Why don't alcohols undergo nucleophilic substitution with ammonia? Alcohols can undergo nucleophilic substitution with $\ce{PCl5, POCl3, HCl}$.  Halogenoalkanes can undergo nucleophilic substitution with $\ce{NH3}$. The $\ce{C-O}$ (in alcohol) and $\ce{C-Cl}$ (in chloroalkanes) bond energies are comparable. Why don't alcohols undergo nucleophilic substitution with $\ce{NH3}$?

To build on ssavec's answer: Nucleophilic substitution reactions require two species: a nucleophile (a Lewis Base) and a substrate with a leaving group. Nucleophiles are often generically represented as $\ce{Nu}$ and leavings groups as $\ce{LG}$.

All leaving groups are not created equal. The leaving group ability of a leaving group is defined as the relative rates of a particular analogous series of substitutions. For example, the relative leaving group ability might be based on the following reaction, with rates determined for various leaving groups:

$$\ce{EtOH + PhCH(LG)CH3 -> PhCH(OEt)CH3 + H-LG}$$

This is an $\mathrm{S_N1}$ substitution, so the first (and rate determining) step of the mechanism is loss of the leaving group (and is independent of the nucleophile):

$$\ce{PhCH(LG)CH3 -> PhCH+CH3 + LG-}$$

The relative rates of this reaction are influenced by the stability of the $\ce{LG-}$ anion (see the Hammond Postulate, which proposes that the transition state of an endothermic process resembles the products).

The stability of the $\ce{LG-}$ anion can be predicted from the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid of $\ce{LG-}$, which is controlled by the same ability to stabilize a negative charge.

The conjugate acid of $\ce{Cl-}$ is $\ce{HCl}$, which is a strong acid. Using some Hess's Law trickery, the $\mathrm{p}K_\mathrm{a}$ of $\ce{HCl}$ has been predicted to be around $-8$.

The conjugate acid of $\ce{OH-}$ is $\ce{H2O}$, which has a $\mathrm{p}K_\mathrm{a}$ around $+16$.

$\ce{H2O}$ is a much weaker acid than $\ce{HCl}$, which means that the negative charge on  $\ce{OH-}$ is less stabilized than $\ce{Cl-}$. Thus, $\ce{OH-}$ is a poorer leaving group than $\ce{Cl-}$ (by a lot).

Another good discussion of leaving groups (and the $\mathrm{p}K_\mathrm{a}$ data used above) come from this site.

In your example reaction (ammonia + ethanol), the product of the reaction has a better leaving group ($\ce{NH3}$, conjugate base of $\ce{NH4+}$, which has a $\mathrm{p}K_\mathrm{a}$ of $+9.75$) than the $\ce{OH-}$ leaving group in the reactant, so the reaction will also run in reverse, and the equilibrium will strongly favor the reactants.

$$\ce{CH3CH2OH + NH3 <=> CH3CH2}\color{red}{\ce{NH3+}}\ce{+ OH-}\ \ K_\mathrm{a} \ll 1$$

As for your confusion about bond energies...

Bond dissociation energies are defined for homolytic bond cleavage, in which each atom keeps one electron when the bond breaks (see image).

Substitution reactions involve heterolytic bond cleavage, in which one atom gets both electrons:

Simply because ammonia is a great base(due to the presence of 2 lone pairs) and hence the basic character always dominates over the nucleophilic character.


Acid-base titration: Unknown base in burette or flask? While finding the concentration of $\ce{NaOH}$ via titration with $\pu{0.05M}$ Oxalic acid solution (in our lab) we put NaOH in the burette whereas added phenolphthalein to oxalic acid in titration flask. But in this video the demonstrator does just opposite ( i.e., $\ce{NaOH}$ in flak with phenolphthalein). I am confused in deciding which one might be correct. My view is that either would do the job but I need some clarification.

The base should be in the burette, and the acid in the flask. With this technique, the point of equivalence is obtained with a rather high degree of accuracy.


Is paper aging(oxidation) and paper burning(rapid oxidation) the same thing? I woke up today with a question of what is fire; looked up the definition and the first response was that fire was rapid oxidation.

I was curious what the difference was between rapid oxidation and regular oxidation that we have all around us as seen with metals rusting, book pages yellowing, artificial aging, paint drying, apples browning due to air exposure.

I cannot find an answer to what the difference between rapid and regular oxidation is, other than the speed of the reaction, and the visibility of the reaction when it becomes fast enough.

I am curious whether there is any difference at all between metals "rusting", paper "burning"/"aging", paint "drying", or if these are all different names for the same process at different rates.

Is it reasonable to say that there is fire everywhere there is oxidation occuring, but it's invisible until the reaction is fast enough?

Rapid oxidation- burning - leads to two main products carbon dioxide and water. You cannot keep any records if paper is burned.
In case of slow oxidation some of paper components are oxidized which changes the colour of paper, but just that. Old manuscripts can be stored for hundred of years if no none burns them.


How to find the mass of dissolved AgCl when placed in a solution of NaCl? I think the following problem in my textbook is malformed:

  Solid $\ce{AgCl}$ is placed in $1\,\mathrm{L}$ of $0.55\,\mathrm{M}$ $\ce{NaCl}$. Find the mass of $\ce{AgCl}$ which dissolves.

I know the method to solve the problem. I think the problem is malformed: When we get an answer, say $x$ grams $\ce{AgCl}$ dissolves, but if we placed less than $x$ grams solid $\ce{AgCl}$ in the solution initially, then it would be impossible for $x$ grams to dissolve.
So we obviously have to make certain assumptions, but I am looking at the solution in my textbook and I don't see in which part of the solution is the assumption used (implicitly). And by the way, how do we solve the problem if we have place less than $x$ grams solid $\ce{AgCl}$?

The question is most likely badly worded and should have been

  What is the maximum mass of $\ce{AgCl}$ that can be dissolved?

As we all know, $\ce{AgCl}$ is damn insouble in aquaeous solutions — and due to the equilibrium constant it doesn’t get better at all if you add chloride ions (reducing the silver ion concentration). So without having done the maths, I’m going to assume that a small spec of nothingness is all that will dissolve. That spec is so tiny that you will have a hard time weighing less so you can actually perform the experiment you suggested.

Because it’s just a flick of dust, and because the authors likely

knew that, too
assumend you knew that, too

they may have gone with a less obvious wording. I mean it’s clear: You can’t dissolve what you didn’t put in, and to-be-dissolved silver ions aren’t going to magically appear with Alakazam.

Since the already present sodium chloride highly ionic, it's dissociation can be considered 100% and thus the contribution of chloride form silver chloride would be thus negligible (in comparison to rock salt's) or suppressed. Thus we can treat the final chloride concentration as $0.55\rm M$. Now the solubility product is given as:
$$K_{\rm sp,AgCl}=\ce{[Ag+][Cl- ]}$$
Now we can find out $\ce {[Ag+]=\frac{K_{\rm sp,AgCl}}{[Cl- ]}}$. This would actually be all from silver chloride and hence the solubility of it.


CO2 removal from mushroom growing tent I'm trying to grow oyster mushrooms that like high temperatures. They look great in an indoor terrarium, and can eventually yield several tasty meals :)

However oyster mushrooms are hard to grow indoors, because they need low $\ce{CO2}$ levels (<1000 ppm according to Paul Stamets), AND high humidity (>95%).

Warm climate oysters tend to grow fast, and can generate quite a lot of $\ce{CO2}$ fast.

If I vent the air in tent about 10 times a day, humidity will suffer (and I can't do that all the time).

If I close the tent, mushrooms will be stunted and club-like, and yields will be next to nothing.

I'm thinking about putting a tray with $\ce{NaOH}$ crytals on it, probably with a fan to slightly move the humid air around a bit.

An other method would be to use an aquarium pump to bubble air through NaOH solution.

My questions are:

Does this look like something feasible? Can some $\ce{NaOH}$ remove enough $\ce{CO2}$ from the air to keep it under 1000ppm?
Which of the above methods seem to be the better one given the aquarium pump can bubble about 100-200 litres of air in every hour and the tent is about 500 L ?

Please stay away from sodium hydroxide, it's way too nasty to use in non-lab setup. You can handle it safely in dry conditions, but in contact with water, it becomes sticky and causes bad burns.

As a substitute, I suggest $\ce{Ca(OH)_2}$, readily available as slaked lime. You can structure it into relatively high-surface sculptures and it solidifies into common plaster upon binding enough $\ce{CO_2}$. It is often used for this purpose due to its low price, low toxicity and ease of handling.

In your situation, I would just put some vertically fixed mesh soaked in very viscous suspension of calcium hydroxide. In the high humidity conditions you'd have no problems with normal drying, so once it solidifies it has done its job and you can just hit the mesh against the ground to crumble the plaster away and put on new layer. It is advisable to use gloves and goggles, but it is not as necessary as it would be in the case of sodium hydroxide (look at all the masons around).

Stratification
Best of all, use no chemicals.  If you have an air circulating fan in your chamber, use it intermittently.  When not running, ambient air enters small hole(s) at the top of your chamber, and carbon dioxide will escape out small hole(s) at the bottom of your chamber.  CO2 at standard temp/press is invisible, but it definitely will sink beneath the unmoving oxygen and nitrogen of ambient air.


Solubility of compounds in water Why is $\ce{NaCl}$ more soluble in water than $\ce{NaOH}$? Well, this is a question from Chemical Equilibrium and the question hints for explanation in terms of Le Chatelier's principle.

It is not. $\ce{NaOH}$ is more soluble ($>100~\mathrm{g}/100~\mathrm{ml}$) then $\ce{NaCl}$ ($<30 ~\mathrm{g}/100~\mathrm{ml}$). $\ce{NaOH}$ when dissolved produces $\ce{OH^-}$ anion, that has relatively small size and forms specific hydrogen bonds (where it is bonded to relatively small positively charged hydrogen) very well. In solid $\ce{NaOH}$ it has to form hydrogen bonds with quite big $\ce{Na^+}$ cation.

The Water molecules move around the $\ce{Na}$ and $\ce{Cl}$ and break the salt apart into ions. The polarity of the water molecule allows this by the more negative oxygen being attracted to the positive Na ions and the positive hydrogen being attracted to the negative chlorine ions. In the case of NaOH, I am assume that bond between $\ce{Na+}$ and $\ce{OH-}$ is stronger than the bond in $\ce{NaCl}$. Also I don't see a case for the use of an equilibrium in this case because $\ce{NaOH}$ almost completely dissociates into its ions $\ce{Na+~~ OH-}$.


Can fluorine act as the central atom in interhalogen compounds? Why can't fluorine be the central atom in inter-halogen compounds?

A $\ce{F-F}$ bond is weaker than a $\ce{F-X}$ bond and thus, fluorine should be happy to form inter-halogen compounds. But, why doesn't it act as the central atom?

In my textbook, the reason given is because of the high electronegativity of fluorine atoms. But, how does that affect fluorine being the central atom or not?

If fluorine is the central atom, it can draw electrons from other less electronegative halogens and be happy. Does this lead to instability of the compound?

Why can't fluorine be the central atom in inter-halogen compounds?

First off, fluorine can be at the "center of things."  Examples would include the strongly hydrogen bonded hydrofluoric acid

and the very relevant example of the trifluoride anion

$$\ce{[F-F-F]^-}$$

The trifluoride anion example is critical as it demonstrates that fluorine can be hypercoordinated (hypervalent).  Of course, in this last example we've "tricked" the fluorine into occupying the middle position, so your question remains (slightly altered), why doesn't fluorine occupy a central position more often?

In order to answer this question, two general cases need to be considered,

The anionic situation $\ce{[X-F-X]^-}$ vs.  $\ce{[X-X-F]^-}$
The cationic situation $\ce{[X-F-X]^+}$ vs. $\ce{[X-X-F]^+}$

where X is Cl, Br or I.

In the case of the anion, Lewis dot structures suggest a negative charge on the central atom.

However, computational analysis of the electron distribution in the triiodide anion (see p. 156) suggests that the central atom is positively charged (+0.27 electron), while the terminal atoms are negatively charged (-0.63 electrons on each terminal atom).  Assuming that the same general trend would be applicable to mixed trihalide anions, then in the case of $\ce{[X-F-X]^-}$ we would be placing positive charge on highly electronegative fluorine atom instead of the larger, more polarizable (e.g. more capable of bearing a positive charge) bromine, chlorine or iodine atom.

In the case of the cation, similar arguments come into play.  Again, calculations for a variety of trihalide cations (mixed and homogenous) suggest substantial positive charge on the central atom (this time in agreement with Lewis Dot structures). The same arguments made above would operate here to destabilize structures with a central fluorine atom. This analysis also found extremely long (less stabilizing) bonds resulted when fluorine (or the most electronegative of the 3 atoms) occupied the central position.

Arguments that don't explain the infrequent occurrence of fluorine at a central position in inter-halogen compounds include:

the lack of "d-orbitals"; as other's have commented, this is an old, invalid argument; the existence of $\ce{F3^-}$ demonstrates that fluorine is capable of hypercoordinate bonding without the need for d-orbital involvement
inefficient 2p-3p overlap; $\ce{FCl}$ has a higher bond strength and lower heat of formation than $\ce{Cl2}$

fluorine is high electronegativity so when it combine with other halogen which is less electronegativity then fluorine can take electron from other atom due to this compound become unstable that's why  fluorine can't be central atom in interhalogen compound.


How do we predict the order of melting points of group 1 and group 2 halides? I get really confused by this:

The order of melting points of group 2 chlorides is: $$\ce{BaCl2 > SrCl2 > CaCl2 > MgCl2 > BeCl2}$$ This is completely understandable by Fajan's Rules.

But, the order for group 1 halides is: $$\ce{NaCl > KCl > RbCl > CsCl > FrCl > LiCl}$$

This is what I don't get! Considering Fajan's Rules, $\ce{NaCl}$ is the most covalent, least cationic size, so shouldn"t it have the lowest melting point?

For group 1 halides polarisation is not a dominating factor(except for lithium) since the size of cations is relatively larger than any other cation therefore we must consider lattice energy for determining melting point. As interionic distance increases lattice energy decreases hence melting point decreases down the group.
Lithium is an exception because of its very small size that brings high degree of polarisation. you may want to refer to lother meyer's curve on atomic volumes of elements to get relative idea about the size of the elements.

Melting point depends on Lattice energy. NaCl has similar sized ions which make the lattice stronger. LiCl comes last since it has a lot of covalent character.


How do you explain pKa to non-professional? Let's say we have substance $\ce{A}$, which is mixed with substance $\ce{B}$ to improve shelf-life because $\mathrm{p}K_\mathrm{a}$ of the substance $\ce{A}$ is $7.9$ and in mix the $\mathrm{pH}$ is $5.2$.

Does this mean that in the solution with $\mathrm{pH}$ near $8$ the substance $\ce{A}$ has multiple molecules in neutral state and not dissociate, thus it forms precipitates?

Does $\mathrm{p}K_\mathrm{a}$ mean that most of the molecules of the substance $\ce{A}$ are in nondissociated state?
 Does lowering $\mathrm{pH}$ in this case causes substance $\ce{A}$ to dissociate?

Are bold parts right? Please explain in plain English without the use of external links.

pKa and pH are related concepts but often confused. pKa is a property of a compound that tells us how acidic it is. The lower the pKa, the stronger the acid. pH is a property of a particular solution that depends on the concentrations and identities of the components.

For this discussion, I'm going to use the terms protonated and deprotonated to mean that a compound is associated or dissociated with a proton.

Based on the relationship between the pKa of a compound and the pH of a solution, we can predict whether a compound will be protonated or deprotonated. If the pH is lower than the pKa, then the compound will be protonated. If the pH is higher than the pKa, then the compound will be deprotonated.

A further consideration is the charge on the compound. Acids are neutral when protonated and negatively charged (ionized) when deprotonated. Bases are neutral when deprotonated and positively charged (ionized) when protonated.

Given the information you provided, if compound A (with pKa 7.9) is in a solution of pH 5.2, compound A will be in the protonated state.

Without knowing anything about the identities of A and B, the following is speculation. Most likely compound A is a base and compound B is an acid. The protonated version of A (its "conjugate acid") is more stable for long term storage than A itself.

Another possibility is that compound A is an acid and B is a stronger acid. If the deprotonated version of A is unstable, then compound B is added to ensure that if the mixture came in contact with a base, something more acidic than A would be present to react with the base.

EDIT BASED ON NEW INFORMATION:
As Mateus B said, lidocaine will be protonated by HCl at the amine nitrogen. This is likely done for two reasons. First, as its hydrochloride, it will be more water soluble. In solution, the compound will dissociate into protonated lidocaine cation and chloride anion. Second, as the nitrogen is protonated, it is less reactive to the environment. Over time, amines can react with oxygen to form the corresponding N-oxide and/or can absorb carbon dioxide.

By specifying that A is lidocaine I could say that by adding HCl to it you'd be forming a diferent compound, lidocaine hydrocloride.


How do steric effects affect inductive effect in compounds with tetravalent nitrogen? I want to compare the electron density on methyl group between tri‐tert‐butyl(methyl)azanium (1) and triisopropyl(methyl)azanium (2), i.e. essentially the inductive effects of $\ce{-\overset{+}{N}R3}$ groups.
$$\underset{\large\textbf{1}}{\ce{Me-\overset{+}{N}(t-Bu)3}}\qquad\underset{\large\textbf{2}}{\ce{Me-\overset{+}{N}(i-Pr)3}}$$
The only difference I see in these compounds is that due to steric effects the bond angle would be greater in compound 1 than in 2. Based on this info alone I am not able to judge which one would have greater electronegativity on $\ce{\overset{+}{N}},$ or so as to say greater inductive effect due to the $\ce{-\overset{+}{N}R3}$ group.
How should I go about comparing the electron density on methyl group in both of these species?

Since you are asking for a theoretical explanation, I believe a theoretical calculation might be helpful. It is possible to calculate the partial charges on atoms using electronic structure programs. I have calculated the partial charges at $\text{B3LYP/pcseg-1//GFN2-xTB}$ level (with GAMESS and xTB).
$\ce{Me-\overset{+}{N}(^iPr)3}$
We only need to look at the partial charges on the Me group to know the electron-withdrawing capacity of $\ce{-N+(^iPr)_3}$ group.
\begin{array}{cc} \hline
\mathrm{atom} & \mathrm{partial\;charge\;(e)} \\ \hline
\ce{C} & -0.4367\\
\ce{H} & \;\;\, 0.1954\\
\ce{H} & \;\;\, 0.1957\\
\ce{H} & \;\;\, 0.1966\\ \hline
\end{array}
Total charge on $\ce{Me} = 0.151$.
$\ce{Me-\overset{+}{N}(^tBu)3}$
Again, we only need to look at the Me group.
\begin{array}{cc} \hline
\mathrm{atom} & \mathrm{partial\;charge\;(e)} \\ \hline
\ce{C} & -0.3087\\
\ce{H} & \;\;\, 0.1436\\
\ce{H} & \;\;\, 0.1430\\
\ce{H} & \;\;\, 0.1414\\ \hline
\end{array}
Total charge on $\ce{Me} = 0.1193$
So, the partial charges would indicate that $\ce{-N+(^iPr)3}$ is more electron withdrawing than $\ce{-N+(^tBu)3}$.
Explanation?
I am not sure I have a definitive explanation for this. But it's possible to look at the differences between $\ce{^iPr}$ and $\ce{^tBu}$—

The cone angle formed by the three R groups for $\ce{-N+R3}$ would be larger for $\ce{R=^tBu}$ obviously, due to the steric bulk of that group. When the angles are the ideal $109.5^\circ$ each bond can be assumed to form from $\mathrm{N\;sp^3}$ orbitals. When N is $\mathrm{sp^2}$, the three bonds would be $120^\circ$ to each other. So, when the cone angle increases, the amound of p-character in those three N-R bonds can be assumed to go down. Conversely the amount of p-character in the 4th bond goes up which would mean less electron withdrawing capacity (p-orbitals are further away from nucleus). So, $\ce{-N+(^tBu)3}$ should be weaker EWG.

As there is no empty p-orbitals on N, so there should be no hyperconjugative effects for the compounds drawn. However, if instead of $\ce{Me}$, there is some other group that has a $\pi$ system attached to N, then there is a possibility of hyperconjugation.

The $\ce{CH3}$ groups have a very weak electron donating effect, so $\ce{-N+(^tBu)3}$ will be slightly weaker EWG than $\ce{-N+(^iPr)3}$. *

Notice that points 1 and 3 predict the same trend, which matches with the calculated trend. However, it is difficult to tell which factor is the most important (and whether these factors actually mean anything).

* Curiously, the partial charge on $\ce{C}$ of the side chain goes up when its $\ce{H}$ is replaced by $\ce{CH3}$, which would suggest that $\ce{CH3}$ is actually electron-withdrawing. This does not match with the standard notion of methyl groups being electron donating. I am not really sure if there is any explanation for this discrepancy. But note that there have been some evidence that $\ce{-CH3}$ groups can act as EWG's in some circumstances. The linked paper specifically mentions in the abstract—"That methyl groups attached to carbon atoms are electron donors must not be generally assumed."

Both N are sp3 hybridized. Both N are quaternary nitrogens with a positive formal charge. The tBu group is more electron donating (think vectors of the N-C-C geometries) than the iPr group, but there is more of a steric penalty due to its bulk; on the other hand, the N-C-H linkage in the iPr ammonium cation will also have a slightly larger dipole than the N-C-C. So, in answer to your question about electron density, I would say the [MeN(tBu)3]+ again because of the EDG effect of tBu > iPr.
Does this make sense? :)


Why is it necessary to heat an esterification reaction? Why do we heat the mixture acid and alcohol?

The reaction of esterification is exothermic so it is not affected by increasing temperature, then what is the use of heat ?

Basically the use of heat is to liberate the water molecule from the mixture of acid and alcohol that is $\ce{OH-}$ ion from alcohol and $\ce{H+}$ from acid that results in formation of ester.Moreover it helps in evaporation of water since esterification reaction forms an equilibrium.The removal of water from mixture forces the reaction to move forward.It helps us to get a large amount of product.Alternative to heating some dehydrating agents like conc. $\ce{H2SO4}$ can also be added.


Why do Electrons leave the Zinc in a Galvanic Cell What induces  electrons to leave the atoms around the part of the zinc metal in a solution to travel up along that zinc metal and above the solution, through the wire and to the copper metal and reduce the copper ions in that solution.

Is it a result of the positive metal ions in a 'sea of electrons' being attracted to the negative ions of the solution making the electrons surrounding them leave, are the electrons repelled by the electrons in the solution or do these electrons 'sense' the copper ions on the other side waiting to be reduced and flows towards them?

I'm just not sure what prompts them to leave in the first place if the positive copper ions are not directly in contact with the zinc.

First, let's think about what happens if you dip a single zinc electrode into some electrolyte solution (for now, it's not going to have copper or anything in it). This situation is subject to this equilibrium: $$\ce{Zn_{(s)}<=>Zn^2+_{(aq)} +2e-}$$

At first, there is no barrier to some of the zinc at the surface dissolving. The standard reduction potential is −0.7618 V (this is for the reverse of the reaction written above) so the oxidation process is energetically favourable. However, though the zinc ion can diffuse through the solution, there is nowhere for the two electrons to go, so they are trapped within the electrode. This excess charge opposes further oxidation—it becomes more and more difficult to force more charge into the electrode. For this reason, zinc electrodes do basically nothing in neutral solutions. N.B. It is possible for these electrons to be used to reduce the hydronium ion into hydrogen gas, but zinc adsorbs hydronium very poorly so this process is rather slow except in concentrated acid.

But what happens if we give these electrons somewhere to go? If we connect the zinc electrode to another electrode that is less easily oxidized than zinc (copper is often used because it's cheap, but it could be siver, gold, etc.), these electrons can flow into the other electrode and participate in a reduction reaction. In the case of a copper electrode, because there are no copper ions in our electrolyte, the reduction reaction that occurs (unless there is a more easily reduced substance in the electrolyte) is: $$\ce{2H3O+ + 2e- <=> H_{2(g)} +2H2O}$$
The two reactions balance each other out so we have no charge buildup to oppose the dissolution of the zinc.

To go back to the core of your question,

I'm just not sure what prompts them to leave in the first place if the positive copper
ions are not directly in contact with the zinc.

If there were any copper ions in the solution, electrons could be transferred directly from the zinc to the copper ions. If you dip a piece of zinc into a copper solution, this is exactly what happens—the zinc dissolves and small beads of copper are reduced on its surface, no second electrode required. Trying to make a zinc-copper cell with copper ion in the solution just makes the cell work worse as a good part of the zinc is lost through direct reduction of copper ions at the surface. The way to prevent this is to separate the two half-reactions with a salt bridge.

The reason the electrons leave in the first place (why the oxidation reaction above occurs) is that it is more energetically favourable for the zinc to be oxidized in the solution than to remain in its metallic state. Another way to think about is that zinc is more easily oxidized than copper so it must have electrons that are more easily lost (have a greater electrochemical potential) and these electrons will flow towards a lower electrochemical potential. Therefore, when you connect the two electrodes together, electrons flow from zinc to copper until an equilibrium is reached. Copper's electrons are not normally high enough energy to reduce the hydronium ion, but because of the electrons coming from the zinc, its electrochemical potential is raised enough for the reaction to occur.

This paper is a great reference for learning more about the lemon battery and other single solution cells (if you have ACS journal access, unfortunately).

First of all, both electrodes (the copper and the zinc, in this case) are dissolving in the electrolyte. However, normally the dissolved electrode material comes out of solution just as fast. It's like trying to dig a hole in the dirt when someone else is filling it up at the exact same speed; nothing happens. Nothing happening is also called an equilibrium.

Second of all, when it dissolves, it becomes an ion, that is, it either gains or loses an electron. Now, some materials want to ionize different amounts than others. This "want" is measured in volts.

Taking into account these two factors, when you put any two electrodes together in an electrolyte, you can tell from a "Standard Electrode Potential Table" what will happen, if anything. In the case of a lemon battery, the ionic zinc wants to lose two electrons more than the copper wants to gain them. Note that one electrode must gain the electrons and one must lose them. So we have these two electrodes giving off ions into solution and also taking them back in. However, when a copper ion by chance touches the zinc metal, it transfers two electrons to the copper making metallic copper and ionic zinc. After the copper adheres to the zinc metal, it can't dissolve anymore because if it, by chance, does dissolve, the zinc metal around it instantly changes it back to metallic copper by giving it two more electrons and dissolving another zinc atom in the copper's place. Likewise, the zinc can't adhere to the copper because it would have to steal electrons from the copper, but the copper doesn't want to.

Some things to keep in mind:

This transfer of electrons is what generates the electricity in the first place.
Not all electrode pairs react.
If you want more in depth information, look up "standard reduction potential table" in Google, as well as "how to calculate the voltage of a galvanic cell".


Nuclear fusion reactions violate law of conservation of mass, but it has not yet been taken into account. Why? According to the law of conservation of mass, during any physical or chemical changes, the total mass of the products is equal to the total mass of the reactants. But in nuclear fusion reaction if considered an example, energy that sun emits in its core is due to collision of hydrogen nuclei and formation of helium nuclei. Here conservation of mass is not obeyed as certain part of mass is converted into energy. So, law of conservation of mass is violated here. Then law of conservation must be redefined as: during any physical or chemical change, the total mass of the products is equal to the total mass of reactants provided mass has not undergone conversion into energy.

Nuclear reactions appear to violate both the Laws of Conservation of Mass and Energy because mass is converted into energy or vice versa.

However, the concept of mass-energy equivalence that emerges as a consequence of Einstein's General Theory of Relativity makes the Laws of Conservation of Mass and Energy special limiting cases of the Law of Conservation of Mass-Energy. Since mass is now a form of energy, we can convert back and forth as long as there is no net loss between the two.

The equivalence of mass and energy comes from Einstein's famous $E=mc^2$, where $c$ is the speed of light (in a vacuum).

I dont really think it violates the concepts of mass and energy conservation as dE=dmc2 and the speed of alpha beta and gamma rays are almost as equivalent as the speed of light hence taking in consideration that einstein stated that energy can be converted from mass if the body is in speed of light or nearly the speed of light ,the statement is correct.
.To be short and non-lengthy i can state that the basic concepts of mass conservation and energy convervation are violated but  the einstein's theory of relativity proves the above statement.


Ionization energy and Electron Affnity Which of the following statements correctly interprets the relationship between ionization energy, atomic radius  Electron affinity when comparing two atoms.

The correct choice as per my instructor is that when atomic radius increase ionization energy decrease and electron affinity becomes less positive.

I do understand that as the atomic radius increase it becomes easier to remove and an electron due to shielding from multiple orbitals. Hence Ionization energy is decreased as the atomic radius increase. Additionally, electron affinity is an exothermic process.I am not sure how  electron affinity (addition of an electron) related to ionization energy and atomic radius. Please clarify. Thank you.

For example ,when a person drops a book. When he or she lifts a book, he or she gives potential energy to the book (energy absorbed). However, once the he or she drops the book, the potential energy converts itself to kinetic energy and comes in the form of sound once it hits the ground (energy released).

Electron affinity is defined as the change in energy (in kJ/mol) of a neutral atom (in the gaseous phase) when an electron is added to the atom to form a negative ion. In other words, the neutral atom's likelihood of gaining an electron.

When an electron is generally added to a neutral atom, i.e first electron affinity, energy is released; thus, the first electron affinities are negative. However, when an electron is added to a negative ion, i.e. second electron affinity, more energy is required. Thus, more energy is released to add the electron into an ion because the negative ion has to force the electron to go into its electron orbital; thus, the second electron affinities are positive.

First Electron Affinity (negative energy because energy released):

$\ce{X(g) + e- -> X- (g)}$

Second Electron Affinity (positive energy because energy needed is more than gained):

$\ce{X- (g) + e- -> X^2- (g)}$

This is because you are practically trying to forcefully add an electron to it.

Coming to your question: Remember that greater the distance, the less of an attraction; thus, less energy is released when an electron is added to the outside orbital. In addition, the more valence electrons an element has, the more likely it is to gain electrons to form a stable octet. The less valence electrons an atom has, the least likely it will gain electrons.

One might think that since the number of valence electrons increase going down the group, the element should be more stable and have higher electron affinity. One fails to account for the shielding affect. As one goes down the period, the shielding effect increases, thus repulsion occurs between the electrons. This is why the attraction between the electron and the nucleus decreases as one goes down the group in the periodic table.

As you go down the group, first electron affinities become less (in the sense that less energy is evolved when the negative ions are formed). Fluorine breaks that pattern, and will have to be accounted for separately.)

The electron affinity is a measure of the attraction between the incoming electron and the nucleus — the stronger the attraction, the more energy is released. The factors which affect this attraction are exactly the same as those relating to ionization energies — nuclear charge, distance and screening.

The increased nuclear charge as you go down the group is offset by extra screening electrons. Each outer electron in effect feels a pull of 7+ from the center of the atom, irrespective of which element you are talking about.

So,

as atomic radius increases, the nuclear attraction on outtermost electron decreases — this leads to decrease in IE — but the shielding also increases — less energy is released when an electron is added to the outside orbital — decreasing the EA.


What is a "Strong" Acid? If a reaction requires a "strong" acid - what does this mean? Does it mean any acid of a specific concentration or pH? Are only certain acids classed as "strong"? I have read on Wikipedia that it needs to ionise completely in an aqueous solution - in this case, is a strong acid (for example, HCl) still classed as a strong acid if it is in a 0.00001M solution?

The strength or weakness of an acid is independent of its concentration. A strong acid dissociates completely in solution, so $\ce{HCl}$ at any molarity would be considered strong, and $\ce{HCOOH}$ (for example) at any molarity would be weak. It is more enlightening to consider $K_a$ values, which are essentially the equilibrium constants for the ionization of an acid $\ce{HA}$.
$$K_a=\frac{[\ce{H}^+][\ce{A}^-]}{[\ce{HA}]}$$
This value will either be very small (<<1) or very large (>>1000). Acids with very small $K_a$ are considered weak, and acids with very large $K_a$ are considered strong (the $K_a$ of strong acids is usually not stated and assumed to be infinite).

Addendum: The $K_a$ value is concentration-independent because you are only interested in how well the acid dissociates from its molecular form. If you had $0.0000001\ \text{mol}\cdot\text{dm}^{-3}\ \ce{HCl}$, practically every molecule of $\ce{H-Cl}$ would dissociate into $\ce{H^+}$ and $\ce{Cl^-}$ no matter how many there are. But if you had $12 \ \text{mol}\cdot\text{dm}^{-3}\ \ce{HCOOH}$ (methanoic acid), only a small proportion of the $\ce{HCOOH}$ would dissociate into  $\ce{H^+}$ and $\ce{HCOO}^-$, irrespective of how many you had to start.

What makes an acid strong is also dependent on the context. E.g. lactic acid is considered strong in body fluids because lactate is completely ionized in body pH 7.4.


In rovibrational spectra, when exactly is the Q branch forbidden? I'm aware that the Q branch ($\Delta j = 0$) is usually not observed in rotational-vibrational spectra of diatomic molecules in the gas state, however, I've heard different things about when exactly this transition is forbidden.

I've sometimes been told that the transition is not allowed in general, for any diatomic molecule, but I've sometimes heard that this is only true if the electronic state has a $\Lambda \neq0$, so which is the correct selection rule here?

I'd love to have this clarified for me. If it is possible I would also like to know when is a Q branch observed in vibronic spectra of heteronuclear and homonuclear diatomics.

I would like tell you what I know about the answer to this question.

So, it's known that the selection rules for a fine vibrational spectrum (rotational-vibrational spectrum) for diatomic molecules make $\Delta J = 0$ or $\pm 1$ transitions allowed between the rotational levels of two vibrational states (usually $V''=0$ to $V'=1$, where the symbols have their usual meaning). But one must be careful in realizing that the $\Delta J = 0$ transition is forbidden for all parallel vibrations in the oscillator and only allowed for perpendicular vibrations.

Now, let's see what are parallel and perpendicular vibrations!

Any vibrational mode in infrared active only when the change in the dipole moment vector is sinusoidal w.r.t. time and thus can interact with the electric field of the electromagnetic radiation. If this change in the dipole moment vector is parallel to the major axis of the molecule, then the vibrational is labelled as a parallel mode of vibration. If the change in the dipole moment is perpendicular w.r.t. the major molecular axis, it is a perpendicular mode of vibration.

Let's look at the case of diatomic molecule, as the question is focused on that.

The major molecular axis for it is the bond axis (containing both the atoms). Since a hetero-nuclear diatomic molecule has only one vibrational mode, viz., stretching, it can only occur parallel to the bond axis. Therefore, $\Delta J = 0$ transition is forbidden in the case of parallel vibrations. Thus, on observing the fine infrared spectrum of hetero-nuclear diatomic molecules, we only observe $P$ and $R$ branches. $Q$ branches are not observed.

References:

Fundamentals of Molecular Spectroscopy - Banwell and McCash
Introduction to Molecular Spectroscopy - Gordon M. Barrow
Atomic and Molecular Spectroscopy - Rita Kakkar


Why are geminal diols unstable? The most stable of:

(a) $\ce{(CH3)2C(OH)2}$

(b) $\ce{CH3 C Cl (OH)2}$

(c) $\ce{CH3CN}$

is $\ce{CH3CN}$, because the other two are geminal diols. How can we tell that geminal diols are unstable?

Formation Of Gemdiol

Carbonyl compounds react with water to give gem diols. This reaction is catalysed by acid. The reaction is reversible reaction.

Gem diols are highly unstable compounds hence equilibrium favours the backward direction. The extent to which an aldehyde or ketone is hydrated depends on the stability of gem diol.

Stability of gem diols depend on the following factors:

(i) Steric hindrance by +I group around α-carbon decreases the stability of gem diols. +I group decreases stability of gem diol and hence decreases extent of hydration.

(ii) Stability of gem diols mainly depends on the presence of –I group on α-carbon. More is the –I power of the group more will be stability of gem diols.

(iii) Intramolecular hydrogen bonding increases stability of gem diols. –I groups present on carbon having gem diol group increases strength of hydrogen bond.

More is the strength of hydrogen bond more will be the stability of gem diol.

Overall stability : (CH3)2C(OH)2 < CH3CCl(OH)2 < CH3CN

Geminal diols are unstable due to repulsion between the two oxygens present on the same carbon.If a compound is able to disperse the electron pair of oxygens by phenomenon like hydrogen bonding,resonance,-H effect(hyperconjugation) etc,their stability will increase.


Why is Vapor Pressure of Liquid Equal to Ambient Pressure at Boiling? What does phase change from liquid to gas have to do with equality of vapor pressure and ambient pressure? I understand that at higher pressures more kinetic energy is needed for molecules in liquid to spread out to form bubbles of vapor since vapor has lower density than liquid which means that higher temperature is needed. But why should equality of vapor pressure and ambient pressure determine when phase change takes place?

When the vapor pressure equals the applied ambient pressure the liquid can form vapor in the bulk liquid hence "boiling".  To accurately measure this care must be taken to mix the liquid well and to prevent overheating. At lower vapor pressures evaporation is only from the exposed surfaces.


Why does dimethyl dichlorosilane undergo hydrolysis to form siloxane polymers, but its carbon analog does not? Dimethyl dichlorosilane will undergo hydrolysis to form siloxane polymers. But why would its carbon analog dimethyl dichlorocarbon not be able to undergo a similar hydrolysis reaction to form polymers?

My guess is that $\ce{Si-Cl}$ bond being much weaker than $\ce{C-Cl}$ bond would be easier to break and undergo hydrolysis, and that the geminal diol is unstable, hence it is less likely to form.

I'm unsure of my answer though. Could anyone please explain and elaborate?

Actually a $\ce{Si-Cl}$ bond is much stronger than a $\ce{C-Cl}$ bond.  It takes about 90 kcal/mole to break a $\ce{Si-Cl}$ bond, but only around 81 kcal/mole to break a $\ce{C-Cl}$ bond.  The strengths of the corresponding $\ce{Si-O}$ (110 kcal/mole) and $\ce{C-O}$ (85 kcal/mole) bonds also need to be considered.

In the silicon reaction we need 90 kcal/mole to break the $\ce{Si-Cl}$ bond, but we get 110 kcal/mole back when we form the $\ce{Si-O}$; the reaction is exothermic by roughly 20 kcal/mole - a large amount.  In the analogous carbon case, we need 81 kcal/mole to break a $\ce{C-Cl}$ bond, and we only get 85 kcal/mole back when we form the $\ce{C-O}$ bond; the reaction is exothermic by only 4 kcal/mole - not much.  There is quite a difference in the overall driving force for the two reactions with the silicon case being energetically much more favorable than the carbon case!

Your guess is correct. The geminal diol is unstable because carbon atom can't hold more than one $\ce{-OH}$ group. Where as Si atom can hold three $\ce{-OH}$  groups. It is this property of silicon that makes the formation of organosilicon polymers possible.


Why don’t all objects melt and liquefy when heated sufficiently? Certain objects such as metals, glass, plastic, sugar etc. melt when heated to a certain temperature but other materials such as a piece of wood, paper, a piece of rock etc. do not. Why is that?

The problem with melting most substances is that you would be doing it in a non-innocent atmosphere and that sometimes even the functional groups inside a compound are non-innocent.

In a perfectly innocent world, everything could be molten given sufficient temperature. Given real world paper, remember that reaction $(1)$ is exothermic and the only reason why it does not spontaneously occur is the lack of activiation energy.

$$\ce{C_nH_mO_l + \frac{$2n + m - l$}{2} O2 -> n CO2 + \frac{$m$}{2} H2O}\tag{1}$$

Once you supply sufficient heat, the activation energy will be overcome and the reaction suddenly becomes spontaneous; for paper, I am told that this happens at around $451~\mathrm{^\circ F}$, which is $233~\mathrm{^\circ C}$ in proper units.

Now you could suggest melting things in an inert argon atmosphere — or maybe even take helium for additional inertness. But then still you have potential reactions that can occur with organic materials. What then will happen might be called smouldering. Technically, this is no more than burning without oxygen.

Everything has a liquid state, so your initial assumption is technically incorrect. But your question is a good one, so let me list the things needed to clarify.

Why does some compounds/solids not turn into liquids at atmospheric pressure? You have to factor in to your question that the states of matter are governed by both temperature AND pressure. Some compounds will not become liquids at room temperature no matter what you do to the pressure, and some compounds will not become liquids at 1 atmosphere, no matter what you do to the temperature. (They will certainly become "fluids" at some point though, but that is an unnecessary complication). If you plot the temperature and the pressure on y and x scale of a chart, the graph explaining the behaviour of a material will look like a slightly right leaning lowercase "y" for most compunds.
Check that out @ Wikipedia

Now, some of the things you have listed do become liquids, but at higher temperatures than what people generally see. Stones for instance, melt.

Also, as others have pointed out, some states are more reactive than others. In general reaction kinetics make reactions speed up from solids->liquids->gases so something that is idle and stable as a solid can vanish as a liquid, due to reaction. Solid states of matter, for instance, generally react so slowly that you can safely assume that solids do not burn*). Which begs alot of questions, with interesting answers. =)

*) Naturally excepting some of the more interesting reactions revolving around Cl3F and the likes


How do I prevent oxygen interfering when I prepare a FeCl2 solution from anhydrous FeCl2 powder? I have a bottle of anhydrous $\ce{FeCl2}$ powder (from here; it is 98% pure and was kept in a glovebox after opening it for the first time to minimize reactions with oxygen) and want to make this into a solution. I work in a glovebox, but what I have done prior is that I prepared the reagents outside of the glovebox and then put them inside. I let nitrogen gas bubble through water and then add this water to a certain amount of the anhydrous $\ce{FeCl2}$ powder. After this, I put the solution into the glovebox.
However, I get nowhere near to the green color of the $\ce{FeCl2}$ solutions that I find online. My solution has a slight green hue, and after some time I find a little bit of orange precipitate at the bottom of my reagent. So I fear that there is some reaction between the $\ce{FeCl2}$ and $\ce{O2}$. What can I change about my method to prevent this?
The purpose of my research is to produce the mineral tochilinite ($\ce{6Fe_{0.9}S\cdot5[Mg,Fe](OH)_2}$). I mix the $\ce{FeCl2}$ reagent with $\ce{NaOH}$, $\ce{MgCl2}$ and $\ce{Na2S}$. I do not want any $\ce{Fe^{3+}}$ in my $\ce{FeCl2}$ reagent, because it is also not present in the mineral tochilinite. If it is present in the reagent, other minerals might start to form (such as magnetite).

See if your group or an other (e.g., organic / organic catalysis) has a Schlenk line.  With some training, this allows you to reduce the interference of oxygen greatly, because you may evacuate your reaction flasks and to replace the air by an inert gas such as nitrogen or argon.  Nitrogen is relatively cheap because it equally is used for welding (as gas) or for cooling (think e.g., NMR spectrometers).  Argon gas, on the other hand has the advantage of a density greater than the one of air, thus you may cover and protect your substance under a blanket of argon.
Then, search for Schlenk flasks.  You identify these quickly for their connector to said Schlenk line. Besside round bottom flasks and tubes (a gallery, a second one), there equally are Schlenk flasks to perform a pressure filtration while working under argon enclosing a filter plate (example).
Then, degas your solvent.  In HPLC, bottled solvents for example are placed in the basket of the ultrasound bath while purging them with a steady, slow flow of argon to remove dissolved oxygen.  Do this for your deionized water.
Weigh in $\ce{FeCl2}$ in your dry Schlenk flask, and add the deionized water prepared under the inert gas.

My variation on the suggested synthesis is not the direct addition of, say, the metal iron (has already been suggested), but instead the partitioned introduction into the reaction chamber of an oxygen scavenger.
For example in one possible embodiment, in place of the direct intervention of Fe in the reactive solution mix, just place a small amount of freshly prepared fine powdered magnesium metal in the chamber, but not in direct solution contact.
My logic is that magnesium metal acting as an oxygen scavenger will anodically corrode (to MgO) with oxygen, thus mitigating the further possible action of oxygen on ferrous ions.
The procedure is then mixing of the FeCl2 reagent directly with NaOH, MgCl2 and Na2S, in the presence of a potential separated side surface reaction between Mg and any present oxygen.
If this fails, try a layer of Mg powder on the reactive mix. However, as in the case of suggested employment of Fe, the resulting chemistry may or may not favor the targeted formation of the mineral tochilinite ($\ce{6Fe_{0.9}S⋅5[Mg,Fe](OH)2}$).


Reaction of a diketone What is the product when ammonium bicarbonate reacts with hexa-2,5-dione?

I thought of formation of 2 moles of NH3, water and carbon dioxide. And then the usual addition followed by elimination to the carbonyl group. But is it the case? Or will some cyclic compound result?

Ketones react with NH3.

R2C=O -> R2C=NH. Hexa-2,5-dione has two carbonyl groups so it leads to cyclic compound. Check Paal–Knorr synthesis in Wikipedia to know the answer.


How can an ionic bond have partial covalent in character? I was going through the topic of 'ionic bond' and read this:

  No bond is 100% ionic in character. It has some percentage of covalent character.

I didn't understand how an ionic bond can be partially covalent in character. Please explain along with an example.

One way is Polarization. When a small cation approaches a large  anion it distorts the charge density towards itself thereby forming a type of bond which has directional characteristic i.e from ** (-) ----> (+) ** which is characteristic of a covalent bond.

This polarization has some rules given by **Fajan **

Smaller the cation better distortion/polarization.
More the amount of positive charge more is the polarization.
or more the charge : radius ratio of cation more is the polarization.

So in short every cation has the power of polarization and therefore no bond is 100% ionic.


Can artificial sea water be prepared by mixing sea salt with water? I am looking for ways to easily prepare artificial sea water. The chemical composition doesn't have to exactly match that of natural sea water. On-line I can find recipes to prepare sea water. But I was wondering, can I just buy natural sea salt and dissolve it in fresh water (Until I get the required salinity ).

The commercial shrimp farms in my region, that use natural sea water cause enormous environmental damage. That is the primary motivation behind my interest in artificial sea water. I want to explore if artificial sea water can be a eco-friendly sustainable alternative.

My intention is not to breed shrimp, but just to grow the shrimp that I get from hatcheries. And yes, I am planning to get the help of people with relevant expertise in this. I just want to gain some prior knowledge and experience before I go for their help.

Some of my present concerns are similar to those of oceanarium though on a much much smaller scale :-) Like them I also want to prepare sea water, which I can control to my exact needs. Locally we get these "raw sea salt" used for industrial purposes, which are claimed to be very similar to natural salts found in sea. They do not contain iodine, caking agents or any other additives etc. I am guessing probably the sea salts that oceanarium imports might be same as these (probably of much better quality).

Yes, it is possible though not easy to make artificial seawater by mixing sea salt and fresh water. An interesting example of this happens at the oceanarium in Lisbon, Portugal, which is one of the largest in Europe. After considerable investigation, it was found that the seawater present in massive quantities just outside the building could not be economically filtered, purified and adjusted to attend the surprisingly demanding and delicate needs of the marine organisms to be displayed. The chemical composition of seawater is actually quite complex, varying significantly with the location and time of collection. As with all life, marine life can be very sensitive to small changes in concentration of compounds, not to mention possible biological incompatibilities between the micro- and macrobiota present in seawater from different regions. The solution chosen for the Lisbon Oceanarium was to import some 16.5 metric tons of sea salt monthly from the Red Sea in Israel and mix it with local fresh water in order to replace the tank water. [1] [2] (Couldn't find a reference in English)

Having said that, you really should consult people experienced in setting up seawater aquariums before trying anything yourself. Seawater aquariums have significantly tougher upkeep and maintenance requirements compared to fresh water aquariums; you can't just mix any old cooking sea salt into water, it has to be in the right concentration, and replacement of the tank water has to be more gradual and frequent, not to mention that the salty water will corrode the actual aquarium more quickly. A good aquarium supplies store should be able to better inform you.

Edit: Now I realize you may actually be interested in cultivating shrimp rather than putting them on display. That's a whole different matter, many many times more difficult. You really need to contact some experts in marine cultivation.

Yes. If you do however, that would likely unsaturate the solution. I don't know if certain animals need certain amounts of salt, but doing this without adding salt might kill them.


How to calculate the concentrations of the species in the carbonate equilibrium from a titration with hydrochloric acid? Lake water contains dissolved sodium carbonate and sodium hydrogen carbonate. The following equilibrium exists:
  \begin{align}
\ce{HCO3- &<=> H+ + CO3^{2-}}&
\frac{\ce{[CO3^{2-}]}}{\ce{[HCO3-]}}  &= 0.958
\end{align}
  When $\pu{10 cm^3}$ of lake water were titrated with  $\pu{0.2 mol/dm^3}$ $\ce{HCl}$, $\pu{22 cm^3}$ of acid were required to neutralise all the carbonate and hydrogen carbonate ions according to the following equations:
  \begin{align}
  \ce{H+ + HCO3- &-> H2O + CO2}\\
  \ce{2H+ + CO3^{2-} &-> H2O + CO2}\\
\end{align}
  Calculate the total amount of substance of acid used, and thus, by using the ratio quoted, calculate $\ce{[CO3^{2-}]}$ and $\ce{[HCO3-]}$ in the lake.

I was able to find the amount of substance of acid $\ce{HCl}$, which is $\pu{4.4E-3 mol}.$ How do I go on?

You have correctly identified the amount of substance $\ce{HCl}$ which were necessary to neutralize all $\ce{HCO3-}$ and $\ce{CO3^2-}$ in the sample. With the given sample volume of $\pu{10 cm^3}$, we can calculate the concentration of both ions in the sample:
$$
\frac{\pu{4.4E-3 mol}}{\pu{10 cm^3}}
= \pu{4.4E-4 mol//cm^3}
= [\ce{HCO3-}]+[\ce{CO3^{2-}}]
$$

Using the given ratio of $0.958$, we can express $[\ce{CO3^{2-}}]$ in terms of $[\ce{HCO3^{-}}]$:
\begin{gather}
[\ce{CO3^2-}]
  = 0.958\times[\ce{HCO3-}]\\
\pu{4.4E-4 mol//cm^3}
  = [\ce{HCO3-}] + [\ce{CO3^2-}]
  = [\ce{HCO3-}] + 0.958\times[\ce{HCO3-}]
  = 1.958[\ce{HCO3-}]\\
[\ce{HCO3-}]
  = \pu{2.25E-4 mol//cm3}
\end{gather}

We then use the calculated hydrogen carbonate concentration and the ratio to find $[\ce{CO3^2-}]$:
$$[\ce{CO3^2-}] = 0.958\times[\ce{HCO3-}] = \pu{2.15E-4 mol//cm3}$$

I would do it like follows: By using the ratio of 0.958 you can simply state the ratio of amount of substances. Then by using the balanced equations you can find the ratio of amount of substances of HCl required for each. Having both the ratio and total amount, you can get the the HCl for each; and then you can find the the required concentraion of those two ions from the balanced equation and total volume.


Cahn–Ingold–Prelog priority of carboxylic group versus ortho carboxylic acid Should the C=C double bond below be labelled as (E) or (Z)?

When using the Cahn–Ingold–Prelog (CIP) priority system and the ‘Sequence Rules’, double bonds are split into two bonds and the doubly bonded atoms are duplicated.

Thus

is treated as

where (O) and (C) are duplicate representations of the atoms at the other end of the double bond;

whereas

is treated as

Therefore, the name of the structure that is given in the question is (2E)-3-bromo-3-iodo-2-(trihydroxymethyl)prop-2-enoic acid.

Simple. The rule is that at the tie..(carbon)...there are three ligands of equivalent priority...however, since their not both acids, the rule is that sp2 hybridization beats sp3.

Also, you'll never isolate that geminal (or triplet) alcohol, someone called it an acid hydrate..(even hydrates of aldehydes are difficult to isolate). Only as an orthoester can such structures be isolated.


How to measure chlorine content using ORP sensor? In my project I need to measure pH, turbidity, and chlorine level of treated drinking water. I found sensors for measuring pH, turbidity, and also chlorine.
The actual problem begins is that the cost of chlorine sensor is too high, so I googled for an alternative solution, which led me to use an oxidation-reduction potential (ORP) sensor.
Can anyone tell me about the relationship between pH, chlorine, and ORP value in millivolts and how to measure chlorine content in treated water using ORP sensor?

For a school project like yours, you can use iodometric titration. Iodometric titration is one of standard methods of water examination. The amount of free chlorine in tap water is really small on the order of < 1 parts per million therefore you need a lot of sample. One would typically take 500 mL of tap water (or more), add potassium iodide, and titrate with 0.01 M sodium thiosulfate. You can use an ORP and plot a potentiometric titration curve.
Consult your teacher for this option. He/she should explain to you the details.
There are spectrophotometric methods but the reagents may not be available in your school.

Based on the comment by M. Farooq with respect to the oxidation-reduction potential (ORP) sensor, I would suggest a pure chemical test.
In an inverted container containing the subject treated water, to test for dissolved chlorine and hypochlorous acid, simply inject a volume of stabilized hydrogen peroxide.
The following reactions should govern the system:
$\ce{ Cl2 (d) + H2O (l) <=> HCl (aq) + HOCl (aq)}$
$\ce{ HOCl (aq) + H2O2 (aq) <=> HCl (aq) + H2O (l) + O2 (g)}$
The last is a fast reaction and I would rapidly note (by marking) the volume of oxygen rapidly released. Subsequent elevation in released oxygen could be produced by other catalytic agents acting on the H2O2 itself.
I would recommend performing a trial run on distilled water to which is added a known amount of Sodium hypochlorite. Repeat with boiled tap water plus NaOCl and observe any growth in oxygen release with time.


Confusion in IUPAC Nomenclature I was trying to name the following compound but I got stuck in naming the aldehydic part of the compound.

I looked for the answer and found it as 6-bromo-2-oxocyclohexanecarbeldehyde. I had come to know earlier that carbeldehyde suffix is used when there are more than two aldehydic group in the compound but why carbeldehyde is used here? I am not able to understand this. If someone can explain which rule is used here. I will be very grateful for this.

Why …carbaldehyde?
It's similar to $\ce{C6H11-COOH}$ cyclohexanecarboxylic acid (not cyclohexanoic acid). It's the case where the functional group's carbon is not included in the parent structure. (Such case are also nitriles/carbonitriles, $\ce{R-CN}$)
Why not …hexyl… or rather …cyclohexyl… ?
This (wrong name) would be constructed similarly to a different, non-preferred type of nomenclature, a radical name, which names e.g. ethanol as ethyl alcohol.
There are more than two aldehydic group in the compound.
They are not. Alhedyde group is ${\ce{-CHO}}$. The other one, $\ce{>C=O}$ is an oxo group (ketone). The group for suffix is chosen by the pre-defined priority, ${\ce{-CHO}}$ > $\ce{>C=O}$.
Why 2-bromo-6-oxo-… and not 6-bromo-2-oxo-… ?
The groups named by prefixes (after highest priority - suffix group is assigned lowest locant possible) are not numbered by their priority, but such that the assigned locant set is lowest possible. If there are more equivalent possibilities, then they are numbered in the alphabetically sorted order (which is used in the name).

Carbaldehyde is not particulary used when there are two aldehydic groups in a molecule.
To describe an aldehyde you either use "-al" or "-carbaldehyde". The difference is that when you use "-carbaldehyde" you also count the carbon, meaning that carbaldehyde stands for "CHO" group including the carbon.

(That is why it is 2-oxo.)


Double Replacement Reactions form Precipitates Why do most all double replacement reactions result in a precipitate and an aqueous molecule? Is it because the individual solute molecules always form stronger bonds such as how in single replacement the molecule with more energy bonds with the molecule of the opposite charge?

Precipitation of silver chloride is a common example of a double replacement reaction. e.g. $$\ce{AgNO3~_{(aq)} + NaCl_{(aq)} -> AgCl_{(s)} + NaNO3~_{(aq)}}$$

Initially all four ions are present in solution. When the solutions are mixed the silver and chloride ions bond together and precipitate, leaving the sodium and nitrate ions remaining in solution along with a small number of silver and chloride ions because the precipitation is an equilibrium so a small number of dissolved ions will still be present (thanks to @Martin for pointing this out). The reaction happens because the precipitation of silver chloride removes these ions from the solution equilibrium, driving the reaction to completion.

Silver chloride is insoluble because the $\ce{Ag-Cl}$ bond has a high degree of covalent character due to the high polarizing ability of the silver cation and the high polarizability of the chloride anion and the similar size of both ions which allows a shorter bond length and so a stronger bond requiring more energy to break apart in order to dissolve the solid. See this question for more detail on this point.

Precipitation helps drive the reaction in one direction by removing some of the reactants from solution.

Another example of double displacement is $\ce{FeS + H2SO4 -> FeHSO4 + H2S}$ where the reaction is driven in the forward direction by the loss of gaseous $\ce{H2S}$.


Can 1 gram hyaluronic acid really hold 6L of water? According to this study (https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3970829/), they make an (uncited) claim about hyaluronic acid: "One of its most important properties is that it can attach and hold large amounts of moisture; approximately 6L of water in just lg ."

Now, the MW of HA is $776.651$ g/mol, so there are about $\pu{0.00128865979}$ moles of HA in a gram. And about $\pu{7.757732e20}$ molecules.

Using google, 6L of water has $\pu{2.007e26}$ molecules.

So this is essentially telling me that roughly 1 million water molecules are able to hydrogen bond with 1 molecule of HA...

Firstly, is this fact true? And if so, how is this possible??

Various figures can be found “on the internet”. E.g. ‘Hyaluronic Hydro Gel Has 1000 Times Water Holding Capacity Perfect for Hair to Heel’ (commercial link intentionally omitted).

Gel liquid phase absorption ability measure seems to be a ‘swelling capacity’ (ratio).

For hyaluronic acid (HA), it can be increased by additional crosslinking. In one paper from 1992, usual swelling ratio was found to be in the order of magnitude 1 to 10, the maximum figure was 120 (for crosslinking ratio much higher than 2.1; 25–30 for the ratio 2.1).[1]

In another paper, another guys are using biphasic nanoparticles (some composite of crosslinked and non-crosslinked HA), they state they have achieved for two different particle types ‘… swelling ratio 1270, 1000 %, respectively’[2] (note the percent sign). Which is, in both cases, approximately, not 1000× capacity, but rather 10× capacity, I'd say.

However, there are polymers with the claimed property – superabsorbents:[3]

  3.27 superabsorbent polymer

  Polymer that can absorb and retain extremely large amounts of a liquid relative to its own mass.

  Note 1: The liquid absorbed can be water or an organic liquid.
  Note 2: The swelling ratio of a superabsorbent polymer can reach the order of 1000:1.
  Note 3: Superabsorbent polymers for water are frequently polyelectrolytes.

References:

Shah, Chirag B., and Stanley M. Barnett. "Swelling behavior of hyaluronic acid gels." Journal of applied polymer science 45.2 (1992): 293-298.
Chun, Cheolbyong, et al. "Effect of molecular weight of hyaluronic acid (HA) on viscoelasticity and particle texturing feel of HA dermal biphasic fillers." Biomaterials research 20.1 (2016): 24.
Compendium of Polymer Terminology and Nomenclature (IUPAC Recommendations 2008)

Attach and Hold - a bit of a misnomer; it forms a 3 D Lattice matrix containing water; given its inherent viscosity and inherent elasticity, it can be compressed and upon release of pressure spring back to its original dimensions thus the visco-elastic/shock absorbing properties advantages medically (knees) and cosmetically (skin fulfillment/youthful appearance etc).


Why does silver oxide form a coordination complex when treated with ammonia? Part of the preparation of Tollens' reagent occurs according to the formula:

$$\ce{Ag2O (s) + 4 NH3 + 2 NaNO3 + H2O → 2 [Ag(NH3)2]NO3 + 2 NaOH}$$

I wonder how exactly the oxygen gets "unstuck" from the silver under the influence of ammonia. I've read about coordination complexes, so I understand what $\ce{[Ag(NH3)2]NO3}$ is, but I'm puzzled at what exactly happens to pull that oxygen apart from $\ce{Ag2O}$

I've done very little exploration of the issue, so this may be too basic a question - in which case just please give a hint on what to read up.

I found this (now deleted) comment:

  $\ce{NH3}$ dissolves silver oxide because the nitrogen has a pair of electrons that it can share with the silver forming a diammine complex. It has nothing to do with pH adjustment. – A.K. Jan 20 '16 at 17:17

Chemguide states this:

  This is made from silver(I) nitrate solution. You add a drop of sodium hydroxide solution to give a precipitate of silver(I) oxide, and then add just enough dilute ammonia solution to redissolve the precipitate.

It's quite elementary (perhaps). Silver (and gold) are known to show two main coordination numbers in its compounds, namely, $2$ and $4$. To start, we can consider $\ce{[Ag(OH)2]-}$ to be our initial complex as in aqueous solutions it can be thought of to be existing in an equilibrium with $\ce{Ag2O}$. See footnote [1].

A quick glance at the order for the strength of ligands tells us that $\ce{NH3}$ is a stronger ligand than $\ce{OH-}$. What I now intend to do is to compare the thermodynamic stability of $\ce{[Ag(OH)2]-}$ with that of the diammine complex $\ce{[Ag(NH3)2]+}$.

The strength of ligands is directly proportional to the thermal stability of a complex so yes, the diammine complex is more stable.

A similar case is seen with copper(I). Cyanide complexes of copper(I) are known to be extremely stable and hence it prefers forming a complex with cyanide over other ligands.

[1] Nevertheless, saying this may be highly controversial because Ellingham diagrams tell us that $\ce{Ag(OH)}$ being unstable gets converted into $\ce{Ag2O}$ which is a much more stable compound for silver to be in. However in aqueous solutions, an equilibrium of such a kind can always exist.

Actually both comments perhaps, to quote:

"because Ellingham Diagrams tell us that Ag(OH) being unstable gets converted into Ag2O which is a much more stable compound for silver to be in. However in aqueous solutions, an equilibrium of such a kind can always exist"

"NH3  is a stronger ligand than OH−"

Assuming a starting Ag2O presence from the reaction, some Ag+ ions from limited solubility, and noting, "a ppt of AgOH appears and dissolves in excess ammonia",  see https://books.google.com/books?id=uwJbSFRIqyAC&pg=PA37&lpg=PA37&dq=radical+attack+on+silver+Ag+%2B+%E2%80%A2OH++--%3E+AgOH&source=bl&ots=jG9-2NG9x3&sig=V9nVefopAXXKP3lFnxQYnJriG14&hl=en&sa=X&ved=2ahUKEwjasar_g7ffAhWlhOAKHQEDA_4Q6AEwGHoECAcQAQ#v=onepage&q&f=false :

2 AgOH = Ag2O + H2O

both comments may be right.

Further, in the presence of visible light leading to the photocatalyst Ag2O/Ag, which per research out of P.R. China (published 2011 in the ‘Chemistry, a European Journal’), which has been cited as a powerful stable visible light photocatalyst capable of generating electrons (e-) and holes (h+) (see https://onlinelibrary.wiley.com/doi/abs/10.1002/chem.201101032 ):

Ag2O/Ag + hv --> e− + h+

With electrons:

Ag+ + e- --> Ag

And, in the presence of active electron holes, any OH- created via:

XOH = X+ + OH-   (where X = H or NH3 or Ag)

could be converted into •OH radical:

OH- (aq) + h+ --> •OH (aq)

which may attack Silver metal directly (see https://pubs.acs.org/doi/abs/10.1021/acs.jpca.7b08081?src=recsys&journalCode=jpcafh and superacidity related redox couples at http://beta.chem.uw.edu.pl/people/WGrochala/Ag2+_mechanism.pdf, and more generally radical attacks on metals in their lower valence states at https://srd.nist.gov/NSRDS/NSRDS-NBS-65.pdf , where one could postulate the half cell reactions: Ag = Ag+ + e- and •OH + e- = OH- with the net reaction below):

Ag + •OH (aq) --> AgOH (aq)

2 AgOH = Ag2O + H2O

Or: AgOH (aq) = Ag+ (aq) + OH- (aq)  (based on some limited solubility of Ag2O)

producing the oxygen free Silver ion to complex with NH3 (or OH-, depending on the respect ion concentration) in a photo cyclic reaction. Note: the hydroxyl radical can be largely scavenged at high ammonia concentration (creating •NH2), so this path is pH sensitive.


Removing aluminum oxide from impure silica We have an assay for a mine. Our goal is to have only silica. Is there a cost effective way to extract the contaminants? The product is a fine powder called diatomaceous earth, or fossil shell flour. Our goal is to use the product to help poor pacific islanders increase their silica uptake and assist in heavy metal detox. Many members of the pacific island nations suffer from heavy metal poisoning. The composition of the powder is:

> 84.7% $\ce{SiO2}$
< 4% $\ce{Al2O3}$
< 2% $\ce{Fe2O3}$
< 0.33% $\ce{CaO}$
< 0.36% $\ce{MgO}$
< 0.5% "other"

We aren't concerned with the calcium, iron or magnesium oxides, just the aluminum oxide.

Is there a simple process to reduce the aluminum oxide levels?
Also, what is the easiest process to convert the silica to orthosilicic acid for increased bio-availability?

You may try the famous Bayer process, which is used for extracting alumina form bauxite. When treated at $160°$C and high pressure with $\ce{NaOH}$, a mixture like yours dissolves $\ce{SiO2}$ and $\ce{Al2O3}$, and lets the other compounds insoluble. This is a first purification. $$\ce{SiO2 + 2 NaOH ->  Na2SiO3  +  H2O}$$ $$\ce{Al2O3  +  2 NaOH + 3 H2O -> 2 Na[Al(OH)4]}$$Later on, bubbling $\ce{CO2}$ in the solution makes $\ce{Al(OH)3}$ to precipitate. $$\ce{Na[Al(OH)4] + CO2  -> Al(OH)3(s) + NaHCO3}$$ Of course $\ce{NaHCO3}$ is immediately transformed into $\ce{Na2CO3}$ by $\ce{NaOH}$. $\ce{Al(OH)3}$ can be eliminated by filtration. And the final solution contains only sodium silicate, apart from the excess of $\ce{NaOH}$. Silica can then be obtained pure by adding  enough $\ce{HCl}$ to destroy $\ce{Na2SiO3}$ and the NaOH in excess.$$\ce{Na2SiO3 + 2 HCl + H2O ->  Si(OH)4 + 2 NaCl}$$ It is advisable to highly dilute the $\ce{NaOH + Na2SiO3}$ solution before adding $\ce{HCl}$. Because the precipitate of $\ce{Si(OH)4}$ is extremely voluminous, and fills up the entire solution.

ore or mineral should be treated with some flux like CaO, which react with silica to form Ca-silicate. and which can be separated after filtration process.


Arrhenius equation and rate constant I am a high-school student and I am currently taking a course on chemical kinetics. Towards, what seems, end of the course at highschool level, the Arrhenius equation came in. Now, I'm not sure if what I'm presented is just a general form of Arrhenius equation and there's more to it, or this is it. Well, anyways, one of the things I am being told to learn is the plot between graph of rate constant $k$ vs temperature $T$. $$k=A\mathrm e^{-E_\mathrm a/(RT)}$$
The graph is sort of looks like with increasing $T$, the value of $k$ is skyrocketing (kinda like $y=x^2$ in first quadrant) but if I'm not mistaken, the rate constant $k$ should equal $A$ as temperature tends to infinity. Now I believe there must be some other factor in the equation that is also a function of temperature and is causing the graph to be the way it is. Can someone help me understand the same?
(In case you didn't get, because I am unable to attach an image, the looks like $y=x^2$ instead of $x=y^2$, which basically interrupts my intuition of Temperature on $x$ axis ever approaching infinity. $k$ on $y$ and $T$ on $x$ axis.)

This is how the graph between the rate constant and temperature looks like:

This might not seem intuitive looking at the Arrhenius equation which says:
$$k=Ae^{\frac{-E_a}{RT}}$$
After using some mathematical prowess someone might say that as $T\rightarrow \infty, ~k\rightarrow A$ and so the graph should look something more like:

Which on first glance seems fair but doesn't match the aforementioned curve.
To understand what is causing the problem we must check everything that might change during a change in temperature.
Upon examination we can find that only two things might cause this discrepancy, $E_a$ and the pre-exponential "constant", $A$.
The activation energy of a system is also independent of the temperature and thus the only possible problem point left is the pre-exponential "constant". Turns out that this constant isn't a constant in true sense, which isn't that disturbing of an idea if you know what that constant signifies.
The pre-exponential constant shows how many effective collisions can take place in a reacting mixture. It is defined as,
$$A=pZ$$
Where,
$A$ is the pre-exponential constant.
$p$ is the steric factor, which takes into account the steric effects during an effective collision.
$Z$ is the collision frequency.
Now as it might seem intuitive, the collision frequency is related to temperature and increases with increase in temperature according to the following relation(using equations for collisions between similar molecules to prevent unnecessary complexity):
$$Z=4nr\sqrt{\frac{\pi RT}{M}}$$
Where,
$Z$ is collision frequency.
$n$ is molecules per unit volume.
$r$ is the radius of molecule.
$R$ is the universal gas constant.
$T$ is temperature.
$M$ is the molar mass of the gas.
Thus the collision frequency and corresponding the pre-exponential constant tend to infinity when $T\rightarrow \infty$, which matches with our observation of $k\rightarrow \infty$ when $T\rightarrow \infty$.
Reference:
The Effect of Temperature on Reaction Rates https://chem.libretexts.org/@go/page/24271 (accessed May 14, 2021).

Surely, as T tends to infinity k tends to A. The graph in this case will be exponential and not $y= x^2$ kind.
You can plot the graph by differentiating the equation (twice to be particular).

The graph looks like $y=x^2$ but only upto $T=\dfrac E{2R}$.
But don't misinterpret the same.


Equivalence point of titration Why is $\ce{[HIn] = [In-]}$ at equivalence point of a titration? I know that at equivalence point moles of known solution is equal to the moles of unknown solution, but I'm not able to relate that fact with the above conclusion. Perhaps I'm don't know something.

Ideally, the concentration of indicator is much, much lower than the concentration of the material to be titrated and of that used to titrate. Therefore, less than a drop of standard solution is needed to make the indicator entirely cross the equivalent point.

For most titrations, the equivalence point of the indicator is not the end point. It’s just somewhere near. And the only thing you can actually see and record is said indicator’s colour change. So it’s the only thing you can resort to.

The important choice in a titration is choosing the correct indicator. If you are titrating a strong acid with a strong base (so the titration’s equivalence point is approximately $7$), you want an indicator whose equivalence point is just behind that (at ca. $8$), so that the tiniest of drops added additionally will cause the colour change.

As you might already know, $\ce{[In^-]}$ is the concentration of the indicator that was added to the solution, which basically shows whether the solution is basic or acidic or neutral. The equivalence point of a titration is when the titrant has reacted with the unknown chemical solution so that it reaches an endpoint of the indicator. The equalization of the concentration $\ce{[HIn] = [In^-]}$ is when the color is at its neutral state: neither colors of the indicator is dominant over the other.


Do electrons in covalent bonds always obey the octet rule? Do electrons in covalent bonds always necessarily obey the octet rule? Always?

The elements in borons row do not always follow the octet rule as in the case of boron fluoride as boron is in the center of the periodic table and will usually only bond to fill up its outer shell with 6 electrons. Additionally, elements with can form expanded octets which can share more than 8 electrons such as phosphorous pentachloride.  Of course hydrogen only has 2 electrons to bond with so it is also an exception to the octet rule.


Except pure alloys, are there any compounds with more metal elements proportion of atoms than nonmetal elements in proportion of atoms? Except pure alloys, are there any compounds with more metal elements in proportion of atoms than nonmetal elements in proportion of atoms?

For example, Aluminium oxide has 2 metal elements & 3 nonmetal elements. Do we have a compound with X metal elements & Y nonmetal elements, where Y is more then X.

There are some rare but very interesting cases in which two metals react when mixed instead of forming an alloy, creating a non-metallic compound made only with metal atoms. Examples are caesium auride ($\ce{CsAu}$), caesium platinide ($\ce{Cs2Pt}$) and several barium platinides ($\ce{BaPt}$, $\ce{Ba2Pt}$, $\ce{Ba3Pt2}$, and possibly others), all of which best described as ionic salts rather than alloys. These are a remarkable consequence of relativistic effects in heavy atoms.

I'm not sure if I undestood you compeletly, but for example compounds called mixed oxides can have cations of many metals in their structure, like $\ce{Y3Al5O12}$ - consisting of yttrium and aluminum cations, and oxide anions.


Rate of electrophilic bromination of toluene and ethylbenzene My book says that electrophilic bromination of toluene (methylbenzene) is faster than that of ethylbenzene. How is that possible?

I am doubting it because ethyl is a better $+\ce{I}$ group than methyl, so ethyl would increase the electron density in the benzene ring more than methyl. As a result, bromine, which is an electrophile, should react with ethylbenzene faster than methylbenzene.

So, why is the bromination of methylbenzene faster than ethylbenzene and please give the reaction mechanism, if possible?

First, I think it's probably important to note that the difference in $\text{+I}$ effect of a methyl vs. ethyl is very small. In principle, the ethyl is more electron-rich, however the strength of the inductive effect decays very rapidly with increasing distance. Ultimately, I suspect variance in steric factors, solvation, and electron-donating ability to the $\pi$-system of the ring by hyperconjugative overlap are the decisive factors, not the difference in induction.

If we consider sterics, the ethyl is obviously the bulkier group, and that may partially explain the difference in reactivity. The ethyl, moreso than the methyl, will hinder the approach of the electrophile, and may potentially raise the energy of the transition state and $\sigma$-complex intermediate slightly by:

(a) electrostatic repulsion toward the electrophile, particularly in the case of ortho- attack, and

(b) inhibit potential stabilization of the $\sigma$-complex intermediate by solvent molecules.

From the standpoint of electronic effects, I suspect the differences are marginal. To the extent of my understanding, the energies of $\ce{C-H}$ and $\ce{C-C}$ bonding $\sigma$ orbitals are very similar, and the efficiency of hyperconjugative overlap should consequently be similar as well. I've occasionally encountered claims that $\ce{C-H}$ bonds are slightly better $\pi$-donors, but I'm not aware of any conclusive evidence to that effect. If that is true, however, it might contribute to the observed disparities in reactivity.

For a quantitative approach, you may want to take a look at the substituent constants derived for various alkyl groups from the Hammett equation. The equation was originally developed with the aim of analyzing substituent effects on the thermodynamic equilibria of benzoic acid derivatives. However, since the reactivity of substituted benzoic acid is also affected by the mesomeric and inductive properties of ring substituents (albeit in the opposite sense from a benzene ring in an EAS reaction), I think the Hammett equation furnishes a reasonable proxy measure of at least the relative strengths of substituents as either donors or acceptors of electron density. I found one table of said substituent constants. (Keep in mind that those table values are for ionization of benzoic acid, so a negative value is actually indicative of an electron-donating effect, which is favorable for an EAS reaction but destabilizing to an acid.)

Finally, as to the reaction mechanism, I'm not going to reproduce it here. Every EAS reaction follows a very similar, schematic mechanism. For the mechanisms, take a look, for example, at the Wikipedia articles on electrophilic halogenation and electrophilic aromatic substitution more generally.

Reactivity of alkylbenzenes towards electrophillic substitution reactions is decided only on the basis of steric hinderence that alkyl groups provides to aromatic ring.if steric hinderence increases reactivity of aromatic ring decreases .steric hinderence by alkyl groups increases in the order -CH3 < 1¤ < 2¤ < 3¤ . Therefore reactivity decreases in the order CH3 > 1¤ > 2¤ > 3¤ .
E.g. toluene > ethylbenzene


Does presence of alkyl groups on alpha carbon increase yield of alkane in Kolbe's Decarboxylation? We know that, during the synthesis of alkanes by Kolbe's Decarboxylation reaction,
the reaction proceeds via a free radical $\ce{R.}$ at the anode.

So, an alkyl group, say methyl, when substituted at the alpha carbon will increase the yield of alkane since the radical is stabilized by hyperconjugation. But my book states that the presence of alkyl group on alpha carbon decreases the yield of alkane. Why is it so?

I have done some electrosynthesis work using the Kolbe reaction, what I found was with straight chain carboxylic acids that my platinum anode was quickly fouled.

When I had some branching at the beta site, I had much less electrode fouling and I got a good yield of my alkane.


Which of the following atoms has the largest first ionization energy? Of the following atoms, which has the largest first ionization energy?

$\ce{Br}$
$\ce{O}$
$\ce{C}$
$\ce{P}$
$\ce{I}$

I got confused between $\ce{O}$ and $\ce{Br}$. The answer given is $\ce{O}$ although in the periodic table $\ce{Br}$ is closer to Group 18 (noble gases). Why is the answer $\ce{O}$ then?

I actually think this is an unfair question, unless you were given some additional information, or you were specifically instructed to memorize specific ionization energies, rather than learning the trends.  The trend for ionization energies is that they increase from left to right across rows, and from bottom to top in columns.  Based on only that information, I can't see how one should distinguish between $O$ and $Br$ ... yes, it happens to be true that $O$ has a higher IP than $Br$, but how are you supposed to tell? Suppose the example had used $S$ instead of $O$, and $I$ instead of $Br$?  Those elements have the same relative positions in the table, but in that case, it would be $I$ that has the larger ionization energy.

So, in this case I think your confusion was quite understandable .. I would never construct a test question like this unless the student had some additional information.  It's not even true that you can always use the rule about them increasing from left to right along rows .. in the first row, $N$ actually has a slightly higher IP than $O$, because there is a penalty for disrupting the half-filled p-shell in $N$, which is anomalously stable.  You can see the periodic trends here.

Ok so I thought about the question to and I thought that Bromine should have a higher first ionization energy but then if you really think about it, the answer would be oxygen. Look at it this way. Oxygen desperately needs two electrons. And Bromine needs only one more. So if you look at it, it would be harder to lose another electron when you already lost two and it would be easier to lose one more if you are Bromine. And also if you look at a basic ionization energy trend, it goes to the top right corner.

Imagine you were broke, would you be willing to pay to get something even though you know that you need money? No like Oxygen.

Think of Bromine as an investor who needs a little bit more $ for an achievement, Bromine would be more likely to sacrifice its electron or to invest in something like an investor.

Really bad analogies but hopefully you get the idea.


How does one tell if a specific molecule is acidic or basic? Let's take the $\ce{KOH}$ molecule into account. I know it is a base from literature, but how would one go about determining if a molecule is acidic or basic simply based on the structure of the molecule? How about amphoteric?

Also, I understand that there are two complementary systems - Lewis and Brønsted-Lowry theories. How do they work and how do they fit together?

It is sometimes challenging to determine if a molecule is going to be acidic or basic if the system in which it is reacting is not considered.  An important point to consider when dealing with acids and bases is that acid/base strength is inherently tied to the solvent.  For this answer, I'm going to limit the discussion to acids and bases in an aqueous environment.
It is helpful to consider the terms acid and base as a means to classify substances.  This way, Chemists can explain chemical reactivity and structure-function relationships of substances.  Very early classification systems depended on our senses (acids are sour, bases are slippery to the touch) and more recent classification systems utilize structural characterization tools such as NMR or crystallography.  Many classification systems have been proposed over the years, and only a few of them have found sufficiently widespread use to end up in textbooks used in the standard Chemistry curriculum.  Below are a few systems, taken from Miessler & Tarr's Inorganic Chemistry textbook.  The 2nd through 4th entries are those in common use today.

NAME (YEAR) acid definition [example]; base definition [example]
Liebig (~1776) Acid: an oxide of N, P, S [$\ce{SO_3}$]; Base: Reacts with
acid [$\ce{NaOH}$]
Arrhenius (1894) Acid: Forms hydronium ion [$\ce{HNO_3}$]; Base: Forms hydroxide ion [$\ce{NaOH}$]
Brønsted (1923) Acid: Proton donor [$\ce{HCl}$]; Base: Proton acceptor [$\ce{NaOH}$]
Lewis (1923) Acid: Electron-pair acceptor [$\ce{Ag^+}$]; Base: Electron-pair donor [$\ce{NH_3}$]
Ingold-Robinson (1932) Acid: Electrophile [$\ce{BF_3}$]; Base: Nucleophile [$\ce{NH_3}$]
Lux-Flood (1939) Acid: Oxide ion acceptor [$\ce{SiO_2}$]; Base: Oxide ion donor [$\ce{CaO}$]
Usanovich (1939) Acid: Electron acceptor [$\ce{Cl_2}$]; Base: Electron donor [$\ce{Na}$]
Solvent system (1950s) Acid: Solvent cation [$\ce{BrF_2^+}$]; Base: Solvent anion [$\ce{BrF_4^-}$]
Frontier Orbitals (1960s) Acid: LUMO of acceptor [$\ce{BrF_3}$]; Base: HOMO of donor [$\ce{NH_3}$]

I find the various acid/base systems very enlightening.  Note how From Arrhenius through Lewis there was a broadening of the acid/base classification system; Arrhenius can't be used to describe non-aqueous acids and bases and Brønsted can't be used with aprotic substances.  Yet, the Lewis definition incorporates the previous two (a Brønsted acid is also a Lewis acid; an Arrhenius base is also a Lewis base).  One then may ask, what's up with the Lux-Flood definition then?  This definition is counter to the trend of broadening the classification system, and yet it is useful in describing anhydrous solid-state chemistry and is used to describe geochemical reactions as well as the chemistry of high-temperature melts.  My point being: classification of substances into acids and bases is only meaningful if it helps explain chemical phenomena.
Which brings us to the Usanovich definition, which essentially states that every reaction is an acid-base reaction.  Such a broad definition is not overly helpful, and to those of us with an affinity towards electrochemistry (ahem), is somewhat arrogant :-)
I do believe that determining if a substance will behave as an acid or a base requires a bit of chemical intuition (or a Socratic method).  For example, you may have performed an experiment in which $\ce{KOH}$ served as a base; you know that, like potassium, sodium is an alkali metal; therefore you presume that $\ce{NaOH}$ would be a base as well.
Back to the question at hand
So how do I suggest one use this information to predict whether a substance will behave as an acid or a base using its structure alone?  Personally, I find the Lewis theory as the most useful classification system in answering this type of question.  If the structure of a compound is set before me and I am to predict its acid/base chemistry, I will ask two questions:

Are there any lone pair electrons that can be donated?
Are there any electron-deficient atoms that could serve as electron pair acceptors?

If the answer to question 1 is yes, then the molecule can behave as a base.  If the answer to question 2 is yes, then the molecule is an acid.  If both are true, then I have an amphoteric substance.
In proofreading this answer, I realize I said I would restrict myself to aqueous systems, in which case using the Brønsted system may be more helpful.  In this case, the questions become:

Is there a hydrogen that can be donated? (You'll be right more often than you are wrong if you rephrase this question as "Is there a hydrogen that is attached to something other than carbon?").
Is there a lone pair that can accept a proton?

In organic chemistry, whether a compound is acidic or basic can be predicted by looking at the molecular structure and can quite easily be explained by applying the resonance theory.  In essence, if the compound can ionize to form a proton and the resulting anion is stable, then the ionization equilibria will be shifted to the right and the compound is acidic.  The more stable the anion, the more acidic the compound is.
Consider the following examples:

Sulfuric acid   $\ce{H2SO4 <=> 2H+ + SO4^2-}$   the sulfate anion is stabilized by 4 resonant structures, therefore sulfuric acid is very acidic.
Nitric acid  $\ce{HNO3 <=> H+ + NO3-}$   the nitrate anion is stabilized by 3 resonant structures, therefore nitric acid is acidic.
Carboxylic acids   $\ce{RCOOH <=> H+ + RCOO-}$   the carboxylate anion is stabilized by 2 resonant structures, therefore carboxylic acids are acidic.
Phenols versus alcohols.  $\ce{C6H5OH <=> H+ + C6H5O-}$   the phenoxide anion has 4 resonant structures while the alkoxide anion has none.  Therefore, phenols are more likely to ionize to produce a proton than alcohols.  That is, phenols are more acidic than alcohols.
Ketones versus alcohols.    $\ce{R1COCHR2R3 <=> H+ + R1COC-R2R3}$   the resulting anion has 2 resonant structures whereas alcohols have none.  Therefore the α-hydrogen to the carbonyl group is acidic because the enoxide ion is more stable than the alkoxide ion (keto-enol tautomerism).
Amides   $\ce{RCONH2 <=> H+  + RCONH-}$  the resulting anion has 2 resonant structures, but they are not equivalent because one has the negative charge on the oxygen, the other one has it on the nitrogen.  Therefore the anion is not particularly stable so amides are only weakly acidic.  The lone pair of electrons on the nitrogen is only slightly available for bonding because they are withdrawn by the more electronegative oxygen.  Therefore amides are only weakly basic. They are practically neutral.
Sulfonamides  $\ce{RS(O)2NH2 <=> H+  + RS(O)2NH-}$   the resulting anion has 3 resonant structures, 2 of which have the negative charge on the more electronegative oxygen atom.  Therefore sulfonamides are a little more acidic than amides and can form salt with metals (e.g. silver sulfadiazine, sodium sulfacetamide).
Urea  $\ce{H2NCONH2}$ the lone pair of electrons on one nitrogen are withdrawn by the more electronegative oxygen (see amide above), but the lone pair on the other nitrogen is available for bonding.  Therefore urea is basic.
Imides  $\ce{R1CONHCOR2 <=> H+ +  R1CON-COR2}$   the resulting anion has 3 resonant structures, 2 of which have the negative charge on the more electronegative oxygen, therefore imides are acidic (for example phthalimide).


Stability of cyclic radicals How do we arrange the following in increasing order of stability?

The answer given is: $5<3<1<2<4$ (numbering according to their order in the image).

My Approach:

I can figure that in the second and the fourth structure, resonance stabilization will be applicable and the fourth structure will be more stable than the second.

But I am not able to figure out how do I arrange the remaining structures.

Radicals are electron-deficient compounds. Therefore, according to Bent’s rule a radical should always populate an atomic orbital with a maximised $\mathrm p$ contribution. Ignoring 2 and 4 for a second, the radical occupies an $\mathrm{sp^2}$ orbital in 3 and 5 while it occupies an $\mathrm{sp^3}$ orbital in 1.[1] This already gives us $3\approx5<1$.

In a next step, let’s consider 2 and 4. These, as you noticed, are adjacent to a π bond, so the radical can occupy a $\mathrm p$ type orbital and participate in π resonance. This is the main way to differentiate 1, 2 and 4, since all three could essentially adopt a planar configuration with the radical confined to a $\mathrm p$ orbital — yet in 2 it can also resonance with the allyl system and in 4 it can resonate across two double bonds. Thus, we now intermediately have $3\approx5<1<2<4$.

All that remains is to determine the difference between 3 and 5, which is not trivial. In 3, you have a single double bond. This structure can distort a little bit, increasing the bond angle on the radical-bearing carbon towards something closer to $\mathrm{sp}$. The freely rotatable single bonds making up most of the ring allow this distortion. The phenyl radical can distort much less, since its geometry is basically already defined from the six-membered aromatic ring. Thus, we may assume 3 to be more stable than 5. Therefore: $5<3<1<2<4$.

Note:

[1]: In fact, rehybridisation is also possible in 1, giving the radical a pure $\mathrm p$ orbital and making the arrangement of substituents planar around the radical centre.

While checking stability of Carbon free radical it is treated as positive charge but we don't see resonance or mesomeric effect because for these effects we have to break bonds Homolatically, which is possible only at high temperature. So we see only H effect or I effect.
Now 5 and 3 are sp2 which are least stable and + I of sp3 carbon chain is more so 3 is more stable than 5.
In 4,2,1 we see H effect so 6,5,4 conjugated hydrogen atoms are there respectively in 4,2,1 so stability order is
4 > 2 > 1 > 3 > 5