking bonds.

If you would like a more in-depth discussion on the true quantum nature of electrons and their interactions, you should check out Mulliken's Nobel Lecture from 1966. It is a great introduction.


How does Gaussian 09 normalize the excitation coefficients? When I did excited state calculations with Gaussian 09, I was asking me on how those values are normalized because a simple sum over the squared of the coefficients did not yield 1.

So ... how are those coefficients normalized then?


Tollen's Test with Salicylaldehyde I have read that Tollens' test is given by aldehydes. The aldehyde is oxidised to a carboxylate ion and silver is deposited.
My teacher had given me notes which pointed that salicylaldehyde does not give this test. However, I cannot find any explanation for the same. I have tried to search online and through books but I did not get anything.
I think it could be related to resonance from the −OH group as something similar is mentioned in my notes, but it is not elaborated. Can anyone please explain this?

Of course, the answer is resonance.  As described by Benet et al., a basic medium (pH>10) favours the Tollen's reaction. On the other hand, if we want to dissolve phenol in water, we will need OH- to generate phenolate. It means the hydroxyl group of salicylaldehyde is sensitive to the high pH, and this hinders attacking the other hydroxides to the carbonyl group by further stabilizing it.


Removing aluminum oxide from impure silica We have an assay for a mine. Our goal is to have only silica. Is there a cost effective way to extract the contaminants? The product is a fine powder called diatomaceous earth, or fossil shell flour. Our goal is to use the product to help poor pacific islanders increase their silica uptake and assist in heavy metal detox. Many members of the pacific island nations suffer from heavy metal poisoning. The composition of the powder is:

> 84.7% $\ce{SiO2}$
< 4% $\ce{Al2O3}$
< 2% $\ce{Fe2O3}$
< 0.33% $\ce{CaO}$
< 0.36% $\ce{MgO}$
< 0.5% "other"

We aren't concerned with the calcium, iron or magnesium oxides, just the aluminum oxide.

Is there a simple process to reduce the aluminum oxide levels?
Also, what is the easiest process to convert the silica to orthosilicic acid for increased bio-availability?

You may try the famous Bayer process, which is used for extracting alumina form bauxite. When treated at $160°$C and high pressure with $\ce{NaOH}$, a mixture like yours dissolves $\ce{SiO2}$ and $\ce{Al2O3}$, and lets the other compounds insoluble. This is a first purification. $$\ce{SiO2 + 2 NaOH ->  Na2SiO3  +  H2O}$$ $$\ce{Al2O3  +  2 NaOH + 3 H2O -> 2 Na[Al(OH)4]}$$Later on, bubbling $\ce{CO2}$ in the solution makes $\ce{Al(OH)3}$ to precipitate. $$\ce{Na[Al(OH)4] + CO2  -> Al(OH)3(s) + NaHCO3}$$ Of course $\ce{NaHCO3}$ is immediately transformed into $\ce{Na2CO3}$ by $\ce{NaOH}$. $\ce{Al(OH)3}$ can be eliminated by filtration. And the final solution contains only sodium silicate, apart from the excess of $\ce{NaOH}$. Silica can then be obtained pure by adding  enough $\ce{HCl}$ to destroy $\ce{Na2SiO3}$ and the NaOH in excess.$$\ce{Na2SiO3 + 2 HCl + H2O ->  Si(OH)4 + 2 NaCl}$$ It is advisable to highly dilute the $\ce{NaOH + Na2SiO3}$ solution before adding $\ce{HCl}$. Because the precipitate of $\ce{Si(OH)4}$ is extremely voluminous, and fills up the entire solution.


What actually happens to the volume of a gas in an isolated system? In an isolated system with adiabatic walls in which gas is filled what exactly happens? How do we define the volume of a gas here? Is volume of 1 mole of gas  present in that inelastic container the same as the volume of say 5 moles of the same gas in that container?

The volume of a gas is based on the pressure, temperature and number of moles.
According to the ideal gas law :

$PV = nRT$

Where P is the pressure, V is the volume, n is the number of moles, R is a gas constant and T is the temperature.

In your case, the volume of both containers is the same, thus the volume of both gases are equal. However, assuming the temperature of the system is constant, the pressure that the gases exist must be different :

V, T, R = Constants -> $ n \propto P$

Therefore, the pressure of the container with 5 moles is five times of that of the one with 1 mole.

To actually make a fair comparison of the volumes of both moles of gases, the pressure that they are in need to be the same. So, if 5 moles of the gas is placed in a container that has 5 times the volume of the one with one mole, the pressure would be the same but the volume would be different (5:1).

Here is a helpful video for you to understand the concept: https://www.youtube.com/watch?v=BxUS1K7xu30


Amount of Energy Needed for Boiling a Litre of Water What is the amount of energy (J or Wh) needed to boil one litre of tap-water in a regular kettle starting at room temperature (20°C)?

The calculation itself assuming ideal and standard prerequisites would be interesting as well as factoring in the non-purity of tap-water and a typical kettle performance factor.

It is simple. Density of water is $1\;\frac{\text{g}}{\text{ml}}$ as we have 1 kg water.

Now according to the laws of calorimetry the amount of heat required

  $Q = c \cdot m \cdot \left( {100{\text{ }^\circ\text{C}} - T} \right) + \Delta {h_{\text{vap}}} \cdot m$

Here

$c$ = specific heat capacity of water, which is 1 cal/(g °C)
$m$ = mass of water; i.e. 1 kg
$T$ = the room temperature in °C
$\Delta {h_{\text{vap}}}$ = specific enthalpy of vaporization of water; 540 cal/g

So you will get the amount of energy needed (in cal) to boil a litre of water

To get answer in Joule multiply the answer in cal by 4.184 J/cal. And be careful with units!


Is it wrong to allege that lemons and limes are alkaline? Please ELI5; I've never studied chemistry. I read about  alkalinity v. basicity here and on Sciencing.com.
Lemons and limes are doubtless acidic.

Lime: The pH range of lime is 2.00 to 2.35.
Lemon: The pH range of lemon is 2.00 to 2.60. This makes lemon slightly less acidic than lime.

'The pH levels of citrus fruits can cause damage to tooth enamel. Typically, the lower the pH, the more damage is done to the teeth. Oranges have a higher pH level, at 4.35, and are less damaging to teeth than lemons at 2.75, limes at 2.88.'

So isn't it wrong for Yuri Elkaim BPHE, CK, RHN

and this unidentified author to allege lemons and limes to be alkaline?

Grouping of food items into various classes is a very old idea of Ancient Greek, Alchemists & Ancient Indian School of Medicine (Aryuveda). These ideas are perhaps thousands of years old. From a scientific perspective these classifications don't mean anything today. For example, in those schools of thought, fish has a very "hot" effect on the body but yogurt is "cold". Similarly, citrus fruits are "cold" but dried dates are "hot". Now nobody can define what hot or cold means and what is the criterion?

Similarly, the chart shown in the original posts are all wrong. It is just a modern version of these ancient schools of medicine. I think the major source of confusion in calling limes and lemons is that citrates (salt of citric acid) are indeed alkaline. Many antacid concoctions use solid sodium bicarbonate and citric acid. The moment you add a spoon to water, it starts to fizz vigorously. The pH of such solutions are slightly alkaline. Perhaps whoever made those charts had these ideas in mind.


Syn-addition or anti-addition to fumaric acid? In my textbook, the addition of alkaline $\ce{KMnO4}$ to fumaric acid (trans) is stated as a syn-addition even though $\ce{OH}$ is present on opposite sides in the product.

But,the defination of syn-addition is:

  Addition of electrophile and nucleophile on the available sites on the same side of molecule.

Isn’t this contradictory?

I don’t know what molecule you ended up with, but the way I draw it, I end up with (2S,3S)-2,3-dihydroxysuccinic acid as shown below.

The syn-addition step would technically be the addition of the electrophile permanganate to the double bond. From there, the oxygens are in place and are not moved. If you reorder the structure to the identical zig-zag representation, you should arrive at a clear syn-orientation.

Note that there are two rotations in this: the first step includes a rotation of fumaric acid by $90^\circ$ so that I can draw the five-membered ring in a plain. After cleaving off $\ce{MnO4^3-}$ (which further reacts to $\ce{MnO4^2-}$ in basic conditions) I arrive at the syn-product.

With maleic acid, everything would be the same, except for the orientation of the carboxylic groups in the second and third structures. That would lead to the hydroxyl groups pointing in different directions even though we are still dealing with a syn-addition.


Solubility order for alcohols Order for solubility of isomeric alcohols in a book was given as:

Primary alcohol >Secondary alcohol >Tertiary alcohol
After this I found a question as :

1) Which is more soluble in water?

(a) 3-ethyl-3-hexanol

(b) 2-octanol

Ans:- (a) & the reason was that (a) has more compact alkyl portion than (b). But as (a) is a tertiary alcohol while (b) is a secondary one ,so according to the given order answer should be reverse.

Also while comparing

2-methyl-2-propanol & 1-butanol

Here 1-butanol had higher solubility which matched the given order. I am not understanding what's wrong about the first case. Please explain me.

The reasons that you cite are both important for the solubility of an alcohol. Generally speaking, if the molecule is branched its (hydrophobic) surface exposed to water is less than in a linear molecule. This effect is more pronounced when the number of atoms is bigger. On the other hand the effect of the degree of substitution on the solubility is less pronounced when the molecular size rise.

This would partially justify your second question, but the problem here is that your second question is wrong, 2-methyl-2-propanol is soluble with water in every ratio, while n-butanol is not.


Is cyclic ether oxidised by periodic acid? Is the three member ring of cyclic ether oxidised by periodic acid? As this also has oxygen attached to adjacent carbon so I think that cyclic ether of three members should be oxidised by periodic acid.

I know the mechanism of oxidation of 1,2 diols by periodic acid but ain't able to relate to this question. Please help me out here.

Cyclic ethers usually do not get oxidized by periodic acid. However, epoxides would do only if they were acidified first or use acidified periodic acid in the reaction (Ref.1). In general, acidified periodic acid has been used as a qualitative test for the presence of epoxide moiety in organic compounds (Ref.1). The procedure is a modification of the qualitative test for  glycols (Ref.2). Hence, epoxide must be converted to 1,2-diols first by acid and oxidation by periodic acid must be followed. The test is described as follows:

  Exactly 2 drops of concentrated nitric acid is added to 2 ml. of a 0.5% solution of periodic acid, and 1 or 2 drops of the unknown is added. Water-insoluble unknowns should first be dissolved in 2 ml. of dioxane or acetic acid. The mixture is shaken and 1 or 2 drops of 5% silver nitrate is added at room temperature. A positive test is the appearance of a white precipitate of silver iodate which usually forms immediately, but may require up to 5 minutes for complete precipitation. In all cases a blank should be run for comparison. The test is negative for simple aldehydes, ketones, and alcohols.

References:

R. Fuchs, R. C. Waters, C. A. Vanderwerf, “General Qualitative Test for Epoxides,” Anal. Chem. 1952, 24(9), 1514 (https://doi.org/10.1021/ac60069a047).
R. L. Shriner, R. C. Fuson, Identification of Organic Compounds; 3rd ed., John Wiley & Sons: New York, NY, 1948, pp. 115-116.


Number of water molecules in a box to simulate liquid water I have seen that depending on the simulation approach, different number of water molecules are employed for studying liquid water properties.

32 in early DFT calculations
The electronic structure of liquid water within density-functional theory
The Journal of Chemical Physics 123, 014501 (2005); https://doi.org/10.1063/1.1940612

216 in early classical molecular dynamic simulations

Molecular Dynamics Study of Liquid Water The Journal of Chemical Physics 55, 3336 (1971); https://doi.org/10.1063/1.1676585

256 I'm not sure if this number of water molecules is just for computational reasons $2^8$= 256

So I'm wondering, What is the reasoning behind them?,

Edit:
In other words, If I want to obtain the ambient density of the system and I don't know anything about it (number of  molecules or the size of the simulation box, assuming cubic), what should I do?

As a starting point, the only thing I know is that the size of the simulation box should be chosen considering a number density that corresponds to the experimental density of the system.

Generally speaking, you use as many molecules in a simulation as you possibly can. There are a couple of caveats. For instance, if you wish to do a simulation using an expensive method (relative to a classical potential) like DFT, you will need to use fewer molecules. This has the obvious drawback that when you use periodic boundary conditions, each molecule will interact with its own reflections much more than we would like (which is not at all). Nowadays, we can certainly do simulations with more than 32 water molecules using DFT.

Also, there is no reason to just keep using larger and larger numbers because it is known that interactions between molecules in the liquid die off reasonably quickly with distance, so at some point you stop gaining anything by adding more molecules. All of the long range effects can be handled by using periodic boundary conditions, which works very well if each molecule is far enough from its own image that a molecule does not have short-range interactions with itself.

Another consideration is that the number of molecules you use depends a lot on what property you are interested in studying. Conventional wisdom about the number of molecules needed is usually associated with the number of molecules that give converged average energies. If you are interested in something like diffusion coefficients, however, you may need quite a few more molecules before convergence is achieved. I have read that up to 512, or even 768, can be needed to get good diffusion coefficients around room temperature for water.

It is true that sometimes a specific number of molecules is chosen to match a specific density you want to do simulations at. This, however, isn't usually too big of a deal as you can always change your box size to match the density you are trying to achieve.

There is no physical reason that people choose powers of 2 for the number of molecules other than that computational scientists tend to use powers of 2. The number 216 is probably used because some seminal paper used this number and got good results, so people tend to just stick with the trend.


Predict the order of ionic radius We have arrange the followig sequence  for decreasing order of ionic radius .

$Se^{-2},I^-,Br^{-},O^{-2},F^{-}$

In my book the answer given as
$I^{-},Se^{-2},Br^{-},O^{-2},F^{-}$

But according to me $O^{-2}$ should be greater than F$^-$

As oxygen has two more electrons , so its zeff should decrease more in comparism to fluorine

You're right about the expected order. From largest to smallest the expected order would be:

$\ce{_{53}I^-}\quad$ = [Kr] $4d^{10}\text{ }5s^2\text{ }5p^6$ = [Xe]

$\ce{_{34}Se^{2-}}$ = [Ar] $3d^{10}\text{ }4s^2\text{ }4p^6$ = [Kr]

$\ce{_{35}Br^-}\text{  }$ = [Ar] $3d^{10}\text{ }4s^2\text{ }4p^6$ = [Kr]

$\ce{_8O^{2-}}\text{   }\text{ }$ = [He] $2s^2\text{ }2p^6$ = [Ne]

$\ce{_9F^-}\quad$ = [He] $2s^2\text{ }2p^6$ = [Ne]

You expressed the idea poorly though. $\ce{_8O^{2-}}$ and $\ce{_9F^-}$ have the same number of electrons. However $\ce{_9F^-}$ has an extra proton which should draw the electrons in closer.

There are a lot of ways to measure atomic/ionic size. No idea why the book would have $\ce{_8O^{2-}}$ and $\ce{_9F^-}$ flipped.


Why does phosphorus give a free electron in silicon doping? Extrinsic semiconductors are created by doping an intrinsic semiconductor; typically silicon is doped with phosphorus to create free electrons or boron to create "holes".

In the case of phosphorus doping, phosphorus takes the place of a silicon atom in the silicon lattice, and its fifth electron becomes a free electron.

Here is what I am struggling to understand with my basic knowledge. Phosphorus has the configuration $\ce{[Ne] 3s^2 3p_x^1 3p_y^1 3p_z^1}$.

This page gives an description of the promotion of a 3s electron to a 3d shell, allowing five bonds to form in the case of $\ce{PCl5}$. I would assume that a similar process results in four bonds forming in the silicon lattice and leaving a free electron.

But that same page later on mentions a 2007 paper which essentially says that that model is not accurate, because such a state is less stable than other ways of modeling the bond (assuming I'm understanding correctly. The problem is described here, I have not read the actual paper.

What causes phosphorus to form four bonds with silicon and leave a free electron rather than, say, forming three bonds and leaving a silicon with an unpaired electron in the lattice, or bonding with another "free" electron? Is it promoting an electron to form the fourth bond and freeing the last?

In short: Why does phosphorus form four Si bonds and leave a free electron rather than forming three Si bonds when doping silicon?

First, don't forget that the phosphorus is sitting on a diamond cubic silicon lattice site, which pretty much defines the symmetry of the potential around it - the P cannot change that, but must respond to it. So, you should not think in terms of atomic orbitals.

Second, the fifth electron is not initially a free electron - it occupies a state that happens to be in the band gap.  Now, that state is really close to the conduction band, only 45 meV below it at 300 K, so at room temperature there is a really good chance that it will jump into and roam about the conduction band.  Nor is that the only state in the gap for P - there is another state 72 meV above the valence band (i.e. it is slightly easier to ionize a 4-fold coordinated P than a Si).

Why doesn't the P only bond with 3 Si? Well, that would leave one unhappy Si atom, and one not much happier P atom.  Again, this is occurring in a crystal, not an isolated molecule. The P will occupy the lowest free energy state. Note that there are other impurities that preferentially occupy various interstitial sites. It is just that P can be coerced into pretending to be pretty much a Si atom - peer pressure, perhaps...


Carbon monoxide treatment When CO is the by product of certain reactions( burning of coal under limited air supply), how is it treated before releasing in the atmosphere or is it converted to some other form? Can that byproduct be used to synthesize some other compound ?

AFAIK, production scale processes are highly optimized, producing just traces of $\ce{CO}$. There traces are either not managed, or can by passed through catalyzators.
On the other hand, burning coke with limited air or oxygen $\ce{2 C(s) + O2(g) -> 2 CO(s)}$ or passing water vapour through burning coke $\ce{C(s) + H2O(g) -> CO(g) + H2(g)}$, are used intentionally, to produce $\ce{CO}$, eventually with hydrogen, for further synthesis.


How closely do GCMS spectra typically agree to conclude a match? I have two unknown peaks on my GC (both with the same MW or rather highest M/Z ratio peak = 98) and I am trying to confirm their identity.

Unknown-1

Unknown-2

Based on the expected chemistry I have identified two potential candidate molecules from a library of GCMS spectra (see images below; just click on the images please if you cannot read the smaller font numbers).

Compound-A

Compound-B

Is it safe to think these are matches or not really? The MS was run in EI+ mode.

Unknown-1 is Compund-B?
Unknown-2 is Compound-A?

Both likely? Neither likely? Any one likely?

Is having the same set of M/z fragments good enough (e.g. 98-83-55 similar between Unknown-1 & Compound-B) or ought the abundance ratios also match mostly?

I would never recommend making an assignment of identity based on Mass Spectrum alone - usually you'd want to get things like IR or NMR Spectra to further corroborate your identification but supposing that the underlying chemistry of the reaction is simple and that all you can do is mass spectroscopy you can use this to identify compounds.

  Unknown-1 is Compound B

Plausible, but the base peak of Compound B is 83, but the base peak of Unkonwn-1 is 55. Need more information, but reasonable guess.

  Unknown-2 is Compound A

Unlikely as the base peak of Compound A is 69, which is distinctly missing from the spectrum of Unknown-2. Unknown-2 also has significant peaks at 55-57 which are insignificant in Compound A's Spectrum.

  Is having the same set of M/Z fragments good enough or ought the abundance ratios also match mostly?

You would have a much stronger case if the abundance ratios also matched. Remember that the peaks are formed by the breaking up of the compound and the relative stability of each fragment. If the compound is the same and run in similar conditions, it should, in principle, break up in the same/similar way.


Can lead-acid batteries release CO? Or can a CO sensor detect gasses other than CO? I recently found myself troubleshooting a CO alarm on a houseboat. Inside the cabin, a Fluke CO-220 was reading 40 ppm. In the generator engine compartment, the CO level was zero (while the generator was running). We eventually found the source of the CO to be a large lead-acid "house" battery which was warm and leaking:

The OL on the display of the CO-220 indicates a value of >1000 ppm CO. I watched it count up through 600, 800, 900 before indicating OL.

Mechanics replaced the batteries, and ensured the battery box was properly sealed and vented. After this, the CO levels were at zero.

Can lead-acid batteries leak carbon monoxide?

This article discusses the chemical reactions of a lead-acid battery, and as far as I can tell there is no carbon involved whatsoever. So I believe the answer here to be "no".

Can a Fluke CO-220 detect gasses other than carbon monoxide?

The CO-220 manual says nothing about any gasses other than CO. In the specifications:

  Sensor type | Stabilized electrochemical Gas-specific (CO)

I contacted Fluke and they responded with:

  I do not know what gas batteries would put off and the CO-220 has a senor [sic] for CO only. Hope this helps.

I assume the answer here is also "no".

So then, what was going on?

You're probably picking up hydrogen gas, which is produced when lead-acid batteries are overcharged at high charging voltages (a danger in its own right). This article details a situation similar to yours: charging a lead acid battery in a golf cart (in a confined space) sets off a $\ce{CO}$ alarm, and typical sensors are activated by $\ce{CO}$ at levels of 150 ppm for 30 minutes and/or 300 ppm $\ce{H2}$ for 30 minutes. Because $\ce{CO}$ is not produced by battery charging, the offending gas is thus identified as hydrogen. After venting, the levels recede.
Also see this post from Hawk Environmental, which details how hydrogen gas can create false positives for carbon monoxide. This post goes into a little more detail about the hydrogen/carbon monoxide sensor issue, and the author states that:

Lead-acid batteries produce Hydrogen when charging.
Carbon Monoxide detectors use something called a "Metal Oxide Semiconductor (MOS)" sensor, which detects a variety of gases including Hydrogen.
A MOS sensor calibrated for CO will give a false positive in the presence of Hydrogen gas at ~10% of the actual value. So, that ~85ppm reported of CO was in reality ~850ppm of Hydrogen gas.

Finally, in an article in the September 1996 issue of Fire Engineering, entitled "Other gases may set off $\ce{CO}$ detectors" (here), it is stated that

CO detectors are electrochemical sensors, which detect a variety of specific gases. In most cases, the gas of interest reacts with the chemical solution in the sensing electrode to produce ions (charged particles) that move through the solution to an electron-accepting electrode, completing an electric circuit. The number of ions produced and the magnitude of the current developed are proportional to the concentration of the gas being measured.

and that

Interfering gases of similar molecular size and chemical reactivity may produce false positive response. Major interferents that can affect $\ce{CO}$ detectors are acetylene, dimethyl sulfide, ethyl alcohol, ethylene, hydrogen cyanide, hydrogen sulfide, isopropyl alcohol, mercaptan, methyl alcohol, propane, nitrogen dioxide, and sulfur dioxide.

I suggest you confirm/deny the above by contacting Fluke and asking them specifically if the sensor does not detect hydrogen gas at all and, if so, at what level. I strongly suspect that the well-documented behavior I've found online and summarized here is the cause of what you have observed in this case.


Why don't proteins form branched polymers? My reference book (Princeton Review for SAT Chemistry Subject Test) mentions that:

Proteins and carbohydrates are both polymers; however, only
carbohydrates commonly form branched polymers. Glycogen and cellulose
are both carbohydrate polymers made up of glucose monomers; glycogen
is a highly branched polymer while cellulose is primarily
straight-chained.

But I don't get why proteins cannot form branched polymers. Proteins have both an amino group and a carboxylic acid group, and I understand that the two groups join together with the elimination of a water molecule during polymerization.
Branched polymers are formed when a hydrogen atom (or any other substituent) is replaced by another monomer unit, and I don't see why this can't take place in the case of proteins. Could someone please explain?

Note that branching is not unheard of, for instance glutathione is an abundant short branched peptide formed by condensation of the carboxylic group on a glutamic acid side chain to the main chain amino group of a cysteine aminoacid. This requires a dedicated enzyme both for synthesis (glutamate cysteine ligase) and degradation of the gamma peptide bond. Such gamma peptide bonds could serve as the basis for branching but afaik occur only in special cases.
An important example of branching also occurs during ubiquitylation, in which ubiquitin is attached to a protein's C-terminus via lysine side chains to target the protein for degradation. A variety of polyubiquitin chain arrangements (including multiple branching) have been identified.
I suppose one can only speculate from its absence (or non-ubiquity) that branching either (1) does not generally confer a significant evolutionary advantage or that (2) there are evolutionary barriers that have impeded development of advantageous biochemical machinery for generation and regulation of branching.


What do the numerals on the top right corner of the cells in the periodic table represent? On the table published here (link) each cell has one to five numerals arranged vertically in its top right corner. The sum of those numbers is always the relative atomic mass of the element. I suspect that, in most cases, two of those numerals represent the number of protons and the number of neutrons, respectively. However, I have no idea what the other numerals represent.

What do the numerals  on the top right corner of the cells in the periodic table represent?

The numbers give you the basic ground state electron configuration of the atoms by telling you how many electrons are in each shell.

An electron's 'position' in an atom can be specified by four quantum numbers:

Principal quantum number ($n$). This tells you which shell the electron is in and can take values $1,~2,~3,~4~...$. The maximum number of electrons in each shell is given by $2n^2$ so the first shell can hold 2 electrons, the second 8, the third 18, the fourth 32 etc. The numbers given in your periodic table are the total number of electrons in each shell in the ground state (lowest energy state) of that element.
Azimuthal quantum number ($l$).
Magnetic quantum number ($m_e$).
Spin quantum number ($m_s$).

See this wikipedia article for more detail on the meaning of quantum numbers.

Electron configurations can be written in more detail using $s,p,d,f$ notation. See this question for more detail.


Formation of polymers - addition polymerisation In addition polymerisation, a double bond in the monomer is broken and a long chain of polymers is formed. How is this bond broken (where is the energy input) and how is the long chain formed?

(something to do with free bonds?)

During radical polymerization, we use molecules called initiators $I$ for forming free radicals. The molecules of initiators contain  at least one fragile bond that can easily dissociate upon heating, and yield free radicals.
$$I\rightarrow2A^.$$
The as-formed radicals  have enough energy to attack the monomer molecule $M$ and transfer the radical to the monomer molecule:$$A^.+ M\rightarrow  AM^.$$
More precisely, one electron of the double bond in the monomer forms a covalent bond with the free electron of the radical $A^.$. The other electron of the double bond is now free. We have  a radical  on the monomer  $AM^.$.

The radical monomer  attacks another monomer molecule, and the polymerization reaction propagates:
$$  AM^. + M \rightarrow AMM^.  $$
In the above reaction, one electron of the double bond in the monomer molecule forms a covalent bond with the free electron of the radical $AM^.$.The other electron of the double bond is now free $AMM^.  $.
and so on..


The product of the reaction of chloroiodoalkane and cyanide ion in acetone What would be the product of the following reaction $\ce{Cl-CH2-CH2-CH2-I + KCN} $ ($\pu{1mol}$ each) in acetone?

I think the  reaction should follow the second order nucleophilic substitution mechanism due to primary halide and strong nucleophile. Thus, iodine atom should be replaced giving $\ce{Cl-CH2-CH2-CH2-CN}$ as product but the book says the product is $\ce{I-CH2-CH2-CH2-CN}$. Where am I wrong?

Iodide ion is recycled in the system. $\ce{KI}$ is quite soluble in acetone $(\pu{1.31 g}/\pu{100 ml}),$ $\ce{KCl}$ is essentially insoluble.

You are quite correct that the first reaction is $\ce{CN-}$ displacing iodide giving $\ce{Cl-CH2-CH2-CH2-CN}$ and $\ce{KI}.$ The soluble $\ce{I-}$ will then do a nucleophilic attack on the $\ce{CH2-Cl}$ to give $\ce{CH2-I}$ and the $\ce{KCl}$ produced precipitates out making the reaction irreversible.


What causes the difference in heat energy released by combustion for the butanol isomers? I am currently working on an investigation where I am burning isomers of butanol to test for the heat of combustion values to see which fuel is most efficient. The fuels I am testing are 1-butanol, 2-butanol, 2-methylpropan-1-ol (isobutanol) and 2-methylpopan-2-ol (tert-butyl alcohol).

I have researched that the heat of combustion values for these fuels are as follows:

1-butanol: −2671 kJ/mol
2-butanol: −2661.1 kJ/mol
Isobutanol: −2662.6 kJ/mol
Tert-butyl alcohol: −2644.8 kJ/mol

So evidently, my experimental results should also show similarity in the ordering of greatest to lowest heat of combustion values.

However, I am wondering why isomers of butanol have different heat of combustions. I know it has to do with the shape of the molecules and whatnot, but how does this affect the energy released? What about bond angles?

The heat of combustion (or any reaction) is the energy that is set free when you make the products ($\ce{CO2}$ and $\ce{H2O}$ here, in their respective stochiomeric amounts) out of single atoms (this is the same for all isomers), minus the energy needed to break all bonds in the educts. The latter is different for the different isomers, because the energy in a chemical bond not only depends on the two partners but also on their neighbours (the "shape" of the molecules).

The usual procedure for this calculation is to not go over single atoms but the elements (and compounds) in their most stable form at standard conditions. These energies a called standard enthalpies of formation. There are long tables listing them for all possible compounds.


What are azeotropic mixtures? I’m not able to get the qualitative idea of azeotropic mixture.

What does it mean in simple language?

If you have two components in a mixture, say water and ethanol, and you heat the liquid to the boiling point of that mixture (at the given pressure), you may find that the composition of the resulting vapor and the remaining liquid are different. In that case, you have a non-azeotropic mixture and the composition of vapor and liquid will change as you continue boiling.

For the case of water/ethanol, at some point, the composition of vapor and liquid will match (and no longer change, no matter how long you boil, until all liquid has evaporated). That composition/mixture is called an azeotrope.

The liquid mixture of oxygen/nitrogen is an example for a system that does not show an azeotrope.