Regarding the mechanism for biological activity of tyramine, amphetamine and ephedrine Graham Patrick explains the mechanism for the activity of tyramine, amphetamine and ephedrine in his book Introduction to Medicinal Chemistry as follows:

  Some amines such as tyramine , amphetamine , and ephedrine closely resemble noradrenaline in structure and are transported into the nerve cell by noradrenaline’s transport proteins.
  Once in the cell, they are taken up into the vesicles. Because these
  amines are competing with noradrenaline for transport proteins,
  noradrenaline is more slowly reabsorbed into the nerve cells.
  Moreover, as the foreign amines are transported into the nerve cell,
  noradrenaline is transported out by those same transport proteins.
  Both of these facts mean that more noradrenaline is available to
  interact with its receptors. Therefore, amphetamines and similar
  amines have an indirect agonist effect on the adrenergic system.

I have two question concerning this explanation:

When these foreign amines have been taken up into the vesicles how do they proceed? Do they re-enter the synaptic cleft by exocytosis, so that they can re-exercise their negative influence on noradrenaline re-uptake? How are they eventually cleared from the synaptic area?
Why do the noradrenaline re-uptake proteins transport noradrenaline out of the axonal terminal under the influence of these foreign amines?

Question: When these foreign amines have been taken up into the vesicles how do they proceed?...Do they re-enter the synaptic cleft by exocytosis...How are they eventually cleared from the synaptic area?
Here is the mode of action of these agents (indirectly acting sympathomimetics):

The most
important drugs in the indirectly acting sympathomimetic amine
category are tyramine, amphetamine and ephedrine, which are
structurally related to noradrenaline.
Amphetamine enters the nerve terminal via the noradrenaline
transporter (NET) and enters synaptic vesicles via the vesicular
monoamine transporter (VMAT), in exchange for noradrenaline
(NA), which accumulates in the cytosol.
Some of the NA is degraded
by monoamine oxidase (MAO) within the nerve terminal and some escapes,
in exchange for amphetamine via the noradrenaline transporter, to act
on postsynaptic receptors. Amphetamine also reduces NA reuptake via
the transporter, so enhancing the action of the released NA.

Exocytosis is not involved in the release process, so their actions do not require the presence of Ca2+. They are not completely specific in their actions, and act partly by a direct effect on adrenoceptors, partly by inhibiting NET (thereby enhancing the effect of the released noradrenaline) and partly by inhibiting MAO.
These agents are cleared by metabolism via oxidative deamination with monoamine oxidase(MAO) enzyme, however  are poor substrates for MAO and, thus, show a prolonged duration of action.
Amphetamine is an example of a drug metabolized through oxidative deamination:

Question: Why do the noradrenaline re-uptake proteins transport noradrenaline out of the axonal terminal under the influence of these foreign amines?
I am not really sure about the “why” part but instead the “how” part AFAICT, this is more of a displacement mechanism:
These drugs, do block reuptake 1(NET) releasing stored transmitter from noradrenergic nerve endings and they displace vesicular norepinephrine into the cytoplasm. The norepinephrine is then transported out of the nerve terminal by the reverse action of reuptake 1 and activates adrenoceptors.
So essentially the drugs increase the availability of noradrenaline to stimulate adrenergic receptors by:
i. releasing or displacing NE (noradrenaline) from sympathetic nerve endings,
ii. blocking the transport of NE into sympathetic neurons or
iii. blocking the metabolizing enzymes, MAO or COMT.
How this helps
References

Rang and Dale’s Pharmacology 8th Edition. Section 2: Chemical Mediators

Medical Pharmacology at a Glance M. JNeal: Drugs acting on the sympathetic system

Lippincott Illustrated Reviews: Pharmacology 6th Edition: Adrenergic Agonists

Goodman and Gilman Manual of Pharmacology and Therapeutics: Agents Acting at the Neuromuscular Junction and Autonomic Ganglia

Principles of Clinical Pharmacology 2nd Ed Atkinson etal
Basic and Clinical Pharmacology B. Katsung et al.


Where did the acac come from in Cr(acac)₃ synthesis by CrCl₃ reagent? Those are the reactions:

$$\ce{(H2N)2CO + 3 H2O -> CO2 + 2 NH4+ + 2OH-}$$
$$\ce{CH3COCH2COCH3 + OH- -> acac- + H2O}$$
$$\ce{CrCl3.6H2O -> Cr3+ + 3Cl- + 6H2O}$$
$$\ce{Cr3+ + 3acac- -> Cr(acac)}3$$

I don't get from where comes the $\ce{C5H8O2}$ in reaction #2!

$\ce{CH3COCH2COCH3}$ is the condensed formula of pentane-2,4-dione, commonly known as acetylacetone. The $\ce{acac^-}$ is just a shorthand for the acetylacetonate ligand, i.e., $\ce{C5H7O2^-}$, which is just the deprotonated form of the acetylacetone used in reaction two.


Can a Birch Reduction be done using gaseous anhydrous ammonia? Instead of generating anhydrous ammonia as a liquid using either pressure or cryogenic freezing, I would like to avoid the hassle entirely by using anhydrous ammonia generated as a gas. Is there a way to use a gaseous form of ammonia for the birch reduction that doesn't generate sodium amide? I'm a hobbyist chemist so I don't have access to a ton of glassware or specialty equipment.

And before someone asks me if I'm making meth- No. I like my freedom and this is just a hobby of mine. Mostly why I'm trying the birch is because I have heard the color of the solvated electron is quite beautiful. Chemistry is a wonderful science and I wish simply showing an interest in it wasn't considered so unusual!

A word of caution: pure ammonia, in either its gaseous or liquified form is extremely dangerous to handle, especially in the context of your 'hobby' where, in your own words you "don't have access to a ton of glassware or specialty equipment." Any ammonia that escapes from your reaction and isn't ventilated properly will cause severe burns to anything it comes into contact with, including completely trashing the lining of your nose and lungs etc. In an attempt to relieve your curiosity, a photograph of a birch reduction showing the characteristic blue is shown below:

That said, the question is an interesting one in the context of laboratory chemistry.

  Is there a way to use a gaseous form of ammonia for the birch reduction that doesn't generate sodium amide?

Yes and No. The most common way of performing these reactions on a small scale in a lab context is to essentially pass the gas over your reaction as illustrated below:

Practically, you're not actually making liquid ammonia and then transferring it to your reaction, but rather your passing it over your reaction flask which is fitted with a condenser which will cause in-situ liquefaction of the gaseous ammonia. Even a simple dry ice condenser is sufficient to do this, and the formation of the active reagent is easily observed by the appearance of the characteristic blue/green colour.  Any gas not liquified is then passed to the back of the fume cupboard, or, ideally, bubbled through a solution of something to neutralise it and hence stop it being released.


Reaction of alkyl bromide in presence of piperidine What are the products of this reaction? I believe there'll be E2 elimination to give the products. Not sure. Help would be greatly appreciated.

The above is the product which I thought might be possible due to E2 elimination

In case there is carbocation formation, this might be the intermediate:

Could there be nucleophilic substitution?

@User6376297 raised a question:

  "Could the cyclopropane ring behave like a double bond, making this the analogue of an allyl bromide, and yielding $\ce{R2NCH2CH2CH=CH-tBu}$ by attack on a cyclopropyl $\ce{CH2}$ and ring opening?" Especially given the steric hindrance near the carbon bearing the leaving group.

Indeed, that is the process going on in this case when the reaction was taken place in piperidine either at refluxing conditions or at room temperature. When (1-bromo-2,2-dimethylpropyl)cyclopropane (the alkyl bromide in question) was reacted in refluxing piperidine, it has yielded $88\%$ of (E)-1-(5,5-dimethylhex-3-en-1-yl)piperidine as a major product, which corresponds to the homoallylic analogue of the well-known SN2 process observed with allylic halide.

In addition, $8\%$ of 1-(1-cyclopropyl-2,2-dimethylpropyl)piperidine was also yielded as a minor product (Ref. 1 & 2). The authors claimed the formation of the minor product was not compatible with simple SN2 displacement due to the great steric hindrance imposed by the neighboring $\ce{t-Bu}$ and cyclopropyl groups, but the tight ion-pair mechanism (not separated by solvent) may explain the presence of SN2 product very well.

The percent yields of SN2' and SN2 products for the reaction at $\pu{25^\circ C}$ had been $73\%$ and $8\%$, respectively.

References:

Reaction of amines with cyclopropylcarbinyl halides: SN2' or
solvolysis?: J. Org. Chem., 1985, 50(24), 4815–4821 (https://doi.org/10.1021/jo00224a033).
Homoallylic substitution reaction of piperidine with 1-bromo-1-cyclopropylalkanes: Tetrahedron Lett., 1983, 24(46), 5031–5034 (https://doi.org/10.1016/S0040-4039(00)94033-4).


How to arrange carboxylic acids (phenylacetic acid and acetic acid) by increasing acidity? Why is $\ce{Ph-CH2-COOH}$ more acidic than $\ce{CH3COOH}$ although the equilibrium of ionization lies mainly backward for both of them?

Attempt:
In the ionized form of $\ce{Ph-CH2-COOH}$ we have a $\ce{Ph-CH2}$-group donating its electron to carbonyl carbon while in case of acetic acid we have a methyl group donating its electron to the carbonyl carbon.
But which one of them is stabilized more?

The $\mathrm pK_\mathrm a$ difference between phenylacetic acid and acetic acid is around 0.5 and the trends become more obvious when other arylalkanoic acids are included in the comparison. Numbers in red are the $\mathrm pK_\mathrm a$ values.

Apparently, phenylacetic acid is a stronger acid than acetic acid. Adding a further phenyl subsituent even increases the acid strength.
The effect levels off with the distance of the phenyl substituent and the carboxylate. Note that 4-phenylbutyric acid is just as strong as acetic acid. It's as if there isn't any phenyl group at all.

Would you think that there is a $-I$ effect of the phenyl substituent on the $\ce{-COOH}$ group?


Nucleophilicity Of Amino Acid Side Chains Out of the side chains of aspartic acid, methionine, threonine, and tyrosine, which is the least nucleophilic?

Aspartic acid < Tyrosine < Threonine < Methionine is the order of nucleophilicity of the amino acid side chains. So, Aspartic acid (neutral form) is the answer. Every side chain has a Oxygen or Sulphur (Methionine) group available for electron pair donation, but in Aspartic acid the Oxygen group has lesser inclination for donation of electron density (due to resonance). Moreover, the solvation layer surrounding the acid group of Aspartic acid makes the job of nucleophilic attachment even more tougher.


How to compare the rate of decarboxylation? For the following acids compare the rate of decarboxylation on heating
I $\ce{Ph-CO-CH2COOH}$
II $\ce{Ph-CO-COOH}$
III $\ce{CH3-CH2-COOH}$
IV $\ce{HOOC-CH2-COOH}$

I tried this question by forming carbanions and comparing their stability. But I am stuck at comparing I and IV.

First thing to see is to see carbanion stability.

Now the question arises that between  carbonyl and COOH  groups which can stabilize better. For that you can look at the COOH 's already possible resonance,within its molecule, which decreases it electron withdrawal  tendency.
Hence order would be: I>IV>III>II.


Factors controlling enzymatic phosphorylation of D-ribulose 5-phosphate Considering the Metabolic Pathway:
http://www.genome.jp/kegg-bin/show_module?M00166

Consider the reaction R01523(the very first reaction). How is it possible to know that the compound C00199(D-Ribulose 5-phosphate) will get transformed to C01182(D-Ribulose 1,5-bisphosphate)? What are the factors on which this transformation depends? One factor I know is structure but how does structure effect this transformation? Does two similar structured compound means that one of the compound can get transformed to the other one?
Note: It ia a part of biology but a chemical reaction or transformation is basically a part of chemistry, so asked this here

After looking at the website I found the reaction is based off of an enzyme called: phosphoribulokinase. In biological systems many reactions occur under the auspice of enzymes or proteins. These are incredibly large molecules made of an extremely long chain of amino acids which are a set of 26 or so small molecules that can 'hook' into the chain structure of a protein. Each amino acid has different properties and thus experiences different forces pulling it towards and away from other amino acids on the chain. What this means is the whole chain folds up into an extremely complex structure. For a protein to allow a reaction to occur it has a small region that the reacting molecule can hook into. There are then sets of atoms and molecules that act as scaffolds that allow atoms in the reacting molecule to be moved without the need for nearly as much energy as would be required without the scaffold. Now this is how it changes structure. Now whilst the protein can just as easily allow the molecule produced to be converted back to the original, this may be less energetically favourable which brings us to equilibrium chemistry involving dynamic forward and backward reactions. What makes this more complicated is that proteins often have separate 'hooks' for molecules to disable them. That is if a lot of substance A is around then the protein will be disabled and no longer convert B into A until more A is consumed and thus stops inhibiting the protein.


Derivation of the Heisenberg uncertainty principle The Heisenberg uncertainty principle states that

$$\Delta x \Delta p \geq \frac{\hbar}{2}$$

where
$\Delta x$ is the uncertainty in the position,
$\Delta p$ is the uncertainty in linear momentum, and
$\hbar = 1.054571800(13) \times 10^{-34}\ \mathrm{J\ s}$[source] is the reduced Planck constant

This means that, regardless of what quantum mechanical state the particle is in, we cannot simultaneously measure its position and momentum with perfect precision. I read that this is intrinsically linked to the fact that the position and momentum operators do not commute: $[\hat{x},\hat{p}] = \mathrm{i}\hbar$.

How can I derive the uncertainty principle, as given above?

The proof I will use is taken from Griffiths, Introduction to Quantum Mechanics, 2nd ed., pp 110-111.

Defining "uncertainty"

Let's assume that the normalised state $|\psi\rangle$ of a particle can be expanded as a linear combination of energy eigenstates $|n\rangle$, with $\hat{H}|n\rangle = E_n |n\rangle$.

$$| \psi \rangle = \sum_n c_n |n\rangle \tag{1}$$

The expectation value (the "mean") of a quantity, such as energy, is given by

$$\begin{align}
\langle E\rangle &= \langle \psi | H | \psi \rangle \tag{2}
\end{align}$$

and the variance of the energy can be defined analogously to that used in statistics, which for a continuous variable $x$ is simply the expectation value of $(x - \bar{x})^2$:

$$\sigma_E^2 = \left\langle (E - \langle E\rangle)^2 \right\rangle \tag{3}$$

The standard deviation is the square root of the variance, and the "uncertainty" refers to the standard deviation. It's more proper to use $\sigma$ as the symbol, instead of $\Delta$, and this is what you will see in most "proper" texts.

$$\sigma_E = \sqrt{\left\langle (E - \langle E\rangle)^2 \right\rangle} \tag{4}$$

However, it's much easier to stick to the variance in the proof. Let's generalise this now to any generic observable, $A$, which is necessarily represented by a hermitian operator, $\hat{A}$. The expectation value of $A$ is merely a number, so let's use the small letter $a$ to refer to it. With that, we have

$$\begin{align}
\sigma_A^2 &= \left\langle (A - a)^2 \right\rangle  \tag{5} \\
&= \left\langle \psi \middle| (\hat{A} - a)^2 \middle| \psi \right\rangle \tag{6} \\
&= \left\langle \psi \middle| (\hat{A} - a) \middle| (\hat{A} - a)\psi \right\rangle \tag{7} \\
&= \left\langle (\hat{A} - a)\psi \middle| (\hat{A} - a) \middle| \psi \right\rangle \tag{8} \\
&= \left\langle (\hat{A} - a)\psi \middle| (\hat{A} - a)\psi \right\rangle \tag{9}
\end{align}$$

where, in going from $(7)$ to $(8)$, I have invoked the hermiticity of $(\hat{A} - a)$ (since $\hat{A}$ is hermitian and $a$ is only a constant). Likewise, for a second observable $B$ with $\langle B \rangle = b$,

$$\sigma_B^2 = \left\langle (\hat{B} - b)\psi \middle| (\hat{B} - b)\psi \right\rangle \tag{10}$$

The Cauchy-Schwarz inequality

...states that, for all vectors $f$ and $g$ belonging to an inner product space (suffice it to say that functions in quantum mechanics satisfy this condition),

$$\langle f | f \rangle \langle g | g \rangle \geq |\langle f | g \rangle|^2 \tag{11}$$

In general, $\langle f | g \rangle$ is a complex number, which is why we need to take the modulus. By the definition of the inner product,

$$\langle f | g \rangle = \langle g | f \rangle^* \tag{12}$$

For a generic complex number $z = x + \mathrm{i}y$, we have

$$|z|^2 = x^2 + y^2 \geq y^2 \qquad \qquad \text{(since }x^2 \geq 0\text{)} \tag{13}$$

But $z^* = x - \mathrm{i}y$ means that

$$\begin{align}
y &= \frac{z - z^*}{2\mathrm{i}} \tag{14} \\
|z|^2 &\geq \left(\frac{z - z^*}{2\mathrm{i}}\right)^2 \tag{15}
\end{align}$$

and plugging $z = \langle f | g \rangle$ into equation $(15)$, we get

$$|\langle f | g \rangle|^2 \geq \left[\frac{1}{2\mathrm{i}}(\langle f | g \rangle - \langle g | f \rangle) \right]^2 \tag{16}$$

Now, if we let $| f \rangle = | (\hat{A} - a)\psi \rangle$ and $| g \rangle = | (\hat{B} - B)\psi \rangle$, we can combine equations $(9)$, $(10)$, $(11)$, and $(16)$ to get:

$$\begin{align}
\sigma_A^2 \sigma_B^2 &= \langle f | f \rangle \langle g | g \rangle \tag{17} \\
&\geq |\langle f | g \rangle|^2 \tag{18} \\
&\geq \left[\frac{1}{2\mathrm{i}}(\langle f | g \rangle - \langle g | f \rangle) \right]^2 \tag{19}
\end{align}$$

Expanding the brackets

If you've made it this far - great job - take a breather before you continue, because there's more maths coming.

We have1

$$\begin{align}
\langle f | g \rangle &= \left\langle (\hat{A} - a)\psi \middle| (\hat{B} - b)\psi \right\rangle \tag{20} \\
&= \langle \hat{A}\psi |\hat{B}\psi \rangle - \langle a\psi |\hat{B}\psi \rangle - \langle \hat{A}\psi | b\psi \rangle + \langle a\psi |b\psi \rangle \tag{21} \\
&= \langle \psi |\hat{A}\hat{B}|\psi \rangle - a\langle \psi |\hat{B}\psi \rangle - b\langle \hat{A}\psi | \psi \rangle + ab\langle \psi |\psi \rangle \tag{22} \\
&= \langle \psi |\hat{A}\hat{B}|\psi \rangle - ab - ab + ab \tag{23} \\
&= \langle \psi |\hat{A}\hat{B}|\psi \rangle - ab \tag{24}
\end{align}$$

Likewise,

$$\langle g | f \rangle = \langle \psi |\hat{B}\hat{A}|\psi \rangle - ab \tag{25}$$

So, substituting $(24)$ and $(25)$ into $(19)$,

$$\begin{align}
\sigma_A^2 \sigma_B^2 &\geq \left[\frac{1}{2\mathrm{i}}(\langle\psi |\hat{A}\hat{B}|\psi \rangle - \langle \psi |\hat{B}\hat{A}|\psi\rangle) \right]^2 \tag{26} \\
&= \left[\frac{1}{2\mathrm{i}}(\langle\psi |\hat{A}\hat{B} - \hat{B}\hat{A}|\psi \rangle ) \right]^2 \tag{27}
\end{align}$$

The commutator of two operators is defined as

$$[\hat{A},\hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A} \tag{28}$$

So, the term in parentheses in equation $(27)$ is simply the expectation value of the commutator, and we have reached the Robertson uncertainty relation:

$$\sigma_A^2 \sigma_B^2 \geq \left(\frac{1}{2\mathrm{i}}\langle[\hat{A},\hat{B} ]\rangle \right)^2 \tag{29}$$

This inequality can be applied to any pair of observables $A$ and $B$.2

The Heisenberg uncertainty principle

Simply substituting in $A = x$ and $B = p$ gives us

$$\sigma_x^2 \sigma_p^2 \geq \left(\frac{1}{2\mathrm{i}}\langle[\hat{x},\hat{p} ]\rangle \right)^2 \tag{30}$$

The commutator of $\hat{x}$ and $\hat{p}$ is famously $\mathrm{i}\hbar$,3 and the expectation value of $\mathrm{i}\hbar$ is of course none other than $\mathrm{i}\hbar$. This completes the proof:

$$\begin{align}
\sigma_x^2 \sigma_p^2 &\geq \left(\frac{1}{2\mathrm{i}}\cdot\mathrm{i}\hbar \right)^2 \tag{31} \\
&= \left(\frac{\hbar}{2}\right)^2 \tag{32} \\
\sigma_x \sigma_p &\geq \frac{\hbar}{2} \tag{33}
\end{align}$$

where we have simply "removed the square" on both sides because as standard deviations, $\sigma_x$ and $\sigma_p$ are always positive.

Notes

1 I have skipped some stuff. Namely, $\langle \hat{A}\psi |\hat{B}\psi \rangle = \langle \psi |\hat{A}\hat{B}|\psi \rangle$ which is quite straightforward to prove using the hermiticity of both operators; $\langle \psi |\hat{A}|\psi \rangle = a$; $\langle \psi |\hat{B}|\psi \rangle = b$; and $a = a^*$ since it is the expectation value of a physical observable, which must be real.

2 This does not apply to, and cannot be used to derive, the energy-time uncertainty principle. There is no time operator in quantum mechanics, and time is not a measurable observable, it is only a parameter.

3 Technically, it is a postulate of quantum mechanics. (If I am not wrong, it derives from the Schrodinger equation, which is itself a postulate.)


Phase rule and degrees of freedom In the formation of 3-bromo-1-butene and 1-bromo-1-butene, from $\ce{HBr}$ and $\ce{C_4H_6}$, in a gas phase:

$\ce{HBr + C_4H_6 \to 1BrC_4H_7}$

$\ce{HBr + C_4H_6 \to 3BrC_4H_7}$

the number of degrees of freedom is given by: $F=C-P+2$

I'm trying to rationalise this but finding it quite confusing. For example, can we state that there is only 1 phase, i.e. a gas phase?

And the components, is the isomer of one independent of another? I'm taking there to be 3 individual components.

My answer is then 4 degrees of freedom, which seems okay, given $p$, $t$, $X$, $V$ could vary?

I'd appreciate your help in this, thanks.

Let's do it step by step.

$C = 2$. The reason is that you have two components: $\ce{HBr}$ and $\ce{C4H6}$. The third compound, $\ce{BrC4H7}$, is simply a combination of both of them. Remember what is $C$: It's the minimal number of components needed to describe all phases in the system. Since $\ce{BrC4H7}$ can de described as $1\times\ce{HBr} + 1\times\ce{C4H6}$, it is not an independent component.
$P=1$ or $P=2$. It depends on how much the reactions has progressed. Basically, you have one phase on each side of the reaction. On the left, you have one gas phase that has a mixture of both $\ce{HBr}$ and $\ce{C4H6}$. On the right, you one one liquid (?) phase: $\ce{BrC4H7}$. So here are the possibilities:

The reaction has not started yet. All you have in the one gas phase. In that case, $P=1$ and $F=2-1+2=3$.
The reaction is going as written, but not completed yet. In this case not all gas is consumed yet, but liquid exists. Then, $P=2$ and $F=2-2+2=2$.
The reaction is finished. All gas is now liquid. Then we're back to $P=1$ and $F=2-1+2=3$.

So what happened in stage 2? Why did you lose a degree of freedom? Well, it depends on what what the driving force for the reaction. For example: Why did gaseous $\ce{HBr}$ and $\ce{C4H6}$ combine to form liquid $\ce{BrC4H7}$? Was it because of cooling or heating? In this case, the lost degree of freedom is $t$. Was it pressure? Then the lost degree of freedom is $P$.

$X$ is a bit problematic. There is not enough information given in the question. In my opinion, $X$ was fixed to begin with: I assume that once the reaction is complete, none of the reactants remain. Thus their ratio is fixed to 1:1 and $X$ is constrained. Had it not been the case, you would also get differing $P$ in stage three because you would have two phases, not one.


How can solids undergo combustion if it only occurs in the gas phase? According to the smoldering Wikipedia article, flaming combustion occurs in the gaseous phase. I know butane undergoes flaming combustion in a Bunsen burner because it's a gas, but how can wood undergo flaming combustion at all instead of just smouldering? What gas is there to react with oxygen in the flames? Does it smoulder to create carbon monoxide then the carbon monoxide undergoes flaming combustion? Can charcoal also undergo flaming combustion? I've never seen a piece of charcoal without wood inside it undergo flaming combustion, only smoldering.

Also, when oil undergoes flaming combustion, is it its vapour that reacts with oxygen to produce a flame?

Combustion is a gaseous phase phenomenon. Oil and gasoline have a high enough vapor pressure at ambient temperatures to produce a gaseous phase of fuel above the liquid. In contrast, hold a lighter up to a piece of wood and try to get it to light. It won't—at least not for quite some time. This is because solid fuels must first undergo endothermic pyrolysis before real combustion can occur. This produces a slew of products which are what truly undergo combustion when solids are burned. Because combustion is exothermic, once a high enough temperature is reached the solid will autopyrolyze, making combustion a self-sustaining reaction.


How is change in entropy of universe related to the energy released by system that can be used to do work? What does change in entropy ($\Delta S$) represent ? How does change in entropy of a universe ($\Delta S_{univ}$) give us information about energy available to do work.

I would like to know the physical significance of entropy of universe and how it is related to energy energy released by system that can be used to do work (Not in terms of equation. I would want to know its physical significance).

Well a reaction which is spontaneous has negative value of Gibbs or Helmholtz free energy. This free energy is the energy which is made available to be used for work.

How does that relate to your question? Both Gibbs and Helmholtz free energy are derived from the argument that change in entropy of the universe is greater than zero. Physically, that is an argument that no chemical system will react in such a way that decreases the number of available microstates unless it increases the number of available microstates for the surroundings by a greater amount.


Name for a spin state with a multiplicity of 13 If I want to name a spin state with 12 unpaired electrons,
what is the correct name?
Tredecim is the latin name for thirteen, but tredecimet sounds odd.

I have seen multiplicity 13 referred to as simply a 13-let. This #-let nomenclature is sometimes used for high multiplicities (e.g. higher than 10).


Bent molecular geometry of water Could you please explain how the bent molecular geometry of water is due to hydrogen bonding? I was under the impression that it was because of the lone pairs (VSEPR Theory).

Your understanding is correct. The usual explanation given in introductory classes is VESPR theory. The oxygen atom has four bond-like "things" coming out of it. Two of those are the covalent bonds with the hydrogen bonds. The other two are the two lone pairs. VESPR theory calls this a steric number of four.

The slightly more advanced explanation comes from molecular orbital theory. Oxygen's valence electrons are all $2\text{p}$ electrons; hydrogen's electron is $1\text{s}$. The only way to mix those orbitals to get enough bonding orbitals is by using two $\text{sp}^3$ hybrid orbitals to generate the two bonds. The $\text{sp}^3$ orbitals are arranged in a tetragonal geometry.


How can diamond have a fixed refractive index? I was doing a lesson on solid states. There were given two kinds of solids namely amorphous solids and crystalline solids. Some differences between these two solids were given in my book. One of the difference was that crystalline solids are anisotropic in nature whereas amorphous solids are isotropic in nature.
Following is from my book

Crystalline solids are anisotropic in nature, that is, some of their physical properties like electrical resistance or refractive index show different values when measured along different directions in the same crystals.

Diamond is a crystalline solid. That means it is anisotropic in nature. This means that measuring the refractive index of diamond from different direction will give different values of refractive index. Then why we have fixed value of refractive index of diamond.
Not just about diamonds, many other crystalline solids have fixed refractive index.

Your textbook is wrong. Some crystalline solids are optically anisotropic. Many crystalline solids are indeed optically isotropic in nature. In solids like this the light beam experiences the same electron density regardless of the direction the crystal is oriented. Some examples of that are (obviously) diamond, halite and sylvite (sodium and potassium chlorides), garnets, spinels and more. Note that this is a feature derived from the electronic structure and it has nothing to do with the actual shape of the crystal.

Even in the case of anisotropic crystals, some are so weakly anisotropic they may considered isotropic in practice. Zeolites are a perfect example of those.


Why does brown sugar have a strong smell but white sugar doesn't? I noticed something while cooking and was wondering what the chemical explanation is for this. White sugar has a smell, but it's different, and much fainter than that of brown sugar.

Brown sugar is unrefined, so does this have something to do with why odorous chemicals are present?

What are these chemicals and why are they present in one, but not the other?

Sugar is made by repeatedly boiling and cooling cane syrup (or sugar beet syrup).  After each cooling, the solution becomes supersaturated with respect to sucrose, causing sucrose to crystallize out of the "mother liquor" (the industrial term for the liquid solution from which crystals form). These crystals of "raw" sugar are heavily processed (washed in syrup, dissolved, treated with carbonate or phosphate minerals, recrystallized, and subject to ion exchange or charcoal color removal) to result in white sugar.

Nowadays, brown sugar is usually made by mixing refined white sugar with blackstrap molasses. In former times, the brown sugar was simply unrefined or unwashed crystal fraction from the 2nd or 3rd crystallization of cane juice. But precise control over the ratio of "contaminating" liquid to white crystalline sugar was difficult, which is why industry moved towards reconstituting brown sugar by mixing molasses and white sugar.

Blackstrap molasses is simply the spent "mother liquor" remaining after the third cycle of boiling and cooling to produce sugar crystals.

The smell of brown sugar thus comes primarily from molasses.  The smell of molasses is very complex.  Here is the abstract of a relevant paper:

  The characteristic sweet-aroma components in refinery final molasses were isolated by using a combination of sensory evaluation, column chromatography and gas chromatography, and were identified by infrared spectrometry, mass spectrometry and color reactions. Identified were ethyl n-hexanoate, ethyl phenylacetate, phenylacetic acid, n-butyl benzoate, isopropyl benzoate, l(+)-pantolactone, benzoic acid, o-toluic acid, m-toluic acid, β-phenylpropionic acid, succinic anhydride, maltol, isomaltol, and 2-methyl-5-hydroxy-6-ethyl-γ-pyrone.

  The aroma of the mixture of these compounds and vanillin and vanillic acid was similar to that of refinery molasses by sensory evaluation. The results indicate that these compounds are important in producing of sweet molasses aroma.


Is there a characteristic that characterizes all, and only, metals? The Wikipedia article on metals describes them as elements whose polyatomic configurations* exhibit any one of a number of patterns. I could not discern any property that characterized all the patterns.

To the best of my knowledge, there are two main categories of elements: metals, non-metals. There's a third category, metalloids, whose members are those that satisfy some of the criteria for inclusion in both main categories.

It would seem very odd if there were nothing that characterized all, and only, metals. In fact it would seem that metals are better conceived of as an arbitrary grouping of what is actually a number of other categories, each of which would include all, and only, the elements that exhibit a single type of polyatomic configuration*.

Is there a characteristic that characterizes all, and only, metals?

*(I'm using the phrase 'polyatomic configuration' in order to express 'the way that atoms of a particular element stick together when you have a number of atoms of that element connected together'. Is there a standard term that expresses that idea?)

The Universe is full of continuous and smoothly-transitioning properties, yet which we often stubbornly try to assign a discrete yes/no answer in order to make the world easier to cope with. Metallicity is one of them. The Van Arkel-Ketelaar triangle is a diagram which attempts to show the transition between covalent, ionic and metallic types of bonding. Here is one populated by a variety of compounds:

The closer a substance is to one of the vertices, the more that vertex is important in describing the nature of bonding in the material. As you can see, the entire triangle is populated quite uniformly. There are certainly many more materials which would fit in the gaps between the materials shown, and even simply varying the ambient conditions (pressure, temperature, etc), a same substance can be put in different regions of the triangle.

Given this behaviour, it should be clear that any attempt to rigorously define a line dividing substances between metals and non-metals will be fraught with difficulty and likely not useful in the first place. No matter where the line is drawn, there will be some compounds that will look strange for "barely being a metal", while others will be strange for "barely being a non-metal".


Can we determine apriori the volume of solvent in an electrolyte Suppose I would like to create a one molar solution of NaCl in water. The molar concentration is defined as

$c_i=\frac{n_i}{V} $

$n_i=1$ in this case. I need to take 58.5 grams of NaCl (atomic weight of Na=23 & Cl=35.5) and dissolve it in water until the total solvent volume is 1 litre. The total volume of the water added is less than 1 litre. But what exactly is that amount of water and how can I determine it apriori from thermodynamics?

You can calculate it using partial molar volumes.

The partial molar volume is defined as the differential change in volume with respect to change in mole fraction, and is determined experimentally. From the definition of partial molar properties, you can calculate the amount of one component needed to achieve a certain volume, if you know all of the partial molar volumes and pure component molar volumes.

There may be some complications to that for this system because you are talking about mixing a solid-phase ionic compound with water, instead of two liquids directly.

In practice, it is much easier to just use a volumetric flask so that you know the total volume is 1 L. If you measure the amount of water you actually used you will find that it is very close to 1 L.


Can sucralose form a syrup? (Almost) Everyone is familiar with making simple syrup from table sugar and water.

I was wondering if pure sucralose is also capable of forming a syrup.

I wanted to do an experiment myself, but to my consternation,

Splenda is not pure sucralose.
pure sucralose is not readily available to me

So, I would like to hear if anyone has tried this experiment, or reasons why sucralose would not form a syrup if such is the case. Pointers to references would be nice.

P.S. I do not have a deep chemistry background, so simple explanations (or even pictures) will be very much appreciated.

Syrup is just a very concentrated solution of sugar.
Because of the high concentration (for instance 65% w/w for USP Syrup, the concentration that avoids bacterial growth) the viscosity increases, and the solution gets syrupy.

In the case of sucralose, the solubility is much lower than that of sugar.
That is normally not an issue, since the sweetening potency is about 600 times higher, at moderate concentrations (80 ppm sucralose have the same potency as 5% sugar).
But to make a syrup, you need a very high concentration.
With a solubility of 28.3 g/100 mL, the viscosity will, probably, be much lower than that of sucrose syrup.

To compare, note that the viscosity of a 30% sugar solution is 3.21 mPa•s, while that is a 60% sugar solution is 56.76 mPa•s.
Take into account that sucralose differs from sucrose only in the exchange of three $\ce{-OH}$ for $\ce{-Cl}$.


What is the Difference Between the Conjugate Effect and the Inductive Effect? As the title suggests, what is the difference between the conjugate and the inductive effect in organic chemistry.

Conjugate effect (or delocalization) is an effect in which molecular orbitals (MOs) are conjugated to new molecular orbitals that are more delocalized and therefore generally lower in energy (the amount of MOs stays the same of course). The electrons can move freely in these new extended orbitals. As an example consider this MO diagram of butadiene.

Note that the lowest energy MO is the MO with the least amount of nodes.

Inductive effect on the contrary does not increase delocalization but just pulls electrons towards the inductive (polarized) atom, this, for example, makes lone pairs less available decreasing the nucleophility of a molecule and increase its acidity. A lot of the time this inductive effect will be achieved trough a $\ce{C-X}$ (fluorine and chlorine) bond and you can observe its inductive effect simply by looking at the acidity of a series of more and less substituted molecules. Compare for example trifluoroacetic acid (pKa of 0.23) with regular acetic acid (pKa of 4.76).

I'm not exactly sure though what effect induction has on the size of a MO (but it won't match the effects of conjugation for sure). It is clear however, that the electron distribution will shift toward the spatial direction of the inductive atom.


Assigning locants when both unsaturation and substituents are present in a hydrocarbon For the following compound:

I understand that the longest chain is of five carbon atoms. However, I am having trouble assigning the locants here.

Going bottom to top we get the locant set as: 223.
Going top to bottom we get the locant set as: 244.

Obviously, there are no functional groups in this compound so the first point of difference rule should govern the numbering of locants. Hence, according to me, bottom-to-top locant assignment is correct here.

Unfortunately, ChemSketch uses top-to-bottom assignment :(

  What rule am I missing here?

I figure out the unsaturation is being given preference here, but I do not know any rule regarding it.

PS:
My IUPAC name: 2-bromo-2-methylpent-3-ene.
ChemSketch's name: 4-bromo-4-methylpent-2-ene

It is correct that the principal characteristic group (the characteristic group chosen for citation at the end of a name by means of a suffix or a class name, or implied by a trivial name) has a higher priority for numbering than any other substituents (expressed as prefixes). It is also correct that the compound that is given in the question does not have any principal characteristic group.

However, you are missing the priority for unsaturation (i.e. multiple bonds), which is between the priority of the principal characteristic group. The most important simplified criteria for the numbering of simple acyclic compounds are:

lower locants for suffixes
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

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

  P-14.4 NUMBERING

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

  (…)

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

  (…)

  (e) saturation/unsaturation:

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

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

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

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

  (…)

Since the compound that is given in the question does not have any principal characteristic group, the first criterion (greater number of suffixes) given in P-14.4 (c) is not relevant in this case.

According to (e), the correct name is (2E)-4-bromo-4-methylpent-2-ene and not (3E)-2-bromo-2-methylpent-3-ene since the locant ‘2’ for the double bond is lower than ‘3’.


Can terminal enols be reduced by using LiAlH4? According to the solution given in my textbook, no reaction takes place in this case.

But I am not quite sure about it. I feel that the enol can easily tautomerize to aldehyde and we know that aldehydes are reduced by $\ce{LiAlH4}$ to give alcohols.

So, will the above reaction take place? What will be the final product after reduction occurs?

Not only can that enol easily tautomerize to the aldehyde, the aldehyde is the major component at equilibrium at room temperature (by a lot!). The equilibrium constant for the similar system of ethanal/vinyl alcohol (one carbon shorter than yours) is $K \approx 3\times 10^{-7}$ (source: Wikipedia page on keto-enol tautomerism):

$$\ce{ CH3CHO <=> CH2=CHOH}\\
K = \dfrac{[\ce{CH2=CHOH}]}{[\ce{CH3CHO}]}\approx 3\times 10^{-7}$$

Thus, for every molecule of enol in this system, there are $\sim 3 \times 10^6$ molecules of aldehyde. $\ce{LiAlH4}$ will reduce the aldehyde and then LeChâtelier's principle will drag the enol along for the ride.


Iodine clock with moving bands Has anyone seen a version of the iodine clock reaction where instead of the system oscillating between clear and blue it shows blue bands travelling down the vessel?

I ask because someone not unknown to this site has made a version of the experiment with such moving bands and I was wondering if it was worth sending a write up of it to the Journal of Chemical Education or some similar journal.

Clarification:

Apologies if I was unclear, I am asking the phenomenon has been published before or if it is unpublished but already well known.

I know the BZ reaction can produce all sorts of patterns, but I have never seen travelling bands in the iodine clock. A quick Google has found no mention of the phenomenon. I am not asking "is it worth publishing", which I believe is against site rules anyway. I'm just asking about prior art.

Whilst not specifically an iodine clock reaction, Belousov-Zhabotinski reactions are oscillatory and can produce moving bands, spirals and concentric circles of colour. This works best in thin films (limiting convection) as the top commenter on the linked video points out.

There's a great thesis by Peter Hantz which looks at these kinds of reactions in detail (the first bit is in french, the rest is in english) and specifically covers Liesegang banding, where an oscillating chemical reaction in a mobile phase causes bands of precipitation in a gel column. As Hantz points out on page 22, the gel is required to prevent convection from destroying the bands, however the viscosity of the gel also freezes the precipitate in place as it forms. This would seem to be close to what you are interested in, however.

Here's a picture of Liesegang bands in test tubes from wikipedia:


What happens when a semipermeable membrane allows only one solute to pass through? I'm essentially a medical student where we deal a lot with osmosis. But when we are taught, it is done generally with only a single solute in consideration.

What if two different solutes are used on either side? The membrane being permeable only to one, not the other but also free for water to move.

Lets assume a beaker of base 10cm x 20cm, of any height. Add 2 litres of water to it, it would come to a height of 10cm. Dividing the beaker equally into two with the semipermeable membrane of the specifications, we'll be left with 1 litre on each side. Now add 1 osmole glucose on one side and 1 osmole NaCl on the other (you can't add only Na+ ions). Where glucose and water can move but NaCl wont.

I guess it'll be safe to assume that $\ x$ moles of glucose moves to the other side resulting in a change in concentration on both sides. Now the water would also move causing not only a change in concentration due to the change in volume distribution but also a change in hydrostatic pressure.
Let the change in height be $\ y $ cm. Now the only constraint I can think of is that the pressure on either side should be equal at the base. By pressure i mean the sum of both hydrostatic and oncotic pressure. Oncotic pressure is given by $\ P_o = icRT$ and hydrostatic is by $\ P_h = h\rho g$.

Now I write $$\ P_{o1} + P_{h1} = P_{o2} + P_{h2} $$

That'd give $$\ (h+y)\rho g + \frac{n+x}{(h+y)A}RT = (h-y)\rho g + \frac{n-x}{(h-y)A}RT$$

Where $\ h$ is the height 10cm , $\ n$ is the no. of moles which we fixed as 1 mole (or rather, osmole) So rest all are constants and the only variables left here is $\ y $ and  $\ x $ which we need.
So rest all are constants and the only variables left here is $\ y $ and  $\ x $ which we need.
 On solving, the equation comes to a cubic in $\ y$ which is actually not so desirable!! Also we need another equation, since it is an equation with two unknowns. So I showed all this since this is the way I would proceed. But I am now stuck.

P.S This was a post on the biology forum and then the physics forum too, a discussion of which I extended here.

I am afraid you are stuck because you considered glucose as osmotically active.
But you assumed that glucose can freely move across the membrane, which is the opposite of osmotic activity.
So, short-short answer : ignore the glucose, it is an osmolyte, but it is not osmotically active on the membrane you are thinking of; its concentration is going to equilibrate on each side of the membrane :
$$[\mathrm{C_6H_{12}O_6}] = \frac{1 \mathrm{mol}}{2 \mathrm{L}} = 0.5\, \mathrm{mol\, L^{-1}}$$
This solution of glucose is simply the solvent.
Also, you may have inverted terms in the equilibrium equation.

A longer answer :
First we assume that the volume and density of the solution are unchanged by the solutes added (ideal solutions).
We are left with $1 \,\mathrm{osmol\cdot L^{-1}}$ NaCl on one side of the membrane, and no osmolyte on the other side.

The equilibrium between hydrostatic and osmotic pressure ($i c R T $) then writes :
$$ 2 \rho g y = i \frac{n}{(h+y)A} RT  $$
Solving for $y$, we get
$$ 2y = \sqrt{\frac{2inRT}{\rho g A}+h^2}-h$$
Using the following values :  $i=2$, $n=0.5 \,\mathrm{mol}$, $R=8.3\,\mathrm{kg\cdot  m^2\cdot s^{-2}\cdot K^{-1}\cdot mol^{-1}} $, $T=3\cdot 10^2\,\mathrm{K}$, $\rho=10^3\,\mathrm{kg\cdot m^{-3}}$, $g=9.8\,\mathrm{m\cdot s^{-2}}$ ,  $A=10^{-2}\, \mathrm{m^{-2}} $ and $h = 10^{-1} \mathrm{m} $, we obtain :
$$ 2y \simeq 7\, \mathrm{m} $$
This value is actually unreachable given the volume of solvent and the geometry of your beaker. All the glucosed water would be drawn on one side, and equilibrium could not be reached.

The osmotic pressure of a solution containing $1\,\mathrm{osmol\cdot L^{-1}}=10^3 \,\mathrm{osmol\cdot m^{-3}} $ is approximately $\Pi = 2.5\, \mathrm{MPa} = 25 \, \mathrm{bar}$. To equilibrate such a pressure with hydrostatic pressure, you need a column of water of depth $z = \frac{\Pi}{\rho g} \simeq 260\,\mathrm{m}$.

The osmolarity of sea water is about $1\,\mathrm{osmol\cdot L^{-1}}$. In a desalination plant, sea water is purified by reverse osmosis using pressures from 50 to 80 bar. Osmotic equilibrium is of utmost importance in physiology, but this you must know better than me. Unless there is active transport, diffusion of solutes, or extraordinary high applied pressures (which may be relevant for deep-sea creatures), water is always going from the least to the most concentrated side of the membranes, so as to even out the osmolarities. Hope it helped.


What is the exact nomenclature of ionization? I'm a little confused as to what reactions are considered ionizations. Is there any requirement for the educts, for example?

Some references suggest that it's only considered an ionization if you produce additional charged atoms/molecules from the educts (and some processes are dubbed detachment, which intuitively look like ionizations to me). Then "double ionization" wouldn't be possible, strictly speaking. To what extend have electrons to be produced. At the moment I think that it's legal to call most processes that look like ionization ionization.

There is a remark in the wikipedia article regarding the distiction from dissociation, but I don't really understand it.

The reference for chemical terminology is the IUPAC Gold Book, or IUPAC Compendium of Chemical Terminology. The entry for ionization is as follows:

  The generation of one or more ions. It may occur, e.g. by loss of an electron from a neutral molecular entity, by the unimolecular heterolysis of such an entity into two or more ions, or by a heterolytic substitution reaction involving neutral molecules […] The loss of an electron from a singly, doubly, etc. charged cation is called second, third, etc. ionization. This terminology is used especially in mass spectroscopy.

It includes examples that you might find instructive. It confirms your intuition that the term has a broad understanding, covering a wide variety of reactions.


Side chain oxidation with KMnO4/H+ I know that in side chain oxidation with the given reagent follows oxidation of benzylic carbon (if atleast one benzylic hydrogen is present). So my answer would be

But in my answer key only it is correct for nitro substituted anthracene. But why? What is the reason?
Thanks for clearing my doubt!

In each case the permanganate oxidises the most electron rich ring.

In the nitro-substituted case, that is the unsubstituted ring, probably first at the position alpha to the nitro group, forming a phenol, then para to that to a quinone but permanganate keeps on chewing until the phthalic acid is formed.

In the aniline case the most electron rich ring is the one with the NH2 attached. Oxidation of anilines by acid permanganate give a polymer called "aniline black" Wiki link. Though I can't find a reaction specific to 1-Naphthylamine other than one that documents the destruction of it by sulfuric/permanaganate without identifying the products here. The site of first oxidation is p to the aniline. This creates a reactive intermediate which is attacked by a second aniline as the first step of polymer formation.


Lattice vectors calculation How can I find the crystal lattice vectors, given the crystal parameters?
For example, consider a crystal of rhombohedral symmetry, where a=b and c are known, and alpha=beta=90, gamma=120. How can the lattice vectors be found?

You can find this using elementary geometry. Also, since the unit cell may be positioned in different ways with respect to the coordinate system, there can be many different answers, all equally valid.

One possible answer implies alignment of the cell vectors $\vec a$ and $\vec c$ with the coordinate axes X and Z, which leads to the following:
$$\left\{\begin{align}
\vec{\bf a} =& \Big(|a|,0,0\Big)\\
\vec{\bf b} =& \left(-{1\over2}|a|,{\sqrt3\over2}|a|,0\right)\\
\vec{\bf c} =& \Big(0,0,|c|\Big)
\end{align}\right.
$$


Bond angles of OCl2 and OBr2 I have looked around and most of the sites suggest that bond angle of $\ce{OBr2}$ is greater than $\ce{OCl2}$ and it is explained by stating the electronegativity difference and looking at electron repulsion (VSEPR). However none of the websites actually give any experimental data so i am unsatisfied. I think that extent of back bonding would be more for $\ce{OCl2}$ and hence its bond angle should be greater.

Can someone please explain why i may be wrong and give experimental data on bond angle along with it?

I think steric effects may play a role in that chlorine atoms are smaller than bromine atoms. The bromine atomic radius is about 0.15 Angstroms larger than the chlorine atomic radius. The larger bond angle can be in part explained by the need to accommodate the larger atoms.


Solid state, closed packed structures, BCC lattice How many next nearest neighborhood respectively each potassium has in bcc lattice?

Second nearest neighbors are the neighbors of the first neighbors. So for BCC let's consider the atom at the body centre, for this atom the atom at the corner are nearest and for the atoms at the corners the atom at body centres of other cubes are nearest. Little imagination(there are 6 body centred atoms surrounding the atom we are considering) and counting gives the answer as six.


When 2-pentene reacts with HBr, what will be the major product? If we consider the first step (formation of carbocation by the attack of the proton).

A couple of textbooks I referred say that when the substituents on the $\ce{C=C}$ are different, the direction of shift of the electron is decided by the inductive effect of the substituents. For example, in this image:

$$
\begin{align}
\ce{\underset{1}{C}H3-\underset{2}{C}H=\underset{3}{C}H-\underset{4}{C}H2-\underset{5}{C}H3 &-> \underset{1}{C}H3-\underset{2}{\overset{\large\ominus}{C}}H-\underset{3}{\overset{\large\oplus}{C}}H-\underset{4}{C}H2-\underset{5}{C}H3} \label{rxn:a}\tag{a} \\
\ce{\underset{1}{C}H3-\underset{2}{C}H=\underset{3}{C}H-\underset{4}{C}H2-\underset{5}{C}H3 &-> \underset{1}{C}H3-\underset{2}{\overset{\large\oplus}{C}}H-\underset{3}{\overset{\large\ominus}{C}}H-\underset{4}{C}H2-\underset{5}{C}H3} \label{rxn:b}\tag{b}
\end{align}
$$

Assuming that a proton $\ce{H+}$ is approaching, both the textbooks say that the electronic shift happens as in the case \eqref{rxn:a} because \eqref{rxn:a} has 2 ethyl groups to stabilize the carbocation by inductive effect. But isn't hyperconjugation more important than inductive effect in determining the stability?

By hyperconjugation, case \eqref{rxn:b} should be the preferred pathway as there are 5 alpha-hydrogens (including the $\ce{H}$ that is attacking and will join at the $\ce{C-}$) and case \eqref{rxn:a} has only 4? By this logic 2-bromopentane should dominate. Am I missing anything?

This sort of reaction isn't investigated seriously anymore. But 80 years ago, Kharasch et al. (J. Am. Chem. Soc. 1939, 61 (6), 1559–1564) wrote that

  Both results agreed within the limit of error (3%) and indicated an equimolar mixture of 2- and 3-bromopentanes [...]

So, for all intents and purposes, there is no major product. It is not always possible to get a definitive answer based on simplified "rules" and concepts. All we can genuinely say in this case is that both intermediates are very similar in stability, and so both products will be formed in significant amounts. In fact, the carbocations may even interconvert rapidly by means of hydride shifts, because when pentan-2-ol or pentan-3-ol is treated with HBr, you get a mixture of the bromides (J. Am. Chem. Soc. 1930, 52 (6), 2440–2451).

This lack of selectivity is precisely why nobody investigates these seriously anymore. Maybe with modern analytical methods we could accurately measure it and find that the product ratio is 52:48, but what's the point? That's just as useless as a 50:50 mixture.


Aetiology of the word "fullerene" in Chemistry I am a mathematician and not a chemist, and I am trying to understand the historical relationship and current usage of the word "fullerene" across Mathematics and Chemistry.

I apologise in advance if I say chemically-ridiculous things, but I am genuinely interested in understanding this.

I will assert some things that loosely describe my understanding of the situation, but which may be chemically wrong. Constructive correction is welcomed.

MY UNDERSTANDING FROM MATHEMATICS

(1) In graph theory (my area of Mathematics) a "fullerene graph" or simply "fullerene" is usually defined to be a 3-regular planar graph, where by "graph" we mean a combinatorial graph with vertices and edges (i.e. abstractions of atoms and bonds).

(2) The standard example of a fullerene is a 60-vertex planar graph, whose faces are arranged as in a standard soccer ball, and of course this is the famous geodesic dome structure or buckyball.

[Actually, mathematicians will also use the dodecahedron as a standard example of a fullerene graph because it has just 12 pentagonal faces and no hexagonal faces. This is fine for a mathematician because 0 hexagons is a perfectly sensible number of hexagons but perhaps it is chemically nonsensical.]

(3) This structure and its name arises from the carbon molecule $C_{60}$ which was first synthesised by Kroto et al., which has this spherical shape, and which they named buckminsterfullerene after Buckminster Fuller.

(4) Mathematics writings on fullerene graphs that attempt to explain the relationship to chemical fullerenes then usually say something like:

an actual physical fullerene must have a molecular structure (atoms and bonds) that is a 3-valent planar graph with pentagonal and hexagonal faces
there are lots of mathematically possible fullerenes, but only a minuscule fraction actually physically exist.
to determine whether or not a mathematical structure might be physically realised, we compute the spectrum of the adjacency matrix and magically believe it relates to a mysterious chemistry thing called the HOMO-LUMO gap.

[In addition, some authors insist that to qualify as a fullerene graph, the pentagons must be separated from each other.]

[I guess that as carbon atoms are 4-valent, there must be a pile of double bonds floating around that us mathematicians have just totally ignored!]

MY QUESTIONS REGARDING THE CHEMISTRY

(1) When did the very specific term "buckminsterfullerene" get relaxed to "fullerene" and by whom?

(2) To exactly what class of carbon molecules does the term "fullerene" apply?

[I have learned that other forms of carbon include diamond where all atoms are 4-valent, then graphite, then fullerenes but I am hazy on the precise mathematical or chemical distinction.  Then I am further confused by the fact that some sources distinguish tube structures (nano-tubes) from spherical structures and only call the latter fullerenes]

Sorry for the ridiculously long first question, but all clarification is gratefully received!

Fullerenes are distinct (meaning every fullerene type has a well defined structure) carbon-only molecules, where all atoms have three bonds ($sp^2$, or nearly $sp^2$, as they're not exactly planar).

Leaving out very freakish topologies (which are unlikely to be stable and not fullerenes), this leads to a closed shell-like structure where all surface elements are pentagons or hexagons. (Which are not perfect, bond lengths differ slightly if the edge is shared by two pentagons, two hexagons, or one-and-one.)

The simpler, smaller forms are spherical (C60) or prolate (C70, ...), larger ones can probably have a surface that is not completely convex. (anyone seen oblate fullerene structures?)

A carbon nanotube with a well defined, closed end structure can be a fullerene, but they usually haven't got one.


Are there any colorful, odorless gases? Gases are frequently described as "colorless, odorless, and tasteless."  I am wondering if this is redundant, since I can't think of many gases that have a detectable color to begin with, much less ones that have a color but no odor.

Are there any gases that are colorful and odorless at standard temperature and pressure?  In particular, gases that impart enough of a color to white light that one could look into a room filled with the gas and detect its presence visually?

(In the limit I suppose one could argue that a standard atmosphere is "colorful" since over many miles one can see blue scattering, but the question here is whether it is ever useful or helpful to describe a gas as "colorless.")

There are plenty that are smelly and colourless, or are odourless and colourless but I would agree that there are none that are coloured and odourless. Surely it is just coincidental if any of these pairs occur because the chemistry involved in vision and smell are totally different. (Whatever the chemistry of smell is it does not involve photons!) If you were less restrictive in your conditions then most gases will be 'coloured' in the sense that they have electronic absorption in the near ultra-violet and also infra-red at longer wavelengths. Perhaps some insects would see gases such as N$_2$O or O$_3$ as coloured. Also if you were to allow vapours then many compounds can be considered if heated, for example azulene (an isomer of naphthalene) is blue and has a small but measurable vapour pressure if warmed slightly. I don't think I want to smell any, however. The ideal of filling ones nose and lungs with aromatic or dye vapours that then condense is not an attractive one!


kWh of radioactive elements database? I recently been studying different radioactive elements. The most difficult part is find the output of these elements in the form of kWh.

Is there a database or chart of these elements' output?

And most importantly does anyone know the kWh of $1~ kg$ of $\ce{^99Tc}$?

$\ce{^{99m}Tc}$ is a nuclear isomer of $\ce{^{99}Tc}$, which is excited by 142.6 keV. Therefore, when one atom decays to $\ce{^{99}Tc}$, it releases 142.6 keV, mostly as gamma radiation, but also some internal conversion occurs. Therefore, for 1 kg of $\ce{^{99m}Tc}$ (10.1 mol):

$$
(142.6\ \mathrm{keV})(10.1\ \mathrm{mol})N_A = 8.673\times 10^{29}\ \mathrm{eV} = 139\ \mathrm{GJ} = 38.6\ \mathrm{MWh}
$$

In general, the total energy released by a nuclear reaction in a given mass is simply the decay energy (commonly known as the Q value) multiplied by the number of atoms. The Q value is simply the difference in mass between the starting material and final product converted to energy:

$$Q = (m_\mathrm{i}-m_\mathrm{f})c^2$$

Q values are easy to just look up, though. (Wolfram|Alpha is a good place)

Of course, depending on the application, all this energy is not accessible or useful. If one were trying to make an RTG, an enormous amount of shielding would be needed to turn the gamma rays into heat, since they penetrate matter very well, where things like Pu-238 work better because they release mostly alpha particles, which are easily absorbed, delivering their kinetic energy as heat.


How do you dry a Graham condenser? I have a Graham condenser and tested it out by distilling water from salt water then I washed it and now I can't get either the coils or the jacket completely dry.

Any advice?  If there isn't a better way I was thinking of popping it in the oven for a while to see if that'd dry it out the rest of the way.

There are two ways I can think of to get rid of the remaining water:

As per LDC3's answer, flush the system with ethanol which will hopefully not have the problem of slow dry time.
Put the glassware in an oven, which will also nicely sterilize the system.


What primary standard acid can be used to standardise NaOH solutions by titration? I've done an experiment to standardise $\ce{NaOH}$ using oxalic acid.

What other primary standard acid can be used to standardise $\ce{NaOH}$?

Can $\ce{HCl, H2SO4, HNO3}$ be used?

I read up and found that:

$\ce{HCl}$ and $\ce{H2SO4}$ are easily diluted, however, the concentration of the "concentrated" solution is not accurately known.

$\ce{HNO3}$ contains a little nitrous acid, $\ce{HNO2}$, which has a destructive action on many acid-base indicators.

Therefore I only found potassium hydrogen phthalate. Is there other primary standard acid that can standardise $\ce{NaOH}$?

Oxalic acid dihydrate and KHP are the most comfortable acids for standardization, as their content per a unit of measurement (in this case it's obviously grams, as they are solid) hardly changes. When you have liquid acids, besides that what you've already mentioned, they can absorb water and other compounds from air. As they are liquid, people usually measure amount of those compounds by volume, and volume is a variable which depends on temperature, pressure.


During formation and breaking of bonds, can the terms potential energy and bond energy be used interchangeably? My chemistry professor uses these terms interchangeably and I was wondering if they refer to the same thing. From what I understand, bond energy is the energy required to break one mole of covalently bonded gaseous molecules. While potential energy refers to the energy gained or lost during bond formation which can be further characterised into endothermic or exothermic.

The answer is a qualified yes. Specifically, the "bond dissociation energy" (often shortened to "bond energy") is a positive number that is the amount of energy one must add to the system to completely break a mole of equivalent bonds. A completely broken bond by definition has 0 potential energy, so bond formation results in a negative potential energy. Thus, the potential energy of the bond is equal in magnitude but opposite in sign to the bond dissociation energy. If we are only talking about the magnitude (which is often the case), they're interchangeable.

In addition to the chemical bonds, a system can have other sources of potential energy such as gravitational potential energy or electrostatic potential energy, so in that sense potential energy is not synonymous with bond energy, but generally it's pretty clear from context when we mean only the potential energy of bonding.


Proper method to electroplate aluminum? I'm looking to plate some aluminum with silver using a basic solution.
My first attempt was to use sodium carbonate to make silver acetate. I sped up the process using hydrogen peroxide. I used the silver acetate as the ion source.
This simply did not work to plate any aluminum.
I concluded that this is because the aluminum was oxidized, preventing a plating from sticking.
I've come to the conclusion that the main problem with plating into aluminum is that it's able to rapidly form a highly resilient oxide layer.
Are there any common ways to prevent this oxide layer from forming? More to the point, are there any common methods to plate aluminum, specifically with silver?

No ! There are no method for plating aluminum. Thousands of people have tried. In vain. It is impossible to prevent the formation of a thin layer of aluminum oxide which is waterproof at the surface of the metal. The only way of preventing it is to dip a piece of aluminum in a solution of mercuric chloride. This produces a reaction $$\ce{2 Al + 3 HgCl2 ->  3 Hg + 2 AlCl_3}$$ and the mercury makes an alloy (or amalgam) with not yet oxidized aluminum, and this prevents the formation of the continuous oxide layer. But this alloy does not prevent the oxidation of Al in contact with air. The oxidation of the amalgamated aluminum proceeds with such a speed that the growth of the aluminum oxide filaments can be seen by the naked eye. In a couple of minutes, white filaments of oxide grow to become one or two centimeters long. This extraordinary phenomena stops when the mercury has been removed from the surface by the growing oxide filaments. Of course, today, it is forbidden to carry out such experiments, because of the mercury toxicity. Fifty years ago, it was allowed. I do remember.


Liquids that keeps the lowest temperature for the longer time after being frozen I am working on a personal project, however I know nothing about chemistry and I would like to ask you guys for a quick question. I need to use a liquid for my project, which after being put into a freezer will freeze, and will be kept frozen for the most time when taken out of the freezer into room temperatures.

Ideally I would like to know if there is a liquid which can freeze after a couple of hours, and when taken out from the refrigerator it can sustain a temperature of -2 -3 degrees for the longer period, in normal room temperatures. And most importantly it can me used again and again.

Thank you so much guys, I will be looking forward to an answer on this one.

To maintain a stable temperature while heat is entering a system implies some phase change: that is, the substance is either melting or boiling, keeping the temperature the same, while some part is changed form solid to liquid or form liquid to gas.

Pure water freezes at 0 °C, but if another substance is added, the freezing point (f.p.) may be lowered. Ordinary table salt, $\ce{NaCl}$, can lower the f.p. to -20 °C. Calcium chloride, $\ce{CaCl2}$, is even more effective. Both are used for melting ice from roads, with $\ce{CaCl2}$ preferred in cold weather.

You might test various concentration to see what best meets your needs. Ideally, the solution f.p. should be a bit above the temperature of your freezer.


Do commuting operators always share the same eigenfunctions? Do commuting operators always share the same eigenfunctions? Could you kindly explain to me if this is true or false and examples for this, as I am a little confused.

If two operators commute, then there exists a basis for the space that is simultaneously an eigenbasis for both operators.

However, if one of the operators has two eigenvectors with the same eigenvalue, any linear combination of those two eigenvectors is also an eigenvector of that operator, but that linear combination might not be an eigenvector of the second operator.

As an example consider the states $|l,m\rangle = |1,1\rangle$ and $|l,m\rangle = |1,-1\rangle$. These are both eigenvectors of $\hat{L}^2$ and $\hat{L}_z$ (the angular momentum operators).

The eigenvalues of $\hat{L}_z$ are $\hbar$ and $-\hbar$, respectively, but the state
$$|1,1\rangle + |1,-1\rangle$$
is clearly not an eigenvector of $\hat{L}_z$, but it is still an eigenvector of $\hat{L}^2$. This is probably why we use the term "mutual" rather than "same". Finally, we form linear combinations of the $|1,m\rangle$ states to get the $l=1$ states that are eigenvectors of, say, $\hat{L}_y$.

As a straight-forward example, consider the following two matrices:
$$L = \left[ {\begin{array}{ccc}
0 & 0 & 0 \\    0 & 1 & 0 \\    0 & 0 & 1
\end{array}}
\right]$$
and
$$Z = \left[ {\begin{array}{ccc}
0 & 0 & 0 \\    0 & 1 & 0 \\    0 & 0 & -1
\end{array}}
\right].$$
The vectors $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ are eigenvectors of both operators. The latter two are eigenvectors of $L$ with the same eigenvalue (namely $1$), but they are eigenvectors of $Z$ with different eigenvalues (namely, $1$ and $-1$). If we instead use the vectors $(0,1,1)$ and $(0,1,-1)$, these are still eigenvectors of $L$ with eigenvalue $1$, but they are no longer eigenvectors of $Z$, as you can verify.


How to determine Delta H from the heat of combustion? The heat of combustion of 2-propanol at 298.15 K, determined in a bomb calorimeter, is -33.41 kJ/g. For the combustion of one mole of 2-propanol, determine (a) $\Delta U$ and (b) $\Delta H$

So I managed to determine $\Delta U$ by first finding the amount grams in one of 2-propanol and then using that number with 33.41 kJ/g...

I'm not able to determine Delta H as I'm not sure what to do... Please help and guide me.

Thanks!

Are you supposed to determine $\Delta_rH$ for the reaction (which is $\Delta_cH$ for the combustion of 2-propanol) or are you supposed to determine $\Delta_fH$ for the formation of 2-propanol?

  If the first case, it might be hard.

The change in internal energy $U$ is equal to the sum of the heat transferred $q$ and the work done $w$. If we assume that only pressure volume work is done, then:

$$\Delta U=q+w = q-p\Delta V $$

The negative sign in front of $p\Delta V$ is to ensure that a decrease in volume is work being done on the system (system gains energy).

As a thermodynamic potential, internal energy is only easy to use if the pressure is constant (and $\Delta V$ is measurable), which is not true in a bomb calorimeter, or the volume is constant (so $\Delta V =0$), which is true in a bomb calorimeter.

At constant pressure: $\Delta U=q$.

Enthalpy is defined such that
$$\Delta H = \Delta U + \Delta(pV) = q - p\Delta V + p\Delta V + V\Delta p $$
$$\Delta H = q + V\Delta p$$

Thus, enthalpy change is only easy to use if the volume is constant (true in a bomb calorimeter) and the pressure change is knowable (you were not given that data) or if the pressure does not change (so that $\Delta p=0$).

  The second case is simpler.

If you were asked to determine $\Delta_f H$ for the formation of 2-propanol, you probably are encouraged to assume that for the combustion reaction $\Delta_rH\approx \Delta_cU$ and go from there. Then, you will need a balanced equation (2-propanol + oxygen forms what?) and some standard enthalpies of formation for the products.

$$\Delta_rH \approx \sum_{products}{\Delta_fH}-\sum_{reactants}{\Delta_fH}$$


Action of hot alkaline KMnO4 on tert butyl benzene Side chain oxidation of aryl alkanes takes place if there is at least one alpha hydrogen present by all accounts I presumed that there would be no reaction. However the answer given was trimethyl acetic acid. Is the answer key wrong?

A similar reaction with chromium oxide in alkaline solution has been documented.

The acidic chromium oxide serves the same role as alkaline KMnO4 i.e a strong oxidizing agent. As pointed out by @Waylander in the comments permanganate oxidizes double bonds even in aromatic compounds.
Reference:
$\large{•}$ Chromium Oxidations in Organic Chemistry; By G. Cainelli, G. Cardillo


Find the stoichiometric coefficient of a reactant given Kp, Kc, and the stoichiometric coefficient of the product Have an hypothetical equilibrium reaction:

  $$\ce{2 A (s) + x B (g) <=> 3 C (g)}$$

  With the constants

  \begin{align}
K_\mathrm{p} &= 0.0105 \\
K_\mathrm{c} &= 0.45
\end{align}

  both at $\pu{250 ^\circ C}$. What is the value of $x$?

Since $\ce{A}$ is solid, I guess we can ignore it, right?

$$\ce{x B(g) <=> 3 C (g)}$$

Using $K_\mathrm{c}$:

$$0.0105 = \frac{\left[\ce{C}\right]^3}{\left[\ce{B}\right]^x}$$

Then

$$\left[\ce{B}\right]^x = \frac{\left[\ce{C}\right]^3}{0.0105}$$

But to solve this, I need the concentration of $\ce{C}$. How can I find it out?

Using the relation of $K_\mathrm{p}$ to $K_\mathrm{c}$:

$$\frac{K_\mathrm{p}}{K_\mathrm{c}} = (RT)^{\Delta n}$$

where $\Delta n$ is the change in the number of moles of gas in the reaction (here, $\Delta n = 3 - x$). Plugging in (using $R = \pu{0.0821 L atm mol-1 K-1}$ and $T = \pu{523.15 K}$) we get:

$$\frac{0.0105}{0.45} = 0.0233 = 42.950615^{(3-x)}$$

To solve we take the natural logarithm of both sides:

$$\ln{(0.0233)} = (3-x) \ln{(42.950615)}$$
$$\frac{\ln{(0.0233)}}{\ln{(42.950615)}} = -1 = (3 - x) \implies x = 4$$

And so $x = 4$. We can check to see if our answer is correct by plugging the calculated value for $x$ into the second expression from the top of this post:

$$0.0233 \buildrel{?}\over{=} 42.950615^{-1}$$

which does turn out to be correct.


Is it possible to build a non-degrading battery? I am not a chemist, nor do I know anything about chemistry. What I am interested in are batteries. I don't really care about the whole "Charge in minutes, last for days" types of batteries (the famously hyped graphene battery) But is it possible to have a battery that will never degrade and always keep its performance without upkeep/maintenance?

I just didn't know if there were some types of laws in chemistry that prevents this from happening. I would ask in electrical engineering but this sounds to me more of a chemistry question.

But is it possible to have a battery that will never degrade and always keep its performance without upkeep/maintenance?

Well, consider the dry pile powering the Oxford Electric Bell, which has been running since 1840. However, that is an example of a nonrechargeable pile.

The question is, can a rechargeable battery be designed that retains it's charging/discharging current profile, reflecting a conserved structural integrity, during an unlimited number of cycles?

While in practice there are many reasons this is difficult to do (loss of electrolyte, occurrence of irreversible passivating reactions, etc) in principle there seems to be no reason why this should not be the case, at least over time on the order of a human lifetime. We manufacture many things which last an awful long time, and batteries would last even longer if designed or treated in an exceptional way, shielded from the environment and charged/discharged in a highly regulated manner. Some battery designs such as thin film lithium-ion allow an exceptional amount of cycling.

  I just didn't know if there were some types of laws in chemistry that prevents this from happening.

The second law of thermodynamics places limits on what you can expect to achieve given the constraints of a particular choice of battery chemistry. This reveals itself for instance in Peukert's law, particularly applicable to lead-acid batteries operating at constant temperature, which says that the capacity of a battery is restricted by its discharge rate. Discharging quickly reduces the total work you can do with the battery.

The second law however does not say we cannot use work to recharge a battery to the same original state a very large number of times. But it does make it very hard! It begs the question, "is such a  battery useful?" Probably not, except theoretically. The design and operation of a battery involves tradeoffs between charge density, discharge and recharging rate, and preservation of structural integrity during charging/discharging cycles.

By the way, you could generalize the question to include other portable high-capacity sources of electrical power. Think of probes that are sent into outer space and continue to operate after, well, a long time. Or consider the quantum battery.


Which affects acidity more, in relation to induction? (Distance vs. quantity) I already know that the answer is III. I'm just looking for further explanation on how the others compare to each other, mainly IV vs V.
I know that iodine is less electronegative than chlorine, making I the least acidic.
Only having one chlorine atom, while the rest have two, puts II next in line.
III is the most acidic, due to the two chlorine atoms being on the same carbon, and closest to the sulfur group.

My question comes in for the comparison of IV and V. While V has two chlorine atoms on the same carbon, they are both 3 carbons away from the sulfur group. Meanwhile, IV has one chlorine atom closer to the sulfur group, but the other isn't on the same carbon atom.
Which one of these is more effected by induction? Proximity to the sulfur group (IV) or being grouped together on a single carbon (V)?

Inductive effects decrease strongly with increasing distance. Thus, the further away an inductively acting group is, the lower its influence will be. For practical intents and purposes, the otherwise very comprehensive Brodwell $\mathrm{p}K_\mathrm{a}$ database does not even list any β-halo compounds; only the difference between α-hydrogen, carbon and halogen is typically given.

Thus, we can expect compound IV to have practically the same acidity as compound II, and that compound V’s acidity is comparable to the unhalogenated pentan-4-thiol. Considering that all chlorines are very far away from the acidic proton, the effects would be smaller still.


Products of HMTD decomposition Hexamethylene triperoxide diamine, "Hex" for short, is an explosive organic compound.

I want to know the products of this reaction:

$$\ce{C_6H_{12}N_2O_6 -> ? + Energy}$$

Answer

This article provide two reactions for the thermal decomposition of HMTD, depending on reaction conditions.

  (at or below 150°C, air or vacuum)

$$\ce{C_6H_{12}N_2O_6 -> 3CO_2 + N(CH_3)_3 + NH_3}$$

  (160°C to 180°C, vacuum)

$$\ce{C_6H_{12}N_2O_6 -> 3CO + N(CH_3)_3 + 1/2 N_2 + 3/2 H_2O + 3/4 O_2}$$

References

Oxley, J.C., J.L. Smith, H. Chen, and Eugene Cioffi. “Decomposition of Multi-peroxidic Compounds: Part II. Hexamethylene Triperoxide Diamine (HMTD).” Thermochimica Acta 388, no. 1–2 (June 18, 2002): 215–225.


Is sugar in mead different from "regular sugar"? My understanding of the process of making mead is that you combine honey, yeast and water in an anaerobic environment. The yeast-bacteria ferments the sugar in the honey and produces ethanol (among other things?). If this is true, how come labels for mead still list e.g. 90g of sugars per litre? If I'm consuming mead, am I really consuming 90g of regular sugars as found in fruit, candy etc or is this some variant of sugar that is chemically different (sugar alcohol?) in a significant way such that traditional advise "eat less sugar" do not apply?

Honey usually contains mainly naturally occuring glucose + fructose, with ratio typically 100:104-170. Sacharose occurs as well, but in minority.

Fermentation consumes just part of sugars, but that is not limited to mead. It applies to beer,wine, and bases for spirits as well.

E.g. typical "future beer" before fermentation has sugar content  $\pu{100-120 g/L}$.


Oxidation of water — what is the half reaction for water? During the electrolysis of a solution of copper sulfate, copper is reduced to form a solid on the inert electrode while water is oxidised at the anode. What is the half equation for water?

\begin{align}
\text{Is it}&&
\ce{4 OH−(aq) &-> O2(g) + 2 H2O(l) + 4 e−}\\
\text{or}&&
\ce{2H2O(l) &-> 4H+(aq) + O2(g) + 4e–}?
\end{align}

What is the difference? And what is the standard reduction potential in an electrolytic cell?

Both your reactions are equivalent. Here’s why. We start from $(1)$.

$$\ce{4 OH- -> O2 + 2 H2O + 4 e-}\tag{1}$$

(I have omitted the phase descriptors for clarity.) In any chemical reaction, we can add or subtract spectators as we see fit; this is akin to mathematical equations where we can add or substract expressions as we see fit as long as we do the same to both sides. For example, I want to add four protons to $(1)$ to give $(2)$.

$$\ce{4 OH- + 4 H+ -> O2 + 2 H2O + 4 e- + 4 H+}\tag{2}$$

If this were now a mathematical equation, I could combine (or divide) parts as I see fit. In a chemical reaction, we need to be a bit more careful. But as long as our focus is not on proton transfers, we can combine anything basic enough with acidic protons. Here, you can combine the four hydroxides and four protons on the reactant side of $(2)$ to give $(3)$.

$$\ce{4 H2O -> O2 + 2 H2O + 4 e- + 4 H+}\tag{3}$$

Now, we notice that we again have spectators. Two water molecules turn up both on the reactant and on the product side (an additional two are unique to the reactant side). We can remove them like we did the protons before, giving us $(4)$. I will also perform some rearranging.

$$\ce{2 H2O -> 4 H+ + O2 + 4 e-}\tag{4}$$

It so turns out that $(4)$ is your second suggested reaction. It is the same equation as $(1)$ as I have just shown; the difference is merely in the type of water molecules present.

Now which one should you use? Notice that one reaction consumes hydroxide while the other produces protons (formally! The consumption of hydroxide is equivalent to the production of protons). Thus, if your reaction medium is basic, you should use $(1)$ for your redox reaction. If, however, your reaction mixture is acidic, $(4)$ is more appropriate. In short, choose the one with the same charged particle as is used in the other half-reaction to make your life simpler.


What determines the high degree of selectivity in free radical brominations? In my organic chemistry class, we were asked what determines the relatively high degree of selectivity in the reaction of a bromine radical with primary, secondary, and tertiary C–H bonds.

I said it is the overall free energy of the reaction because according to Hammond's postulate, endothermic (endergonic) reactions have transition states closely resembling the product side, making the difference between primary, secondary, and tertiary radicals more important in the transition state.

Would this logic be wrong?

First, let’s take a look at the radical chain reaction that leads to the substitution in question:

$$\begin{align}\ce{Br^. + H-R &-> HBr + R^.}\tag{1}\\[0.4em]
\ce{R^. + Br-Br &-> R-Br + Br^.}\tag{2}\end{align}$$

Obviously, the selectivity is determined in the first step as long as no radical recombinations take place. If we break down the process in the first step into sub-processes, we get:

cleavage of the $\ce{C-H}$ bond;
formation of the $\ce{H-Br}$ bond.

Again, the second is the same no matter which radical is formed; therefore, it must be the cleavage of the $\ce{C-H}$ bond that leads to selectivity. Indeed, you can imagine radicals as being electron-deficient compounds and therefore behave similar to cations. Being similar to cations means that you can assume hyperconjugation to be a viable stabilisation mechanism for radicals; therefore, the energy lost by homolytically cleaving a tertiary $\ce{C-H}$ bond is slightly less (a less unstable product) than that lost cleaving a primary $\ce{C-H}$ bond. Exploiting this energy difference is the basis of selectivity.

This selectivity is higher if bromine is used than if chlorine is used. The reason is precisely the Hammond postulate you cited: the reaction with bromine is predicted to be more endothermic and thus the transition state more product-sided. A layman’s explanation would be that bromine can approach slowly and try out whether it feels happy about reacting with this hydrogen — think of playing Jenga and carefully testing if a brick is moveable or not. Chlorine on the other hand, is faster to react and thus has a less product-sided transition state.

There is a second factor that reduces selectivity. Assume 2-methylbutane; this compound conveniently has primary, secondary and tertiary hydrogens. Probabilities of bromine radicals to react with each of these hydrogens have been calculated; since I don’t have actual numbers on me right now, I am going to ballpark the following: assume, that a secondary hydrogen has a five times as high probability of reacting with bromine when compared to a primary one and that the tertiary hydrogen has a 15 times as high probability. Naïvely, we could assume that the reaction yields products with a ratio of $1:5:15$ — but we need to remember how many hydrogens of a given type there is.

The possible products are 1-bromo-3-methylbutane, 1-bromo-2-methylbutane (both primary attacks), 2-bromo-3-methylbutane (secondary attack) and 2-bromo-2-methylbutane (tertiary attack). For each of these, the following number of hydrogens can react to form the product in question:

1-bromo-3-methylbutane: 3
1-bromo-2-methylbutane: 6
2-bromo-3-methylbutane: 2
2-bromo-2-methylbutane: 1

Thus, to accurately calculate the ratios, we need to multiply the preference factor with the number of hydrogens, resulting in a product ratio of $3:6:10:15$ in the same order as presented — a lower fraction of tertiary bromide than presumed.


Which pH indicator will give a dramatic color change with small changes in the amount of carbon dioxide dissolved in water? I want to do a demonstration for students where they take a capped syringe and change the volume above a solution of seltzer water.  The amount of carbon dioxide dissolved will vary, hopefully changing pH enough to cause an indicator to change color, the more dramatic the color change, the better.

I have tried universal, bromcresol green, and methyl red with little success.  I can add substances to the seltzer water or change the temperature of the water.  I'd like a SAFE system I can create that will change color upon moving the plunger.

Try neutral red. It's transition region is from 6.8 to 8.0 > red to yellow

Also check this wikipage for other options.


The intrinsic electron-withdrawing nature of alkyl groups What you are about to read may be a very mind-boggling paragraph but please do not regard it as nonsense. Please think through it thoroughly.

In Chapter 6 of Organic Chemistry (4th ed.) by Maitland Jones Jr. and Steven A. Fleming, the following is written (p. 237):

  In 1970, Professor John Brauman (b. 1937) and his co-workers at Stanford university showed that in the gas phase the opposite acidity order obtained. The intrinsic acidity of the four alcohols of Table 6.6 is exactly opposite to that found in solution. The acidity order measured in solution reflects a powerful effect of the solvent, not the natural acidities of the alcohols themselves. Organic ions are almost all unstable species, and the formation of the alkoxide anions depends critically on how easy it is to stabilize them through interaction with solvent molecules, a process called solvation. tert-Butyl alcohol is a weaker acid in solution than methyl alcohol because the large tert-butyl alkoxide ion is difficult to solvate. The more alkyl groups, the more difficult it is for the stabilizing solvent molecules to approach (Fig. 6.22). Of course, in the gas phase where solvation is impossible, the natural acidity order is observed.

Let me provide some context. The author is discussing the acidity of alcohols in the gas phase and in the aqueous phase. In Table 6.6 as shown above, the pKa values are provided to show the decreasing acidity of the alcohols as the alkyl portion increases in size. The author explains this by saying that it is not really due to the electron-donating effect of the alkyl groups, concentrating the negative charge on the oxygen atom. Rather, he argues that it is due to the decreased extent of solvation due to the steric hindrance provided by the bulkiness of the alkyl groups. This is a reasonable rationalisation.

However, the authors then go on to discuss the acidity of the alcohols in the gas phase and mention that the acidity trend is reversed in this situation. The alkyl groups now increase the acidity of the alcohol, rather than decrease it. The authors then state that it is due to the electron-withdrawing nature of the alkyl substituents. They even provide a qualitative molecular orbital picture of this "electron-withdrawing" interaction, as shown in the following paragraph (p. 238):

  Alkyl groups have both filled and empty molecular orbitals (see the problems at the end of Chapter 1 for several examples). A pair of electrons adjacent to an alkyl group can be stabilized through overlap with the LUMO of the alkyl group. (Similarly, an alkyl group stabilizes an adjacent empty orbital through overlap with the alkyl HOMO.)

I would like to ask if anyone has ever heard of such an unconventional rationalisation and whether there is a sound theoretical basis for the postulations made above by the authors. If the postulations are indeed inconsistent with current chemical understanding, please provide a clear response detailing the areas of conflict.

I wonder if the pKa for water (15.7) is a typo. Usually we think it to be 14.0 (at 25C, anyway). And I wonder how much the gas phase acidity changes over this series.

Acidity in the gas phase means ROH must dissociate to RO- and H+ but doesn't put the proton onto anything else. Since the proton isn't solvated, it must have been knocked loose, say in a mass spectrometer, where the fragmentation potential can be measured. That just seems quite different from our usual idea of "acidity". But OK, it's an observation of a characteristic of a series of related compounds.

The tendency to designate specific places for electrons (LUMO, HOMO) is understandable, but sometimes this description is uncertain and unnecessary. In a gas phase RO- ion, the negative charge is not 100% around the nucleus of the oxygen atom. Surely it will become unsymmetrical and drift where it might find a little positivity, like on the carbons and the hydrogens. Alkyl groups are polarizable and stabilize both anions and cations (https://onlinelibrary.wiley.com/doi/abs/10.1002/ejoc.200700004).

It's simpler to explain the acidity changes in water and in the gas phase by referring to the size-dependent polarizability of the alkyl groups, but using molecular orbitals may have seemed better for publication.


How do I find the expiration time of drug in the following example? An optically active drug has one chiral center and only dextrorotatory isomer is effective. Moreover, it becomes ineffective when original activity is reduced to 35 % of original. The rate constant is  $1 \cdot 10^{-8}\ \mathrm s^{-1}$. Find the expiration time of drug in years.
I considered the reaction as $$\ce{A -> B}$$ $$a-x \rightarrow x $$
so the, $\left(a-2x\right)r = 0.35$ and then I got $x$ and then I used
$$ t=\frac{2.303}{k}\log\frac{a}{a-x} $$ I got answer 3.65 years. But the answer given is 1.37 years.

Your general approach to the calculation looks great.  My only question is, what is the meaning of $x$ in your calculation?

Here's how I would do the problem.

$$\ce{A->B}$$

$$A = A_0 e^{-kt}$$

$$\frac{A}{A_0} = e^{-kt}$$

$$0.35 = e^{-(~{\mathrm 1 \times 10^{-8} {\mathrm s^{-1}}})t}$$

$$ \ln{0.35} = -(~{\mathrm 1 \times 10^{-8} {\mathrm s^{-1}}})t$$

$$ t = \frac{-\ln{0.35}}{(~{\mathrm 1 \times 10^{-8} {\mathrm s^{-1}}})}$$

If I did the units conversion from $\mathrm{s^{-1}}$ right, then the rate constant in units of $\mathrm{{yr}^{-1}}$ is 0.316.  (I just multiplied by 60×60×24×365.25).

$$ t = \frac{-\ln{0.35}}{(~{\mathrm 0.316~{\mathrm {yr}^{-1}}})}$$

When I plug in the numbers, I get 3.2 years.  It's possible I did something wrong, but my answer is way closer to yours than to the given answer.

Here are some possible mistakes that the original writer of the question could have made to give the wrong, lower, answer.

The question has a bunch of superfluous information about the chirality of the drug.  Mistakenly thinking that this is important to the question, perhaps somebody divided by 2 to "adjust" for chirality.  But this is not the correct approach.  We're given the activity of the drug and the decay rate of activity, so the structure of the molecule is completely irrelevant to the problem.
Another possible error is misinterpreting "is reduced to 35 % of original" mistakenly as "is reduced by 35%".  That leads to the calculation for $t$ being $\frac{-\ln{0.65}}{0.316}$ and gives the expected (but wrong) answer.

The bottom line is I think you mostly did the calculation correctly and the given answer seems incorrect, at least given the wording of the question as you supplied it.


Why are glycosidic bonds always 1->4 in polysaccharides? I know 1->6 bond exists too, in amylose for example, but that one seems logical, since the 6th carbon is easier to access.

I understand why the bonds originate from the hemi-acetal carbon, but I don't understand why they are 1->4.

Why not 1->3 or 1->2 ? Since there are hydroxyles on every carbons in the ring?

While I was writing this, I just had an illumination, maybe someone can confirm:

Could it be that, the formation of glucose chains in plants and animals always make use of enzymes and that it is these enzymes which, by the shape of their active site, dictates what the bonds will be?

Why are glycosidic bonds always 1->4 in polysaccharides?

The premise of the question is false.

  Why not 1->3 or 1->2 ?

1,3-linked glycosidic bonds certainly exist.  In fact there's a whole class of molecules called beta glucans of which the 1,3-linked variety forms a major part.

  Could it be that, the formation of glucose chains in plants and animals always make use of enzymes and that it is these enzymes which, by the shape of their active site, dictates what the bonds will be?

I think what you say here is true, but you should keep in mind that there are many different enzymes, and some will specifically form one linkage, and others will form alternate linkages.


Nitrogen inversion barriers of trimethylamine and N,N-diisopropylmethylamine Which would have a smaller nitrogen inversion barrier: $\ce{NMe3}$ or $\ce{N(i-Pr)2Me}?$

I think that $\ce{NMe3}$ should have a smaller inversion barrier as it is less bulky, but data shows otherwise. Are there some other effects that contribute to the inversion barrier that I might not know of?

So I thought about this for a bit and I think I have an answer. Be warned, I might be wrong.

The inversion process involves a conversion from $\mathrm{sp^2}$ to $\mathrm{sp^3}$. So $\mathrm{sp^2}$ hybridization increases angle between these bulky groups, which would be in fact favorable. Hence the low activation energy.


In E1 reactions, does carbocation stability have impact on the final product? In this question, I believe there is a mistake for iii). I think that there the tertiary carbocation would be formed (more stable), which would lead to the tetra substituted alkene (even more stable). Am I wrong?

The answer above only shows one of the two possible isomers that could be formed by elimination from the tertiary carbocation intermediate.

I suspect that both isomeric cyclohexenes with tri-substituted double bonds would probably be produced in roughly equal amounts, along with a smaller amount of the exo-methylene compound.

Edit:  Thanks to Bon and Yolo123 for their comments

Elimination towards the isopropyl group is also a possibility, but two factors argue against this being the major elimination pathway.  First there is a statistical factor;  there is only proton on the central isopropyl carbon available for elimination, whereas the are 4 (2+2) protons available alpha to the carbocation center in the 6-membered ring.  Second, if you look at a model and consider that the proton to be eliminated must be lined up with the carbocation p-orbital you see that each methylene group adjacent to the carbocation has one hydrogen well lined up for elimination, whereas to properly line up the isopropyl proton results in some adverse steric interactions between the isopropyl methyl hydrogens and the ring methylene hydrogens.

My guess is that the gem-dimethyl group in the cyclohexane ring will not favor one in-ring elimination pathway over the other - my model doesn't show them exerting any adverse steric effect one way or the other.