Effects of Ka on the Half-Equivalence Point I'm new to studying chemistry at a university level and am currently reviewing acid-base titrations. I am wondering about the relationship between your ionization constant for a weak acid ($K_\mathrm{a}$) and both the volume of strong base needed to reach the half equivalence point and the $\mathrm{pH}$ at the half-equivalence point.

Assuming these two weak acids are of the same concentration, if you had a higher $K_\mathrm{a}$ value for one of them, my reasoning is that your $\mathrm{pH}$ at the half equivalence would be lower than that of the weak acid with a lower $K_\mathrm{a}$ value because given the equation

$$\mathrm{pH} = -\log[\ce{H3O+}],$$

a greater hydronium ion concentration (from being present in larger amounts as a product given the larger $K_\mathrm{a}$ value) would result in a lower $\mathrm{pH}$ overall.

However, I am not entirely sure if this is correct, and I am also unsure as to how the volume of strong base titrant needed to reach the half equivalence point would be affected.

If we consider $\ce{HA}$ as a weak acid, then at the half equivalence point, $$\mathrm{p}H = \mathrm{p}K_\mathrm{a}$$ As $$\mathrm{p}H = \mathrm{p}K_\mathrm{a} + \mathrm{p[\ce{HA}]} -\mathrm{p[\ce{A-}]}$$ and for the half equivalence point,
$$\mathrm{p[\ce{HA}]} =\mathrm{p[\ce{A-}]}$$
So the higher the $\mathrm{p}K_\mathrm{a}$ is, the higher is $\mathrm{p}H$ at the half equivalence.


Why is fluorine more reactive than chlorine? Why is fluorine more reactive than chlorine despite chlorine having a higher electron affinity?

It depends on what reaction you are talking about.

In the context of organic SN1 reactions, the reactivity is $\ce{R-F}$ (least) $\ce{< R-Cl < R-Br < R-I}$ (most). [1] This reflects the leaving group ability, with larger atoms better able to accommodate the extra electron.
In the context of radical homolysis of $\ce{X-X}$ the order of reactivity is $\ce{F2 < Cl2 < Br2 < I2}$ as the sigma bonds become weaker as you move down the period.[2]
In polar aprotic solvents (DMSO) nucleophilicity follows $\ce{F- > Cl- > Br- > I-}$. This order reverses in polar protic solvents due to solvation: J. Chem. Educ., 1970, 47 (6), p 473
Transition metal $\ce{MX_n}$ bond enthalpies have been measured empirically and calculated: J. Chem. Phys., 1957, 26 (6), 1644–1647.. It would be inappropriate to over-generalize, as any discussion here needs to also account for metal oxidation states, HSAB theory, electronic configurations, and ligand-field theory

References:

Clayden, Greeves and Warren Organic Chemistry, p. 348.
Clayden, Greeves and Warren Organic Chemistry, p. 971.


Why are alkanes also called paraffins? Why are alkanes also called as paraffins?   Is it because they are mainly found from petroleum and natural gas?

The origin of the word is historical, and has to do with the observed properties of the class of hydrocarbons we know refer to as alkanes. The name originated in the Latin for "little affinity."

From M. P. Crosland's text Historical Studies in the Language of Chemistry:

  The term paraffin was coined by Reichenbach (from Lat. parum + affinis = little affinity) to denote a particular substance, a mixture of hydrocarbons (Journal fur Chemie und Physik, 59 (1830), 455). The term was extended to cover any hydrocarbons of the series $\ce{C_{\rm n}H_{\rm 2n + 2}}$ by Watts (Dictionary of Chemistry, 1868, vol. iv, p. 344 and Supplement, 1872, p. 705).

Reichenbach, referred to above, is Carl Reichenbach, a multifaceted scientist who did pioneering work in isolating (and thus naming) hydrocarbons in the early 1830's.


Is this a valid method to find percentage of pure ethanol in an impure sample? To generalize the problem, I will just say:

  5 gram of impure ethanol (or any other type of combustible material) is placed inside a bomb calorimeter where volume is constant. If the combustion produces a certain amount of energy, how should I find the mass in gram of pure ethanol contained in that quantity of impure ethanol?

Here is what I will do:

find the standard molar enthalpy of ethanol.
write a combustion equation.
find the energy produced in the combustion of one mole of ethanol.
then divide that amount of ethanol by the amount of energy stated in problem.

To find how many mole of pure ethanol, convert the mole into gram.

But there are two things that make me think that this way of solving the problem is not correct:

the quantity of impure ethanol is a mixture of pure ethanol and another unknown material. What if this unknown material produce part of the energy stated in the problem?
the standard molar enthalpy is measure under standard conditions (1 atm, 25 °C). However, the condition inside the bomb calorimeter is unknown.

What if this unknown material produce part of the energy stated in the problem?

It would be stated in the problem that the amount of heat that the unknown material produced.

  The standard molar enthalpy is measured under standard conditions (1 atm, 25°C). However, the condition inside the bomb calorimeter is unknown.

We are given some information, the final temperature of the system. But you are right in that we do not know the pressure, and the specific heat of the remaining items is difficult to calculate.


Why does d/l-2,3-dibromobutane on reaction with NaI/acetone give cis-2-butene? This is what I have read/know about the dehalogenation of vicinal dihalides by NaI/acetone:
The two halides which are leaving must be on opposite sides (from the mechanism).
In case they are present in the same side, we must rotate the C-C bond first and then proceed with the reaction.
In both cases we end by with a trans alkene.
But now I came across this:

meso-2,3-dibromobutane on reaction with NaI/acetone gives trans-2-butene.
d/l-2,3-dibromobutane on reaction with NaI/acetone gives
cis-2-butene.

I have no problem regarding the first statement. But regarding the second, from what I have learnt, the product again must be trans-2-butene.
I drew the Fisher projection and tried to figure out the reaction. For the 1st reaction, meso-2,3-dibromobutane must undergo C-C bond rotation, (performing two "swaps") and then we would end up with d/l-2,3-dibromobutane leading both 1st and 2nd reactions to trans-2-butene.
Where is my mistake? Is my understanding of the mechanism of the dehalogenation of NaI/acetone wrong? Any good sources to read regarding the same (I couldn't find it in any of my books) are appreciated, too.

The process of elimination of two bromides which you are thinking is not the correct way how the reaction occurs in this case. First there will be a nucleophilic substitution (preferably SN2) by $\ce{I-}$ on the two carbon atoms containing bromine. As iodide is a better nucleophile than bromide, this substitution is majorly driven forward, and after the formation of vicinal di-iodide, the compound now undergoes bond rotations if required and make the two iodine atoms orient themselves anti-periplanar for a E2 elimination.

So, essentially not only elimination, but substitution also occurs prior to elimination, and thus meso-2,3-dibromobutane never actually gets converted into (d/l)-2,3-dibromobutane during the reaction. Also, going by Sawhorse Projection formula is particularly useful in this case rather than Fisher projection. The former gives you a better visualisation of the syn or anti-periplanarity.

Here is the mechanism of the reaction with the meso-compound as starting material:

Similarly, you can draw the mechanism with the optically active 2,3-dibromobutane as your initial compound:

Now you can easily see how the corresponding alkenes are formed depending on the substrates.


What is the meaning of parentheses in "NAD(P)H"? If someone writes "NAD(P)H is oxidized by ...", does that mean they are referring to both NADH and NADPH?

If so doesn't that lead to confusion when thinking quantitatively since the reactions may proceed at different rates, NAD may be present in a different concentration than NADPH, etc?

If not, what do the parentheses around P mean?

If an enzyme is called $\mathrm{NAD(P)H}$-dependent, that means it is using both $\mathrm{NAD^+/NADH}$ and $\mathrm{NADP^+/NADPH}$ as its cofactors. For example, most oxidoreductases are $\mathrm{NAD(P)H}$-dependent, which are able to oxidize a relevant substrate by transferring a hydride ($\ce{H−}$) group to a nicotinamide adenine dinucleotide cofactor (either $\mathrm{NAD^+}$ or $\mathrm{NADP^+}$), resulting in their reduced forms, $\mathrm{NADH}$ or $\mathrm{NADPH}$ (Ref.1):

You can find a list of $\mathrm{NAD(P)H}$-dependent oxidoreductases within Reference 1.

Reference:

Lara Sellés Vidal, Ciarán L. Kelly, Paweł M. Mordaka, John T. Heap, "Review of NAD(P)H-dependent oxidoreductases: Properties, engineering and application," Biochimica et Biophysica Acta (BBA) - Proteins and Proteomics 2018, 1866(2), 327-347 (https://doi.org/10.1016/j.bbapap.2017.11.005).


Geometry of the transition metal complex in solid dipotassium bis(oxalato) copper(II) tetrahydrate I was thinking of the geometry of the transition metal complex in solid $\ce{K2[Cu(ox)2] · 4 H2O}$ solid around the $\ce{Cu(II)}$ ion.
At first I thought square planar, but is it possible or is there any chance that it could be octahedral with two water molecules since it is a hydrate?
Or do brackets always indicate what is directly complexed in this case?

Yes, square brackets are used to denote a coordination sphere, but they may sometimes be used incorrectly. Ideally you want to check the crystal structure, luckily there is one [1]:

The $\ce{Cu(II)}$ atom has square-planar geometry with two bidentate oxalate ligands. Each $\ce{K+}$ cation is coordinated by eight oxygen atoms from one bidentate and two unidentate oxalate groups and four water molecules.

Fig. 1. (a) Crystal structures of $\ce{[K2Cu(ox)2 · 4 H2O]}$ (1) and (b) $\ce{[K2Cu(ox)2 · 2 H2O]}$ (2) with atom numbering scheme. The thermal ellipsoids are drawn at 50% probability.

Reference

Fan, J.; Sun, W.-Y.; Okamura, T.; Yu, K.-B.; Ueyama, N. The X-Ray Crystal Structural Characterization of Dipotassium Bisoxalato Copper(II) Tetrahydrate, $\ce{[K2Cu(Ox)2·4 H2O]}$ $(\ce{Ox}$ = oxalate Dianion). Inorganica Chimica Acta 2001, 319 (1), 240–246. DOI: 10.1016/S0020-1693(01)00466-2.


Pair of analogous periods on the periodic table On the periodic table, period 2 and 3, 4 and 5, 6 and 7 and so on have similar blocks, identical length and groups. What are such pairs of analogous periods called?

In a bit of a web search, there does not seem to be any particular names for families of periods in the Periodic Table.  However, according to WebElements,

  The f-block elements are assigned to Periods 6 (lanthanoids) and 7 (actinoids) since that is where they are located in the full, or extended, version of the periodic table.

However, these, according to the image below, only apply to the Lanthanides and Actinides, not all of periods 6 and 7.

Image source


Why is a titration necessary? Why can't we just compare the moles of the acid and base? For the following question:

$\mathrm{1g}$ of $\ce{Mg(OH)2}$  is completely dissolved in $\mathrm{25}$ mL of $\mathrm{0.5}$ M $\ce{HCl}$. This solution is then titrated with $\mathrm{0.1}$ M $\ce{NaOH}$, and it takes $\mathrm{40}$ mL of $\ce{NaOH}$ to reach the end point. What volume of $\mathrm{0.5}$ M $\ce{H2SO4}$ would neutralise 1g of $\ce{Mg(OH)2}$ ?

This is solved by the following steps:

moles=concentration*volume=$\mathrm{0.5}$ M * $\mathrm{25 mL}$ $\ce{HCl}$=$\mathrm{12.5}$ mmol $\ce{HCl}$ which is equivalent to $\mathrm{12.5}$ mmol  $\ce{H+}$.
moles=concentration*volume=$\mathrm{0.1}$ M * $\mathrm{40 mL}$ $\ce{NaOH}$=$\mathrm{4}$ mmol $\ce{HCl}$ which is equivalent to $\mathrm{4}$ mmol  $\ce{OH-}$.
Difference = $\mathrm{12.5-4=8.5 }$ mmol of $\ce{H+}$. That is, $\mathrm{8.5}$ mmol of $\ce{H+}$ was used in the reaction with the $\ce{Mg(OH)2}$.
Finding out how much $\mathrm{0.5}$ M $\ce{H2SO4}$ is required to give $\mathrm{8.5}$ mmol of $\ce{H+}$ :  $\mathrm{1}$ mole of $\ce{H2SO4}$ gives $\mathrm{2}$ mole of $\ce{H+}$, so we need $\mathrm{8.5/2=4.25}$ moles of $\ce{H2SO4}$. Concentration=moles/volume. Volume = moles/concentration=$\mathrm{4.25}$ mmol/$\mathrm{0.5}$ M=8.5 ml $\ce{H2SO4}$ required.

Why are all these steps necessary? Can't we directly compare the number of moles of $\ce{Mg(OH)2}$ and $\ce{H2SO4}$? I.e. as follows:

Molar mass of $\ce{Mg(OH)2}$=$\mathrm{24.3+16*2=56.3}$
moles=mass/molar mass=1g of $\ce{Mg(OH)2}$/$\mathrm{56.3}$=$\mathrm{0.018}$ moles of $\ce{Mg(OH)2}$, which is equivalent to $\mathrm{0.018*2=0.036}$ or $\mathrm{36}$ mmol of $\ce{OH-}$.
So we need $\mathrm{36}$ mmol of $\ce{H+}$ to neutralise $\mathrm{36}$ mmol of $\ce{OH-}$. $\mathrm{1}$ mol of $\ce{H2SO4}$ contains  $\mathrm{2}$ mol of $\ce{H+}$. So we need $\mathrm{36/2=18}$ mmol of $\ce{H2SO4}$.
Concentration = moles/volume = $\mathrm{0.5}$ M = $\mathrm{18}$ mmol/Volume. So Volume = $\mathrm{18/0.5=36}$ mL of  $\ce{H2SO4}$.

My second attempt yields the wrong answer. So why can't the problem be solved by comparing the number of moles of hydrogen ions and hydroxide ions? Why is the titration with other molecules required to solve this problem?

Note: I have attempted to summarise the question here. But if you prefer to read it in its original form:

Man! you must have got impure sample of Mg(OH)2
Moreover, Mg(OH)2 being sparingly soluble may not be finding a way for direct titration, so the method suggested in the problem is a process we call as BACK-TITRATION. This you have to follow i all such situations, where direct titration might not, for any of the reasons, be possible.

Because all 1 g Mg(OH)2 is not pure, you can expect incorrect answer if you are directly comparing the moles of H+ & OH- ions.

Would this be fine?


What are the conditions for making amines from alkyl halides? From what I understand, to make an amine from an alkyl halide you need to heat the alkyl halide with a mixture of ammonia and ethanol. I have two questions:

Do both ammonia and ethanol need to be concentrated? Because I think I read somewhere that you use a 'concentrated mixture of ethanol and ammonia', although I'm not sure whether the 'concentrated' part refers to the ethanol or the ammonia. I'm guessing it refers to the ammonia.
What is the role of the ethanol? Does it act a a catalyst, or just a solvent, or both?

Ethanol is just a solvent often used for this reaction because 1) the reactants and products are soluble in it and 2) the reaction proceeds at a reasonable rate in boiling ethanol.  If your trying to prepare the corresponding primary amine, then a large excess of ammonia should be used.  Otherwise, if the concentration of the initial product (the primary amine) becomes comparable to the concentration of ammonia, then the primary amine could react with the starting alkyl halide and a complicated mixture of products could ensue (see this article for a nice description of all the possible products that could be formed if the ammonia concentration is not high enough).

The reaction is typically run at reflux (boiling ethanol).  Since ammonia is a gas at room temperature it would quickly escape from the reaction and we wouldn't have a high concentration of ammonia for long.  Therefore, the ammonia is condensed into a solution of the ethanol and alkyl halide (usually an alkyl bromide) in a glass tube at low temperature.  When the desired concentration of ammonia is obtained, the tube is sealed and then heated to allow the reaction to take place.


Can a flame start on water (video included)? Can a flame a start on a wet piece of paper immersed in water?

The video is shown below (not the full video):

https://youtu.be/Ey3z8z4Hxtc?t=1h27m29s

This is not in English so you might not understand anything. In the video they simply say it is demons or whatever that starts the fire, which is silly to me, of course.

My questions:

What kind of compounds can we use (or the fraud used in the video)?
What do they do in movies, tv shows, magic shows, or even the olympics, to start flames on water?
What changes the color of flame. For example it was pink in the video?

Watch this video from 1:50 onwards to see the reactions of sodium and potassium with water. https://www.youtube.com/watch?v=H6ZDiRIvc2E

This video, 0:56, shows  gasoline (petrol) burning on water. (Note that the batteries have nothing to do with it). https://www.youtube.com/watch?v=Bs13jBWqFjA


Why are free radicals so reactive? Why are free radicals are so reactive? They can break almost any bond, including $\ce{C-H}$ bonds, which are fairly stable. Don’t they have an activation energy or something? Do they not also experience steric effects when colliding with reactants ?

Radicals are so reactive because they require so much energy to form. When we speak about radical reactivity, "more reactive" generally means a more exothermic hydrogen atom abstraction step. The more exothermic this step, the more the transition state resembles the reactants (Hammond's postulate), with a correspondingly lower character of the carbon-centered radical product. This makes the reaction less sensitive to the stability of the carbon-centered radical.

This being said, some radicals do have an activation energy, and some don't. The $E_a$ is essentially zero for all hydrogen atom abstractions by fluorine atom, and the free energy barrier arises solely from the pre-exponential factor of the Arrhenius equation ($k=Ae^{-E_{a}/(RT)}$), which is near 13 for the halogenations listed below.

The activation energies for abstraction by chlorine atoms are also exceedingly small, but there is a slight preference for more highly substituted $\ce{C-H}$ bonds. Lastly, the activation energies for abstraction by bromine atoms are substantial, clearly showing a preference for abstraction of 3°, 2°, and 1° $\ce{C-H}$ bonds in that order. It is because of these differences in activation energies that we observe the varying regioselectivities of the different halogenation reactions$^{[1]}$.

$[1]$ Anslyn, E. V.; Dougherty, D. A. Modern Physical Organic Chemistry; University Science: Sausalito, CA, 2006.


Semipinacol-type rearrangement leading to ring expansion What will be the major product in the following reaction?

I tried to work out the mechanism, which starts with protonation of the alkene, and I got C to be the answer (which is the answer in the book as well). But, I've heard that three-membered rings generally don't expand due to their exceptional stability. Is this true? And between the alkene and alcohol functional groups, which is more likely to get protonated?

Three-membered rings are not exceptionally stable, on the contrary, they suffer
a lot of ring strain and readily take part in ring-opening reactions.
Exceptions include the Hückel-aromatic cyclopropenyl cation, and the
resonance-stabilized cyclopropylmethyl cation.

Your assumption that the alkene is protonated is correct, as this generates a
substituted cyclopropylethyl cation. The regioselectivity of the protonation occurs to give the more stable secondary carbocation instead of a a primary carbocation. A 1,2-alkyl shift then occurs, which is in this case assisted by the lone pair on oxygen. Deprotonation of the carbonyl then yields the four-membered ring ketone C (2-methylcyclobutanone). [The carbons are numbered in blue to make it easier to see what's going on: the C1-C3 bond is broken and C3-C4 bond is formed.]

This is similar to the semipinacol rearrangement; in this case, the carbocation is generated via protonation of an alkene instead of by loss of a leaving group.

The formation of products A or B would require protonation of the alcohol group, followed by loss of water in an E1 reaction. In either case, the final deprotonation yields a three-membered ring with an endocyclic or exocyclic double bond, which is more strained than a saturated three-membered ring. This reaction pathway is therefore less favorable than ring expansion. Protonation of the alcohol can certainly happen, but since it does not lead to the formation of any stable products, it is said to be unproductive.


Why is the tropylium anion non-aromatic? Recently I came across a question asking which compounds are anti-aromatic. This was one of the options:

I’m confused as this is not an answer. I think the lone pair electrons on the carbanion participate in resonance and so there are 8π electrons which, follows the 4nπ electrons rule for anti aromaticity.
So why isn't it anti-aromatic?

The rules in determining the aromaticity of a compound are as follows:

The system must have 4n+2 π electrons
The system must be planar
The system must be completely conjugated
The molecule must be cyclic.

The cycloheptatriene anion does not follow two out of these four rules. Namely,

The anion system has 4n electrons (whereas it should have had 4n+2 electrons in it π system)
The anion system has an sp3 carbon and is hence non-planar (where it should have been planar)

Due to these two issues (second issue mainly), there is no ring current in the complete ring and so no conditions need to be considered whether the compound is antiaromatic or aromatic in nature.
Cyclopentadienyl anion also has an sp3 carbon as an anion. However the anion formed is aromatic if planar and so this overcomes the ring strain and so retains aromaticity.


What goes on the NFPA 704 for multiple chemicals? If you have a chemical that is classified (Health, Fire, Reactivity) 2, 1, 2. And another chemical that is 1, 3, 2. If these two chemicals are stored together, what is put on the storage container/building? What is the rule for this; the largest in each category is displayed, 2, 3, 2. And the specific hazards are listed?

According to Section 4.2.3.3 of NFPA 704:

  Where more than one chemical is present in a building or specific
  area, professional judgment shall be exercised to indicate ratings
  using the following methods:

  (1) Composite Method. Where many chemicals are present, a single sign
  shall summarize the maximum ratings contributed by the material(s) in
  each category and the special hazard category for the building and/or
  the area.

  (2) Individual Method.  Where only a few chemicals are present or
  where only a few chemicals are of concern to emergency responders
  (taking into account factors including physical form, hazard rating,
  and quantity), individual signs shall be displayed. The chemical name
  shall be displayed below each sign.

  (3) Composite–Individual Combined Method. A single sign shall be used
  to summarize the ratings via the Composite Method for buildings or
  other areas containing numerous chemicals. Signs based on the
  Individual Method shall be used for rooms or smaller areas within the
  building containing small numbers of chemicals.


Why is methanol toxic? There are two points of view for the answer of this question:

The biological view, the only one that I faced during my research, states that since it can trigger perilous conditions like metabolic acidosis, cause a neurologic sequelae, etc. methanol is toxic.
The chemical point of view in an answer, which is the one I seek, is the reactions that occur in the body with methanol as a reactant.

Why should it be a reaction? Because I assume something will be toxic for humans when it reacts and disables some of their bodies' vital compounds; e.g.: Hydrogen cyanide reacts with enzymes, renders their active sites useless, and finally causes heavy reduction in bio-chemical reactions that results in quick death.

So, the question is: What reactions use methanol in the body that makes it toxic for us?

Methanol isn't particularly toxic in and of itself, although it's no walk in the park.

If methanol flowed through the body without being broken down, it would cause roughly the same kind of harm as ethanol, i.e. intoxication.

The real culprit is one of its metabolic products, methanoic acid, also known as formic acid.

To understand how formic acid, present as the formate ion, is toxic, we look to Wikipedia:

  Formate is toxic because it inhibits mitochondrial cytochrome c oxidase, causing the symptoms of hypoxia at the cellular level, and also causing metabolic acidosis, among a variety of other metabolic disturbances.

Edit: As DavePhD points out, an intermediate product in this process is formaldehyde, or methanal. While formaldehyde is also toxic, it is rapidly metabolized to methanoic acid.

Reedit: The deeper, more historical reason that this happens is that methanol isn't readily available in nature, meaning that few species have developed biochemical tools to deal with it. There simply hasn't existed an evolutionary pressure to deal with methanol.


Why do electrons in sigma bonds have zero orbital angular momentum about the internuclear axis? As per the title, why do electrons in σ-bonds have zero orbital angular momentum about the internuclear axis?

As far as I know, the outermost electrons participate in bonding. Suppose that the 3p electrons are the outermost and form a σ-bond. Initially, their individual orbital angular momenta are

$$|\vec{l}| = \sqrt{l(l+1)}\hbar = \sqrt{3}\hbar$$

since they have the orbital angular momentum quantum number $l = 1$. How can this become zero after forming a σ-bond?

My quantum is slightly rusty, but I think this is how it goes (I would appreciate any corrections or clarifications). We're talking about diatomic molecules here. As you no doubt know, σ-orbitals are invariant about rotation around the internuclear axis: that is to say, you can rotate the molecule as much as you like, but the sigma orbital will still look the same. This is true whether the sigma orbital is made from s, p, or d orbitals.

If we define the $z$-axis to be the internuclear axis and $\phi$ to be the angle around this axis (the azimuthal angle), then we can also say that σ-orbitals have no dependence on the angle $\phi$.

In a diatomic molecule, because the orbital angular momentum (OAM) operator $\hat{L}$ does not commute with the Hamiltonian, the quantum number $l$ is no longer a 'good quantum number': that is to say, the eigenstates of the Hamiltonian (which are the MOs) are no longer simultaneously eigenstates of $\hat{L}$, so the MOs do not have a specific value of $l$ that can be assigned to them. However, for a one-electron molecule (e.g. $\ce{H2+}$) it turns out that $\hat{L}_z$ still commutes with the Hamiltonian, where the $z$-axis is the internuclear axis (or molecular axis). This operator represents the projection of the OAM onto the internuclear axis, and the corresponding quantum number is denoted $m_L$ or $\lambda$ (depending on where you read).

Now, let's denote the wavefunction of the σ-orbital by $\psi$. We know that the eigenvalues of $\hat{L}_z$ are unit multiples of $\hbar$, or in other words, $\hat{L}_z\psi = m_L\hbar\psi$. [Note that this is no different to $\hat{L_z}\psi = m_l\hbar\psi$ in an atom, except that we use a capital L because the context is different, i.e. we are talking about a diatomic molecule.]

It turns out that in spherical coordinates, $\hat{L}_z$ is equal to $-\mathrm i \hbar (\partial/\partial \phi)$. Since we've established that $\psi$ has no dependence on $\phi$, the partial derivative $(\partial \psi/\partial \phi)$ is zero and hence $\hat{L}_z\psi = 0$, meaning that the quantum number $m_L = 0$, as desired.

It doesn't matter what the constituent atomic orbitals are. However, I thought a bit of clarification would be useful. For the atomic p-orbitals, you are correct that these possess $l = 1$. However, note that $l$ is a quantum number which characterises the magnitude of the angular momentum, whereas $\lambda$, or $m_L$, is a projection quantum number. That is to say, it is possible to "add" two atomic orbitals with $l = 1$ and have the projection cancel out.

If you think about the projection of the orbital angular momenta in the original p-orbitals, they are of course $m_l = (+1,0,-1)$. However, these are the complex p-orbitals which are obtained as eigenstates of $\hat{L}_z$. The real p-orbitals, i.e. the dumbbell-shaped p-orbitals that you and I think about when constructing sigma bonds, are either $m_l = 0$ (i.e. the $\mathrm p_z$ orbital) or are linear combinations of $m_l = \pm 1$ (i.e. the $\mathrm p_x$ and $\mathrm p_y$ orbitals). So, it's not at all counterintuitive that when you add these together to form sigma bonds, the projection $m_L$ is simply zero.


Discrepancies between specific heat of vaporization of water and enthalpy of formation of water and steam According to Wikipedia, the enthalpy of formation of water is $-285.8~\mathrm{kJ/mol}$ while the enthalpy of formation of steam is $-241.818~\mathrm{kJ/mol}$, implying the following:

$$\ce{H2O(l) -> H2O(g)}\qquad (\Delta H=44.0~\mathrm{kJ/mol)}$$

Let us derive the enthalpy of the above reaction using the specific heat capacity of water and the specific heat of vaporization of water instead.

$1~\mathrm{mol}$ of water weighs $0.018~\mathrm{kg}$.

To raise $1~\mathrm{mol}$ of water from $25~\mathrm{^\circ C}$ to its boiling temperature requires $(0.018~\mathrm{kg}) \times (4200~\mathrm{J~kg^{-1}~K^{-1}}) \times (75~^\circ\mathrm{C}) = 5.67~\mathrm{kJ}$.

To turn that amount of water to steam requires $(0.018~\mathrm{kg}) \times (2258~\mathrm{kJ~kg^{-1}}) = 40.644~\mathrm{kJ}$.

Adding these two terms give $46.314~\mathrm{kJ}$, implying the following:

$$\ce{H2O(l) -> H2O(g)}\qquad (\Delta H=46.314~\mathrm{kJ/mol)}$$

Is the discrepancies between the two thermochemical equations simply due to measurement inaccuracies?

According to “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL, the standard molar enthalpy of formation at a temperature of $T=298.15\ \mathrm K$ is

$\Delta_\mathrm fH_\mathrm m^\circ=-285.8\ \mathrm{kJ\ mol^{-1}}$ for liquid water,

and

$\Delta_\mathrm fH_\mathrm m^\circ=-241.8\ \mathrm{kJ\ mol^{-1}}$ for gaseous water.

(These values are in agreement with the values that you have found on Wikipedia.)

Both values refer to a temperature of $T=298.15\ \mathrm K$; therefore, you do not have to raise the temperature of the liquid water to its boiling point. Nevertheless, you may do this; however, then you would have to cool down the steam back to $T=298.15\ \mathrm K$ later.

Instead, you can directly use the enthalpy of vaporization at the given temperature. Note that the enthalpy of vaporization depends on temperature, and that your value for the specifc enthalpy of vaporization of $\Delta_\text{vap}h=2258\ \mathrm{kJ\ kg^{-1}}$ refers to the boiling point at normal pressure, i.e. to a temperature of approximately $T=100\ \mathrm{^\circ C}$.

According to REFPROP – NIST Standard Reference Database 23, Version 9.0, the molar enthalpy of vaporization at a temperature of $T=298.15\ \mathrm K$ is

$\Delta_\text{vap}H_\mathrm m=43.987\ \mathrm{kJ\ mol^{-1}}$,

which is in good agreement with the difference of the values for the molar enthalpy of formation:

$-241.8\ \mathrm{kJ\ mol^{-1}}-\left(-285.8\ \mathrm{kJ\ mol^{-1}}\right)=44.0\ \mathrm{kJ\ mol^{-1}}$


What does chemical bonding look like at the quantum level? How does chemical bonding look like on the quantum scale per se since the electron is a wave, and a particle?

I believe that generally in chemistry the wave characteristics of the electron are far more prevalent/relevant that the particle characteristics. So it's usually easiest just to think of it as a wave (or I think "region of space" is easier).

A chemical bond looks like the electron wave or area spread out between two or more atoms. This shape of the wave has lower energy than any of the orbitals it could occupy on just one atom. Because it has lower energy, that means it needs energy from somewhere for the electrons & atoms to change back to a different configuration. So unless it gets some more energy it cannot change away from the bonded state. That explains why chemical bonds are stable and hold atoms together.


kJ per mole or kJ per 2 moles? From Chemguide:

Is the measurement unit $\ce{kJ*mol^{-1}}$ correct here? Shouldn't it be measured simply in kJ, because the $-572$ figure is for the whole reaction, i.e. for 2 moles of the product?

the whole reaction, i.e. for 2 moles of the product?

This is your mistake.

The balanced equation $\ce{2H2 + O2 -> 2H2O}$ does not mean that 2 moles of the product are being formed.

The numbers, called stoichiometric coefficients, only tell you about the ratio in which the reactants react and form products.

That is to say, if I had a system in which I reacted 1 mol of $\ce{H2}$ with 0.5 mol of $\ce{O2}$ to form 1 mol of $\ce{H2O}$, the above equation would still be correct. This is because the ratio of $\ce{H2}:\ce{O2}:\ce{H2O}$ is still $2:1:2$. As long as the ratio is $2:1:2$, the equation will still be correct and applicable.

What $\Delta_\mathrm{r}H^\circ$ tells you is: how much the enthalpy change would be, if the number of moles reacting were the same as the stoichiometric coefficients. Therefore for the above reaction, saying that $\Delta_\mathrm{r}H^\circ = -572~\mathrm{kJ~mol^{-1}}$ means that

If 2 mol of hydrogen were to react with 1 mol of oxygen to form 2 mol of water, the enthalpy change would be -572 kJ.

That means that if you multiplied the equation by 2, $\Delta_\mathrm{r}H^\circ$ would also be multiplied by 2.

$$\ce{4H2 + 2O2 -> 4H2O} \qquad \Delta_\mathrm{r}H^\circ = -1144~\mathrm{kJ~mol^{-1}}$$

This line tells you: if 4 mol of hydrogen were to react with 2 mol of oxygen to form 4 mol of water, the enthalpy change would be -1174 kJ. That's obviously true because you're reacting twice as much reactants to get twice as many products, so the enthalpy change must be twice as much.

  So what is the per mol for

As chemguide said, it is not per mol of anything. The per mol arises from the definition of the reaction enthalpy change, $\Delta_\mathrm{r}H^\circ$. I have another answer describing it in the context of Gibbs free energy here.

However, one way to look at it is to remember that the reaction does not stipulate the actual amounts of reactants and products. That is to say, the value $\Delta_\mathrm{r}H^\circ$ on its own cannot help you find an enthalpy change, unless you tell it exactly how much of the reactants and products you have. If you don't specify whether you start with 2 mol or 1 mol of water, you cannot find $\Delta H$ for the process.

That means you need to multiply $\Delta_\mathrm{r}H^\circ$ by an amount of substance (units mol) to get $\Delta H$ (units kJ). So, $\Delta_\mathrm{r}H^\circ$ must have units of kJ/mol.

What is the actual amount of substance that you are multiplying by then? It is simply the amount of substance of any species reacting, divided by its stoichiometric coefficient in the equation.

Therefore, if you have 6 moles of oxygen reacting, the "amount of substance" that you multiply by would be $6~\mathrm{mol}/1$. The enthalpy change would be

$$\Delta H = (-572~\mathrm{kJ~mol^{-1}})\left(\frac{6~\mathrm{mol}}{1}\right) = -3432~\mathrm{kJ}$$

You have to divide by the stoichiometric coefficient. This is necessary for consistency. For example, if you were only given this $\Delta_\mathrm{r}H^\circ$:

$$\ce{4H2 + 2O2 -> 4H2O} \qquad \Delta_\mathrm{r}H^\circ = -1144~\mathrm{kJ~mol^{-1}}$$

then the "amount of substance" you would multiply by is $6~\mathrm{mol}/2$. You would have

$$\Delta H = (-1144~\mathrm{kJ~mol^{-1}})\left(\frac{6~\mathrm{mol}}{2}\right) = -3432~\mathrm{kJ}$$

Note how the answer comes out to be the same.

But it doesn't have to be for oxygen either. That's what Chemguide means when it says it's not for any particular species in the reaction, because it's applicable to all of them. For example, if I told you that 44 mol of water was formed, then you would use

$$\Delta H = (-1144~\mathrm{kJ~mol^{-1}})\left(\frac{44~\mathrm{mol}}{4}\right) = -12584~\mathrm{kJ}$$


How do I excel in Chemistry? I wanted to ask an online group of experts/community about how I could improve in chemistry. I study in 10th grade currently and don't get very good grades. Most of the time its a B or C.

I have to select science stream, in which physics, chemistry and maths are compulsory. I am good in physics and maths, but what should I do in chemistry?

I have a hard time grasping concepts like Ionisation Energy, structure of benzene etc. and these are only going to get 100 times tougher next year!

Chemistry needs a devoting reader. You should try to love chemistry as much as you want! That'll be the ignition for lots of reading and stuff. My pointers:

Love chemistry.
Learn in an organized matter. (For example, do not jump from studying about covalent bonds' basics to reaction mechanism.
Take notes of the most important points of what you learn.
Look for open-source online tutorials and not only the books.
Some chemistry concepts are related to each other. Learn the first one's basics and then read the other one's. This will build up your understanding, literally. For example, first learn about electronegativity and then study the covalent bonds' basics. This is the exception for rule 2.
Try to differentiate between what's taught in schools to be true and what really is. For example, look for exceptions of the periodic trends that are taught in schools.
Be imaginatively practical. Don't just say enough for the lewis structure of water, but do it about $\ce{SF6}$.
Always try to go relatively higher than what your textbook has written. We have a proverb in my mother language meaning "If you have 100, you do have 90."
Ask what misunderstandings you have in here!
Never look to doing a research as a waste of time.
When approaching to solve a chemical problem, do it step by step.
When approaching to solve a problem, don't forget to write units of measurement.
When you reached an answer to the problem, ask yourself: "Is this possible?" If you've calculated 5 grams of water to be $\frac{18}{5}$ mol, and you know that a mole of water is 18 grams, shouldn't something be wrong here?
For any definitions, uses, notions, and practical experiments you encounter, ask yourself: "What is this for, either in real life, experimental sciences or theoretical chemistry?"
Look for online exercises, free of charge, but usually important.
Follow the latest news that are being discussed everywhere in the world of chemistry.
For specific concepts in chemistry, e.g.: Periodic table, doing a research about the history is very helpful.

This is as far as I can count. You can ask "why"s in comments.


How does Walter White make pure crystal meth using a non-stereospecific reaction? In the fantastic TV series Breaking Bad, Walter White, the dying chemistry teacher, takes to making crystal meth (d-methamphetamine) by several routes.

Initially, he uses the common small-scale route starting with pseudo-ephedrine which uses a stereospecific starring material to make a stereospecific end product (d-methamphetamine is the active compound). But doing this on a large scale is limited because it is hard to get large quantities of the pseudo-ephedrine.

Walter uses his knowledge of chemistry to move to an alternative route starting with phenyl-acetone(or P2P as it often called):

There are several ways of doing the reduction to produce the final product (White first seems to use a Thorium dioxide dehydrocarboxylation in a tube furnace but later seems to favour the aluminium amalgam reduction).

But all the P2P methods produce racemic products. White clearly knows this as the issue of getting the right enantiomer is raised more than once in the show. His product is considered to be exceptionally pure.

So how does he get an enantiomerically pure product? Or perhaps, since the show might not want to tell us the answer, what possible methods could he have used?

Intriguing question.

First, the best yield would be achieved by selectively producing one enantiomer instead of the other. In this case, White wants D-methamphetamine (powerful psychoactive drug), not L-methamphetamine (Vicks Vapor Inhaler). Reaction processes designed to do this are known as "asymmetric synthesis" reactions, because they favor production of one enantiomer over the other.

The pseudoephedrine method for methamphetamine employs one of the more common methods of asymmetric synthesis, called "chiral pool resolution". As you state, starting with an enantiomerically-pure sample of a chiral reagent (pseudoephedrine) as the starting point allows you to preserve the chirality of the finished product, provided the chiral point is not part of any "leaving group" during the reaction. However, again as you show, phenylacetone is achiral, and so the P2P process cannot take advantage of this method.

There are other methods of asymmetric synthesis, however none of them seem applicable to the chemistry shown or described on TV either; none of the reagents or catalysts mentioned would work as chiral catalysts, nor are they bio- or organocatalysts. Metal complexes with chiral ligands can be used to selectively catalyze production of one enantiomer, however the aluminum-mercury amalgam is again achiral. I don't remember any mention of using organocatalysis or biocatalysis, but these are possible.

The remaining route, then, is chiral resolution; let the reaction produce the 50-50 split, then separate the two enantiomers by some means of reactionary and/or physical chemistry. This seems to be the way it works in the real world. The advantage is that most of the methods are pretty cheap and easy; the disadvantage is that your maximum possible yield is 50% (unless you can then run a racemization reaction on the undesireable half to "reshuffle" the chirality of that half; then your yield increases by 50% of the last increase each time you run this step on the undesirable product).

In the case of methamphetamine, this resolution is among the easiest, because methamphetamine forms a "racemic conglomerate" when crystallized. This means, for the non-chemists, that each enantiomer molecule prefers to crystallize with others of the same chiral species, so as the solution cools and the solvent is evaporated off, the D-methamphetamine will form one set of homogeneous crystals and the L-methamphetamine will form another set. This means that all White has to do is slow the evaporation of solvent and subsequent cooling of the pan, letting the largest possible crystals form. Then, the only remaining trick is identifying which crystals have which enantiomer (and as these crystals are translucent and "optically active", observing the polarization pattern of light shone through the crystals will identify which are which).


calculate the final concentration of this equation A student mixed 260ml of 1.2 M lead(II) nitrate with 300ml of 1.90M potassium iodide. What is the final concentration of $\ce{NO3^{-}}$ ?

the answer is 1.11M , but he used the moles of lead(II) nitrate to find out the moles of $\ce{NO3^{-}}$ ,, and I think he should had used the moles of potassium iodide since it is the limiting reactant.

could anyone explain ?

[…] he should had used the moles of potassium iodide since it is the limiting reactant.

Did any reaction take place upon mixing solutions of $\ce{Pb(NO3)2}$ and $\ce{KI}$?
If so, did the reaction change the number of moles of $\ce{NO3-}$ in solution by precipitation, oxidation, reduction, etc.?
Would the result of the calculation be different if just 300 mL of water were added to the $\ce{Pb(NO3)2}$ solution?

Update I

Now for the rocket science (solubilities at 20 °C):

\begin{array}{cr}
\mathrm{compound} & \mathrm{solubility\ [g\cdot L^{-1}]} \\ \hline
\ce{Pb(NO3)2} & 522\\
\ce{KI} & 1430\\
\ce{KNO3} & 316\\
\ce{PbI2} & 0.76\\
\end{array}

The rather low solubility of $\ce{PbI2}$ suggests that on combination of the two solutions, the following may happen:

$$\ce{Pb^{2+} + 2 NO3- + 2 K+ + 2 I- -> 2 K+ + 2 NO3- + PbI2 v}$$

So again:

Does this in any way change the absolute amount of  $\ce{NO3-}$ in solution?
What does change for the concentration of $\ce{NO3-}$ when 300 mL of another solution is added to 260ml of 1.2 M $\ce{Pb(NO3)2}$ solution?

Update II

Initial $\ce{Pb(NO3)2}$ solution

   260 mL of a 1.2M solution of $\ce{Pb(NO3)2}$ contain $1.2\ \mathrm{\frac{mol}{L}}\ \cdot\ 2\cdot\ 0.26\ \mathrm{L} = 0.624\ \mathrm{mol}\ \ce{NO3-}$

After mixing

   Addition of 300 mL of the $\ce{KI}$ solution results in the precipitation of $\ce{PbI2}$, but does not change the absolute amount of  $\ce{NO3-}$ in solution. However, it changes the total volume and hence the concentration, which calculates to $\frac{0.624\ \mathrm{mol}}{(0.26 + 0.3) \mathrm{L}} = 1.11 \ M$


Why does caffeine powder cause apple cider (or any carbonated drink?) to foam? Something I noticed today, when adding some pure caffeine powder to apple cider.

Why does caffeine powder cause apple cider (or any carbonated drink?) to foam?

It seems to foam (with around 150mg caffeine added) slowly, but constantly so that it almost seems that it will eventually make all the carbon dioxide fizz to air.

Is this really related to caffeine or the powdered form in general?

I don't need to know why you're mixing uppers (caffeine) with downers (alcohol) but if the powdered caffeine is removing the fizz from your cider then you might try dissolving it in hot water first and then adding the hot caffeine solution to your cider.  In dissolved form, the caffeine cannot provide the nucleation sites for the formation of CO2 bubbles that MaxW and vapid pointed out.

The solubility of caffeine in water is 200mg/ml at 80 oC so you should be able to dissolve an adequate dose of caffeine in a small enough quantity of hot water to avoid both excessive warming of your cider and precipitation of the caffeine upon cooling and dilution.

It behooves me to mention that recommended caffeine administration should not exceed 200mg every 3-4 hours in adults, but since you're adding it to alcoholic beverages, I doubt you give a monkey's.  Here's the drugs.com link anyway:

https://www.drugs.com/monograph/caffeine-caffeine-and-sodium-benzoate-injection-caffeine-citrate.html

And solubility data here:

http://www.sigmaaldrich.com/content/dam/sigma-aldrich/docs/Sigma-Aldrich/Product_Information_Sheet/c0750pis.pdf


For what is reaction enthalpy more relevant than reaction free energy? I know enthalpy is a component of the Gibb's Free Energy (along with entropy). For practical purposes, though, isn't Gibb's Free Energy what a chemist is actually interested in, as that determines the overall equilibrium of a reaction? What important detail does enthalpy inform us of and what applicable function does that detail serve?

I am asking this because I have a DFT calculation that yields both the enthalpy and free energy, and I see a paper citing the enthalpy, so I am wondering why it is being preferred over the free energy.

While a detailed answer to this question could go on at length, the major practical distinction is the following:

The enthalpy change of a reaction/process indicates how much heat input or output is involved with the reaction, with a negative $\Delta H_\mathrm{rxn}$ indicating heat output.
As you note, the Gibbs free energy change of a reaction/process indicates whether a reaction is thermodynamically favored or disfavored, with a negative $\Delta G_\mathrm{rxn}$ indicating a favorable reaction.

Note that, e.g., a reaction/process can have $\Delta H_\mathrm{rxn} > 0$ (endothermic), but also $\Delta G_\mathrm{rxn} < 0$ (thermodynamically favored).  That is, despite the reaction requiring heat input, it is still favored to occur.  This is because the change in entropy of the process, $\Delta S_\mathrm{rxn}$, is sufficiently large that the $T\Delta S_\mathrm{rxn}$ term is able to overcome the positive $\Delta H_\mathrm{rxn}$.  A common example is the endothermic dissolution of various salts, such as ammonium nitrate $\left(\ce{NH4NO3}\right)$.

Thus, the paper probably used the enthalpy results because they were comparing their calculations to experimentally-measured enthalpies (heats) of their process(es) of interest.


Calculating Vmax (or Kcat) given specific activity of an enzyme over length of time incubated with an activator Is it possible to figure out the $V_\mathrm{max}$ (or $k_\mathrm{cat}$) and/or $K_\mathrm m$ from a plot of enzyme activity expressed as specific activity over time (i.e. $\mathrm{U/mg}$ on the $Y$ and time on the $x$ axis). I've figured out how to express specific activity as turnover but I'm not sure this helps with calculating $V_\mathrm{max}$ or $k_\mathrm{cat}$.

The reaction:
p38 phosphorylates MSK1 with Michaelis–Menten kinetics

Experimental conditions:
$1\ \mathrm{U/mL}$ of p38 was added at time $t=0\ \mathrm s$ to a solution containing MSK1. They don't say how much MSK1 but express its activity as $\mathrm{U/mg}$ (which is analogous to $\mathrm{(\mu mol/min)/mL}$, otherwise known as specific activity)

The Data:

Time         MSK1 activity
(s)          ((µmol/min)/mg)
__________________________________________________
   0               0
 300             175
 600             350
1200             450
1800             475
2400             500

Calculated turnover:

Time         MSK1 turnover
(s)          (1/(nmol s))
__________________________________________________
 18000          300
 36000          600
 72000         1200
108000         1800
144000         2400
     0            0

In actual fact, I though $k_{\rm cat}$ was the turnover and that this was a constant, specific for each enzyme. This data however suggests that as the reaction progresses more MSK1 is produced the turnover number increases, which makes sense and therefore calls into question the two definitions of $k_{\rm cat}$.

The enzyme catalysis reaction following MM kinetics can be depicted as:

$$\Large\ce{E + S<=>[k_f][k_r]ES->[k_{cat}]E + P }$$

Note that $V_{max} = k_{cat}.[E_0]$ i.e. conversion rate constant (turnover number) times the total amount of enzyme.

You should note that the enzyme here is p38 and substrate is MSK1 (which in turn is an enzyme). When MSK1 gets phosphorylated, it assumes an active form. MSK1 activity (and specific activity) would reflect the concentration of phosphorylated MSK1 (MSK1-P) i.e. the product. Note that the $k^{^{MSK1-P}}_{cat}$ is its intrinsic property and it does not change; only the $[E^{^{MSK1-P}}_0]$ changes.

I hope your doubt is clear.


What is the solid crud that appears inside my kettle? Almost every share-house I've lived in has had some nasty looking solids collecting inside.  I always regret looking inside the kettle because a cup of tea never tastes quite as good after you've seen this:

What is that stuff?  What's the physical process that causes it to collect and why is it usually concentrated around the heating element?  Is it related to calcium in tap water, or is there any trick to prevent it appearing in your kettle?  Is it dangerous to ingest and should be removed, or is it harmless?

This is limescale and you have it because your tap water is rich in minerals (mainly, calcium, as you note). You could remove these minerals with an ion exchanger, but you wouldn't to it with a tap water since these minerals give it a better taste. Limescale is not dangerous to ingest, but certainly not pleasant either (and I see on the photo that there is a slot for a mesh that should filter scales). Thick layer of limescale is also bad for the heating element – it hinders heat transfer to water and leads to overheating.

It is a good idea to remove limescale once in a while. You can dissolve limescale with any weak acid, like vinegar, but vinegar smells, so I suggest buying a descaling agent.


Regioselectivity in addition of hydrogen iodide to vinyl chloride I encountered a question which was stated as:

  The reaction of vinyl chloride with hydrogen iodide to give 1-chloro-1-iodoethane is an example of anti-Markovnikov addition. Is this true or false?

The way I could see it, vinyl chloride has a dipole moment as shown below, since chlorine is more electronegative than the carbons in the double bond.

Therefore the carbon on the left would have a greater partial positive charge than the carbon on the right. I therefore concluded that in a Markovnikov addition, the iodine atom would add to the CH2 carbon.

Since the question states that the product is 1-chloro-1-iodoethane, with the iodine atom adding to the CHCl carbon, that would thus be an example of anti-Markovnikov addition, according to the argument above.

I was told, however, that the reaction shown above was indeed Markovnikov addition. How is this so?

As far as I can see, that reaction follows Markovnikov's rule. The hydrogen goes where there is more hydrogen and the halide goes where there is less hydrogen.
If you are confused as to why the product of that reaction is 1-chloro-1-iodoethane, it's because the electromeric effect the chloride atom exhibits trumps the inductive effect. We need to take the electromeric effects into consideration because the chloride atom is attached to a $\mathrm{sp^2}$ carbon. The inductive effect makes it attract electrons to itself, but the electromeric effect makes it push them away, like a $\ce{CH3-}$ would.

It's the reason why $\ce{CH3-}$ and $\ce{Cl-}$, when attached to an aromatic nucleus, orient new substituents in the same positions, ortho and para (though $\ce{CH3-}$ has a stronger effect than $\ce{Cl-}$).


Why does nose grease knock down the head on beer and soda? Freshman year, my first semester, I was taught many things.  One important trick I learned was when I was passed a beer in a 12 oz. plastic cup, I could eliminate the fresh foamy head by wiping my nose and touching it to the top of the foam.

I thought to do this recently with soda (as it seems soda head is more persistent).  The trick works there too.

What is the chemistry behind this?  Is it the salinity?  Could I use something less revolting than nose grease to do the same trick?

Any oil should work.

Naturally occurring surfactants, such as lecithin, make bubbles or foam because one end of the detergent molecule is hydrophilic while the opposing end is hydrophobic (not rabietic, though). Without oil, the hydrophobic end prefers air to water, encapsulating the air. With oil added, though, the water-repellent end is attracted to the oil, forming micelles, instead.

You could put vegetable oil or mineral oil on your finger, but then the server might wonder why you stick your finger in the beer.

You might also try making a beer-powered boat! Tell your server, and ask for a free beer to prove that this works.


Why do heavier transition metals show higher oxidation states? In p-block elements, higher oxidation states are less stable down the group due to the inert pair effect. This is not the case for transition metals.

Why do heavier transition metals show higher oxidation states than lighter ones? Is the inert pair effect not valid for transition metals also?

The inert pair effect is based on the fact that main group elements’ oxidation states depend on s and p orbitals (and only them). When going down the periodic table, the energy difference between s and p orbitals changes leading to some elements losing their valence s electrons more easily than others.

Transition metals’ chemistry happens in the d orbitals primarily — unless you count the copper and zinc groups wherein a significant part of the chemistry is in fact only s orbital chemistry. The d orbitals — at first approximation of the free ion — are degenerate, i.e. they all have the same energy. Thus, these electrons are typically much more accessable. Furthermore, going down the periodic table increases the number of electrons counted as core electrons meaning that the outermost valene electrons experience a weaker effective nuclear attraction. It is therefore easier (i.e. requires less energy) to remove valence electrons and higher oxidation states are much more accessible.

If it weren’t for the inert pair effect, this would also be visible in main group chemistry — except it is, if you compare the relative stabilities of iodate and chlorate.


Would osmosis occur between two solutions of equal moles of water but unequal moles of a nonpolar molecule? Earlier I learned that osmosis was, like diffusion, a result of the laws of entropy, kinetic energy, and random movement. However, if you have two solutions of equal moles of water, then net movement should be zero, right?

In the case above with equal moles of water, if you now have a semi-permeable membrane (allowing only water molecules to cross) and solutes dissolved in the water (at difference concentrations), then there would still be net movement of water molecules towards the concentrated solution. There was an explanation in my AP Bio textbook that the water molecules hydrating the solutes were unable to pass through the membrane, so net movement occurs because of a difference in free water concentration.

Now I am just wondering what would occur if you had a nonpolar molecule instead of a solute. Would the arrangement simply remain at this?

I just wanted to see if I truly understood this concept.

Your example with the no-polar substance wouldn't qualify in a discussion of osmosis. There needs to be a solution on at least one side of the membrane. Since a typical non-polar substance would be insoluble in water, osmosis would not apply.

However, in your diagram, water would move from left to right due to a concept called pressure potential. You will eventually learn about this in AP Bio (I taught it for 24 years.) Pressure potential and osmotic potential work together, in a concept called water potential, to calculate which way water will move across membranes. You will probably perform a lab utilizing this concept this year. Pressure potential, in your example, is just a fancy term stating that because both the atmosphere and the non-polar substance on top of the left side are greater than just the atmosphere pushing down on the right, water will move to the right.

One of the things you should be careful about is that it is not the number of moles that are important, it is the concentration of substances. Get rid of your non-polar substance. Move the membrane over to the left so there is much more water on the right. Imagine the water levels are equal on both sides so we don't have any pressure potential to deal with. In this scenario, there would be no net movement of water even though there's far more total water on the right. This is because the concentration of water on both sides is the same, 100%


Why does Soda Pop after Shaking Why does shaking a soda can and opening it make the soda jump out of the can?

Soda drinks are carbonated meaning that carbon dioxide is dissolve into it. When carbon dioxide dissolves, the following reaction occurs: $$\ce{CO2(g) + H2O(l)~ \leftrightharpoons ~H2CO3(aq)}$$ As you can see, carbonic acid (a weak acid) is formed. It this which is responsible for the 'fizziness' of the drink.

Note that this reaction is in equilibrium, meaning that not all of the carbon dioxide dissolve to form carbonic acid, but only a certain percentage actually dissolves. Usually, the percentage of carbon dioxide that is dissolved is constant, however this can be changed by doing certain things to the system.

For example, by decreasing the pressure the equilibrium will shift to the left, meaning that the percentage of carbon dioxide that is dissolved will decrease. This will result in an increase of gaseous carbon dioxide in the bottle. This reason for this is explained by the Le Chatelier's principle. You can also imagine it as a piston pushing the gas into the liquid as shown in the below image. Image (a) is at low pressure while image (b) is at high pressure.

Normally in soda bottles, the pressure inside the bottle is relatively high to ensure that more carbon dioxide dissolves into the drink. So when you open a soda can you decrease the pressure. This results in more carbon dioxide gas to produced. It is the release of this gas that causes that pfffftt sound.

Now when you shake the bottle, due to turbulence, small bubbles containing carbon dioxide form.The formation of these small bubbles allow even more carbon dioxide gas to form. So if you were to shake a soda can and then open it soon after, the soda will jump up quite violently as there is now a lot of carbon dioxide being released.

However if you were to shake a soda can and then wait a while before opening it, it most likely just create the normal pfffftt sound. This is because you give the system some time to go back to its original equilibrium position, meaning that some of the carbon dioxide gas will redissolve back into the solution .


How to remove nitrose gasses from ozone gas mixture? I produce ozone gas with an ozone generator based on corona discharge principle.
These kind of ozone generators have the disadvantage that if $\ce{N2}$ is present, a bit of $\ce{NO}$ and $\ce{NO2}$ will be produced too.
To produce oxygen I use a oxygen concentrator, with an output of about $90\%$ $\ce{O2}$ and about $10\%$ $\ce{N2}$.
For that reason I have a little quantity of $\ce{N2}$ entering the ozone generator, which subsequently lead to unwanted side reactions like $$\ce{N2 -> NO -> NO2}.$$

The $\ce{NO2}$, even in that small quantity as it is, does influence the following processes in an negative way.

The easiest way to circumnavigate this issue would be to take pure oxygen – however for cost reasons I cant use LOX, so I have to stay with the concentrator.
I am searching for an reagent or an catalyst, which is not effected by the ozone but does absorb or disintegrate the $\ce{NO}$, and especially the $\ce{NO2}$.

To stripe $\ce{NO2}$ from a normal gas is not so difficult, but the ozone does make trouble.
I thought about dry $\ce{NaOH}$, soda lime, or dry $\ce{KOH}$, but I have the feeling that these chemicals will react with $\ce{O3}$.

Any suggestions?

There is no need to use caustic since $\ce{NO2}$ reacts with water itself. As per [1] it is even better absorbed in water than in $\ce{NaOH}$.

Ozone is soluble in water depending on the concentration of $\ce{O3}$ in the gas phase. It is much more soluble than oxygen so you will loose some, but it will have much lower concentration of $\ce{NO2}$. Of course, the washed ozone will be moist. If you need it dry, you can use a column of molecular sieves or some other desiccant.

[1] Absorption of Nitrogen Dioxide by Aqueous Solutions,
F. S. Chambers Jr., T. K. Sherwood,
Ind. Eng. Chem., 1937, 29 (12), pp 1415–1422


Olfactory Receptors Bonding with Organic Material How strong is the bond between the olfactory receptors and an organic material, like a perfume?

Is there a chemical that is safe for inhalation that would weaken this bond and prepare the olfactory receptors to receive new scents?

I know that materials like charcoal and zeolite absorb these molecules, but they are carcinogenic when inhaled, so I am searching for an alternative. Any and all advice is greatly appreciated!

Thanks

The so-called "bond" you are talking about is relatively weak. Binding between most odiferous compounds (most of them are organic compounds, but there are some inorganic ones like ammonia and hydrogen sulfide) and olfactory receptors (think proteins with a groove that fits the shape of the odor chemical) is rapid and reversible. There are no covalent bonds formed- just weak intermolecular forces like London forces and maybe hydrogen bonding. You need that rapid reversibility in order to tell which way a smell is coming from  or to track something by smell.

For sure, high doses of odiferous compounds can temporarily overload your olfactory receptors - the result being that your sense of smell is degraded or qualitatively changed for minutes or perhaps hours. I do not know for sure but I suspect that some chemical substances (probably only if they are designed to do so) can irreversibly bind to receptors (similar to so-called "suicide" substrates that inactivate an enzyme) and permanently inactivate them, in which case the olfactory cells would have to produce more receptors (hours to days?) in order to restore your sense of smell.

The question I think you need to consider is: is there a chemical substance that would purge the olfactory receptors of old odor compounds and render them freshly able to detect new odors? I think the answer may be no. Even if you could place a zeolite or activated carbon particle next to an olfactory receptor, the odor chemical that is bound to it would still have to diffuse out (on its own) in order to bind with the particle.

But lets see if others out there have any ideas about this. And give my regards to your grandmother!


How is molecular orbital theory used in drug research? I was reading up on MO theory and came across a particular section:

  Application: Computational Chemistry In Drug Design

  While the descriptions of bonding described in this chapter involve many
  theoretical concepts, they also have many practical, real-world applications. For example, drug design is an important field that uses our understanding of chemical bonding to develop pharmaceuticals. This interdisciplinary area of study uses biology (understanding diseases and how they operate) to identify specific targets, such as a binding site that is involved in a disease pathway. By modeling the structures of the binding site and potential drugs, computational chemists can predict which structures can fit together and how effectively they will bind.

Whilst I can see from the idea of in-phase and antiphase combinations of wave functions that you could potentially 'match' two atomic orbitals together but I am confused as to how this happens over a large molecule and the binding site of the target?

Common drug design approach does not involve much QC. There are two general approaches - QSAR and docking

QSAR

The idea is that a family of chemicals (training set) with some having desired biological activity, is used to train a pattern recognition algorithm. The trained algorithm is then used on a library of candidates and once recognised as active are then synthesised and tested in a lab.

Docking

The idea is to construct a molecule strongly interacting with a biological important target (a receptor or an enzyme). Since intermolecular interaction are weak, their calculation from first principles is very computationally expensive and is not practical in most cases. Since biologically relevant molecules are usually fairly trivial, even if complex molecules (For example, there is usually no need to consider interactions involving continuous conjugated systems), the intermolecular interactions are easily described using approach of molecular mechanics. So, in this approach a set of molecules is generated and then tested for interaction with a protein target. The tight spot is to find a way two molecules can bind, typically involving search over extremely large conformer space. Billions of potential candidates for ligand-protein complex must be tested, so the computationally cheapest solution (molecular mechanics) is used.

========

As you can see, neither approach normally uses orbital theory or any complex QC, and even if such complex QC is involved, it is clearly subservient. Instead machine learning algorithms and ways to efficiently search conformation space for low-energy conformers are the focus of research.


Halogenation of 4‐methylcyclohex‐1‐ene My problem is with the methyl group. If it wasn't there, I had no problem.
When the halonium bridge is made does the methyl group shift to give a more stable transition state or not?

In order for a methyl shift to occur, there would need to be an empty orbital resulting from a carbocation intermediate for the methyl group to occupy.

Note that in the bromonium intermediate shown above, neither carbon in the three-membered ring has an entire empty orbital available. Hence methyl shift doesn't occur and the mechanism continues as expected, resulting in halohydrin formation.


Controlling the reaction of alcohols with acid to either ether synthesis or elimination Let’s say you take ethanol: $\ce{CH3-CH2-OH}$ and you add an acid and heat. You can get water, the acid, and $\ce{CH2=CH2}$ (a dehydration reaction). However, you can also get $\ce{CH3-CH2-O-CH2-CH3}$, water, and an acid (symmetric ether synthesis). How would you make one reaction preferable over another?

By modifying the conditions, in this case: the concentration.

The first and rapid step of the mechanism is the protonation of the alcohol:

$$\ce{HA + CH3CH2-OH <=> A- + CH3CH2-OH2+}$$

From then on, it depends what happens:

If the molecule is left alone, i.e. its concentration is low, something that has a just enough basicity might at the correct time be close enough to a methyl hydrogen to allow the elimination to happen. $\mathrm{E1}$ is not an option since the primary carbocation is extremely unstable and Wagner-Meerwein rearrangements for enhanced stability are not possible, thus the elimination must preceed by the $\mathrm{E2}$ mechanism. If $\ce{A-}$ is the only basic compound present, it must accept the proton.
$$\ce{A- + H-CH2-CH2-OH2+ -> A-H + H2C=CH2 + H2O}$$
If the molecule happens to find one if its kind before the former can take place, the two of them could engage in a substitution reaction by the $\mathrm{S_N2}$ mechanism. If the molecule’s concentration is high, this gets more likely. And if it happens, the second molecule’s hydroxy group is nucleophilic enough (even though alcohols are generally bad nucleophiles) to displace the extremely good leaving group $\ce{-OH2+}$ and liberate water. The extraneous proton can later be removed as the ether sees fit.
$$\ce{CH3CH2-OH2+ + HO-CH2CH3 -> CH3CH2-(OH+)-CH2CH3 + H2O}$$
$$\ce{CH3CH2-(OH+)-CH2CH3 + A- <=> CH3CH2-O-CH2CH3 + HA}$$

Thus, the first reaction will always happen but be rather slow. Its rate is close enough to being independent of the molecule’s concentration. The second reaction’s rate is directly dependent on the molecule’s concentration and will get faster the higher the concentration is. High concentrations (and low acid concentrations) can thus favour the substitution mechanism over the elimination one.

This will hold true more or less for all cases where two reactions compete against one another. It is always possible to modify the conditions to prefer one or the other. However, it is often not as simple as in this simple case.


Predicting product of a Hofmann elimination with a cyclic amine What is the final product of this reaction?

Here's what I have:

Does this look correct? I'm concerned because I'm used to seeing the nitrogen removed from the product (in the cases of primary amines).

With excess of methylamine, I wouldn't exclude additional methylation, yielding a trimethylanilinium iodide.

However, the reaction should stop there since a further hydrogen abstraction from a $\beta$-position isn't possible.


Why are catabolic reactions exergonic and anabolic reactions endergonic? Catabolism breaks apart large molecules and anabolism creates large molecules. Catabolism is supposed to be exergonic, while anabolism is supposed to be energonic. Id on't really understand this.

By definition a chemical bond is at an energy minimum. So if catabolism is taking apart large molecules with many bonds, which are at an energy minimum, how is this supposed to release energy (be exergonic) if the bonds are already at an energy minimum?

By contrast why is it that anabolism, which is putting creating a large molecule (which is a minimum energy state) from many atoms is enderognic and thus requires energy? I mean you're creating a minimum energy state, why would this require energy?

In most chemical reactions, including those of catabolism and anabolism, some bonds break and others are formed. For example, if you hydrolize a biopolymer such as a polysaccharide or protein into its building blocks, you break bonds in the polymer and in water, but form new bonds between parts of the water and parts of the polymer.

Making a polymer from building blocks sometimes is endergonic (like some anabolic reactions) and sometimes is exergonic (like making polyethylene from ethylene). There is no inherent rule that says more complex molecules have higher or lower Gibbs free energy than the small molecules they are made from.

You could say there are three reasons why we eat food: To directly use the substances we ingest (e.g. vitamins), to use the substances we ingest to make new substances we need (e.g. amino acids in proteins), and to oxidize substances (in humans with $\ce{O_2}$ as the terminal electron acceptor) and harvest the Gibbs free energy. We choose food accordingly (e.g. we don't eat sand). Also, we don't eat Grignard reagents or other highly reactive substances because they are usually not available and we are not equipped to handle very exothermic reactions in our bodies. So the food we eat is fairly stable, and often it requires Gibbs free energy to drive biosynthetic reactions (explaining why many anabolic reactions are endergonic)

Molecules that can be hydrolyzed in the digestive tract using exergonic reactions are good as food. It is difficult to run endergonic reactions in the digestive tract (the reactants often used to drive endergonic reactions, most notably ATP, are found inside cells, not in the stomach or intestine). For those reasons, any large molecules we digest can be broken down to smaller molecules in exergonic reactions (these smaller molecules then enter cells, where further reactions could be exergonic or endergonic).

Specifically for hydrolysis (breaking biopolymers into building blocks) and condensation reactions (making biopolymers from building blocks), the hydrolysis reactions tend to be exergonic and the condensation reactions tend to be endergonic. This might be related to the high concentration of water throughout the body.


Can substituted acetone undergo Cannizzaro reaction? I know the Cannizzaro reaction takes place in carboxyl compounds without $\alpha$-H with the regular examples being formaldehyde and benzaldehyde.

But can this reaction take place with say diphenylmethanone or di ter-butyl ketone (2,2,4,4-tetramethylpentan-3-one) or 2,2-dimethylpropanal ?

I couldn't find any examples of these compounds showing the cannizzaro reaction.

As far as I'm aware, the Cannizzaro reaction can't occur with ketones, only aldehydes. The reaction mechanism involves the transfer of hydride from the former carbonyl carbon (of the tetrahedral intermediate resulting from hydroxide's nucleophilic addition to the carbonyl carbon), during which step the carbonyl reforms, yielding a carboxylic acid. Since a ketone, by definition, has no hydrogens bonded to the carbonyl carbon, it obviously can't undergo such a reaction. I would expect ketones with no $\alpha$-hydrogens to be essentially unreactive with hydroxide, since aldol reactions and keto-enol tautomerism are impossible with no $\alpha$-hydrogens, and most hydrates are unstable.


Reduction of 2-cyclohexenone to cyclohexanol Is $\ce{NaBH4}$ a suitable reducing agent for the conversion of 2-cyclohexenone to cyclohexanol?

In Clayden (page 506, second edition), I have found the reduction of cyclopentenone to cyclopentanol using $\ce{NaBH4}$ in $\ce{Et-OH}$. (99% yield!). First NaBH4 performs 1,4 addition and then 1,2 addition to give the product.

However, I am unable to find a reference for reduction of cyclohexenone to cyclohexanol.

I found this in Smith:

I don't understand why it can't reduce the double bond through Michael addition in case of cyclohex-2-enone as well. What is the reason behind it?

According to Johnson, M. R.; Rickborn B. Sodium borohydride reduction of conjugated aldehydes and ketones J. Org. Chem., 1970, 35, 1041.
DOI: 10.1021/jo00829a039:

2-Cyclohexenone can certainly be reduced fully. It has very similar reactivity to cyclopentenone. In this report, mixtures of products were observed. The experimental indicates the researchers used 0.5 eq of $\ce{NaBH4}$ per carbonyl.

Furthermore, the use of Luche conditions can again help you favor only 1,2 reduction:


Writing Ionic Equation According to the rules of writing ionic equation, only compounds in aqueous state can be split or dissociated into their ions, whereas elements or compounds in solid, liquid or gas state remain as they are. But what about molten sodium chloride which is NaCl(l) in liquid state, isn't it dissociated into ions when melted? So can we split it into its respective ions when writing ionic equations?

The statement, "only compounds in aqueous state can be split or dissociated into their ions," is false in a number of ways.

As you state, molten salts are ionic, and there are room-temperature ionic liquids, such as ethylammonium nitrate , $\ce{(C2H5)NH3+·NO3-}$.
In addition to water, there are other ionic solvents, such as nitromethane, $\ce{CH3NO2}$. Salts in these solvents break up to form ions.
Further, not everything dissolved in water is ionized. Sugar (sucrose), for example, is not ionized.


Gaussian vs ABINIT for solids I see that Gaussian has feature to set periodic boundary conditions specifying the parameter Tv in the input file. Does it do it via the plane wave basis set? Also, the question is whether it is efficient to use Gaussian to compute the relaxed lattice constant, band structure and DOS for solids or it is better to use programs like ABINIT.

Even if it is possible to calculate periodic structures in Gaussian, I would not suggest it.
Better option is to use codes devoted to calculation of solids, among others the ABINIT you mentioned, or e.g. Quantum Espresso.

Gaussian does not use plane wave basis set, will be slower and in general less powerful than codes devoted to solids.


How to determine the mass of ammonium nitrate needed and enthalpy of solution? I am trying to figure out how to determine the mass of ammonium nitrate needed to reduce the temperature of solution from $\pu{23 ^\circ C}$ (measured room temperature at start) to $\pu{3 ^\circ C}$, once it dissolves in water.

I have the formula:

$$Q=ms\Delta T$$

I looked up the enthalpy of solution for ammonium nitrate and found it to be $\pu{25.41 kJ/mol}$.

I also have data that if I add $\pu{15 g}$ of $\ce{NH4NO3}$ the temperature will decrease by $\pu{25 ^\circ C}$, if I add $\pu{10 g}$ the temperature will decrease by $\pu{20 ^\circ C}$, and if I add $\pu{5 g}$ the temperature will decrease $\pu{13 ^\circ C}$.

Clearly then, to get my solution to be $\pu{3 ^\circ C}$, I would need to add a mass somewhere in between $10$ and $\pu{15 g}$.  In the lab I could just add random amounts between until the solution settled on $\pu{3 ^\circ C}$. Shouldn't there be a mathematical way to solve this rather than by random guessing?

I tried to use $Q=ms\Delta T$ by substituting the enthalpy of solution (in J) for $Q$, the specific heat of water for $s$, and $\pu{20 ^\circ C}$ for $\Delta T$ (since I am going from $23$ to $\pu{3 ^\circ C}$). However, solving $$\pu{25410 J} = m (\pu{4.186 J// ^\circ C})(\pu{20 ^\circ C})$$ gives me $\pu{303.66 g}$ which is clearly nowhere near the $\pu{10 - 1 g}$ that is sufficient to change the temperature.

How can I accurately determine the mass using a formula?

I am trying to figure out how to determine the mass of ammonium nitrate needed to reduce the temperature of solution from $\pu{23 ^\circ C}$ (measured room temperature at start) to $\pu{3 ^\circ C}$, once it dissolves in water.

The result will depend on the amount of water where you're dissolving the salt. For instance, the data you present refers to different solvent mass. And by the way, for your temperature difference ($\pu{20^\circ C}$) the problem is already solved: a $\pu{20^\circ C}$ temperature decrease, will require $\pu{10 g}$ of the salt.

It is important to distinguish the mass of the salt from the mass of the solvent.
\begin{align}
Q &= m_\mathrm{salt} \times \Delta_\mathrm{solution} H^\circ_\mathrm{salt}\\
Q &= m_\mathrm{water} \times C_{p(\mathrm{water})} \times \Delta T\\
\therefore m_\mathrm{salt} &= \frac{m_\mathrm{water} \times C_{p(\mathrm{water})} \times \Delta T}{\Delta_\mathrm{solution} H^\circ_\mathrm{salt}}
\end{align}

Hope this helps.


Why isn't nitrous acid a product of nitrite and water? Question
  Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the equations
  $$\ce{NO2- + H2O -> ?}$$

I thought it would be $\ce{NO2- + H2O -> HNO2 + OH-}$, but my textbook says that the correct products are $\ce{HNO3}$ and $\ce{OH-}$. What am I doing wrong here? The answer that they give doesn't even seem to balance.

It should form nitrous acid... If it was $NO_3^{-}$, nitrate ion, then it would presumably form nitric acid $HNO_3$.

Nitrous acid reacts with water to form nitrite and hydronium:

$$\ce{HNO_2 + H_2O -> NO_2^- + H_3O^+} $$

Water is an amphoteric species. Assuming we just have water the nitrite ion, the nitrite ion accepts the hydrogen forming nitrous acid:

$$\ce{NO_2^- + H_2O -> HNO_2 + OH^-} $$

This reaction is interesting as well since nitrous acid will also dissociate again into nitrite ions and water as well:

$$\ce{HNO_2 + H_2O -> NO_2^- + H_3O^+} $$


Interpretation of the following mass spectrum I'm given the above spectrum. $M^+ = 102$, probably $\ce{C_6H_14O}$ or $\ce{C_5H_10O_2}$.
I'm thinking of ethyl propionate. (that picture also shows the fragments I expect). I can account for the following peaks:
m/z = 73: Loss of ethyl from acyl side (expected lowish intensity)
m/z = 57: Loss of alkoxy to form $\ce{CH3CH2CO+}$ (moderate intensity)
m/z = 45: Alkoxy ion (weak)
m/z = 29: ethyl carbocation, moderate intensity.
I can't account for 74, since according to my book the alkoxy side would not proceed through normal Mclafferty, but the Mclafferty + 1 rearrangement... That should yield m/z=75.
I proposed methyl butyrate, and of course that's not right. My instructor wrote to pay attention to the m/z = 57 (M-45) and m/z = 85 (M-17). This fits well with carboxylic acids, but again cannot account for the m/z = 74. McLafferty of carboxylic acids would result in a fragment of m/z=60...
Can someone help me out, I've come to a dead end for hours now.

I was on to the wrong structure. My main issue with the carboxylic acid was the fragment m/z=60 that it would produce by McLafferty. I didn't have that, so I thought that left me with t-Bu-COOH, which was the only structure with no $\gamma$-hydrogens. Besides I couldn't explain the m/z=74 peak with an acid. On the other hand I couldn't explain the M-17 peak by ethylpropanoate.

But of course, if the carboxylic acid has a methyl substituent at the $\alpha$-carbon, then that methyl group would also be part of the McLafferty fragment, therefore resulting in m/z=74!

The spectre is of: 2-methyl-butanoic acid

M-17 and M-45:

Alpha cleavage of carbonyl => loss of hydroxyl radical or carboxyl radical

m/z = 74:

Loss of ethene by McLafferty rearrangement

m/z = 45:

$\ce{COOH^+}$ fragment

M-15 and M-29:

Loss of methyl radical or ethyl radical

m/z = 29:

Ethyl carbocation


4-Bromo, 2,4-di-methylhexane and 3-Bromo, 3,5-di-methylhexane are the same thing? Came across the following question. The answer only says that option 3 is incorrect which is I agree to but dont you think option 4 is also incorrect. As per IUPAC rules isnt it the Halide that should get lowest possible number (in this case number 3 instead of 4) or is it both are same ? I am confused please help.

Both names, ‘4-bromo-2,4-dimethylhexane’ and ‘3-bromo-3,5-dimethylhexane’ are unambiguous and describe the same compound. However, only ‘4-bromo-2,4-dimethylhexane’ is in accordance with IUPAC recommendations.
First of all, the seniority order of functional classes, which is used to choose the functional class (here: …hexane), does not apply to numbering of locants within the same compound class.
Concerning numbering of locants within the same compound class, 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:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(…)

Furthermore,

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with
other locant sets, each cited in order of increasing value, has the lowest term at the first
point of difference; (…)

Therefore, the example is correctly named as 4-bromo-2,4-dimethylhexane rather than 3-bromo-3,5-dimethylhexane since the locant set ‘2,4,4’ is lower than ‘3,3,5’.


How does turbidity affect dissolved oxygen? I have to do a EEI for school about turbiduty of water and finding the most cost effective method for purifying a sample of water. My group was planning to do a standard test where we use materials such as filter paper, cloth, sand, gravel, etc. and run the turbid water through it. As our school has no turbidometers we wanted a way to have some quantitative data. To do this we were planning on conducting a dissolved oxygen test (wrinklers method) as we found some sources linking turbidity to oxygen levels. Would this be a valid test? Would the filtered water need to sit overnight, for example, before we tested the DO?

And help is appreciated!!!

According to this document, turbidity and oxygen content can be correlated because the more microbes in the water, the more turbid the water is, and the less oxygen is left (because they breathe it).

But a correlation is no cause-effect relationship. Both are linked in the particular case of microbes because they affect both by their presence.

Not a reliable method at all. More oxygen in the water does not mean it is potable. If you let sit a bucket of water contaminated with microbes, the oxygen levels will drop over time, but if you stir the water, you re-dissolve some up to saturation levels, but you still have the microbes present.

You may want to evaluate the oxygen demand, or the biomass (absorption at 600nm), centrigugate and evaluate visually, etc.


Chemical equilibrium — why multiplication, rather than addition? Why do we multiply (and not take the sum) the concentration of products and reactants when more than one entity is present in either side and rise the power to the coefficient term?

Suppose either

$$\ce{aA + bB <=> cC +dD}$$

described by

$$K_{\mathrm{eq}} = \frac{[\ce{C}]^c \cdot [\ce{D}]^d} {[\ce{A}]^a \cdot [\ce{B}]^b}$$

where we multiply the concentrations and rise the power to the coefficient. Or the dissociation of

$$\ce{A_aB_b <=> aA+ + bB-}$$

described by
$$K_{\mathrm{sp}} = [\ce{A}]^a \cdot [\ce{B}]^b$$

again based on a product, and not the sum.

This seeks a clarification of math based on chemistry. How to explain it to an 8th grader?

For the 8th grader, hmmm...

Let suppose this funny equilibrium equation:

$$\ce{2 S <=> H}$$

Where $\ce{S}$ are ice skaters on a giant skating ground and $\ce{H}$ are "collision heaps" of 2 skaters laying on the ice.

Now, what is the frequency of collisions leading to a heap, if there is "skater concentration" $\ce{ [S] }$ ?  It should be $k_1 \ce{[S]}$ as frequency should be proportional to $\ce{[S]}$, correct ?

But wait ! When there is more skaters, each skater would collide with  other skaters more frequently,  proportionally to $\ce{[S]}$ as well !

So collision frequency would be like $k_1 \ce{[S]}^2$

Now the heap "recovery":

It should be proportional to $\ce{[H]}$, as more heaps, more skaters would be  raising at the same time. So the heap recovery rate would be like $k_2 \ce{[H]}$.

Now imagine the equilibrium on the skating ground, where the rate of collisions is equal to the recovery rate
and number of collision heaps is constant.

Then $k_1 \ce{[S]}^2= k_2 \ce{ [H]}$

If we express the equilibrium constant $K=k_1/k_2$, then $$ K = \frac{\ce{[H]}}{\ce{[S]}^2}$$

Do you see ? No sum, but multiplication. Twice as many skaters leads to 4 times more heaps.

You can create similar system, where just for fun collide just boys and girls together.


Relationship between the water ionization constant and ionic strength I read in a textbook that

  the water ionization constant ($K_\mathrm{w}$) increases as the ionic strength of the solution increases

and it confused me. If the ionic strength increased, then the activity of each of the ions $\ce{H3O+}$ and $\ce{OH-}$ would decrease, leading to a decrease in the value of $K_\mathrm{w}$.

Can someone please explain the mistake in my way of thinking?

The equilibrium constant is given though $$K_x = \prod_{\ce{B}} x_{\ce{B}}^{\nu_{\ce{B}}},$$
where $x$ is a specific quantity. We can choose that to be the activity $a$.

For the auto-ionisation
$$\ce{H2O (aq) <=>[H2O] H+ (aq) + OH- (aq)}\tag{1}\label{water}$$
we find therefore
$$K_a = \frac{a(\ce{H+})\cdot a(\ce{OH-})}{a(\ce{H2O})}.$$

Assuming the equilibrium $\eqref{water}$ is largely on the left side and therefore there is an excess of water, a reasonable simplification is to treat $a(\ce{H2O})$ as constant.
$$K_a' = K_a\cdot a(\ce{H2O}) = a(\ce{H+})\cdot a(\ce{OH-})$$

In ideally diluted solutions we also find
$$\lim_{c\to0}\gamma_i = \frac{a_i}{c_i}=1,$$
and therefore
$$K_c = c(\ce{H+})\cdot c(\ce{OH-}) \overset{\mathrm{def}}{=} K_\mathrm{w}.$$

The ionic strength is defined as
$$I_c = \frac12\sum_{\ce{B}}c_{\ce{B}}z_{\ce{B}}^2,$$
or for our case
$$I_{c,\mathrm{w}} = \frac12 \bigg[c(\ce{H+}) + c(\ce{OH-})\bigg].$$

Because of $\eqref{water}$ we can write
$$I_{c,\mathrm{w}} = c(\ce{H+})$$
and
$$K_\mathrm{w} = c(\ce{H+})^2, $$
therefore
$$I_{c,\mathrm{w}} = \sqrt{K_\mathrm{w}}.$$

From this relation you can see, that when $K_\mathrm{w}$ increases, the ionic strength also increases, vice versa.

  If the ionic strength increased, then the activity of each of the ions $\ce{H3O+}$ and $\ce{OH−}$ would decrease, leading to a decrease in the value of $K_\mathrm{w}$.

The first part is correct, the activity of each ion would decrease. However, there are more ions, so the activity of the ensemble increases more (given ideal dilution), therefore also leading to an increase in $K_\mathrm{w}$.


Substances - A Macroscopic Concept Does the term “substance” refer to any sample of matter that is tangible? In other words is “substance” a macroscopic concept or can we also refer to individual particles such as atoms, molecules, etc. as substances?

It depends on the context

"Substance" is not a tightly defined term so the argument is largely linguistic not about chemistry. Unless, that is, you are talking about observable chemical properties and use the word. Then, the context of use matters and you might need to be careful in how the word is used.

With gases it doesn't much matter how much stuff you have as the individual molecules don't interact much. But for solids and liquids when you talk about "substance" the apparent chemical properties may depend on how many molecules or atoms you have. So you might need to be careful how many of the components are present. The reason why I say this is that some collections of small numbers of components have different properties to larger collections.

Nanoparticles of gold, for example, have different properties to bulk metallic gold. Some show much stronger colours that metallic gold (eg small colloidal gold clusters are used to impart a red colour to some specialty glass products). So the small clusters have different properties to bulk gold and you don't see the bulk properties until the amount of gold is large enough.

So although the substance of both bulk gold and colloidal clusters of gold atoms might be said to be the same, it is probably better to be more specific and talk about colloidal clusters as being a different "substance" to bulk gold. But this isn't universally agreed as usage of substance is not precisely defined.

In short, if you are talking purely about what something is made from then the amount doesn't matter and you can say substance for any amount but if you are talkingg about the properties of the material it can be better to be specific as the properties do change with the amount of the components involved.


Is there a direct relationship between the adsorption energy and the catalytic activty I have seen in several papers, where people correlate adsorption energy with the catalytic activity; the greater the adsorption energy, the greater would be the catalytic activity.

I understand that in order for a catalytic reaction to take place, the adsorbates should get adsorbed on the surface of the catalyst. But what if the adsorbates are too strongly adsorbed on the catalyst. This is unfavorable for the reaction. Isn't it?

Therefore, I don't understand how one can correlate higher adsorption energies with greater catalytic activity.

Adsorption energy is indeed related to catalytic activity, but not "greater energy means greater activity".

As you were guessing, there is an optimum window in adsorption energies for which catalytic activity is maximum, this is explained by the Sabatier principle:

  The Sabatier principle is a qualitative concept in chemical catalysis named after the French chemist Paul Sabatier. It states that the interactions between the catalyst and the substrate should be "just right"; that is, neither too strong nor too weak. If the interaction is too weak, the substrate will fail to bind to the catalyst and no reaction will take place. On the other hand, if the interaction is too strong, the product fails to dissociate.

There is a nice recent paper in which you can find much more mechanistic details.


What's the reason for discrepancy between DFT calculated image and NC-AFM-acquired one in this article? Looking at the images in this article, I've noticed that the molecule electron density images showed as calculated via DFT look very symmetric, while those obtained by NC-AFM appear somewhat distorted. Namely, the ring at the left (the one with $\ce{OH}$ group) looks elongated, while the one at right is, on the contrary, compressed in horizontal direction, and the $\ce{OH}$ group seems to be tilted to the $\ce N$ atom. Also, the calculated image seems much more smeared, while the measured one gives some highly localized density "rods" for the bonds.

I've made a gif-combination of these two images so that the difference was easier to see:

Here the B is DFT-calculated version, while F is one measured by NC-AFM.

I can suppose one or several of the following might be among the reasons for this:

The substrate of $\ce{Cu(111)}$ wasn't taken into account when calculating(?), while in the measurement it appears to distort the molecule
DFT, being an approximate method, gives too simplified results
Smear vs thin rods difference might because in the DFT case the picture is of electron density while in AFM case the image shows AFM frequency shifts, which is not necessarily the same.
The probe of the microscope distorts the molecule

What are the true reasons for the differences?

1) Most definitely. Adsorbed molecules often are very distorted, up to chemical reaction with support, especially when reactive surface is used. Despite common misconception, even noble metal surfaces are quite reactive.

2) DFT, while being indeed inexact (as any calclation with limited size of basis set), catches major features of electronic density very well.

3) AFM stands for atomic force microscopy, which is not connected to electronic density directly. Now, naphthalene molecule features large $\pi$-system, which is very polarizable. Since nonvalent interatomic forces beyond electrostatic interactions usually increases with polarizability of interacting particles, I would expect $\pi$-systems  to be highlighted on the NC-AFM images.

4) I'm not an expert on atomic-resolved microscopy, but I never read about this possibility in the articles I worked with, though the works I usually works with used STM. and SEM mostly.