text:
The carbon-13 NMR for 1,1,1-trichloroethane
I was working the following spectroscopy problem from Clayden 2e that asks us to identify the structure of a compound from information about its carbon-13 NMR peaks.

Problem 3.5: The solvent formerly used in some correcting fluids is a single compound $\ce{C2H3Cl3}$, having 13C NMR peaks at 45.1 and 95.0 ppm. What is its structure? How would you confirm it spectroscopically?

I initially thought that the structure would be 1,1,1-trichloroethane (shown below) since there will be two different carbon chemical shifts: one bonded to three $\ce{H}$ atoms and one $\ce{C}$ atom in the 0-50ppm region (corresponding to the 45.1 ppm peak) and another peak for a carbon bonded to three $\ce{Cl}$ atoms in the 50-100ppm region (corresponding to the 95.0ppm peak).

Although it considers 1,1,1-trichloroethane, the solutions manual to the book concludes that the correct structure is 1,1,2-trichloroethane (shown below), the other possible structure with formula $\ce{C2H3Cl3}$.

However, this compound would have both peaks close to eachother in the 50-100ppm range.
Wouldn't the correct molecule be 1,1,1-trichloroethane, and wouldn't the solutions manual be incorrect? The NMR specs below that I found online corroborate my thinking.

References
Clayden, J., Greeves, N., Warren, S. Organic chemistry, 2nd ed.; Oxford University Press: New York, 2012.
https://www.chemicalbook.com/SpectrumEN_71-55-6_13CNMR.htm
https://spectrabase.com/spectrum/5WUteTf4M4u

I don't have a copy of the textbook so can't confirm, but the spectra predictions you post are accurate. The wikipedia (infinite in its wisdom) writes of 1,1,1-trichloroethane as a solvent, but not of 1,1,2-TCE. A blooper?

Most likely was a blooper on the part of the authors. I just wanted to make sure that my explanations for the two NMRs I found online were correct.

So probably you should write a very short answer so the Q can be closed

You can find another reliable 13C-NMR on this site: http://www.hmdb.ca/spectra/nmr_one_d/3289, which also confirms you are right about 1,1,1-trichloroethane (can find in: https://pubchem.ncbi.nlm.nih.gov/compound/1_1_1-trichloroethane#section=1D-NMR-Spectra).

SDBS: https://sdbs.db.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi confirms your conclusion.

I fully agree with your assessment and assume that this is a mistake in the book. I wonder what exactly the authors suggested as spectroscopic verification. I personally would suggest using 1H-NMR as a verification method. In 1,1,1-trichloroethane, only one hydrogen peak should be observed as the three protons of a methyl group are always homotopic. In 1,1,2-trichloroethane, two signals in a $1:2$ ratio should be observed as a triplet and dublet, respectively. Depending on what the book suggests as verification, it should be easy to confirm the nomenclature error.

The authors did suggest using 1H-NMR as the method to "confirm it spectroscopically". This is what they provided as the suggested method of spectroscopic confirmation: "The second structure (1,1,2-trichloroethane) looks better but it would be easily confirmed by proton NMR as the first structure (1,1,1-trichloroethane) would have one peak only but the second would have two peaks for different CHs. The solvent is indeed the second structure 1,1,2-trichloroethane."

@Ethiopius Yup, that sounds like the authors were having a major derp moment.


text:
Which is more reactive, chlorine radical or chloride ion?
Ions are very reactive for obvious reasons (i.e. Coulomb force); $\ce{Cl-}$ ions will be very quick to form an ionic bond with positive ions (unless inhibited, e.g. through being in a solution; let's consider both $\ce{Cl}$ and $\ce{Cl^-}$ to be in gaseous phase).
Chlorine atoms without reduction ($\ce{Cl}$), aren't charged, yet tend to combine into $\ce{Cl2}$, utilising a covalent bond.
Thus, both $\ce{Cl}$ and $\ce{Cl-}$ tend to form bonds. My confusion lies in the following:

Noble gases are chemically inert, $\ce{Cl-}$ has the electron configuration of $\ce{Ar}$, yet it's very chemically reactive.
Neutral atoms aren't as attracted to each other as ions, and could be considered more "chemically stable" than ions with higher charge. However, even though $\ce{Cl}$ has no charge, it still seems eager to react and get to the electron configuration of a noble gas.

In both cases, information seems conflicting, and perhaps a more clear explanation will make me understand the situation. Is one more prone to react than the other?
(N.B.: My knowledge on chemistry, maths, physics etc … stretches to (but doesn't include) quantum physics, so feel free to go all-out.)

In addition to the answers you got it is evident fron your text that you are considering ions to be reactive as the same of to form ionic bond. Altough they reactivity is dictated by being charged, they  can react in different ways and end up in a covalent bond.

The apparent example of chloride ions rapidly forming ionic bonds isn't really a chemical reaction but a change of state. The ionic solid still contains the (stable) chloride but in a solid crystal with counterbalancing ions. The stability of that solid drives the physical change of state not any reaction of the chloride.

@matt_black I disagree with that statement. The combination of positive and negative ions to form an ionic solid can certainly be considered as a chemical reaction simply because there is a chemical change occurring. Ionic bonds (i.e. electrostatic forces of attraction) are being formed, with the release of heat energy. The potential energy of the ions experience an overall decrease as the ions experience the mutual attraction and repulsion forces, with net attraction dominating.

@TanYongBoon But that makes the freezing of water a chemical reaction too. Few would agree with that.

@matt_black In that case, what do you take to be the distinguishing features of a chemical reaction?

I think we can also approach this from an energetics perspective. Reactivity can also be seen to be related to thermodynamic stability in the following manner: The more thermodynamically stable a chemical species is, the less reactive it is. With this in mind, we can look at the enthalpies of formation of $\ce {Cl.}$ and $\ce {Cl^-}$.
Your question is slightly vague as you did not state the specified state of the substance. This can critically affect the answer. The reactivity of a heavily solvated $\ce {Cl^-}$ ion is certainly different from that of a free $\ce {Cl^-}$ ion in the gaseous phase.
By definition, the enthalpy change of formation, $\Delta H_f$, of an element in its standard state (e.g. $\ce {Cl2 (g)}$) is $\ce {0}$. $\ce {2 Cl.}$ are produced from the homolytic cleavage of the $\ce {Cl-Cl}$ bond in $\ce {Cl2}$. Knowing the bond dissociation energy of this bond to be $\ce {+242 kJ/mol}$, we can approximate $\Delta H_f$ of $\ce {Cl.}$ to be $\ce {+121 kJ/mol}$.
Similarly, the gaseous chloride ion can be produced from the gain of an electron by $\ce {Cl.}$. The first electron affinity, $\ce {1st E_a}$, of chlorine is the energy change when one mole of electrons is acquired by one mole of these gaseous chlorine atoms to give one mole of singly, negatively-charged gaseous $\ce {Cl-}$ ions. $\Delta H_f$ of $\ce {Cl- (g)}$ can thus be approximated as $\Delta H_f$ of $\ce {Cl.}$ + $\ce {1st E_a}$ of chlorine . Since this $\ce {1st E_a}$ is $\ce {-349 kJ/mol}$, this enthalpy change of formation for $\ce {Cl^- (g)}$ can be approximated as $\ce {-228 kJ/mol}$.
As can be seen from energetics data, $\ce {Cl^-}$ is more stable than $\ce {Cl.}$ in the gaseous phase. Thus, we may say that the $\ce {Cl.}$ is more reactive in the gaseous phase. Furthermore, being charged, $\ce {Cl^-}$ is solvated by water molecules in aqueous solution, further increasing its stability as it has ion-dipole interactions with the neighbouring water molecules. Electrically-neutral chlorine radicals would not experience this solvation effect, thus they would also be more reactive in the aqueous phase. In non-polar solvents, $\ce {Cl^-}$ and $\ce {Cl.}$ would have similar reactivity as in the gaseous phase since they do not experience starkly different solvation effects. Overall, based on energetics, we may say that $\ce {Cl.}$ is generally more reactive than $\ce {Cl.}$.
Additionally, it is important to point out the difference in the type of reactivity of the two different chemical species. Being negatively-charged, chloride ions participate in ionic reactions, involving charged species. These reactions include nucleophilic substitutions, as well as nucleophilic additions. On the other hand, chlorine radicals mostly participate in non-polar reactions, involving mostly radical species, such as free radical substitutions.

A quantitative approach indeed seems to be appropriate for an answer, and based on the reasoning through electron affinity, I'm assuming that qualitatively, one may conclude that the noble gas configuration is the least energetic state an element could reach, thus the most stable?
N.B.: As you pointed out, I was in fact referring to gaseous states, so I've edited the question to make that more clear.

@Mew That is not true at all. The idea of the noble gas configuration as the state of the lowest energy is very wrong, if you are referring to the chemical species in the gaseous phase. For example, to make Mg2+ (g) from Mg (s) in standard state, the enthalpy change is very large and endothermic. What makes this ionisation of Mg an energetically favourable process is the bonding interactions that the Mg2+ ion can undergo (e.g. ion-dipole interactions with water molecules, ionic bonds with Cl- ions etc.)... I hope I have made you aware of how important it is to specify the physical state...

Then why is the noble gas configuration so commonly observed in compounds if it is not the most stable configuration energetically? The reason is simply this coincidentally allows the chemical maximise the energy loss overall from the exothermic bonding interactions and the endothermic electron affinity/ionisation/atomisation processes.

Ions actually aren’t particularly reactive (chemically speaking). Furthermore, most of the reactions they participate in are either full-blown redox reactions (i.e. the product will be a completely different ion or nothing at all) or simple ligand exchange reactions. There are some examples in organic chemistry, especially concerning nucleophilic substitutions, where ions will react in a non-redox fashion to form a new stable compound—but these don’t often lead to high reaction enthalpies.
Radicals, on the other hand, are particularly reactive. The most obvious indication of this is a chlorine radical being able to abstract a hydrogen atom from a $\ce{C-H}$ bond in radical chain reactions:
$$\ce{R-H + Cl. -> R. + H-Cl}\tag{1}$$
The bond dissociation enthalpy of a typical $\ce{C-H}$ bond is in the area of $\pu{420 kJ/mol}$ which is quite a lot of energy. You will be hard-pressed to find an example of an anion abstracting a hydrogen from a hydrocarbon unless you are using a very, very strong base or have a particularly acidic carbon atom. Most notably, the chloride anion is such a weak base that it will never deprotonate any neutral, unactivated hydrocarbon.

You've more or less walked all around the answer. Think about this partial statement of yours:

Noble gases are chemically inert, [and] $\ce{Cl^-}$ has the electron configuration of $\ce{Ar}$.

The free chlorine atom is a radical with one unpaired electron. This arrangement is highly unstable and the chlorine radical either wants to either extract an electron from some atom to become a $\ce{Cl^-}$ anion, or to form a covalent bond somehow.
So, in general, the free chlorine atom is more reactive than the chloride ion. However which is more reactive in a particular reaction, depends on the reaction. For instance in interstellar space the $\ce{Cl^-}$ anion would react more with $\ce{H^+}$ cations because of the coulombic attraction.

Using chlorine as an example, $\ce{Cl-}$ is reactive for the reasons you would think - it is negative so it is drawn to other species with positive charges or electron deficient areas. $\ce{Cl-}$ does have a stable electron configuration, but that doesn't mean it not reactive. $\ce{Ar}$ is only not reactive like $\ce{Cl-}$ as $\ce{Ar}$ has no charge meaning it is not drawn to species other than by van de Waal's forces.
$\ce{Cl}$ atoms are a little bit of a different case. chlorine atoms will reactive together to form $\ce{Cl2}$ very rapidly giving both atoms a stable electron configuration. But one individual chlorine atom is very very reactive as it has an unpaired electron - a so called chlorine free radical. Any free radical is very reactive as the unpaired electron results in a less stable atomic orbital - so it reacts very quickly to increase the atom's stability.  when chlorine atoms bond together to form $\ce{Cl2}$ they are still reactive but no where near as reactive as a chlorine free radical - such so you cannot just find chlorine free radicals they are only formed during reactions and act as very reactive intermediates.
So to answer to your original question individual $\ce{Cl}$ atoms are much more reactive that $\ce{Cl-}$ but as we don;'t get chlorine free radicals outside of reactions a better question would be is $\ce{Cl2}$ or $\ce{Cl-}$ more reactive? and the answer to this depends on what environment they are in as the two reactive in different ways.

So then, if I understand correctly, an atom's orbital instability (the absence of filled or semi-filled subshells) is a greater motivator for a reaction to happen than an instability - if we may call it that - in charge? As a follow-up: What kind of force is at the base of radicals being so violent? I can't see how, for example, the Coulomb force has any direct role in this.

a semi-filled sub-shell ISN'T unstable to have that clear. But each sub-shell is made of orbitals, of which each can hold two electrons and it is unstable when an orbital only has one electron in it. the reason why free radicals react so violently is that they want to fill the orbital as they are thermodynamically unstable with it only having one electron in it. There isn't really a force that explains their reactivity as they are neutrally charged. It is only due to stability of the atom

I feel like that is explaining what instability is, but not which physical phenomenon underlies the effect. Based off of Tan Yong Boon's response, it seems that this is part of the second law of thermodynamics, but I might be overlooking a specific actor (i.e., it'd seem strange that $Cl$ bonds to $Cl$ because of its radical state, but without any force making the two approach each other, based on their neutral state).

source: Stack Exchange

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

sort: votes


text:
Why is boron or carbon solids and chlorine a gas at room temperature
I know different compounds have different states at room temperature because of inter-molecular forces ( dipole-dipole force, hydrogen bonding, and London dispersion forces).
But how do they act on elements? Is there any other inter-moleculer force.

The reason why boron and carbon are solids at room temperature because of their structure.
Both carbon and boron form giant covalent structures, meaning a lot of energy has to be put into break the bonds between atoms and turn into a liquid. A example of this would be diamond - here each carbon atom is covalently bonded to 4 others meaning no inter-molecular  force exist between the atoms are these are not small simple molecules.
Chlorine however is a small simple molecule $\ce{Cl2}$. Here only van der Waals forces hold the molecules together meaning that they can easily be broken - hence why its a gas at room temp.
So elements boiling point is always dictated by inter-molecular forces as it is dependent on its structure. some structures will allow inter-molecular forces to arise giving rise to relatively low bp and mp, while others have giant structures that doesn't give rise to inter-molecular forces only intra-molecular forces.

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


text:
What is a simple formula to approximate the change in relative humidity as temperature rises/falls?
The moisture-carrying ability of air goes up as the temperature rises. So if I take a volume of air and heat it or cool it (allowing it to expand/contract so the pressure remains constant) how can I calculate the (approximate) relative humidity of the warmer/cooler air that contains the same amount of water vapor?
Say I blow 30°c air at 20% relative humidity over a cool object (but not cool enough to condense water out of the air.)  The air is cooled to 20°. What would the relative humidity of the cooled air be if it contained the same amount of water vapor? How do I calculate that?
I am building an Arduino-based control for a warm air dryer for ski boots and bike gear. I will measure the temperature/humidity of the heated air as it comes out of the dryer, and then measure it again as it escapes the gear that I am drying. While the gear is still damp, the waste air should pick up additional moisture, and so its relative humidity should be higher that it would be if I simply cooled it down. When my gear is completely dry, the cooler waste air should contain the same amount of water vapor that it started with, so the relative humidity should be the amount calculated for the measured change in temperature.
I want a "good enough" solution, not one that's mathematically pure. If it gets me answers that are within a percent but involves less computation, that's better than an ideal solution that involves lots of time-consuming calculations. The Arduino microcontroller is fairly slow and cannot do high speed floating point calculations.
(Would this question be better on an engineering forum? I could see arguments either way.)

I meant see this: https://physics.stackexchange.com/questions/337440/dependence-of-relative-humidity-on-temperature-and-pressure?rq=1

You'll need the temperature and RH of the intake air and the temperature and RH of the exiting air. Then it is about what kind of math the microprocessor can do. I can't see why "slow" calculations would be a problem, even a minutes of extra drying shouldn't matter. In fact you'd probably want to add some extra time just to be sure that the items are really dry.

You'll of course need to know the vapor pressure of water as a function of temperature (which gives you 100% RH). See Wikipedia for several different formulas and their comparison.

If you want to convert the %RH from one temperature to another you can use one of various parameterizations, for instance the Antoine equation:

range: 273. - 303 K,
  Bridgeman and Aldrich, 1964,
https://webbook.nist.gov/cgi/inchi?ID=C7732185&Mask=4#Thermo-Phase
A = 5.40221;
  B = -1838.675;
  C =   -31.737;
log(P) = A + B/(C+T);

Then obtaining the $RH_2$ at $T_2$ from that at $T_1$ is as simple as
$RH_2 = RH_1 P_1/P_2$, where $P_1$ and $P_2$ are the saturation vapor pressures of water computed with the equation above, assuming the ambient pressure (and therefore the partial pressure of water) is constant.

That looks like what I want. Is that log base 10, base 2, or natural log?

Wikipedia uses A - B/(C+T) gives slightly different coefficients, but good over range from 1 to 99 C.

Cool. That seems to work. At first I didn't realize the temps were in Kelvin, so I was getting partial pressures of infinity (floating point overflow from crazy-big powers of 10).

That's natural log (base e).


text:
How to check leaving group ability of cations?
In the reaction of an ether with a hydrogen halide (HX), the positive hydrogen ion replaces one of the alkyl groups attatched to the oxygen; but why does this happen, isn't the hydrogen ion a better leaving group than the carbocation?

The leaving group takes the electron pair with it when the colavent bond breaks. Neither the proton nor the carbocation are the leaving group. The protonated alkoxy oxygen is the leaving group.

@Ben Norris I just wanted to know whether a hydrogen ion replacing alkyl group is more feasible or an alkyl group replacing a hydrogen is more feasible.

Yes it is, but that protonation is entirely reversible; the H+ comes off and goes on all the time. Occasionally the alkyl group comes off, that is not reversible and drives the reaction forward.


text:
Do We change sign of electrode potential when reversing an equation?
I have a cell equation as:
$$\ce{2Cr^0 + 3Cd^2+ -> 2Cr^3+ + 3Cd^0}$$
I am provided the standard reduction potentials as:
$$\begin{align}\ce{Cd^2+ + 2e- &-> Cd} &&    E= \pu{-0.4V}\\
\ce{Cr^3+ + 3e- &-> Cr} &&      E= \pu{-0.74V}\end{align}$$
Now do I use chromium’s reduction potential as $\pu{-0.74V}$ or do I flip the equation and use $\pu{+0.74V}$ (as it is getting oxidised and I have the equation for reduction)?
This is using the formula $E(\text{cell}) = E(\text{cathode}) -E(\text{anode})$, all reduction potentials.

$$\ce{Cr^{3+} + 3e^-  -> Cr}\quad\quad\quad\text{E = -0.74}$$ $$\ce{Cr  -> Cr^{3+} + 3e^- }\quad\quad\quad\text{E = +0.74}$$

The reduction potentials are a special case of standard reaction enthalpy, because you can transform $\Delta E$ into $\Delta H$ just by calculating with a few constants.
Standard reaction enthalpy and all other thermodynamic values related to reactions change sign when the direction of reaction is reversed.

The short answer is just throw the numbers into the equation.
In all other thermodynamic equations, everything is "added," and if there are subtractions, it comes from the fact that values themselves are negative. In this case, technically, the potential for the reaction of chromium is $\pu{+0.74 V}$ (because it is the reverse reaction of the tabulated value).
However, there exists a convention to keep the value and thus the sign of the standard reduction potential. The sign change is already accounted for by the minus sign in the equation:
$$
\begin{align}
E(\text{cell}) &= E(\text{cathode}) - E(\text{anode}) \\
  &= \pu{-0.4 V} - (\pu{-0.74 V}) \\
  &= \pu{0.34 V}.
\end{align}
$$
In the end, whether or not you change the sign partly depends on who looks at your work. As long as you get the right answer, some people will not care if you stuck to convention or not, but others might. Some do not even know the convention.

Note that the formal potential of a redox reaction $E = -nF\Delta G$ and the measurable  electrode potential are two different things.
You change the sign of the former when reverting the equation, but you cannot change the sign of the latter. The electrode keeps just one sign and the reaction occurs there in both directions.
E.g. the potential of $\ce{Zn/Zn^2+}$ electrode (not reaction!) is always negative, with ongoing bidirectional $\ce{Zn <=> Zn^2+ + 2e-}.$


text:
Do non-classical carbanions exist?
There are some non-classical carbocations, such as the 2-norbornyl cation, in which the positive charge is heavily delocalised. Have non-classical carbanions also been discovered? I have not seen any non-classical carbanions in books. If there are any non-classical carbanions, please provide some examples.

If an answer has helped solve your question, please consider upvoting and/or accepting one (by clicking on the tick next to the voting buttons). That's the Stack Exchange way of saying thank you! See also: https://chemistry.stackexchange.com/help/someone-answers

In addition to the species mentioned in the answer above, I found another one in Organic Chemistry by Morrison and Boyd(Seventh Edition)

Please give me some references

Interesting question. It is much less studied and reported on than the case of non-classical carbocations, but I did find a few papers. Brown and Occolowitz (Ref.1) reported that deuterated bicyclo[3.2.1]octa-2,6-diene 1b, below, undergoes base-catalysed de-deuteration (potassium tert-butoxide in DMSO) to give 1a much faster (ca. $3 \times 10^4$) than the bicyclo[3.2.1]octa-2-ene counterpart, i.e. 2b $\rightarrow$ 2a. They suggested this is evidence for the intermediacy of a 6π-electron, delocalised (non-classical) carbanion 3.

Much more recently, Brown reviewed and provided additional DFT computational evidence for stabilisation and charge delocalisation in these species (Ref.2).
References:

J. M. Brown, J. L. Occolowitz, “A non-classical carbanion,” Chem. Commun. (London) 1965, (16), 376-377 (https://doi.org/10.1039/C19650000376).
John M. Brown, “Origins of Stabilization and Evidence for Charge Delocalization in the Bicyclo[3.2.1]octadienyl Anion and Related Species,” Australian Journal of Chemistry 2014, 67(9), 1296-1300 (https://doi.org/10.1071/CH14244).

Can one unselect an accepted answer??


text:
Taking a procedure for ICP and adapting it to use with AAS
I was reading a paper regarding metal detection and they gave their procedure for an acid digest of a soil sample which was then analysed by ICP-OES.
They used nitric acid, hydrogen peroxide and sulfuric acid.
I want to do a similar analysis, but I only have AAS available to use.
Can I simply dilute the acid digest in a volumetric flask and analyse with the AAS?
Or are there other steps to consider?
Are the choices of acids of concern? Can they cause interference in the flame ionisation?

In short - it depends. For most metals no problem in using the same digestion protocol for ICP (OES or MS) or AAS. Are you using flame or graphite furnace?  If you are using flame AAS -just prepare a blank digest (same ratio of acids but no sample) use that as you blank.  If you are looking to quantify Na, K or Ca there can be significant background issues from soil samples - most trace metals are not a problem - but you need a good blank to qualify.  The LOD (limit of detection) will vary depending upon the matrix S/N (signal to noise).
Hope that helps.  If using graphite AAS you are typically limited as to the amount (volume wise) you can add to  the furnace.  You can try multiple dry downs before atomization.  Again a good plank is paramount.


text:
Clarification in the mechanism for Molisch's test for glucose
Wikipedia says

Molisch's test is a sensitive chemical test, named after Austrian botanist Hans Molisch, for the presence of carbohydrates, based on the dehydration of the carbohydrate by sulfuric acid or hydrochloric acid to produce an aldehyde, which condenses with two molecules of a phenol (usually α-naphthol, though other phenols such as resorcinol and thymol) also give colored products), resulting in a red- or purple-coloured compound.

Can somebody provide an arrow pushing diagram to illustrate the intermediate steps of these reactions? (especially for the dehydration of glucose)
Also, for the reaction with of 5-(hydroxymethyl)furfural with 2 moles of phenol, it seems that the acyl group is acting like an electrophile and naphthol is engaged in an electrophilic aromatic substitution, but then again, I am not sure how 2 moles of naphthol combined there. Any help would be appreciated.

The route from glucose to 5-HMF is complex. It is detailed here: https://reader.elsevier.com/reader/sd/pii/S2095495617307362?token=40D8718D0F9B852A01A1E7C29B634152BC21F2582E9F7B6D22B4BA877A039281E7220B1DB197F582E3D70AFA4025F9A2. The reaction with phenol or alpha-naphthol are electrophilic, aromatic substitutions. The test would be applicable to other aldohexoses and aromatic aldehydes.

@user55119 I am unable to understand how the electrophilic aromatic substitution took place with two moles of alpha-naphthol. Could you please provide the mechanism scheme from 5-HMF to the purple-coloured dye?(as in the oxidation part, I don't understand how the H+ was removed)

I'll work on it.

YUSUF HASAN: 5-Hydroxymethyl furfural (5-HMF) 1 when protonated on the carbonyl oxygen becomes a reactive electrophile. Addition of phenol 2 at the reactive para-position affords 3 which deprotonates at the para-position, rearomatizing the ring to a phenol and liberating a proton. Secondary alcohol 4 is protonated on oxygen with loss of water to form stabilized cation 5. A second equivalent of phenol adds to the cation as previously described leading to the triarylmethane 6. The methane hydrogen is susceptible to oxidation by atomspheric oxygen, which is in a triplet state (unpaired electrons behaving as a free radical). Resonance stabilized radical 8 is formed along with the hydroperoxy radical 7, which can abstract the hydrogen from triarylmethane 6 forming more of radical 8 and hydrogen peroxide. Oxidation of radical 8 with oxygen gives the resonance stabilized carbocation 9 and superoxide anion, aka, superoxide radical anion 10. This species can be protonated by the the strong acid conditions of the Molisch test to form more of the hydroperoxy radical 7.


text:
Chain elongation with a substituent
Can this compound exist in a stable form. If so, can it be used to add the chloroethyl group to a carbonyl compound

No, this is not a stable species. It will immediately eliminate to give ethene.
If you want to add a C2 unit you're better off with the commercially available vinyl Grignard.


text:
Water as a "Harmful Acid"
Apologies for the question which is far below the brilliance of the minds on this page. I have an A-Level in Organic Chemistry but I'm tying myself in knots with this one.
I'm a magician looking to create a Russian Roulette effect with water and acid (sort of!) the actual secret is that both solutions are water.
Is there a readily available metal; or substance that looks like metal, that would make the water behave like an acid to convince an onlooker that it is a potentially life threatening acid.
As a bonus question – is there a way to cheat the litmus test so that one solution (also water!) turns it red, the other reacting neutrally (that's the easy one!)?

I wonder if a low dilution of something like acetic acid or citric acid would do the trick. It won’t give you a pH 1 solution, but with the typical universal indicators on pH paper you would be well into the acidic range.

Thanks Jan, that's not a bad idea. I'm leaning towards using a classic magicians switch to substitute a piece of litmus paper with a pre-dyed red strip, so that the water just moistens the paper to create a more convincing effect. Obviously alkali metals are not readily available so thinking an effervescent tablet made to look metallic might be  good substitute.

How are you going to convince your auditorium that this "acid" is dangerous? You cannot dissolve an earthworm or a bank note in water. ;-)

Magicians are expert liars :D

It is possible to obtain a small sample of sodium or potassium metal from a university  - the reaction is violent in water. For the acid part,  you could use acetone and packing peanuts dissolve easily in them to show that everything will dissolve in 'acid'!

If you are using water alone, then sodium lead alloy reacts as sodium would but, with the lead added, it's much safer.  But you can have a wider variety of options if, as the comments suggest, you use a dilute acid like vinegar (reacting that, for instance, with zinc).  Wear eye protection no matter what you do.


text:
Can ZnO act as a reducing agent for noble metals like Pd?
Palladium ions (e.g., $\ce{Pd^2+}$) can be reduced to $\ce{Pd}$ by more reactive metals such as zinc. Metal oxides such as $\ce{ZnO}$ are very stable but can its atoms in the oxide structure still be displaced by Pd?
I know the answer is probably "No" but in one of my samples $\ce{ZnO}$ nanoparticles were removed from the solution after I added $\ce{Pd^2+}$ solution to the system and $\ce{Pd}$ nanoparticles were formed. The solvent was methanol. After washing the solution and drying I did XPS and no trace of zinc was observed but saw $\ce{Pd}$ peaks.

The metals aren’t connected to the Swedish entrepreneur Alfred Nobel but to the English adjective to nobility: noble. As for the actual question, I sadly cannot give you an answer.

Did you measure the size of the nano-particles? According to Wikipedia, at most, the XPS instrument will only probe $\pu{20 nm}$ into a sample. Thus, I wonder, $\ce{Pd}$ has been deposited on $\ce{ZnO}$ surface. "Palladium Deposition on $\ce{Pt}$ Surface by Reduction of $\ce{Pd^2+}$ Ions with Organic Substances" has been studied before (https://link.springer.com/article/10.1023/A:1010314807520).

Zinc chemistry is overwhelmingly that of the +2 oxidation state, it does not go higher. For ZnO to reduce Pd+2, the Zn has to go up from +2 which it cannot. If you are getting reduction of Pd+2, Zn is not responsible.

Perhaps ZnO is catalyzing the reduction of Pd(II) to Pd(0) with methanol as the reducing agent (oxidized to formaldehyde). I presume the Pd(II) is stable in methanol.


text:
Evidence for a chemical reaction
I know that the emission of light is evidence for a chemical reaction.  However, are there any instances where emission of light can also be considered a physical reaction?

Synchrotron radiation? Czerenkov radiation? Bremsstrahlung? What you are searching for?

An Edison-style light bulb? LEDs? ;-) Actually chemical rections emitting light are rather rare, esp. if you take out those that simple become very hot. If you don't live in a rural area (fireflies) or have had a good chemistry teacher, you probably have never even seen a light that would qualify as "chemical".

Most light doesn't result from chemical reactions
While there are many chemical reactions that can emit light and many related reactions where the emission is related to chemistry (for example the light from some flames is caused by electronic transitions), most light you see isn't.
An incandescent light bulb emit light because of black-body radiation produced because it is hot (this is not normally though of as having anything to do with chemistry). Fluorescent tubes emit light because of electronic transitions in gases or phosphors (both of which might be considered closer to chemistry because they involve electronic transitions but not the sort that indicate a reaction as they are entirely reversible). LEDs are similar.
A well set up Bunsen burner emits a small amount of light from the chemical reactions occurring in the flame but most well-regulated flames don't emit much light. A candle, on the other hand, has a relatively bright flame. but most of the light comes not from chemical emissions but from the incandescent emission from small soot particles produced from the (deliberately) inefficient burning. Ultimately that heat come from chemistry but most of the light comes from the basic black-body physics of hot objects.
So, in short, most light is physical not chemical.

I was initially thinking about the light bulb as an example of light emission as a physical change. Because I know that electricity passing through the filament in a light bulb causes a physical change.  Heat and light are produced, yet no chemical change. But the Introductory Chemistry textbook I have presents emission of light as evidence of a chemical reaction (example of glow sticks and fire flies) and no mention that it could also be physical change. But the light bulb made me think otherwise and, thus, I wanted to reach out to others for additional input/confirmation.

There are a number of places where it might be appropriate to write "visible light" rather than just light.

@user72414 Can you give the name of this book, so others might be warned about it?

Introductory Chemistry (6th Ed) by Nivaldo J. Tro (pg. 209-210).  It's interesting that the author does point out how gas evolution is not always an indication of a chemical change (physical change of liquid water to gas for example).  The book also stresses that the changes only provide evidence but not definitive evidence (which is true).  But I think it would have been less confusing had the author included something about how emission of light can be from chemical changes AND physical changes.  I guess that will come about in the 7th edition?;-)


text:
How does the concentration of a gas affect the entropy of a system in a chemical equilibrium?
The question asks about factors that increase the entropy of the system. The mark-scheme states that "decreasing the concentration of a gas without changing the volume of the container" is the answer. I do not understand how that is possible.
The chemical equilibrium of the question is the decomposition of calcium carbonate:
CaCO3(s)  CaO(s) + CO2(g)
My thoughts are that in order to decrease the concentration of the gas without changing the volume, the equilibrium would have to shift to the left. But that would correspond to a lesser amount of gas, so a lower entropy.
My next thought was that the mark-scheme I had was wrong. I did some digging and found that the question had 3 "G2" complaints, meaning that 3 IB chem teachers had complained about the question during the examination period. But yet the IB determined that the question was fair and kept it, and so here I am trying to wrap my mind around the topic. Alas, I am in need of some help.

What is the question statement?

Ahh. I completely forgot to add what the question was asking for! Thanks for the reminder.

I agree with your line of thought. Can you add a link to those complaints or other specifics? Perhaps more detail regarding the nature of the question too.

That helps to know that I'm not missing something glaring. I've added what I have about the complaints. I'm now thinking that the "concentration decreasing" in this question might be referring to diffusion. If it is, then what a terrible question!

I think "decreasing the concentration of the gas" is supposed to imply "adding an inert gas to the container", which would certainly increase the entropy. (This process would be equivalent to a fraction, say half, of the carbon dioxide molecules suddenly becoming nitrogen molecules, and then having the two gases mix. The entropy of mixing is positive.)
This is indeed poor wording.


text:
Does an airborne chemical stimulant exist?
I recently read the creepypasta entitled "The Russian Sleep Experiment". In this story they lock five test subjects in a room with the intention of keeping them awake for 30 days by releasing a gaseous stimulant into the room constantly. I'll let you read the rest, it's an incredible horror story, but not very credible. Anyways I wanted to know if in real life such a gaseous stimulant exists.
During the Moscow theater hostage crisis Russian forces "pumped an aerosol anaesthetic, later stated by Russian Health Minister Yuri Shevchenko to be based on fentanyl". So it's possible for an opiate-based depressant can be made gaseous (technically aerosol).
So is there an airborne stimulant that actually exists?

Gaseous: no. Aerosol: yes.

Regardless of whether the stimulant is actually gaseous at standard temperature and pressure, it can be made airborne as an aerosol simply by spraying. If the room to be filled with vapour is sufficiently small then it is no problem to just spray a solution of your stimulant and pump it inside. But even gym hall sizes should be fine with adequate technique applied.


text:
How to find equilibrium constant in reactions which have a limiting reagent?

If in the following reaction
$$\ce{H2 + I2 <=> 2HI}$$
the initial concentrations of $\ce{H2}$ and $\ce{I2}$ are $\pu{1 M}$ and $\pu{5 M}$, respectively, and at equilibrium $x$ moles of each have combined. Then what will be the $K_\mathrm{c}$ for the reaction?

What I think is that the $K_\mathrm{c}$ will not change because it does not depend on the initial concentrations of reactants and depends only on the stoichiometric equation, but the one thing I can't reason out is that those extra $4$ moles of $\ce{I2}$ should also have an affect on its equilibrium concentration (since those $4$ moles are still present in the vessel) and thus affect $K_\mathrm{c}$.
Please help me find the correct explanation for the aforesaid question.

can you just clear up what you mean by  "x moles of each have combined" do you mean the total moles at equilibrium is x or at equilibrium there is x moles of each species.

Lookup the definition of equilibrium constant and fill in the equilibrium concentrations. Then you'll get the value of the equilibrium constant.

Just how the very word "change" came about? To speak of a change, you must have some "before" and some "after". Here you don't have any "before".

If $x$ moles have combined, then there must be $2x$ moles of $\ce{HI}$, $1-x$ moles of $\ce{H2}$ and $5-x$ moles of $\ce{I2}$, so therefore $K_c$ is ...

@IvanNeretin By "change" I mean the deviation from the ideal condition where moles of $\ce{H2}$ and $\ce{I2}$ are given according to the balanced equation ( 1 each in this case).

@CurtF.'s interpretation of my question is absolutely correct

There is no such thing as ideal conditions. $K_c$ is still the same constant, no matter what is the ratio of reagents.


text:
Constraints on rate coefficients in chiral catalytic system
Consider some reversible chemical reaction in a closed system:
$$\ce{A <=> B + D} \tag{1}$$
where $\ce{A}$ is some input reagent and $\ce{B}$, $\ce{D}$ are some output reagents. Let's denote forward and backward rates as $k_f$ and $k_b$ correspondingly.
Now, consider the same reaction in the presence of some catalyst $\ce{C}$:
$$\ce{A + C <=> B + D + C} \tag{2}$$
Due thermodynamic constraints the equilibrium state cannot change, which means that forward and backward rates of catalyzed reaction ($c_f$ and $c_b$) must be equal to the rates of uncatalyzed reaction multiplied by the same multiplier, $k$:
$$
    \begin{matrix}
    c_f = k \, k_f \\
    c_b = k \, k_b \\
    \end{matrix}
 \tag{3}
$$
Now, consider the same scenario if the reagents and catalysts are chiral so that each of the reagents has an enantiomer:
$$\hat{E}(A) = a, \hat{E}(B) = b, \hat{E}(C) = c, \hat{E}(D) = d \tag{4}$$
where $\hat{E}$ denotes an enantiomer.
Then the first reaction splits into two:
$$
    \begin{matrix}
    \ce{A <=> B + D} \\
    \ce{a <=> b + d} \\
    \end{matrix}
 \tag{5}
$$
with the same forward and backward rates $k_f$ and $k_b$ and catalytic version splits into 4:
$$
    \begin{matrix}
    \ce{A + C <=> B + D + C} \\
    \ce{a + c <=> b + d + c} \\
    \end{matrix}
 \tag{6}
$$
and
$$
    \begin{matrix}
    \ce{A + c <=> B + D + c} \\
    \ce{a + C <=> b + d + C} \\
    \end{matrix}
 \tag{7}
$$
where (6) is just the same as the catalytic reaction (2) above but now expressed for both enantiomers and (7) accounts for the fact that chiral catalysts are not 100% enantioselective: if chiral reaction is catalyzed by a chiral catalyst, then it is also catalyzed by enantiomer of that catalyst (though with different rate coefficients).
To consider thermodynamic equilibrium in such a system we must also take inter-conversion of enantiomers (racemization) into account: the concentration of both enantiomers must be equal in equilibrium for each pair of enantiomers. This leads to the following constraints:
$$
    \begin{matrix}
    c_f + c^e_f = k \, k_f \\
    c_b + c^e_b = k \, k_b \\
    \end{matrix}
 \tag{8}
$$
where $c_f$, $c_b$ are forward and backward rate coefficients of reactions (6) and $c^e_f$, $c^e_b$ are forward and backward rate coefficients of reactions (7).
So, thermodynamics imposes two constraints on four parameters. If we consider many catalysts for reaction (5), then all of them will obey equation (8) with some value of $k$.
The question is: are there any other constraints, which can be imposed on coefficients of reactions (6) and (7)?

Why bother with species D, it just makes things more complicated.  Also if you do not add reactions $A\leftrightharpoons a$ and similarly for the other species, you effectively have two separate reactions, one for species $A$ and another for $a$.

The considerations without D species and without reactions like $\ce{A <=> a}$ results in wrong thermodynamic considerations and, subsequently, in erroneous conclusion that chiral catalyst would result in the same multiplier for forward and backward rates of chiral reaction, thus not changing even local equilibrium state. But it does! So, both D species and racemization are required :(

As it happens I disagree with your conclusion, $A\leftrightharpoons a$ etc is the racemisation step.

@porphyrin Great! That's why I am here. If we put a racemase and just one of the species (L or D), then what will we observe after a while? Will it be a racemic mixture of L + D regardless where we start OR will it be just mostly one of the species? Of course, we will need enough cofactor and maybe even some inflow of energy. But, ... what will be the outcome? I can't find the answer on the web :(  This is the key underlying question.


text:
What is the use to know that a compound optically active or not?
What is the use to know that a compound optically active or not? Is it important to know that?

As it stands now, your question is too broad to get an answer more meaningful than the following: It is no more important than any other property of a compound, that is, it is important in some circumstances and unimportant in others.

In regards to??

Yes, it is important.
For reasonably enantiopure chiral molecules, it is expected to record and publish a specific rotation when it is first described. For the person recording these, they are a little bit annoying, especially if you have a polarimeter as old as the one I had to work with during my PhD. Therefore, you want to cut time as much as possible. By immediately determining that a compound is achiral, you eliminate the need to record a specific rotation because theory predicts it is zero.
Conversely, if you were to simply record all specific rotations of your compounds and then just remove the ones that end up as zero you would likely miss the chiral molecules you synthesised as a racemate. If, however, you take a look at the molecule on paper, are able to determine it should be chiral—given the methods and chemicals employed—but still find a specific rotation of zero you have something that must be discussed in the final paper or thesis.


text:
Confusion between moles and molarity units
I am a total newbie in the field and i would like some kind of guidance.
So, I have been recently given 2 datasets with measurements regarding several phospholipid fatty acids of plasma HDL. Both datasets consist of the same classes of FAs but with different units of measurement (μΜ and mol%).
Since I have no previous experience with that kind of data, can someone give me a hint about how (and if) it is possible to convert from one unit to the other?
Thanks in advance.

yes it will be possible as $M$ is $mol dm^{-3}$

$\ce{M}$ is a shortcut for $\frac{\ce{mol}}{\ce{l}}$. $\ce{mol}\%$ on the other side is defined as $\frac{n}{n_{\ce{tot}}}$. In order to convert between the two you need to know the density $\rho$ and the molar mass $M$ (don't confuse this with the unit) of your solution/mixture. You can then do
$$
c_{\ce{mol}\%} = \frac{c_\ce{M} \cdot V_{\ce{mix}}}{V_{\ce{mix}} \cdot \rho \cdot M^{-1}} = \frac{c_\ce{M}}{\rho \cdot M^{-1}}
$$
$c_{\ce{mol}\%}$ is the concentration in $\ce{mol}\%$ and $c_M$ the one in $\ce{M}$.


text:
Stability of alkene
Sorry for this silly doubt but why is CR2=CH2 alkene more stable than CHR=CHR alkene?
Both have same number of alpha hydrogens. I thought that the latter is more stable as in its trans isomer, the R groups are most seperated (least sterric hinderance) but in my book it is mentioned the other way.

I don't think this is universally true.

Perhaps because the sp2 CH2 in the first alkene is more electronegative than the CHR in the second one, so when electron density decreases more at the CR2 side of first alkene, it gets stabilized more by electronic releasing effect of the two R's as compared to CHR=CHR

I think that one reason is in the tags list you have chosen

Do you mean to say that in the latter case, the flow of electrons of the two R groups interfere with each other during hyper conjugation?

I think That's true only if the -R is an alkyl group without any other substituents and if -R is not a tert- group.

Unfortunatly, I can not comment yet, so maybe someone can move this to the comment section.
I don't think, this statement is generally true. It depends on the inductiv and mesomeric effects of the side groups (+-I/+-M).
You also have to choose, what you mean with stability. The energy that is needed to break the doublebond or the reactivity of the double bond.
If a doublebond has four substituents (which are not Hydrogen of course) any reaction on the double-bound is stericly hindered. When there are still Hydrogens on it, the doublebond might be more likly to react. That doesn't mean, that it requires less energy to break the bond.


text:
When to use internal energy and enthalpy in balances
I am confused as to when you use ΔU and when to use ΔH, I don’t really understand the difference. Is it that ΔU is used for closed systems whilst ΔH is used for open systems?

Is it that ΔU is used for closed systems whilst ΔH is used for open systems?

That's not quite correct.
Consider the case of a simple system: a volume of gas. The problem is that when processes take place, the gas may expand or contract. Since the gas pushes back on the external pressure, some amount of work may be done the system ($W = p\Delta V$). The internal energy change $\Delta U$ does not automatically factor in this work.
Enthalpy, on the other hand, is a quantity which is defined ($U + pV$) such that it does account for the pressure-volume work of the system, so $\Delta H$ is the change the energy of the system and the pressure-volme work done by the system.
So, the summary is that if you have a constant volume process (where pressure volume work is by definition zero), use $U$. If you have a constant pressure system, use $H$.
Even the free energies are this way as well. There's Helmholtz free energy ($A$ or $F$) for constant volume processes and Gibbs free energy ($G$) for constant pressure processes.
In chemistry, most processes are done in open air (unless you're using a bomb calorimeter), in which case all processes are simply subject to atmospheric pressure, so these processes would all be best described using enthalpy and Gibbs free energy.

You would use $\Delta U$ along with Q-W for a closed system.  For an open system (involving mass entering and leaving a control volume) operating at steady state, you would use $\Delta H$ along with $Q-W_s$, where $W_s$ is the shaft work per unit mass passing through the system (which does not include the work to push mass into and out of the system) and Q is the heat added to the control volume per unit mass passing through the system.
The open system version of the first law of thermodynamics (neglecting kinetic- and potential energy effects) is
$$\dot{U}=\dot{Q}-\dot{W_s}+\dot{m}_{in}h_{in}-\dot{m}_{out}h_{out}$$where h is the enthalpy per unit mass of entering and leaving streams to the control volume.  At steady state, $\dot{U}=0$ and $\dot{m}_{out}=\dot{m}_{in}=\dot{m}$.  So, for steady state operation, $$\dot{Q}-\dot{W_s}+\dot{m}(h_{in}-h_{out})=0$$or equivalently, $$\Delta h=\frac{\dot{Q}}{\dot{m}}-\frac{\dot{W_s}}{\dot{m}}$$


text:
Why does covalent nature in ionic solids increase the lattice energy?
I have observed in my textbooks that ionic solid that have a greater covalent character tend to have higher lattice energy. Can this be generalised as fact?

(First of all I have to note that I'm just a student, not a specialist in solid body physics or chemistry)
Let's consider chemical bond in general, starting with diatomic molecule (our findings for this case we will generalize to crystalline structure later):
When we say that chemical bond is formed, we mean that for electrons of each atom by atomic interaction (interaction of atomic orbitals of each atom to each other) become available to be situated on molecular orbitals(MO) (defining as superposition with different contributions of atomic orbitals(AO)). Now we must say that depending from atomic orbital energy and its symmetry, result of interaction with another AO may be different.
If we have two interacting AOs with big (bigger than stated value) difference in energies it leads to formation of two MOs: one of them, bonding MO, must have energy closer to such AO that is lower by energy, another relatively must have energy closer to another AO (see picture). In this case we have low energy of stabilization (system weakly feel energy changing before and after interaction).
If difference as big as whole system looks like we give both electrons to one of AOs it means that electron density of bond is shifted to one of atoms. In this case we say that we deal with IONIC bond.
Now COVALENT bond: energies of interacting orbitals are close. This way we have big stabilization energy (see picture). Because of it we can say that covalent bond is stronger then ionic.
Simplifying: ionic bond is provided by Coulomb interactions (in approximation), but covalent bond is provided not only by them, and this is the reason why it's stronger.
Generalization to crystalline lattice: now we must consider group of interacting orbitals, but main findings remain the same: bond forming by AOs with same energy is stronger.
And, as lattice energy is energy of interacting particles forming lattice, it's clearly determined by MO's energy.
That's why I think yes, we may generalize your finding as a fact.

Welcome to Chemistry.SE!  To help you become aquainted with out site, I would just like to suggest if you haven't already, please take a minute to look over the help center and tour page to better understand our guidelines and question policies.

It depends on what is meant in the framing of the question—the lattice energy of a crystal is defined as the molar energy of sublimation for some ionic solid where it is sublimated into its constituent ions. For some ionic solid $A_mX_n$, the ionization is the change in internal energy $\Delta U_{lattice}$ of the following reaction:
$$ A_mX_n (s) \rightarrow m A^{n+} (g) + n X^{m-} (g) $$
You could approximate an ionic solid as only bound from Coulombic (i.e. charge/ionic) attraction and calculate what the change in energy would be of separating two ions by using Coulomb's law. If that's sufficient enough for you to understand, then you can skip this next math. If you would like to see the math, here's what that calculation would look like:
$$ N_{A} \int_{B_{l}}^{\infty}{\frac{-mnQ}{4\pi \epsilon _0r^2}}dr $$
where $N_{A}$ is Avogadro's number, $Q$ is the charge of the proton, $B_l$ is the bond length between the two ions, and $\epsilon _0$ is the vacuum permittivity.
When comparing this calculated energy (or an energy calculated from a similar ionic-only model), often the energy is greater than you would expect it to be. That's because there is some extra character that comes not just from Coulombic attraction but from covalent attraction.
Covalence is the stabilization of a bond through the quantum mechanical mixing of two atomic orbitals into a bonding orbital of low energy and an antibonding orbital of high energy. This covalent character further increases the stabilization of the lattice, and thus the energy required to break apart the ions will be greater.
The Coulombic attraction between two ions at the measured bond length distance doesn't change even if some of that bond length is determined by covalent character in the bond—if we measure the bond length and apply the ionic-only model, we underestimate the extra contribution of stabilization from the covalent character. However, covalent character won't completely determine trends in lattice energy; only with every other factor being equal could you compare two ionic solids and predict lattice energy from the covalent character.


text:
Chemical composition of seawater
Is it true that the sea water is composed of about $86\%$ oxygen, $11\%$ hydrogen and $3\%$ of minerals? The chemical formula of water is $\ce{H2O}$ (two hydrogen and one oxgen) that shows that the number of hydrogen is greater than that of oxygen.
If the number of hydrogen is greater, then why does the sea water consist of $11\%$ hydrogen and $86\%$ oxygen, which is lesser than the oxygen?
The book which I am reading says which is confusing me:

... Seawater is composed of about $86\%$ oxygen, $11\%$ hydrogen and $3\%$ of minerals, consisting mainly of sodium and chlorine.

The book that you're reading is measuring by mass.
If you have pure water then you would expect oxygen to make up $\frac{16}{16 + 2}\times 100\% \approx 89 \% $ by mass. Likewise, hydrogen would make up $\frac{2}{16 + 2}\times 100\% \approx 11 \% $ by mass.

As it has already been mentioned, these are mass percentages, which are a bit imprecise. For the reference, CRC handbook of chemistry and physics [1, p. 14-17] lists elements' abundances in seawater near the surface in $\pu{mg L-1}$, alphabetically. I took 15 (starting with 16th element $\omega < 10^{-4}~\%$) most abundant elements from that array and calculated mass percentage for each:
\begin{array}{llr}
\hline
\text{Element} & \rho_i/\pu{mg L-1} & \omega/\% \\
\hline
\ce{O} & \pu{8.57E5} & 85.8443 \\
\ce{H} & \pu{1.08E5} & 10.8182 \\
\ce{Cl} & \pu{1.94E4} & 1.9433 \\
\ce{Na} & \pu{1.08E4} & 1.0818 \\
\ce{Mg} & \pu{1.29E3} & 0.1292 \\
\ce{S} & \pu{9.05E2} & 0.0907 \\
\ce{Ca} & \pu{4.12E2} & 0.0413 \\
\ce{K} & \pu{3.99E2} & 0.0400 \\
\ce{Br} & \pu{6.73E1} & 0.0067 \\
\ce{C} & \pu{2.80E1} & 0.0028 \\
\ce{Sr} & \pu{7.9}     & 0.0008 \\
\ce{B} & \pu{4.44}     & 0.0004 \\
\ce{Si} & \pu{2.2}     & 0.0002 \\
\ce{F} & \pu{1.3}     & 0.0001 \\
\ce{N} & \pu{5.00E-1} & 0.0001 \\
\hline
\end{array}
References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; 2017; Vol. 97.

You say the numbers are a bit imprecise, but they have to be, because not all seawater is the same!


text:
Why CO2 has 4 vibrational degrees of freedom?
If I have a diatomic molecule, so a linear molecule, $3N-5=6-5=1$ holds, since this vibrational degree of freedom corresponds to the stretching coordinate beetwen the two nuclei.
If I consider water for example, I have a nonlinear triatomic molecule, so the formula says $3N-6=9-6=3$ vibrational degrees of freedom. Also in this case it makes sense since they correspond to the bending angle and to the $2$ stretching lenghts.
While, If I consider $\ce{CO2}$, in this case I have a triatomic linear molecule, so $3N-5=9-5=4$ vibrational degrees of freedom. Of course $2$ of these correspond to the two stretching lenghts, but I really don't understand what the other two correspond to. Since the molecule is linear, there shouldn't be angle to consider, otherwise it would not be linear anymore.

The molecule is linear in the ground state. There are two bending motions in addition to the symmetric stretch and anti-symmetric stretch.

Molecular vibrations can certainly occur by changing the O-C-O bond angle away from 180º. When we call the molecule linear we are saying that the geometric minimum is a linear configuration; however, vibrational modes can (and do) break that symmetry. What's important is that the average nuclear position is a linear molecule (or if you're doing a geometry calculation you would say that it's the 0 K positions).
So the two vibrational modes you're missing are the two O-C-O bends at 90º from each other.
Also as a note: you seem to be suggesting that the two stretches correspond with the two bonds stretching. That's not quite true. One of them corresponds to both C-O bonds stretching outward at the same time (in phase) and the other will correspond one compressing while the other stretches (out of phase).
For a full derivation of the normal modes of CO$_2$ there are many sites, but this one looked good: http://www.colby.edu/chemistry/PChem/notes/NormalModesText.pdf


text:
The carbon-13 chemical shift of acyl chlorides vs ketones
In lectures, we were taught that ketones and aldehydes tend to have chemical shifts of around 200 ppm, while acid derivatives (carboxylic acids esters etc) tend to have values around 160–170 ppm.
The reasoning given by my supervisor was that despite acid derivatives having more electron withdrawing groups, due to conjugation, the net effect was that these additional groups donated electron density to the carbonyl carbon, hence shielding it more than the carbonyl carbons in ketones/aldehydes (i.e. the conjugative effect dominates over the inductive effect). This all makes sense for me except for acyl chlorides.
The reason why acyl chlorides are not adding up for me is that in the very next topic, we were taught that the CO double bond of an acyl chloride is stronger than that of a ketone/aldehyde! Since carbonyl bonds are strengthened by anything that withdraws electrons, I can only conclude that the Cl has a net electron withdrawing effect (i.e. its inductive effect dominates the conjugative effect).
Since the Cl group appears to actually withdraw electron density, going back to NMR, shouldn't the carbonyl carbon in acyl chlorides resonate at a higher frequency than the carbonyl carbon in a ketone/aldehyde?

Relevant, but I'm too tired to read thoroughly now: https://doi.org/10.1063/1.1703231 The issue is more complicated than just electron density, as 13C usually doesn't play by the same rules as 1H (see https://chemistry.stackexchange.com/a/62565/16683).

@HunnyBunch Actually, I am not actually convinced by the reasoning presented in your 2nd para regarding resonance donation outweighing inductive withdrawal. If we were to draw the resonance structures showing delocalisation of the lone pair into the carbonyl group, we necessarily would have to show the C-O $\pi$ bond breaking. Thus, even though there is resonance donation of the lone pair, increasing electron density around the carbonyl carbon, the effect of C-O $\pi$ bond breaking actually reduces electron density around the carbon atom. Thus, the reasoning fails to explain the phenomena.

@orthocresol I am not entirely convinced by the explanation they have suggested in their discussion. The reason for my disagreement with it is given in my previous comment.

Halogens $(\ce{Cl},$ $\ce{Br},$ $\ce{I})$ also act as conjugative donors. Not very good ones but you still have to take that into account especially because, as you said, conjugative effect is larger than inductive effect (and it is very different in $\ce{^1H}$- or $\ce{^13C}$-NMR).
The $\ce{^13C}$ chemical shift in $\ce{CH2Cl2}$ is $\pu{53 ppm}$ but the one for $\ce{CH2I2}$ is $\pu{–62 ppm}$ (yes, with a minus sign).

Thank you for your response, but how would you rationalise the second part of my post?

Well in the 2nd part of your post, you actually consider only one bond (C=O) out of the 3 possible bonds (C=O, C-Cl, C-C).

Hmm, I'm not altogether convinced... Indeed 1H and 13C effects are very different, but I don't see how conjugation applies to 13C shielding in CH2I2? There is a known shielding effect with heavy atoms, see e.g. https://chemistry.stackexchange.com/a/73685/16683 https://onlinelibrary.wiley.com/doi/full/10.1002/mrc.4720

I guess conjugation effects are usually how the relative IR stretches are rationalised, but I'm not sure if NMR is the same: the picture is rather complicated as there are both diamagnetic and paramagnetic shielding effects.


text:
ChemDraw temporary label for chemnum LaTeX
Currently I'm searching for a nice way to implement my ChemDraw graphics into $\mathrm{\LaTeX}$.
There is a package called chemnum which will auto-label the compounds in the eps-graphic which sounds nice. The package describes, that I need to put "temporary markers" inside the document where the number should be.
How do I get these markers? I guess it is not a normal text area, isn't it? I'm using ChemDraw 16.0.1.4.

It's better to ask $\mathrm{\LaTeX}$-related questions on TeX.SE. There is already a plethora of similar questions there, check them out – probably you'll find an answer.

This is not about LateX. It is about chemdraw. Thats why the question is in here but I will check out the Questions in the Tex.Stackexchange

I'm voting to close this question as off-topic because it's software issue.

Just put tmp1 etc below the compounds you want to label, then let the latex macro replace those with the actual numbers from the aux-files. This requires at least two runs, and you are limited to using epslatex. There are certain drawbacks to this.

The ChemNum package documentation that can be found on CTAN specifies:

Create the scheme and place temporary tags like TMP1, TMP2 and so on where you want the compound labels to be.

A question on TeX.SX specifies:

When you create a text box and type into it, it will store it as a string correctly in the EPS. However, it was (sometimes) automatically putting it into chemical formula mode - so that in TMP1, the 1 was subscript (TMP1). Even if you fix this manually, it now saves the TMP and 1 separately in the EPS, meaning CHemNum fails to find TMP1.

Thus, it seems all you need to do is create textboxes with the appropriate labels.

And for creating textboxes with the appropriate labels in ChemDraw see also this question: https://chemistry.stackexchange.com/q/45055/7951


text:
Why does using NaI as a nucleophile and acetone as solvent lead to an SN2 reaction mechanism instead of SN1? Is I- not considered a weak nucleophile?
In a reaction with a secondary leaving group, why does using $\ce{NaI}$ with acetone as a solvent lead to an $\mathrm{S_N2}$ reaction mechanism instead of $\mathrm{S_N1}$? I thought that halogens were weak nucleophiles, and only really led to $\mathrm{S_N1}$ reactions?

Students seem to like hard and fast rules like ‘$\ce{\text{iodide} -> \text{weak nucleophile} -> \mathrm{S_N1}}$’ where reality allows for grey tones, in between scenarios and others.

@Jan could you explain why this is not a hard and fast rule then? In most of the sources I read about this, they all just say NaI is good example of an SN2 reaction, so it seems like that is the case in most scenarios, and I'm just trying to understand why.

If it’s a nucleophile, it’s capable of $\mathrm{S_N1}$ and $\mathrm{S_N2}$ pathways. The ratio of $\mathrm{S_N1}$ to $\mathrm{S_N2}$ depends on factors such as the leaving group, steric influence of the electrophile, the solvent, the temperature and the strength of the nucleophile. That’s all there is to it. Saying ‘x will only do y because z’ is wrong most of the time.

No, iodide is an amazing nucleophile. It also also happens to be a great leaving group, so substituting for iodide and having it stay on the product is hard.

As Jan wrote, there isn't a hard and fast rule encompassing every mechanism, so there may not be a definitive answer to the question. Nevertheless, I found an interesting site that talked a little bit about the Finkelstein reaction:
$$\ce{NaI + RX -> NaX + RI}\qquad\ce{X} = \ce{Cl}, \ce{Br}$$
This is in acetone solvent (I don't know how to format it along the arrow).
If you go by Henry Rzepa's blog, the solvation of sodium halides by acetone leads to a solvent shell that isn't well "fit" for a chloride or bromide anion. The writer discusses how the gaps between solvated sodium ions is just large enough to fit an iodide anion, which means the anion has minimal distance to the sodium atoms around it, and therefore a minimized potential energy. Bromide and chloride are too small, leaving a gap of empty space around them, and the distance to sodium cations is larger than optimal. It becomes energetically more favorable for the sodium and halide ions to simply fall out of solution than arrange themselves in a solvated cage. This drives the reaction forward in accordance with Le Chatelier's principle.
While this relates to the solubility of the halide salts, maybe it can also serve to speculate on the mechanism of the Finkelstein reaction. In the vicinity of the solvated $\ce{NaI}$ cage, there isn't a favorable space to release bromide or chloride into, so the iodine attack is forced to proceed through $\mathrm{S_N2}$.
This is far from rigorous research, so I'd take it with a grain of "salt" as a correct answer, but thought it might be interesting to think about.

NaI in acetone is the reagent used in Finkelstien reaction. Alkyl halides are treated with the reagent to give alkyl iodides.  The important aspect of this reaction is that the reaction is driven forward because of the lower solubility of the products NaCl and NaBr in acetone even though I- is a weak nucleophile.

This in itself doesn’t explain why one nucleophilic mechanism should dominate over another.

Note :
When does SN2 attack occur. Let's analyse this and you will get your answer in end.
First of all the steric hindrance around the carbon atom should be least i.e if RX is 1° (primary) then SN2 attack always (mostly) occurs.
If RX is 3° (tertiary) then SN1 attack always (mostly) occurs, as steric hindrance is very high.
Now if RX is 2° (secondary), the mechanism can be SN1 or SN2, depending on the Nucleophile and the solvent.
If solvent is aprotic  and nucleophile is fairly good (strength): like DMF, acetone etc., then SN2 attack is favoured (I am not explaining why it occurs, but you can refer to the mechanism)
If solvent is protic and nucleophile is not so strong: then SN1 mechanism is favoured.
In your question, $\ce{I-}$ (iodide ion) is mild nucleophile and acetone is aprotic solvent. Thus SN2 mechanism is favoured more.


text:
Thought process behind Rutherford's gold foil experiment
In Rutherford's gold foil experiment, if he at first thought the atom looked like the plum pudding (as Thomson had said), why did he think the alpha particles would not be repelled by the positive "pudding" and that they would go straight through?  Is it because of the high energy of the emission of the alpha particles?

Yes high speed of the alpha particles and relatively small electric field at the foil/atoms due to spreading of charge in the plum pudding model.

They already knew that alpha particles propagated through air, although the beam did spread out as one would expect from low-energy-transfer collisions (some did actually backscatter but were not detected in those experiments). To backscatter you need to obey the standard kinematic formulas - what you hit has to be heavier than you are, and in the plum pudding 'model' there were no compact high mass objects that would lead to backscattering. Forward scattering leading to broadening of the beam, as observed in air, was expected.

Rutherford conducted his gold foil in order to prove Thomson's plum pudding model but the result was contradicting.
He expected that most of the Alpha particle will show very minor deflections  because the positive charge was evenly spread throughout the atom according to the plum pudding model.

To my knowledge, this statement is not true. Thomson's model was nowhere nearly universally accepted.


text:
How could you accomplish the following transformation in sufficient yield?
I've considered Riley oxidation but the major product would have a hydroxyl group at the benzylic carbon to the left of the double bond instead of the methyl group as shown. Is there a way to make the reaction favor oxidation of the methyl group?

This is really not straightforward. Any conditions you choose are likely to react at one of the benzylic sites and aromatise the top ring.

The structure shown is similar to dronabinol which has an allylic group and a benzopyran ring. There is another drug called nabilone in which the allylic methyl group is oxiidized to the cyclohexanone. Search for these 2 compounds in the literature may give you some idea that will inspire you.

I'm thinking a Pd catalysed allylic acetoxylation might work as the methyl group is the least hindered site for attack. I don't think it would go exclusively at that point though.

As has been stated by @Beerhunter those benzylic hydrogens are a serious concern . Nonetheless, here is a different offering. To solubilize carboxylic acid 1 both the phenolic group and the acid should be protected. Here, both of the groups are protected as methyl groups. [An ester group and silylated (TBS?) phenol would also suffice.] The double bond of 2 is epoxided to 3. The critical step is the isomerization of the epoxide to an exo-methylene allylic alcohol. There are several reagents that do this transformation. Yamamoto1 has shown that methyl groups are more reactive than methylene groups with the epoxide rearranging reagent diethyl aluminum 2,2,6,6-tetramethyl piperidide although no benzylic methylene examples were provided.
Acetylation of the allylic alcohol affords allylic acetate 4. Grieco2 has reported that bis-acetonitrile PdCl2 isomerizes allylic acetates to their tautomer in what is an equilibrium reaction favoring the more stable isomer. In this instance, the trisubstituted, endocyclic double bond compound 5 is expected. The remaining steps involve removal of the protecting groups of choice to afford the target 6.

1) S. Tanaka, A. Yasuda, H. Yamamoto and H. Nozaki, J. Am. Chem. Soc., 1975, 97, 3252.
2) P. A. Grieco, T. Takigawa, S. L. Bongers and H. Tanaka, J. Am. Chem. Soc., 1980, 102, 7587.

I like this answer, but fear for the atom economy of the sequence.

@Beerhunter: At this point, I don't think atom economy is an issue.

The methods required are generally outside of the mainstream lab chemistry,  but that does not mean they can't be used in lab setups.
As Waylander states, use of oxygen with acetic acid can be catalysed to acetyloxylate the molecule. Then the acetate ester is simply cleaved.
The main issue in this molecule is the activated benzylic position. These are usually prone to oxidation without trying. So most methods suggested may also functionalise this benzylic position too:
Radical bromination with NBS and an initiator; biocatalysts; singlet oxygen.
If bromination is used,  one could deliberately exhaustively react until the allylic position is functionalise, then selectively reduce the benzylic position - this is possible using catalytic hydrogenation or transfer hydrogenation through careful catalyst and condition selection.
A bit more fiddly is using singlet oxygen.  The best method to use this is through flow rather than batch methods (also safer). However,  formation of allylic hydroperoxides from allylic functionality is well documented. A reductive step then needs to take place to yield the alcohol. Sulfides and triphenylphosphine are usually cited but other methods are possible.
All the methods above are at least two steps though apart from biocatalysis.

I'm not an organic chemist, but upon looking for similar systems I found a nice collection of various papers that all use selenium dioxide, so allylic oxidation as you mentioned it with various other functional groups. I copied the sections from the papers (therefore sorry for the quality and the Chinese in one paper) together with the preparations.
To me this looks like the most promising route here. According to that mechanism it should favor the allylic position here selectively and therefore and also due to sterics hopefully not oxidize the longer alkyl chain.

Riley oxidation is SeO2 oxidation.

None of the examples you quote have the adjacent aromatic ring which makes things much more difficult.

How would that ring influence it? I can't really see any delocalization of the benzene ring into the other ring. So it probably only causes some minor effects on the electron density in the double bond.

Did you read question properly? There are 3 allylic postiton - 2 of which are also benzylic, and OP wants OH in the one which isn't. Probably undoable in single step.

@Justanotherchemist  Look at the starting structure; it is a dihydronaphthalene. Show it an oxidising agent and it will aromatise. That is what makes this challenging. If I was tasked with making the target structure I would not start from here!


text:
Does the hydrohelium dication exist?
It is well-known that there exists the $\ce{He2^2+}$ dication, and also $\ce{HHe+}$ and $\ce{H2+}$ cations. But I could not find any definite information about the existence or inexistence of the $\ce{HHe^2+}$ cation.
It would be isoelectronic to $\ce{H2+}$, but would have a weaker bonding due to greater helium nucleus charge. It looks unlikely to be stable, but did anyone prove it by calculation or experiment?

There is a paper by S. K. Knudson, Semiclassical energies of low-lying states of one-electron diatomics. By skipping through the paper I found some calculations on your compound. But I didn't have time to read through yet. So perhaps it's not helpful at all to you.

Just in case you want to look at the paper though: https://doi.org/10.1016/S0301-0104(00)00006-9 In my first skim I found the following quote (tl;dr): 'The only bound ground state is that for $\ce{H2+}$; the other cations have repulsive ground states. '

@Martin-マーチン, however they might have stable excited states. Surely there must be at least one. This is also the case for the He2 dimer.

I am not sure how to elaborate much beyond saying that yes, there are bound states of $\ce{HHe^{2+}}$. I cite three papers below which give numerical calculations of various states, ref. 1 gives calculations for the lowest 20 states of the system.
It appears to be that the 1s$\sigma$ is not a bound state in this system, but the 2p$\sigma$ state is bound. I would expect there are some other bound states as well.

References:
[1] Winter, T. G., Duncan, M. D., & Lane, N. F. (1977). Exact eigenvalues, electronic wavefunctions and their derivatives with respect to the internuclear separation for the lowest 20 states of the HeH2+ molecule. Journal of Physics B: Atomic and Molecular Physics, 10(2), 285.
[2] Laaksonen, L., Pyykkö, P., & Sundholm, D. (1983). Two‐dimensional fully numerical solutions of molecular Schrödinger equations. I. One‐electron molecules. International Journal of Quantum Chemistry, 23(1), 309-317.
[3] Buehler, E. J., Gooch, E. E., Dial, J. L., & Knudson, S. K. (2000). Semiclassical energies of low-lying states of one-electron diatomics. Chemical Physics, 253(2-3), 219-230.

Well, I'm pretty sure there are various ways to elaborate. Right now your refs are longer then the rest :D What's for example the energy gap between these states, or half-life?


text:
Thermodynamics of a spontaneous isothermal chemical reaction.

Which of the following is/are correct for spontaneous isothermal
  chemical reaction?
(A) $\Delta H = 0$,  because $\Delta T = 0$
(B) $\Delta S = 0$
(C) $\Delta U = 0$,  because $\Delta T = 0$
(D) $\Delta G < 0$

My answer: $\mathrm{A, C, D}$
Reasoning:
$U$ is a state function and only depends on the initial and final states and their temperature. Since $\Delta T = 0$, $\Delta U = 0$
$\Delta H = \Delta U + \Delta(PV)$, but $PV = \text{constant}$, so $\Delta H = 0$
$\Delta G <0$ - the requirement for any reaction to be spontaneous.
But answer given is only option D.
I think the flaw in my $\Delta H = 0$  argument is that reactants can be solids and liquids as well and we cannot apply $PV = \text{constant}$ to them. Am I right?
But I  still can't figure out why $\mathrm C$ is wrong.

For ideal gases U is a function of T only, but not generally. Even for ideal gases $\Delta H = \Delta U + RT \Delta n$ and you are considering a reaction where $\Delta n$ may not be 0.

I want to say this is a duplicate of https://chemistry.stackexchange.com/q/39645/16683...

The internal energy and enthalpy of a pure ideal gas depends only on temperature.  But, the internal energy and enthalpy of an ideal gas mixture depends both on temperature and on the amount of each chemical species that is present.  To be more precise, it is equal to the internal energy and enthalpy per mole of each component as if it were a pure species times the number of moles of that species present, summed over all the species in the mixture:  $$U=\sum{n_iu_i}$$and$$H=\sum{n_ih_i}$$So, even at constant temperature, $$\Delta U=\sum{(\Delta n_i)u_i}$$and$$\Delta H=\sum{(\Delta n_i)h_i}$$You should have been taught this when they were covering the thermodynamics of mixtures and chemical reactions.

I'm confused why your answer is about ideal gases when the question doesn't say anything about the nature of the reactants and products.  Does the question assume they are gases, much less ideal ones?

No.  My answer just uses ideal gases as an example.  If one of the given choices is not true for an ideal gas (mixture), it certainly isn't true for a chemical reaction in a mixture of real gases (which exhibits ideal gas behavior in the limit of low pressures) or for a liquid phase reaction.  Incidentally, OPs analysis was strictly valid also only for ideal gases.  In real gases, the internal energy and enthalpy are functions of pressure, and PV is not constant even if temperature is constant.  What I didn't want to do in my discussion was to start introducing partial molar properties.

That makes complete sense.  (If you add the third sentence of your comment to the answer, you'll get a +1 from me, if such things matter to you.)

The correct answer is (D).
Systems can be held at constant temperature by being placed in a heat bath.  This lets heat flow into and out of the system as needed to ensure that the temperature stays constant.
A. $\Delta H=0$ is not necessarily true, because an exothermic (or endothermic) reaction releases enthalpy.  For example, consider the reaction $\ce{C6H12O6 + 6 O2 -> 6 H2O + 6 CO2}$.  This is the reaction that your body uses to get energy from eating sugar, and it happens at nearly isothermal temperatures (37 °C).  But enthalpy is released, and used to power your body's metabolism.
B. $\Delta S=0$ is not necessarily true, because systems held in a heat bath can exchange heat with their surroundings, leading to entropy changes.  The glucose oxidation reaction above releases enthalpy, and without heat flow to the heat bath, it would result in a temperature increase.  But with heat flow, the change in entropy cannot be zero.
C. $\Delta U=0$ is not necessarily true, because (among other reasons) $U = H - PV$, and we have already established that $\Delta H$ is not necessarily zero.
D. $\Delta G < 0$ is true because that's essentially the definition of a spontaneous reaction.  If it weren't true your body couldn't get energy from metabolizing sugar.


text:
Why are lyophillic sols uncharged and lyophobic sols charged?
Gums, resins, glue, gelatin starch are uncharged and lyophillic where as metal sulphides and hydroxides are charged and lyophobic.
Right after studying o-chem this seemed quite contradictory as polar solvents are supposed to dissolve polar solutes and non polar solvents non polar solutes.
Why does this not work for sols, why are polar/charged compounds lyophobic but uncharged lyophillic?

I am late but can u try adding more tags ? I have the same doubt


text:
NaOH or lye. Would it work to remove ice snow or water from a paved surface?
Thinking of ways to clear ice using chemical reaction.

Yes, it probably will; it is terrible for the environment and the roads though.

All you are producing is a salt solution with a lower freezing point than pure water. There is no chemical reaction.

Lye is dangerously corrosive, so even if it worked, it would be a very bad choice as a deicing agent.

Hope everybody avoids skin contact with the paved surface after you've applied NaOH to it.

The grass probably wont grow back for a while, and parents may be upset that kids got caustic burns on their skin, you may or may not also get fined by authorities, but yes lye will melt ice.

If you insist on a chemical reaction, combine a little salt (on concrete, calcium chloride is better) with magnesia (the unhydrated oxide, not the hydroxide).  The salt melts the ice and then the magnesia, a basic anhydride, reacts with the water.  Then the salt is free to melt more water, it's acting as a catalyst while the magnesia is the actual reagent.  The product of this reaction is precipitated instead of destroying the concrete/asphalt/grass, but that also means you're left with a mess.
Or just use salt alone and rely on the usual freezing point lowering.  :-)

Use CaCl2 instead of NaCl on concrete.


text:
Why does diatomic oxygen have five filled molecular orbitals, but is considered to have two bonds?

So I'm taking an introductory chemical physics course and we are covering LCAO-MO.
We learn about the molecular orbitals that arise out of a combination of the two atoms atomic orbitals. This totally makes sense to me, but then we start learning about electron configurations and orbital diagrams. When we get to oxygen the textbook shows that oxygen has 8 electrons each, and when they combine they fill up the σ1s σ2s σ2pz π2px π2py bonding orbitals and the σ*1s σ*2s antibonding orbitals (with the π*2px π*2py antibonding orbitals each half filled). My question is:
Why does this result as a double bond?
I see it is partially addressed here: Molecular Orbital Theory and No. of bonds
But what they don't speak to is why there are two filled π2p bonding orbitals but only one pi bond. I understand how to calculate bond order but in my understanding thats our approach at reasoning out bond strength, it returns an integer value - it does not really tell me much about the actual bonds themselves. Is the pi bond that results some sort of hybrid between the π2px and π2py orbital? If there are two electrons each in the π2px and π2py orbitals, how does one electron in each of their respective antibonding orbitals play out in the formation of the bond?
Thank you ChemSE community!
*editted to further clarify I am aware of bond order calculations


text:
Why does Carnot Engine have the curves as isothermal and adiabatic?
Why exactly does the Carnot engine have two isothermal and two adiabatic curves? (My book says that “a process is quasi-static if the temperature difference between the system and the reservoir is infinitesimal. Hence it is isothermal. For the connecting two isotherms ;only reversible adiabatic is suitable. If(say) it were isochoric then we shall need a series of reservoirs in the temperature range defined by the two reservoir and the sink” .I don’t get it. Please help.

To maximize the efficiency of a heat engine, you need to have work done by the engine to be maximum, whereas the heat input to be minimum. If you considering two heat reservoirs between which the engine is operating, you need to have isothermal processes, as the temperature doesn't change, but to satisfy the above mentioned criterion, you will find out that adiabetic processes are the best ones as there is no heat input, but still the engine does work. If you consider any other process instead of adiabetic process, you can't maximise the ratio of work done to heat input to the engine.

I'm voting to close this question as off-topic because this question is not about chemistry. The carnot engine is a set of physical processes.

Well then you may as well be in favour of shutting thermodynamics out of chemistry.

@Jan I think the question might be homework, but I disagree with closing the question on the grounds that it is about physics. The OP has a point that it would seem to say that much of thermodynamics such be off-topic. On the other hand, I agree that the OP might get a better answer to this on Physis SE (once they add a little more of their own thoughts on the problem they are having)

@Tyberius Not all of thermodynamics but I do see the extreme end as being practically physics only. Not that I come from pretty much the opposite side of the spectrum being an organic chemist.

@Jan I think the line between the two is blurring more as time goes on. I would agree this is probably closer to physics than chemistry, but its a very commonly covered topic in chemical thermodynamics courses. For a lot of physical chemistry, you do have to make detours through topics that are almost entirely just physics or math to make sense of certain useful concepts (e.g. isotherms, adiabatic, engines, and heat sinks)

By definition, a Carnot cycle is comprised of 2 reversible isotherms and 2 reversible adiabats.
Regarding your question about isochoric heating or cooling, if you are going to carry this out reversibly, then you can't do it with just one constant temperature reservoir.  For the heating or cooling to be reversible, the temperature difference between the system and the reservoir(s) must be tiny over the entire path.  To do this, you need a sequence of constant temperature reservoirs running from the starting temperature to the ending temperature.  After the system has equilibrated with each reservoir in the sequence, you move on to the next reservoir.  That way, the temperature difference between the system and the reservoir is tiny along each step of the process path.

I think I meant why would it be that only those specific curves imply maximum efficiency for which the Carnot cycle is known for?

You can represent any reversible cycle as a superposition of mini Carnot cycles.  A diagram of this is in virtually every thermodynamics book. The mini Carnot cycles have different temperature extremes, so each of the mini cycles has its own efficiency.  The overall efficiency you the cycle is a weighted average of these mini-cycle efficiencies.  But it will be less than the efficiency calculated from the pair of most extreme temperatures.


text:
Why nitrogen monoxide does not dimerise
This question has already been asked link Why doesn't nitrogen monoxide dimerize?
but explination is not clear. I would like it to be explained in simple manner using
 molecular orbital diagram is included please explain in simple term using molecular orbital theory.
Why this is not the case
In NO2, when drawing it's Lewis structure, you'll see that valency of nitrogen isn't satisfied, there is and odd electron left on it making it somewhat unstable.
enter image description here
Odd electron get fulfilled when it dimerises.
enter image description here

If you think the answers to the question aren't sufficient, try gaining some rep and offering a bounty.

I have already put a link saying that its already been asked

Nitric oxide (nitrogen monoxide) has one more electron than the N2 molecule. N2 has a sigma framework (sp hybridized) with 2 electrons in the outward lone pair on each nitrogen and one electron from each nitrogen in the sigma bond. The remaining 2 electrons on each nitrogen are in the two pi bonds, so that each nitrogen has a total of 8 electrons: 2 in the lone pair, 2 in the sigma bond and 4 in the two pi bonds. N2 has a bond order of 3.
The extra electron in NO goes into an antibonding pi molecular orbital. The resulting bond order in NO is 2.5. A high-energy antibonding pi electron is not well designed for bonding to anything, and yet, solid NO has been found to consist of loose dimers with a rectangular structure: the tightly bonded N-O bond is 1.10 Angstroms long and it is bonded N to O and O to N, with a separation of 2.38 A between the monomer NO molecules.
Reference is Advanced Inorganic Chemistry, Cotton and Wilkinson, 1966


text:
pKa Conceptual Questions
If anyone could clear up some of this conceptual confusion, that would be much appreciated.

Is $\mathrm pK_\mathrm a$ an intrinsic property of a molecule? Or extrinsic? A property is extrinsic if it depends on the environment (e.g. weight), and intrinsic if it is independent of the environment (density). Source: Thoughtco: Intrinsic Property Def.
Doesn't $\mathrm pK_\mathrm a$ change according to environment/conditions the molecule is in? According to the definition, $\mathrm K_a = [\ce{H^+}][\ce{A^-}]/[\ce{HA}]$. Then, given a $\mathrm pK_\mathrm a$ table, what are all the $\mathrm pK_\mathrm a$ values calculated with reference to? What are the standard conditions in which scientists calculate their $\mathrm pK_\mathrm a$ values.
Is there a way to calculate $\mathrm pK_\mathrm a$ from known intrinsic properties of a molecule? E.g. if I give you one molecule of a compound, and only one molecule, is it possible to calculate $pK_a$ based off the intrinsic structure of that molecule, given no other information?

It's not a property of molecule at all. Just some random equilibrium constant of highly limited use.

Well, given that it's related to the equilibrium constant of a reaction the molecule takes part in, I would say it's as much of an 'intrinsic' property as equilibrium constants in general are (they depend on the temperature and a few other things). In this case, the other reactant is water, present in excess as the solvent; perhaps that's why we tend to think of $pKa's$ as properties 'of the molecule'. They are in fact properties of the acid+water system within the specified conditions.

The tables you find in the literature are just experimentally determined values of $pKa's$ in water; I don't see what 'reference' you mean. The temperature is usually mentioned.

As for the 'prediction' of $pKa's$, there are in silico tools that are able to calculate them, usually based on QSAR models. In essence, a lot of $pKa$ data were examined, and the structural features of the molecules were associated with their contribution to the $pKa$ value. This was the 'training' of the model. Then the model becomes like a function taking as input the structural features, and producing as output a prediction of the $pKa$, with various degrees of accuracy depending on how good the training set and the model itself were. So yes, it is possible.

BTW, I am fairly shocked by the above statement that $pKa's$ are of 'highly limited' use. In my day-to-day work (computational chemistry, drug discovery), $pKa's$, either predicted or experimental, are essential to many of our activities, and have an enormous impact on many properties of biopharmaceutical interest. Not to mention the application in synthetic chemistry, where the choice of a base is fundamentally down to its ability to deprotonate certain groups (sometimes selectively).

@user6376297 Why highly limited? Your constant is for highly diluted water solution  (or other specific solvent) and specific temp. I wouldn't be surprised if an experimental value wouldn't be good enough even in certain in vivo situations. If one looks from broader perspective - how many solvents are used in organic synthesis? What about gas phase? What about highly concentrated solutions? There are good reasons why Hammett scales are used.

Thank you @user6376297 for your answers. I wasn't aware of QSAR models, and I assumed the pKa's were calculated using water, but I wasn't sure. You've answered all of my questions, but now that this question is on hold, how do I mark this post as resolved?

@Mithoron On a practical scale, I haven’t heard of the Hammett scales in my day-to-day lab work at all but $\mathrm pK_\mathrm a$ values are frequently referenced. The water (sometimes DMSO but usually water) values are definitely ‘good enough’ for all our chemical considerations.


text:
Titration curves
If I were to model the titration of $30 \ \text{cm}^{3}$ of $ 3 \ \text{mol dm}^{-3}$ $\ce{CH3COOH_{(aq)}}$ with $3 \ \text{mol dm}^{-3}$ $\ce{NaOH_{(aq)}} $ , what would the equation look like (graphing volume of base added vs. pH of the resulting system)? I am able to obtain a graph that looks like a titration curve by using the Henderson-Hasselbalch equation [more details below], but according to Wikipedia, the equation fails when

a strong acid or base is involved
strong concentrations $> 1 \hspace{1mm} \text{mol dm}^{-3}$ are involved.

These conditions apply to this situation. Nevertheless, the equation I have found (it's very slightly not continuous) is
$$f(v) =
\begin{cases}
4.76 + \log_{10}{\left(\dfrac{3v}{0.09 - 3v} \right)},  & v \in ( 0 , 0.03) \\
8.86 & v = 0.03 \\
14 - \log_{10}{\left(\dfrac{3v - 3}{0.03 - v} \right)}, & v \in (0.03 , \infty)
\end{cases}$$
I would like to know how to fix this issue (if there is any way to better represent this titration). Thank you.

Before the equivalence point
$$\text{moles of $\ce{H+}$}: 30 \ \text{cm}^{3} \cdot {1 \ \text{dm}^{3} \over 1000 \ \text{cm}^3} \cdot 3 \ \text{mol dm}^{-3} = 0.09 \ \text{mol} $$
$$\text{moles of $\ce{OH-}$}: v \cdot 3 \ \text{mol dm}^{-3} = 3v \ \text{mol}$$
calculating the acid and its conjugate base
using an ICE chart,
$$\ce{CH3COOH + OH- <=>CH3COO- + H2O}$$
$$\text{moles of $\ce{CH3COO-}$}: 3v \ \text{mol}$$
$$\text{moles of $\ce{CH3COOH}$}: 0.09 - 3v \ \text{mol}$$
$$[\ce{CH3COO-}] = {3v \over 0.09 + v} \ \text{mol dm}^{-3}$$
$$[\ce{CH3COOH}] = {0.09 - 3v \over 0.09 + v} \ \text{mol dm}^{-3}$$
using Hendersen-Hasselbalch
if $f(v)$ is pH,
$$f (v) = 4.76 + \log_{10}{\left(\dfrac{3v}{0.09 - 3v} \right)}$$
At the equivalence point
using an ICE chart,
$$\ce{CH3COO- + H2O <=> CH3COOH + OH-}$$
$$K_{b} = 10^{-9.24} = {v^{2} \over 0.09 - v}$$
$$0 = v^2 + 10^{-0.24} x - 0.09 \left(10^{-9.24} \right)$$
$$v = 8.86$$
After the equivalence point
To be honest, I think just screwed around with this graph until I got a similar shape. I knew that that
$$\lim_{v \to \infty} f (v) = 14$$
at standard conditions, so I subtracted the log function from 14.

I don't think there is any need for piecewise functions. The equation relating added titrant and pH is analytically solvable and continuous. See https://chemistry.stackexchange.com/questions/73448/finding-ph-after-dilution/73554#73554 . It all boils down to a system of mass balances and equilibrium constants, really. See also: https://chemistry.stackexchange.com/questions/87644/nahco3-required-to-raise-ph-of-weak-hcl-solution-from-3-5-to-7-2-doubting-my-ca/87670#87670


text:
Differences in Glycerol Nomenclature
Could someone differentiate between the terms glycerol, glycerin, glyceryl and glyceride?

Google is your friend! https://en.wikipedia.org/wiki/Glycerol

There is no difference between glycerol and glycerin; they both refer to propane-1,2,3-triol (already answered here).

Glyceryls refer to glycerol derivatives in which a hydrogen has been replaced (e.g. glyceryl trinitrate).

Glycerides represent a specific example of glyceryls and are fatty acid esters derived from glycerol (e.g. triacetin).