text:
Cr(II) and Mn(III) - their oxidizing and reducing properties?
My textbook states that $\ce{Cr^2+}$ is a reducing agent while $\ce{Mn^3+}$ is an oxidizing agent in spite of both having $\ce{d^4}$ configuration. The explanation states that when $\ce{Cr^2+}$ gets oxidized to $\ce{Cr^3+}$, it attains a stable half-filled $\mathrm{t_{2g}}$ orbitals but when $\ce{Mn^3+}$ gets reduced to $\ce{Mn^2+}$, it attains a stable half-filled $\mathrm{d}$ orbitals.
First, is it a reasonable explanation? If this is the case, why can't $\ce{Cr^2+}$ get reduced to $\ce{Cr+}$ to attain a stable half-filled $\mathrm{d}$ orbitals and $\ce{Mn^3+}$ get oxidized to $\ce{Mn^4+}$ to attain a stable half-filled $\mathrm{t_{2g}}$ orbitals, i.e. in-short is the vice-versa true?

Related question with same answer but in a different context of the 4f block: Why don't we see these lanthanide species?
You have a misconception regarding the stability of oxidation states. The factors you have listed are honestly not very important in determining the stability of a certain oxidation state. They will tip a delicate balance in favour of one oxidation state or another, but they are hardly the sole determining factors.
If you are going from $\ce{M+}$ to $\ce{M^2+}$, you need to consider two main things.

You need to ionise the second electron from $\ce{M+}$, denoted $I_2$. That represents an input of energy.
$\ce{M^2+}$ is smaller and more highly charged than $\ce{M+}$, so you can gain some energy back on the basis that the $\ce{M^2+}$ ion is more strongly solvated or forms stronger ionic/covalent bonds with anions.

Loosely speaking, if the energy you get out of (2) is enough to compensate for the energy you lose in (1), then $\ce{M+}$ will act as a reducing agent. And the converse is true as well. It is for this reason that one does not observe Mg(I) compounds (apart from some esoteric organometallics). The $I_2$ of Mg is so small that it can be recouped extremely easily. Likewise, the $I_2$ of K is so large that there's no way any stronger bonding can compensate for that.
For the cases you have described:

$\ce{Cr^2+}$ simply does not act as an oxidising agent because $I_2$ of Cr is rather small. The same is true of the whole 3d block. How many 3d compounds do you know of with the metal in a +1 oxidation state? Without resorting to organometallic compounds, all I can suggest is copper(I). The reason why none of the other transition metals exist in the +1 oxidation state is precisely because their $I_2$ values are fairly small.
Similarly, $\ce{Mn^3+}$ does not act as a reducing agent because $I_4$ is way too large. Remember that successive ionisation is extremely difficult.

What your textbook should have written is:

The $I_3$ of Mn is anomalously large because you are "losing the stability of a half-filled 3d subshell". Hence, $\ce{Mn^3+}$ is a good oxidising agent.
The hydration enthalpy of $\ce{Cr^3+}$ is anomalously large (for a 3+ ion) because there is a large ligand-field stabilisation energy associated with the $(t_{2g})^3$ configuration. (This is factor number 2, not factor number 1). Hence, $\ce{Cr^2+}$ is a reducing agent.

These factors are small compared to the raw size of the ionisation energies. Yes, the $d^3$ ion $\ce{Mn^4+}$ also enjoys a large LFSE, and its hydration enthalpy would be larger than expected. However, that 4th ionisation energy is just too big, and that simply cannot be compensated for by the comparatively tiny increase in LFSE.

Did you mean to write that the $I_3$ of Mn is anomalously high, not anomalously low?

@AshishAhuja Thanks, good catch! I fixed it.


text:
Mechanism of substitution reaction with no change in stereochemistry

When (2S,3R)-3-iodobutan-2-ol undergoes a substitution reaction with sodium azide the only organic product from the reaction is (2S,3R)-3-azidobutan-2-ol. Give a mechanism for the reaction.

My Attempt
Now I know this is not a normal $\mathrm{S_N1}$ or $\mathrm{S_N2}$ reaction since the stereochemistry remains the same and only one of the enantiomers are formed. I am guessing that at first there is some intramolecular reaction where the iodine atom attacks the other carbon bonded with the hydroxyl group. Then the azide substitutes the iodine atom.

I can explain the reason why the configuration does not change by using neighboring group participation of the hydroxyl group. This is the mechanism that I propose:

The hydroxyl group attacks C-3 and $\ce{I-}$ leaves.
Then $\ce{N3-}$, the nucleophile, attacks C-3.
Since two successive inversions occur at C-3, the net result is retention.

As you can see, the configuration has not changed.

@bon brought up a point: The formation of a mixture of products when an epoxide opening takes place. If the nucleophile were to attack C-2 instead,

C-3 will still have the S configuration.

As noted by @Jan in the comments, the epoxide intermediate is symmetrical and both epoxide carbons are equivalent. This can be shown by a $C_2$ rotation:

Epoxide formation under neutral conditions? Also, one would expect a mixture of products from the epoxide opening.

@bon I am sorry if this is wrong.  I read about retention of configuration because of neighboring group participation in jerry march's book. So I thought it would be the reason here.

I'm not saying it's definitely wrong but I'm just surprised (and intrigued).

@bon: if the nucleophile were to attack at the third carbon, then the product will also have retention configuration. So Nu attack on 2nd and third carbon will produce the same compound. I just drew the possible products on my book.

Oh yeah my mistake.

Upon some research it appears this is correct. I learnt something new today!

The intermediate epoxide is $C_2$ symmetric. This renders the carbons homotopic and means that both attacks give the same product.

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text:
Bromination of a benzene based molecule

Using two equivalents of $\ce{Br2}$ and catalytic $\ce{FeBr3}$, compound A can be converted into B. In the same reaction vessel but in multiple steps, the conversion of B into C ($\ce{C11H12Br3I}$) can be obtained using $\ce{H2SO4, NaNO2}$ and $\ce{NaI}$.
The molecule A is below:

My Attempt
What I am struggling with is determining what double bonds the $\ce{Br2}$ is added on. I am pretty sure that the double that is not on the benzene ring will get brominated, but I am not sure what other double bond will also get brominated.
I know the mechanism for the conversion for B into C, which is basically the sandmeyer reaction followed by substitution of iodine. However what further confuses me is that the end product only has 3 bromine atoms, not 4 bromine which is what I expected (since bromination occurs twice).

No other double bond gets brominated. Electrophilic substitution in the ring, that's what happens.

@IvanNeretin Thanks. How do I know which carbon in  the ring gets substituted with the bromine atom?

The one in para position to the hydrocarbon tail, I'd say.

I think you are likely to get a mix, but ortho to NH2/para to alkyl seems likely dominant.

Related on the topic of selectivity.

You can deduce the end product from considering the change in the molecular formula from start to finish. Aromatic bromination results in the loss of one hydrogen and gain of one bromine, bromination of the double bond results in no loss of hydrogen but gain of two bromines and the Sandmeyer reaction results in loss of two hydrogens. Therefore the intermediate must have fourteen hydrogens which means only one aromatic bromination has taken place.

The aromatic bromination likely produces a major product with the bromine at the ortho position shown. The amine group strongly activates the positions ortho and para to it so substitution will occur at one of these three positions. I suspect the one shown is most likely due to the bulky nature of the other attatched group, which stericically hinders attack at the other two positions.

It might probably be worth mentioning that, while in this particular case the electrophilic addition does take place on the pi bond (as justified by the change in molecular formula), BUT in general, with the reagent $\ce{Br2 / FeBr3}$, only EAS can take place.

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text:
Distilling small volume
I have 50 ml of a compound I would like to purify. Since the impurities presents have close boiling points (b.p.), I have to use fractional distillation. Am I going to lose significant quantity of my product volume because of loss related to the distillation itself?
If so, what can I do to minimize the loss?
I thought about adding a higher b.p. solvent to the distillation flask so I can distill all the compound without leaving some in the flask. What do think about this method?
Are there any other strategies?

Adding a chaser is certainly necessary to maximize yield.

In order to both achieve sufficient separation and minimize loss during fractional distillation, you want

a large inner column surface for optimal contact of the vapour phase (upstream) and the liquid phase (downstream)
a rather small inner column volume

Often, these destillation are performed using a Spaltrohr distillation apparatus. Here, the vertical column consists of two concentrical, threaded glas tubes with a gap of 1-2 mm  in between.

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What is Brown's Gas
I've recently heard references to something called "Brown's Gas" which can be used to cut/weld metal but is regarded as 'a cold flame'. Does it actually exist and if so, why is it not widely used?

Might get some traction on Skeptics.SE on this one.

I guess you could consider the flame "cold", but only relative to things like acetylene or dicyanoacetylene.  The adiabatic flame temperature is 3200 $\unicode{x2103}$.

http://chemistry.stackexchange.com/questions/4018/differences-between-h-h-o-and-h-o-h

Brown's gas is actually hydrogen-knallgas, a 2:1 mixture of hydrogen and oxygen.
See https://en.wikipedia.org/wiki/Oxyhydrogen.


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Why does trihaloacetaldehyde formed during the haloform reaction not undergo a Cannizzaro reaction?
We know trihaloacetaldehyde is formed during the haloform reaction of acetaldehyde with a halogen in strong base. Since it has no alpha hydrogens, I expected that it might undergo a Cannizzaro reaction, as we are using a strong base.   Why not?

https://en.wikipedia.org/wiki/Haloform_reaction is for ketones not aldehydes.

@Mithoron Please go through the same link, you will find: The only primary alcohol and aldehyde to undergo this reaction are ethanol and ethanal,

Och, didn't thought about them in this context... Well then it happens for the same reason all haloform reactions happen: CX3 anion is leaving group good enough to be substituted by OH-. Much better then H- for that matter.

Albeit haloform reaction is primarily for methyloketons, aldehydes which have relatively good leaving groups connected with carbonyl (like chloral), or get it during halogenation, indeed also react.
Here we have another factor limiting usage of Cannizzaro reactions, as although intermediate tetrahedral anion (6) could in principle undergo this reaction if $\ce{R = H-}$, it doesn't if there's a better leaving group then $\ce{H-}$. The point of haloform reactions is to make them if they aren't present already.

(source of picture)


text:
Entropy as "extent of energy dispersal" vs. entropy as defined by the Clausius equation
Currently, the concept of entropy as "energy dispersal" has been proved to be more easy to understand than the concept of "disorder". While the first can be applied successfully to explain the expansion of a gas through vacuum, I don't get how the temperature influences the "energy spreading" according to the Clausius equation (although this is simple to understand with the concept of "disorder"):
$$dS = \frac{dq_{rev}}{T}$$
If the temperature is low, the energy added to the system ($dq_{rev}$) increase its entropy ($dS$) more than if the temperature is high. How this can be explained applying the "energy dispersal" interpretation?

You don't understand a concept.  Ok.  What is the question?

@Lighthart, why do you think I don't understand entropy?

I thought that is what you stated. My apologies for misconstruing you words.


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How many vibrations are expected for butatriene and cyclobutadiene?
I have butatriene and cyclobutadiene, the formula for finding the number of vibrations expected are

Linear: $3N-5$
Nonlinear: $3N-6$

where $N$ is the number of atoms.
Both of these molecules are $\ce{C4H4}$ so they both have 8 atoms (because I think you include the $\ce{H}$ atoms).
Butatriene is nonlinear so the vibrations expected would be
$$3N-5 = 3(8)-6 = 18,$$
and Cyclobutadiene is nonlinear so
$$3N-6 = 3(8)-6 = 18.$$
Is this correct? Or do you only count $C$ atoms when you do this calculation?
Would all nonlinear $\ce{C4H4}$ isomers have the same number of vibrations expected?

Butatriene is not linear - the hydrogen atoms are not on the axis formed by the four carbon atoms.

@snurden Oh your right! I didn't think about that, thank you! do you happen to know whether or not H's are included when calculating the vibrations expected?

Yes, they should be included (just as you already did) since they are just regular atoms as all the others are. (Or does water have no vibrations? ;))

Perhaps an explanation of where these mysterious equations come from will help you understand how to apply them properly.
I only need you to make one, single (correct) assumption: all atoms in a molecule are capable of moving in each of the three Cartesian coordinate directions.
That is, each of those little hydrogens can wiggle around in the x, y, and z coordinates as much as it pleases.
That means that we have a total $3N$ ways that the whole molecule can move because each of $N$ atoms can move in three directions.
Now, let's consider how molecules move. If every one of those atoms happened to be moving in the same direction, what would that look like?
Well, that would be translation of the molecule (i.e. moving). Any molecule can do this in three directions.
So, excluding translation we have $3N-3$ ways of moving around.
Now, another thing we know molecules do is rotate. By the same reasoning, we can say that a molecule can rotate in all three coordinate directions.
This leaves us with $3N-6$ ways to move around.
But, what if that molecule was linear? Well, then that molecule can only rotate in two unique directions because spinning around like a cylinder is known to be a zero energy process. Another way of saying that is that it doesn't happen.
So for a linear molecule there $3N-5$ ways of moving around still because we only subtracted out two rotational modes.
Of course, now we see that the only other way our molecules can move is by vibrating.
Thus, for a linear molecule, $$Number\ of\ normal\ modes=3N-5$$and for a nonlinear polyatomic molecule,$$Number\ of\ normal\ modes=3N-6$$
So, using these equations which now hopefully make some more sense, it's quite obvious that we do indeed include the hydrogens in our $N$ and find there are 18 normal modes of vibration for both molecules.
And, if there were an isomer of the formula $\ce{C4H4}$ which was linear then there would be 19 normal modes, but there is no such molecule so, yes all molecules with the molecular formula $\ce{C4H4}$ have 18 vibrational modes.


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What chemicals are mixed together to cause chemoluminescence?
My brother is supposed to do a science experiment at school for an open house. We found an interesting experiment in which you pour several chemicals together, causing the resulting concoction to glow brightly. You can make several colors in this way. Then you can dump them together, to demonstrate how white light is made up of several colors mixed together. Several questions.
1. Which chemicals are used?
2. How do you get the colors?
3. Where can you buy the chemicals you need?

"Where can you buy the chemicals you need?" Looks off-topic

I second what @Mithoron said. Also, as written, it's pretty broad. If you can narrow your query - what colors? what procedure(s) are you interested in investigating? - that will help us help you.

related http://chemistry.stackexchange.com/questions/35527/how-do-things-glow-in-the-dark https://chemistry.stackexchange.com/questions/14241/is-there-anything-funny-i-can-do-with-chemicals-in-glowsticks

If you want different colours, you might want to give the peroxyoxalate reaction, known from glow sticks, a try.
Typically, you will need three components:

a diaryl oxalate
hydrogen peroxide (to oxidize the diaryl oxalate, probably to a 1,2-dioxetanedione)
a fluorescent dye, which determines the final colour.

Wilhelm J. Baader has examined the reaction in great detail, and you'll find a lot of references on his website.


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How big is the influence of basis set and functional on ZPE and thermal corrections?
Most quantum chemistry packages are capable of computing ZPE and thermal corrections for a given temperature. But how big is the influence of the basis set chosen and the functional used?
Is it sufficient to calculate this at a lower level than the corresponding final high-level single point calculation? What about differences between the energies - if there is an error, is it systematic so that energy differences are not affected?

The short answer is without benchmarking you will not know. The slightly longer answer is, that these corrections only make sense for well converged optimised structures. It is not too uncommon, to use these corrective terms and add it to a single point calculation of higher level. Computing frequencies at the SP level should give you (almost) the same corrections and is a waste of resources.

I agree on Martin's comment - that's what I usually read in (benchmark) papers.

I agree with all the comments, but I want to add: Please also consider what kind of method you are using. A DFT calculation on a transition metal complexes has error bars where I wouldn't even consider ZPE corrections at all without a very good reason. A small molecule where you want the heat of formation as accurately as possible? Do whatever you can afford. I am surprised that you did not mention a method or system to be honest.

This is more of a comment, apologies.
Consider this paper (Ruth L. Jacobsen, Russell D. Johnson III, Karl K. Irikura, Raghu N. Kacker: Anharmonic Vibrational Frequency Calculations Are Not Worthwhile for Small Basis Sets, J. Chem. Theory Comput., 2013, 9 (2), pp 951-954). The magnitude of the computational-level-dependent scaling factors for the frequencies reported therein should give you a sense of the magnitude (I would also think that at least some other studies of the topic are cited).


text:
Can dry ice explode?
Can dry ice be used as a explosive chemical? I am intrigued by this chemical dry ice and seen various experiments with it, and seems to have potential to be a   explosive.
So can dry ice explode?

I've seen (no reference at hand) dry ice in a coca-cola plastic bottle explode. But there's a difference between deflagration and explosion. The former one is subsonic, the later one supersonic. Dry ice doesn't come anywhere near supersonic.

And beside, it wouldn't be a chemical reaction.

I've done this myself. The bottle gets ripped open with great force (DON'T TRY IT AT HOME!), but as @GyroGearloose correctly noted, this is certainly not a chemical explosion.

@GyroGearloose - is sublimation, a phase transition, not chemical? I guess one could argue that it is physics, but the thermodynamics of phase transitions would seem to straddle the two disciplines.

I toned this back a bit as to avoid any appearance of malfeasance :)

@JonCuster Gyro is right, a phase transition is considered a physical change.

Jo wrote explosive chemical not chemical explosion- that is different. Jo is not trying that the nature of the reaction is chemical.  Chemical reactions are only a little semantically different from physical reactions which have bonding changes just different from "chemical" reactions.

Chemicals used as explosives are those which, by some reaction/state change, can produce a large pressure (and thus usually volume) change very quickly.
While dry ice (solid carbon dioxide) is much more dense than in its gaseous phase, at room temperature the change happens very slowly, and is just unstable rather than stable but reactive, so can't be left and triggered later.
This makes it pretty useless as a working explosive.
Since it can exert a significant pressure while it sublimates, however, it can make other things explosively fail -- like the dry ice in the soda bottle demonstration. A quick search suggests that soda bottles are rated to withstand a pressure difference of around 100 psi (~6.8 atm, 0.69 MPa), and the increasing pressure won't stop the carbon dioxide subliming/boiling (since at that point it'll be a liquid) until ten times that.
Now, there is a market for things that slowly expand with a lot of force ("non-explosive demolition agents"), but they're solids that you mix into a slurry and pour into a cavity, since these can trivially exert orders of magnitude more pressure -- a common brand-name one, Dexpan, is rated at 18000 psi -- and they also turn from a slurry or solid into another solid. If you tried this operation with something that turned into a gas, you'd need to seal the cavity to stop the gas just escaping, whereas with solid-to-solid reactions, gravity and the solid itself will do the job for you.

Actually, you are close.  While technically not an explosive (and really, it's liquid CO2 vs solid CO2), charges containing liquid CO2 that are heated can be used as blasting agents for mining and tunneling.  They make up for relatively low shock/energy output by being very safe and also breaking material via tension instead of compression.  A well-known example is Cardox, which is used in coal mining.  Or see this.


text:
Is 98%+ purity Reagent Grade Ca-EDTA powder as safe to use as USP Grade in the body?
Is $98\%+$ purity reagent grade $\ce{Ca}$-EDTA powder as safe to use as USP grade in the body?  Or, does the reagent grade have unsafe impurities that USP grade would not have even though they are the same purity?

Definitely not as safe. There's a much higher chance of lead and other common mineral impurities.

I wouldn't say it safe or unsafe but rather that it is risky.  The reason you shouldn't use reagent grade materials for internal use is because Reagent grade is based on purity alone.  USP grade is based somewhat on purity, but is more concerned with a lack of toxic impurities. your Calcium EDTA at 98% can be reagent grade but the balance 2% can be magnesium, sodium, nickel (though not likely), etc...

While normally health-related questions such as this one are considered off-topic, this one is sufficiently cut-and-dried that it probably merits an answer.
NO!!!! In general, reagent-grade chemicals are NOT safe to use internally like USP grades.  The entire point of the USP specifications is to ensure, to the extent practical, that the relevant chemicals are prepared from suitable starting materials and using suitable methods such that toxic or hazardous materials are absent from the final product. This is a big reason why the USP grades are often more expensive than the lab grades.
It's certainly possible that a given reagent-grade chemical may actually be safe to ingest, but there is no real way to know for sure without conducting extensive analysis of the impurities.

I'd agree that reagent grade shouldn't be used for ingestion. However I'd disagree about relative price. I think which is more expensive would depend on the chemical. I can buy table salt at the grocery much cheaper than reagent grade sodium chloride.

@MaxW Table salt isn't USP grade, it's food grade. (At least, I'm pretty sure there's a difference there.)

@MaxW $\ce{NaCl}$ isn't the best example... take instead BHT, the preservative. Sigma has food grade listed at US $274 per 10kg and USP grade listed at US $326.50 per 500mg.

@MaxW I'm pretty sure the key difference is that food grades are only approved for ingestion, while USP grades are approved for ingestion or inclusion in pharmaceutical formulations.


text:
Comparison of enol content

How can we find the order of percentage of enol form
of the compounds in the following equilibria?

Percentage of enol form in x is highest as due to aromaticity. But how can we compare that for y and z. The compound z is non-aromatic but compound y has resonance.

You may also think of hydrogen bonding here. See that in case of y, intramolecular hydrogen bonding is possible and hence greater percentage of enol content is of y than z. You may note that this is a general factor of esters (in this case a cyclic ester) in determining enol content percentage and that this factor may not always be the 'winning' one (as you can compare between x and y).

Do you mean intermolecular hydrogen bonding?

I may be wrong, but isn't the O serving to increase electron density in the C=O π* orbital, and thereby decreasing chances of enol formation?

The ring formed due to intramolecular H bonding wd be 4 membered thus destabilizing the enol.

2nd compound seems to have more stable keto as well as more stable enol than the third compound due to conjugation. Either I am missing something or it wd be hard to compare without experimental data.

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text:
Does (chloromethoxy)ethane react in a bimolecular or monomolecular mechanism?
It is known that (chloromethoxy)ethane forms stable carbonium ions based on Nobel prize Chemist George Olah research Olah's method. This would indicate that (chloromethoxy)ethane reacts through $\mathrm{S_N1}$ mechanism. However, $\mathrm{S_N2}$ mechanism is also very feasible for highly electrophile carbon.

Which one of the mechanisms is dominant?

Organic chemistry books in this case are not very clear. For example Clayden speaks about Olah's method and neighbouring group participation (pp. 419, 970 1st ed) but also gives relative rate for $\mathrm{S_N2}$ mechanism (pp. 425 1st ed)

Probably both mechanisms can happen and it would require experimental data from reaction kinetics and computational calculations to determine which one of the mechanism is domintant.
One example of determined mechanism about neighboring group effect, also called anchimeric assistance is mustard gas (chloromethine) alkylation reaction with DNA.

Effect of chloromethine is based on its ability to block DNA replication by alkylating guanine, and after the war it has been also used as cancer drug.


text:
Can an acid or base have a pH less than 0 or more than 14?
This question has been asked earlier, but I could not understand it.
Please, someone explain it in an easier way.

Also http://chemistry.stackexchange.com/questions/25155/can-a-negative-ph-exist and http://chemistry.stackexchange.com/questions/6997/ph-range-outside-convetional-0-14

It was asked many times, you should read more answers and if you still don't get it, tell what's your problem otherwise I doubt you could get answer useful for you.


text:
Conformation of six-membered ring constituting a bridged compound
What is the conformation of the cyclohexane ring constituting [2.2.1]-bicycloheptane?
Is it boat or twisted boat? Is that conformation abundant in other bicyclic systems containing a cyclohexane ring?

EDIT: From Modern Organic Synthesis by Zweifel I took the following image:

which is in contradiction with current answers. Can anyone account for this observation?

Please be so kind and mark the edits in the question, so it is obvious how our argumentation builds up.

@ssavec No, definitely not. Please never mark any edits whatsoever, that is what the edit history is for. A question should always stand as is, not as a chronological build up. The extension that is now introduced does hardly make any sense to me, as I don't know what are the current answers, and I don't know how it even relates to the first question.

@Martin-マーチン You got the point. The added question does not relate to the original one at all. By substantially modyfing the question, OP rendered the existing answer (mine) look very incomplete. I don't think the question should become more complicated after it was answered.

The question was: Conformation of six-membered ring constituting a bridged compound. I provided [2.2.1]-bicycloheptane as an example with the follow-up questions: Is it boat or twisted boat? Is that conformation abundant in OTHER bicyclic systems containing a cyclohexane ring?" I've got the answer for [2.2.1]-bicycloheptane, it is boat, but what about other systems. I provided an example where boat isn't the most stable conformation. The correct conformation for  just [2.2.1]-bicycloheptane doesn't answer my question.

@ssavec I agree with that. Once an answer is given, it is better to ask a new question and provide a link for context.

The question: "Is that conformation abundant in OTHER bicyclic systems containing a cyclohexane ring?" wasn't added later and when I first asked the question there wasn't any problem with it. Xumuk gave an answer that all systems adopt a boat conformation and to disprove it I edited the question. A complete answer is still missing.

According to structure determined by electron diffraction, norbornane is present in bridged boat conformation. http://pubs.acs.org/doi/abs/10.1021/ja01014a032, see Figure.

Norbornane and its derivatives are often used for studying effect of equatorial and axial substitution of cyclohexane and is considered rigid.
Therefore, I don't expect significant population of the second conformer you sketched.
Edit:
For the [2.2.2]bicyclooctane (not the lactone mentioned in question), both conformers are known in crystalographic database.
Bear in mind, that bicyclooctane constist of three hypothetical cyclohexanes, possibly equivalent by symmetry.
For any compound of interest, you should find enough data in crystallographic database, but I'd expect all reasonable conformations to exist.

That's not the full answer. Is that conformation abundant in other bicyclic systems containing a cyclohexane ring like bicyclo[2.2.2]octane?

Which one is lower in energy?

I need solid evidence because there are contradictory answers here.

@Marko If you need solid evidence, than find it yourself.

I need for one of these answers so I can accept it. Geoff said that the most stable conformation for bicyclo[2.2.2]octane is twist-boat, XuMuk said that it is boat and you didn't give a definite answer on that matter.

Norbornane, as mentioned by the other answers forces the cyclohexane ring into a prototypical "boat" conformation. The bridging carbon creates two five-membered rings in the low-energy envelope or "pucker" conformation:

Put simply, the 1-carbon bridging system creates a very rigid system. While there are many eclipsed C-H interactions, changing conformation would require stretching bonds or angles - distortions that are much higher in energy.
In larger bicycloalkanes, like bicyclo[2.2.2]octane, there is now more conformational flexibility to adopt a twist-boat conformation:

Notice that from the bridgehead carbons, there are many eclipsing interactions. In Avogadro, I can watch this all-boat conformation twist into the twist-boat:

While it's important to do conformational analysis for every molecule (small interactions can change the low-energy conformer), I would guess most of the larger bicycloalkanes to adopt twist-boat configurations.

So which conformation is the more stable one for bicyclo[2.2.2]octane?

It's twist-boat - you can see the twist in the final image (of bicycle[2.2.2]octane).

Relative energy of hydrocarbon conformers is reasonably tricky question, involving among others long range weak interactions, dynamic effects and symmetry. That's why I (as computational chemist) tried to avoid answer based on calculations. BTW, how large is the energy difference of those?

Your molecule is "bicyclo[2.2.1]heptane", also called "norbornane".
And it looks like:

Coordinates:
C    -0.336900000000      0.382800000000      0.560000000000
C     1.186600000000      0.194300000000      0.678100000000
C     1.458300000000     -0.983300000000     -0.277900000000
C     0.054500000000     -1.313600000000     -0.817600000000
C    -0.397400000000     -0.196100000000     -1.775400000000
C    -0.669200000000      0.981300000000     -0.819300000000
C    -0.819800000000     -1.054000000000      0.403500000000
H    -0.777400000000      0.930600000000      1.394100000000
H     1.468700000000     -0.067200000000      1.698900000000
H     1.738100000000      1.093400000000      0.400200000000
H     1.874300000000     -1.831700000000      0.267200000000
H     2.157900000000     -0.722600000000     -1.072900000000
H    -0.028300000000     -2.315100000000     -1.241800000000
H    -1.309100000000     -0.480900000000     -2.302600000000
H     0.360200000000      0.039800000000     -2.523700000000
H    -1.716400000000      1.283000000000     -0.870400000000
H    -0.059000000000      1.855400000000     -1.049900000000
H    -1.889000000000     -1.133000000000      0.198000000000
H    -0.570700000000     -1.688700000000      1.255400000000

Could you draw the 3d of twisted boat? (it definitely is not the minimal energy conformer)
I can try to calculate the energy of the conformers by quantum-chemistry, but there is a problem - minimisation leads to the boat conformation.
This is MO of norbornane:

This is 3d structure of 2-Oxabicyclo(2.2.2)octan-3-one ((1s,4s)-2-oxabicyclo[2.2.2]octan-3-one):

and xyz coordinates:
C     0.863100000000      0.683400000000      0.628700000000
C     1.100200000000     -0.847300000000      0.586700000000
C    -0.188300000000     -1.567200000000      0.104800000000
C    -1.274300000000     -0.508200000000     -0.211400000000
C    -0.760200000000      0.456300000000     -1.309800000000
C     0.533400000000      1.154500000000     -0.810100000000
C    -0.391200000000      0.927800000000      1.487000000000
O    -0.385000000000      1.631600000000      2.492800000000
O    -1.472000000000      0.267900000000      0.979400000000
H     1.721200000000      1.214600000000      1.041700000000
H     1.383200000000     -1.208500000000      1.576700000000
H     1.934600000000     -1.081400000000     -0.074900000000
H    -0.549400000000     -2.249900000000      0.875300000000
H     0.019000000000     -2.174800000000     -0.776900000000
H    -2.210400000000     -0.981200000000     -0.509400000000
H    -1.525500000000      1.199400000000     -1.538800000000
H    -0.568800000000     -0.088500000000     -2.235000000000
H     0.404800000000      2.237900000000     -0.827600000000
H     1.365600000000      0.933900000000     -1.479200000000

I don't know what authors call for "twist boat", but for compounds, you are asking about, there is definitely only one conformer!
For visualisation these geometries you can use a lot of software - chemcraft, VMD, avogadro, pymol...

That's not the full answer. Is that conformation abundant in other bicyclic systems containing a cyclohexane ring like bicyclo[2.2.2]octane?

@Marko This conformation is very solid, as said ssavec - population of other conformations is almost zero.

I will edit my question to prove my point.

@Marko It is on page 35, before 2.2. "The lactone moeity acts as a stereochemical bridge, maintaining the hydroxyl and
carboxylic acid groups in a cis relationship.  I think there is little mistake in what to call "twist boat".

It is drawn so, but it could be a mistake.

@Marko Please, try to draw it in 3d or make in any chemical constructor. You will see, twisted almost impossible.

Here: http://www.bluffton.edu/~bergerd/models/newman3.html I don't see what it would be so improbable.

@Marko your link is totally irrelevant for the lactone issue.

@Marko I completely agree with you - it is possible for simple C6H12 (Cyclohexane). But I insist in building the model of bicyclo[2.2.2]octane or 2-Oxabicyclo(2.2.2)octan-3-one! Simple ring - another situation.

I've put the link because one can easily compare the boat and twist-boat conformations of simple cyclohexane and to prove that there shouldn't be that much difference as you suggested.

@Marko What kind of proof do you need? Citation from another book will be enough?

Book, publication, calculation...

@Marko Ok http://www.sciencedirect.com/science/article/pii/002228609185011Q (X-Ray study of hetero ring flexibility in norbornane, norbornene-fused 1,3-oxazin-2-thiones Structure of 5,8-methano-r-4-phenyl-c-4a, c-5,6,7,c-8,c- 8a-hexahydro-4H- 1 ,3-benzoxazin-2 (3H)-thione) there are x-ray data: https://summary.ccdc.cam.ac.uk/structure-summary?sid=Elsevier&id=doi:10.1016/0022-2860%2891%2985011-Q There is a very little twist in cycle. But better have a look at: A statistical approach to the geometry of norbornane from X-ray data of substituted derivatives (10.1107/S0108768185002117)

Okay, norbornane has the boat conformation for the 6-membered ring, but bicyclo[2.2.2]octane is the disputable one. Any data about it's conformation: boat or twist-boat?

@Marko, whait, 2-Oxabicyclo(2.2.2)octan-3-one or bicyclo[2.2.2]octane?! Off-course the bicyclo[2.2.2]octane can be in twist-boat. But you in question asking about 2-Oxabicyclo(2.2.2)octan-3-one.

I am asking in general for bridged bicyclic systems. I've just found the lactone example in a textbook to prove that not all bridged bicyclic systems with 6-membered rings must adopt a boat conformation.

what software you people use to draw these molecules ?

@Lucian09474 I have used two softwares: chemcraft, it can be obtained very cheap for academic, also, there is in the Internet very old version, but without any limitations (chemcraft freeware now is very limited, but for fast and smooth pictures is on of the best and easiest). MO were calculated by Jaguar - programme of the Schrodinger -. It costs not a little money. Also, there are a lot of other, not so bad drawing/rendering software: VMD, vega zz, Mercury...


text:
Energy profile of cyclohexene conformations
The energy profile for cyclohexane flipping looks like this:

How does the corresponding diagram for cyclohexene look like?

I found the answer in the book: "Practical Aspects of Computational Chemistry"
. In chapter 19: Unusual Properties of Usual Molecules. Conformational Analysis of Cyclohexene, Its Derivatives and Heterocyclic Analogues.
The value for ring inversion of cyclohexene is smaller than the one of cyclohexane (which is near 10 kcal/mol):

Value of ring inversion barrier is within 4.2–
  7.9 kcal/mol for molecular mechanics and 5.2–6.6 kcal/mol for quantum chemical
  methods.

The inversion proceeds as follows:

The ring inversion process in cyclohexene includes three stage. The first
  stage involves the transition from an equilibrium half-chair conformation to a twist-boat conformation with a significant increase in energy. The second stage
  may be described as a very easy transformation of one twist-boat conformation to
  another via boat almost without any changes in energy. The third stage includes the
  transition from a twist-boat conformation to a half-chair conformation accompanied
  by a significant decrease in energy. The difference in energy between the twist-boat and the boat
  conformation is less than 0.01 kcal/mol. The boat conformation
  is just central point of this plateau and it is only the most convenient point for a description of the saddle point for the ring inversion
  .

The interconversion diagram should be something like this:


text:
Reaction of sodium with terminal alkyne
Can sodium (just sodium, not with liquid ammonia) react with terminal alkynes in a simple acid-base reaction to form sodium alkynide and dihydrogen (just as it reacts with $\ce{H2O}$ to form sodium hydroxide and $\ce{H2}$)?

"University Chemistry, Volume 2" by C. Parameshwara Murthy, p 164 says it can. No conditions are listed, though.

@DGS- but that is in the presence of liq NH3 ,which would react with sodium to form NaNH2. The terminal alkyne then reacts with NaNH2 to liberate H2. I want to know if sodium can directly react with the alkyne

I guess you may find your proof sooner or later, but you should understand that as sodium is solid and ethyn gaseous, there's not much point in reacting them like this.

Take a liquid alkyne. Also since they are mildly acidic, the metal must get solvated

The one I saw didn't mention ammonia, which is kind of crucial to that reaction if that is the mechanism. I'm assuming it wasn't there.

Sry, you were right @DGS. There isn't ammonia. I saw another reaction on page 161 and thought you had meant that. Thanks, my question is answered


text:
Migratory aptitude in pinacol pinacolon rearrangement
Out of ortho-methoxyphenyl and para-methoxyphenyl which one has more migratory aptitude in pinacol-pinacolon rearrangement?

Groups with higher electron density have a greater tendency to migrate. So, o-methoxy phenyl should be the answer but due to steric factors p-methoxy phenyl can always be an equally likely answer.  It can only be found out accurately by experimentation and considering other reaction conditions.


text:
How can semi-permeable membrane be so smart?
According to the definition, a semi-permeable membrane is a membrane that allows only solvent molecules to pass through it; solute molecules are blocked.  Solvent molecules flow from the membrane side that has low solute concentration (i.e. high solvent concentration) to high solute concentration (i.e. low solvent concentration).
Lets take a case:

I take oil as a solvent in one side and water as a solvent in other side. I take the same solute in both sides.
Initially, the water side is filled to height $h$ and the oil side to height $2h$. The amount of solute is same in both the sides. So the solute concentration is more in water than in the oil.  Accordingly, oil will flow from the oil side to water side. Now, I add more water on the water side to reduce the solute concentration on the oil side.  Accordingly, water should now flow towards the oil side.

So what will happen? Will this happen? What makes the semi-permeable membrane so smart or intelligent?

Water and oil? Do you understand how solubilities work?

@Mithoron I know water and oil are immiscible.. If that was what you meant... I was just trying to give an example .. Please correct or edit my question if you can to explain it better :)

A semi-permeable membrane is not ‘smart’. If you wish, it is actually rather dumb. All it can do is perform a simple selection of ‘I’m letting you through’ versus ‘I’m not letting you through.’ Most membranes will work along the lines of size, polarisability or both.
Size is easily explained: The membrane’s pores have a certain size and what is larger than that cannot diffuse through. Most classical solvents are rather small molecules — water molecules are very small. So it is easy to conceive a membrane that can let water through but has pores too small for large biomolecules or maybe even ions. Remember that dissolved ions are usually surrounded by a solvation sphere, so their size is substantially larger that it may look.
Polarisability is also a rather simple concept but it requires a certain structure of the membrane. The membrane could, for example, consist of a very hydrophobic surface. In that case, ions would likely not want to go near it. It’s a little bit harder to explain, but it may also allow the solvent molecules to go through while the solutes cannot.
Of course, as I said, the membrane is dumb and cannot distinguish solvent from solute. Say you had one that can let water through but not glycerin; maybe due to glycerin’s size. And say you have a $1~\%$ solution of water in glycerin, so $1~\mathrm{g}$ water in $99~\mathrm{g}$ glycerin. The latter is clearly the solvent. However, the membrane would still only let water pass since it is ‘dumb’.
‘Smart’ and ‘dumb’ are bad words altogether, though, since this all is based on physical properties which have nothing to do with smartness (even though smart is a rather frequent keyword in many scientific papers).

Most of the time what you say about semi-permeable membranes is correct.

a semi-permeable membrane is a membrane that allows only solvent molecules to pass between the membrane; solute molecules are blocked. Solvent molecules flow from the membrane side that has from low solute concentration (i.e. high solvent concentration) to high solute concentration (i.e. low solvent concentration).

However, you have to keep in mind several assumptions that are inherent in the definition as you have given it.  The most important of these is that thermodynamic activity differences in the membrane-permeable solvents determine the membrane behavior.  When phases are immiscible, such as the water/oil system in your example, it is essential to think in terms of activities, not in terms of concentration.

I take oil as a solvent in one side and water as a solvent in other side. I take the same solute in both sides.

So we are imagining a solute that can be dissolved by both water and by oil.  What do we know about the solubility of the solute in these two solvents?  What is the partition coefficient?  And what can we say about the thermodynamic activity difference across the membrane for the three species?  Let's go step by step:

Water.  Water is not soluble in oil to an appreciable degree.  In equilibrated phases, the thermodynamic activity of each chemical species is the same in each phase.  Therefore in an oil-water mixture, the thermodynamic activity of water in the aqueous phase is the same as the thermodynamic activity of water in the oil phase. The extremely low solubility of water in oil means that if there were to be a difference in thermodynamic activities across the membrane, only a few molecules would need to cross over for the condition of equal water activities to be satisfied.
Oil.  Oil is not soluble in water to any appreciable agree.  Therefore the same argument applies.
Solute.  The solute could have a different concentration on each side of the membrane, but because we don't know the partition coefficient, we don't know what the equilibrium state is.  Suppose the solute were membrane permeable.  Even if it was twice as concentrated in water than in oil, if it's partition coefficient favored an equilibrium where the oil-to-water concentration ratio was 10, then it would flow from the water to the oil.  But, in your example the solute is not permeable to the membrane.

Initially, the water side is filled to height $h$ and the oil side to height $2h$. The amount of solute is same in both the sides. So the solute concentration is more in water than in the oil.

Now hopefully the discussion above makes you realize that there aren't many chemiosmotic forces in your example.  The only driving force left is hydrostatic pressure.  So the oil will cross the membrane, not the water, just because of hydrostatic pressure differences.  This is regardless of the solute concentration.

Now, I add more water on the water side to reduce the solute concentration on the oil side. Accordingly, water should now flow towards the oil side.

Well adding more water raises the height $h$ to some higher height and thus increases the hydrostatic pressure.  So the increased pressure forces water back through the membrane.

Semi-permeable membranes such as cell membranes use active transport to move things AGAINST a concentration gradient (Or through a membrane that would otherwise be impermeable), using energy to do so. You are right, for many materials, things simply balance out like your example. However, in active transport, only some materials are taken in.
Here's an article about active transport: https://en.wikipedia.org/wiki/Active_transport


text:
What is intramolecular energy?
I got this term while reading a portion of specific heat capacity in Ideal gas chapter. What does it mean? Does it mean internal energy?

yes I think that it must do. Inter-molecular is between molecules, intra- within a molecule. It is usually used in the context of intra-molecular energy transfer, such as occurs after absorbing energy from a photon.


text:
High temperature bath chemicals
I have an electric mantle that I want to bathe reaction vessels at temperatures up to $\mathrm{750^{\circ}C}$.  Obviously that rules out oils.  What substances are used as bath media at such temperatures?
I was thinking of using lead or Rose's alloy, but wondering if there are more standard high-temperature bath chemicals?  I suppose an ideal one would be both inert to lab steel, glass, and a standard atmosphere, and liquid down to room temperature.

What about sand?

@ToddMinehardt You did beat me by seconds :D. Sand is an excellent option, both in terms of good heat transfer and lab safety.

@KlausWarzecha and Todd - Good suggestion.  The reason I was thinking of liquids is that the electric mantle doesn't apply heat very evenly, but natural convection in a liquid would maintain even temperature.

Gallium may be nice

@Mithoron - good one.  Reminded me of the highest range between melting and boiling point question.  Galinstan was also mentioned.  But sounds like gallium would likely attack the electric mantle.

@feetwet And gallium would probably also steadily form oxides at any air-exposed surface; not so good for long experiments.

Different types of solder can give a very high operating temperature, are cheap and offer excellent conduction and heat contact when molten. We have used lead and cadmium free solder to 500 degrees C with no problem. Your issue may be the melting point (ours was 230C), but the right mix may bring this down for you. If you have the time and inclination, you could modify a commercial mix yourself. Otherwise,  sand would get my vote.


text:
Molar heat capacity and heat capacity
Is molar heat capacity and heat capacity

at constant volume both represented by $C_v$?
at constant pressure both represented by $C_p$?

Yes.  Unfortunately, we use usually use the same symbol for both.  The context of the problem and solution determines which one is being used.

The symbols, which are standardized in ISO/IEC 80000 Quantities and units (14 parts) and recommended by IUPAC in Quantities, Units and Symbols in Physical Chemistry (Green Book), distinguish the various related quantities:
$$\small
\begin{array}{llll}
\hline
\text{Quantity} & \text{Quantity} & \text{Unit} & \text{Unit} \\
\text{name} & \text{symbol} & \text{name} & \text{symbol} \\
\hline
\text{heat capacity} & C & \text{joule per kelvin} & \mathrm{J/K} \\
\text{heat capacity at constant pressure} & C_p & \text{joule per kelvin} & \mathrm{J/K} \\
\text{heat capacity at constant volume} & C_V & \text{joule per kelvin} & \mathrm{J/K} \\
\text{specific heat capacity} & c & \text{joule per kilogram kelvin} & \mathrm{J/( kg \cdot K )} \\
\text{specific heat capacity at constant pressure} & c_p & \text{joule per kilogram kelvin} & \mathrm{J/( kg \cdot K )} \\
\text{specific heat capacity at constant volume} & c_V & \text{joule per kilogram kelvin} & \mathrm{J/( kg \cdot K )} \\
\text{molar heat capacity} & C_\mathrm m & \text{joule per mole kelvin} & \mathrm{J/( mol \cdot K )} \\
\text{molar heat capacity at constant pressure} & C_{\mathrm m, p} & \text{joule per mole kelvin} & \mathrm{J/( mol \cdot K )} \\
\text{molar heat capacity at constant volume} & C_{\mathrm m, V} & \text{joule per mole kelvin} & \mathrm{J/( mol \cdot K )} \\
\hline
\end{array}$$
However, in contrast to symbols for units, symbols for quantities are only recommendations. Unfortunately, quantity symbols are used inconsistently in the literature, sometimes in a sloppy way. In principle, a given symbol can indicate different quantities.


text:
Perkin reaction for aliphatic aldehydes
Why isn't Perkin reaction possible with aliphatic aldehydes? Since the carbon atom of carbonyl group of aliphatic aldehydes is more electrophilic than that of aromatic ones, the reaction must be more feasible in case of aliphatic compounds. Please explain.

Which proton is most likely to be abstracted when heating acetic anhydride and an enolisable aliphatic andehyde with a base?

What if there are no alpha hydogens?

In Perkins reaction, condensation occurs between aromatic aldehydes (which cannot undergo self condensation) with an acid anhydirde in presence of sodium or potassium salt of corresponding acid.

Consider Perkins reaction of benzaldehye with acetic anhydride to form cinnamic acid. The first two steps occur in weakly basic medium (sodium acetate). The last two occur in acidic medium (protonation and subsequent dehydration). Notice that the last step is particularly favoured owing to formation of a conjugated product.
This wouldn't be the case for aliphatic aldehydes (with no alpha Hydrogen , else self condensation may occur). Also, if the anhydride used has only one alpha Hydrogen (here acetic anhydride has 3 alpha Hydrogen), then the reaction barely proceeds to completion.Instead an aldol type product is obtained. This shows the role conjugation plays in making the last step fairly irreversible.
Furthermore, this reaction occurs with vinylogs, hetrocyclic aldehydes and even pthalic anhydrides. Notice that in all cases , final product is conjugated.
It would be worthwhile to mention another alternative mechanism, which has been mentioned in Peter Stykes, Wikipedia as well.

Some Support for this mechanism comes from fact that anhydrides with only 1 alpha hydrogen donot give this reaction. Again, here the base promoted dehydration step would be largely irreversible incase of aromatic aldehydes due to conjugation.
Hope this helps.
References:

Perkin, W. H. J. Chem. Soc. 1868, 21, 181
XXIII.—On the hydride of aceto-salicyl.DOI: 10.1039/js8682100181
Peter Stykes. Organic mechanism. Sixth Edition

3.https://en.wikipedia.org/wiki/Perkin_reaction

SN Sanyal: Reactions,Rearrangements and Reagents.

5.http://www.name-reaction.com/perkin-reaction

-So from what I understand, the main reason is that most of the steps are reversible, and the final step (in which the product has a double bond conjugated with the phenyl ring, which is not possible in case of aliphatic compounds without alpha hydrogens) is highly favorable and shifts the overall reaction to completion. Am I right?
By the way I can't help noticing,  you seem like an IITJEE guy

Yes, you got it right. I was just trying to validate this for other (possible) mechanisms. The second mechanism in that sense would be more probable... although I should research more on that line. And to your second question: yes.


text:
Lowering Freezing Point of Amines
What is most efficient method of lowering freezing point of amines without greatly effecting properties? Method must involve incorporating an additive at small percentage.

Protonate the amine?

Hmm... That's backward, I guess.

Add another amine. It is better to choose an amine with just few more carbons.

Can I use a plastisizer or even something like salt similar to reducing freezing point in water.

May be there is some additive that can help inihibit mechanism that promotes freezing?

If another amine is added then how much more needs to be added to inhibit the freezing? If its too high an amount then properties could be effected. Essentially I have a problem that this amine crystallises or freezes below a certain temperature and I then have to warm which is time consuming.

What about adding dimethylformamide?

Sounds like you're performing a reaction or preparation neat. Without knowing anything else, I would think about the more inert solvents, such as an ether, MTBE? As little as 1-4% (sometimes less) of an additive can affect melting points. e.g. I have used 4% NMP in sulfolane to keep the latter at liquid at 20 degrees C. Just experiment!


text:
Partial Coverage During Zinc Plating Bath
I am experiencing partial coverage of a part during a DIY home zinc plating operation and was hoping someone here may be able to identify the cause of this.
I am using a solution of household vinegar and epsom salt to zinc plate some pieces during a motorcycle restoration but I am having difficulty getting the zinc to deposit on an area of a specific part.
On accompanying picture of a footpeg mount it is evident where the plating has not adhered.
I have done multiple runs and cleaned the part thoroughly between each run and even oriented it in different directions to the zinc anode and even connected my cathode connection point directly to the area not accepting any zinc all to no avail and was wondering if anyone has any ideas on what may be happening here and how I may resolve the issue. (My guess is there is an electrical or magnet field being generated by the "sides" of the part and preventing plating in this area as all other areas not visible have zinc adhered from the first run)

have you used any stirring?

Sorry I didn't mention that. Yes, I have stirred the solution before and during use and the part has been fully submerged in the solution and during I also use an aquarium bubbler to keep the solution moving. Although I have also tried it without the bubbler after noticing this problem and have even reoriented the mount (pictured) in different orientations to  the anode still to no avail

If you look close at the image, it will be evident that not only the surface facing you is not plated, the inner sides are not plated either and it appears the absence of plating follows the contour of the sides which lead me to believe there is possibly an electrical issue (magnetism or other) between the sides preventing the plating of this area. (just a guess)

Ions of zinc move following electrical gradient. Between the inner sides the gradien is very low if exists at all. I'd guess that either very intensive stirring or moving the anode close into the inner space should work. Otherwise, if the part is not very large, you can try to put it into melted zinc. Its mp is rather low and zinc can be easily smelted using propane torch or even common wood flame. Hell, I even melted it once using common alcohol flame.

@permeakra thanks for commenting. I'd like to extend our conversation by adding that this piece is relativetly small (would fit in  the palm of your hand) and I have positioned it directly in front of the anode and it plated the edges facing the  anode (pictured) as well as the back side which is against the paper in the photo but I can't get it to adhere to this inner area and the sharp line where the plating ends makes me believe something else is at work

As Epsom salt is MgSO4 . 7 H2O, I wonder how is done zinc electroplating without any zinc in electrolyte. Unless it uses a zinc anode and waits until some zinc dissolves.

It is hard to imagine this would be due to an electric field or a magnetic field.
I would say this has to be a different material or coating in the region where you are having bad coverage.
Would it be possible to sandpaper this piece to get rid of any potential coating? Even leftover glue from some sticky on top would cause an issue.
You could also try to electrochemically clean the surface immediately prior to the plating. Applying a highly cathodic potential should lead to some electrolysis that would evolve hydrogen gas which would knock any impurities off the surface.
Either method could roughen the surface somewhat, I don't know enough about motorcycles to say if this would be a dealbreaker.

I thank you immensely for your contribution. I'd like to add that I have wondered if there was a difference in materials at play but the piece appears to be made of the same material throughout. Regarding sandpapering the piece; I have done similar by aggressively running it into a spinning wire wheel and it appears to be bare metal across it's entire surface which I've then cleaned multiple times with Kerosene - > Brake Clean -> MEK and then Distilled Water before submerging in the solution and after plating the unplated area even develops light rust as can be seen in picture

I do like your suggestion regarding electrochemically cleaning the surfaces before plating to increase their cathodic potential and I may visit this in the near future. Thanks

I think your problem is called "throwing power" in the industry. That is a reason professional platers have chemists to modify bath chemistry . And maybe electrode design.

Would you care to elaborate? An answer of this length is pretty much just a comment.

I was was trying to identify or name the cause. I have zinc plated test coupons but have no technology background in plating.

Throwing power is how far from an anode you can plate a cathode uniformly. https://www.corrosionpedia.com/definition/1090/throwing-power or, "Why would the zinc cations prefer to bypass the nearby metal to go deeply into the hole?" Perhaps you could put the zinc anode in the space between the sides. But then the outside won't get plated... Maybe do the plating in two sessions; perfect uniformity is not as important as complete coverage. +1 on the answer!


text:
Do IUPAC names for haloalkanes require numerals?
I'm new one of chemistry. I want to know the IUPAC nomenclature rules. That is if we write alkane with halogen must be indicate the numerals? Are the following names right or wrong?

chloropentane
trichloropentane

Yes, you must use the numerals to avoid ambiguity.  1-chloropentane is a different molecule with different properties than 3-chloropentane.  Similarly 2,2,3-trichloropentane is a different molecule with different properties than 1,1,1-trichloropentane.

Exceptions, where the locants are not included, would be halogenated methane, monohalogenated ethane, hexahalogenated ethane (and all other perhalogenated alkanes). Theoretically also e.g. pentahalogenated ethane would be unambiguous, but the preferred name should contain the locants.

You must use the numbers unless your substance in a mixture of isomers; otherwise your molecules structure is ambiguous.  For chloropentane this name alone does not specify weather the chlorine is on the 1,2,or 3 position which is important as the position of a chlorine can dictated physical and chemical properties as well as reactivity in various organic reactions.  Same is true for trichloropentane.  there are there chlorine but without numbers I dont know if the chlorines are all on one carbon, if they are in the 2,3,4; 1,3,5; 1,2,5; 1,2,3; 2,3,5; 1,1,2; 1,3,3 1,1,4 positions or any of the many more possibilities.  all of these molecules give different products in various organic reactions.
Again though if is a mixture of isomers (as is typical of hexane solvents) it is not important to give number


text:
What is the boiling point and density of LSD?
I have tried to do a lot of research and didn't succeed to find the boiling point and density of LSD. I would want to know this information. In addition, I can't use Clausius–Clapeyron equation because I don't have the data that I would need to use that equation.

Bioorganic molecules usually don't have boiling points.

Why can't you use the Clausius-Clapeyron equation?

@CurtF. Because I don't have the data that the equation needs.

I think your question would be better if you explained that detail in the text of the question itself.

@CurtF. Good point, I edited it.

You want to distill it? That would be very impractical, use chromatography for purification. ☺

Lysergic acid diethylamide doesn't boil. It decomposes. The density is $1.2 \pm 0.1~\mathrm{g~cm^{-3}}$.
Data can be found at  ChemSpider. There's a predicted boiling point of 541°C. However, lysergic acid decomposes at its melting point of about 240°C (CRC Handbook of Chemistry ad Physics, 64th ed.). This suggests that lysergic acid diethylamide does not have a real boiling point - at least at normal pressure.

Please cite your sources.

The statement about the boiling point is mainly common sense, given the structure and molecular weight of lysergic acid diethylamide. An additional argument is that lysergic acid decomposes at its melting point of about 240°C (CRC Handbook of Chemistry ad Physics, 64th ed.). The predicted boiling point of LSD is above 500°C[1] what is far above this temperature. The densitiy can also be found at[1].
[1] http://www.chemspider.com/Chemical-Structure.5558.html

Concur, the decomposition is common sense, but supporting the argument with detail is a good idea. I recommend editing your answer to include the data and references of your comment.

… that's for lysergic acid. LSD itself (lysergide) has m.p. 80–85 °C (tartrate: 198–200 °C), no b.p. (ever measured?), as expected (Merck Index).


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Can someone explain the differences in decomposition rates for when two molecules which have 2 different substituents?
Just having trouble on why these two have such different decomposition rates, I would like to think its because NO2 is electron withdrawing but I cannot see why this would change the rate...some please explain?

What is the decomposition pathway?

Would it first decompose at the O-O bond? then the C-C bond next to the C=O, is that correct?

Uhhuh, that seems pretty legitimate. So how would the NO2 group stabilise the radical that you form?

By withdrawing electrons away from the ring? I assume this would make rate decrease due to the radical becoming less able to react further on? I know that NO2 is an electron withdrawing group so this would occur where as in the methyl group, it is electron activating and thus would make the radical more unstable? thus increasing the decomposition rate


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What material should I mix with graphite to make a brick?
I want to make flake graphite into a brick but I don't know what material will work best. Would mixing it with clay work?

There's probably a large amount of research into graphite/clay blends as they are commercially important for the production of pencil leads.

Clay would work, yes. It is, after all, dirt. Adding graphite to it would only be changing the composition of that dirt.
Typically, you'd only need any sort of binding agent: If you're willing to make a weak brick, you can add some cornstarch and literally press it into a brick. You could also try things like epoxy and acrylic compounds, or things like plaster.
If you wanted to keep the composition of the brick to be mostly carbon, you could consider using chemicals like bitumen and tar, although these bricks would be more flexible (depending on the graphite to whatever ratio).

Yes, I recently did this myself actually. Clay is what is mixed with pencil lead graphite as well, so it makes sense that it works.


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Naming compound containing a ring (cycloalkane) side chain?
Why is the correct name 5-cyclopentyl-2-methylheptane rather than 2-methyl-5-cyclopentylheptane? I am using the 2nd edition Hornback Organic Chemistry textbook. I got this answer from the solutions manual.
I have searched the textbook and the internet for the explanation and am still very confused. I thought the order of the side chain in the name was alphabetical, not including the prefixes? So by this logic, "methyl" should be before "pentyl," right?

In view of the release date of your book (2005), the used names for organic compounds are probably in accordance with the 1993 Guide to IUPAC Nomenclature of Organic Compounds. However, note that the current version of the recommendations can be found in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book).
According to the 2013 recommendations, two methods are possible for naming systems composed of rings and chains: either the ring or the chain can have seniority. To construct the preferred IUPAC name (PIN), the ring is always selected as the parent structure. Therefore, the PIN for the compound given in the question is (6-methylheptan-3-yl)cyclopentane since the cyclopentane ring has seniority over the heptane chain.

In general nomenclature, however, the context may favour the chain, so that the parent structure has the greater number of atoms in the principal chain. The corresponding systematic name for the compound given in the question is 5-cyclopentyl-2-methylheptane.
According to the 1979 recommendations, a hydrocarbon containing a small ring attached to a long chain was generally named as a derivative of the acyclic hydrocarbon. The 1993 recommendations do not explicitly mention this rule but refer to the 1979 recommendations. Hence, it is not surprising that your book still uses such a name, although the 1998 Guide to IUPAC Recommendations already explains that the ring is always selected to be the parent structure.
Now coming to your actual question:
According to the 2013 recommendations, the prefixes ‘cyclopentyl’ and ‘methyl’ in the name 5-cyclopentyl-2-methylheptane are arranged alphabetically.

P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.
P-14.5.2 The name of a prefix for a substituent is considered to begin with the first letter of its complete name.

Therefore, ‘cyclopentyl’ begins with ‘c’ and ‘methyl’ begins with ‘m’. Note that, for example, ‘2,3-dimethyl’ is considered to begin with ‘m’ since the multiplicative prefix ‘di’ does not alter the alphabetical order. Whereas, for example, the compound substituent name ‘(1,2-difluorobutyl)’ in 7-(1,2-difluorobutyl)-5-ethyltridecane is considered to begin with ‘d’.


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Is the mixture of alkyl dimethyl benzyl ammonium chloride and vinegar dangerous?
So I was trying to clean cat pee out of my carpet and I scrubbed it with a Lysol wipe that says the active ingredient is alkyl dimethyl benzyl ammonium chloride and without thinking then sprayed white vinegar on the area. Now I'm afraid that made a bad chemical mixture? Do you know if those 2 ingredients are dangerous?

Please note that I worked in medical research, but I am not a medical doctor!
I think that you're safe, and I cannot imagine a reaction between acetic acid, the main component in vinegar, and a quarternary ammonium salt that would release harmful products.
Benzalkonium chloride is a long known disinfectant and has been used as a biocide and preservant in medical products. However, it may cause problems. The European Medicines Acency has published a Q&A on Benzalkonium chloride.
In general, it is not the worst idea to have a look at MSDS (Material Safety Data Sheets) for compounds and products, in order to be aware of what to do if something spills on your skin, gets into your eyes, is swallowed, or inhaled. Here is an example of a MSDS for benzalkonium chloride.
To sum it up:

Mixing household chemicals can release harmful products, as known for the reaction of bleach and acid, from which chlorine gas is released.
I seriously doubt that you have created a harmful mixture in your case.
Manufacturers are obligated to provide MSDS, wich have to include first aid measures! Take a look at this vital information.


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Naming a ring with an alcohol group?
Is it necessary to include the "1" in the name to indicate that the OH group is at the 1st carbon? Isn't it kind of assumed?
Is it necessary to name a compound "2-(3-methylbutyl)-1-cyclopentanol"?
Or is "2-(3-methylbutyl)cyclopentanol" sufficient?
Thanks!

If my response answered your question, please mark it as "accepted" - thank you!

It depends on the priority of the substituents on the ring. If the highest-priority group is the alcohol group, then the numbering of the carbons in the ring starts there - so the 1 is implicit, and you do not need to state it. See 2-(3-Methylbutyl)cyclohexanol as an example.

Note that the alkyl group position is assigned a number relative to the position of the -OH group. However, if there is a higher-priority group than -OH then the position of the alcohol will be assigned a number relative to this higher-priority group; see 2-hydroxy-3-chlorobenzoic acid.

In your example, the 1 is not necessary.


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How does increasing the ionic strength of an aqueous solution decrease the activity coefficient?
Activity is a new concept for me, and I am having a little bit of trouble understanding it.
For the equation:
$$\ce{aA + bB<=>cC + dD}$$
$$K_c= \frac{[\ce{A}]^a[\ce{B}]^b}{[\ce{C}]^c[\ce{D}]^d}$$
As I understand it, this is not the true equilibrium constant expression for the chemical equation above. My textbook calls $K_c$ in this instance the "concentration quotient." The technically correct equilibrium constant expression is:
$$K=\frac{(a_\ce{A})^a(a_\ce{B})^b}{(a_\ce{C})^c(a_\ce{D})^d}$$
Where $K$, in this equation, is the "true" equilibrium constant determined by the activity, $a$, of each species in equilibrium with each other. The activity is represented by the equation: $a_\ce{A}=\gamma_\ce{A} [\ce{A}]$, where $\gamma$ is the activity coefficient.
My textbook Quantitative Chemical Analysis (Ninth Edition, by Daniel C. Harris), states that "the equilibrium 'constant' is not really constant." And then it goes into how increasing ionic strength increases the solubility of ionic compounds in aqueous solution. It then later states "the activity coefficients must decrease with increasing ionic strength."
This would seem to contradict the idea that increasing ionic strength increases the solubility of an ionic compound, would it not? If the ionic strength increases the solubility of an ionic compound, then wouldn't that mean that the concentrations of the ions that make up that ionic compound would increase, and thus the activity coefficient, therefore, must increase to show an increase in activity, otherwise known as the "effective concentration?"
Or is the activity independent of the ionic strength such that in order to keep the activity the same under higher concentrations, the activity coefficient must be smaller? This would then bring up the question: is the equilibrium constant, $K$, (determined by the activities of the components in the equilibrium expression) constant even under different ionic strengths, while the "concentration coefficient" (determined merely by the concentrations of the components in equilibrium with one another) does vary under different ionic strengths? I'm not sure I'm getting this very well.

Thermodynamic activity is essentially the same the chemical potential.  At the solubility limit of a compound, the dissolved species is in equilibrium with the solid species.  For example, a saturated solution of sodium chloride is in equilibrium with solid sodium chloride crystals.  In such equilibria, the chemical potential of the dissolved sodium chloride is equal to the chemical potential of the solid sodium chloride crystals.

If the ionic strength increases the solubility of an ionic compound, then wouldn't that mean that the concentrations of the ions that make up that ionic compound would increase, and thus the activity coefficient, therefore, must increase to show an increase in activity, otherwise known as the "effective concentration?"

Let's assume that increasing the ionic strength doesn't alter the crystals of solid sodium chloride that dissolved sodium chloride is in equilibrium with.  That means the chemical potential of the crystal is unchanged.  At the solubility limit, the dissolved sodium chloride is still in equilibrium with the crystals, and as before, this equilibrium requires that the crystals have the same chemical potential (aka thermodynamic activity) of the dissolved species.
If there is more of the dissolved species (a higher concentation), the only way the activity can stay the same is if the activity coefficient goes down.
So you were right when you wrote:

...the activity [is] independent of the ionic strength such that in order to keep the activity the same under higher concentrations, the activity coefficient must be smaller...
This would then bring up the question: is the equilibrium constant, $K$, (determined by the activities of the components in the equilibrium expression) constant even under different ionic strengths, while the "concentration coefficient" (determined merely by the concentrations of the components in equilibrium with one another) does vary under different ionic strengths?

The equilibrium constant is a ratio of thermodynamic activities.  The "concentration coefficient" can vary and will not be constant any time that activity coefficients are not 1.0.  Ionic strength is one of many variables that can affect activity coefficients.  So to conclude:

Yes, the equilibrium constant $K$ is determined by the activities of the components in the equilibrium expression.
The "concentration coefficient" $K_c$ can vary as a function of ionic strength (or other variables).

So, the activity must remain the same even under different ionic strengths?

At the solubility limit, yes.  So if the solubility of $\cf{MgSO4}$ under ionic strength A is 27 g/L, and the solubility under ionic strength B is 40 g/L, then (a) the activity of $\cf{MgSO4}$ at 27 g/L under condition A is the same as the activity of $\cf{MgSO4}$ at 40 g/L under condition B, and (b) the activity coefficients are obviously different.  This is because the activity of crystalline magnesium sulfate does not change as in equilibrium with dissolved $\cf{MgSO4}$ under both conditions.

Okay, so because the magnesium sulfate's activity would remain the same, the activities of the magnesium cation and the sulfate anion would also remain the same, meaning the equilibrium constant/solubility product constant (as determined by those activities) would also remain the same under different ionic strengths, right?


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Organic chemistry polymerisation
I have a doubt regarding the polymerisation of ethene, in the case of polymerisation, we basically are arranging the molecules in a systematic order and thus decreasing its entropy so shouldn't we be providing a low temperature in order to keep the process spontaneous...
Whereas, in reality the conditions for the reaction are high temperature and pressure?
Please help...

For a process to be spontaneous, the overral entropy of the univesrse should increase. Thus, the change in entropy of the system can be negative for a spontaneous process. At elevated temperatures, the spontaneity condition is still being met and additionally you are providing enough energy for bonds to break and reform between reactants.

Rate of a reaction depends on the temperature conditions. So to make a reaction proceed significantly, we have to take the reactants up to the level of activation energy. One more thing to note here is that low temperature is not a fixed temperature, for example - consider 2 reactions A and B, now if low temperature for reaction A is 400 Kelvin, then it is certainly possible that for reaction B low temperature is say 800 Kelvin.

Spontaneous has a very specific meaning in thermodynamics - it is  process where the entropy change of the universe is positive. It is not related to the activation energy or the rate of reaction.

Why did you copy the question into a comment?

Please help me with this question. I am a bit doubtful about my previous answer.

Your answer is OK, removed dubious spontaneity. It's kinetic vs thermodynamic misunderstanding in question. Your kinetic info is ok


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What test can be used to differ amide and amine?
For aliphatic amide and amine. I know we can use $\ce{NaOH}$ because amide will give off $\ce{NH3}$, but what will amine yield if we test it with $\ce{NaOH}$?

Amines don't react with NaOH

In order to identify amines, you might want to search for the Hinsberg reaction.

Pfft, if you have simply choice between amine and amide, then make solution and use basicity indicator.

I do not know how to distinguish chemically between an amide and an amine based on the reaction with $\ce{NaOH}$ but I can tell you a different way to distinguish between the two kinds of compounds.
An amide on acidic hydrolysis will yield the corresponding carboxylic acid (i.e. with the same number of carbons) as follows. $$\ce{RCONH2->[\ce{H3O+}][\Delta] RCOOH}$$
The carboxylic acid produced can be tested for using the sodium hydrogencarbonate test, in which adding a pinch of $\ce{NaHCO3}$ produces a brisk effervescence due to the evolution of carbon dioxide gas.
$$\ce{RCOOH + NaHCO3->RCOONa + H2O + CO2 ^}$$
So basically, out of the two given compounds, the one that gives you a brisk effervescence when you first hydrolyse it in an acidic medium and then follow it with a pinch of $\ce{NaHCO3}$, is the amide, and the other is the amine.

You will have to be careful doing this and make sure that you remove all of the acid used for hydrolysis, before adding the hydrogencarbonate, otherwise you will get false positives.

Yes, that is true. I did not see that earlier.

From Wikipedia:

While the conjugate acid of an amine has a pKa of about 9.5, the conjugate acid of an amide has a pKa around −0.5. Therefore, amides don't have as clearly noticeable acid-base properties in water.

Therefore, if you are dealing trying to distinguish a water-soluble amide from a water-soluble amine, and neither molecule has other functional groups, you could simply dissolve them in water and then measure the pH.  If it's alkaline, it's an amine.  If the pH doesn't change much, it's an amide.

Easiest way is to use TLC(thin layer chromatography) plate and stain with ninhydrin. Ninhydrin colors primary and secondary amines and amides only with vigorous heating when amide bond breaks.


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If Mendelev hadn't put the periodic table in its current order, would it still have been made popular?
What I mean is, does the current model make the most sense? The QWERTY keyboard was made popular because someone won a typing contest with it, and is widely used today, even though other keyboard designs might have made more sense. Is our current periodic table also just popular for an arbitrary reason, or is it actually the best design possible? If things had played out differently, is there a high chance we would use another table?
Thank u

Mendelev's periodic table actually wasn't in the typical form now used.

You are also wrong about the keyboard. In the olden days typewriters were used that had the letters on levers that swung up an impacted a ribbon which left an impression of the letter on the paper. If you struck two keys at almost the same time then the keys would jam when both levers were up at the same time. So the keys were arranged by engineers to prevent jams. There was no contest. https://en.wikipedia.org/wiki/QWERTY

The periodic table is very different to things like the QWERTY keyboard layout which is essentially arbitrary.
The structure is determined by the fundamental chemical properties of the elements. When the table was first developed Mendeleev was wise enough to arrange the table with both gaps and with some elements in chemical property order rather than the order indicated by know physical properties such as atomic mass (which usually gives the same order as chemical properties but gets it wrong a small number of times).
These properties are fundamental to the elements and not a matter where human choice can have any impact at all. Sure, there are alternative ways to get the graphic design right, but none of these work if the violate the original principles of grouping similar elements together. Therefore, no alternative choices by Mendeleev could work since they would fail to achieve this.

Just like your keyboard analogy, the periodic table could have developed in a different way but it turned out to be so because since the beginning of classification of elements, all attempts of classifying the element are being made to classify elements on the basis of fundamental properties like atomic mass, electronic configuration, atomic number etc. The trend of the development of the periodic table of elements was headed in a particular direction from the very beginning, leading to the modern periodic table.

As for wether it is the best possible design or not, I believe that there is scope of improvement in every human design ever made. For example, the modern periodic table also has its own flaws such as not being able to properly classify hydrogen on the basis of its electronic configuration and not being able to decide wether to put hydrogen in halogens or alkali metals. Also the transition metals have properties that do not correspond to all elements in their particular group. So, yes if things had played out differently, we would definitely be using another periodic table of elements.

If someone could come up with a better periodic table design based on a pro/con analysis, then I do believe scientists would switch. The fundamental problem is that the "periodic properties" aren't actually rigorously  periodic. So you can think of the periodic table as a good first approximation.


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Isonitrile to nitrile conversion
If we are given $\ce{CH3NC}$ and told to convert it into $\ce{CH3CN}$ by doing some chemical reactions.
My attempt
According to my textbook there are no chemical reaction of isocyanide conpounds.  I tried reduction of isocyanide giving a secondary amine. Then I don't know how to convert a secondary amine to cyanide compound.
Hints are too appreciated.
Edit:
I have got a solution but I don't know whether the reaction are spontaneous.

I have written these reaction on the basis of theoretical knowledge I don't know whether they are spontaneous or not. Correct me if I am wrong somewhere.

N-Methylformamide is not unstable! It's just a polar solvent, that is bottled and shipped. To my knowledge, neither ADR, nor IATA regulations classify the compound as a particularly dangerous good.

Practically is it possible to carry out this reaction ?

What do you mean by "spontaneous" in this context? BTW from dumb org. chem homework POV it's mostly OK, but has no sense in practice. As Klaus wrote, simply heating it would be real, practical thing.

@Mithoron spontaneous means whether the reactants on reacting gives the product in sufficient amount or not. Or in other words spontaneous can be said as that reaction occur on its own .

You should learn proper words for synthesis. Don't mix thermodynamics here.  One asks if reactions are viable or have good yields.

Just heat it up: Isonitriles can be thermally rearranged to nitriles.
$$\ce{R-NC ->[\Delta] R-CN}$$
Please have a look at some references, such as:
Michael J. S. Dewar, M. C. Kohn, Ground states of $\sigma$-bonded molecules. XVI. Rearrangement of methyl isocyanide to acetonitrile, J. Am. Chem. Soc., 1972, 94, 2704-2706 DOI
Michael Meier, Barbara Mueller, Christoph Ruechardt, The isonitrile-nitrile rearrangement. A reaction without a structure-reactivity relationship, J. Org. Chem., 1987, 52, 648-652
DOI

UPDATE
As far as the sequence outlined in the question is concerned:

The synthesis of methyl formamide from methyl isonitrile is pretty much the opposite of how things are done in real life. Isonitriles, used in the Passerini and the Ugi reaction, have shown to be available by dehydration of formamides, e.g. using $\ce{POCl3}$. For an example with cyclohexyl isonitrile, see Org. Synth., 1961, 41, 13.
N-Methyl formamide is far from being unstable! Release of methyl amine, which, by the way, has a BP around -6 °C might be possible under alkaline conditions.
Making methanol from methylamine, again, is a rather pointless mind game. In real life, methylamine is produced by reacting methanol with ammonia at hight temperatures in the presence of acidic catalysts.
You definitely want to make methyl chloride in the lab! With a BP around -24 °C it's very easy to handle, and the GHS classification with H351 (Suspected of causing cancer) and H373 (May cause damage to organs through prolonged or repeated exposure), this is a real gem. Yes, I'm kidding! Don't do it!
For a nucleophilic substitution with cyanide, the high volatility of methyl chloride again probably doesn't make your life easier, but just shorter.

In summary, my suggestion is simple: Forget about the proposed synthesis! It is laborious, difficult to handle in the lab, and an economical nightmare.

I have updated my question by putting a solution but I don't know whether the reaction are spontaneous or not.

It's too simple guys first you have to convert isocyanide into primary amine. It can be done by hydrolysis of isocyanide, you'll get primary amine and formic acid.
Then after getting primary Aimne, on reaction with Cu/500°C or Ni/500°C you will get Cyanide.

The OP wanted a method which gives a good yield, and a far better way for that is via rearrangment as highlighted in the answer above


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Why are fires smokey?
Combustion, in chemical terms, is a reaction with a certain molecule and oxygen, and it produces, energy, $\ce{CO2}$, and $\ce{H2O}$ (That's how I learned it at least). So what makes a fire'smokey'? I had always thought that smoke was simply $\ce{CO2}$, but read that it was actually the result of incompletely 'combusted' material, yet I don't fully understand what this means. Is smoke made from material that starts to combust but stops before it finishes, or does it never even begin combustion? What would the chemical reaction be of a wood (Or other similar) fire that produces a lot of smoke? And finally, is there a certain element or compound that gives smoke its 'smokey' smell?

Related http://chemistry.stackexchange.com/questions/6954/why-does-a-wood-fire-create-benzene

Smoke is essentially aerated ash, unburned organics (including charcoal) and steam.  You have to remember wood is not a pile of cellulose, but the remains of a dead organism. All the essential minerals that the plant needed for life are still in the wood such as calcium, phosphorous, sodium and potassium (namely potassium).  When the organic matter is fully combusted, water and carbon dioxide are produced and the other elements are left as ash which can become aerated.  When the water vapor reaches the cool air it condenses into a visible steam.  However in real life wood is not fully combusted, but rather many unburned organic compounds due to a limited amount of oxygen, which give it its smell and fine aerated particles of charcoal which is scentless but very dark.  This presence of unburned hydrocarbons in smoke is the cause of backdrafts in firefighting and can be demonstrated by lighting the smoke of a candle that has been blown out.

Thanks @A.K. I guess this is as close as I can get to a chemical formula :)
Also, thanks for the edit on those symbols, I'm still figuring out this stack exchange so I'm not sure how to do that yet :)

Fires need oxygen to burn properly. By that I mean if there isn't sufficient oxygen supplied to the fire, the material that is burning does not completely burn up. They come to the middle of the combustion reaction but they run out of oxygen. These bits of incomplete combustion are called soot or smoke. $\ce{CO2}$ is another by-product of combustion. There are particular woods that give off certain smells and flavours like hickory, but smokey smells are probably only caused by carbon.

Carbon does not have a smell at all. The smell must be caused by some mixture of organic substances, too complicated (and too carcinogenic) to look into.

@IvanNeretin I'm sure it's not the only one, but guaiacol is one of the major odor compounds in wood smoke. The purified stuff has a remarkably potent and persistent aroma.


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Why does H2O2 have a gauche structure?
The gauche structure of $\ce{H_2O2}$ is quite well known. It has the two $\ce{OH}$ groups in different planes at right angles. Why does it adopt this structure?
I can understand that it might have something to do with the repulsions between the lone pairs on the oxygens. But in that case a staggered structure would make more sense since that minimises the repulsion even further. Maybe another factor is at play here that I am missing.
EDIT: My main question in this post is why the gauche structure arises in the first place. However, the other post is mainly focussed on the variation of the dihedral angle in different phases. Although the OP asks why hydrogen peroxide adopts a gauche structure, a satisfactory answer has not been provided. (Note that the OP has accepted the provided answer. Since that does not answer my main question, I think a new question is justified in this regard)

Related: Dihedral angle of gaseous and crystalline HOOH

https://en.wikipedia.org/wiki/Gauche_effect

I think this question should be reopened. As OP has noted, the "related" question is not about the cause of the gauche structure of hydrogen peroxide.

In liquid or gas and even in solid hydrogen peroxide has a gauche structure. One reason for this is to reduce electronic repulsion and the other is to give a well packed structure in solid phase.also the interactions between different molecules leads to a gauche structure in liquid and gas.


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Formose reaction - mechanism of first step
The following image was taken from wikkipedia's article about the Formose reaction. What is the mechanism of the first step of the reaction sequence? How do the two formaldehyde molecules condense?

Wikipedia says: 'The reaction is catalyzed by a base and a divalent metal such as calcium.' So I'd guess first complexation, the proton abstraction, then C-C bond formation.

Would it help to know that 1. under standard conditions, formaldehyde is a gas, 2. the reaction is performed in aqueous solution, 3. in aqueous solution, formaldehyde mostly exists as a hydrate $\ce{H2C(OH)2}$?

There are many articles on the topic. I am far too lazy to read them and write up an answer but here are some: Tetrahedron Lett., 1959, 1 (21), 22-26, J. Org. Chem., 2006, 71 (25), 9503–9505, J. Am. Chem. Soc., 2014, 136 (10), 3720–3723

@orthocresol In none of these articles is the first reaction considered.

@Klaus I don't understand. There must be some deprotonation step but it seems so improbable in any case.

It is. There are two pathways for the first reaction - one occurs without catalysis and the mechanism seems to be unknown (the last reference writes that it may involve radiation), and the other is catalysed ostensibly by the products 2, 3 and 4. Most of the discussion is focused on the catalysed reaction.

So there isn't a definite answer about the mechanism?

@Marko I find your question very interesting! I think that we can definitely rule out direct deprotonation of formaldehyde in aqueous alkaline medium. However, deprotonation of methanediol, formadehyde exists as a hydrate in  aqueous solution, seems possible, and it might be further facilitated by the presence of larger cations. But till now, I haven't figured out how an umpolung, rather than a Cannizzaro reaction might take place under these conditions.

Only a radical mechanism seems plausible: homolytical cleavage of C=O bond, addition of the radical to the other formaldehyde molecules and hydrogen abstraction by O*. That would fit early Earth's conditions, but I don't know whether this reaction works in the lab and how, so I can't be sure.


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Why NaHCO3 added to the water used (Ion Chromatography Suppression)
I'm studying HPLC detection techniques and stuck on the conductometric suppression technique. It is describing how buffer ions (in my case $\ce{Na+}$ and $\ce{HCO3-}$) are being either adsorbed ($\ce{Na+}$) or protonated ($\ce{H2CO3}$) which makes them electrolytically inactive and leaves just water in the column.
Great! I understand that background conductivity is a problem. What I do not get is why was $\ce{NaHCO3}$ added to the water in the first place?

Are buffers used to dissolve the sample?

Ion chromatography usually means chromatography where aqueous (and usually inorganic) ions are the analytes.  They are separated from each other by the use of ion exchange resins.
If your analytes are anions, then they will be separated on an anion exchange resin.  These resins have a crosslinked organic polymer that contains permanent positive charges.  The negative charges that neutralize these resin-bound positive charges come from aqueous solution.  Different aqueous anions will have slightly different propensities to bind to and temporarily neutralize the solid-phase cations.
If you don't use an anion-containing eluent such as bicarbonate or hydroxide, the analytes that you would like to eventually detect through conductometric detection will permanently stick there!  So adding e.g. $\ce{NaHCO3}$ as an eluent will force them off of the column and ensure that your analytes eventually flow to the detector.
A decent diagram of the process is found in this PDF from Thermo (formerly Dionex). $%edit$


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Neutralization equation
If I have cider vinegar and want to calculate the total moles of acid found in it, how can I do it? I have the moles of NaOH and I know that it includes both acetic and malic acid but how can I combine them?