text:
Why does the acid in hops make beer taste bitter?
For years, I've learned that acids taste sour and bases taste bitter.  Recently, I've read in "The Beer Bible" and various beer sites that the hops in beer, which are acidic, cause beer to have a bitter taste.
So my question is, is it true that the acid in hops make beer bitter?  And if true, does this mean that (Arrhenius) acids aren't necessarily sour tasting, but sometimes bitter?
Alternatively, is there some kind of chemical reaction between hops and beer that make beer more basic, hence the bitter taste?

While there is certainly quite a bit of chemistry involved here, you might get a more general and tailored answer at [beer.se]. It might be a bit too broad here, but - as a non-specialist - I would be hesitant to vote on that reason.

It doesn't. Then again, acidity and sour taste are pretty much the same thing; basicity and bitter taste, not so much. There are many compounds which are neither very acidic nor very basic, yet taste way more bitter than any base.

Well that all acids taste sour should be taken with a grain salt. As for not all salt are tasting salty :)

Beer has a moderately acid pH, 4.0-5.0 would be typical. Sour is based on pH not total acid content, many acids are weak acids, or are soluble in small amounts (like fatty acids), and some like lewis acids can have a pKa above 7.0 (meaning they buffer to a basic pH) because acids are defined by their effect on protons and hydroxide ions not by pH.
Alpha-acids are compounds in hops that are isomerized by boiling for an extended time, one hour is a typical, cold soaking hops does not add alpha-acid related bitterness.


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First point of difference rule 1,6-dimethylcyclohex-1-ene

According to the first point of difference rule, second one should be correct right?

The name on the right is incorrect because it gives the locant 2 to the first substituent, whereas the one on the left gives locant 1 to the first substituent.

Please note that the ‘lowest sum rule’ that is mentioned in the picture does not exist. See also: https://chemistry.stackexchange.com/a/28009/7951

First of all, the ‘lowest sum rule’ mentioned in the picture in the question does not exist in IUPAC nomenclature.
As already explained in a related answer, the most important simplified criteria for the numbering in such cases are:

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

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The compound that is given in the question doesn’t have any principal characteristic group that could be expressed as a suffix. Therefore, Rule (c) is not relevant in this case.
Thus, a low locant is assgined first to the double bond according to Rule (e). Therefore, the name of the parent structure without any further substituents is cyclohex-1-ene. (Without further substituents, however, the locant ‘1’ would be omitted.)
Finally, low locants are assigned to the remaining substituents as a set according to Rule (f).

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.

Thus, the correct name is 1,6-dimethylcyclohex-1-ene rather than 2,3-dimethylcyclohex-1-ene since the locant set ‘1,6’ is lower than ‘2,3’.

@ Loong +1 for elaboration.In several of your posts ,you have quoted Blue Book .Would it be possible to obtain pdf for the same.If possible a link would help.

The highest priority is double bond for this compound. And then consider to give lowest sum for substituents on double bond. Here is only one substituent present on double bond (methyl group), hence it should get the lowest number. Therefore, 1,6-dimethylcyclohex-1-ene is the correct name (1,6-dimethyl-1-cyclohexene is also correct). In either way, note that there should not have a space between 1,6-dimethyl- and cyclohex-1-ene.

No, the right one is incorrect as the numbering starts at the double bond, it is suppose to start at the first substituent.

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What is the purpose of a boiling stick when heating a reaction mixture?
What is the purpose of a boiling stick when heating a reaction mixture?
I read some stuff online and still don't understand how it works. Does it just ensure that the mixture will boil evenly? If so, then what's wrong when a reaction doesn't boil evenly?
This is the LibreTexts page I was reading information from. I don't understand the terms "bump" and "superheated". If the liquid is above the boiling point, why hasn't it turned into a gas then?

Why the page you have linked does not satisfies you? For the final question: thermodynamically it should be a gas. That gas had not time, not yet, to form a bubble and move freely to the surface. This can happen suddenly causing kind of "explosions" spilling liquid around or even disjoint the glassware.

That last explanation was what I was wondering about. Thanks for that.

In undergrad chem lab I can attest to it pretty much violently emptying a test-tube once.

Boiling chips and boiling sticks are practical solutions to problems chemists face in the lab.
When boiling liquids they often observe the the boiling does not happen evenly. Instead of a gentle warming giving gentle bubbling we find that nothing happens for a while until, suddenly, a great deal of bubbling happens and this causes the liquid to splash overwhelming the capacity of the vessel and throwing the liquid into places we don't want it to be (during distillations, for example, the splash may throw liquid to the top of the distillation column, contaminating the product with raw liquid; during reactions the product can be expelled from the vessel entirely which is rarely what you want).
Boiling chips or sticks ensure the boiling happens evenly,  keeping things under control.
The problem they solve is that, in a smooth vessel, there are often not many rough patches to initiate bubbles. Bubbles are much easier to initiate if there is a small rough patch on the vessel wall. Chips or sticks provide such a rough patch. In the absence of a rough patch some liquids will superheat and may spontaneously vaporise but suddenly and violently. Roughness ensures bubbles start to form evenly and in many places avoiding superheating and making the vaporisation much more even.
The idea that things won't happen this way as "If the liquid is above the boiling point, why hasn't it turned into a gas then?" is a mistake from applying the ideas of thermodynamic equilibrium to a situation that is clearly not in equilibrium. If you are heating something it is clearly not in equilibrium and you have to worry about the practical effects not the ideal thermodynamic view. In practice, vaporisation is encouraged by nucleation sites (rough patches on the glass, for example). If there are not enough of those, it may not happen evenly.

The boiling point of a substance is defined as the temperature at which the vapor pressure of a liquid or solution is equal to atmospheric pressure; it does not mean that gas has started precipitating out of solution.  The key here is to focus on the fact that boiling is gas precipitating out of solution in much the same way that a solid precipitates out of a supersaturated solution.  Have you seen the precipitation demonstrations where a supersaturated solution has a crystal of the solute introduced and the precipitate immediately starts growing in solution? The boiling chip is serving a similar purpose to the solute being introduced to a supersaturated solution.
In order for the precipitation to begin happening, it requires enough of the precipitating material to overcome the viscosity of the solvent and join together into its own unit.  Boiling chips provide an uneven surface upon which gases can form bubbles and escape the solution; lab glassware typically is too smooth.
As for bumping, you can think of the superheated liquid as being a supersaturated solution where any disturbance can provide the nucleation site. Once the liquid is disturbed, you get precipitation just like in the sodium acetate demos, except this precipitation is a gas which occupies a much greater volume than a solid or liquid, and it splashes all over the place.

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Will ferric nitrate or nitric acid dissolve polylactic acid?
Will ferric nitrate dissolve 3D printed items? I wish to etch some sterling silver, for which we use either ferric nitrate or nitric acid, and wonder about 3d printing the design straight onto the silver to form the block against the acid.

Hi Julie, welcome to Chem.SE! Please take the tour to familiarize yourself with our site. Thanks! Btw, your question title does not match the content? Were you trying to dissolve polyacetic acid or 3D printed items?

@Gaurang That's because PLA(polylactic acid) is used in 3D printing. I did a project on that.


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What is it meant by two different groups in stereoisomerism?
I understand there are some properties in an organic compound that allow stereoisomerism.

double bond can't rotate
two double bonded carbon atoms each have two different atoms or groups attached

I'm unsure by the meaning of two different atoms/groups around on each side of the double bond.

With respect to but-2-ene. From my understanding, both sides of the carbon atom have $\ce{-CH3}$ on either side of the double bond so how does this comply with the rule that "Two different atoms or groups attached around each carbon of the double bond"

Related: Aren't cis and trans isomers geometric?

Stereoisomers are different due to the spatial arrangement of their atoms.
The picture that you gave has two $\ce{CH3}$ groups on one side. That is called cis - as they are on the same side. There is another isomer, called trans. That is when one of the $\ce{CH3}$ is on the top and the other $\ce{CH3}$ is at the bottom. The arrangements of the $\ce{CH3}$ and $\ce{H}$ is different in cis and trans.
This link explains it well: Chem.Libretexts


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Why do objects gain electrons?
I have read that objects gain electrons to become negatively charged and also that they lose electrons to become positively charged. I have also read that same charges repel each other and non-same charges attract each other. My question is how can objects gain electrons when the electrons should repel each  other.
For example: an element has 17 electrons, so its electronic configuration will be:
K L M
2 8 7

So, it will gain 1 electron. But how can this happen? I mean, shouldn't the electrons repel each other? Moreover, if it gains an electron how will it become stable? I mean I agree that there will be 18 electrons but to become stable shouldn't there be 18 protons? So from where will the extra proton come from?

The extra proton will not come at all. Then again, in the middle there is a nucleus with 17 protons, and it attracts the extra electron with the same force as the existing 17 electrons repel it, or even a bit stronger.

but don't the 17 protons cancel out the 17 electrons?? So what attracts the extra electron??

@ProNeon they don't cancel out

aren't electrons and protons opposite of each other??Doesn't electrons carry -1 charge and protons carry +1 charge?? So don't they cancel each other??

They would, if they all were located precisely at one point, which they aren't.

@IvanNeretin Does that mean the protons and electrons have different charges?? I mean does that mean protons have more than +1 charge and neutrons have less than -1 charge??

No, they have exactly opposite charges.

@IvanNeretin So what does exactly causes the attraction??The position??Or something else??And why does position matter??Aren't the electrons revolving around fixed orbits and aren't the protons present in the nucleus??

I guess you know that the electrostatic force depends not only on the charge, but also on the distance, right?

A clear answer would have the shape and size of a textbook on quantum mechanics. Electrons do not revolve, nor do they really have a position.

or maybe https://chemistry.stackexchange.com/questions/18857/electron-affinity would be a better dupe

@Mithoron I fail to see how either of the linked questions answers the OP's question: "shouldn't the electrons repel each other?" Do you mind explaining? Thank you.


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How does pH change during the electrolysis of water?
How does the pH change during electrolysis of a water and  magnesium sulphate solution? Also, will this pH change happen every time? Does the pH change differently when there is just water in the anode and cathode?

If u electrolyse a solution. pH change is uniform no demarcation between anode and cathode pH as the electrolyte is the same.

The pH will increase or decrease or stay the same, depending on where and when and how you look.
At the cathode, $\ce{Mg^{2+}}$ ions will be reduced to magnesium metal. The magnesium  will immediately react with water to evolve $\ce{H2}$ gas and form $\ce{Mg(OH)2}$, which is a strong base (though not very soluble). The pH increases near the cathode. The $\ce{H+}$ concentration due to the slight acidity of $\ce{MgSO4}$ is negligible but any $\ce{H+}$ which is removed by being turned into $\ce{H2}$ gas results in the same effect: the pH increases at the cathode.
At the anode, $\ce{SO4^{2-}}$ anions are oxidized, $\ce{O2}$ is evolved and the sulfate ion is converted into $\ce{SO3}$ and hydrates to $\ce{H2SO4}$. The pH at the anode decreases as sulfuric acid is formed.
If you stir the solution and then take a pH measurement, the pH will not have changed, because the $\ce{H2SO4}$ will react with the $\ce{Mg(OH)2}$ to produce $\ce{MgSO4}$ again, and the only change will have been that $\ce{H2O}$ was converted into $\ce{H2 and O2}$.

Hi James! I see that you’ve been here for quite a while. Please use MathJax to format questions and answers. Thanks!.

The pH will decrease, because magnesium sulphate is acidic.
As the electrolysis proceeds, the magnesium sulphate becomes more concentrated.

Your electrolysis will split water into it's compounds, you just have to look at redox potentials.
$\ce{2H+  + 2e− ⇌ H2}$   E° = 0 V
$\ce{Mg^2+  + 2e− ⇌ Mg}$   E° = −2,362 V
The redox potential of hydrogen is higher, thus it will be reducted before Magnesium.
It is similar with the sulphate and oxygen, in this case the oxygen will be oxidated.
Edit:
DavePhD is right!
The pH will drop a little bit until the solubility limit has been reached, then it should stay constant until there is no water left; no water no pH.
In addition to that Magnesium sulphate has a good solubility in water, the pH will decrease for a longer time until precipitation.


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Why is potassium cation harder than copper?
According to HSAB theory, the smaller radius and the less polarizable an acid is, the harder it is. According to Wikipedia, the ionic radii of $\ce{ K+}$ is much larger than $\ce{Cu+}$, which makes sense because d-electrons are poor shielders. Also, $\ce{Cu+}$ has larger electron density but it has a higher effective nuclear charge in comparison to potassium ion because of the poor shielding ability of d-electrons. Yet why is $\ce{K+}$  ion considered hard while $\ce{Cu+}$ ions are considered soft?

I have misspelt harder in the title can any privileged user/OP take a look.

$\ce{Cu+}$ has a kernel of $\ce{K+}$, with an extra 10 protons in the nucleus and a $\mathrm{3d^{10}}$ electron shell around it. $\ce{Cu+}$ is smaller because it has more positive charge pulling those electrons in. But size isn't everything!
The first ionization potential of potassium ($\pu{4.34 V}$) is much lower than the first ionization potential of copper ($\pu{7.73 V}$). But from there on, the situation changes: Second ionization potential: $\pu{31.63 V}$ for $\ce{K}$ vs $\pu{20.29 V}$ for $\ce{Cu}$; third ionization potential: $\pu{45.72 V}$ for $\ce{K}$ vs $\pu{36.83 V}$ for $\ce{Cu}$; fourth ionization potential: $\pu{60.91 V}$ for $\ce{K}$ vs $\pu{55.2 V}$ for $\ce{Cu}$; and fifth ionization potential: $\pu{82.66 V}$ for $\ce{K}$ vs $\pu{79.9 V}$ for $\ce{Cu}$. It is not till the sixth ionization potential that potassium is again easier to ionize than copper: $\pu{100.0 V}$ for $\ce{K}$ vs $\pu{103 V}$ for $\ce{Cu}$.
So, although the smaller size of the $\ce{Cu+}$ ion suggests that it would be harder (after all, it's stuffed with more electrons!), it turns out that the $\ce{Cu+}$ ion has a softer (i.e. more easily ionized, therefore more easily polarized) outer skin.
Perhaps this is not predictable unless you take into account more than just ionic radii, but at least it is explainable.


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Reactivity of Benzaldehyde between aldehydes
Are Benzaldehydes less reactive than aldehydes due to the -R Effect of the -CHO group present on the benzene ring?

For a carbonyl group to show reactivity toward nucleophilic reagents there should be positive charge on the carbon of the carbonyl group. The existence of positive charge on the carbon allows the reagent to donate electrons much easily.

Alternatively, A compound with more stabilized positive charge will show less reactivity towards nucleophilic reagents. The positive charge could either be stabilized by resonance or electron donating groups. In case of benzaldehyde, the positive charge is highly stabilized (due to resonance) than in other simpler aliphatic aldehydes.

I'd like to add that steric factor is equally responsible for its low reactivity at the carbonyl group.


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Heating p-Bromobenzenesulfonic acid in the presence of dilute HCl
How do I predict the outcome of this reaction?

I expected this to be an acid base reaction leading to the formation of the weak acid para-chlorobenzenesulfonic acid. However, the answer given in the worksheet was surprising $\ce{-SO3H}$ was eliminated and the product stated was $\ce{C6H5Br}$. I would like to be acquainted with the mechanism of this unusual  reaction. I couldn't find it in Clayden or Solomon Frhyle.

Electrophilic sulfonation of aromatic rings is reversible.  The firt paragraph here may be helpful.

I think @ron has hit the bull's eye in the comments! I would like to elaborate upon that:

Since each step in the sulfonation of benzene is an equilbirum, sulfonation is a reversible reaction. According to Le-Chatlier's principle addition of $\ce{H+}$ ions in the solution would shift the equilibrium to the left in the last step of sulfonation where arenium is deprotonated, hence addition of $\ce{HCl(aq)}$ accompanied with heating for kinetic requirements results in the production of 4-bromobenzenesulfonic acid in the given reaction.  This process is called protodesulphonylation.

Well said!  I'll add one more point, since aromatic sulfonation is reversible it can be synthetically useful. You can use the group to block and direct (see here for an example) and then remove the sulfonic acid group afterwards.


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Numerical on change in degree of dissociation wrt pressure

The degree of dissociation of $\ce{PCl5}$ is 0.4 at 2 atm pressure and T kelvin. At what pressure will degree of dissociation of $\ce{PCl5}$ be 0.6?

Taking the degree of dissociation as 0.4, I got the partial pressure of $\ce{PCl5}$ to be $\frac{2}{2.33}$. I then thought of taking the degree of dissociation as 0.6 and multiplying the mole fraction of $\ce{PCl5}$ with the total unknown pressure and equating it to the partial pressure of $\ce{PCl5}$. But I then realized that wouldn't the partial pressure change with change in total pressure?
I'm not able to figure this question out. Any help would be greatly appreciated!

The dissociation of $\ce{PCl5}$ is given by the equation:
$$\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}$$
with $\displaystyle K_\text{p} = \frac{p_{\ce{Cl2}}p_{\ce{PCl3}}}{p_{\ce{PCl5}}}$. Putting this in terms of total pressure, $\displaystyle K_\text{p} = \frac{x_{\ce{Cl2}}x_{\ce{PCl3}}}{x_{\ce{PCl5}}}p_\text{T}$.
In terms of the degree of dissociation, $\alpha$, assuming an initial amount $y$ of moles of $\ce{PCl5}$, at equilibrium, we have $n_{\ce{PCl5}} = y - \alpha y\,\text{mol}$, $n_{\ce{PCl3}} = \alpha y\,\text{mol}$, and $n_{\ce{Cl2}} =  \alpha y\,\text{mol}$. Thus, their molar fractions are $x_{\ce{PCl5}} = \frac{1 - \alpha}{1 + \alpha}$, $x_{\ce{PCl3}} = \frac{\alpha}{1 + \alpha}$, and $x_{\ce{Cl2}} =  \frac{\alpha}{1 + \alpha}$, so:
$$K_\text{p} = \frac{\left(\frac{\alpha}{1 + \alpha}\right)^2}{\left(\frac{1-\alpha}{1+\alpha}\right)}p_\text{T} = \frac{\alpha^2}{1-\alpha^2}p_\text{T}$$
Since at $p_\text{T} = \pu{2 atm}$, $\alpha = 0.4$, we get $K_\text{p} = \pu{0.381 atm}$. Hence, for $\alpha = 0.6$, since $K_\text{p}$ doesn't change (i.e. assuming it's an isothermic system), we get $p_\text{T} = \pu{0.677 atm}$. This is in agreement with Le Chatelier's principle: reducing the total pressure will push the equilibrium towards the side with more gaseous moles, so the dissociation constant in this case will be higher.

Thank you so much! I was confused why Kp would remain same but then realised since Kp=Kc(RT) ^deltaN, since Kc and T are constant and R and delta N are fixed values, Kp must remain the same. Thanks for the help!


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The chemical reaction between lemon juice and polystyrene foam cups
I have noticed that when I have lemon juice on my fingers and I touch a polystyrene foam cup (I will refrain from using the term 'styrofoam'), the cup begins to break down, i.e., melt. I can leave fingerprints on the cup in this manner. What is it about lemon juice and foam cups? What is happening, other than what I see; the cup melting.
I have never tried replicating this using other citric acids, such as an orange. So, I don't know yet if it's just lemon juice that will do this.

It is not any acid in the juice, but limonene, the natural, fragrant, oil found in citrus fruit, and similar terpenes that are good organic solvents.
In fact, these oils are sold for the purpose of cleaning. Note, though limonene itself is considered comparatively nontoxic, after it has dissolved the polystyrene, it might be a slightly more dangerous, since any unreacted styrene monomer in solution is somewhat more toxic.

Is it possible that the fresh juice from the lemon is causing a chemical reaction within the cup when used with a beverage? Or when the juice is added to a beverage, is the dilution enough to prevent a reaction?  Cold tea/fresh lemon vs. hot tea/fresh lemon.

Are you saying that limonene will break down polystyrene? Or that traces of styrene from manufacturing would leech into solution?

I'm not sure about depolymerization, but unreacted monomer might be present, as well as various plasticizers, such as phthalates. Though high temperatures will depolymerize polystyrene, I have noted a strong styrene-like odor when the plastic is dissolved, or even shredded, at room temperature.

I did up voted too early as dissolving polystyrene, or anything, does not involve the breaking of covalent bonds. I doubt for the sudden release of monomer traces as they should be negligible or absent in a cup which is likely thought to contains drinks. Good point for limonene, tough.


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Assignment of the 13C NMR spectrum of 1-(ferrocenyl)ethanol
The molecule in question has the following structure:

(source: sigmaaldrich.com)
The 13C NMR is given below:

My assignments are:
23.700 ppm, -CH3
65.569 ppm, CH adjacent to the -OH
66.138/66.183 ppm; 67.924/67.977 ppm, substituted Cp carbons with protons attached
68.344 ppm; Cp carbons
94.857 ppm; Quaternary C
Are these assignments correct? The Cp carbons usually appear between 80-95 ppm but they are shifted down for some reason (I'm not sure why).
Additionally, why are the substituted Cp carbons (non-quaternary) doublets? I thought it may be due to the presence of staggered/eclipsed forms of the molecule but this would affect all the carbon environments.

Because of the chiral centre outside your ring, C2/C5 on the top Cp ring are diastereotopic (and C3/C4 too). Based on a tiny bit of internet digging, I think the shifts around 65–80 ppm are perfectly normal for ferrocenes - it surprises me more that C1 on the top Cp has such a high shift.

@orthocresol: I agree. For example, substituted $\ce C_p$ carbons in this reference (!http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.475.8371&rep=rep1&type=pdf) are reported as: $\delta$ $68.6, 69.4, 83.6$.

Thanks, I completely forgot about the diastereotopic effect. One more question - in this case, I assigned the 68.344 ppm peak to the Cp protons purely because it had a much larger peak than the others (i.e. the 65.599 ppm peak). Is this a valid way to assign this peak even though 13C NMR integration isn't really a thing?

I think it is perfectly fair, especially in the absence of other data. You can't do 13C integration quantitatively, unless you modify the usual pulse sequence, but qualitatively there is still a link: more carbons = stronger signal. In particular it is the rate of relaxation that messes up the integration (faster relaxation = stronger signal) but between the top and bottom Cp rings I personally wouldn't expect there to be significant differences, except for the quaternary carbon of course. [Also some NOE enhancement due to decoupling but again I don't expect any real differences.]

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Enthalpy of neutralisation for weak acids and weak bases at standard state
I recently read about why for weak acids and bases the enthalpy of neutralisation is less as compared to the strong ones. This was because the acids and bases being weak, they dissociate incompletely and even in this they absorb energy.
I have two questions regarding this:

At standard state, for a solute, we take the hypothetical case of unit molality behaving as one which is infinitely diluted, then shouldn’t even weak acids or bases act as strong acids/bases (because they are, at least in theory assumed to behave as if they were infinitely diluted)?
It's mentioned that during dissociation they (weak acids/bases) absorb heat. Wouldn't even strong ones absorb at least some heat to effect their dissociation.

I find the first question to be more important in terms of conceptual value.

Q1, well you are correct to a certain level. standard state of a solute in a solution of any specified solvent is the state having solution under pressure exactly equal to standard pressure 1 bar,composition of solute equal to 'a standard composition', solution must be an 'ideal solution (perfect solution)', and finally solution must show 'properties of an infinitely diluted solution'. here standard composition refers to either, standard amount concentration (1 mol/dm3), standard molality (1 mol/kg) or standard mole fraction (1). note that temperature is not a part of definition of standard state. this is a hypothetical state for any temperature we use or what standard composition you may chose. here you have chosen standard molality,its ok.(note that we can prove that for an ideal solution showing properties of infinite dilution at a given pressure and temperature molar enthalpy of solute is a constant what ever its composition may be.plug into somewhere that defines you exactly the terms 'perfect solution',infinite dilution and properties of infinite dilution to understand this, its easy). infinite dilution means amount concentration or equivalently any other composition term of solute approaching zero. however acid  composition is molality =1mol/kg as you prefer or any other standard composition, but yet it is not infinitely diluted although it shows properties of infinite dilution. key thing here to understand is that infinite dilution is one thing and showing properties of infinite dilution is another. this is not the only hypothetical state you are dealing with, standard state of a gas existing in pure form , as a solute or in a mixture is the pure form of gas at pressure 1 bar and gas is ideal or equivalently compressibility factor =1. going back to your question, we can talk about strong and weak acids or bases. in fact Ka and Kb values are based on this.  Q2; strong and weak are relative terms. the better word is strength. even strong acids or bases absorbs heat .

The answer is difficult to read (no paragraphs, not fully formulated sentences) and does not address the question of the magnitude of enthalpies of dissociation for strong vs. weak acids.


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Does CHT=CHT have a geometric isomerism because of tritium atoms?
Is the radioisotope of hydrogen considered as a new atom? Can you detect it with an H NMR?

Just to address what the answers don't answer - yes, you should be able to pick this up the 1H or 3H NMR as the couplings are different in these two different species.

Yes isotopes can induce geometric isomerism.  You can induce chirality by using different isotopes of atoms; the molecule $\ce{CHDClF}$ is chiral with the deuterium atom (you do not need radioactivity) while ordinarily $\ce{CH_2ClF}$ is not.  Chirality won't work for ethylene (that molecule is still planar and thus mirror-symmetric about its own plane no matter what iostopes are used for any atoms), but cis-trans isomerism with two deuterium atoms still can work.

So compounds like $\ce{CHDTCl}$ rotate the plane-polarized light or in other words show optical activity. Could you please tell whether my conclusion from your answer is correct, as I think optical activity (due to difference in the counter-rotating circularly polarizing light) is because of chemical properties (attributed to valence electrons, spatial orientation, etc.,) and not physical properties (like different atomic masses of isotopes) ?

Neutrons are made up of quarks which are charged, so can interact with electromagnetic radiation.  Differences in number of neutrons can thus show up as rotation of polarized light.

Actually we define geometric isomerism to just differ between compounds. As the cis and trans forms of that compound are different they will have geometric isomerism.


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Comparing the basicity of aniline and ortho anisidine
In  my worksheet there is a question that asks for the comparison of basicities of the three anisidine isomers and aniline:

After my analysis using:
$\text{Basicity} \propto \text{+M effect (except for meta annisidine)} \propto \dfrac{1}{\text{-I effect}}$,
I could make the following conclusions:

m-anisidine should be the least basic
p-anisidine should be more basic than aniline.

But the main problem is comparing the basicity of aniline and o-anisidine and comparing them with p-anisidine. I doubt the presence of ortho effect in ortho isomer after seeing the ball and stick model which makes the interaction of lone pair of nitrogen with hydrogen atoms of $\ce{-OCH3}$ improbable.
The answer given is:

 p-anisidine> aniline > o-anisidine > m-anisidine (I have verified it with $pKa$ values from PubChem)

I would say SIP.

But I want to think of it in terms of "tendency to donate electrons" @AvnishKabaj

Steric inhibition of protonation reduces the "tendency to donate electrons"

Why you considering the methyl group on $\ce{-OCH3}$ to be "locked" in position, facing away from the $\ce{-NH2}$ group? It is free to rotate in all directions, and it will come at a position next to the $\ce{-NH2}$ group, where the ortho effect will apply.
For a visualization, here is a similar image:

(taken from Martin's excellent post)

PS: I believe that Steric Inhibition of Resonance should not be invoked here, because both sides of the $\ce{-NH2}$ group are not blocked by the $\ce{-OCH3}$ group, and $\ce{-OCH3}$ is certainly not as bulky a group as t-butyl.

Nice! So basically in ortho effect here, hydrogen bond will be formed between the hydrogen of NH2 and O?

Nope, that way the lone pair of nitrogen is still free. The interaction will be between N of NH2 and H of OCH3 (also see)


text:
What acid did I accidentally create, and how?
So, I wasn't trying to do a chemistry experiment this evening; it was entirely unintentional!
I was mixing up a batch of fertilizer for my plants (no nitrates involved, so I wasn't worried about potentially dangerous reactions).  The main goal was to add some sulfur (which over time lowers the pH) and potassium (in the form of potassium sulfate); to balance it out there also were lesser amounts of calcium superphosphate and magnesium sulfate, a little iron sulfate, and tiny amounts of trace elements (boric acid and some simple compounds of manganese, copper, zinc, and molybdenum)
The potassium sulfate tends to come in big chunks because it's hygroscopic, and I know that hot water helps dissolve it (and that the sulfur doesn't dissolve in water, even hot water), so I heated the mixture (in water) in the microwave, then stirred with a plastic utensil.  I noticed a surprising amount of offensive-smelling offgassing coming off of it.  So I put in my analog pH meter, and it pegged instantly, hard (its limit is 3,5, so it must have been much lower).   Even after adding some calcium oxide (which instantly fizzed and started gelling), the pH was still pegging hard.  And after 1 dilution (maybe 7:1?), it was still pegging, and only after the second dilution did it get stop pegging (but still showed strongly acidic).  There was heavy brown discoloring in the bottom of the plastic container, and it looked like the sulfur might have been reacting as well.
I assume I created sulfuric acid (and/or possibly phosphoric?) somehow, but I'm not sure what the reaction was!  I wasn't expecting this at all.  Normally in horticulture, sulfur takes ages to acidify soil, as it's decomposed by bacteria. I know that combusted sulfur yields sulfur dioxide, which - oxidized over a vanadium pentoxide (or other) catalyst - yields sulfur trioxide, and thus (with water) sulfuric acid... but that's clearly not what happened here.    Any clue what reaction we're looking at?

Perhaps anhydrous calcium chloride could work? Sigma Aldrich sells it with a 99% purity, with the remaining impurities being traces of metals basis.

Dissolve some of the potassium sulfate in distilled water and measure its pH. If the resulting solution is highly acidic it is possible that you have potassium hydrogensulfate instead of potassium sulfate.

Potassium sulfate isn't hygroscopic (contrary to the description in the question); however, potassium hydrogen sulfate is.

Probably you are looking at two different phenomena:
Acidity
If the potassium sulfate is hygroscopic, then as seen inthe comments it may have been potassium hydrogen sulfate, $\ce{KHSO4}$ instead of the normal sulfate $\ce{K2SO4}$. The $\ce{HSO4^-}$ ion is a fairly strong acid with $\pu{pK_a}\approx2$, so reasonably solutions of salts with this ion and no normal sulfate would be expected to have $\pu{pH}<2$ even at concentraions as low as 0.1 molar. So they require much dilution to get the $\pu{pH}$ within your reading range. Thus your accidental acid may be potassium hydrogen sulfate instead of normal potassium sulfate.
Gassing
There is no indication that sulfur reacts with sulfate or bisulfate salts to give sulfur dioxide. But it can react with air, especially when heated (probably your microwaving destroyed microbes). It's supposed to go to sulfuric acid, but sulfur dioxide can form as an intermediate product and potentially gas out. A neutral or basic solution might convert the intermediate to sulfite or bisulfite ions, but an acid solution (see above) would promote formation of the gas.

SO2 is not easily oxidized to SO3 or sulfuric acid.

But it could be formed. Neutral or basic solutions containing sulfite ions would be easier to oxidize to sulfate.


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What is the function of acetic acid as a solvent in EAS reactions?
I saw acetic acid being used in two reactions:

1.

2.

I was asked to predict the outcome of the first reaction but I couldn't, because nitration requires generation of $\ce{NO2+}$ which in turn requires an acid stronger than nitric acid. (acetic acid is weaker)
I would like to know how exactly is $\ce{AcOH}$ participating in these two reactions. Is it supporting the generation of electrophile? If yes, then how? I couldn't find any relevant information on the net. The second reaction is from Clayden's book and it doesn't highlight the role of $\ce{AcOH}$ there.


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Does S use its p-orbitals for bonding?
This question was inspired from the following questioN:

I thought the answer would be either A (maybe accounting for larger sizes?) or B (Bent's rule) but apparently the answer is D. Can someone explain why this is so?
This question is #54 on the 2010 USNCO Part 1

A screenshot or picture of an exercise is not searchable. Please consider rewriting it, so that it can be of help for future visitors.

I have edited it for the source

The reason for (D) is pretty much given here, and in short, s-p separation is large in sulfur, therefore no hybrid orbitals. Which also excludes Bent's rule. While (A) is true, a longer bond would not affect the bond angle at all, so that cannot be the reason. (B) is a bit difficult to judge, because there are many factors affecting polarity, however, polarity does not necessarily influence the geometry that much. And (D) is just plain wrong, wrong, wrong, wrong. No d orbitals for sulfur.

I mean, according to VSEPR theory sulfur can access d-orbitals for bonding. I know that this is false now but why is it false? Also isn't the s and p separation smaller in the n = 3 subshell than in the n = 2 subshell?

VSEPR is taught more often wrong than it is taught right. "according to VSEPR theory sulfur can access d-orbitals for bonding" is such a common, but completely false statement. There have been multiple studies, showing that octet expansion is a wrong concept and that d-orbitals will not be used for bonding at all. This is due to the fact that energy levels become more spread out with higher n. Which also answers your last inquiry.

Why do subshell energy separations increase with increasing energy shells? I feel like that's counter-intuitive to the Rydberg equation / quantum mechanical systems such as particle in a box, which places energy levels closer together with increasing radius? Is it because "n" increases in $n^2 h^2 / 8mL^2$?


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Why do potassium, rubidium, cesium all have lower standard potentials than sodium?

Why do potassium, rubidium, cesium all have lower standard potentials than sodium?

Sodium has a higher charge density than any of them, so its enthalpy of hydration should be larger, right? The argument applies for why lithium has a more negative standard reduction potential than the other three, but why doesn't this argument apply for sodium?

I think lithium is the outlier in this series due to the enthalpy of hydration. The trend from sodium to caesium is linear and consistent with ionization energies, meaning the enthalpies of hydration are not big enough to account for the difference in ionization entropies, thus the trend.


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Is it possible to melt HDPE to a point where it has similar viscosity to engine oil?
I have a few pieces of HDPE that I want to melt and cast into a cube of around  $\pu{5 cm}$ each side, similar to what one would do with cast resin.
I tried putting it into an oven and heating it to $\pu{250°C}$, where it melted, but never enough to pour it somewhere.
Is there a viable way to melt HDPE scraps to a point where it becomes fluid, with a viscosity similar to engine oil, and retains this state long enough to pour it?

HDPE oxidizes rapidly at those temperatures. You'd need to at least do it an inert environment I would guess. Try looking into HDPE rotomolding to see what conditions they use.

You may see a small reduction in viscosity with increasing temperature, but at elevated temperature the polymer will start to oxidize and/or undergo thermolysis. If I were trying to reduce viscosity I'd try blending in some low molecular weight polyethylene (LDPE). Not sure if you have the capability of doing this or if you could tolerate the resulting changes in property (loss of tensile strength, etc.). This reference:
https://doi.org/10.1080/03602550600609663
suggests you may need to go as high as 40%+ to get a significant change in viscosity.

I also have LDPE scraps lying around, would those alone be easier to melt? Mechanical properties are not an issue.

Yes, LDPE generally has a lower melting point than HDPE. That's a better bet if you're not concerned with mechanical properties.


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What are "tensides"?
I was reading about chemicals that control pore size of polyurethane foams and there it was mentioned that "tensides" decrease pore size. What are tensides? I assume it's the opposite of a surfactant, that is, it increases surface tension, but that is just my guess. I have no idea what it actually is.
I tried to Google it but strangely I just got bunch of "surfactant" results even through I wrote it just tensides. Some German site when translated from German to English changed the word tenside to surfactant, so I don't know if it is different class of chemicals or just a German word for surfactant. From the original article where I saw the word, it was written in a way as to suggest that they both are different from each other.

See also: What is the opposite of a surfactant?

I don't remember to have seen the term tensid outside the context of the German language.
The term is typically applied to agents that decrease the surface tension (hence tensid) of water. Note that the logic behind the English equivalent surfactant is just the same.
So, in summary tensid and surfactant mean the same.

I have seen the term "tensider" (likely meaning "surfactants") on labels in Swedish too, this is probably because Swedish is a Germanic language. All sources I could find seem to confirm that it is just a word for surfactants. I have seen the word used in German as well.


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QM modelling of charge separation reactions
I have a general question regarding the modelling of reactions in which neutral reactants form charged species/ions (in solution). How can one get the energetics approximately right on reactions like thes? Examples would be very simple acid/base reactions like
$$\ce{NH3 + HCl -> NH4+ + Cl-}$$
but also more complex reactions like
$$\ce{Ru(CO)H2(PPh3)3 + HCl -> Ru(CO)H(PPh3)3+ + H2 + Cl-}$$
Important contributions that I can think of, but do not seem to be sufficient in many cases, are:

Solvation. Some correction must be applied to account for solvation effects of the ions. I usually try to obtain corrections to $\Delta$G for this from COSMO-RS, and the solvation energies are a lot higher than for neutral species, but still seem to low. It is hard to judge whether these corrections are good, since for most of the interesting cases no data on solvation enthalpies in certain solvents etc. exists.
Diffuse functions. For better description of anions, diffuse functions in the basis set can make a difference. However that change also seems to be small when the basis set is already large (say def2-QZVPP to def2-QZVPPD), and for very large basis sets, the computational cost is also considerable.
Coordination sphere of ions. Many (smaller) ions probably have a special coordination sphere, which also depends on the solvent, and may form aggregates or oligomers. For new species it's quite likely that solution structure is not known, but even for seemingly simple cases (for example $\ce{NaOMe}$ in $\ce{MeOH}$) it is hard to find data.

Are there other things that should be considered? What is a good "standard" protocol to deal with these kinds of reactions?

It is not clear how much is approximately right for you, especially if you are asking a rather wide range of reactions, systems and solvent.

What about explicit solvation?


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Why does a mixture of nitrous acid with concentrated nitric acid cause nitration of benzene instead of nitrosation?
A question came up in our worksheet to predict the reaction product of nitrous acid and concentrated nitric acid with benzene.
I quickly recognized it to be similar to the nitration reaction. Following this ChemGuide page, which gives the reaction of nitration as:

$$\ce{HNO3 + 2H2SO4 -> NO2+ + 2HSO4- + H3O+}$$

I wrote this reaction:
$$\ce{HNO2 + 2HNO3 -> NO+ + 2NO3- + H3O+}\label{a}\tag{1}$$
and thought that this electrophilic $\ce{NO+}$ would then perform EAS on the benzene to give nitrosobenzene (aka nitrosation reaction).
However, the answer given in my worksheet is nitrobenzene instead. I am unsure if the following reaction (for the generation of the electrophile $\ce{NO2+}$):
$$\ce{HNO3 + 2HNO2 -> NO2+ + NO2- + H3O+}\label{b}\tag{2}$$
is even valid. To me, reaction $\ref{a}$ seems more favorable because it involves the formation of the more stable $\ce{NO3-}$. Then why is reaction $\ref{b}$ favored instead? Why is nitrobenzene the product of this reaction, and not nitrosobenzene?

You were absolutely correct when you said that nitrosation takes place.
What actually happens is that after nitrosation occurs, concentrated $\ce{HNO3}$ oxidises the nitroso group to form nitrobenzene.
Source
A Guidebook To Mechanism In Organic Chemistry; Chapter 6, section 6.2;Peter Sykes


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What is a benzene ring minus a hydrogen atom called?
I thought it was Phenyl but the teacher said it wasnt and that it started with "A" I cant seem to figure it out

It is phenyl all right. Then again, there is a broader term: aryl.

What is the difference between aryl and arylene? She said it was close to arylene but that arylene was a more general name

Is it that the latter had a hydrogen atom removed from two benzene rings? If so, is there any other case where a benzene ring or more have one h atom removed for each(i could only find the other name for arylene that is arenediyl)

It had to be where the hydrogen atom is removed not substituted

You could be referring to "aryl".  "Phenyl" is specific to $\ce{C_6H_5}$, "aryl" is generic to monovalent benzenoid  aromatic ring groups.


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How many products are possible when chlorination of cyclohexane takes place?
Since each atom is identical and equally likely to get attacked in cyclohexane therefore according to me only one product will be formed.Our sir made us write this answer

Mono chlorination?

Yes monochlorination

Should be one only, where did you get the question from.

Polychlorination will give more than three products, even if we limit "poly" to 2.

Exactly how many products and what will be their structure?

I count seven .

This question is honestly very vague, and lacks lots of details. "Our sir hasnt specified" isn't a valid excuse for a poorly framed question, and neither can we help you with understanding your teacher's intent. I'd advise you to carefully decide what is the concept that's bothering you, and not consider it a "homework" that you're giving us.

For cyclohexane, you're correct that since all carbon atoms are exactly the same, only one monochloro derivative exists.
The answer could be 2 if we did the monochlorination of cyclohexene instead.
Allowing dichlorination gives more than just three products - due to both geometrical and optical isomerism.
Therefore, to obtain the answer 3, the reactant ring must be different. Currently, as it stands, the answer 3 is wrong.


text:
What is it called when a benzene ring loses a hydrogen atom?
Does not get substituted by another group it just gets removed  (so not aryl) what is it? Like when two benzene rings lose one H atom each we get arylene? Whats another name for it? I just need the name for when the atom is removed and NOT subtituted
Thank you
Edit: thats not the answer I asked for
I asked what it was called when it lost the hydrogen atom and you say its aryl but thats when a group is in place of the lost atom

The name of the benzene minus a hydrogen is called a phenyl group (C6H5). https://en.wikipedia.org/wiki/Phenyl_group Look this page up for more information. The question you've asked is very similar to the one that @Gaurang has commented, so just be careful when you ask questions on here. Hope this helps

Is that when it doesnt get subtituted by a functional group?

Yes, it says in the wiki page as well.

But wheb I read it it says that the "ar" refers to the group

Ar refers to aryl, which is a broader denotation of the group. Phenyl I believe is the correct/ more used term.


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Does the pH of Water affect how Acidic or Basic it is?
I've noticed a lot of people have been recently uploading videos in which they test the pH of multiple brands of bottled water and claim that the more acidic ones (5.5 < pH < 6.5) are worse for your health than the more basic ones (7.5 < pH < 8.5).
However, considering that water is made up of both H+ and OH- ions, wouldn't that mean that the concentration of both ions would be equal regardless of the pH that a sample of water has? If this is true, would it mean that two different brands of bottled water would still be neutral even though they have different pH values?

It surely would, if there were absolutely nothing else in the water. This, however, is not the case.

pH is the measure of water basicity or acidity. If the pH is 7, the water is neutral, if it is less than 7 it is acidic, if it is greater than 7 it is basic. So to answer your question, if the pH is not 7, the water is not neutral.
Absolutely clean water has pH 7. However, absolutely clean water is extremely rare. Most waters contain various dissolved substances and some of those substances influence the pH. Even if you leave absolutely clean water in contact with air, the air gases start to dissolve in the water immediately and pH goes down due to dissolved carbon dioxide. So absolutely clean water, even common distilled water, in contact with air is acidic.
Regarding the health effect, I think it is an oversimplification. pH itself is not so important, the important thing is, what is dissolved in the water. You cannot judge that knowing only the pH.


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Is 1,2- Hydride shift possible in ketonic cations?
The question in my worksheet asks me to predict the result of the reaction between benezene and $\ce{(CH3CH2CH2CO)2O}$ in presence of anhydrous $\ce{AlCl3}$.
My attempt:
After complex formation with $\ce{AlCl3}$ we get this carbocation: $\ce{CH3CH2CH2CO+}$ that has two resonance structures.
But, if we consider hydride shift we get $\ce{CH3CH2C^+HC=O}$ which is not only stabilised by resonance but also by the $\text{+I}$ effect of ethyl group.
However, the answer is $\ce{PhCOCH2CH2CH3}$ which isn't formed by the second carbocation with hydride shift .
So I'd like to know if hydride shift is possible in ketonic carbocations or not? And if it is, is the answer wrong?

If hydride transfer were to occur, the resultant carbonyl  would be stabilizing a cation. A definite NO_NO! Read up on the the Friedel-Crafts reaction.

The carbocation doesn't rearrange since oxygen donates it's lone pair and the resulting carbocation is quite stable since the octet of all the atoms is complete.


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Acid Catalysed Ring Expansion – Mechanism?
Ring expansion of vinyl cyclobutane -

Could someone please help me with a detailed mechanism for the above conversion?
It is the acid-catalyzed rearrangement of 4,7,7-trimethylbicyclo[3.2.0]hept-3-en-6-one to 3,3-dimethyl-6-methylidenebicyclo[2.2.1]heptan-2-one.
I couldn't find the mechanism anywhere, and couldn't propose a good one either.
Please help.
P.S.
As one of the users in the comments is interested in seeing my best effort, I'll explain my thought process-
As the carbonyl oxygen is the most nucleophilic, it is the first to be protonated in acidic medium. Furthermore, this species has a resonating structure (less contributing, but the only one that can lead to a ring expansion, I believe) – one obtained by shifting the π bond towards oxygen – a carbocation.
After this, I'm clueless. I think, that the expansion could have something to do with the hydrogen at alpha position to the sp² hybridised carbon atom in the five membered ring – as the product has an exocyclic double bond.

You are correct about protonating the carbonyl group. [The first of your products is the same as the reactant.] The cyclobutanone 2 was prepared from a mixture of geranic acids. (J. J. Beereboom, J. Org. Chem., 1965, 30, 4230). The structures of 2, 4 and 5 were determined by Beereboom. I am providing the most reasonable mechanism. Cyclobutanone 2 must have a more stable cis ring juncture owing to its method of formation or exposure to p-TsOH. The trans isomer of 2 cannot give the trans isomer of 3 because of strain. Tertiary carbocation 3 arises by addition of the double bond to the protonated carbonyl. Collapse of the carbocation produces bicyclic ketone 4, which in the presence of p-TsOH isomerizes to the exocyclic isomer 5. In general, 1,1-disubstituted double bonds are less stable than tri-substituted ones. In this case, ring strain may be present in the endocyclic alkene 4.


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Ring expansion of two fused rings to a larger ring
I saw the following reaction mechanism in paper Tetrahedron Lett. 1976, 17 (33), 2869–2872.:

I'm not able to understand the following parts:

How did the conversion of 27 to 29 take place? I've never seen that kind of a ring expansion, and a detailed explanation would help - maybe including the necessary transition state(s) and why it happens.
27 to 28 and 30 to 31 aren't clear as well; why should such rearrangements happen and how do they happen? What is the mechanism?

A key thing to remember is that 4-membered rings are highly strained and not very stable.

Alright, I was aware of that. What's next?

@schrodinger_16 The bridge C-C bond of 27 has just shifted to the right and created that big ring. It is interesting to note though, that there are two ways for it to shift: one creates this 8-member ring, the other creates two five member rings; I'm curious how we can even predict theoretically which combination will have lesser strain.

I think the possible mechanism for the intermidiate conversions can be as following.Here thr $\ce{R}$ is actually $\ce{-OCHO}$.

What's the reason for that hydride shift? Clearly, it decreases the carbocation's stability by decreasing the number of hyperconjugative structures.


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Does mass equal to moles times molar mass take into account coefficients?
I have an equation $$\ce{ 2NaN3 -> 2Na + 3N2}$$ I've calculated there are $\pu{4.4 moles}$ of $\ce{N2}$ (using the data given in the question). I then did $\large{\frac{4.4}{3 \cdot 2}}$ to get number of moles required of $\ce{NaN3}$ (although if this is wrong then correct me). And now I want to calculate the mass of $\ce{NaN3}$ used (knowing there are $\pu{3 moles}$).
Where I have $\pu{ 3 moles}$ of $\ce{2NaN3}$ and I want to calculate mass do I use the $\mathrm{M_r }$of $\ce{M_r(NaN3)\cdot 2}$ or just a single mole of it.
Eg:
 $\text{mass = moles} \cdot  \mathrm{M_r} = 3 \cdot 65 $ or $\text{mass = }~3 \cdot 130$?

Apparently you are confusing relative molecular mass $M_\mathrm r$ and molar mass $M$. Relative molecular mass $M_\mathrm r$ is a dimensionless quantity.

Furthermore, the quantity ‘amount of substance’ $n$ shall not be called ‘moles’; i.e. don’t confuse the quantity and the unit.

What part of my question confuses them? Is mass = moles * M?

What do you mean by $\pu{3 moles}$ of $\ce{ 2NaN3}$? The coefficient (the 2) is the number of moles of that molecule that you have.
If you're looking at an equation with $\ce{ 2NaN3}$ written in it, and you want that 3 times, then you're going to have $\pu{6 moles}$ of $\ce{NaN3}$, so $\text{mass}~= ~6 \cdot 65$ .

I have an equation 2NaN3 --> 2Na + 3N2. I've calculated there are 4.4 moles of N2 (using the data given in the question). I then did 4.4/3 * 2 to get number of moles required of NaN3 (although if this is wrong then correct me). And now I want to calculate the mass of NaN3 used (knowing there is 3 moles). Hope this has clarified.

I've edited my question to put this in to help other try and understand where I am getting confused

Ah I see. The coefficients in the chemical equation simply show the ratio between the number of moles of each compound that are involved in the reaction. So for every 2 moles of NaN3 you form 2 moles of Na and 3 moles of N2.
This is distinct from the actual number of moles you have of each chemical. You have calculated that if 4.4 moles of N2 are formed, then 2/3 as many moles of NaN3 must have reacted (ie 3 moles). Therefore mass = 3 * 65.


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Why does iron(III) form the trithiocyanate complex instead of the triisothiocyanate one?

As far as I know, $\ce{Fe^{3+}}$ is a hard acid, and thus shouldn't it favor complexation with the nitrogen's lone pair rather than the softer S?

How hard is the nitrogen?  Doesn't pi bonding have something to do with hardness/softness?


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Is it correct to term boron trifluoride a "catalyst" in electrophilic aromatic substitution?

In Friedel Craft Alkylation of aromatic compounds using $\ce{BF3}$ :

$\ce{BF3}$ acts as  a lewis acid

It is used as a catalyst

Amount of $\ce{BF3}$ used in the reaction is almost equal to that of amount of benzene used.

$\ce{BF3}$ acts as a nucleophile.

I am sure about options $1$ and $3$ being correct and $4$ th being incorrect but I am hesitant about option $2$.
Catalyst is a substance that lowers the activation energy and enhances the rate of the reaction, in the end remaining unconsumed.
Now, what Lewis Acids do is that they make halogens stronger electrophiles by forming  complexes with them. Taking this further we can say that yes they serve as "catalysts" in the reaction because otherwise the reaction would have been lot slower.
However, the given answer is $1,3$. Is my reasoning correct and the answer wrong? Is it functioning as a catalyst in the reaction?

On the formal definition, I think $\ce{BF3}$ may fail the remains unconsumed part. It will form $\ce{[BF3X]-}$, which would need to revert to $\ce{BF3 + X^-}$ in order to complete the catalytic cycle. I don't think the $K_\mathrm{d}$ of this complex is particularly high, so this is not likely to happen, hence it won't be a proper catalyst. This is also why you need a stoichiometric ammount of $\ce{BF3}$ for the reaction to proceed to completion.

Tbh, in school, we were always taught that this complex (or any such catalytic complex) dissociated completely during the reaction. So, unless we know the fact you stated, it is tough to arrive at the answer.

It's an assumption based on the fact that you need stoichiometric amounts of the compound (thus you don't get it back during catalysis), and the fact that $\ce{BF3}$ is quite a strong Lewis Acid since it lacks an octet.


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Why do nitrogen molecules not act as ligands in haemoglobin?
Nitrogen molecules $(\ce{N2})$ have lone pairs, which, as far as I know, is the property of oxygen molecules $(\ce{O2})$ that allows them to act as ligands bonding to iron in haemoglobin in the blood. Why then is atmospheric nitrogen unable to act as a ligand and form dative bonds to iron in the blood as oxygen is able to?

N2 is even worse as ligand then oxygen, its complexes are rare.

It can, at least at high pressure, see e.g. https://www.ncbi.nlm.nih.gov/pubmed/3417671

@Mithoron Okay, but why is N2 a worse ligand than O2?

https://chemistry.stackexchange.com/questions/118773/why-co-is-a-stronger-and-more-common-ligand-than-n2 I have a similar question but still no answer

Related: https://chemistry.stackexchange.com/questions/14434/how-many-ligands-in-heme/14448


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How to write the rates for a bimolecular collisional dissociation?
I have questions concerning two species involved in a reaction, where one species is dissociated completely by collision with another, fairly inert species. The reaction looks like this:
NO  + N$_2$ $\xrightarrow{k_f}$ N + O + N$_2$
My understanding is that the reaction requires the N$_2$ molecule in order for the reaction to proceed. So, for the rate:
$\frac{d[NO]}{dt}$ = $-k_f [NO][N_2]$
$\frac{d[N_2]}{dt}$ = 0
Is the above expression for the rate of change for NO correct?
Second, if I want to consider there being a reverse process, what I believe I need is an equilibrium constant such that:
$K_{eq} = \frac{k_f}{k_r} = exp(-\frac{h_p-h_r}{RT}) $
Where $h$ is the enthalpy of formation, R is the universal gas constant ($8.314 [\frac{J}{molK}]$, and T is the temperature of the gas. the subscripts $p$ and $r$ refer to the products and reactants, respectively. For "naturally" occuring species, the enthalpy of formation is 0.
For this particular reaction, I looked up the enthalpy of formation for the species involved:
$h_{N_2} = 0 [\frac{J}{mol}]$
$h_N = 472.68\times 10^3 [\frac{J}{mol}]$
$h_O = 249\times 10^3 [\frac{J}{mol}]$
$h_{NO} = 90\times 10^3 [\frac{J}{mol}]$
Once the equilbrium constant has been solved for, I determined the reverse reaction rate $k_r$ by dividing my known rate $k_f$ by $K_eq$. The resulting rate then, including reverse reactions would look something like:
$\frac{d[NO]}{dt}$ = $-k_f [NO][N_2] + k_r[N][O][N_2]$
$\frac{d[N_2]}{dt}$ = 0
Is the above formulation for the reverse reaction rate, and the expression for the time rate of change of the gas correct?
Thank you

It seems that for one, I am incorrectly calculating the equilibrium constant $K_{eq}$. I should be using the expression $\Delta G = H - T\Delta S$ rather than $h_p, h_r$ in my calculation for the equilibrium constant. Making the change, the equation should instead look like $exp(-\frac{\Delta G_p - \Delta G_r}{RT})$ where the subscripts $p,r$ still mean products and reactants, respectively.


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NMR spectrum of 3-hexylthiophene: why is the methyl group not a triplet?
I have an example spectrum, and it shows which peaks correspond to which H, but I don't understand why the spectrum looks like that
e.g. I would expect the peak at 0.88 to be a triplet based on the (n+1) rule, why is it such a large multiplet?

The reason why the terminal methyl group is not a simple triplet is because of strong coupling, which is the scenario whereby the coupling constant between two nuclei $J$ is comparable to the difference in resonance frequency $\Delta \nu$.
As labelled in your question, protons B and C have almost the same resonance frequency (0.01 ppm difference corresponds to a 3 Hz difference on a 300 MHz spectrometer). The coupling constant between these protons, though, is probably on the order of 7 Hz or so. Hence, protons B and C are considered to be strongly coupled.
What does this have to do with protons A, which your question is asking about? Well, by virtue of the fact that protons A are coupled to protons B, which are in turn strongly coupled to protons C, there will be some "extra" coupling observed in the peak corresponding to proton A. The term for this is virtual coupling.
It turns out that this is the case no matter what nuclei they are or exactly how they are connected in the molecule - as long as X and Y are strongly coupled, any nucleus A that is coupled to either X or Y will have some degree of virtual coupling. The reason for this is fundamentally quantum mechanical in nature. Suffice it to say, though, that the n+1 rule will not predict multiplet shapes accurately under such conditions.
Further reading on Hans Reich's excellent website. In particular you may want to take a look at the "Methyl Region" section which is pretty similar to your example (although in your case it is actually even more complicated).


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Can a molecular sieve be used increase the saturation of a fructose solution?
The Problem: Let's say we have an experiment where we are attempting to grow rock candy crystals by letting a solution of sugar crystalize on a wooden dowel. However, an error may have been made in that there was too little sugar added for the volume of water to reach saturation and thereby grow the crystals. The solution is to remove some of the water so that the solution becomes saturated.
But that's a problem.
The experimental setup is in an environment were letting it evaporate away is not allowed. It cannot be moved outside or really leave the lab it's in at all. Our access to the setup is very limited and getting resources to it is a logistical and financial nightmare. The test setup may as well be in space. The test apparatus is also not very conducive to modification in that we can't modify it at all and the only access to the solution is through a 3 mm inner dia tube. Repeating the experiment is out of the question because of a whole host of constraints, again assume it's in space.
So my question is, can we send up a molecular sieve to be inserted into the container to absorb some of the water and cause the sugar to reach saturation and thereby start to crystalize? If not, is there something else that can be done to the solution?
Other suggestions must be as harmless as possible. They should also not cause a pressure increase in the container, nor interfere with the growth of the sugar crystals. Any form of density separation will not work.

Molecular sieves release a lot of heat when exposed to water, if that's a problem then it might not be a good choice.

I was unaware of this property. How much is a lot of heat though? The experimental set up can tolerate high heat (~100 C ) and there is plenty of water to serve as thermal mass.

That's a very good question, I can't tell with numbers. I just know that a way to test if the MS work is to add a few drops of water while keeping them on your hand and if they get really hot, then they're good.

A more fundamental problem is that even if the molecular sieve idea worked, the molecular sieves would act as nucleation points for the sugar. So you would likely end up with sugar coated molecular sieves.

@MaxW that's more preferable than no crystal growth at all.


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Is hydronium an electrophile, nucleophile, both or neither?
Is hydronium an electrophile or a nucleophile? I expected $\ce{H3O+}$ to be a nucleophile due to the presence of electrons in the $\ce{2p}$ orbitals of $\ce{O}$. But the answer key of the test states that its none of them.
I can rationalise it this way: If $\ce{O}$ donates its 2 electrons then there would be 2 positive charges on $\ce{O}$ leading to instability.
But I am not too sure of my reasoning either. Isn't hydronium serving as a nucleophile in its reaction with $\ce{OH-}$ ? Why is it then neither an electrophile nor a nucleophile?
Google books results show that its mentioned as an electrophile in some books. While there's disagreement on other sides on the internet?  What's the true classification?

$\ce{H^{+}}$ (or equivalently $\ce{H3O^{+}}$) is an electrophile.  If you react $\ce{C6H6}$ with $\ce{D3O^{+}}$ you will get $\ce{C6H5D}$ as the initial product, formed by the standard electrophilic aromatic substitution mechanism.

@ron But I think there $\ce{D+}$ serves as the electrophile not $\ce{D3O+}$

$\ce{H^{+}}$ is shorthand for $\ce{H3O^{+}}$ which is shorthand for the Zundel and Eigen cations.  See here.

@ron I understand that $\ce{H+}$ is just shorthand for $\ce{H3O+}$. However, how do you explain that $\ce{H3O+}$ is an electrophile? We can explain that $\ce{H+}$ is an electrophile as it has an empty duplet, but we can't say the same for $\ce{H3O+}$. What other reasoning do we have then?

@GaurangTandon  Like I mentioned above, $\ce{H3O^{+}}$ doesn't really exist either.  You actually just have a solvated proton that still needs 2 electrons.

@ron Correct me if I am wrong, but the linked answer says that $\ce{H5O2+}$ and $\ce{H7O4+}$ are the main ions present in solution. Then doesn't it contradict your statement "You actually just have a solvated proton that still needs 2 electrons."?

@GaurangTandon I see no contradiction.

@ron Ok, I think get it. You're trying to suggest that $\ce{H5O2+ = 2H2O + H+}$ so it still only is a solvated $\ce{H+}$ ion. Am I right?

@GaurangTandon Yes, that's exactly right.

As I mentioned earlier H3O+ can be further protonated in strongest superacidic solutions, so it can still act as nucleophile. Almost anything can be one or another depending on reaction.

So both an electro as well as nucleophile

https://chemistry.stackexchange.com/questions/93809/is-hydronium-an-electrophile-nucleophile-both-or-neither

Firstly, I would like to point out that, when we talk about electrophiles (E) and nucleophiles (Nu), we may be calling the entire molecule an electrophile or a nucleophile, but in reality we are referring to a single atom or functional group that serves as the E/Nu. For example, when we call hydroxide (OH-) a nucleophile, we really mean that the oxygen anion is nucleophilic.
Now for hydronium. It should be noted that there are no 2p orbitals on the oxygen atom; it is sp3 hybridized. Given the positive charge on oxygen, it is already electron-deficient, meaning that it is highly unfavorable to lose more electron density by sharing it with another atom through a bond (as it would do if the oxygen acted as a nucleophile).
Instead, one off the hydrogen atoms acts as an electrophile. Once again, the positively-charged oxygen atom is electron-hungry, weakening the H-O bond and making the σ* orbital easier to donate electron density into. As a result, one of the hydrogen atoms of hydronium is quite easy to strip off, giving it its strong acidity (pKa~ -1.74). Since no other parts of hydronium react with other species, the only reactive component is the hydrogen atom, which serves as an electrophile and gives hydronium its categorization as an electrophile. Below is an arrow-pushing mechanism depicting the deprotonation of hydronium.

*Note that the oxygen of α-Methylbenzyl alcohol is acting as the nucleophile to attack a hydrogen atom of hydronium, the electrophile.


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Compounds with high electron density
If $d$ is the ratio of the total number of electrons in a compound and the average radius(of all anions and cations, or covalent radii), are there known compounds (inorganic or organic) where $d$ is high?
Are there any rules or analytic techniques to design such compounds that have stable chemistry?
Please don't confuse the electron density in the title to this one
Edit: Assume the constraint that we want common elements and no exotic, radioactive ones.

Your d is high where the atomic number is high.

@IvanNeretin: Edited. Consider the effect of average radius as well.

I still stand by my statement.

Mercury has 80 electrons and a covalent radius of 132 picometers.


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Which Alpha-H will be extracted in intramolecular condensation rxn between ester and ketone
In this problem, which hydrogen will be extracted - the one near the ketone group (Marked 2) or the one near the ester (Marked 1)?
My approach:- In intramolecular reactions between aldehydes and ketones, the base extracts $\ce{H}$ from the carbon adjacent to ketone (making it a nucleophile) as the carbonyl carbon of aldehyde has more positive character, thereby rendering attack by the ketonic nucleophile easy.
But here case is between ester and ketone.
One can argue that esteric $\ce{H}$ (Marked 1) should be extracted by the same logic as the ketonic carbon has more plus character.
Another factor is that if the $\ce{H}$ near ketone is extracted, then it can remove the $\ce{C2H5O^-}$ group(as it is a better leaving group).
So, which effect dominates? I am unable to decide.

Your immediate concern seems to be which hydrogen is abstracted, 1 or 2, in structure 1 (see below). Examination of pKa’s is useful. pKa’s: EtOH, 16; alkyl ketone, ~20; aliphatic ester, ~25; and β-diketone, ~9; aliphatic carboxylic acid, ~5. Clearly, alkyl ketones are more acidic than aliphatic esters. The methyl ketone hydrogen 3 in structure 1 is slightly more acidic than the methylene hydrogen 2 as evidenced by the haloform reaction. But this subtle point is immaterial here. More importantly, the question is what is the structure of A and how is it formed? [A appears to be an intermediate but you have not shown the whole question.] For the sake of discussion, let’s assume that we are dealing with concentrated NaOEt/EtOH. All of these enolates are undergoing proton exchange under the reaction conditions and the concentrations of the enolates are relatively low.
There are two possible reaction pathways to consider. The enolate 2 adds to the ester carbonyl as in intermediate 3. This step is reversible (blue arrows) but elimination of ethoxide leads to β-diketone 4, which, given its low pKa, is irreversibly deprotonated to enolate 5. Acidification of enolate 5 provides β-diketone 4, aka A.) The ring closure (2 → 3) occurs through a 5-membered ring transition state that is more favorable than a 7-membered ring transition state resulting from the enolate of the methyl ketone of 1. This reaction must be conducted with at least stoichiometric base because the alkoxide is consumed in the reaction 4 → 5.

The alternative pathway (1 → 6 → 7 → 8 → 9) invokes the ester enolate adding to the ketone.  All of these steps are reversible except perhaps 9 → 8. The irreversible formation of β-diketone channels the reaction irreversibly in the direction of 5. If the formation of unsaturated ester 9 is irreversible, then pathway 1 → 4 is faster than 1 → 9. Can one argue that the pathway 1 → 4 is preferred because there is more ketone enolate than ester enolate? Not necessarily!
Consider the reaction of cyclohexanone (10) with diethylsuccinate (11) under a variety of basic conditions (Stobbe condensation). The product of this reaction is the carboxylate salt 15. Here it is the ester enolate that adds to the ketone! The favorable 5-membered ring lactone formation (12 → 13) leads to the irreversible elimination 13 → 15. Alternatively, the addition of the cyclohexanone enolate to the ester group of diethyl succinate (11) would lead to the irreversible formation of β-diketone enolate 17. In spite of the fact that there is expected to be a preponderance of ketone enolate in this route, it does not prevail!

Another query if you may.If we replace the ester by an aldehyde group,then will aldehyde enolate be favoured or ketonic enolate?Although Aldehydic adjacent carbon H should be more acidic(due to fact that it has 1 less +Inductive effect of methyl as compared to ketone),My book says attack will take place by Ketone enolate on Aldehydic carbon since Aldehydic carbonyl has more positive character(hence rxn faster).Here are we to ignore acidity since there is not much difference between pka of aldehyde and ketone?Thanks in advance

You are still too focused on the which hydrogen is more acidic. This is an issue when making kinetic enolates (LDA, aprotic solvent). If the ester is replaced with an aldehydic group, you are looking at an intramolecular aldol condensation. If the aldehyde is the electrophile, you would get 1-(cyclopent-1-en-1-yl)ethan-1-one (1-cyclopentenyl methyl ketone). If the ketone is the electrophile, then the product is 2-methylcyclopent-1-ene-1-carbaldehyde. These products are under thermodynamic control and are generally conducted in aqueous base.

TL;DR
The hydrogen near the ketone will be plucked.
Comparing positive charge characters
Ketones and esters both have carbonyls. The oxygen of the carbonyl pulls electrons away from the carbonyl carbon. Thus, both carbons (in the ester and the ketone) have partial positive charge.
Ketones have 2 alkyl groups bonded to the carbonyl. The partial positive charge is partially releived by the +I effect due to these groups.
Esters have an alkyl group and an alkoxy group attached to the carbonyl. The alkyl group is responsible for a +I effect and the alkoxy group provides a +M (positive mesomeric) effect, which is stronger than the +I effect.

Thus, there is more partial positive charge on the carbonyl carbon of the ketone.
Comparing enolates
The enolate formed by the ketone is more stable than that formed by the ester.

The lone pairs on the oxygen of the alkoxy group further destabilise the enolate.


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Is planarity really necessary for conjugation?
If we look at the $\ce{ClO4-}$ anion, the chlorine atom in it is sp3 hybridised. Hence, it has a tetrahedral structure in which conjugation shouldn't be possible, but while explaining the acidic strength of $\ce{HClO4}$, I've seen many people claim that it is very acidic because the $\ce{ClO4-}$ anion is stable due to the conjugation of charge between the four oxygen atoms.
So, all of this boils down to: Is planarity really necessary for conjugation?

Comments are not for extended discussion; this conversation has been moved to chat.

https://chemistry.stackexchange.com/questions/48113/why-is-hclo4-the-strongest-oxyacid

this is a rather good question several students have actually asked me about it. it is not any exeption or contradiction .i have written the explanation below

Could you type your answer out please? The content seems to be very nice. Thanks

I can't read that, please type it out. I would then even consider removing my down-vote.


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Acidity of aldehydes and ketones and enolate formation
In most academic courses you'd learn that aldehydes are more acidic (lower $\text{p}K_\text{a}$) than ketones due to the lower electron donating effect of the proton compared to the alkyl group of the ketone.
However, when talking about condensations, for example, it's not a weird choice to form a ketone enol over an aldehyde enol on the basis of "thermodynamic stability due to a more substituted alkene under thermodynamic control". Yet, $\text{p}K_\text{a}$ is the measurement of such stability, and thermodynamically the aldehyde enol is more stable. Adding to this, a following argument is that "aldehydes react much quicker towards nucleophilic attack than ketones". I agree with this, but weren't we just discussing thermodynamic control?
The questions are then:

Is it valid to use the "more substituted alkene" to argue a step on a mechanism under claims of thermodynamic control?
Can this type of argument be coupled with a kinetic one like the example provided as well? I.e. is it valid to mix thermodynamic and kinetic arguments one in the same? (e.g. this intermediate forms because it's more thermodynamically favored, and the next one forms because it is kinetically favored).

(I realize question 2 may be broader than this, but I'm trying to keep this in the sense of mixed aldol condensations)

I am probably wrong, but does the word "condensation" refer to simply aldol condensations (self and cross) or some other reactions as well?

Since the whole post is about aldehydes, ketones and their formation of enolates, I think it is safe to assume I'm talking about aldol condensations.

I would like to first comment on the reasoning given as to why aldehydes are more acidic than ketones. The usually provided reasoning is, as stated here:

H atoms are regarded as having no electronic effect : they don't withdraw or donate electrons.
Alkyl groups are weakly electron donating, they tend to destabilise anions (you should recall that they stabilise carbocations). This is because they will be "pushing" electrons towards a negative system which is unfavourable electrostatically. Hence, the anion of a ketone, where there are extra alkyl groups is less stable than that of an aldehyde, and so, a ketone is less acidic.

We need to be clear of what we are measuring when we measure the acidity of an aqueous solution of an aldehyde. It is important that we recall what is mentioned in this post, on the acidity of benzaldehyde, as it relates to the acidity of aldehydes in aqueous solution. Most aldehydes exist in aqueous solution with a significant proportion of their hydrated forms, which are geminal diols. Hence, it is very likely that the $\mathrm pK_\mathrm{a}$ values we have obtained are for the $\ce {-OH}$ protons of the hydrate and not for the $\ce {C-H}$ protons at the $\alpha$ position of the aldehyde.
Let's take a look at acetaldehyde. Ethanol has $\mathrm pK_\mathrm{a}$ value of $15.9$ and the hydration equilibrium constant for acetaldehyde is $\ce {1}$, as mentioned here. Based on the guideline that geminal diols have $\mathrm pK_\mathrm{a}$ $2.5$ units less than that of the monohydric alcohol, the predicted $\mathrm pK_\mathrm{a}$ value for acetaldehyde is $13.4$. The experimental measurement gives $13.57$, as seen from here, which is very close to the value determined via calculation with the guidelines. This closeness of the values suggests that the hydrate is indeed formed and the acidity that we are measuring for the aqueous aldehyde is indeed that of the hydrate.
If this is what we are measuring, then clearly the reasoning highlighted in the box may not even be correct. In fact, we can't even be sure that the $\ce {\alpha}$ $\ce {C-H}$ proton of the aldehyde is more acidic than that of a ketone since we won't be able to measure their acidities, at least not in water.
With the clarification I have made, I hope that there is no longer any inconsistency in the issues you have mentioned. In particular, it is correct to say that under thermodynamic control, the intermediate with greater stability (e.g. the enolate with the more substituted alkene) is formed. Regarding your 2nd question, I am unsure of what you mean but I will just say that kinetic and thermodynamic control refer to different reaction conditions resulting in the reaction undergoing different pathways to thus form different products.
As an extension, we can also consider if the $\mathrm pK_\mathrm{a}$ values given for the ketones are representative of the acidity of the $\alpha$ protons or that of the $\ce {-OH}$ protons of the hydrate. Let us consider acetone, with $\mathrm pK_\mathrm{a}$ given as $\ce {19.16}$ here. The hydration constant for acetone is $1.4 \times 10^{-3}$ and the $\mathrm pK_\mathrm{a}$ of 2-propanol is $16.5$ (as seen here). Based on the same sequence of calculations and the same guideline, the predicted $\mathrm pK_\mathrm{a}$ value is $16.9$. The significant difference between the predicted value and the experimentally measured value suggests that the acidity of the acetone is not just due to that of the hydrate $\ce {-OH}$ protons and that there may be significant deprotonation of the $\alpha$ protons taking place.
An important question that we would like to know the answer to is what then is the $\mathrm pK_\mathrm{a}$ value of an $\alpha$ proton of an aldehyde. Well... It seems like due to hydration of the aldehyde, we would not be able to directly measure the acidity of this proton. However, one would reasonably estimate that it is close to that of the ketone $\alpha$ hydrogens. Again, that is something we may not be able to measure directly for the same reason. An educated guess would be in the range of $18-22$ as it needs to be more lower than the $\mathrm pK_\mathrm{a}$ of esters but higher than that of alcohols.


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How did the atomic number and atomic mass of elements discovered?
Elements are small for our naked eyes.so,how did the discovery of its atomic number and mass number took place?

http://scienceline.ucsb.edu/getkey.php?key=408


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Is an antiaromatic compound more stable than a non-aromatic compound because of more conjugation?

(image from a LibreTexts page (archive link))
It is shown that the antiaromatic compound is more stable than the non aromatic compounds 2 and 3 because of a more conjugated system. Here's the exact statement:

In the first structure, delocalization of the positive charge and the π bonds occurs over the entire ring. This becomes apparent when we look at all the possible resonance structures as shown below.

  In the second structure, delocalization is only possible over three carbon atoms. This is demonstrated by writing all the possible resonance forms below, which now number only two.

  Finally, the third structure has no delocalization of charge or electrons because no resonance forms are possible. Therefore, it is the least stable of the three.

Is this correct? I thought that the conjugation in an antiaromatic compound destabilized it more than a non aromatic compound.

Good question! I highly doubt if this is true. But, I do not have the relevant data to post an answer. That said, I have also archived that page and added a link and the relevant paragraphs to your question, just in case the linked webpage changes and makes your question appear obsolete for future visitors.

It is often weird to compare the stability of species that differ by their formula anyway. The case is not as to compare a hypothetical localised cycloexatriene to benzene. ...but somehow to compare cyclohexane and benzene merely on demoralisation.  Saying that Cyclopentadienyl cation is anti aromatic means it does not get planar. The non planar cation could be more stable than the other(s). But even in the case the explanation by the link is not correct. Delocalisation as depicted does not occur.

Demoralisation reads as delocalisation. T9 :)) And even non planar cyclopentadienyl cation is very unstable.

To be accurate, the stability of a compound depends on the exact configuration of the atoms. We have numerous theories telling a different story about the compound and all hold good for some and terribly wrong for some other. Chemistry is still a developing science and to look at it from the existing theories may lead to inaccurate results.

Sometimes, only the delocalisation of electrons is not sufficient to conclude that the compound is more stable. We should also look at which molecular orbital the electron pair is delocalising. But a clear picture of these kind of $\pi$-conjugated systems can be got when it is approached quantum-mechanically, with the help of Huckel-$\pi$ Molecular Orbital Theory.  According to this theory, to get the $\pi$-electronic energy of the system, we have to solve the Secular Determinant, which is $$\ce{det(H_{ij}- ES_{ij}) =0  ,where i,j varies from 1 to n}$$ in a $n$ Carbon System.  In Huckel Theory, for solving the determinants two approximations are taken,$1$. $\ce{H_{ii} = $\alpha$ for all i, and H_{ij} = $\beta$ for |i -j|=1.}$ This approximation tells that the effective Hamiltonian for the whole conjugation is negligible if the carbons are not close to each other(i.e. adjacent).The two different values are obvious because of the in Hamiltonian, the Columb integral, the Exchange integral interferes. $2$. $\ce{S_{ii} =1; and S_{ij}= 0 else}$. This tells the Overlap integral(Note that, this is an integral, and clear overlap in the picture doesn't mean integral should be large) are negligible for any two different carbons.  Taking these two approximations, if you solve the determinant, you will get energy levels for $5$ carbon systems,$$\ce{E = $\alpha$ + 2$\beta$;  $\alpha$; $\alpha$ - $\beta$}$$Typically, the values of $\beta$ are negative and as you increase in $\alpha$ or decrease in $\beta$, the energy is increased and system gets destabilisation. .So, the lowest energy level is $\alpha$ + $2$$\beta$, and then the second and after that the third one.Your first compound has four $\pi$ electrons. So, two electrons will be in the level $\alpha$ +$2$$\beta$, and other two is in $\alpha$ only. So,net energy, is $4$$\alpha$ + $4$$\beta$ in mostly conjugated system $1$st compound.  But in the second case there are only two $\pi$ electrons. That two will go into $\alpha$ + $2$$\beta$.So, net energy is $2$$\alpha$ + $4$$\beta$ in second one.  So, second system has lower energy than the  first as coefficient of $\alpha$ is lesser in second one, though second compound is less conjugated. So, extent of delocalisation doesn't always determine stability.


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Addition of a methyl group to a heterobicyclic ring

I am battling to understand how this reaction works. It should be a carbonyl addition reaction but I am not 100% sure how to reach each step. Could someone explain the mechanism to me?

Here, the lone pair on $\ce{N}$ is in conjugation with the double bond.So, during formation of the first product, the lone pair delocalises and as a canonical form, a carbanion is formed which acts like a Nucleophile and performs $\ce{S_N2}$ with $\ce{CH_3I}$.  For the formation of the next two products, $\ce{K_2CO_3}$ acts like a base and abstracts a proton from the methyl substitued carbon. That abstraction forms a carbanion which is also in conjugation with a double bond and hence it's stable. The next two products are actually both $\ce{S_N2}$ products, but the nucleophilic centre are change due to that negative charge delocalisation.The mechanism is depicted as follows,

Thank you! This makes it a lot clearer. Champion answer!

Please upvote or mark as correct answer if you like it..

It is mostly NOT Carbonyl addition. K2CO3 is used only for a weak basic medium.
I think that as the lone-pair of nitrogen is in resonance, it will delocalise resulting in negative charge on the carbon atom.
This atom can act as nucleophile for Sn reacion with MeI being the target.
The first of your three products is obtained this way.
Now, Nitrogen takes back its lone pair leading to carbocation formation.
Elemination takes place.
Then Nitrogen attacks MeI due to presence of lone pair.
This gives your third product.
Similar mechanisms may result in the second product.


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What does the neutralisation of acids and bases mean?
From what I found in textbooks and google is that neutralisation is when acids and bases react to produce salt and water. Most examples I have learnt do so, however a reaction like:
$$\ce{NH3 + HCl -> NH4Cl }$$
It does not produce water.
So is neutralisation defined as only when salt and water produce a neutral salt and water or is it just produce any salt and water. And if it is any then how can does the example above work?

Neutralisation is when something acidic plus something basic produces something more or less neutral. Water may be a part of the picture, or it may not; this is irrelevant.

There are many theories put forward to explain acid-base reaction.
According to Arrhenius acid-base theory. All acid liberate protons (H+ ions) when dissolved in water. All bases liberate Hydroxyl ions (OH- ions) in water.
When the reaction takes place H+ and OH- combine to form water, and the remainder of the ions (what are left after liberating H+ and OH- ions) form what is called a salt.
In such reactions, Water is always formed.
What you have shown is an example of a Bronsted acid-base reaction.
Bronsted acids are proton donors (have ionisable hydrogen ion).
Bronsted bases have the capability to accept them (either via hydroxyl ion liberation or by their lone-pair as in NH3)
Clearly, the formation of water is not a necessity in a Bronsted acid-base reaction.

You can read more here :https://en.wikipedia.org/wiki/Br%C3%B8nsted%E2%80%93Lowry_acid%E2%80%93base_theory