text:
Is PbO ionic compound?
In IIT-JAM 2018, There was one question: "which one of the following oxides are ionic?" I have selected $\ce{PbO}$ but according to the official answer key, $\ce{PbO}$ isn't ionic.
Although I know this distinction between ionic and covalent isn't defined sharply. But lead is a metal and oxygen is a non-metal. So according to me, it should be ionic. I tried to find on the internet I couldn't able to find out about $\ce{PbO}$, although $\ce{PbO2}$ is defined in between of ionic and covalent according to this paper. $\ce{PbO}$ should be more ionic than $\ce{PbO2}$ since oxidation state of the lead is lower, it will be less electronegative.
Even $\ce{PbCl2}$  is considered as Ionic according to many books, where chlorine is less electronegative and much bigger in size as compared to oxygen, So at least $\ce{PbO}$ should have more ionic character than $\ce{PbCl2}$ according to the Fazan's rule.
So should I challenge the official answer key? If yes, then on what basis? Which reference should I send to them?

Iono-Covalent Character of the Metal–Oxygen Bonds in Oxides: A Comparison of Experimental and Theoretical Data

Be careful about punctuation, there shouldn't be any spaces before commas. As far as question is concerned, there shouldn't be such option as this isn't clear case, but I'm afraid that you're wrong and they're right.

@Mithoron but according to DavePhD PbO is ionic. So how is it wrong ?

Haha, this is getting silly, both this question and such distinction are pretty much pointless. With one calculation you might get, say 47% of ionic character, with another mayyybe 53%, who knows? And what difference it would make? Would you  actually know more about this compound? See for ex. https://chemistry.stackexchange.com/a/17072/9961

If you want a reference to challenge the question, try page 248 of Advanced Study Guide Chemistry.  There is a table on that page that says the PbO is "ionic".

There are lots of different thoughts about qualitative assignment of whether $\ce{PbO}$ is ionic or covalent.
So, I think it will be better to approach it quantitatively and for that we can take help of Hannay-Smith's equation for calculation of ionic character in any binuclear compound.
% Ionic Character = [ 16 ($ \Delta\ $E.N.) + 3.5 ($ \Delta\ $E.N.)$^2$ ]
So, according to Pauling scale of E.N., we have E.N. of $\ce{Pb}$ = 1.87 (thanks to @Mithoron for correcting me) and E.N. of $\ce{O}$ = 3.44.
Now, putting the values we will get the % ionic character of $\ce{PbO}$ to be 33.75%, which, unfortunately for you and me (I also appeared in this year IIT-JAM and submitted $\ce{PbO}$ to be ionic),indicates that $\ce{PbO}$ is predominantly covalent in character.

Pauling's e. value is much lower for Pb (+2), 1,87 according to https://en.wikipedia.org/wiki/Lead

@Mithoron Yes, but in Hannay Smith equation, the e.n. of 'neutral atoms' are considered and not of them with their oxidation states. And I know you will now ask for ref. which unfortunately I can't get at the moment, but I think the example in this link ( https://www.google.co.in/search?q=hannay+smith+equation&oq=Hannay+Smith&aqs=chrome.1.69i57j0l2.5906j0j7&client=ms-android-lenovo&sourceid=chrome-mobile&ie=UTF-8#imgrc=JLKAPEQwCLjpBM%3A ) will help you understand that the e.n. of neutral atoms are under consideration!

sigh Electronegativity depends on oxidation state for lead on +2 oxidation state it's 1,87, for Pb on +4 it's higher. Bare non-molecular ions have no electroneg. whatsoever.

@Mithoron Obviously, that was too ignorant of me! I just got carried away with this ( https://www.google.co.in/search?client=ms-android-lenovo&ei=uuSWWoHIB4TivASOvJmoDg&q=electronegativity+value+of+Pb&oq=electronegativity+value+of+Pb&gs_l=mobile-gws-serp.3..33i160l2.10762.18738..19340...2....248.3647.2-16..........1j5..mobile-gws-wiz-serp.....3..0j35i39j0i67j0i70i251j0i22i10i30j0i22i30.WT4yQ9yYzRk%3D ) and tried to pile a logic upon that! I am correcting my answer. Thank you!  But I doubt where that value 2.33 came from?

There are many different electronegativity scales. If I try to use Hannay-Smith's equation using Pauling electronegativity scale then Al2O3 is also covalent. Al(III) has an electronegativity of  1.61 on Pauling scale. PIC of Al2O3 is 41 %.  But according to IITB Al2O3 is ionic but PbO isn't. How unfair !!
Moreover, the electronegativity value of Al(III) and Pb(II) is almost same on Allred Rochow electronegativity scale, i.e. 1.47 and 1.50 respectively.

Yes, but I think (though I am not at all sure) Hannay-Smith's equation is only applicable to binuclear 'diatomic' species as you can see in all the examples given in the link I provided earlier.

Moreover, PbCl2  is considered as Ionic according to many books, where Cl is less electronegative and much bigger in size as compared to O, So At least PbO should have more ionic character than PbCl2  according to the Fazan's rule. If PbCl2 is considered as Ionic compound, then I see no reason not to call PbO ionic.

I don't understand the purpose or the advantage of using an empirical equation, which is itself based on another empirical property (electronegativity), to try to obtain a precise quantification of percent ionic character.

One of the greatest errors in all of science is the well-propagated notion that there are "ionic" and "covalent" bonds, as if a bond were always one or the other, and that we can tell them apart by electronegativity difference.
Neither notion is true.

Bonds always have a mixture of ionic and covalent contributions, even where they are between equivalent atoms like $\ce{H2}$.  Nonpolar bonds as in molecular hydrogen have small, opposing ionic contributions that cancel each other out.

Bonds can show much greater ionic character than would be expected from the electronegativity differential. In many cases, a strong ionic bonding character is driven  by molecular-orbital structure rather than electronegativity. For example, in magnesium diboride, the boron forms macromolecular structures with low-energy vacant orbitals and thus tends to become anionic (see here).

My answer to this question is to get a better textbook when one is available.


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One-pot synthesis of pyrazole
I am trying to follow a published paper. It is a one-pot synthesis of pyrazole from a ketone and and acid chloride via a diketone intermediate.
Here is the produre from the paper:
General Procedure: Ketone (2 mmol) was dissolved in 5 mL dry toluene in a screw cap vial (with septum) and then the solution was cooled to 0°C under nitrogen. LiHMDS(2.1 mL, 1.0 M in THF, 2.1 mmol) was added quickly via syringe with agitation stirring and the formed anion was allowed to sit for approximately 1 minute before the addition via syringe of acid chloride (1 mmol) in one portion with stirring. The vial was then removed from the ice bath and allowed to stand for 1 minute, then 2 mL AcOH was added with stirring. 10 mL EtOH and 5 mL THF was added to form a homogeneous
mixture, and then hydrazine hydrate (2 mL, 1.1 g, 34.3 mmol) or substituted hydrazine(5 mmol) was added and the mixture was allowed to auto-reflux and was held at that temperature for 5 minutes, when LCMS showed that all diketone had reacted. The resulting solution was added to 1.0 M NaOH solution and extracted with EtOAc. The organic fraction was then washed with brine, dried over Na2SO4, and evaporated under reduced pressure.See specific compounds for purification details.
My questions are:

What is the purpose of excess hydrazine hydrate, acetic acid, and enolate in producing the pyrazole?
How many molar equivalents of sodium hydroxide should I use at the end of the reaction?

It would be helpful if you copied the text of the exact procedure. As is, it's impossible to tell when or what is added.

Good general question, but please narrow it down to specifics. As it stands, it's too broad and a bit unclear (to me).

Excess hydrazine hydrate to neutralise the AcOH and ensure complete reaction of the diketone.
Excess AcOH to protonate all anions arising from the LiHMDS, mostly to completely protonate the anion of the diketone.
Excess enolate (and LiHMDS) to ensure all the enolate reacts. The diketone product is more acidic than the enolate and will hence protonate the enolate, so you need 2 eqs or more of base/enolate to get complete reaction.
How much NaOH for the wash at the end? More than eqs of AcOH. Don't see why they specify a base wash as everything you want to remove into the aq phase would be aq soluble anyway.

Making the solution basic helps keepimg the pyrazole in the organic phase.

Can pyrazole be deprotonated then? Also, I am assuming hydrazine hydrate replaces the OH of AcOH. Does that make the solution less acidic, hence fewer equivalents of NaOH needed?

Pyrazole is a relatively weak base (pKb 11.5). It can be deprotonated but that requires strong base. Hydrazine is a strong base so will be fully protonated by AcOH, but the free base is extensively water soluble.

One last question. How would a basic solution keep the pyrazole in the organic layer? Why would I need the equivalents of NaOH to be more equivalents of AcOH?

You need more eqs of NaOH than AcOH to ensure the aq is basic. It is possible to protonate pyrazoles and the protonated form is more aq soluble

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Can hyperconjugation take place between lone pairs of an atom attached to an sp2 carbon?
When $\ce{HBr}$ is added to 2-chloro-2-butene, the proton preferentially bonds to the $sp^2$ carbon that's not bonded to chlorine. Does this mean that hyperconjugation takes place between the lone pairs of chlorine and the positively charged $C3$? I would assume that since there is still a free rotation about the $\ce{C-Cl}$ bond and that chlorine is in a tetrahedral formation this would be the reason behind the regioselectivity.

The positive charge (in the intermediate) formed on C2 is stabilized by resonance with the lone pairs of chlorine.

@ApoorvPotnis: isn't the inductive effect stronger for halogens? Wouldn't it be a net destabilizing effect that chlorine induces?

@harry I am seeing this comment 3 years after it was posted and honestly, I don't know the answer. I have forgotten all chemistry I knew years ago.

@ApoorvPotnis I'm the person who asked the question. I now work in a completely different field and I don't understand anything from my own question XD

Hyperconjuation is the interaction of electrons in sigma orbitals (in $C-H$ or $C-C$) with an adjacent empty p orbital or $\pi$ orbital. What you are describing is not hyperconjugation, but resonance; more specifically, the chlorine shows the $+R$ effect.
Now if you draw the resonance structures of this molecule you will realize that the $C_3$ has a negative charge on it in one canonical form. This means, in the actual molecule, that $C_3$ carbon has a partial negative charge, which is why the hydrogen from $HBr$ preferentially bonds to that carbon.
A more acceptable explanation would be that, when the hydrogen bonds to $C_3$, the resultant intermediate carbocation (positive charge on $C_2$) is stabilized by the $+R$ effect of chlorine, resulting in a racemic mixture of $2-bromo-2-chlorobutane$ to form as the major product.

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How are the van der waal constants 'a' and 'b' related to molecular size?
'a' represents the measure of attractive forces between the gas molecules so how is it related to the molecular size? if a1> a2 then what does it tell us about the molecular sizes of 1 and 2?

a is not related to molecular sizes, b is.

$a$ is included in the pressure correction term ,$b$ is included in the volume correction term. Higher value of  $a$ means more attraction and hence , lesser size.

vdW equation is just a model. While the textbook derivation of the equation may sound intuitive, 'a' and 'b' are not actual molecular parameters, just parametrization.

The real gas equation or van der Waals’ equation of state is
$$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT $$
where the constants a and b are known as van der Waals’ constants.
$a$ is a measure of van der waals forces of attraction existing between the molecules of a given gas. Greater the value of $a$ greater is the strength of the forces. It doesn’t affect molecular size.
On the other hand, $b$ is the incompressible volume per mole for a gas. It is directly proportional to the size of gas molecules.

Note that for example, water and ammonia have almost similar $b$, but their $a$ differs significantly. Source

But their molecular sizes are similar. So we can say $a$ doesn’t affect molecular sizes right?


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Can we classify all the d-block elements as transition metals?
I thought that properties of d-block elements are transitional between those of s-block and p-block elements, and that is the reason for calling them transition metal.
My textbook says that not all d block elements are called transition metals.
My doubts:

What are these non-transition metals that are in d-block?
Why are these specific elements not called transition metals?

The idea that the properties of d-block elements are transitional between those of s-block and p-block elements, and that is the reason for calling them transition element. It is absolutely correct, but the thing is transition metals are defined as slightly different manner.
Definition of a transition metal/Criteria for an element to be a transition metal

A transition metal is one which forms one or more stable ions which have incompletely filled d orbitals

Appreciate that the general electronic configuration (EC) for d-block elements are slightly different for those of transition metals.
EC of d-block elements: $\mathrm{(n-1)d^{1-10}ns^{1-2}}$
EC of transition metal/ion: $\mathrm{(n-1)d^{1-9}ns^{1-2}}$
Some examples of elements which are in d-block but not a transition metal

Scandium has the electronic structure $\mathrm{[Ar] 3d^{1} 4s^2}$. When it forms ions, it always loses the 3 outer electrons and ends up with an argon structure. The $\ce{Sc^{3+}}$ ion has no d-electrons and so does not meet the definition.
Zinc has the electronic structure $\mathrm{[Ar] 3d^{10} 4s^2}$. When it forms ions, it always loses the two $\mathrm{4s}$ electrons to give a 2+ ion with the electronic structure $\mathrm{[Ar] 3d^{10}}$. The zinc ion has full d-levels and does not meet the definition either.

There's a small thing that one must take into account that some elements are capable of forming multiple metal ions. In those cases we must consider more common ion (the most stable one).

The pefect example for this case is $\ce{Cu}$ $\left(\mathrm{[Ar] 3d^{10} 4s^1}\right)$, $\ce{Cu}$ is capable of forming two ions:

$\ce{Cu^{1+}}$ $\left(\mathrm{[Ar] 3d^{10}}\right)$ (not the most common ion of $\ce{Cu}$)
$\ce{Cu^{2+}}$ $\left(\mathrm{[Ar] 3d^{9}}\right)$  (most common ion formed by $\ce{Cu}$)
Hence, copper is definitely a transition metal because the $\ce{Cu^{2+}}$ ion has an incomplete d-subshell.
Using this idea we can also say  $\ce{Zn}$, $\ce{Cd}$, and $\ce{Hg}$ are not considered as transition elements.
How are these metals different from rest of the d block elements?
$\ce{Zn}$, $\ce{Cd}$, and $\ce{Hg}$ are not hard metals (as compared to other d-block elements), which can be attributed to the fact that they have no unpaired electrons which makes their metallic bonding weak.

Technically, scandium could exhibit oxidation states lower than +3.  But the cited examples either appear to retain metallic bonding or, as seen with elements more commonly rendered as "transition metals", they are organometallic.

You might also want to add that the properties of zinc, cadmium, mercury differ from the characteristic properties of the transition elements. Like formation of colored salts. Or the fact that Zn/Cd/Hg are not at all very hard.

According to IUPAC gold book, "An element whose atom has an incomplete d sub-shell, or which can give rise to cations with an incomplete d sub-shell." So, Sc should also be a transition element.

@freakylearner we may be seeing the definition evolve to favor including scandium as a transition metal.  See this question.

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What is the senior parent chain in the following compound?

Will the IUPAC name be (2-hydroxymethyl) but-3-ynoic acid (or) 2-ethynyl-3-hydroxy propanoic acid?

Please note that the triple bond shall be depicted in a linear fashion.

The most important simplified criteria for the choice of a principal chain are:

greater number of suffixes
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds
greater number of prefixes
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding wording of the rules taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) is as follows.

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)
(j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);
(…)
P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.
(…)

You have correctly identified the suffix (“oic acid”) for the principal characteristic group. The next criterion for the principal chain is the greater number of skeletal atoms (i.e. the longest chain). Thus, the principal chain corresponds to the but-3-ynoic acid part and not to the 3-hydroxypropanoic acid part since but-3-ynoic acid has a longer chain than propanoic acid. Therefore, the correct name is 2-(hydroxymethyl)but-3-ynoic acid.

"greater number of suffixes" I don't understand this. How can there ever be more than one suffixes? Do you have an example?

@GaurangTandon As you can see from the quoted rules, ‘greater number of suffixes’ is my simplified short version of ‘The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix)’. For example the principical chain in 1-[3-(3-hydroxypropyl)phenyl]ethane-1,2-diol is ‘ethane-1,2-diol’ (two substitutents corresponding to the principal characteristic group, i.e. suffix ‘ol’) and not the longer ‘propan-1-ol’ (only one substituent corresponding to the principal characteristic group).

Oh, thanks! :) That helped. Though you might want to lengthen that "simplified short version" to something like "greater count of the principal functional group" to make it more unambiguous. I am not sure if I am the first person to be confused by that...

Honestly though, this simplified criteria is so immensely useful. Throughout the past year ever since I've been searching for decent IUPAC nomenclature I've never seen it before. You should really make a nice introductory page sort of thing for it. Now that I have fully understood it I really repent not knowing it earlier.


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Coloured salts for silica garden
I recently thought about making a silica garden. I have copper sulphate and nickel sulphate right now. What other salts would be the best to put along with these two? Also should I buy sodium silicate directly or should I use silica gel and sodium hydroxide? Any help would be appreciated! Thanks.

I would suggest Cobalt Nitrate (reddish-brown) , Iron(III) Chloride (orange) and Zinc Sulphate or Calcium Chloride (both white) as they look good with blue and green (Copper Sulphate and Nickel Sulphate respectively).
If you’re doing this at home and don’t have lab like apparatus, I would suggest buying Sodium Silicate as you are required to heat Sodium Hydroxide in the making of Sodium Silicate, which can be dangerous if proper care is not taken.


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Effect of temperature change in an equilibrium reaction
Talking about an exothermic reaction in equilibrium it is said that heat is on the product side, so according to Le Chatelier principle, the reaction should move backward to counteract the increase in the number of products.
But it is also stated that in an exothermic reaction when the temperature increases, the value of equilibrium constant decreases, which means the product is less than reactant, then shouldn't the reaction move in the forward direction?
Please help me with wherever I'm going wrong.

"Equilibrium constant decreases" means "reaction moves backward".

As per Le Chatelier's principle, when the temperature is increased, for an exothermic reaction , the reaction shifts in the backward direction. This is clear from the van't Hoff's equation:
$$ \log\left( \frac{K'_\mathrm{p}}{K_\mathrm{p}}\right) = \frac{\Delta H^0}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$
where $T_1$ corresponds to $K_p$ and $T_2$ corresponds to $K'_p$ . Also $T_2 > T_1$.
This is because $K'_\mathrm{p}$ becomes less than $K_\mathrm{p}$.  Hence the reaction goes in the reverse direction.


text:
What acid will dissolve iron carbonate?
Issue: removing through dissolution a thin (2-4 mm) iron carbonate crust covering fossilized bone without damaging the bone.  Microblasting with conventional baking soda ineffective; removal of iron carbonate with mini-pneumatic air chisel also removes top layer of bone with the iron carbonate crust. Any solutions would be appreciated.

The mineral composition of the bone itself is required to make a good suggestion. Without it, my first bet would be potassium cyanide and my second be would be HF+H2O2

My first two bets are the same, only in a different order. Also, it would be not exactly HF, but a solution with a little H+ and a lot of F-. I suggest that you turn this into an answer, because both existing answers seem to be missing the point.

Two issues arise immediately: 1) dissolving a variable thickness of iron carbonate (probably also of varying composition) and 2) not affecting the fossilized bone (which also probably has varying composition). A third issue might be that the iron carbonate seems to be either very tightly bonded to the bone, or perhaps even part of it by being part of the first substitution layer.
If the fossil is known to be mostly silica, acetic acid (vinegar) could work. Best method would be to scrub the fossil under the surface of the solution with frequent examination to see where additional scrubbing is appropriate, and where not.
If the fossil is largely calcium carbonate, recourse could be to oxidize the ferrous to ferric using a) calcium nitrite solution or b) hydrogen peroxide (H2O2). Calcium nitrite is a commercial corrosion inhibitor for rebar steel in concrete, and functions to oxidize ferrous to ferric at pH 11-12. The product at high pH is ferric oxide (hydrated). The aim in this process is to chemically attack the iron carbonate and turn it into something else that is not attached to the fossil. Scrubbing under the liquid surface (using rubber gloves!!) could turn the iron carbonate to a powdery product that is not attached to the fossil, at high enough pH to avoid dissolving calcium compounds. The chemical reaction is:
2 Fe++  +  2 NO2-  +  2 OH-  -->  Fe2O3  +  H2O  +  2 NO.
The NO will oxidize in air to NO2 which forms nitric and nitrous acids in water, and is toxic, but so is the nitrite, so you have to be very careful not to breathe the vapors or leave the oxidizing solution on your skin.
Perhaps a safer method would be to use hydrogen peroxide. This reaction is well-known, but is usually done in acid; a pH of 5 should be safe enough for the fossil, even if it is carbonate-based. Doing the reaction in acid keeps the ferric iron soluble, but hydrogen peroxide is also reactive in alkaline solution and a high pH might be more suitable for carbonate substrates. The reaction is:
H2O2  + 2 H+  +  2 Fe++  -->  2 H2O  +  2 Fe+++.
There is the added advantage that using excess peroxide causes foaming which could help remove particles from the fossil. Foaming causes mist which should not be breathed, and rubber gloves should be worn because H2O2 will eat your skin if you leave it on you for any significant period.


text:
What is the best method to prepare meta-nitrobenzoic acid?
Question:

Which is the best method for carrying out the following reaction?

I know that nitro group is a meta directive group in benzene nucleus and so is the carboxyl group. Option (b) cannot be the right answer since it first does the Friedel Crafts alkylation and then it does the nitration which would bring the Nitro group on either ortho or para position which is not desired.
However by going through options (c) and (d) I am getting the same product which is asked in the problem.
If we go via option (c) we will have:

Alkylation resulting in methylbenzene
Oxidation which will produce benzoic acid
Nitration which will give m-nitrobenzoic acid

And if we go by option (d) we will have:

Nitration - giving us nitrobenzene
Alkylation - this will introduce methyl group at meta position
Oxidation of the methyl group will give us again m-nitrobenzoic acid

According to my book the correct option is given as only (d). And that's where I'm confused. I'm not able to decide which of the following will be the best method.
Source:- "Elementary problems in Organic Chemistry" (S. Balaji publication) Chapter - Aldehydes, Ketones and Carboxylic Acids (page 401) (amazon link)

Remark: This paper suggests that option (c) can be used. Though it does not comment if its the best method.

The paper also says “In practice, this nitration reaction can result in the production of quite a bit of the ortho product as well, unless the temperature is kept very cold throughout the reaction“.

@MollyCooL Oh, good catch, I skipped that! Though I don't see any reason why ortho products would also be obtained.. :/

True even I don’t get that :(

The d) method does not work, as nitrobenzenes are usually very unreactive toward Friedel-Crafts Alkylation due to high electron withdrawing effect of -NO2 group.

As an addendum to SRM, nitrobenzene is used as a solvent in Friedel-Crafts reactions! Moreover, option c) is poor since the product toluene is more reactive than benzene. Over alkylation is to be expected. Methyl chloride would have to be the limiting reagent. Option f) benzene  (cat. FeBr3, Br2) --> bromobenzene (Mg, ether) --> phenylMgBr  (CO2) --> benzoic acid (HNO3, H2SO4) --> m-nitrobenzoic acid.

More effective to prepare benzaldehyde from benzene (Gatterman reaction) then nitrate (benzaldehyde less deactivated than benzoic acid) and finally oxidise.

@Waylander: As long as the choices aren't a-e.

None of a-e look like good strategies

I think the answer should be option C, because in both cases the first group obtained i.e. $\ce{-COOH}$ or $\ce{-NO2}$, is meta directing. But $\ce{-NO2}$ is has greater negative mesomeric effect than $\ce{-COOH}$. Hence if $\ce{-NO2}$ is attached first the reaction rate will be slower than if $\ce{-COOH}$ is added first.
You may check the negative mesomeric effect order here.


text:
Friedel craft alkylation halo-carbon bond cleavage

In this reaction, why is the $\ce{C-F}$ bond broken instead of the $\ce{C-Cl}$
bond, even though $\ce{C-F}$ bond is stronger than the $\ce{C-Cl}$ bond?

The boron probably associates much better with the fluoride, which activates the attached carbon towards substitution. After all, tetrafluoroborate is a very stable anion.

@Zhe Could this be because the $\ce{2p}$ orbitals of fluorine have a stronger overlap with the $\ce{2p}$ orbitals of boron, as compared to those with the $\ce{3p}$ orbitals of chlorine? That said however, I am not entirely sure whether the product ratio of 3-chloropropylbenzene to 3-fluropropylbenzene is 40:60 (a minor difference) OR 15:85 (a major difference)

Could you provide a reference for this reaction?

In Friedel-Crafts alkylation, the halogen-carbon bond breaking depends on the strength of the bond formation between halogen atom and the Lewis acid used. The more stronger is the forming bond, the easier it is to generate a carbocation for the reaction (in all cases, the carbocation explicitly needn't have to generate, a strong $\delta ^+$ charge on the electrophilic carbon is sufficient enough to have a $\ce{S_E2}$ or $\ce{S_E1}$ type reaction). Here, the Lewis acid used is  $\ce{BCl3}$ where there is a vacant  $\mathrm{2p_z}$ orbital on the $\ce{B}$ atom (assuming the molecule is oriented in a triangle in the  $\mathrm{xy}$-plane), and the Lewis base is the given reagent 1-chloro-3-fluoropropane, which has two basic centres. But donation through $\ce{F}$ will correspond to a stronger  $\mathrm{2p_\pi-2p_\pi}$ overlap with Lewis acid, rather than weaker  $\mathrm{3p_\pi-2p_\pi}$ overlap in the case of donation through  $\ce{Cl}$ atom. Thus, bonding with $\ce{F}$ will correspond to a very huge amount of $\mathrm{-I}$ effect of the adjacent carbon atom, which will weaken the bond between them by pulling the electron pair towards  $\ce{F}$ and easily favour the attack by the $\pi$-electron cloud of benzene ring. Also, the repulsion between $\pi$-electron cloud and the  large sized $\ce{Cl}$ atom is very less during the formation of  the above compound. But, if it would have bonded through  $\ce{Cl}$, there won't be a great electrophilic nature of the adjacent carbon due to less effective bonding in the case of  $\ce{Cl}$ (corresponding to even lesser  $\mathrm{-I}$ effect), so attacking will correspond to a highly energetic transition state, and also, there will be a slight electrostatic repulsion with the $\ce{F}$ atom of the $\pi$-electron cloud.
Due to such aforementioned phenomena and energetically favoured conditions, bonding through cleavage of  $\ce{C-F}$ bond is more preferred over cleavage of $\ce{C-Cl}$ bond.


text:
Can ethanol be both a Lewis base and a Lewis acid?
I'm asked to determine if ethanol is a Lewis acid, Lewis base, or both.
Since $\ce{O}$ has two lone pairs of electrons I would think that it could act as a base, since they could easily be donated. If it was a Lewis acid, it should have an empty orbital, and I can't find such one. All of the carbons are bonded and oxygen's orbitals are full. Still my book tells me it can act as both.
Is the textbook wrong or could one say that the 1s orbital to $\ce{H}$ (bonded to $\ce{O}$) acts as "empty", since $\ce{O}$ is more electro-negative than $\ce{H}$?

Why is this being down voted, is it a stupid question? Should I delete it?

You're not missing anything. Well, maybe that the textbooks often mistakenly treat Bronsted acid as Lewis ones (donate proton) while proton donors should be treated as adducts, where proton as acid exchanges its ligand for another.

I edited and undownvoted but you should remember that textbooks usually contain some errors. Also you should really change your username, it doesn't do you any good.

@Mithoron My username is a joke about someone trapped inside a website (as you can see him knocking on the screen). Also thank you for the response, but if ethanol cannot act as both, then it should be a base, which doesn't seem right?

Of course it's a base, What's supposed to be not right here?

Lewis acidity and basicity require some amount of context because they can't function in those roles without at least a putative base or acid, respectively. Certainly the oxygen here has a lone pair so it could be basic, and the $\sigma^{*}$ orbital of the O-H bond (on the hydrogen side) can accept an electron pair. What matters more is the context of what acids and bases are present to interact with these moieties. There is no reason why these pieces can't separately fulfill their roles independent of each other.

But hasn't the $\sigma$-bond already got an electron pair? How can it accept another one?

Indeed it can't. If it would, H- ion would need to leave.

The other way. It's a proton, ejecting a alkoxide... @Mithoron

@cansomeonehelpmeout I'm talking about the antibonding orbital. The bonding orbital, as you've noticed, is already full.

He knows this but if you put a lone pair of nucleophile in antibonding orbital it kicks out the one that's already bonding and it would be H- not H+ You'd be oxidating hydrogen here and it still wouldn't be pure acid + base reaction but exchange of ligands of  R-O+ cation.

Oh, I see what's not clear about my answer. I'll edit.

OK, this could lead to proton exchange, but if hydrogen bond would get created (especially symmetric one) the then it could be treated as acid + base reaction.

Ethanol  donate protons to the strong bases forming alkoxide ion. Ethoxide ion is a strong conjugate base which shows that it is a weak acid.
On the other hand, ethanol donate electrons to hydrogen in a reaction with Hydrogen halide and hence it can be considered as strong lewis bases. This is evident from the fact that there are lone pairs on the oxygen atom.

Is there no way that I can know if it is an Lewis acid/base just by looking at the structure?

As far as I know,  electron deficient molecules do act as lewis acids and electron rich molecules act as lewis bases. So from the structure it is possible to identify. But I never mentioned it to be lewis acid, by that reaction I just told it is a weak acid.


text:
Why is neodymium the most paramagnetic lanthanide?
My textbook says that paramagnetism rises to a maximum in neodymium.
I don't understand how gadolinium has 8 unpaired electrons, whereas neodymium has only 4. Shouldn't paramagnetism be higher  in gadolinium?
Has this got anything to do with the magnetic moment associated with orbital angular momentum?

This is from Shriver and Atkins' Inorganic Chemistry (p.586, 2009 ed) :

The magnetic moment of many d-metal ions can be calculated by using the spin-only approximation because the strong ligand field quenches the orbital contribution. But, for the lanthanoids, where the spin orbital coupling is strong, the orbital angular momentum contributes to the magnetic moment, and the ions behave like almost free atoms. Therefore, the magnetic moment must be expressed in terms of the total angular momentum quantum number J:
  $$\mu = g_J{\{J(J+1)\}}^{1/2}\mu_B
$$
  where the Landé g-factor is
  $$g_J=1+\frac{S(S+1)-L(L+1)+J(J+1)}{2J(J+1)}
$$
  and $\mu_B$ is the Bohr Magneton.

Being a high school student, the only thing of matter for you is the key point at the top of the page:

The magnetic moments of lanthanoid compounds arise from both spin and orbital
  contributions.

Old question, but what is your textbook? Gadolinium is ferromagnetic just below room temperature and the most paramagnetic just above. Looking at this table we extract
$$\begin{array}{rl}\text{Metal}&\chi_M/10^{-6}\text{cm}^3\text{mol}^{-1}\\\hline
\text{La}&95.4\\
\text{Ce}&2500\\
\text{Pr}&5530\\
\text{Nd}&5930\\
\text{Pm}&\cdots\\
\text{Sm}&1278\\
\text{Eu}&30900\\
\text{Gd}&185000\\
\text{Tb}&170000\\
\text{Dy}&98000\\
\text{Ho}&72900\\
\text{Er}&48000\\
\text{Tm}&24700\\
\text{Yb}&67\\
\text{Lu}&182.9\\
\end{array}$$
The Gadolinium result is quoted at $350 \text{ K}$ so as not to be too close to its Curie point. So Gd is in fact the most paramagnetic at temperatures where it's not actually ferromagnetic. You should not think of lanthanide metals as atoms because they are most typically $+3$ ions with the $6s$ electrons at least stripped from the atoms. Europium and Ytterbium maybe as $+2$ ions and Ce and Tb perhaps as $+4$ ions to achieve half- or fully-filled $4f$ orbitals.
So your count should be $2$ unpaired electrons for Nd and $7$ for Gd.


text:
Reaction of heated cupro-nickel in isopropanol or methanol
There are a number of youtube videos where someone heats a US quarter (which is copper sandwiched between layers of 75% copper, 25% nickel) to red heat and cooling it in methanol. This results in a coin that appears to be "copper plated". I've replicated this effect at home with rubbing alcohol and a US nickel, which is a solid piece of 75% copper, 25% nickel. The apparent copper coating is somewhat fragile, and there is a layer of black underneath.
What reaction is occurring here? Is the outer coating copper, and what could the black layer be? I've looked for an explanation online, but I've only found wildly inaccurate speculation.


text:
What's the role of sodium carbonate and hydrochloric acid for the neutralization of disaccharides into monosaccharides?
A couple of days ago we conducted the following experiment at school:

Add 5mL of an aqueous solution of sucrose ($\ce{C6H22O11}$) to 1mL of concentrated hydrochloric acid and heat it for 10 minutes. Then, add sodium carbonate ($\ce{Na2CO3}$) until the foaming stops. Then, conduct the Fehling experiment by adding the Fehling reagent to the following solution and heat it.

The results of the Fehling test on the solution was positive, meaning there was a compound with free aldehyde functional groups present.
The explanation to the above experiment was given as well:

$\ce{C12H22O11_{(aq)} + H2O_{(l)} -> C6H12O6_{(aq)} + C6H12O6_{(aq)}}$
Neutralisation:
  $\ce{Na2CO3_{(aq} + 2HCl_{(aq)} -> 2NaCl_{(aq)} + H2O_{(l)} + CO2_{(g)}}$
The neutralised solution performs a redox reaction on Fehling's reagent, so the result is the red precipitate.

First of all, I am interested as to why the heating aqueous solution of sucrose and hydrochloric acid result in the occurrence of foaming. Why does the sodium carbonate stop the process of foaming?
Second of all, I'd like to know why do we need to neutralize the solution in question for the test to be positive? Glucose molecules ($\ce{C6H12O6}$) are aldehydes and they should color the Fehling's reagent all by themselves. Is neutralization therefore obsolete?

You need to read the procedure more closely. The foaming occurs when sodium carbonate is added to the sucrose plus HCl.

Yes, I know that. But why does it occur? Is it because the reaction of sodium carbonate neutralization produces CO2?

Yes! Sodium carbonate plus acid gives carbon dioxide gas. Your system has excess HCl when sodium carbonate is added.

Thank you sir :D but what is the answer to the second question?

Fehling's soln is strongly basic. Possibly the Copper Tartrate complex is not stable in acidic environments and the aldehyde oxidation will not take place.

ratioale cogitande, thank you for the answer :D

by the way, can you please write that as the answer (I feel a little unpriviliged to do it myself since it was you who gor the solution) @Waylander

The foaming occurs due to carbon dioxide generation when you add the sodium carbonate to the sucrose and HCl solution as there is excess HCl. It stops when all the excess HCl is neutralised.
Fehling's solution B is strongly basic and when you add it to Fehling's A it forms the bis(tartrate) complex of Cu(II) which is deep blue. This is reduced  by the aldehydes to Cu2O which is insoluble in neutral and basic aqueous solution - hence the characteristic red ppt. Cu20 is soluble in dilute HCl, so if you did not neutralise you would get no ppt.


text:
Mismatch of plane of symmetry vs Gold Book's definition for a pseudochiral carbon
Here is pentane-2,3,4-triol:

At first glance, without deciding the configurations at each chiral carbon, we can clearly see that this molecule has a plane of symmetry, perpendicular to the plane of paper and passing through the $\ce{-OH}$ and $\ce{-H}$ on the 3rd carbon atom.
However, if we then sit down to compute the R/S configuration at carbons 2 and 4, we find them to be R and S respectively. Thus, going by the definition of a pseudo-chiral carbon, given in Gold Book as:

a tetrahedrally coordinated carbon atom bonded to four different entities, two and only two of which have the same constitution but opposite chirality sense.

we would infer that the 3rd carbon is indeed pseudo-chiral. This is confirmed by ChemSketch which generates the name of the above compound as "(2R,3s,4S)-pentane-2,3,4-triol".

Thus, we have observed that using the plane of symmetry to check chirality, we end up declaring the 3rd carbon to be achiral, rendering the molecule optically inactive. However, if we use the Gold Book's definition, we end up declaring the 3rd carbon to be pseudo-chiral, rendering the molecule optically active.
Another fact confirming that the molecule is optically active is that there exist two isomers, of two opposite configurations of the central carbon atom -  (2R,3s,4S)-pentane-2,3,4-triol and (2R,3r,4S)-pentane-2,3,4-triol. Here are their structures:

I have looked at several questions on pseudochirality here, yet I have been unable to resolve this conundrum. I believe I am using a wrong fact or a misguided inference, but whatever is it, I am unable to grasp it, and thus need help resolving the conundrum.

Yeah, the middle carbon becomes pseudo chiral but that doesn't have an impact on the compound being meso.

The Gold Book definition for a meso compound is

the achiral member(s) of a set of diastereoisomers which also includes one or more chiral members.

So, a meso compound has chiral subsets. Here we would have three such subsets:

Leftmost carbon (2R)
The centermost (pseudochiral 3S)
The rightmost (4S) carbon

Even though  the centermost carbon is pseudo chiral, the plane of symmetry still exists. The compound remains meso, achiral.
Nomenclature is just for naming a compound.

The following two molecules are diastereomers, not enantiomers.

For the examples I've looked at, a small (r) or (s) is used for the pseudochiral carbon instead of (R) or (S). For example, see this section in Wikipedia. That means that this is not a real stereocenter. But we do need to assign something because there is intrinsic stereochemistry because it has relative stereochemistry with the two enantiomeric substituents, and changing the center produces diastereomers. As you noted, the reason it's a pseudocenter is because it cannot be a full stereogenic center by symmetry.

Thanks! Do you mind expanding your answer a bit? As it stands, "because there is intrinsic stereochemistry because it has relative stereochemistry with the two enantiomeric substituents, and changing the center produces diastereomers." makes very less sense me. Perhaps, splitting it into 3-4 sentences might be useful?

@Zhe: do you mean that we can say there're two isomers r and s, but that there isn't a convention to assign priorities to the two chirally-opposite-but-structurally-identical groups, and therefore you can't actually draw the structures and specify which is r and which is s?

@HarryHolmes They are different structures, and you can assign uniquely and unambiguously assign $r$ and $s$ to the middle "stereocenter." But that's the problem. It's not a stereocenter because of symmetry. You can absolutely draw the structures and assign centers. We just did!

(2R,3s,4S)-Pentan-2,3,4-triol is a meso compound owing to the plane of symmetry. C2 and C4 are both chirotopic and stereogenic. Thus the R/S designation. C3 is ACHIROTOPIC and stereogenic (pseudoasymmetric). Hence, the lower case "s" (C3-OH>C2-R>C4-S>H, which are the CIP priorities). Inversion of C3 leads to a new meso diastereomer (2R,3r,4S) that has the same chirotopicity and stereogenicity at the three centers as did the original triol. To invoke enantiomers one needs the pair (2R,4R) and (2S,4S). C3 in each of these enantiomers has no designation (non-stereogenic and chirotopic) because inversion returns the same enantiomer. For a discussion of stereogenicity and chirotopicity in lieu of "pseudoasymmetry", see Mislow, K. and Siegel, J., J. Am. Chem. Soc., 1984, 106, 3319.

Can you contextually explain what you mean by "achirotropic"? The goldbook definition is a bit vague - https://goldbook.iupac.org/html/C/C01065.html Also what does "OH>R>S>H" mean?

@Gaurang Tandon: Sorry for the late reply. Your Goldbook link seems clear to me. Inversion of C3 returns the same structure, i.e., achirotopic. As to the priorities,  O(H) is highest, H the lowest. What about the 2 enantiotopic groups, C2 and C4? Follow CIP rules. R precedes S. Thus, O(H)>R>S>H.

When do we decide the compound has plane of symmetry... By analysjng it's 3D structure before making the isomers?

Take a look here:  http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/RS14272/achiral.html

Pseudoasymmetric centres are stereogenic but achiritopic. Now, the 3rd carbon is pseudoasymmetric centre. It has stereogenicity only when the two groups are  different in configuration. Note that I have used the word stereogenicity, and not chirality, because whatever the configuration of the two $\ce{CH(OH)CH3}$ group may be 3 rd carbon does not induce optical activity or chirality in the molecule. That is why it is a pseudo yet asymmetric centre (In the sense that it has 4 different groups).
So the configuration you have given is obviously meso, but as a whole if you ignore the stereochemistry in the structure then for this compound you have 4 isomers: 2 optically active and two meso (these two meso compound arising from the 3rd carbon being stereogenic)

Thanks for your answer! "Note that I have used the word stereogenicity, and not chirality, because whatever the configuration of the two CH(OH)CH3 group may be it does not induce optical activity or chirality" That was exactly my question actually, as I was unable to justify it through a plane of symmetry. Still, +1 for describing it through words. Though I don't get your second part. How have you counted four isomers?


text:
Activation energy in physical adsorption

Physical adsorption of molecules on surface requires activation energy, because bonds of the adsorbed molecules are broken.

Is this a valid statement?
I have been under the assumption that physical adsorption does not involve any energy of activation as there is no chemical bond formation between the adsorbed molecules and the surface (and hence there is no necessity for bonds of adsorbed molecules to be broken).

The definition given above would more closely describe chemisorption. At large separation from the surface the interaction is zero. In physisorption the interaction becomes attractive and forms a potential well as the surface is approached. There is no activation energy needed to do this. At shorter separations than the potential minimum, the potential becomes repulsive. The minimum is usually at larger separation than the minimum for equivalent species exhibiting chemisorption. Species may physisorp before chemisorption occurs

Comment by porphyrin converted to answer


text:
Would the lone pair be in the equatorial plane or the axial plane for bromine pentafluoride?
I've been learning chemical bonding and I'm confused about the structure of $\ce{BrF5}$.
It has a square bipyramidal geometry but a square pyramidal shape - I got that.
I don't  get why the lone pair gets placed at the top perpendicular to the equatorial plane instead of the equatorial plane, where it's more stable.

And why does the angle change a lot from 90 degrees to 84 degrees? Similarly for $\ce{ICl4-}$- why do the lone pairs go to the axial plane?

There are no axial and equatorial positions here.

Oh! It didn’t show that way, now raajsuriya fixed it i suppose. But wait is there a square biphramidal geometry known? Or is it equivalent to octahedral?

Square bipyramidal geometry means octahedral geometry. All the vertices of an octahedron are equivalent.

@MollyCooL Yes, a square bipyramidal geometry is called an octahedral geometry because of its eight sides!

Okay thank you! Just got a bit confused between the two.

As @IvanNeretin said in the comments, "There are no axial and equatorial positions here". This is a symmetric 3-D structure. Think of this structure as two planes cutting each other perpendicularly and you can rotate it any way you want. This is well reprsented by this image.

So your line "I don't get why the lone pair gets placed at the top perpendicular to the equatorial plane instead of the equatorial plane, where it's more stable." does not make any sense. No matter where you place your lone pair it will still have no effect on the stability of the compound because everytime you think you are shifting your lone pair to a different position, it is forming the same molecule again.
Angle changes from 90 degrees to 84.8 degrees due to the lone-pair present. This creates a lone pair bond-pair repulsion and therefore slightly pushes the 4 Floruines, perpendicular to the lone pair, away from it making an 84.8 angle and not a 90.
Again in the case for $\ce{ICl4-}$ there is no axial or equatorial plane. Here's a diagram that should be helpful:

Keep in mind that (a), (b), (c), (d) can form a plane as well and according to your theory, it will become the equatorial plane. The main key point here is that the lone pairs must be furthest away from each other to be symmetric.


text:
Reaction of sulphur dioxide with solution of potassium permanganate
Solution of $\ce{KMnO4}$ is decolorised by $\ce{SO2}$. Is this due to the reducing nature of $\ce{SO2}$ or acidic nature of $\ce{SO2}$ ?
I encountered following two explanations. While first one appears to be reasonable, I could not understand the second one.
Reducing nature of $\ce{SO2}$: $$\ce{2KMnO4 +5SO2 +2H2O -> K2SO4 + 2MnSO4 +2H2SO4}$$
Acidic nature of $\ce{SO2}$:
$$
\begin{align}
\ce{SO2 + 2H2O &-> H2SO4 + 2H}\\
\ce{2H + O &-> H2O}\\
\ce{2KMnO4 +3H2SO4 &-> K2SO4 + 2MnSO4 +3H2O + 5[O]}\\
\end{align}
$$
Also, it is said that $\ce{H2SO4}$ is a moderately strong oxidizing agent. How can $\ce{H2SO4}$ reduce $\ce{KMnO4}$ to $\ce{MnSO4}$ (last equation)? Is the second explanation correct?

Your first equation under "Acidic nature" is incorrect. $\ce{SO2 + H2O -> H2SO3 + H2O}$ Since now there's no nascent hydrogen atoms, the second reaction won't occur, and hence the entire "Acidic nature" reasoning has failed.

Potassium permanganate is a very strong oxidising agent (stronger than sulphuric acid). In acidic medium, $\ce{Mn}$ gets reduced from its +7 oxidation state to +2 oxidation state. (n-factor is 5).
$$\ce{2KMnO4 + 3H2SO4 -> K2SO4 + 2MnSO4 + 3H2O + 5[O]}$$
Since $\ce{SO2}$ is a strong reducing agent, it reduces $\ce{Mn}$ as denoted by the equation
$$\ce{5SO2 + 2KMnO4 + 2H2O -> 2H2SO4 + 2MnSO4 + K2SO4}$$
Both the reactions are viable. Because of the formation of $ \ce{MnSO4}$ and the disappearance of $ \ce{KMnO4}$, decolorisation occurs.
But sulphur dioxide reacts with water to produce sulphurous acid and not sulphuric acid.
$$ \ce{ SO2 + H2O —> H2SO3} $$
Hence, the acidic nature reasoning doesn’t hold good.

Thanks. Few clarifications: (1) Is the first equation redox? Mn gets reduced from +7 to +2. Which species is getting oxidised? (2) How is H2SO4 produced if SO2 gives H2SO3? (3) What is the conclusion? SO2 decolorises KMnO4 due to both acidic nature as well as reducing nature?

(1) You can say oxygen is getting oxidised from -2 to 0 because of the formation of nascent oxygen (Although I am not sure about it) (2) $\ce{H2SO4} $is formed by contact process. (https://en.m.wikipedia.org/wiki/Contact_process) and also by reacting$ \ce{SO3}$ with water but the latter is highly exothermic. (3) Read gaurang’s comment on your question. That was why I said it can form only $\ce{H2SO}$  in my answer. I ll add that also.

Reg. (2), my question was not on how H2SO4 is manufactured. Rather, it was in the given context of SO2 reacting with KMnO4 in acidic medium. So, based on your inputs, I understand that SO2 with H2O gives H2SO3, and not H2SO4. So 'acidic nature' explanation does not hold. Thanks.


text:
Is the kinetic gas equation valid for non-cubical containers?
I have seen the derivation of the following equation for a cubical container with $N$ molecules.
$$PV = \frac 13 mNu^2 $$
How is this equation valid even for spherical container or any other non cubical shaped container?
(I prefer rigorous equation-based proof but I'm okay even with analytical proof )

With any non-cubic container, the problem would require some calculus.

And I'm not even sure that it's valid.I was just told so.

How would any of the molecules "know" the container isn't exactly cubic? Slightly more formal, assume the container deviates from a cube by dV in one place and -dV in another. This does not affect the equation, it turns out, regardless of the magnitude of dV.

The equation tells us that energy represented as $pV$=(force/area) . volume = force . distance $\equiv$ Newton . metre = Joule  is equal to the (kinetic) energy of the particles $mu^2$ = mass . velocity = kg . m /s$^2 $ = Joule. So this cannot depend on the shape of the container.


text:
Alternatives to sodium bicarbonate
I am searching for a molecule that is edible for human, reacts to liquid by generating $\ce{CO2}$ or a similar gas and that leaves behind a clear liquid, similar to sodium bicarbonate but without the milky whiteness.Any ideas?

Sodium bitartrate?

Ammonium bicarbonate, "baker's ammonia", is sometimes used as a leavening agent for baking (e.g. cookies) as it decomposes to gaseous ammonia, water, and carbon dioxide which are driven off in the oven heat. You can get it in some food stores. Hope this is helpful.


text:
pKa of different sites of butanone

From what I can tell, deprotonation of the central hydrogen leads to the more thermodynamically stable product (most substituted double bond. i.e. most stabilization by hyperconjugation). Therefore, if butanone is allowed to equilibriate in the gas phase, I would expect the deprotonated form A to be present in greater amounts than B. Therefore, I would conclude that the pKa of the central hydrogen is lower than that of the terminal one, in the gas phase.
My questions:

Is this correct? (please point out any mistakes or omissions from my argument)
How does this change with solvation, and why?

EDIT: To clarify questions: Any data on the acidity of the different sites in solution is greatly appreciated. The gas-phase specification above is simply there to be able to treat the problem solely thermodynamically (I appreciate that data on gas phase equilibration of acids might be hard to come by)

I've seen many other posts about ketone deprotonation, but browsing them I have   been unable to answer my questions. Refer: "Which enolate of butanone is more stable?", "Regioselective enolate formation"

I take it the opposite is true (B is in greater amounts) that's why you asked the question. If so could you cite the source?

Equilibrate in gas phase? Why are you considering that?

@Zhe The gas-phase part is simply to rule out any solvation effects which may (or may not?, i don't know) have an effect.

@AvnishKabaj I am not sure what is correct, thats why i asked the question, and would like the answer to include a credible source.

The more substituted enolate is the thermodynamic product. Gas-phase equilibration is a really weird thing to talk about...

@Zhe Okay,  what does that imply for the pKa?

pKa is directly related to thermodynamic stability, so...


text:
Defects in crystalline structure

Starting from a perfect $\ce{MgO}$ crystal a solid solution of $\ce{Al2O3}$ and $\ce{MgO}$ is prepared with an atomic proportion of $15:85$ of $\ce{Al}:\ce{Mg}$. Knowing the number of vacancies per $\ce{Mg}$ atom is $0.088$, calculate the % variation of density.

My reasoning is as follows:
For every $3~\ce{Mg^{2+}}$ atoms, we have $2~\ce{Al^3+}$ to maintain electric neutrality and 1 vacancy to keep the number of lattice points. Then, in one cell of crystal, which is FCC, there are initially 4 $\ce{Mg^{2+}}$ atoms which will then be redistributed as 2 $\ce{Mg^{2+}}$, $\frac43$ $\ce{Al^{3+}}$ and $\frac23$ vacancies to fulfill stoichiometry.
Then, I calculate the initial density as:
$$\rho_\mathrm o=\frac{(4 \ce{Mg^{2+}}\times\frac{\pu{24.31g}}{N_A \pu{at/mol}})+ (4 \ce{O^{2-}}\times\frac{\pu{16g}}{N_A \pu{at/mol}})}{a^3}$$
And the final density as:
$$\rho_\mathrm f=\frac{(2 \ce{Mg^{2+}}\times\frac{\pu{24.31g}}{N_A \pu{at/mol}})+ (\frac43 \ce{Al^{3+}}\times\frac{\pu{26.98g}}{N_A \pu{at/mol}})+(4 \ce{O^{2-}}\times\frac{\pu{16g}}{N_A \pu{at/mol}})}{a^3}$$
When taking the difference and dividing by the initial density and multiplying by 100%, I get 31.9%, when the solution is 4%. What am I doing wrong?

Well, you forgot several things, but in particular where does the 15:85 ration of Al:Mg figure into your calculation?

Update: I have successfully reached the right solution. Thanks again to @JonCuster

@Bee Could you please post a self-answer?


text:
Are the words "apparatus" and "equipment" synonyms?
Are the terms "apparatus" and "equipment" synonyms, within a chemistry or physics laboratory? If not, what are distinguishing factors?

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

https://i.stack.imgur.com/wYuF6.png

@Mithoron I foresaw this claim and hence the context. It might not seem to be a very technically relevant topic.. but FYI there are students of science who find such questions to be .. well .. intriguing..

@pentavalentcarbon I guess that answers the question in a very general sense including the ‘lab’ scenario.. thank you.

Within a lab they are synonymous. But if you consider in general, apparatus refer to non-machines. Equipment on the other hand can be divided into a couple of sections:

Non-mechanical which can be things like kitchen equipment, measuring cups, knives (all hand tools)
Mechanical machines which are things like sewing machines, mechanical diggers.
Powered machines that are powered by either gas, steam or electricity.


text:
Why is the pressure correction constant 'a' independent of temperature?
In the van der Waals equation of state, 'a' which is the correction value is based on inter-molecular forces of attraction? Then should inter-molecular forces decrease when temperature increases, and thereby cause a change in value of 'a'?
Then why is 'a' independent of temperature?

The parameter $a$ is independent of temperature because we defined it this way. There is no other reason. Indeed, van der Waals equation is very much not a law of nature. It is just an empirical approximation. There are other approximations, maybe a dozen or more. You may create your own, where $a$ would decrease with temperature. Wait, it's been done already - look at the Berthelot equation!


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Equipment Suggestion Small Burette?
Is anyone familiar with a piece of equipment that can be purchased like a small burette? I need to find a way to dispense a liquid at a constant rate, however, the amount of liquid that I need to dispense will probably be less than 5 mL. A syringe is good in that regard, but it requires that I am able to push down with a constant pressure (which I doubt I can do). I've heard that they make burettes that are smaller (i.e around 25mL). Does anyone know if they make them smaller or at least not 2 feet long. Thank you for any useful information.

How about a syringe pump?

Yes, a syringe pump would work. I forgot to mention that the amount of money I can spend is around 50-75 dollars which is much less than most syringe pumps. Thank you for your response. We are considering trying to build a make-shift syringe pump but are still not sure if they could be something else we could purchase that'd be better/faster to get.

any old HPLC pump around?

The usual commercially available burettes are standardizes in the international standard ISO 385:2005 Laboratory glassware – Burettes. This standard includes small burettes with capacities of 1, 2, 5, 10 and 25 ml. The very small versions are typically advertised as “micro burettes”.
$$
\begin{array}{lllll}
\hline
\text{Nominal capacity} & \text{Graduation divisions} & \text{Accuracy limits} & & \text{Max. length}\\
& & \text{Class A and AS} & \text{Class B} \\
\mathrm{ml} & \mathrm{ml} & \pm\ \mathrm{ml} & \pm\ \mathrm{ml} & \mathrm{mm}\\
\hline
1 & 0.01 & 0.006 & 0.01 & 575 \\
2 & 0.01 & 0.01 & 0.02 & 650 \\
5 & 0.01 & 0.01 & 0.02 & 820 \\
5 & 0.02 & 0.01 & 0.02 & 800 \\
10 & 0.02 & 0.02 & 0.05 & 820 \\
10 & 0.05 & 0.03 & 0.05 & 620 \\
25 & 0.05 & 0.03 & 0.05 & 820 \\
25 & 0.10 & 0.05 & 0.10 & 620 \\
50 & 0.10 & 0.05 & 0.10 & 820 \\
100 & 0.20 & 0.10 & 0.20 & 870 \\
\hline
\end{array}$$

@Mflickin you can accept the answer...


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Why is this scintillator component turning into a milky-white emulsion?
I am a chemistry and physics double major doing some chemistry work in a nuclear physics lab. My mentor is a physicist, so his area of expertise is occasionally mismatched with what I need for my chemistry end of the project.
The scintillator fluid we use is commercially made, so I don't know all of the specific components, just that it is primarily diisopropylnapthalene (DIPN). I was trying to replace a LiCl emulsion (a standard scintillator cocktail component) with a lithium carboxylate.
I have tried dissolving my lithium salts into the scintillator, both the standard commercially-made one and a custom one that's the same composition but without the surfactants. Using only fractional molar concentrations (as low as 0.1-0.2 M) it turns milky. LiCl emulsions in our lab were stable between 1M to 10M, at least.
I know that water contamination can cause a milky emulsion to appear in oils. I think I used a strong enough desiccant to remove all the water, and I couldn't find -OH groups in IR spectra of my samples.
Could this be the O- parts of the carboxylates getting emulsified? Or something else? Why would my lithium carboxylates be milky but a LiCl emulsion not milky?
Any ideas would be great. I'm trying to understand the interactions going on here so I can either revise my methodology or choose new molecules to try. Thank you!
Edit: Given my UV-Vis data, the transmission was worse without the surfactants, which I think is because the surfactants made my carboxylates dissolve more easily. But I'm not sure if that has anything to do with the milky appearance.

Could you please mention which cocktail you use (e.g. Ultima Gold AB or similar)? Maybe it's not the solvent but one of the other components.

@Loong It is indeed Ultima Gold AB (and a formulation that is the same but without surfactants)!


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Why does the sodium become transparent after burning?
I know that if we throw a piece of sodium into water then it will melt first then it will explode.
But under different conditions, the above situation will not happen. If we keep a piece of sodium on a wet cloth then it will not explode. Rather, it will burn and turn into a transparent small ball and then the ball will bust. (Video of example.)
What is the reason behind such situation? What reaction actually occurs when the sodium is kept on wet cloth?

Videos/links? Have you tried this (of course with proper safety equipment)?

Exactly the same one...

Yeah same one but no answers there , I think you did not watch full video , he said he even don't know what happens in that situation.

I'd surmise that it is an "optical delusion" due to sodium melting into a near sphere with a very high reflectivity since the surface is pure clean metal with no oxide coating.

17 days passed  I got no answer ?


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orgo- ketone derivatives
In the following reaction, I mix $\pu{0.25g}$ semicarbazide hydrochloride (M.M. $\pu{111.54 g/mol}$), $\pu{0.4g}$ sodium acetate ($\ce{CH3CO2Na}$), $\pu{3mL}$ of $\ce{H2O}$ (M.M of sodium acetate trihydrate is $\pu{136 g/mol}$), and $\pu{0.25 g}$ acetone (M.M $\pu{58.08 g/mol}$) in a large test tube.
My understanding of this reaction is that the sodium acetate deprotonates the semicarbazide hydrochloride to make it a good nuc while simultaneously buffering the aqueous solution to create ideal conditions for the acetone ketone to be protonated, become a good electrophile, and get attacked by the carbazide intermediate. I need help on how to figure out what the balanced equation is, how to find limiting reagants and how to determine what the theoretical yield of my product should be.

You are correct in your description of the reaction so the balanced equation should be obvious to you: 1 eq Semicarbazide.HCl + 1 eq Sodium Acetate + 1 eq Acetone gives the carbazone of acetone (plus NaCl and AcOH).
The limiting reagent and yield is obtained by calculating how many moles of each reactant are present, which is the mass of reagent divided by the molecular weight. e.g. for Acetone it is 0.25/58.08 = 4.3 mmol
If you do this calculation for Sodium Acetate and Semicarbazide.HCl it will become clear which component is the limiting reagent.
The theoretical yield is the number of moles of limiting reagent x MW of the product acetone semicarbazone


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Computational modeling of steric repulsion between two polymer mediated magnetic Nanoparticle?
Can you recommend any modern article or research papers on steric interactions and forces between two or multiple magnetic nanoparticles with polymer brushes in a nonpolar solvents? something with more physics and mathematical formulas not just text and graphs. My goal is to learn how brushes create osmotic pressure, what are the forces, and then create a simulation of two particles interacting together.
I am new to this field and currently reading Israelachvili's Inter-molecular Forces and Flory's Polymers books they are very general and don't have detailed Physics in case of steric repulsion.


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Total irreducible representation of a quantum system - Slater determinant?
I'm beginning to study quantum chemistry and I became pretty confused about the total state of the molecule.
When we describe the single orbitals using irreducible representations like in the picture below, then we can get the total quantum state by direct product
$$(2a_1)^2(1b_2)^2(3a_1)^2(1b_1)^2=(A_1)^2\otimes (B_2)^2\otimes (A_1)^2\otimes (B_1)^2 =A_1,$$
as it is described in this answer. But when I read about multi-particle wavefunctions, I realized, that the wavefunction of the whole system should be antisymmetric, so I should assemble Slater determinant to get the multi-particle wavefunction with the enforced antisymmetry property.

Questions
1) How is Slater determinant connected with the direct product above?
2) Does $(A_1)^2$ mean $A_1 \otimes A_1$ or is it purely the notation for 2 electrons in the orbital?

I haven't read through it in detail, but I think this should help: https://www.google.com/url?sa=t&source=web&rct=j&url=http://www.uwyo.edu/kubelka-chem/mm_notes_10.pdf&ved=2ahUKEwia1cyHqszZAhWE14MKHWdoBZoQFjAFegQIABAB&usg=AOvVaw0LHVQaMyEpqFSff8yvl-u8

(1) It isn't. (2) Yes.

@IvanNeretin Would you mind to respond in more detail? So, why can we find the total irreducible representation using direct product? And by "yes" you mean, that $()^2$ is purely the notation and not power?

The Slater determinant is just a linear combination of these direct products. The symmetry will still be the same.

Two electrons in the orbitals leads to $(A_1)^2 = A_1 \times A_1 = A_1$. So it isn't just notation.

@pentavalentcarbon So, that means that every  IR, e.g. $(B_2)^2$ equals $A_1$?

Yes (extra text to satisfy comment length)

The electronic wavefunction must be anti-symmetric with respect to the exchange of any two electrons, and Slater determinants provide a practical way to do achieve that.


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Why was my soap measured as acidic?
I've always been told soap is basic in pH. A few years back in a high school chemistry class, we measured the pH of various substances with universal pH indicator. We measured some liquid hand soap and it came out as acidic, which puzzled even the teacher. Unfortunately I have no idea what brand it was and I still don't know why it came out as acidic.
I learned yesterday that the fundamental reaction that produces soap happens when a base reacts with an oil chain which breaks down the glycerol backbone. Depending on which exact reaction path is being used to make the soap, some base may be added to ensure no oils remain unreacted. After the reaction is complete, some acid can be added to neutralize the base. Could it be likely that extra acid was added to the soap to neutralize the remaining base and the solution was left acidic? What other possible ideas could explain why the soap was acidic?

Soaps are made to give off slightly acidic solution, one can hear about "pH 5.5 soap" even from commercials, so "I'm surprised you were surprised" ;)

It has been claimed that skin has a pH of about 5.5. Of course, this must be the sweat coming out of your skin, since skin isn't a liquid. So many body washes add citric acid to their formulas to be milder to tender skin.
For washing after poison ivy or car grease, pH 11 is much more effective.


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Readings about chiral resolutions at the undergraduate level
Does anyone know of a resource which clearly explains the scientific theory and experimental methods involved in chiral resolutions?
I have access to the American Chemical Society, Tetrahedron, Royal Society of Chemistry and many other journals and books.
I am an undergraduate student doing laboratory research. I am about to separate enantiomers, which are atropisomers. I am looking for something that I can understand with my limited experience, as I have only taken two semesters of undergraduate organic chemistry. Also, maybe something brief since I will have to do the resolutions in only a few weeks, but that is less important.

Unfortunately, separation of enantiomers is not a generally solved problem. It depends very heavily on the functional groups in your substrate. If you're only looking to separate a small amount, I'd recommend chiral prep HPLC.

I would try and find the book Enantiomers,  Racemates and Resolutions by Jacques, Collet and Wilen


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What is the real gas equation at high pressures?
Is it correct to ignore the $\frac{n^2a}{V^2}$ term at high pressures as is done? I think that, as pressure increases, this term also increases as Volume decreases so it is not safe to ignore it!

https://chemistry.stackexchange.com/questions/16110/the-van-der-waals-equation-high-t-and-high-v-m


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Trends in ionization energy in the second period

Question
  Which element among the following has the highest ionisation energy: fluorine, oxygen, neon.

I know that all these elements belong to period 2 and ionisation energy increases from left to right across a period. But is there going to be an exception for neon? Or is it that Neon will also follow the normal rule?

I down voted because it does not show good research skill, as this answer can be found by simply searching on Google.

Compared with oxygen and fluorine, neon has to ionize an electron from the same subshell (2p) and there is more nuclear charge holding that subshell down.  Neon thus requires the highest energy to ionize.
Now consider the process of accepting an electron.  Oxygen and fluorine can do so in their 2p subshell but in neon that is full, the electron has to go into a higher subshell like 3s.  That limits the electron affinity of neon.
So neon is hard to knock an electron out and yet does not release much energy accepting an electron.  Add that up and you get ... an essentially inert element!  Like ... neon.


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Does the alpha effect apply to third row elements?
Alkylperoxy anions are stronger nucleophiles than alkoxides (the so-called alpha effect), but would an alkyldisulfide anion be a better nucleophile than an alkylthiolate anion?


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Solubility equilibrium

This is a worked example about solubility equilbrium I am new to this subject and I am really confused first of all I know that Q is the product of concentration of Ag and Cl in the solution and ksp is the product of molar solubility of Ag and Cl then why the difference of these two is not amount of precipitation
second: if we are not given ksp value of AgCl and we are given only the last part ( part b) how can we find the ksp value?
P.s: after change the amount of cl become zero and amount of Ag become 99*10^-6 so it means that after mixing two solution precipitate is formed and amount of dissolved Ag ions become 99*10^-6 and all of Cl ion is precipitated am I right?
(Sorry if my questions doesn't make sense for you it is because I am new to subject and I don't have any teacher I really tried hard on this example and I saw lots of videos about this subject but I didn't understand it)

Q is known as the ionic product which is defined as product of concentration of ions of the electrolyte, each raised to the power of their coefficients in the balanced chemical equation in solution at any concentration. $\pu{K_{sp}}$ is the same but it is applicable only for a saturated solution.
To find the solubility for the salt with the given $\pu{K_{sp}}$ value, For $\ce{AgCl}$, we have the equilbrium,
$$ \ce{ AgCl(s) <=> Ag+(aq) + Cl-(aq)}$$
If initally we have $‘a ‘$ moles of AgCl, at equilbrium we will have $’a-s ‘$ and we have $\ce{[Ag+]=[Cl-]=s}$. Hence we get the relation as,
$$ \pu{K_{sp}} = s^2 $$
This is the maximum concentration of the ions present in solution at equilbrium.(because $\pu{K_{sp}}$ = Q).
When Q just crosses $\pu{K_{sp}}$, salt is precipitated out.
Since they are products of ionic concentration, we can’t directly subract Q and $\pu{K_{sp}}$ to find the amount of salt precipitated.
I’m just working backwards to get $\ce{K_{sp}}$.
Let mass of $\ce{AgCl}$ precipitated be $’m ‘$ g.
No. of moles precipitated $n = \frac{m}{M}$.  Let the volume be $’V’$ litres,
$\ce{[AgCl]}$ precipitated $x = \frac{n}{V}$
$$ \ce{Ag+ + Cl- -> AgCl} $$
We know initial concentrations of the ions. Let $\ce{[Ag+]}$ and $\ce{[Cl-]}$ be $’p‘$ and $’q‘$ respectively.
Then solubility product is,
$$ \ce{K_{sp}} = (p-x)(q-x) $$
Hope that makes sense.

And all the $\ce{Cl-}$ is precipitated, what you said is correct.

you mean when Q=Ksp  s=amount of precipitated ions? And also can you please do the calculation to find ksp value from part b? Because when I did the calculation for ksp
It  didn't come up to be 1.6*10^-10 which is the actual ksp value for AgCl

When Q=$\pu{K_{sp}}$, at this point addition of more salt results in precipitation.  Solubility is the maximum moles of salt that can be dissolved per Litre of solution at a given temperature. I ll make my answer more clear. Please list out what we are given exactly and I shall try to find $\pu{K_{sp}}$. Thanks

Suppose that we are given the mass of precipitated Agcl and the volume of solution and we know the initial mole number of Ag and Cl

so only in case that Q>ksp preciption occur then if Q=ksp we don't have any precipitation and hence No equilbrium right?

I tried to find the ksp value this way initial concentration of Ag is 10^-2 and initial concentration of Cl  is 10^-4  and concentration of Agcl is 510^(-5)           ksp=((10^-2)-(510^-5))((10^-4)-(510^-5)) which is equal to 4.97510^-7 but this not equal to 1.6*10^-10 which is the actual ksp of Agcl

can you please check my calculation?

I did! But starting off with those is the $\ce{[AgCl]}$ precipitated correct? I doubt!

You mean that initial concentrations are wrong? but if I find their mole and find their initial concentration in 0.02L solution the answer is not 1.6*10^-10 can you please make it clear I really need to understand this

No, the initial concentrations of the ions can be anything, but corresponding to that, the concentration of $\ce{AgCl}$  should also be correct. You can check that with the known $\pu{K_{sp}}$ value and try and get the concentration of $\ce{AgCl}$. Then use my method to get $\pu{K_{sp}}$.


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What is the definition of a component in a phase diagram?
I have been trying to understand the concept of a component in relation to phase diagrams in my materials class. I have tried searching online and found that the definition is supposed to be that a component is a singular element, however I have also found a water (a 2 element compound) phase diagram is listed as a singular component phase diagram, which contradicts the previous definition of a component.
So is a component all relative to the temperatures/concentrations it operates under? Like, if a liquid is made of two elements and you're looking at it between 20-100 degrees Celsius and it only separates into its individual elements after 100 degrees Celsius, then would it be classified as a singular component phase diagram under 100 degrees Celsius?
So in short I need help understanding what a component is in relation to phase diagrams.

Since you are coming from a materials background, likely the first binary phase diagram you encountered was a diagram with two pure elements on either end of it. This might look something like:

And you probably understand how to parse out the x-axis - one goes from pure silver on the left to pure tin on the right. At any point you can determine the phase(s) stable at a given atomic ratio and temperature.
Now one comes to binary diagrams that are more properly called pseudo-binary diagrams. Here the components are indeed not elements, but some variety of compound(s) that can interact like you are used to on a classic binary diagram. For example:

Here one has pure GdF$_{3}$ on the left, and pure CsF on the right. The composition axis is no longer in atomic fraction (my bad on labeling the axis), but is instead a weighted average of the pure compounds. So, for instance, the purple line representing Cs$_{3}$GdF$_{6}$, is found at a 'composition' of 0.75. This is because it is an amalgamation of 3 CsF units and 1 GdF$_{3}$ unit, so it is (3/4) units towards CsF. Similarly, the CsGd$_{2}$F$_{7}$ is one CsF and two GdF$_{3}$ units, so is (1/3) units towards pure CsF.
The math works out just as for the true binary element case, but you have to think in terms of ratios of the compounds, not specific elements in the compounds.


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Propose a synthesis of indane from benzene
The question is to propose a synthesis of indane from benzene. I started by the following. The black box is what I am not sure about. I know that the stable carbocation should be the middle one, this is why I am not sure. I need help here.

Better to do Friedel–Crafts acylation chemistry, then get rid of the carbonyl groups later, that avoids the problem which you pointed out

Related: Double alkylation of benzene to form tetralin

If you treat your bromo alkyl compound with AIBN/tributyltinhydride it should cyclise to your product by a radical mechanism

@Waylander I'm not seeing it - that would give you a compound with two hydrogens too many, I think

@Waylander https://i.stack.imgur.com/r8xYd.jpg

OK, I see what you're saying. Reflux this product over activated MnO2 to get the required

I still haven't taken this reaction (AIBN) so i don't think i can use it.
I did another way:
1- Fridlecraft with Cl-Ch2-Ch2-Cl
2- Mg/ether to prepare my grignard reagent
3- Co2
4- H+ ; the benzene ring will attack the carboxyl group
5-Reduce by clemmenson
My concern is step 1 and 4

Polyphosphoric acid is reagent of choice for step 4. Instead of Grignard + CO2 attack the chloride with KCN and hydrolyse to get the acid.

Why doesn't grignard work though :( ?

It would, you can after all buy phenethyl magnesium chloride. But as a general principle for more complex molecules the cyanide + hydrolysis is a better option

Just saying: there’s a carbon missing in the product of the first step.

another way would be to (1) Friedel-Craft acylation with succinic anhydride/$AlCl_3$ (2)Clemmenson reduction (3) $H_3PO_4$ (4) Clemmenson reduction.

Succinic anhydride has too many carbon atoms, your route will give tetralin

@Waylander Potentially a naive question from an inorganic chemist: Is such reaction with AIBN/Sn known to happen like that? It looks like you will be breaking the Ph ring aromaticity. Can you provide a reference where someone has done something similar ?  Just being curious, thank you.

Now that I'm looking I can't find a ref, Possibly I have mis-remembered but I no longer have Scifinder access to check.

Do people actually do Clemmensen reductions these days or is this paper chemistry?

@Beerhunter This is what i found interesting with this question. How would you actually do it in the lab. The first step in the reaction is fine (got a ref for it) but for the second i am not sure the Br will survive the highly acidic and reducing conditions of the clemmensen reduction. Or in the case of AIBN (which is a reasonable suggestion btw) not sure if you will get the cyclized product or a dimer or both.

The synthesis proposed could be fine if reordered and a step removed. Acylation occurs as presented. Add some more Lewis acid (notwithstanding a calorimetry check first), then cyclise to give the 1-indanone. Then reduce last using your favourite reduction method. The two I've witnessed industrially are a) TFA with triethylsilane or tetramethyldisiloxane, b) methanesulfonic acid with 5% Rhodium catalyst and hydrogenation at 60 bar. Both work on benzoyl groups really well.

@Beerhunter, I surmise it is simply that old reactions such as the Clemmensen are traditionally taught early on.

To prepare indane (6) from benzene, make the procedure as simple as possible and amenable to scale-up. Propiophenone (2) is readily available by Friedel-Crafts acylation of benzene with propionyl chloride (1). Removal of the carbonyl group of propiophenone and the creation of a carboxyl moiety can be achieved via the Willgerodt reaction to produce amide 3. Vigorous  acid or base cleavage of the amide is possible but nitrous acid affords a mild route to carboxylic acid 4. One can use the AlCl3 F-C route again but hot polyphosphoric acid is a clean method for the intramolecular acylation without having to form the acid chloride. Finally, Clemmensen reduction of indanone (5) gives indane (6). Wolff-Kishner reduction is also a good option for the final step.

Nice one,  this Willgerodt reaction (unknown to me till now) was the missing link to the type of synthesis i was trying to get at as well!

@AMM: I should add that the initially formed product in the formation of amide 3 is the thioamide, which undergoes hydrolysis under reaction conditions.

Below is a suggested route for synthesizing indane starting from benzene and allyl bromide. I appreciate that this may not be a desirable answer for a school question but it is how one could perform the synthesis in real life and is based on high yield reactions that have been previously reported (see references, also note that this may not be the most efficient way):

References:
1) Synlett, 1999, No. 8, 1245.
2) J Org. Chem. Vol.78, No 19, 9772.
3) J Org. Chem. Vol.57, No 23, 1992.
4) Molecules 2014, Vol 19, No 5, 5599.
5) Catalysis letters 2011, Vol 141, No 4. 507.

Be quicker to do a Heck with your bromobenzene and acrylate.

Yes, i was thinking of cross coupling approaches too and specifically converting the benzene to PhB(OH)2 and then suzuki coupling but didn't want to use transition metals or 'uncommon' ligands for this one and keep it more 'organic'. The Heck is a reasonable suggestion though. I assume you would follow with hydrogenation of the double bond and hydrolysis to get the acid afterwards ?

Exactly so, then polyphosphoric acid for the cyclisation

I think the effeicient and better way to synthesize the given compound with greater yield from Benzene may be this,

I dont think this is a good way since you will get a mixture of all possible Friedel Crafts products such as PhCH2CH2CH2Ph etc.

Not if you add benzene into the mixture or use a lanthanide Lewis acid in a flow system with a dichloropropane and benzene feed in the appropriate ratio.

@Beerhunter This doesn't sound very practical and i cant find it in the literature. It does sound very interesting though. Can you provide a reference please as it might be of use to me in general (it doesn't have to be exactly the same reaction)? Thank you.

The concept will have been demonstrated, but not necessarily on exactly the same compounds. The literature is just what other people have done, which doesn't always make it the best way to do something. The methodology of using something like scandium triflate has been demonstrated.  Using flow chemistry to control stoichiometry has been also.

Yes, this is why i asked for something similar and not the same so I could have a look into it. Indeed, literature is not necessarily the best way to do something but it is in many cases the fastest and safest way compared to re-inventing the wheel. The fact that something makes sense as a concept is very far from actually working in the lab. There has been plenty occasions where people in the lab tried to take a 'shortcut' but ended up spending much more time compared to performing a few more, but well established, reactions. As far as i am concerned, what you suggest might or might not work.

This route is problematic: 1) alkylation of benzene via F-C gives products that are more reactive than benzene; 2) polymerization here is a problem IF the primary carbocation were the reacting species; 3) n-propyl chloride gives isopropylbenzene. Rearrangement of the primary to the the secondary carbocation occurs.  Expect the same problem here.

@AMM I don't disagree in the slightest.  There are a lot of paper reactions thrown around as working, when in fact they don't. We - as a community - should also think outside of the round bottom flask.  It amazes me that unpleasant and hazardous reagents ate still being lectured as the go to ones for simple transformations.

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


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Calculate pressure equilibrium constant from concentration equilibrium constant

$$\ce{N2(g) + 3H2 (g) -> 2NH3(g)}$$
  Calculate $K_\mathrm p$ from $K_\mathrm c$, given that $K_\mathrm c = \pu{4.2 M^-2}$
  and that there is 0.2 moles of $\ce{N2}$, 0.6 moles of $\ce{H2}$ and 0.425 moles of $\ce{NH3}$. Temperature = $\pu{600K}$

My attempt at the question:
$K_\mathrm p= K_\mathrm c(RT)^{\Delta n}$
$K_\mathrm p= 4.2(8.314\times600)^{-2}$
$K_\mathrm p= \pu{1.69\times10^{-7} Pa^-2}$
However, in our answer sheet the value of $K_\mathrm c$ is divided by $10^6$ and the answer is $\pu{1.69x10^-13 Pa^-2}$. Could someone please explain why $K_\mathrm c$ is divded by $10^6$?

If you intend to get an answer in Pascal, which is an SI unit, you need to have your original values in SI units as well.
Note that:
$$K_\mathrm c=\pu{4.2M^-2}=4.2~\left(\frac{\text{moles}}{\text{litre}}\right)^{-2}=4.2~\left(\frac{\text{moles}}{\pu{10^-3m^3}}\right)^{-2}$$
The conversion factor for litre (non-SI) to metre cubed (SI unit) is $10^{-3}$. Since you have $K_\mathrm c$ in units of $\pu{M^-2}$, the final answer would be $10^{-6}$ of the answer you had arrived at.

Elaboration:
Let's do dimensional analysis on the RHS factor-by-factor.

$K_\mathrm c$ has the units $\left(\frac{\text{moles}}{\text{litre}}\right)^{-2}$ as shown above.
$RT$ has the units = $\pu{Jmol^-1T^-1}\cdot\pu{T}=\pu{J\cdot mol^-1}$
$(RT)^{-2}$ has the units = $\pu{J^-2\cdot mol^2}$.
Since one joule $=\pu{Pa\cdot m^3}$, hence $(RT)^{-2}$ has the units = $\pu{Pa^-2\cdot m^-6\cdot mol^2}$
Multiply $K_\mathrm c$ and $(RT)^{-2}$ we get
$$\text{Units of }K_\mathrm p=\left(\frac{\text{moles}}{\text{litre}}\right)^{-2}\cdot\pu{Pa^-2\cdot m^-6\cdot mol^2}=\pu{Pa^-2\cdot m^-6\cdot litre^2}$$

Thus your final answer is not in the units of $\pu{Pa^-2}$ but rather $\pu{Pa^-2\cdot m^-6\cdot litre^2}$. Can you now see why we need the conversion factor?

Please note that unit symbols and unit names shall not be mixed and that unit names are not mathematical entities. Therefore, it is not permissible to write $K_\mathrm c=4.2~\left(\frac{\text{moles}}{10^{-3}\ \mathrm{m^3}}\right)^{-2}$.

@Loong So, I should write $K_\mathrm c=4.2~\left(\frac{\text{mols}}{10^{-3}\ \mathrm{m^3}}\right)^{-2}$ instead?


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The order of aromaticity of benzene, thiophene, pyrrole, furan, and pyridine
I have been taught that the order of aromaticity is: benzene > thiophene > pyrrole > furan. However, I have been unable to deduce any logical explanation for that order. Also, I wish to compare pyridine's aromaticity with these, but my professor does not know about it.
According to me, all the rings are aromatic and satisfy the Huckel's rule of aromaticity with $6\pi$ resonating electrons and a planar ring. So, that point is useless for doing any comparison of aromaticity.
But, we note that $\ce{2p\pi-2p\pi}$ overlap is most favorable in conjugated systems. Hence, thiophene should be the least aromatic (since it has a $\ce{3p\pi-2p\pi}$ overlap instead).
I agree that benzene should be the most aromatic because both its resonating structures are equivalent and have no charge separation. Using exactly the same reasons, pyridine should be the second best aromatic ring (lesser than benzene because the N atom is an electron withdrawing group and is reducing the electron density in the ring)
That said, every resonating structure (except the uncharged one) for pyrrole and furan will create a formal positive charge on the hetero-atom. Since N is less electronegative than O, it will be slightly more stable than O with that positive charge. Hence, pyrrole will be more aromatic than furan.
Therefore, according to me, the aromaticity order should be: benzene > pyridine > pyrrole > furan > thiophene.  But it is obviously wrong.
Hence, my question is: what is the order of aromaticity of benzene, thiophene, pyrrole, furan, and pyridine. And why?

It's not trivial to establish a quantitative degree of aromaticity, actually. Phil Baran writes benzene > pyridine > thiophene > pyrrole > furan, but I don't know where the values come from. Empirically, though, pyrrole and furan undergo dearomatisation fairly easily because they're very electron-rich; thiophene has more benzene-like reactivity because it's less electron-rich (sulfur 3p poorer overlap with carbon 2p).

@orthocresol Did you mean pyrrole (0.039) > thiophene (0.032) > furan (0.007) instead?  (i'm referring to the $\beta$-values in the "Heterocylic Aromaticity values" in the pdf you linked)

I was looking at the aromaticity percentages. But I don’t know where the values come from.

This isn’t a full answer so I’m commenting it. Regarding thiophene, furan and pyrrole, in thiophene due to the lesser electronegativity of sulphur, the lone pair delocalises easily, and so aromatisation is better. I have no idea about how to include pyrridine among these. They(pyrridine and the other three) seem to be different systems so I’m having trouble comparing them.

I looked up Aromaticity percentages and found this: https://pubs.acs.org/doi/abs/10.1021/jp0657686. It says here that "In the present work concept, the ΔΔHH2 value of benzene defines the completely aromatic character (+100%), and the closed shell of the singlet cyclobutadiene represents maximum antiaromaticity (−100%)." So it's a reference scale, defined with respect to Benzene.

What standard should be used? Ease of hydrogenation is one. Perhaps competitive electrophilic "aromatic" substitution such as Friedel-Crafts reaction.

Strictly speaking Hückel's rules do not really apply to these compounds except for benzene, as they are not derived for these cases. Also aromaticity is not a well enough defined concept to actually derive an order. You can evaluate their reactivity with respect to different mechanisms, but then you are comparing exactly that, and you are only deducing properties towards aromaticity. How could you quantify something that you can't really measure?

@Martin-マーチン  Your comment, and the previous comments by AbhigyanC and user55119 make it seem that these "orders of aromaticity" stated by various books are basically pointless i.e. there are many reasonings to define aromaticity (hydrogenation/EAS/etc.) and they all could lead to different orders. Unless explicitly stated by the book the reasoning they've used, we can have more than one valid guesses. Is that true?

@GaurangTandon That is somewhat more of a philosophical question. When you compare hydrogenation among those compounds, you do exactly that, you don't actually measure aromaticity. So with respect to aromaticity, these are all guesses; with respect to reactivity it is data, and that is in principle what you want to know. If you know which compound is "more aromatic", but it doesn't let you deduce which reactions it will undergo more than the other, then that is basically a pointless characteristic, designed to generalise. We already know that these approaches might lead to wrong conclusions.

@Martin-マーチン "it doesn't let you deduce which reactions it will undergo more than the other, then that basically a pointless characteristic, designed to generalise." Yep, it does seem to be so now. It's surprising how popular this order is yet it is basically pointless :-O

Thiophene and pyrole being "more aromatic" than furan do not undergo Diels Alder reaction

The main criteria of aromaticity is resonant or delocalization energy.
It is difference between energy contained in bonds if they were localized like in Kekule's formula for benzene and actual energy.
For benzene if is about 150 kJ/mol.
The more this energy the more stable molecule is.
The approaches for calculating this energy may differ.
As for β-values mentioned in paper by @orthocresol it looks like it is Dewar's resonance energy per one π-electron calculated based on modified Hückel's MO method as mentioned in GSE article.
There is a table in this article with the same β-values:

In English it is

Benzene                            0.065     Pentalene                     -0.018
Naphthalene                        0.055     Azulene                        0.023
Anthracene                         0.047     Fulvene                       -0.002
Tetracene                          0.042     Pyridine                       0.058
Cyclooctadecanonaene([18]annulene) 0.012     Pyrrole                        0.039
Cyclobutadiene                    -0.268     Furan                          0.007
Azete (Azacyclobutadiene)         -0.160     Thiophene                      0.032


text:
Surprising values of intermolecular forces of ethane, oxygen and xenon
I know that gases with stronger intermolecular forces should have a larger value of the van der Waals constant $a$. So I looked up a table on van der Waals constants and tried to reason about the relative values of that constant. Though I was able to explain most of the values, there was one thing I found absurd. Here:
$$
\begin{array}{cc}
\hline
\text{Substance} & a~(\pu{atm {dm}^6 mol^{-2}}) \\
\hline
\text{Oxygen} & 1.364 \\
\text{Xenon} & 4.137 \\
\text{Ammonia} & 4.225 \\
\text{Ethane} & 5.507 \\
\hline
\end{array}
$$
My intution says that dioxygen, which has a slightly higher molar mass (32 vs. 30) than ethane, but significantly less molecular size or volume, should have a only slightly lower value of $a$ than that of ethane, but here we find a significant difference! Another example is xenon; its molar mass (131) is much larger than ethane (30), but its value of $a$ is smaller. Does it have to do with the dominance of molecular size over mass? Or atomicity or atomic dipoles? I suspect that it's probably the large and branched structure of ethane that's causing this.
Why does the molecular size/volume seem to play such a dominating role in the value of the constant $a$? I thought it was primarily influenced by molecular mass.

Van der Waals forces are weak and depend on polarizability. There is a correspondence between the boiling points of oxygen (−183 °C), xenon (−108 °C), and ethane (−88.5 °C) and their constants (1.364, 4.137, 5.507). These gases are polarizable to varying extents, but not polar with charge separation. The size of the molecule and its mass will affect polarizability - perhaps a way to look at this is to say electron density in the outer regions of the molecule will determine how easily dipole-dipole interactions can evolve.
But ammonia stands out with a boiling point (−33 °C) which is too high to be explained by van der Waals forces alone. Of all the gases listed, ammonia is the only one with significant hydrogen bonding capability. Perhaps if $\ce{H2O}$ were included, it would be an even greater outlier.

You seem to suggest that vdW forces are independent of molecular size. Am I correct?

van der Waals forces can be calculated from ionization potentials and polarizabilities. They vary approximately as 1/r^6, where r is the separation between two molecules (the space between them, not the distance between centers). Polarizability is related to how "soft" the molecule is, which is not directly related to size, but all other things being equal, a larger molecule would probably be more polarizable.

Ok, so you are suggesting here that the molecular size has a greater effect than the molecular mass of a molecule?

@Guarang Tandon I think you have the right idea. It is more about size than mass. A lot of times as a simplification, van der Waal forces are described as being related to mass, but that's just because for the simplest case of single atoms or small molecules, size is closely correlated with mass.

so let me just wind up the entire discussion- Van der waal forces actually do not depend on molar mass (atleast to a significant extent) but depend greatly on molecular size, since molar mass is generally proportional to molecular size we generally say that vanderwaal force increases with increasing molecular mass but that is not true, molecular size is the only dominating factor. If I am right please add this to your answer as this line alone would answer my question.

OK, I will add a line - but make it more about polarizability: The size of the molecule and its mass will affect polarizability - perhaps a way to look at this is to say electron density in the outer regions of the molecule will determine how easily dipole-dipole interactions can evolve.