Usage of ammine vs amine in nomenclature In the nomenclature of complex salts we use ammine for NH3 instead of amine.

I thought this was to differentiate between ammine ligand and amine in organic ligand (like en).

However, Wikipedia cites the usage has historical reasons. So why exactly do we use ammine instead of amine?

(Will do more research into ammines come February.)

Premise

Ammonia is spelled with two 'm's. The more natural derivative of the word in a linguistic sense would also have the same number 'm's. Thus whichever predates the other, ammine complexes or amines, would claim the throne.

In a structural sense, when awknowledgement of the first ammine complex came it was literally in a solution of ammonia (see more below).

Etymology of ammonia and amine

  Our travels begin in sunny Ancient Egypt. One important deity of the time was Amun, the god of wind and air along with his wife-in-mythology Amaunet. At the beginning of the Middle Kingdom of Egypt (c. $2055-$$1650$ BC), he was annointed patron saint of Thebes.  Later Amon become part of a trinity, along with gods Re and Ptah. This coincided with Thebes restored as capital city due to pharaoh Ahmose I (c. $1539-$$1514$ BC).

  Eventually Amon would become the god of gods. This is when he also goes by Amon-Re (or Ra), after uniting with sun god Re.$^{\text{[a]}}$

  Egypt's glory shone far beyond its borders. Ancient Greeks adopted their own version of Amon. There he would be called Ammon, with two 'm's.

As it so happens, near one of Amun's temples in then-Egypt is a depository of $\ce{NH4Cl}.$ From this háls ammôniakós (Greek for 'salt of Amun') we can easily extract $\ce{NH3}.$

$$\ce{NH4Cl(s) ->[t^\circ] NH3(g) +HCl(g)}.$$

It makes sense for the option with two 'm's to outdate amine. Etymonline agrees (in ~ English):

  ammonia (n.) $-$ $1799$, modern Latin, coined $1782$ by Swedish chemist Torbern Bergman ($1735$$-$$1784$) for gas obtained from sal ammoniac [----];

  amine (n.) $-$ "compound in which one of the hydrogen atoms of ammonia is replaced by a hydrocarbon radical," $1863$, from ammonia + chemical suffix -ine (2).

Note: has been expanded upon under 'Amine in chemical literature' further down.

The why

Why ammine for coordination compounds and not amine? It is probably because the study of metal ammines predates any knowledge of amines. Indeed, in $1798$ Tassaert$^{[1]}$ reported on 'ammoniacal solutions of cobalt(II) chloride [which when] allowed to stand overnight yield an orange-colored crystalline product containing six molecules of ammonia'$^{[2]}$. The compound in question was $\ce{CoCl3*6NH3}$. These and similar compounds were dubbed metal ammines$^{\text{[b]}}$.

Note that at the end of  $18^{\text{th}}$ century organic chemistry had not yet developed. The crucial and accidental evidence against vitalism$^{\text{[c]}}$ came in $1828$ when the German chemist Wöhler artificially synthesised an organic compound $-$ urea $-$ an amide with two amine qroups!

$$\ce{AgNCO + NH4Cl-> AgCl + NH4NCO \\ NH4NCO ->[t^\circ]NH_3 + HNCO <=>\underset{urea}{CO(NH2)2}}$$

First pivotal trace of amines pops up in $1842$, $44$ years after Tassaert's discovery of coordination chemistry. Russian chemist N. N. Zinin uses dihydrogen sulfide to reduce nitrobenzene$^{[3]}$$^{\text{[d]}}$:

$$\ce{C_6H5NO2 + 3H2S ->[NH3]\underset{aniline}{C6H5NH2} + 3S + 2H2O}.$$

On Zinin's research, co-founder of amine chemistry A. W. von Hofmann declares$^{[4]}$,

  If Zinin had done nothing more than to convert nitrobenzene into aniline, even then his name should be inscribed in golden letters in the history of chemistry.

Aliphatic amines were simultaneously and independently synthesised in $1849$ by chemists Wurtz of France and Hofmann of Germany. Both could be credited with the discovery of ethylamine. Hofmann's method allowed for the additional preparation of secondary and tertiary amines.$^{[3]}$

Wurtz treated ethyl isocyanate with potassium hydroxide; Hofmann applied heat to solutions of alkyl halogenides and ammonia.$^{[3]}$

$$\tag{Wurtz}\ce{C2H5N=CO + H2O->[KOH]C2H5NH2 + CO2}$$
$$\tag{Hofmann, I}\ce{R-I + NH3->[t^\circ]R-NH2*HI}$$
$$\tag{Hofmann, II}\ce{2R-I + NH3->[t^\circ]R2-NH*HI + HI}$$
$$\tag{Hofmann, III}\ce{3R-I + NH3->[t^\circ]R-N*HI + 2HI}$$

For quaternary amines, Hofmann used the following scheme:
$$\tag{Hofmann, IV}\ce{R3N + RI->R4-N*I}.^{[3]}$$

But because ammine was already reserved, they went with amine.$^{[5]}$ Furthermore, if you think about it, the naming is rather successful. Well, at least for primary amines.$^{\text{[e]}\ \text{[f]}}$

$$\text{metal ammines contain the complete}\ \ce{NH3->} \text{keep the}\ m\\ \text{pr. amines have one less hydrogen or}\ \ce{-NH2->} \text{lose the}\ m$$

Amine in chemical literature

In $1834$, German chemist Liebig optained a dirty white or buff-coloured amorphous substance by heating ammonium thiocyanate.$^{[5]\ [6]\ [7]\ [8]\ [9]\ [10]\ [11]}$

$$\tag{$\alpha$}\ce{KSCN + NH4Cl ->[t^\circ] NH4SCN + KCl}$$
$$\tag{$\beta$}\ce{NH4SCN <=> \underset{thiourea as thione}{S=C(NH2)2} <=>[t^\circ] \underset{thiourea as thiol}{HS-C(NH)NH2}}$$
$$\tag{$\gamma$}\ce{\underset{thiourea as thiol}{HS-C(NH)NH2} ->[t^\circ] NH3 + HNCS \\ \underset{thiourea as thiol}{HS-C(NH)NH2} ->[t^\circ] H2S\ \ \ + \underset{cyanamide as aminonitrile}{N#C-NH2}}$$
$$\tag{$\delta$}\ce{\underset{cyanamide as aminonitrile}{N#C-NH2} <=> \underset{cyanamide as diimine}{HN=C=NH}}$$

He named the product melam without any admitted etymology other than his own fantasy. Indeed, Liebig mentions in his original paper$^{[5]\ [6]}$,

  [These designations] are, if you like, grasped from the air (i.e. pure inventions), and serve the purpose just as well as if they were derived from the colour or one of the properties.

Liebig proceeded to boil melam with a solution of potassium hydroxide and ammonia. This yielded melamine.$^{[5]\ [6]\ [7]}$

Neither the structure of melam nor melamine were known to Liebig at the time.$^{[5]}$ Melamin(e) was most likely derived by Liebig as melam $+$ -in(e).$^{[5]}$$^\text{[g]}$

A few years before $1849$, Liebig had predicted the existance of amines.$^{[5]}$$^\text{[h]}$ Liebig called them compound-ammonias$^{[5]}$. The word amine itself was used already in $1853$$^{[12]}$ in the works of French chemist Gerhardt. Often times this was put in parentheses as an alternative name. For example$^{[12]}$,

  méthyl-ammoniaque (méthylamine),
  amyl-ammoniaque (amyl-amine).

Acceptance of the word amine was not immediate. Besides having to distinguish between amines and metal ammines, amides were too an obstacle. The restriction to retain amine for ammonia [where any] hydrogen atom [has been] successively replaced by hydrocarbon radicals was generally accepted in $1863$.$^{[5]}$

TL; DR

$$\text{Amun in Greek}+\text{nearby }\ce{NH4Cl}\ \text{depository}\ce{->[Tassaert dibs on ammine in 1798][\text{vitalism until a.l. 1828}]}\text{amine in 1863}$$

Extra information

$\text{[a]}$ According to some texts, the unification as Amon-Ra goes back to the reign of Mentuhotep II ($2055-$$2004$ BC). Amon's (or Amon-Ra's) importance declined, then increased again during Amenemhet I ($1985-$$1956$ BC), fell and rose to national prominence with Ahmose I.$^{[13]}$
$\text{[b]}$ Most of the early work was done with ammonia, and thus the resulting class of complexes were already then known as metal ammines$^{[14]}$. Developments in naming specific compounds are summarised in this diagram:

$$\text{after discoverer}\ce{->[\text{m.c.l.}]}\text{basis on colour}\ce{->[\text{m.c.l.}]}\text{count number of ammonias only}\ce{->[\text{m.c.l.}][Werner's coordination theory 1893]}\text{premodern}$$

where $\text{m.c.l.}$ is short for many compounds later. Some examples of such practice is included the table$^{[14]}$ below.

\begin{array}{|c|c|c|c|}
\hline
\mathbf{Complex} & \mathbf{Colour} & \mathbf{Name} & \mathbf{Present}\ \mathbf{formulation}\\
\hline
\ce{CoCl3*6NH3} & \text{yellow} & \color{gold}{\text{luteo}}\text{cobaltic chloride} & \ce{[Co(NH3)6]Cl3} \\
\ce{CoCl3*5NH3} & \text{purple} & \color{purple}{\text{purpure}}\text{ocobaltic chloride} & \ce{[CoCl(NH3)5]Cl2} \\
\ce{CoCl3*4NH3} & \text{green} & \color{green}{\text{praseo}}\text{cobaltic chloride} & trans\text{-}\ce{[CoCl2(NH3)4]Cl} \\
\ce{CoCl3*4NH3} & \text{violet} & \color{violet}{\text{violeo}}\text{cobaltic chloride} & cis\text{-}\ce{[CoCl2(NH3)4]Cl}  \\
\ce{CoCl3*5NH3*H2O} & \text{red} & \color{pink}{\text{roseo}}\text{cobaltic chloride} & \ce{[Co(NH3)5(OH2)]Cl3} \\
\ce{IrCl3*6NH3} & \text{white} & \color{gold}{\text{luteo}}\text{luteoiridium chloride} & \ce{[Ir(NH3)6]Cl3} \\
\hline
\end{array}

An attentful reader ought to notice that $\ce{CoCl3*6NH3}$ is yellow and $\ce{IrCl3*6NH3}$ is white. But they both have the same prefix -luteo. At first it did indeed mean yellow but later the same scheme was adopted to notate an equal number of ammonias. Crazy but it is true!$^{[14]}$ You can probably see that proper nomenclature could not get there soon enough.
$\text[c]$ In chemistry, vitalism was the idea that compounds found inside living things are fundamentally different from inorganic chemistry. Such organic molecules were believed to be impossible to synthesise outside an organism.

To captivate how little of organic chemistry was understood in the first half of the 18th century, Wöhler stated in $1835$$^{[5]}$,

  [Organic chemistry is] like a primaeval forest of the tropics, full of the most remarkable things.

Proponents of vitalism had a hard time accepting the evidence. They      argued that the process was achievable since carbamide is simply a waste product. However, luckily notable progress followed.$^{[15]}$

M. Berthelot $-$ natural fat ($1854$ doctoral thesis)
H. Kolbe, E. Frankland $-$ acedic acid ($1861$)
A. Butlerov $-$ sugars ($1861$, see Formose reaction)

${\text{[d]}}$ Aniline had been successfully separated in $1826$ by German merchant and chemist O. Underderben. Note, however, that this was not a synthesis $-$ he used dry destillation of indigo.$^{[5]\ [16]}$
${[\text{e}]}$ This was very much intentional. Wurtz recognised in $1849$ that methylamine and ethylamine could be regarded as ammonia in which one equivalent of hydrogen is replaced by methyl or ethyl.$^{[5]}$
$\text{[f]}$ Classification of primary, secondary, and tertiary amines was introduced by Gerhardt in $1856$.$^{[17]}$
$\text{[g]}$ Some authorities$^{[18]}$ claim melamine as having come from mel(am) $+$ amine. This is wrong since amines were unknown at that time.$^{[5]}$
$\text{[h]}$ From Zinin to Liebig, amines were probably grouped as alcaloïdes artificiels (French for artificial alcaloids) because the structure had not yet been proposed.$^{[19]}$

Refrences and bibliography

$[1]$ B. M. Tassaert, Ann chim. phys., 28, 92 (1798).
$[2]$ Fred Basolo, Ralph G. Pearson. Mechanisms of Inorganic Reactions. (1958). (page 2)
$[3]$ Michele Giua. Storia della Chimica. (1962) (pages 342$-$344)
$[4]$ A. W. von Hofmann. Ber. Deut. chem. Ges. (1880). 13,. 449$-$450.
$[5]$ W. E. Flood. The Origins of Chemical Names (1963) (pages xxii, 33$-$35, 37$-$38, 137$-$138)
$[6]$ J. von Liebig. 'Uber einige Stickstoff $-$ Verbindungen'. Justus Liebigs Annalen der Chemie, 10, 1, 1$-$47 (1834).
$[7]$ Bernard Bann, Samuel A. Miller. 'Melamine And Derivatives Of Melamine'. Chemical Reviews, 58 (1), 131$–$172. (1958)
$[8]$ Klaus Bretterbauer, Clemens Schwarzinger. 'Melamine Derivatives – A Review on Synthesis and Application'. Current Organic Synthesis, 9, 342$-$356 (2012).
$[9]$ Michihiro Ohta, Shinji Hirai, Hisanaga Kato, Vladimir V. Sokolov, Vladimir V. Bakovets. 'Thermal Decomposition of $\ce{NH4SCN}$ for Preparation of $\ce{Ln2S3}$ ($\ce{Ln}$ = $\ce{La}$ and $\ce{Gd}$) by Sulfurization'. Materials Transactions, vol. 50, No. 7, 1885$-$1889 (2009).
$[10]$ Zerong Daniel Wang, Motoko Yoshida, Ben George. 'Theoretical study on the thermal decomposition of thiourea'. Computational and Theoretical Chemistry, vol. 1017, 91$–$98 (2013).
$[11]$ A. Kawasaki, Y. Ogata. 'Kinetics of the formation of melamine from dicyandiamide'. Tetrahedron, vol. 22, 1267$-$1274 (1965).
$[12]$ M. Charles Gerhardt. Traité de chimie organique. First edition. (1853) (pages 8, 134, 210$-$211, 277, 396$-$397, 427, 463, 545, 551, 611, 613, 616$-$619)
$[13]$ Seppo Zetterberg. Maailma ajalugu. (2015). (pages 41$-$43)
$[14]$ Fred Basolo, Ronald Johnson. Coordination Chemistry. (1964) (pages 4$-$13)
$[15]$ H. Karik. Üldine keemia. Handbook for students. (1987) (page 223)
$[16]$ Alex Nickon, Ernest F. Silversmith. Organic Chemistry: The Name Game. (1987) (page 305)
$[17]$ M. Charles Gerhardt. Traité de chimie organique. Fourth edition. (1856) (page 592)
$[18]$ The American Heritage Dictionary of the English Language: Fourth Edition. 2000. https://web.archive.org/web/20081201105219/http://www.bartleby.com:80/61/24/M0202400.html (Jan 3, 2017)
$[19]$ Encyclopædia Universalis. 'Amines'. http://www.universalis.fr/encyclopedie/amines/ (Jan 2, 2017)


Ring expansion and carbocation formation If in a reaction the carbocation is formed outside the ring then why does the ring expansion does not happen as follows:

To understand which rearrangement pathway is preferred, we must consider the relative stabilities of the intermediates. The rearrangement that puts the carbocation one carbon away from the the oxygen does not have any stabilizing resonance structures:

$\hspace{5.4cm}$

The rearrangement that places the carbocation adjacent to the oxygen delocalizes the positive charge to two atoms, and is much more stable:

$\hspace{3.7cm}$


Biochemistry for a mathematics student? I am a second year mathematics student. I am mostly interested in statistics. Right now my curriculum doesn't offer me much choice (all second year courses are mandatory) but next semester I will be able to select the courses I want. Among the ones that are offered is a course called Biochemistry. It is actually a second year class for those how study chemistry and chemical engineering but I guess there must be applications of statistics in this field since it is also offered to third year math students.

My problem is that I am a math student: I have no knowledge or whatsoever in chemistry or biology. Granted that I had some in high school but back then I just learned everything by heart in order to pass.

My questions are the following:

With no knowledge of chemistry, is it realistic to undertake a biochemistry class?
The course description lists the following prerequisites: "Basic knowledge in general and organic chemistry". How important are these subjects in biochemistry? What are some good references (books) for self-learning these subjects?
Is Stryer's Biochemistry a good book for self-learning?
Are biochemistry and biostatistics related at all?
Are there many applications of statistics in biochemistry?

I have already searched for some information on the internet but I'd like to have some specific answers.

If it helps here is the information page of this class.

Finally I hope this kind of question is accepted here, if not please point me to a better place. Thanks in advance!

With no knowledge of chemistry, is it realistic to undertake a biochemistry class?

I would say that taking a biochemistry course without any background in chemistry or biology would be very difficult, but it would be possible.

  The course description lists the following prerequisites: "Basic knowledge in general and organic chemistry". How important are these subjects in biochemistry?

The info page link that you gave would not work for me, so I can't tell how in-depth this course would be, but I know that the undergraduate biochemistry courses I have experience with expect you to be familiar with (at a minimum):

Chemistry

Bonding theory
Molecular structure
Nomenclature of organic molecules
Polarity and its relation to hydrophobicity/hydrophilicity
Basic organic reaction chemistry (particularly Lewis acid/base and functional group chemistry)
Basic reaction kinetics
Basic thermodynamics

Biology

Macromolecules (proteins, carbohydrates, fats)
Cell structures
Basic anatomy/physiology

All of these are fundamental to just about every concept in biochemistry, and they will probably expect you to already be familiar with if not good at them.

  What are some good references (books) for self-learning these subjects? Is Stryer's Biochemistry a good book for self-learning?

That I can't answer. I can tell you that Voet & Voet's "Biochemistry" is a good biochem book, but I am not sure if it would be suitable for self-learning. I don't have experience with Stryer's book. Voet & Voet also have a "Fundamentals of Biochemistry" book that might be better - but again, I don't have any experience with that one and can't compare to Stryer. In my personal experience, I have found that reading as many different books as possible is the best way to self-study, but your results may vary.

I can give you some suggestions for general chem and organic books:

General Chemistry

Brown, Lemay, and Bursten's "Chemistry: The Central Science"
Tro's "Chemistry: A Molecular Approach"

These are both very easy to read and clearly explain all of the fundamental concepts, with lots of example problems worked out in detail.

Organic Chemistry

Wade's "Organic Chemistry"

This is the only organic chem book I have much experience with, but I remember it being very easy to understand.

For biology textbooks, I'm afraid I can't be much help.

  Are biochemistry and biostatistics related at all?

Only in the sense that biostatistics is statistical methods applied to problems found in biology, and many problems found in biology are related to biochemistry. You would certainly want to learn some basic biochemistry if you were going to work on bioinformatics problems, for example.

  Are there many applications of statistics in biochemistry?

There are - particularly in areas like design of experiments, genomics, proteomics, and drug screening, and other various assays.


Does benzene vapor react with air in a closed barrel? If we would have benzene in a closed barrel for quite a while at room temperature, some vapor would form fast due to its volatility, right? But let's say that when the barrel was closed, some air got trapped inside. So it's inside at any time. Is there any chance that the benzene would react with the components in the air at room temperature? Or would we simply have benzene vapor and air? Or would we have benzene, some products of reaction and air? I made a sketch for a better understanding of my question.

So which one is the real one? And if it is the second one, would the reaction be similar to combustion, or?Is temperature a major player in this? And if so, how strongly would it affect the reaction?

Combustion is occurring, but the kinetics are very very very slow at room temperature.  So, on a time scale of days, weeks, and months, the system will be closer to the first picture.  On a time scale of decades, centuries, and millenia, the system will be closer to the 2nd picture.


Counterions in adsorption of ions When an ion is adsorbed from an aqueous solution onto a solid, what happens to its counterion? Obviously, the solution does not become charged - does this mean that it too is adsorbed?

(Example: $
\ce{NaF}$ with activated alumina as the adsorbent)

I suppose an alternate explanation would be that the remaining cation hydrolyses , but this does not seem likely in the example above as $\ce{Na^+}$, being the salt of a strong base, will not hydrolyse.

Since this got no answers, I'm posting an answer from Reddit user midnight-cheeseater, which can be found here.

  There are two main possibilities here, depending on how the ions are adsorbed:

  If the solid only adsorbs a type of ion, either positive or negative. Such as the example quoted of alumina adsorbing fluoride ions. The alumina will become negatively charged at the specific points where the fluoride ions stick to it. These negative charges will attract anything nearby which is positively charged. So the $\ce{Na+}$ ions will associate themselves with the negatively charged areas on the alumina.
  As a solid adsorbs one ion, it may release another. This process is commonly seen in ion-exchange resins. Such as those which are used to produce "de-ionized" water. These adsorb positively charged metal ions and release $\ce{H+}$ ions at the same time. They can be regenerated by running fairly concentrated nitric acid through them.

  The counter-ions are typically chloride, sulfate or nitrate, as well as some other halides, carbonate and phosphate. So this essentially converts a dilute solution of metal salts into a dilute solution of hydrochloric, sulfuric and nitric acids. Which explains why "de-ionized" water is rather acidic, as opposed to distilled water which is neutral.

  Another common example found in dishwashers and water softening systems is an ion-exchange resin which adsorbs $\ce{Ca^2+}$ or $\ce{Mg^2+}$ ions while releasing $\ce{Na^+}$ ions at the same time. This converts dissolved calcium or magnesium salts (usually bicarbonates or sulfates) into sodium salts. These are regenerated by running saturated $\ce{NaCl}$ solution through them, which is why dishwashers and water softeners require a supply of salt.


Isothermal compression at critical temperature Suppose I am compressing a gas isothermally at its critical temperature. Initially, it has a large volume. I start compressing it and reach the critical point. At this point, the gas will start to form liquid and both of these phases cannot be distinguished, right? Now if I further compress it isothermally at its critical temperature so that the pressure reaches beyond critical pressure, what will my gas look like? Will it be a liquid or a gas or something in between, I don't know. Is there a video where I can see something of this sort? Just curious

https://youtu.be/RmaJVxafesU?t=242

carbon dioxide has a critical point at 31 degrees celsius.


How to determine age of stones (and other inorganic matter)? I'd like to ask how can one determine /with relative precision/ the age of stones and stone monuments, like statues, columns, stone walls or stone tablets? I know that one determines the age of organic substances by radiocarbon dating, however, this method is inapplicable for rocks, so what else can we try?

Under age I mean the period since a stone was carved into a historical artifact and I'm interested in periods that are accurate up to a historic age, like Antiquity or the Renaissance - I suppose one cannot ask for more, unless there are visual marks on the artifact, which provide clues?

PS I hesitated much whether my question is Chemistry- or Physics- related and decided to post it here, since it's purely practical. If you decide otherwise, I'll move it to Physics.SE ;)

The question being asked, I believe, is the age of the inscription or statue rather than that of the material from which it is made.

The answer, though, is that it is quite difficult to do on the basis of a single stone object. If it is found in situ, then the age of surrounding items may be a guide.

There are some clues to the age of stone objects, though. For example, if a stone object (e.g. the walls of Siloam's Tunnel) has soot on its surface, such as from torches, then radiocarbon dating shows a minimum age (because the soot could have been deposited some time after the tunnel or object was made).

Weathering of a surface can provide information, providing the object has been exposed to known atmospheric conditions; but a buried statue would erode differently from one on the surface.

For stones, or better, for ceramics, that have been heated, thermoluminescence provides an accurate way to measure age (or at least time since last being heated).

One indication for the age of the Egyptian pyramids is their alignment to a calculated geologic reference that changed with time due to the precession of the Earth.


Diethyl Ether reaction mechanism I was trying to determine the reaction mechanism behind the production of diethyl ether. Looking at the ethyl alcohol + sulphuric acid method, I have come to the following conclusions for the reaction steps:

The ethanol nucleophilic oxygen gets protonated by the addition of sulphuric acid. This is more favourable because of the high density of electrons surrounding the oxygen. The bond broken between the carbon and oxygen is also needed to form the bond with hydrogen.

Water is then released and the ethyl alcohol conjugate base is formed (H3C-CH2+). The remaining ethyl alcohol molecules can either react with the HSO4- or the ethyl conjugate... Since HSO4- is more stable than the ethyl conjugate, ethyl alcohol reacts with the ethyl conjugate. Carbon is more electronegative than hydrogen and is an electrophile. It is attracted to the oxygen from the ethyl alcohol and replaces the H, thus forming diethylether, H+ and H2O where the sulphuric acid essentially acts as a catalyst!

Am I correct in presenting the mechanism this way? If the sulphuric acid still remains in the solution, why does it not protonate the diethyl ether? Is it because the stability of the diethyl ether conjugate base is unstable?

Thanks

Your mechanism would create a primary alkyl cation, namely the ethyl cation ($\ce{CH3CH2+}$). Primary carbocations are extremely unstable and only generated if there are stabilising effects present such as in the benzylic or allylic case — and these are then often not considered primary cations but rather allylic or benzylic ones.

Rather, the protonation of ethanol ($\ce{CH3-CH2-OH}$) creates an oxonium ion as you suggested in the first step:

$$\ce{CH3-CH2-OH + H2SO4 -> CH3-CH2-OH2+ + HSO4-}$$

The $\ce{CH2}$ carbon is then attacked from behind in a nucleophilic $\ce{S_{N}2}$ reaction liberating water in the same step. It is important to note that the attack of another ethanol molecule and the liberation of water happen synchronously; there is no first or second.

$$\ce{CH3-CH2-OH2+ + CH3-CH2-OH -> CH3-CH2-OH+-CH2-CH3 + H2O}$$

Finally, the oxygen which is now part of a protonated ether loses the proton to give the final product diethyl ether.

In the case of a nucleophilic substitution on a substrate like tert-butanol, or even better, tert-butyl bromide, one would consider the leaving group to be liberated first and a carbocation to be created; however, tertiary carbocations are orders of magnitude more stable than primary ones.


Redox Reaction Equation with elusive oxygen I've been balancing redox equations for the last couple of weeks and have become quite accustomed to the procedure however I've come across an equation that eludes me every which way when it comes to balancing oxygen. I come to the next to last step to balancing my redox equation but am unable to write my net ionic equation.

  Liquid hydrazine reacts with an aqueous solution of sodium bromate. Nitrogen gas and bromide ions are formed.

I am asked to balance this reaction in an acidic solution.

I first write out my equation:

$$\ce{N2H4(l) + NaBrO3(aq) -> N2(g) + Br- (aq)}$$

I break my equation up into two half equations, one being my oxidant equation and the other my reductant. I balance them both:

$$\ce{N2H4(l) -> N2(g) + 4H+ (aq) + 4e-}$$

The other half of the equation is where I get stuck with balancing the oxygen. I end up balancing the hydrogen on both sides to be 10 but I come up short of one oxygen on my product side.

$$\ce{NaBrO3 (aq) -> Br- (aq)}$$

I use the algebraic method to assign an oxidation number to $\ce{Br}$ on reactant side of half equation.

$\ce{Br + Na + 3O = 0}$

$\ce{Br + 1 + 3(-2) = 0}$

$\ce{Br - 5 = 0}$

$\ce{Br = 5^{+}}$

$\ce{Br}$ on the product side has an oxidation number of -1 equal to its net charge

In order to balance the oxidation charge of both sides, I add $\ce{6e-}$ to the reactant side. This then brings the oxidation charge to -1 on both sides of the half reaction.

$$\ce{NaBrO3(aq) + 6e- -> Br- (aq)}$$

I then compensate net charge of the acidic reaction by adding 5 hydrogen ions to the reactant side to bring the net charge on both sides to 1-.

$$\ce{NaBrO3(aq) + 6e- + 5H+ (aq) -> Br- (aq)}$$

I then must add water to the product side in order to balance Hydrogen and oxygen atoms in the half reaction. I have 5 hydrogen for reactants so I must add 2.5 water molecules on the product side to match hydrogen.

$$\ce{NaBrO3(aq) + 6e- + 5H+(aq)->Br- (aq) + $2.5$H2O(l)}$$

Staying customary to the law of combining volumes and Avogadro's hypothesis I multiply the half equation by 2 to get a whole number for water

$$\ce{2NaBrO3(aq) + 12e- + 10H+(aq) -> 2Br- (aq) + 5H2O(l)}$$

The oxygen atoms here elude me. I can't figure out how to balance them. I figured I took a wrong turn on the way but my troubleshooting comes to no avail thus far. Can anyone fill me in on this little secret?

First of all you need to write equations with real ions and molecules that do exist in solution. Your oxidation agent here is $\ce{BrO3-}$. In neutral media you can use $\ce{H+}$, $\ce{OH-}$ and $\ce{H2O}$. Your half-reaction can be made from this scheme:

$$\ce{BrO3- -> Br-}$$

There is no oxygen in the right part so we need to add $\ce{H2O}$ there, we need 3 molecules.

$$\ce{BrO3- -> Br- + 3H2O}$$

Now there are no H atoms in the left part and 6 H atoms in the right part. Fix this by adding 6 protons, $\ce{H+}$, in the left part:

$$\ce{BrO3- + 6H+ -> Br- + 3H2O}$$

Charge of left part is +5 and charge of right part is -1. Now equalize charges in left and right parts by adding 6 electrons to the left part.

$$\ce{BrO3- + 6H+ + 6e- -> Br- + 3H2O}$$

Half-reaction is ready.


Find empirical formula when individual masses are known I'd like to illustrate my question with an example:

Imagine nickel reacts with fluoride and you start with $0.766\ \mathrm{g}$ of nickel, an unknown amount of fluoride and you end with $1.261\ \mathrm{g}$ of nickelfluoride. The mass of fluoride in the end must then be $1.261\ \mathrm{g} - 0.766\ \mathrm{g} = 0.495\ \mathrm{g}$.

So, in the end there's $0.013$ moles of nickel and $0.026$ moles of fluoride. Now the question is, what is the empirical formula for nickel fluoride? ($\small\ce{NiF2}$). I see that for every mole of nickel you need $2$ moles of fluoride but then why is the empirical formula $\small\ce{Ni2F4}$ incorrect?

An empirical formula is, according to IUPAC’s Gold Book definition:

  Formed by juxtaposition of the atomic symbols with their appropriate subscripts to give the simplest possible formula expressing the composition of a compound

Wikipedia goes even further:

  the simplest positive integer ratio of atoms of each element present in a compound

and then

  An empirical formula makes no reference to isomerism, structure, or absolute number of atoms.

So, NiF2 is an empirical formula, while Ni2F4 simply isn't. Another example is given by, again, Wikipedia:

  For example, formaldehyde, acetic acid and glucose have the same empirical formula, CH2O

Their molecular formulas (which account for the absolute number of atoms in a molecule) differ, though: formaldehyde CH2O, acetic acid C2H4O2, glucose C6H12O6.


Forming a dioxolan with levodopa I've got 2 problems with reaction:

1) Does the nitrogen or the carboxyl group react with the methanal?
   Do I have to use a protecting group?

2) Is there a chance it reacts similar to a marquis reaction?

Before answering the question, you might find it helpful to know that people tend not to go about making the dioxolane, but instead make use of natural products such as safrole to buy in a precursor with the ring already formed. If you work at a University or a research institution, this shouldn't be an issue. If you don't, I strongly advise you against trying this chemistry, as your desired product bears significant resemblance to multiple illegal drugs.

To answer specific parts of your question:

  1) Does the nitrogen or the carboxyl group react with the methanal? Do I have to use a protecting group?

The reactivity of the amino acid portion depends on the ionisations.

In both the protonated and deprotonated form, the carboxylic acid/carboxylate can't react with formaldehyde.

Amines do react with formaldehyde in the presence of an acid catalyst (think about the Mannich reaction). Whether or not your substrate (amino acid) as drawn would react with formaldehyde likely depends upon whether the amine is protonated or not (think about the amino acid and whether or not it would exist in its zwitterionic form under the conditions given).

  2) Is there a chance it reacts similar to a marquis reaction?

The mechanism of the Marquis reaction isn't fully elucidated, however it does give positive results for many alkaloid drugs

As already stated, given the similarity of your compound to several illicit drugs, I wouldn't be at all surprised if it did give a positive result.

Moving on to the actual chemistry you've proposed.

The standard method for making 1,3-benzodioxole (and derivatives) is to take the catechol and treat it with CH2Br2 under basic conditions.

This works better than formaldehyde (I'm not actually convinced the formaldehyde would work at all, but I'll presume you've researched the method) because rather than trying to expel water (which as you point out, you'd have to Dean-Stark to remove the water), you only expel halide ions, which don't interfere under the conditions.

(As another side note, experimentally you also need to add a phase-transfer catalyst as the CH2Br2 isn't miscible with the base/aqueous layer).


The MOL format specification doesn't specify an origin, so presumably any origin can be chosen, as long as the coordinate values are within the allowed range. (The format allows a range of roughly -10,000 to +100,000, but I don't know if all programs support that large a range.) For example, in one of the examples in the specification the atom closest to the origin is over 17.6 units away. Where the origin is actually set would depend on the software used to create the file.

The coordinates for each atom specify the nucleus of the atom. I think you have a misconception about atomic size. The size of an atom, and thus the space it takes up, can be defined in different ways. The representation of an atom as a sphere (with the nucleus at the center) is a convention, only approximating reality, and the size of the sphere is arbitrary (but usually based on measurable properties). The nucleus is the only point in the atom that has a well-defined location, and thus the only part that meaningful coordinates can be defined for.

In MOL files, the presence of hydrogen atoms can be implied from valence and number of other bonds. (They can also be specified explicitly if needed.) In the long set of zeros at the end of each atom line (just after the element), the fourth zero specifies to add however many single bonds to hydrogen atoms are needed for the atom's valence. The sixth zero indicates to use the element's normal valence. For example, the carbon of the $\mathrm{CH_3}$ group has only one explicit bond, so three bonds to hydrogen are implied by carbon's normal valence of 4. How implicit hydrogens are handled depends on the software (e.g., this molecule has four implicit hydrogens that, per the usual convention for skeletal formulas, are not shown in your image).

The specification for the MOL format is available from Accelrys. The this answer is based on the specification and my knowledge of chemistry. Also, it only describes the MOL V2000 format, which is what you sample file is (based on the header); there is a newer V3000 format.


Strontium-89 vs strontium-90 Why is radioactive strontium-89 is considered medically useful for bones (along with other benefits) whereas strontium-90 (also being radioactive) is harmful for human body?

How does the addition of one atomic mass has adverse effects on human body?

The half-life of $\ce{^{89}Sr}$ is 50 days compared to a half-life of 29 years for $\ce{^{90}Sr}$. So a large enough dose of $\ce{^{90}Sr}$ to be useful would leave the patient radioactive for the rest of his/her life.

Edited for use of $\ce{^{89}Sr}$. In a comment Karl pointed out that the isotope is used to treat bone cancer.


What is more reactive towards a cyanohydrin formation an aldehyde or an Ketone? Which is more reactive—aldehydes or ketones—towards the nucleophilic addition by cyanide to form cyanohydrins?

Aldehydes will generally form cyanohydrins more readily than ketones, but some ketones such as trifluoromethyl or acyl ketones may behave more like aldehydes due to the electron deficiency at carbonyl carbon.


How is the numbering in phenylpropenes? Just wondering why the name of this compound is 3-phenylpropene? I understand the concept of phenyl groups but I'm confused where the 3 came from, because if it is referring to the double bond in propene, then shouldn't it be numbered 2?

Obviously, there are three carbons in propene to which a phenyl group might be attached. Let's have a look and name the compounds:

1 with the phenyl group at C-2 of propene is 2-phenylpropene, also known as $\alpha$-methylstyrene.

2 with the phenyl group at C-1 of propene is 1-phenylpropene. To be exact, the compound drawn is (E)-1-phenylpropene. E is for entgegen, that is on opposite sides of the double bond.

3 is 3-phenylpropene, also known as allylbenzene.


Why does the displacement reaction of zinc in copper(II) sulfate solution result in dark metallic copper? It is commonly known that when zinc metal is placed in a solution of copper(II) sulfate, a displacement reaction occurs, and elemental copper is deposited onto the decomposing zinc metal. A practical demonstration of this shows the zinc turning black from the deposited copper.

My question is, why is the deposited copper black? I thought metallic copper was red. Moreover, the dark color can be seen instantaneously, meaning the color cannot result from copper oxidizing.

I recently read a paper that claimed the copper is deposited in the form of plasmonic nanoparticles, and said nanoparticles display excellent solar absorption (and low thermal emittance, in fact). Any ideas?

All metals look black or dark when presented as a sufficiently thin powder. The metals display their color only when the dimension of the grain is greater than the wavelength of the light, namely 400 to 700 nm. This is about 1000 times bigger than the atoms. When prepared from a precipitation reaction, the metal atoms are never arranged neatly in a regular fashion. So they do not reflect the light like a metallic plate.


Does more acidity = better electrolyte? I am trying to figure out what makes an electrolyte better than another. I originally thought it was acidity, but a potato produces more voltage than a lemon. When I tried this same thing with acetic and muriatic acids, I found that acetic acid produced a constant voltage of a little under 1, while the muriatic (hydrochloric) acid started at about .60v and rapidly dropped into a negative voltage. So if acidity is not an indicator of a good electrolyte, what is?

Essentially, 'how good of an electrolyte' something is depends on how many particles the substance dissociates into. Let us discuss acidity. Now, from experimental evidence, it may seem that 'stronger' acids and mixtures, such as HCl, $\ce{H2SO4}$, Muriatic, are all good electrolytes. These are not because of their relative acidities. You will notice that in my last sentence, I put 'stronger' in quotes. This is because when talking about acids, 'stronger' has a different meaning than the one you are used to. A strong acid is one that dissociates virtually 100%. In other words, when you put a strong acid in water, all of it turns into ions, and you have none of the original acid left over. Similarly, a weak acid is one that does not completely dissociate into ions. By the definition of an electrolyte, stronger acids are better electrolytes (more ions in solution). This leads to our next point, ions in solution. As mentioned by Zhe above, the 'Van't Hoff Factor' is important. Essentially, it is just a measure of how many particles a substance dissociates into, for example NaCl has a VHF of 2, whereas $\ce{CaCl2}$ has a VHF of 3. The higher a substances VHF, the better of an electrolyte it is, the reasoning being that with an increase in number, there is an increase in dissolved particles, and therefore an increase in the state of being an electrolyte. Now, going back to acids for a moment, you may be wondering which acids are strong. There are 6, and they are HCl, HBr, HI, $\ce{HClO4},\ \ce{H2SO4},\ \ce{HNO3}$. Weak acids are everything else. These strong acids will have the best electrolytic results if you use them as such, much better than lemon juice (citric acid) or vinegar (acetic acid). In your example, you used muriatic acid (HCl) as one of your ion-producers, which explains the rapid voltage change, occurring from the dissociation of a strong acid.

$$\text{In essence, the answer to your question is:}$$ $${\text{How many particles a substance dissociates into determines how good of an electrolyte it is.}}$$


What is the structure of a lipid and why is this example non-polar? Lipids usually consist of 3 fatty acids and 1 glycerol. I looked up what they look like

So how is it not a polar molecule?  The $\ce{-OH}$ side has negative charge, and the hydrocarbon side chain has a positive charge.

Lipids usually consist of 3 fatty acids and 1 glycerol.

That's wrong. You're giving the definition of fats, which are just a subclass of lipids.

The structure in your question is the skeleton of cholesterol - without the stereochemistry.


Synthetic route for the conversion of ethanol to but-1-yne How do I convert $\ce{CH3CH2OH}$ (ethanol) to $\ce{CH3CH2C#CH}$ (but-1-yne) ?

I'm not too good with reaction mechanisms and conversions are beyond me.

My Attempt:

I've searched through a couple of textbooks and notes and I don't see any direct way to do the reaction.

I think I can use thionyl chloride to to convert from alcohol to chloroalkane:
$$\ce{CH3CH2OH + SOCl2 -> CH3CH2Cl + HCl + SO_2\uparrow}$$
The $\ce{SO2}$ escapes but I'm not sure how I can get rid of the $\ce{HCl}$.

Assuming that I do get rid of it somehow, I can take the resulting chloroalkane and add an alkynyl group to it hence forming the required higher alkane.
$$\ce{CH3CH2Cl + C2HNa + NaCl -> CH3CH2C#CH + \dots}$$

Again, the mechanism of these reactions elude me. I've only figured out the above from randomly stated properties of reagents. A detailed method of the conversion would be helpful.

The first reaction is the conversion of the alcohol to an alkyl chloride using thionyl chloride. The mechanism of the reaction starts with nucleophilic attack of the alcohol oxygen on sulphur, followed by displacement of $\ce{Cl-}$. Chloride then substitutes the alkyl chlorosulfite group in an $\ce{S_{N}2}$ reaction, and the latter decomposes into $\ce{SO2}$ and $\ce{HCl}$.

(Image source)

In the second reaction, chloroethane is alkylated with sodium hydrogen acetylide to yield the desired product.

$$\ce{C2H5-Cl + HC#C^{-} ->~ C2H5-C#CH + Cl-}$$

This is a simple $\ce{S_{N}2}$ reaction, in which $\ce{Cl-}$ is substituted by $\ce{HC#C^{-}}$. Acetylide salts of alkali metals can be prepared by deprotonation of ethyne with a strong base, for example, with butyllithium in THF or sodium amide, which can be generated in situ by dissolving sodium metal in liquid ammonia.


Transesterification of propylene carbonate and methanol I'm going to try out the transesterification of propylene carbonate and methanol. I know that the mechanism is as follows :

-What are the products yielded after the reaction?

-And which catalyst should I use? (reaction is happening at normal air pressure and at room temperature)

I know it is used to obtain  dimethyl carbonate (DMC) I think the others products should vary from which catalyst you choose. Have a look to Transesterification of propylene carbonate with methanol using $\ce{MgAlCO3}$
hydrotalcite as solid base catalyst: Indian Journal of Chemistry
Vol. 49A, September 2010, pp. 1182-1188 to choose the catalyst. I think there are two little typos in Scheme 2 where the mechanism is shown: miss $\ce{^-}$ from $\ce{CH_3O^-}$ (first step) and miss $\ce{C}$ from one part ($\ce{CH_3CO}$) of DMC formula (second step) but is a good article.


Why do we use coefficients for RFM (relative formula mass) in atom economy? Usually when calculating using moles = mass / RFM, the coefficient is not multiplied by a compound's RFM to calculate RFM.

But in calculations for atom economy, the (balanced) coefficient of a compound is used in calculating RFM. Why is this the case?

Before we come to your actual question, we should clarify a few other problems related to quantities and units in your expression:

  moles = mass / RFM

It is essential to distinguish between quantities and units. For example, density is defined as “mass per volume” and not “mass per litre”. In your expression “moles” is a unit name; but it should actually be the quantity “amount of substance”:

  amount of substance =  mass / RFM

Names of quantities (such as “amount of substance” and “mass”) shall not be arranged in the form of an equation. You should use the corresponding quantity symbols:

  n = m / RFM

It is not permissible to use abbreviations (such as “RFM”) for unit symbols (“Mr”) or unit names (“relative molecular mass”).

  n = m / Mr

There shall be a space on both sides of most operators but not for the solidus “/”.

  n = m/Mr

And finally, your equation is wrong, since amount of substance is not given as mass per relative molecular mass. Remember that relative molecular mass is a dimensionless quantity. Actually, amount of substance is given as mass per molar mass:

  n = m/M

Now coming to your actual question.

Atom economy is defined as the quotient of relative molecular mass of the desired product by relative molecular mass of all reactants.

Thus, for a simple reaction

$$\ce{A + B -> Y + W}$$

where $\ce{Y}$ is the desired product and $\ce{W}$ is a by-product that becomes waste, the atom economy would be

$$x=\frac{M_\mathrm r(\ce{Y})}{M_\mathrm r(\ce{A})+M_\mathrm r(\ce{B})}$$

However, for a slightly different reaction

$$\ce{C + D -> 2 Y + V}$$

you would get two molecules of the desired product $\ce{Y}$ for each molecule of the reactant $\ce{C}$ and for each molecule of the reactant $\ce{D}$. The yield of product per used amount of reactants is twice as high; therefore, the atom economy is twice as high:

$$x=\frac{2M_\mathrm r(\ce{Y})}{M_\mathrm r(\ce{C})+M_\mathrm r(\ce{D})}$$

In the same way, any other coefficients in the chemical equation would have to be taken into account.


Isothermal process gives V = A(T)exp(-KP)? Substance with isothermal compressibility K, where $\mathrm{K = -\frac{1}V(∂V/∂P)_T}$
undergoes constant isothermal, mechanically reversible process from state ($P_1$, $V_1$) to ($P_2$, $V_1$) where V is molar volume.

(a) Starting with the definition of K, show that the path of the process is described by: $\mathrm{V = A(T)e^{-KP}}$

I don't know how to get A in terms of T, the furthest I can get is $\mathrm{A = V_1e^{KP_1}}$

$A$ is already an implicit function of $T$, so your answer is essentially correct.

Starting from the definition of $\kappa$ and integrating both sides with respect to $P$, we obtain $$-\int\kappa\,\mathrm{d}P = \int\frac{1}{V}\frac{\partial V}{\partial P}\,\mathrm{d}P,$$ which yields as solution $-\kappa P = \ln V + f(T)$, where $f(T)$ is our "constant" of integration. Rearranging and exponentiating yields $V(P,T) = e^{-f(T)}e^{-\kappa P} = A(T)e^{-\kappa P}$, $A(T) := e^{-f(T)}$, and we are done.


How do I model non-covalent interactions in Gaussian? I'm doing some research about the dyes found in photography, and found Organic Chemistry of Photography
by Shinsaku Fujita, which shows the dyes used. I'm looking at doing an excited states and UV-VIS calculation, of course optimized before with a geometry optimization, with some of the dyes presented in the book. Right now, I'm trying to build number one in the image below, but have come upon a bit of a stumbling block.

The problem is the $\ce{(CH2)3SO2O-}$ and $\ce{(C2H5)3NH+}$ part on the rightmost nitrogen, as I'm not sure how to model the interaction between $\ce{(CH2)3SO2O-}$ and  $\ce{(C2H5)3NH+}$ (triethylamine), as both have no more covalent bonds available.

To visualize the problem, I built out two molecules (everything but triethylamine as one, then triethylamine below). As you can see, both are connected with as many covalent bonds as possible.

How can I model the interaction between the two molecules (pictured above) in Gaussian, for UV/Vis calculations?

As hBy2Py, Greg, and others have pointed out in their comments, the cation is just a counter ion for your benzothiazole-based cyanine dyes.

I suggest to be reasonably lazy and perform the calculations on the cyanine dye only.

Even for calculations in solvents of sufficient polarity (from an experimental photochemical perspective, acetonitrile seems to be a good choice), I would just ignore the cations.


Do enolates get protonated at the carbon or oxygen atom? An enolate has two resonance structures. When it gets protonated, where is the $\ce{H+}$ more likely to go? On the oxygen to produce an enol, which tautomerizes to the keto, or on directly onto the carbon (no enol intermediate). I know that both lead to the same product, but when drawing the mechanism this would be good to know.

Like all ions in solution, the enolate does not exist on its own in free form. It will be solvated with maximum efficiency. Depending on the solvent itself, the counter ion might be further or closer away, but charge balance needs to be maintained in the bigger picture.

Contrary to popular belief, the enolate ion is nothing that truly exists in non-gaseous phase. Therefore the question what gets protonated first turns into the question: Does the oxgen or the carbon have the stronger interaction with the solvent? This is quite hard to judge and very dependent on the whole system. there is certainly no general rule for it and as previously stated by bon, it does not really matter, because keto-enol tautomerism is fast and the ratio is entirely determined by the relative stabilities to the involved species.

There is another issue when talking about protonation. The proton itself is not free, either. In an aqueous solution the smallest model would be the hydronium ion. But this is far from being complete either. A whole branch of computational and theoretical chemists work on describing water clusters with models. It is extensive. One of the reasons for it being hard to comprehend is the high flexibility and mobility of protons even when bound to other species. The high conductivity of an acidic or basic solution is mainly governed by the Grotthuss mechanism. This kind of mechanism can in first approximation also be applied to keto-enol tautomerism.

But before we dig deeper into any mechanism, I would like to look at the properties bon outlined for the enolate. The following results were obtained at the DF-BP86/def2-SVP level of theory for the deprotonated butan-2-one resonance hybrid.
Let's look at the obvious first: resonance. With the help of the natural bond orbital theory (NBO) I performed the analysis. The first and most obvious thing is that it finds a so called hyper-bond between the oxygen and the two carbons with an occupancy of 3.95 electrons. As we will see in the refined analysis, this is in favour of the carbon-carbon bond (59.5/40.5 considering only these three atoms).
A more refined analysis, I chose a cut-off of 5%, include secondary resonance structures, seven in total, gives us the following weights for the major contributors:

From these structures we can support the highlighted arguments in the statement by bon:

  Attack on an enolate by an electrophile (in this case just a proton) is governed by the balance between electrostatic interactions, which are strongest at the oxygen atom due to its greater negative charge, and orbital interactions, which tend to be stronger at the carbon atom because the HOMO of the enolate has a greatest contribution at this atom [carbon].

The predominant contributor is is the enolate, hence in the overall description also its negative charge should be predominant. The π-system is certainly part of the HOMO in any case. Since it is mainly localised over the carbons, the highest contribution should be there.

We can further support that by looking at the natural charges of this model. More prominent is it looking at the charges obtained in terms of the quantum theory of atoms (QTAIM) in parenthesis.

The second part can be cleared up by looking at the HOMO; on this level of theory, the major contribution is C3-p of 56% and C2-p of 32%. The contribution of O1-p is below the threshold of printing, hence it is <10%. The major contribution of O1-p is found in HOMO -2 with 44% and a contribution of C2-p of 35%. It certainly looks a bit different in the picture. The contour value is set to 0.04.

All these calculations are only evaluations of the ion in the gas phase, so they can serve as an example to understand the true nature a bit more. It is in no way intended to be complete, it shall only provide a more concise educated guess of the general nature.

Now that we have a lot of information on the enolate itself, it's time to look towards a mechanism of a solvent-mediated keto-enol tautomerism. I have chosen the most simplest model possible, with one water (hydronium) explicitly included. The result is a six-membered cyclic transition state, where a proton is transferred from the oxygen of the enol to the oxygen of water, and one proton is transferred from water to the carbon of the double bond.
In theory there might well be many more water molecules involved. In principle any protic solvent is capable of doing this. The principle reaction path is the same for a base mediated mechanism, for example with ammonia.

If you look closely enough, I have borrowed heavily from the Grotthuss mechanism. A somewhat simultaneous transfer of protons. It needs to be acknowledged, that this is only an approximation of a stationary system. The animation only shows the transitional vibration mode (imaginary mode). It is not the calculated structural change or reaction path.
Unfortunately the calculations even on this low level of theory take quite some time. Currently I have no barrier height to report as an estimate, how fast this reaction would be approximately. I'll update this as soon as I can.

There is of course the possibility of a non-mediated transition state. This one, however, proceeds through a four-membered cyclic transition state. This one is highly strained, if you like the concept of ring strain, and seems unlikely compared to the above in any other medium than the gas phase.
If you are repulsed by the concept of ring strain, then you might want to look at it from a different angle. The deformation energy is high, since the carbon-carbon-oxygen is forced into a near 90° angle, hence increasing the p-content of the bonding orbitals. Higher p-orbital content means higher energy of the molecule. Therefore the activation energy will also be higher.

For a deeper insight into this, I recommend “Ab Initio Molecular Dynamics Study of the Keto–Enol Tautomerism of Acetone in Solution” (Clotilde S. Cucinotta, Alice Ruini, Alessandra Catellani, András Stirling, Chem. Phys. Chem. 2006, 7 (6), 1229-1234), which is available at researchgate.net.


How does B2Cl4 have two perpendicular c2' axes? I have been looking at the symmetry of $\ce{B2Cl4}$ and was cannot understand how it has two perpendicular $C_2'$ axes?

I understand it has a $C_2(z)$, $S_4(z)$ and two dihedral planes bisecting each of the $\ce{Cl-B-Cl}$ bonds. However my book says it also has an additional two $C_2'$ axes in the $x$ and $y$ planes centred at the middle of the $\ce{B-B}$ bond. I just can't see it. Any help? ( I understand it will be difficult online but I would very much appreciate your assistance.

Diboron tetrachloride has the same symmetry as allene, which I will use to demonstrate the other two $\ce{C_2^{'}}$ axis. The back $\ce{CH_2}$ group is turned 90° in comparison to the front $\ce{CH_2}$ group. This conformation is most stable as it minimizes steric interactions between both groups.

In this case, a picture tells more than I thousand words I think.

The two $\ce{C_2^{'}}$ axis are turned by 45° in comparison to the C-H bonds. It therefore also has two dihedral planes, thus belonging to the $\ce{D_{nd}}$ point group.

If you have difficulties picturing this in your head that is not strange. Many people make mistakes in assigning point groups. You get a better visualization by practicing online.


Why is the ionic radius of hydride so large? The order of ionic radii for halides and hydride is apparently as follows:
$$\ce{F-} < \ce{Cl-} < \ce{Br-} < \ce{H-} < \ce{I-}$$
Why is the hydride ion so large, even larger than bromide which has filled 3d and 4p subshells? Some claim that it is because the ratio of positive to negative charge is 0.5, which is an unusually small ratio. But this explanation doesn't fully convince me.
Nat. Commun. 2014, 5 (1), No. 3515 also makes mention of the unusually large size:

The ionic radius of an $\ce{H-}$ ion is $\pu{134\pm 2 pm}$, which is slightly larger than the radius of $\ce{F-}$(II) ($\pu{128.5 pm}$) and almost the same as the radius of $\ce{O^2-}$(II) ($\pu{135 pm}$). This evaluation agrees with the generally accepted trend that the radius of an H− ion is nearly the same or slightly larger than that of an $\ce{F-}$ ion and increases with increasing electropositivity of the coordinating cation.

It is not.
Presenting the order of ionic radii like that is useless.
Ionic radius of a given ion is not constant and mainly rely on crystal structure (coordination environment, more precisely) and used experimental techniques (crystal structure determination and theoretical calculations).
As a consequence, you can only compare homogeneous data.
Proposing the ordering based on some cherry-picked values from random sources or their averaged values is pointless at best.
This very issue is discussed by Messer [1]:

The effective ionic radius of fluoride ion is given as 1.33 Å by Pauling (8) and 1.36 Å, by Zachariasen (9), for coordination number 6. The effective ionic radius of hydride ion varies from 1.27 to 1.52 Å (10), being much more highly sensitive to the particular cation present and to the particular assumptions used in deriving the radius than the fluoride value. For the alkali metal hydrides other than lithium hydride, with the same face-centered cubic structure as the corresponding fluorides, the effective hydride radius is 1.47 Å for $\ce{Na+},$ and 1.52–1.54 Å for $\ce{K+},$ $\ce{Rb+},$ and $\ce{Cs+}.$ For $\ce{LiH},$ $\ce{CaH},$ and $\ce{BaH},$ the hydride ion radius is 1.34–1.36 Å. The closest similarities between hydrides and fluorides should thus appear here.

You can find the order of ionic radii for group 1 hydrides and halides in a more recent work by Lang and Smith [2]:

$$\ce{Li+} < \ce{H-} < \ce{Na+} < \ce{F-} < \ce{K+} < \ce{Cl-} < \ce{Rb+} < \ce{Br-} < \ce{Cs+} < \ce{I-}$$

Again, note that this doesn't only contradict your order, but it is no universal sequence and is only true for these compounds with soft-sphere model used for calculation.
References

Messer, C. E. Hydrides versus Fluorides: Structural Comparisons. Journal of Solid State Chemistry 1970, 2 (2), 144–155. DOI: 10.1016/0022-4596(70)90062-9.
Lang, P. F.; Smith, B. C. Ionic Radii for Group 1 and Group 2 Halide, Hydride, Fluoride, Oxide, Sulfide, Selenide and Telluride Crystals. Dalton Trans. 2010, 39 (33), 7786. DOI: 10.1039/c0dt00401d.


If the pH is zero, what would be the normality of a sulphuric acid solution? If pH of the solution is zero, what would be the normality of $250 \, \mathrm{ml}$ $\ce{H2SO4}$ solution?

My work:
$\mathrm{pH} = 0$, hence the concentration of $\ce{H+}$ ions is $\pu{1 mol L^-1}$, which implies that the molarity of the solution is $\pu{1M}$ hence the normality of the solution must be $2~\mathrm{N}$.  Is this correct?

$\ce{H2SO4}$ has 2 $\mathrm pK_\mathrm a$s, one that is strong whereby $\ce{H2SO4}$ ionizes to $\ce{HSO4-}$ and $\ce{H+}$ and another that is weak whereby $\ce{HSO4-}$ ionizies to $\ce{SO4^{2-}}$ and $\ce{H+}$

At $\mathrm p[\ce{H+}] = 0, [\ce{H+}] = \pu{1M}$

At $\mathrm {pH} = 0$, sulfuric acid is substantially $\ce{HSO4-}$ and $\ce{H+}$, with little $\ce{H2SO4}$ or  $\ce{SO4^{2-}}$.

Therefore at $\mathrm p[\ce{H+}] = 0$, sulfuric acid concentration is $\pu{1M}$, because each mole of sulfuric acid has released one mole of $\ce{H+}$.

Normality of an acid is the concentration of the acid times the number of acidic protons per molecule.  Sulfuric acid has two acidic protons per molecule.

Normality $= \pu{1M} \times 2 = \pu{2N}$


Calculating partial pressure of carbon monoxide $\ce{ZnO}$ is exposed to pure $\ce{CO}$ at $\pu{1300 K}$ and the equilibrium
$$\ce{ZnO(s) + CO(g) <=> Zn(g) + CO2(g)}$$
is then established at $\pu{1 atm}$ pressure. The density of the gas mixture is $\pu{0.344 g L-1}$ at the same temperature. Calculate the partial pressure of $\ce{CO(g)}$ at equilibrium (atomic weight of $\ce{Zn}$ is $65.4).$

My attempt
Since at equilibrium the amounts of $\ce{CO}$, $\ce{Zn}$ and $\ce{CO2}$ are equal (as no information about initial amounts are given), mole fraction of $\ce{CO}$ is 1/3.
Since pressure is given as $\pu{1 atm},$ the partial pressure of $\ce{CO}$ is 1/3 atm. Where I am wrong?

For simplicity, let's assign numerical indices to the compounds of interest — all gaseous products participating in equilibrium:
$$\ce{ZnO(s) + \underset{1}{CO(g)} <=> \underset{2}{Zn(g)} + \underset{3}{CO2(g)}}$$
Partial pressure of carbon monoxide can be found via its mole fraction $x_1$ and given total pressure $p$:
$$p_1 = x_1p\tag{1}$$
To find unknown mole fraction, two facts can be taken into consideration:

As Karsten also pointed out, the molar ratio between the products is $1:1$ so that $x_2 = x_3$. Which gives
$$x_1 + x_2 + x_3 = 1 \quad\implies\quad x_2 = x_3 = \frac{1 - x_1}{2}\label{eqn:2}\tag{2}$$

All three mole fractions are linked together in the expression for the average molar mass:
$$\bar{M} = x_1M_1 + x_2M_2 + x_3M_3\label{eqn:3}\tag{3}$$

The average molar mass can be found via average density of the gaseous mix $\bar{ρ}$ by applying ideal gas law:
$$pV = nRT$$
$$p = \frac{nRT}{V} = \frac{mRT}{\bar{M}V} = \frac{\bar{ρ}RT}{\bar{M}}$$
$$
\begin{align}
\bar{M} &= \frac{\bar{ρ}RT}{p} \\
  &= \frac{\pu{0.344 g L-1} × \pu{0.082 L atm K-1 mol-1} × \pu{1300 K}}{\pu{1 atm}} \\
  &\approx \pu{36.67 g mol-1}\tag{4}
\end{align}
$$
Now, knowing numerical value for $\bar{M},$ we can combine \eqref{eqn:2} and \eqref{eqn:3}:
$$\bar{M} = x_1M_1 + \frac{1 - x_1}{2}M_2 + \frac{1 - x_1}{2}M_3$$
Grouping the like terms and expressing $x_1$:
$$
\begin{align}
x_1 &= \frac{M_2 + M_3 - 2\bar{M}}{M_2 + M_3 - 2M_1} \\
  &= \frac{\pu{44.01 g mol-1} + \pu{65.38 g mol-1} -2×\pu{36.67 g mol-1}}{\pu{44.01 g mol-1} + \pu{65.38 g mol-1} -2×\pu{28.01 g mol-1}} \\
  &\approx 0.67\tag{5}
\end{align}
$$
Finally, the partial pressure of carbon monoxide:
$$p_1 = 0.67×\pu{1 atm} = \pu{0.67 atm}$$


Will adding up protons and electrons (without neutrons) create a new element? If protons have a positive charge and electrons have a negative charge,
can we add up several protons and electrons together to create a new element, without adding neutrons to hold the nuclei together?

Yes and no.

Elements are defined by the number of protons only. It does not matter if (say) a carbon nucleus has six or seven (or eight) neutrons, they will all react the same.*

With that, to create new elements, you would need to get up to some 115 or so protons fused together. But there is a reason for neutrons: all the positively charged protons in the nucleus repel each other electrostatically so neutrons are there to stabilise the nucleus — you can think of it as cushioning protons apart. For each element there is a set number of neutrons that will create a stable nucleus. E.g. for hydrogen (one proton) it is zero or one, for oxygen it is eight, for carbon six or seven, for tin it can be $62, 64, 65, 66, 67, 68, 69, 70, 72$ or $74$.

Nuclei with other neutron numbers decay radioactively on half-life scales from femtoseconds to millions of years. I would expect no-neutron nuclei (non-hydrogen) to be on the very short edge of that spectrum (although I’m no radiochemist).

*: There are reactions in certain systems that depend a lot on hydrogen being $\ce{^1H}$ rather than $\ce{^2H}$ or $\ce{D}$ because they rely on the tunneling effect whose probability decreases with mass (I think by the factor $m^2$). That’s also why one shouldn’t drink $\ce{D2O}$


Products of Reimer–Tiemann reaction of 4-methylphenol p-cresol (4-methylphenol), undergoes the Reimer–Tiemann reaction upon treatment with chloroform ($\ce{CHCl3}$) in alkaline medium. Which of the following is the major and minor product?

  A. 2-hydroxy-5-methylbenzene-1,3-dicarbaldehyde
  B. 2-hydroxy-5-methylbenzaldehyde
  C. 4-(dichloromethyl)-4-methylcyclohexa-2,5-dien-1-one
  D. 4-(dichloromethyl)-4-methylcyclohexa-2,5-dien-1-ol

The answer given is that B is the major product, and C the minor product.

I have two questions about this:

Why is the mono-aldehyde B the major product, and not the dialdehyde A?
I think both electron-donating groups should greatly increase the reactivity of benzene, allowing the reaction to take place twice. On the other hand, steric factors would oppose the second formylation. I do not know which one predominates.
How can one derive the formation of the minor dearomatised product C?

The Reimer-Tiemann reaction is known as the ortho-formylation of phenols under strongly alkaline conditions.

It's not an addition of a carbene to a double bond, like the Simmons-Smith reaction, the pyrrole to pyridine ring enlargement or the Buchner expansion, which is catalyzed by transition metals.

As ron already pointed out, pH is one crucial requirement, having a phenol (and not just any other arene) is the other.

What we're actually facing here is the addition of a carbanion to a strongly electron-deficient carbene.

As again already described by ron, there is significant charge density at the 4-position of the phenolate, and the initial addition step might indeed happen here too.

The resulting intermediate can't rearomatize other than in a back reaction and that's what it will mainly do! Only a minor part will grab a proton and yield C.

That was close, but no cigar! Let's have a look at the addition at the 2-position.

Through the interplay of dienone, phenolate and phenol we get rid of one chloride and all further steps to the final benzaldehyde are likewise assisted by this tautomerism.

We still haven't answered the question why no 1,3-dialdehyde is formed. Let's have a look at the major product B under alkaline conditions and its resonance structures.

One of them looks like the enolate of a 1,3-dicarbonyl conpound, with additional charge distribution over the $\pi$ system.

This species has a significantly lower localized electron density than the anion of the starting material. So, the party stops here.

Edit In a comment, it was asked whether the geminal dichloride will convert to the corresponding aldehyde under alkaline conditions.

Such a transformation could involve a nucleophilic substitution, leading to a geminal halohydrine which then eliminates HCl to furnish the aldehyde.

In the case of benzylic dihalides, this is possible even without the intramolecular assistance of a phenolate (cf. this example) using secondary amines as a base and in the case of diphenyldichloromethane, the transformation to benzophenone is achieved by heating in the presence of water.

In the case of C, the simple deprotonation of the dihalide might be a dead-end alternative pathway to substitution.


How is Melatonin Synthesised? I should like to know roughly how the hormone melotonin is synthetically made for the pharmaceutical industry (i.e. not how it is made in living beings). I can't seem to find any reference to its manufacture on the web, it is seen to have a fairly simple structure on its Wiki page and biologically it is synthesised from the forerunner amino acid tryptophan. So I believe it is likely to be readily synthesised for use in pharmaceuticals without the use of animal products, but I cannot confirm this.

Specifically, a close relative of mine has been prescribed melotonin and I want to confirm that this substance is not made from animal products, as I live in a country where the authorities in their great wisdom have deregulated the food and meat processing industries. Astoundingly, they did this at exactly the same time that a UK royal commission recommended re-regulation of the same to prevent contamination of food and pharmaceuticals by the Creutzfeld-Jacob prion.

I do not know the exact process in industry, but these are the steps that I thought of (someone likely thought of this before me, it is a very simple synthesis).  The main synthesis of the ring is done in Fischer indole conditions.  Therefore I broke up the molecule into a compound known as 4-acetamidobutanal and p-hydrazinoanisole.  The first compound is an amide that is available from the reaction of 4-aminobutanal and acetyl chloride or acetic anhydride in the presence of a base.  The latter compound is available from the reaction of p-anisidine with sodium nitrite and hydrochloric acid to form the diazonium chloride, and then reduction with sodium sulfite and sodium hydroxide.  This is just an example of a synthesis.  The overall synthesis shouldn't be more than 5 or 6 steps.

I found an article highlighting the syntheses, you can take a look.  Apparently the synthesis that I described above is very similar to the most effective synthesis, a one-pot synthesis done in very similar conditions and having similar starting materials.


How can electron move between two metals without the presence of electric field? Electron can move from one location to another if it feels a force that is if there is a presence of electric field. Now suppose I have a $\ce{Zn}$ rod immersed in $\ce{CuSO4}$ solution. Electrons are transferred from $\ce{Zn}$ to $\ce{Cu}$. What results in this motion of electrons from $\ce{Zn}$ to $\ce{Cu}$. How can electron move without a field?

The wording of the question hides the answer.

Electrons do not move from Zn to Cu. They move from neutral zinc to cupric ion Cu++. The field arises from the mix of H2O, H+, OH-, SO4--, and Cu++ all around the Zn rod. Could electrons be forced onto Zn from OH- or SO4-- ? No.There is no gain in stability; energy would be needed. Or could electrons be attracted to H+; yes, and even more so to Cu++. And the final products (Zn++ and neutral Cu) would be more stable.


To understand logic and relation of biochemical cycles I recenty learned the Kreb cycle, in that I saw many chemical are same.

I want to know if there is logic/threrom/technique in which I can write a chemical  cycle to get a product i.e I want basic reactions and how to put it in sequence to make a cycle ?

tl;dr: You can't reasonably 'propose' a biochemical pathway on paper the same way you might be able to for an organic reaction.

Biochemical pathways are highly complex and interwoven. The Krebs cycle doesn't act in isolation, but rather is part of a far larger network which provides it with the components it needs (enzymes, cofactors, oxidatants/reductants etc.) and the regulation required to ensure it works correctly.

The figure below demonstrates a more elaborated Krebs cycle, complete with all of the relevant interactions with various other pathways:[1]

Given their complexity, the discovery of the true nature of a pathway takes a long time, and uncertainty often remains due to the difficulty in directly observing the reactions taking place (though with modern genomics/sequencing its becoming easier to see what might be taking place).

In the case of the Krebs cycle, the key discovery was made by Gyorgyi who won the 1937 Nobel Prize in Physiology for 'his discoveries in connection with the biological combustion processes, with special reference to vitamin C and the catalysis of fumaric acid'.

Using isolated cells from pigeon breast tissue, Gyorgyi was able to identify the importance of succinic acid and fumaric acid in the pathway, noting that the rate of respiration was increased significantly if the tissue was incubated in their presence (as measured by gas evolution, which was already known at the time to be a consequence of respiration).

Building on this discovery, Krebs published the first full account of cellular respiration in his 1937 paper Metabolism of Ketonic acids in animal tissues.[2] In this paper he outlines all of the key steps needed, but even at this stage many of the important details were missing (many of the other necessary pathways weren't known, so Krebs had no idea where the succinate was coming from in the body, for example).

[1]: Gerhard, M.; Schomburg, D. Biochemical Pathways- An Atlas of Biochemistry and Molecular Biology; Wiley: New Jersey, 2012

[2]: Krebs, H. A.; Johnson, W. A. Biochem. J. 1937, 31, 645


Laboratory setup for the distillation of sulfuric acid Purification of Laboratory Chemicals, 5th Edition, by Armarego, reads (p. 53):

Reagents such as water, ammonia, hydrochloric acid, nitric acid, perchloric acid, and sulfuric acid can be purified via distillation (preferably under reduced pressure and particularly with perchloric acid) using an all-glass still.

as well as (p. 479):

Sulfuric acid, and also 30% fuming $\ce{H2SO4}$, can be distilled in an all-Pyrex system, optionally from potassium persulfate. Also purified by fractional crystn of the monohydrate from the liquid. Dehydrates and attacks skin—wash immediately with $\ce{H2O}$.

I've never seen sulfuric acid fume (only spread with steam when heated), and Wikipedia states a boiling point of $\pu{337 ^\circ C}$, along the fact that it slowly decomposes to $\ce{SO3 + H2O}$ above $\pu{330 ^\circ C}$, so I'm wondering, mostly from a theoretical standpoint:

How far lower can the boiling temperature effectively get under reasonably strong vacuum?

Does the vacuum not also bring down the temperature level at which sulfuric acid degrades?

What would be an ideal laboratory (not industrial!) setup and process for the safe distillation of sulfuric acid in fair enough amounts for laboratory use (say producing in a batch the equivalent of $\pu{5L}$ at 30% concentration)?

Nota Bene: I know it's a better idea to buy sulfuric acid at the desired level of purity, but that is not the question.

Just on the purification of sulphuric acid, well the one he gets has a dark green tinge, and what is in the sulphuric acid is anticorrosive inhibitors, etc, and they are all organic in nature, and the simple heating up a beaker of sulphuric acid, and before it gets to hot, slowly add 100 ml of 3% grocery store bought food grade hydrogen peroxide, and stir whilst slowly heating, well before long the acid will be crystal clear, any fumes are the skeleton structures of the organic additives, and concentrating it back to 98% sulphuric acid is either density determined, temperature, to insure water has been removed hence both will get a much purer, crystal clear sulphuric acid. Anything organic will be consumed by the peroxide and what doesn't will turn into a gaseous product that evaporates of. The product described has served his associates and himself, for many years.


How to synthesize 4‐[(4,4‐dimethylpentyl)oxy]‐2,2,8,8‐tetramethylnonane starting with 4,4‐dimethylpent‐1‐ene? <img src="https://i.stack.imgur.com/FBDDt.jpg" alt="4,4‐dimethylpent‐1‐ene to 4‐[(4,4‐dimethylpentyl)oxy]‐2,2,8,8‐
tetramethylnonane" />

My attempt:

I don’t know how to add another chain. I think I probably need to react with another A molecule, but I don’t know the mechanism. Can anyone help me with this?

Divide your alkene starting material in two portions - one twice the amount of the other.
Take the larger portion and do an anti-Markovnikov addition of HBr to it using peroxide or light reference to give the terminal bromide (1-Bromo-4,4-dimethylpentane).
Take the smaller portion and ozonolyse it, working up with Dimethyl Sulfide to give the aldehyde (3,3-dimethyl butanal).
Take half the bromide material and form the Grignard with Mg metal. Add the Grignard to the aldehyde. Work up and isolate the resulting secondary alcohol. Dissolve the secondary alcohol in DMF at 0c and form the alkoxide with NaH, add the second portion of the bromide to the alkoxide solution. Work up, isolate and purify to obtain the desired product.


Do microwaves heat polar molecules or molecules with polar bonds? IR is only absorbed my molecules with polar bonds
regardless of the overall polarity of the molecule , what about Microwaves ?

It depends on the strength of the incident EMF. Pure $\ce{H2}$ can be ionized by microwave radiation in multi-photon collisions, as well as being heated. Even nonpolar molecules have quantum fluctuations in the electron cloud, so they interact to some extent with n electromagnetic field.


Why do unbonded electrons exist in pairs? Basically the term which we use to refer them is lone pair. In Lewis structure why we represent those unbonded electron in pairs. Like here (structure of SO2)

Here if we assume both the unbonded electrons to be existing in pairs then it's geometry will be different from what we would observe in case when both electrons exists freely (without pair).

So why we basically consider those electrons to be existing in pair? Is it any kind of observation or any theory or both?

A single unpaired electron is called a radical, and usually aren't stable. Each electron orbital can hold two electrons, one spin up, other spin down. So a lone pair is just a full electron orbital, and cannot accept another electron. An orbital with 1 electron will readily react to get another electron and form a pair.

Check this: Free Radical

The first reaction they show is UV light being used to break $\mathrm{Cl}_2$ gas into $2\mathrm{Cl}$ radicals, each with a single lone electron, forming free radicals. These radicals could be used as free radical polymerization initiators, or for other reactions.

In the structure you show, the molecule forms a trigonal planar configuration, with the lone pair forming the third arm. If you could stably achieve two free radicals instead of the lone pair, it would have tetrahedral configuration, with the radicals forming 2 of the 4 arms.

The theory behind it is more quantum mechanics than I can remember, but I know it involves the Pauli exclusion principle and Hund's rule.


Why is it called "carbonation"? Why is it referred to as "carbonation" and we drink "carbonated" beverages when carbonate is $\ce{CO3}$ while $\ce{CO2}$ (carbonite?) is present in carbonation?

Nope. Carbonate is $\ce{CO}_3^\color{red}{2-}$ (an ion), not $\ce{CO3}$. Carbonite is $\ce{CO}_2^\color{red}{2-}$.

Carbonation involves dissolving $\ce{CO2}$ gas in water. It turns out that $\ce{CO2}$ reacts (maybe not the best term to use here*) with water via the following reaction $$\ce{CO2 + H2O -> CO3^{2-} + 2H+}$$

So you have effectively carbonated the soft drink.

When you depressurize the bottle by opening it, the reverse reaction occurs and you get carbon dioxide.

*Such dissociation is normal and integral to dissolution of polar solutes in water, so generally it is considered as part of the dissolution


Convention for numbering of carbons I'm trying to understand how carbons are numbered in organic molecules (i.e. as in the following image).

(source: Principles of Biochemistry by Nelson and Cox, p. 532)

I've read and heard different methods, among which:

the end of the carbon chain with the highest oxidation state is assigned the lowest number.
the end of the carbon chain with the highest total mass is assigned the lowest number. the number of bonds is used as a kind of multiplier (i.e. O=C-OH corresponds to a total mass of O*2+C+O+H). (source: What is the convention of numbering carbon atoms in organic molecules?)
the end of the carbon chain with the substituent of the highest atomic number is assigned the lowest number (source: Principles of Biochemistry by Nelson and Cox, p. 73).
the carbonyl, acetal or hemiacetal carbon is given the lowest possible number (source: Carbon numbering in carbohydrates)

These methods could come to contradictory conclusions, though. One example I could think of:

Phosphorous would be the highest atomic number constituent, while the remaining rules (carbonyl carbon, total mass and oxidation state) would assign the carbonyl carbon the lowest number.

I assume some are just simplifications. What is the correct method, or is there no single general one?

Your initial example showing derivatives of glucose and fructose might be misleading since the nomenclature of natural product parent compounds does not necessarily follow the usual nomenclature rules of general organic chemistry.

In particular,

  P-102.2.2 Numbering parent structures

  The carbon atoms of a monosaccharide are numbered consecutively in such a
  way that:

  (1) a (potential) aldehyde group receives the locant 1 (even if a more senior
  characteristic group is present);

  (…)

(This rule is probably easier to understand if you draw the monosaccharides in their acyclic form.)

Generally, however, nomenclature of organic chemistry has different rules for the numbering. The most important simplified criteria for the numbering are:

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

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

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (e) saturation/unsaturation:

  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

Therefore, the numbering in your own example corresponds to the name 3-phosphinopropanoic acid since the acid group is the principal characteristic group of this compound.


Formation of 2,4-Dichlorophenol by free chlorine mediated oxidation of Triclosan I cannot find much guidance in the literature. This is the reaction between free chlorine in water and triclosan.
I understand ether cleavage and electrophilic aromatic substitution (maybe) involved. Any ideas as to what the mechanism of this reaction might be?

I'm not sure if this qualifies as an answer since it's more of a guess. Catechols (1,2-dihydroxybenzene derivatives) can be oxidized to ortho-quinones. Applying that transformation here (chlorine in water is certainly oxidizing conditions) gives the oxonium of the ortho-quinone, which can be hydrolyzed to 2,4-DCP.


What's the difference between mass spectrometry, mass spectroscopy, and mass spectrography? It seems that the terms mass spectrometry, mass spectroscopy, and mass spectrography are mostly used like synonyms. Is this correct? Or is there a difference?

According to the Encyclopaedia Britannica, mass spectrographs use(d) photographic plates or films for the detection of ions, as in the apparatus thought out by F. W. Aston in 1919.


Equilibrium constant for esterification process Teacher told us that the concentration term for pure liquids and solids is always $1$. But online I found this reaction:

$$\ce{CH3COOH(l) + CH3CH2OH(l) <=> CH3COOCH2CH3(l) + H2O(l)}$$

According to this page on LibreTexts (search for "esterification" to quickly find it),

$$K_c = \frac{[\ce{CH3COOCH2CH3}][\ce{H2O}]}{[\ce{CH3COOH}][\ce{CH3CH2OH}]}$$

Shouldn't $K_c = 1$ as all reactants and products are liquids? The above expression even contains $[\ce{H2O}]$.

The key point is pure liquids. In context of the esterification equilibrium, it considers the reactant concentration in the common solvent - which may or may not be  one of the reactants.

Note that if the solvent is at the same time a reactant or a product, which concentration (activity) can be considered constant, then there can be defined the equilibrium constant for this solvent, with the solvent excluded from the equilibrium.

$$K_{aqua} = \frac{[\ce{CH3COOCH2CH3}]}{[\ce{CH3COOH}][\ce{CH3CH2OH}]}$$

$$K_{ethanol} = \frac{[\ce{CH3COOCH2CH3}][\ce{H2O}]}{[\ce{CH3COOH}]}$$

$$K_{ethylacetate} = \frac{[\ce{H2O}]}{[\ce{CH3COOH}][\ce{CH3CH2OH}]}$$

$$K_{acetic acid} = \frac{[\ce{CH3COOCH2CH3}][\ce{H2O}]}{[\ce{CH3CH2OH}]}$$


How can one find products when one knows the reagents? I'm going to use an example (I could have used a different one);
On the document it is about the study of the chemical balance of an acid-basic solution.
The chemists have written the chemical reaction equation to make a table of the evolution of pH during the reaction.
If I understand with this equation I'll be able to understand others.
$\ce{CH3COOH + H2O ->CH3COO- + H3O+}$

Here is what I already understand:

This is the equation of methanoic acid with water
I understood why the general formula of methanoic acid is $\ce{CH3COOH}$

What I want to know is how do we know that

$\ce{CH3COOH}$ changes into $\ce{CH3COO-}$ ?
$\ce{H2O}$ changes into $\ce{H3O+}$?

In general, how does one find the products when one knows the reagents?
Does reagents molecules always correspond the same ions?

If I went into specifics, this answer would span a few thousand books. Instead, I'll give an overview:
There are many ways of determining the products given the reagents:
Actually reacting the reactants
One can, of course, actually make the reaction take place. Once you've done that, you need to identify the contents of the resulting "soup". A few methods for this  (You may have to separate the products first) :
NMR Spectroscopy
Here, a nifty technique is used to determine the magnetic properties of a compound. From these properties, one can make reasonable guesses at the substructures existing in a molecule, and check against a list of possible products. This is a very powerful technique, and usually gives good results.
Qualitative analysis
Here, we use various chemical reagents on the products, and analyse these products, making conclusions about the nature of the original products.
Microscopy
Here, using an STM, one figures out the structure of the molecule. This isn't used that often, but is an interesting technique.
There are many other such techniques, with varying accuracies and applicabilities.
Predicting the products in the absence of experiment
A lot of chemical research is into improving its (already vast) predictive power. The end-game of most sciences, after all, is to be able to predict stuff. So, in many cases, we can just predict how the reaction proceeds, without actually making the reaction occur in a lab.
Figuring out a reaction mechanism
This generally occurs in organic reactions. Here, you break the reaction into elementary steps. Each step can be explained by looking at electronic attraction/repulsion, as well as the increase in stability (which can generally be estimated by taking into account the inductive, mesomeric, hyperconjugation, and steric effects, though other things may effect the overall stability as well). From these steps, one gets a proposed mechanism, which may be verified.
Knowing solubilities and other numerical values
In ionic/inorganic reactions, the path a reaction takes generally hinges on the solubility of its final products. This is sometimes the case for other parameters as well. Knowing various numerical values lets one compare two possible reaction paths
Computational chemistry
Here, one uses the equations of quantum mechanics (Schrodinger's equation and a few of its cronies), and solves them for the given problem using numerical methods.

The particular reaction you have given
This one can be easily explained by resonance. In $\ce{CH3COOH}$, the $\ce{H}$ is loosely bonded, as the $\ce{C=O}$ bond resonates with the $\ce{C-O}$ bond. This makes it easy for the $\ce{H}$ to float away as $\ce{H+}$, leaving behind a pretty stable $\ce{CH3COO-}$ (made stable by resonance of the negatively charged lone pair over the oxygen atoms).
So far, we know that $$\ce{CH3COOH -> CH3COO- + H+}$$
Now we know that it is behaving like an acid. It is extremely well known that water forms a Hydronium ($\ce{H3O+}$) ion by self-dissociation, and this is accelerated by making the medium acidic (All free $\ce{H+}$s find the nearest $\ce{H2O}$ molecule and form $\ce{H3O+}$). $\ce{H3O+}$ is stable due to resonance, again.
So we now know that $$\ce{H+ + H2O -> H3O+}$$
Put the two reactions together and you get $$\ce{CH3COOH + H2O -> H3O+ + CH3COO-}$$
Though, in this case, the historical method of finding out the reaction may have been different -- acetic acid has been well known and studied for a very long time.


Energy of photons of given frequency and number I’m having trouble with calculating the energy of $\pu{5.00E10}$ photons of frequency $\pu{1.00E9 s-1}$.
I know that to calculate the energy $E = hf$ is to be used. However, I am unaware as to what to do with the $\pu{5.00E10}$ value and its meaning.

Analogy: you have 5 bottles of coke 1 L each; what is the total volume?
The formula you shown $(E = hf)$ is derived for exactly one photon.
Now, since you have 50 billion of photons $(N = \pu{5.00e10}),$ guess what the total energy would be?

 $$E_\mathrm{tot} = Nhf = (\pu{5.00E10})(\pu{6.63E-34 J s})(\pu{1.00E9 s-1}) = \pu{3.32E-14 J}$$


What does a fractional order of reaction mean for the mechanism? I'm doing a high school/sixth form college investigation of the kinetics between magnesium ribbon and hydrochloric acid.

I have obtained a rate order was 1.5 w.r.t [H+] and hence the rate equation is rate = k[H+]1.5.

What would that mean about the mechanism?

I remember my teacher once telling me that a fractional rate order just means that there is an intermediate stage and each step has its own rate order which contributes to the rate equation - is this right?

Below is a pictorial representation of what my teacher told me:

Also, what would this mean for my rate-determining step? If the above is true, what would I choose as the RDS if they both contribute a different amount to the rate order?

Here's an example.

Let's consider a reaction of the generic form:

$$\ce{3A -> 2C}$$

Now, let's say we have some knowledge that this reaction proceeds by the following two-step mechanism:

  Step one is fast and reversible.

$$\ce{A <=>[k_1][k_{-1}] 2B}$$

  Step 2 is slow and irreversible.

$$\ce{A + B ->[k_2] C}$$

The rate of the reaction (expressed as the change in concentration of the product per unit time) can be expressed in terms of the concentrations of A and B.

$$\mathrm{rate}=\dfrac{\Delta [\ce{C}]}{\Delta t}=k_2[\ce{A}][\ce{B}]$$

However, we would prefer to not use intermediates in our rate law. Since the first step is fast and reversible, we can use the law of mass action to create a relationship between the concentrations of A and B:

$$K_1=\dfrac{k_{1}}{k_{-1}}=\dfrac{[\ce{B}]^2}{[\ce{A}]}$$
$$[\ce{B}]^2=K_1[\ce{A}]$$
$$[\ce{B}]=\sqrt{K_1[\ce{A}]}$$

If we substitute into the  rate law:

$$\mathrm{rate}=\dfrac{\Delta [\ce{C}]}{\Delta t}=k_2[\ce{A}]\sqrt{K_1[\ce{A}]} =k_2 K_1^{1/2}[\ce{A}]^{3/2}$$

The term $k_2 K_1^{1/2} =\dfrac{k_2 k_1^{1/2}}{k_{-1}^{1/2}}$ becomes $k_{\mathrm{obs}}$, the observed rate constant of the overall reaction. The fractional order comes from the second step being rate-determining, but the first step being an equilibrium with stoichiometry that generates the fractional tern.


Fast way to isolate 4-nitrophenol from 4-aminophenol I'm working in the lab with a very old (like 30 years old) sample of 4-nitrophenol, wich of course has been reduced with the time to 4-aminophenol. I need to isolate both reagents and work only with the nitrophenol, is there any quick way to do that?

I would use solvent extraction. Dissolve the mixture of phenols in an organic solvent. Next shake the organic solution with dilute hydrochloric acid in a separation funnel. Do this several times to wash all of the 4-hydroxyaniline out of the organic phase.

Next shake the organic phase with concentrated sodium chloride solution (brine) and then add anhydrous sodium sulfate. Shake the mixture up in a sealed conical flask. Now decant or filter it to give you a solution of 4-nitrophenol which you can now evapourate down to give you the phenol.

I would like to know how the nitrophenol was reduced into the amine over the years it was in storage. This is a more interesting question to my mind is how did the nitro compound get converted on storage into the amine.


Rate of Evaporation of solution Why does rate of evaporation of solution decreases on adding a non volatile component?

Also what happens if we add a volatile solute?

The rate of evaporation decreases because the activity of the solvent in the liquid phase decreases, while the activity in the vapor phase (or solid, for that matter) stays the same. This is also the reason for the increase in boiling point and the decrease in melting point of the solvent.

If we add a volatile solute, it will evaporate too, thus changing the solvent activity in the vapor phase as well. The resulting behavior can't be characterized universally; you'll have to consult the binary phase diagram of boiling for the particular system, and these can be pretty diverse.


Where did the INN name of fentanyl come from? Fentanyl is a potent anilidopiperidine opioid. Where did its INN name come from?

Figure 1: Fentanyl

I know that the INN name of paracetamol came from its chemical name para-acetylaminophenol (par-acet-am-ol). Is there a meaningful origin of the INN name of fentanyl?

My guess: the name may be based on two of the immediate precursors to the drug, N-Phenethyl-4-piperidone (FENT, from Phen + ethyl?) and aniline (ANYL):

Definitely not as straight-forward as your para-acetylaminophenol example.


How does salt push out the water from vegetables, meat, etc.? Where can I find out, how does this process work? Have you noticed, that when you salt mushrooms, eggplant... or add sugar to lemon - some "juice" comes out.

Any explanations? Or maybe name that reaction?

It is called "Osmosis" and the applications of this are many. Since you are a self-identified "noob of chemistry" I'll not go too far into the specifics of why this happens but try to explain on a laymans basis.

As you probably know, water is a solvent. It can dissolve (some) solids so that no solid remains, only liquid. This usually has some effect on the solute (water in this case). Lets say you have two glasses of water - one of which has some salt dissolved in it. Nature would very much prefer if they were at the same concentration of salt. Nature wishes for 2 glasses of water with $\frac{some}{2}$ salt in each.

So, when there is an apple, with water inside and some salt outside, nature would like there to be about the same concentration of salt inside and outside. So she (nature) will provide a driving force for taking water out of the apple and into the salt. The skin of the apple is a semi-permeable membrane: only water can pass. So nature will drive the water through the skin to the salt. And she will keep at it until the apple is so empty of water that it is so hard to find water in the apple that the forces are in balance.

The same thing applies to sugar. Sugar is also one of these compounds that can dissolve in water and will affect the water in the same way.

For a more detailed explanation I'd google "Osmosis". This interesting phenomenon is responsible for many things, wonderful and useful. You can actually pump water with it. Nature will "pull" water against the force of gravity in order to balance out the concentrations. (This is basically how a salt power plant works)


What chemical reaction can cause this blue-ish deposit? Recently I noticed that some of our glasses and cutlery had developed a blue-ish deposit as you can see for the two sets of spoons below. For each set the spoon on the left has the blue-ish deposit whereas the spoon on the right is the original.

I traced the cause of this blue-ish deposit to a specific effect. It only occurs when the spoon or glass has been used for a medicine called Questran-A, which is a powder that needs to be dissolved in water and then ingested. And more specifically, it only happens when these glasses/spoons are subsequently washed in our dishwasher.

This leads me to believe that some reaction occurs between the (left-overs) of the medicine and the dishwasher tablets. My question is: does anyone know what reaction between the chemicals in the medicine and those in the dishwasher tablets could cause this blue-ish deposit?

Obviously this will be impossible to say without the ingredients of both, so here is a list of the contents of the dishwasher tablets and the medicine:

Dishwasher tablets (Albert Heijn):

pentasodium triphosphate
sodium carbonate and sodium carbonate peroxide
taed
sodium silicate
PEG-90
alcohols, C16-18, ethoxylated mind 30 EO
acrylic copolymer
PEG-4
maleic acid/acrylic acid copolymer,sodium salt
alcohols, C12-18 ethoxylated propoxylated
zinc sulfate
beta-alanine, N-(2-carboxyethyl)-, N-coco alkyl derivs, disodium salt
mn-complex
benzotriazole
subtilisin
tetrasodium etidronate
alpha-amylase
benzisothiazolinone

Questran-A:

4 mg colestyramine
30 mg Aspartame(E 951) equal to 17 mg Phenylalanine
Citric acid(E 330)
kelcoloïd
Silicon dioxide(E551)
Orange flavoring
Xantham gum (E 415).

Let's speculate a bit ;)

Colestyramine is a twodimensional polymer network, build from styrene and and 1,4-divinylbenzene.
The polymer has a lot of benzyltrimethylammonium groups. That means it can act as an anion exchanger.
It is used to strongly bind bile acids
Neither the polymer itself, nor the adduct with bile acid are supposed to degrade in the bowel. The loaded polymer is supposed to be excreted.

The dishwasher tablets, on the other hand, contain the sodium salt of a maleic acid/acrylic acid copolymer.

That means we have two polymers, one with a lot of $\ce{Ph-CH2N^+(CH3)3}$ groups and another one with lots of $\ce{-COO-}$ groups.

Sounds like a mass wedding of two chain gangs :D

Edit

It is conceivable that this deposits a thin film on the spoons and the glassware.

In order to explain the bluish colour, we have two options:

The polymer aggregate absorbs in the 580 nm range - we observe the complementary colour. Due to lack of conjugation, I don't see how this could happen here.
The colour results from some dispersion effect related to opalescence, as suggested in Michiel's comment.