text:
Where's my experimental error coming from?
I’m teaching chemistry to a home-ed 15 year old.  Occasionally we do some practical work, since I believe very strongly that physical science learning should have some real lab work to bring it alive, and also to teach things like observational skills and attention to detail.
My background is a PhD in organic chemistry, so I’m familiar with hazard assessment, and all the lab work I do with my student is something I’ve done myself already.  Any time she touches the apparatus it’s under close supervision by myself.
So the practical involved taking hydrated copper sulphate, heating it to drive off the water, weighing before and after, and thus calculating the number of water molecules of crystallisation, based on the respective molecular weights of the anhydrous salt and water.
I have a porcelain evaporating dish to contain the hydrate during heating, electronic scales that will weigh to 300 g in increments of 10 mg, and commercially bought copper sulphate that is allegedly 99.5 % pure.
I started with the hydrate, heated it in the evap dish over a flame (using a gas cooker ring) until it had all turned greyish white, then allowed it to cool and weighed again.
The results were:
Starting wt hydrate: 13.03 g
Wt after heating: 8.22 g
Hence water lost: 4.81 g
So that’s 4.81/18 = 0.267 mol water, and
8.22/(63.5 + 32 + 64) = 0.0515 mol anhydrous copper sulphate
Hence 0.267/0.0515 = 5.18 mol water per mol copper sulphate.
So my question: why not 5.0?
The electronic scales (quite cheap) were an obvious first candidate for a source of error.  I’ve compared its readings at weights of about 2, about 20, and about 50 g, with another similar model, and they agree to within 10 mg (one digit in the final decimal place) at all weights.
I could perhaps have not heated the hydrate enough, and not driven off all the water – but then I’d have less than 5, not more, waters per mol.
Or the hydrate, as supplied, could be a little wet.  I haven’t investigated that yet, but I plan to weigh a sample before and after leaving in a warm airing cupboard overnight.
If you’ve read this far, I’m already grateful, but I’d be even more so, if anyone could suggest where the difference between 5 and 5.18 is coming from.
I did it a second time, and got 5.16.
The difference between 5.00 and 5.16 is a weight error of about 200 mg, and I was really careful, so this seems unlikely.
The weight after cooling of the evap dish is constant.
Any ideas?
Edit (2):

There was no sign of $\ce{CuO}$ after the heating.  No trace of black, and indeed we dissolved the anhydrous salt in water (for growing some seed crystals later), and there was no insoluble residue.
I’ve now heated the dish over a hot blue natural gas flame for ten minutes.  Its weight before cooling was 0.02 g (on my scales) less than its empty weight cold.

Edit (3)
I weighed out some of the hydrate into another dry dish, flattened it out as much as possible within the confines of the dish to expose as much surface area as possible, and left it in the airing cupboard for 24 h at 26 degrees Celsius.  Then I re-weighed.
Pre drying wt: 42.68 g
Post drying wt: 41.98 g
Assumed water content: 42.68 − 41.98 = 0.7 g (!!!)
So assuming that 26 degrees is not warm enough to cause partial loss of water of crystallisation, that suggests that at best my hydrate is only 98.36 % pure, not 99.5 as it says on the bottle (typo corrected above).
So repeating my calculation above:
Wt before: 13.03 × 0.9836 = 12.82 g hydrate
Wt after: 8.22 g
Water lost: 4.60 g
Moles water: 4.60/18 = 0.256
Moles $\ce{CuSO4}$: 8.22/159/5 = 0.052
Ratio: 4.96
I’m much happier with that!
Many thanks.

so with this last source of error you edited in (the 0.02 g) my estimate of $\Delta W_e=0.1$g actually seems quite realistic. Do you perhaps have any information about the nature of the .6% impurities in the original sample?

Still working on that - I'm going to leave a sample fresh from the bottle in a warm cupboard overnight tonight, and see if it gets any lighter.  Pending marking your answer as accepted, my grateful thanks in the meantime for your detailed analysis.

I would go with a full fledged error propagation analysis on this one, because without systematic approach you might be guessing forever. So if I look at the calculation process you have the following equation to get from known quantities to the answer:
$$X=\left(\frac{W_0}{W_e}-1\right)\left(\frac{M_{CuSO_4}}{M_{H_2O}}\right) \tag{1}$$
where $W_0$ is the start weight, $W_e$ the end weight and $M_i$ is the respective molecular mass. $X$ is the desired answer.
To find the expected error in the answer, $\Delta X$, assuming no error in the molecular masses, we calculate:
$$\Delta X=\sqrt{\left(\frac{\partial X}{\partial W_0}\Delta W_0\right)^2+\left(\frac{\partial X}{\partial W_e}\Delta W_e\right)^2} \tag{2}$$
which works out to:
$$\Delta X=\frac{\sqrt{(\Delta W_0)^2 W_e^2+(\Delta W_e)^2 W_0^2}}{W_e^2}\left(\frac{M_{CuSO_4}}{M_{H_2O}}\right)  \tag{3}$$
So now the final thing left to do is estimate how big $\Delta W_0$ and $\Delta W_e$ are. For $\Delta W_0$ I think it is safe to say that this is only the inaccuracy of the scale, so perhaps $0.01$ g. For $\Delta W_e$ I would go for the worst case scenario: that all of the 0.6% impurity in the original copper sulfate has evaporated, meaning that the weight will be off by 0.6% times $W_0$ i.e. $0.08$ g.
Now plugging in the numbers in Eq.$(3)$ I find $\Delta X=0.14$ mol water per mol copper sulphate. Which is very close to the actual error you find.
This error is almost completely determined by the error in $W_e$. If, like @brinnb suggests, an additional reduction in $W_e$ could be expected due to adhered water say that the error in $W_e$ is $0.1$ g you already get to $\Delta X=0.17$ mol water per mol copper sulphate.
Conclusion
The error you find falls within the worst-case expected error due to impurity and potentially adhered water.

You did what I thought of doing but didn't get around to do. +1

Did you heat the dish before weighing it to find the tare?  Water adheres to the dish when the dish is at room temperature.  This is probably a small effect.
A larger effect is decomposing copper sulfate to copper oxide and sulfur trioxide.  You can check for this by looking for the telltale brown/black color of copper oxide.


text:
Hydroboration/oxidation vs acid-catalyzed hydration of 1-phenylpropene
I am trying to predict the products of both the hydroboration/oxidation and acid-catalyzed hydration of 1-phenylpropene:

Both C1 and C2 of the double bond are secondary carbons. However, I have the feeling that there are slight differences in the electron density on each carbon, which will affect the addition of the boron electrophile and the stability of the carbocation for the two reactions respectively.
Which regioisomers are predominantly formed from both reactions?

Hydroboration
With the simplest hydroborating agent $\ce{BH3.THF}$, the hydroboration/oxidation of 1-phenylpropene proceeds to give 1-phenylpropan-1-ol as the major product.[1] The isomer of the substrate used was not reported.

There is some ambiguity in the above source, as the labels 8 and 8′ (which the paper uses to refer to the two regioisomers) are consistently swapped; hence, I have taken the names and not the labels to be accurate. However, as additional proof, there are other examples of similar regioselectivity in the hydroboration of styrenes (here, an E/Z mixture of styrenes was used):[2]

Rationalising this selectivity is not very easy, but in one of Herbert Brown's papers where the same regioselectivity is seen with 9-BBN as the hydroborating agent, he writes:[3]

The preferred site of hydroboration in 1-phenylpropene, the site to which boron becomes attached, is the 1-position. This is not what one would expect on the basis of steric effects. One possible explanation for this is the combined effects of phenyl conjugation (-K) and methyl hyperconjugation (+K) which act to decrease the amount of electron density at the 2-position and to increase it at the 1-position.

I'm not totally convinced by this, but I also don't have any better explanation. Further evidence for this explanation is the fact that the (Z)-isomer reacts with poorer regioselectivity than the (E)-isomer. This is because there is allylic strain in the (Z)-isomer which forces the phenyl group to twist out of conjugation with the C=C double bond (depicted below). Consequently, the electron-withdrawing effect of the benzene ring (via conjugation) is reduced, and the formation of the 2-hydroxy isomer is less disfavoured.

Acid-catalysed hydration
Examples of these in the literature are harder to find and I cannot find any generic examples of acidic hydration of $\ce{R^{1}-C=C-R^{2}}$, where $\ce{R^1}$ is an aryl group and $\ce{R^2}$ an alkyl. However, I see no particular reason to go against the conventional wisdom: the more stable carbocation intermediate is formed at the benzylic position, which should lead to the 1-hydroxy compound as the major product.

References

Camacho, C.; Uribe, G.; Contreras, R. Diphenylamine Borane, A New Stable Amine .Borane with Remarkable Hydroborating and Reducing Properties. Synthesis 1982, 1982 (12), 1027–1030. DOI: 10.1055/s-1982-30051.
Allegretti, M.; Bertini, R.; Cesta, M. C.; Bizzarri, C.; Di Bitondo, R.; Di Cioccio, V.; Galliera, E.; Berdini, V.; Topai, A.; Zampella, G.; Russo, V.; Di Bello, N.; Nano, G.; Nicolini, L.; Locati, M.; Fantucci, P.; Florio, S.; Colotta, F. 2-Arylpropionic CXC Chemokine Receptor 1 (CXCR1) Ligands as Novel Noncompetitive CXCL8 Inhibitors. J. Med. Chem. 2005, 48 (13), 4312–4331. DOI: 10.1021/jm049082i.
Brown, H. C.; Nelson, D. J.; Scouten, C. G. Hydroboration. 63. Kinetics and regiospecificity of the hydroboration of isomeric cis- and trans-alkenes via 9-borabicyclo[3.3.1]nonane (9-BBN). Effects of 1,1- and 1,2-dialkyl interactions. J. Org. Chem. 1983, 48 (5), 641–643. DOI: 10.1021/jo00153a003.

Note: $H_3O^+$ looks like  $H_3O^-$ due to thin width of vertical bar in $+$
During HBO, the carbon nearer to the phenyl ring is more sterically hindered than the other one next to it, otherwise these two are similar in many respects exscluding some, this can be understood by imagining the intermediate $R_3B$ by taking R as branched through both carbons one by one, maybe then you'll appreciate the fact of difference in steric hindrance.
In case of acid catalyzed hydration, the carbon nearer to it has more carbocation stabilization than the one next to it, because of the conjugation with the benzene/phenyl ring. Even if the carbocation is formed on the $\beta$ wrt phenyl, it'll rearrange to $\alpha$/allylic position.Thus the product.

Notes:
Herbert C. Brown (1912–) won the Nobel Prize in 1979 for his invention and development of hydroboration, mostly carried out at Purdue University, which made this new chemistry familiar to the practising organic chemist. The prize was shared with Georg Wittig, showing how important organic chemistry with main group elements had become. He vigorously opposed ‘non-classical’ carbocation theory.
For most substrates, the addition in hydroboration is stereospecific and syn,
with attack taking place from the less-hindered side.
Brown, H.C.; Zweifel, G. J. Am. Chem. Soc. 1961, 83, 2544; Bergbreiter, D.E.; Rainville, D.P.
J. Org. Chem. 1976, 41, 3031; Kabalka, G.W.; Newton, Jr., R.J.; Jacobus, J. J. Org. Chem. 1978,
43, 1567.
The only special case known (atleast to me) is the treatment of iPrCHCHMe with borane which
gave 57% of product with boron on the methyl-bearing carbon and 43% of the
otherBrown, H.C.; Zweifel, G. J. Am. Chem. Soc. 1961, 83, 1241.
In all cases, the boron goes to the side of the double bond that has more hydrogens,
whether the substituents are aryl or alkylFor a thorough discussion of the regioselectivity with various types of substrate and hydroborating
agents, see Cragg, G.M.L.Organoboranes in Organic Synthesis Marcel Dekker, NY, 1973, pp.63–84,
137–197. See also, Brown, H.C.; Vara Prasad, J.V.N.; Zee, S. J. Org. Chem. 1986, 51, 439.

Apparently, ordinary hydroboration doesn't have the desired regioselectivity for 1,2-disubstituted alkenes when one of the substituents is a phenyl group (see this reaction scheme). Ordinary borane adds the boron atom predominantly at the benzylic position in those instances. I think it's possible to overcome this by using a sterically hindered borane (e.g., 9-BBN), however.

@GregE. You are right about the expected regioselectivity of the hydroboration. Can you upgrade your comment to an answer as the current one is not correct? I wouldn't think 9-BBN will change considerably this selectivity since it is mainly electronically driven.

@K_P, thanks for the comment. As far as the regioselectivity, I understand that electronic factors are decisive. Admittedly, I'm speculating, but I think with a sufficiently bulky borane (like 9-BBN, disiamylborane, catecholborane, etc.) it might be possible for sterics to override electronics and direct the addition to the less hindered carbon. There is ample literature indicating superior regioselectivity when using bulky boranes, but I can't find any reference specific to phenyl-substituted alkenes, so I can only conjecture. I'd like to research more before I write an answer.

@Aditya, I recognize the text you've added from March's Advanced Organic Chemistry, and nothing in the excerpts you've used specifically addresses this issue because the degree of substitution at the two carbons of the alkene in 1-phenylpropene is the same, and sterics and electronics likely have opposite influences on regiochemistry in this instance. Perhaps this is covered in the primary literature March cited, but I don't have access to it at the moment to check. In the meantime, see scheme (d).

(-1) Regioselectivity of hydroboration is wrong, as has been pointed out in previous comments. By the way, 9-BBN doesn't reverse the regioselectivity (cf. ref 3 in my answer).

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

authors: عبدالله عبدالمعبود

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Isomer Identification Using Condensed Structural Formulae

Which of the following pairs are isomers?
a) $\ce{C5H10}$ and $\ce{C10H20}$
b) $\ce{CH3(CH2)4CH3}$ and $\ce{CH3(CH2)3CH3}$
c) $\ce{CH3CH(CH3)(CH2)2CH3}$ and $\ce{CH3(CH2)2CH(CH3)2}$
d) $\ce{(CH3)3CH}$ and $\ce{CH3CH2CH2CH3}$

My textbook says the answer is C). I beg to differ. I think it is D). C) is just a different way of writing 2-methylpentane.

isomer
One of several species (or molecular entities ) that have the same atomic composition (molecular formula) but different line formulae or different stereochemical formulae and hence different physical and/or chemical properties.

Source:
Pure and Applied Chemistry, 1994, 66, 1077 (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994))
Pure and Applied Chemistry, 1996, 68, 2193 (Basic terminology of stereochemistry (IUPAC Recommendations 1996))
a)
C5H10
and
C10H20
different chemical formula → not isomeric
b)
CH3(CH2)4CH3 = hexane (C6H14)
and
CH3(CH2)3CH3 = pentane (C5H12)
different chemical formula → not isomeric
c)
CH3CH(CH3)(CH2)2CH3 = 2-methylpentane (C6H14)
and
CH3(CH2)2CH(CH3)2 = 2-methylpentane (C6H14)
same chemical formula and same structural formula → identical, not isomeric
d)
(CH3)3CH = isobutane (C4H10)
and
CH3CH2CH2CH3 = butane (C4H10)
same chemical formula and different structural formula → isomeric

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
What is the reason for the exceptional stability of the cyclopropylmethyl carbocation?
Can someone explain this to me by drawing resonance structures for the cyclopropylmethyl carbocation please?
Also one more question, is the tricyclopropylmethyl carbocation more stable than tropylium ion?

Note that benzyl cation $\ne$ tropylium ion. One rearranges to the other, but they are not synonymous.

@Ben Norris Yes Im aware of that. It was a separate question actually. I will edit it to avoid confusion :)

It is commonly said, that a cyclopropane fragment behaves somewhat like a double bond. It can conjugate, and pass mesomeric effect similar to a double bond, but the donor orbital is $\sigma_{\ce{C-C}}$ instead of $\pi_{\ce{C=C}}$. Cyclopropane can be considered as a complex of a carbene and an alkene, where the carbene $\mathrm{p}$ orbital interacts with the $\pi^*_{\ce{C=C}}$ orbital while the carbene $\mathrm{sp^2}$ orbital interacts with the $\pi_{\ce{C=C}}$ orbital, so this 'virtual' double bond behaves somewhat like a normal double bond.
On the other hand, the structure of the cyclopropylmethyl cation is downright weird. It is well known that both cyclopropylmethyl and cyclobutyl derivatives give very similar product mixture under $\mathrm{S_N1}$ hydrolysis conditions, resulting in both cyclopropylmethyl and cyclobutyl derivatives (see, for example, J. Am. Chem. Soc. 1951, 73 (6), 2509–2520). This is commonly described by the conjugation in following manner (here, the 1-cyclopropylethyl cation is depicted):

Here bonding of $\ce{C-4}$ with $\ce{C-1}$ and $\ce{C-2}$ can be roughly described as an interaction of the vacant $\mathrm{p}$-orbital of $\ce{C-4}$ with filled orbital of $\pi$-bond between $\ce{C-1}$ and $\ce{C-2}$. It does not matter much, where the original leaving group was - at  $\ce{C-2}$ or $\ce{C-1}$. Since the positive charge is more or less symmetrically distributed between three atoms and the small ring is somewhat relieved of its geometrical strain (both cyclopropane and cyclobutane are very strained molecules, not only due to angle strain, but also considerable steric interactions between hydrogens), the cation has remarkable stability.
Similar effects are common in the chemistry of small bicyclic systems, with norbornane derivitives being the chosen test subjects for decades with the 2-norbornyl cation being probably the most well-known example. March's Advanced Organic Chemistry, 7th ed., Section 10.C.i discusses such nonclassical carbocations in great detail, with the cyclopropylmethyl system being described on pp 404–406.
With further addition of multiple cyclopropyl groups, however, the full conjugation becomes sterically hindered, so extra groups beyond the first have less of an effect. Of course, the stability of these cations is far below that of the tropylium cation, which has very little strain and also possesses aromatic character, distributing the positive charge over seven(!) carbon atoms.  In fact, the stability of the tropylium system is so high, that even the cyclooctatrienyl cation (also known as the homotropylium cation) adopts a similar structure.

Thank you for answering and a good explanation. You may want to add the reaction between Ethene and Methyl carbene to form Cyclopropane above. :)

Hi, (+1) thanks for your detailed answer! I wish to seek some clarification. My prof taught that cyclopropyl methyl cation is stabilized by "banana bonds" (aka bent bonds) and "dancing resonance". He didn't detail these concepts in as much detail as you did yours, so I don't know their internal details. Though, you did not mention either of these two concepts in your answer. I wish to ask if these two concepts are related to anything you wrote? Or do they not affect the stability of cyclopropyl methyl cation at all?  Thank you!

@GaurangTandon It is common in chemistry to have two or more models that are mathematically equivalent or very similar. Modern chemistry dangles between two such concepts: one of localized bonds, resonance and hybridization and one of molecular orbitals. When the first model works, both models are equivalent despite completely different inner workings.  The first model is simpler to grasp, but struggles to offer a simple explanation in complicated cases. I offered explanation in terms of molecular orbitals, you professor offered explanation in terms of resonance.

Is cyclopropylmethyl carbocation is more stable than triphenylmethyl carbocation?

Cyclopropane is a highly strained molecule with a bond angle of $60^\circ$. The normal tetrahedral bond angle is $\pu{109^\circ{28}'}$, so we'd expect a ring strain of $49.28^\circ$!
For a cyclopropane C atom, the four hybrid orbitals are not equivalent. The two orbitals directed to the outside bonds have more $\ce{s}$- character than a normal $\ce{sp^3}$ orbital, while the involved in ring bonding have less, because the more $\ce{p}$ like they are, the more they resemble the ordinary $\ce{p}$ orbitals whose ordinary bond angle is $90^\circ$. Since the angle strain is a measure of the difference between the preferred angle and the real angle of $60^\circ$, this additional $\ce{p}$ character relieves the strain.  The external bonds have $\sim 33\%$ $\ce{s}$ character, so they are $\sim \ce{sp^2}$ orbitals, while the internal orbitals have $\sim 17\% \ce{~s}$ character so they are $\sim \ce{sp^5}$  orbitals. Thus, each C-C bond is formed by $\ce{sp^5-sp^5}$ overlap.
The bonds in cyclopropane are called bent bonds and they are intermediate in character between $\sigma$ and $\pi$.
Now, coming to the stability of cyclopropyl methyl carbocation, it is symmetrically stabilized by both $\ce{C-C \sigma}$ (2-3 and 2-4) bonds.  You may call it bent bond resonance  with the vacant $\ce{p}$ orbital of the carbocation.

It is worth mentioning that cyclopropyl group stabilizes an adjacent postive charge even better than a phenyl group.

Source: March's Advanced Organic Chemistry

I'm pretty sure the image is incorrect. I don't believe the primary cation exists. And the whole point of resonance is that the structure does not change.

Would a cyclobutyl cation be a valid resonance structure?

In cyclopropyl methyl carbocation, the sigma electron cloud expands outwardly due to angle strain; therefore, the valence electron cloud of $\ce{CH2+}$ is surrounded by or enveloped by the expanded electron cloud making carbocation stable.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Number of diastereomer pairs of 1,3-dichloro-1,2,3-triphenylpropane

How can I determine the number of possible pairs of diastereomers here?
My first guess was two: RR with RS and SS with RS. But what about the potential chirality center middle carbon? There are two identical branches attached to it; however, those two branches can change R/S. So, is it a chirality center?

I basically agree with Ron's answer, but had to draw all of the possible structures to confirm it. The complication is that carbon-2 is non-stereogenic but it may be chirotopic depending on the configuration of the neighboring atoms. The IUPAC Gold Book calls this a pseudo-asymmetric carbon atom. Carbon-2 is not a stereocenter, because it does not have four different "things" attached to it. However, its configuration is important relative to carbons-1 and -3. Below are all of the possible combinations.

Out of the eight possibilities, we see there are only four unique compounds (names generated in ChemDraw):
A: (1R,2s,3S)-1,3-dichloro-1,2,3-triphenylpropane
B: (1S,3S)-1,3-dichloro-1,2,3-triphenylpropane
C: (1R,2r,3S)-1,3-dichloro-1,2,3-triphenylpropane
F: (1R,3R)-1,3-dichloro-1,2,3-triphenylpropane
The small r and s are essentially used to indicate the stereochemistry at carbon-2 relative to carbons-1 and -3. Note that in compounds B and F, the stereochemistry at carbon-2 is inconsequential as in both cases the phenyl group is syn to one chloride and anti to another chloride. I haven't been able to find an adequate explanation of assigning the r/s designations, but I believe that the R configured substituent receives higher priority than the S configured substituent, and Cahn–Ingold–Prelog system is followed from there.

Hmm. The solutions said the answer was two. Like my guess... I conclude that the solutions is mistaken.

Carbon 2 is achirotopic, not chirotopic and it is called a stereocenter as it can have r/s configuration. It is not strictly a chiral center because two attached groups differ only by their configurations.

As far as I can tell (by reading Anslyn/Dougherty section 6.3.3) C-2 is chirotopic only in those isomers which are chiral, i.e. B and F. In the meso compounds A and C, C-2 lies within the plane of symmetry and is therefore achirotopic. So, both of you are correct. :)

As for the r/s notation, future visitors to this question may be interested in: What does lowercase r-s notation mean? and IUPAC name for 1,2,3-trichlorocyclopropane? where the notation is explained in detail with reference to the Blue Book.

@orthocresol I've changed it to reflect that C2 may or may not be in a chiral environment.

CIP-assignments of A and C are incorrect. Actually, they are correct assignments of E and G. I appreciate if you correct them ASAP.

Let's number the 3 aliphatic carbons to make our discussion clearer.

Using the definition of a chiral carbon as a carbon that has 4 different substituents, carbons 1, 2 and 3 are all capable of being chiral. Carbons 1 and 3 are relatively straightforward chiral carbons, it's easy to see the 4 different substituents and assign "R" or "S" configurations.  However carbon 2 is problematic.
If we assign R- and S- configurations to carbons 1 and 3 respectively, then there are 4 different groups attached to carbon 2, but the difference in 2 of the groups is just "R" or "S".  In fact, in this situation there is actually a plane of symmetry bisecting the molecule and the molecule is achiral - it is a meso-isomer.  The plane of symmetry contains carbon 2 and the hydrogen and phenyl group attached to carbon 2, and, as expected, the R- and S-substituents are enantiomeric.  In such cases it is correct to say that carbon 2 is pseudo-asymmetric (thanks to @GaurangTandon for pointing this out!). The Cahn–Ingold–Prelog priority rules can still be used to describe carbon 2, the R substituent is given priority over the S substituent and lowercase "r" and "s" are used to describe carbon 2, the pseudo-asymmetric carbon.
In the case where we assign R- and R- (or S- and S-) configurations to carbons 1 and 3 respectively, then there is no longer a plane of symmetry bisecting the molecule.  Carbon 2 is now prochiral (this means changing one substituent is all that is required to make carbon 2 chiral), but the molecule as a whole is now chiral.
Going in the order carbon 1-2-3 here are the various possibilities:

1R,3R is chiral and it has an enantiomer 1S,3S
1R,2r,3S is an achiral (there is a plane of symmetry bisecting the molecule), meso-compound and a diastereomer of 1R,3R;
1R,2s,3S is a second achiral, meso-compound.

So there are 3 diastereomers; one is chiral and exists as a d,l pair and the two other  diastereomers are achiral, meso-compounds.

Ron, just to be sure, the question was asking for pairs.
So, I'm listing them:

1R,3R and 1R,2r,3S

1R,3R and 1R,2s,3S

1S,3S and 1R,2r,3S

1S,3S and 1R,2s,3S

1R,2s,3S and 1R,2r,3S

Could you please back me up? I found 5 total.

This is actually quite weird. This was a multiple choice question and there not odd numbers :O ! I should use another exercise package. Too many mistakes.

@yolo123  I don't think that is right.  Diastereomers don't come in "pairs".  As I said in my answer, there are 3 diastereomers.  Maybe the question was asking how many pairs of enantiomers?

Oh boy! Why have I not thought of that...? Word for word:
"How many pairs of diastereomers are possible in the following molecule? PhCH(Cl)CH(Ph)CH(Cl)Ph?"

Definitely, this question is very ambiguous... Honestly, thanks a lot jerepierre and ron. Very informative.

Unless I misdrew the molecule from the condensed structural formula.... But I don't think so...

I don't think it is hindered enough to prevent rotation around the central carbon, so I think it only has two stereogenic centres, leading to four possible stereocentre combinations as you have identified: RR, SS, SR, RS
This results in more than two pairs of diastereomers though:
1) RR with SR
2) RR with RS
3) SS with SR
4) SS with RS

This analysis is incomplete because there are two possible compounds with configuration (1R,3S), depending on the relative (syn/anti) configuration at C-2.

@jerepierre Indeed, thanks for posting a better one :)


text:
What new material is D3O?
I am a PhD student in materials engineering. What is D3O? It is not a chemical, but rather a "polyurethane energy-absorbing material" (from Wikipedia). I would wish to know and study the chemical composition of this impact shock absorbing materials.

The inventor, Richard M. Palmer, has filed at number of patents on "energy absorbing materials", one of which is US patent (7,381,460).
On the other hand, there does not seem to be any peer-reviewed publication by Mr Palmer.
There's always the option to ask the manufacturer D30 for further information and/or material samples.
The possible downside of such a cooperation is that you might not be allowed to report on the progress of your experimental work on conferences or in publications due to an NDA previously signed!

The patent says the dilatant material they use is made by Dow Corning. They refer specifically to "Dow Coming under catalogue no. 3179". 3179 Datasheet. The datasheet says primary it is for toys, presumably like Silly Putty, which according to the Silly Putty Wiki, is also a silicone-based polymer.

The patent also has a specific example of the chemistry (see Example 1 therein). Note that the patent recently expired.

D3O is basically polyurethane with polyborodimethylsiloxane, or at least that is what the patent what's you to believe.
Polyborodimethylsiloxane is an expensive chemical and the thickness of the D3O back protector and limb products is on average the same with the thickness of their competitors using "normal" polyurethane.

Regarding "Polyborodimethylsiloxane is an expensive chemical and the thickness of the D3O back protector and limb products is on average the same with the thickness of their competitors using "normal" polyurethane."
Please note that:
1. PBDMS is much different from 'normal' foams such as PU, PS, TPU, EVA, etc.
2. PBDMS is a viscoelastic liquid or non-Newtonian fluid having high Deborah numbers
3. There are covalent bonds within the polymer, but hydrogen bonds between the molecules. It provides ‘Grab-release’ mechanism  with dynamic, reversible network(Si-O-B)

PBDMS is an interfacial nano-adhesive, hydrogen-bonded, self-healing, hydrophobic, super stretchy or flubber-esque, etc. Again, it forms dynamic, reversible network.

In conclusion,
PBDMS is a spider silklike siloxane. The emulsified gels or suspensions with nanoparticles(ex. carbon nanotube, graphene, cellulose nanofibrils, nanoclay, etc.) may find many potential applications. Its features are:
①   spider silk-like,
②   energy absorbing
③   self-healing
④   can be stretched several times its length before it breaks
⑤   stronger than carbon steel
⑥   tougher than carbon fiber and Kevlar
⑦   antimicrobial, fragrant
⑧   waterproofing, nonflammable
Reference

https://pubs.acs.org/cen/whatstuff/stuff/7848scit3.html
https://www.facebook.com/Aussie-Aerogels-552921575091430/?ref=bookmarks
https://omnexus.specialchem.com/tech-library/article/2018-wmf-startup-challenge-innovative-ideas (#4, K-armourgel and Spider Silk Sand by Korea Aerogels)

If you are associated with the company, that is no problem. But you should disclose it (informally, like a line saying "Note I am working for Company Abc"). A nice answer!


text:
Is this Fischer Projection of (3R, 4S)-3-ethyl-3,4-dimethyl-1-hexene wrong?
(3R, 4S)-3-ethyl-3,4-dimethyl-1-hexene
Draw the Fischer Projection of the compound above.
Here is the answer:

Is there a mistake at the upper stereocenter? I think so.

Is there a mistake at the upper stereocenter?

Yes you are thinking correct, the R configuration will be the diastereomeric pair of this compund, the second chiral carbon is in correct configuration. Don't you want to ask why?

How can it be a diastereomer when there is a double bond in the group at the top? I thought diastereomers needed to have a mirror plane on the bond between chiral centers.

@LDC3 see the figure, they're not enantiomers, nor homomers, not structural isomers and so on, and the one thing left is diastereomers


text:
Are these statements about change in enthalpy contradictory?
Just going through my thermodynamics chapter I came across 2 contrasting statements:

Change in enthalpy=enthalpy of (products-reactants) at constant pressure and temperature.

Change in enthalpy=Energy required to break bonds in molecules of reactants -energy
released in the formation of bonds in molecules of products.

So, if I consider these statements individually then these sound correct to me but they are different in the sense that in the first one it is (products-reactants) and in the second one (reactant -products)?
Is the book wrong or am I mistaken in my fundamentals  ?

They are basically same statements. We can use Hess law to elaborate these two statements. For example, we can use the following reaction,
                                 CH4=C+2H2
Here, Delta reaction = Enthalpy of CH4 - Enthalpy of C + 2(Enthalpy of H2).......(1)
We can break the reaction events as,
1) Breaking of 4 C-H bonds
2) Formation of 2 H-H bonds
So, delta H = Energy required to break 4 C-H bonds + Energy required to forming two H-H bonds.......(2)
Now we always take the reactor to be the system and that's why heat input to the reactor is positive and heat going out of the reactor is negative. In the formation of bonds, the energy are released, so the formation energy of H-H is negative with respect to our thermodynamic system.From 2
So,  delta H = Energy required to break 4 C-H bonds - Energy released in the formation of two H-H bonds.......(3)
Now, energy required to break 4 C-H bond= Energy of C + Enrgy of 4 H - energy of CH4
                                        =-Energy of reactants...(4)
Similarly,  energy required to form 2 H-H bond= Energy of 2 H2 - energy of 4 H
                                              =Energy of products (As C is in reference state).........(5)
Adding 4 and 5 yields, equation (1), which proves that both statements are same.

Bond breaking is endothermic, bond making is exothermic. To use bond strengths, sum the strengths of the broken bonds and sum the made bonds. Since the breaking is endothermic and the making is exothermic you can add the positive to the negative and get the enthalpy change for the reaction.
With heats of formation, it’s as if the above process has already been done with reference to the constituent elements of the substances involved. So, subtracting the heats of formation of the reactants from the heats of formation of the products is like doing the above process but with the actual values of the bonds in the substances, which can vary depending upon molecular structure.
Look at water as an example. The strength of a $\ce{O-H}$ bond is about 110 kcal/mol. $\ce{O=O}$ is about 120 and $\ce{H-H}$ is about 105. To make 2 mol of water you have to raise the potential energy of the system by 330 kcal by breaking the oxygen and hydrogen molecules into atoms. You then make 4 mol of $\ce{O-H}$ bonds in making 2 mol of water, lowering the potential energy by 440. This is a net change of −110 kcal for the process. −110 kcal is about 460 kJ. I rounded a lot and there’s always a difference in the bond strengths in a table and those in a molecule. A heat of formation table will tell you that a mol of water is about 240 kJ/mol lower in potential energy than the oxygen and hydrogen it was made from. You can take the 2(−240 kJ/mol) and subtract 1(0) for the oxygen and 2(0) for the hydrogen and end up with −480 kJ/mol, about what we got from the bond strength calculation.

1) change in enthalpy = enthalpy of (products-reactants) at constant pressure and temperature.

$$\Delta H_{p,T} = H_\text{products}-H_\text{reactants}$$

2) change in enthalpy = energy required to break bonds in molecules of reactants − energy released in the formation of bonds in molecules of products

$$\Delta H_{p,T} = (-H_\text{reactants})-(-H_\text{products})= H_\text{products}-H_\text{reactants}$$
because the energy required is negative of enthalphy of formation/ enthalpy of substance.

how can you say that the energy required is negative of enthalpy of substance ?
I am sorry but I didn't understand the last part.

@KaranSingh energy released is taken negative and absorbed positive; convention

ok so according to that-how is Energy
required to break
bonds in
molecules of
reactants negative as you have shown?
we need to provide energy to break bonds hence it should be positive?!?!:


text:
Why does an increase in reduced mass lead to a decrease in frequency?
In terms of IR spectrum, why does an increase in reduced mass lead to a decrease in frequency?

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site. Finally, we have an important policy: your questions (especially homework questions), should show your own work or thinking that you have already done in an initial attempt to answer the question.

Consider a simple molecule $\ce{A-B}$ composed of atoms $\ce{A}$ and $\ce{B}$ and let the bond connecting them be described as a spring.  Then the stretching and contracting of the $\ce{A-B}$ bond (spring) can be considered as a simple harmonic oscillation and is described by Hooke's law.
${\nu = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}}$
where ${\nu}$ is the frequency of the stretch,  ${k}$ is the force constant describing the specific spring and ${m}$ is the mass of the system.  In our $\ce{A-B}$ example, the mass of the system is described by the reduced mass ${\mu}$, where
${\mu = \frac{m_A m_B}{m_A + m_B}}$
So our equation becomes
${\nu = \frac{1}{2 \pi} \sqrt{\frac{k}{\mu}}}$
From this equation we can see that as the reduced mass increase's the stretching frequency will become smaller.

that's not hooke's law, it is a consequence of it used to describe by SHM


text:
Formation of azobenzene from nitrosobenzene and N-hydroxyaniline
I'm trying to create a reasonable reaction mechanism for the reaction of nitrosobenzene and N-hydroxyaniline, which happens as part of the reduction of nitrobenzene to azobenzene using magnesium as reductor and methanol as solvent. Ron was very helpful to give me a general outline of this entire mechanism:

The key step is formation of nitrosobenzene (this is just a reasonable proposal).

Some of the nitrosobenzene can be further reduced to aniline and
  N-phenylhydroxylamine, both of which are nucleophiles and can react
  with the nitroso N=O bond (analogous to a carbonyl). Elimination of
  water in both reactions produces azobenzene and azoxybenzene
  respectively. The azoxybenzene can be further reduced to azobenzene.

I went on to try and create a reasonable mechanism for the reactions of N-hydroxyaniline with nitrosobenzene and aniline with nitrosobenzene (both in the presence of magensium with methanol as solvent):

I wonder whether this is a correct reasonable reaction mechanism. I tried to move radicals as much as possible from positively charged/highly electronegative atoms (e.g. the nitrogen in the nitro functional group / oxygen) to the less electronegative atoms (e.g. carbon). My reasoning was as follows (I have not yet learned about in radical chemistry, but I give it a try): radicals can be removed in two ways: 1) the radical electron leaves the molecule, which results in a positive charge. This is less likely to happen with a more electronegative atom like oxygen in comparison to for example carbon, so that the radical is more stable on oxygen. 2) The radical reacts with an other molecule and obtains an extra electron forming a new covalent bond (in this case the electron is donated by methanols C-H bond), resulting in a neutral molecule with no radicals. Since oxygen is more electronegative than carbon it is more "greedy" to fill the singly occupied orbital, making the radical more stable on carbon.
Furthermore the second way to lose the radical is much more likely than the first, since it leaves a neutral molecule instead of a charged one.
At last, I was told that reactions involving metal surfaces (like magnesium in this case) are often not well understood. Why is that? Because there are too many pathways?

Below is my proposed mechanism. You'll see that we go through some of the same key intermediates, so I think you are on the right track. In your second step, it is much more likely that the proton is removed by acid base reaction instead of through the electron transfer process you've shown. Since the initial intermediate (my second structure) is zwitterionic, we can just say that it will neutralize itself. If the reaction conditions are not acidic (from what I see, it's just methanol and magnesium), then there will not be an appreciable amount of the protonated hydroxylamine. The mechanism doesn't require it, as nitrogen is strong enough to "kick out" hydroxide. The result is essentially an azoxybenzene, so we just need to figure out how to lose the other hydroxyl. As the cationic form, it will want to accept electrons, such as from the magnesium as a free electron. Back to a neutral (radical) compound, the other hydroxyl can be kicked out, giving a radical cation on nitrogen. Filling in a second electron from magnesium results in the azobenzene product.


text:
Wittig reaction with benzaldehyde

Somehow I am told that my product is incorrect. Why is my product incorrect?
As far as I can see this is a Wittig reaction resulting in a two carbon increase with removal of the oxygen.
The only thing that strikes me as odd is that the Wittig is conjugated to a bromide salt. The only way this can happen is if the phosphorous part has an overall positive formal charge. Does this mean that the n-butyllithium will abstract a proton and leave the carbon backbone of the Wittig as $\ce{CH3CH}$?
Even so this doesn't seem to affect my final product ...

The Z isomer is also formed.

@Ron does temperature have anything to do with this?

no there's no general trend bth are formed

Choice of solvents will also influence the E/Z ratio

Below is the preparation of the Wittig reagent you are given. Yes, the intermediate is a phosphonium salt, with a formal positive charge on phosphorus. Treatment with base (BuLi) deprotonates at the adjacent carbon giving a phosphorus ylide.

As for the Wittig reaction itself, one postulated mechanism (depicted in Organic Chemistry by Maitland Jones) is shown below. Nucleophilic attack of the ylide onto the aldehyde gives an open betaine intermediate, which closes to form an oxaphosphatane. The strained 4-membered ring decomposes readily to give an alkene and triphenylphosphine oxide.

The stereochemistry of the Wittig is related to the nature of the ylide used. Unstabilized ylides, such as the one here, predominantly give the Z-configured alkene, while stabilized ylides (usually through conjugation to a carbonyl) give an E-configured alkene as the major product. With substrates similar to the ones here, I obtained a 4:1 ratio of Z:E products, so the selectivity is not perfect, as K_P points out.
The stereochemical outcome is intimately related to the mechanism, and according to my copy of Jerry March's Advanced Organic Chemistry (4th ed), there is no evidence for the betaine intermediate. Instead, the oxaphosphatane is formed directly through a [2+2] cycloaddition mechanism. This mechanism is attractive because the stereochemical demands of the cycloaddition account for the preference of the (less stable) Z-product. However, this does not explain why different ylides give different selectivities.

there's no proof if betaine or oxaphophatane until now as Solomons Fryhle says

For many years it was assumed that a diionic compound, called
a betaine, is an intermediate on the pathway from the starting compoundsto the oxaphosphetane, and in fact it may be so, but there is little evidence
for it. to the oxaphosphetane, and in fact it may be so, but there is little evidence
for it.

I suspect they  want you to specify the stereochemistry too. The expected product should be the cis (predominantly at least). That's the textbook answer, reality is not that ideal.
EDIT: If I remember correctly the Z- isomer is favoured in higher temperatures and in lower temperatures you will get worse selectivity, or even predominantly trans. In any case your answer is not "wrong" maybe incomplete.

Could it be that the Z-isomer is favored at low temperatures? Since it's the less stable product, it must be kinetically formed. That implies that low temperature would favor the Z-product.

@jerepierre This (low temp E/Z selectivity) was in the back of my mind so I'll have to look it up. From a quick search, the wiki article seems to agree but I'll try to look further - at home now and missing my copy of March too. (as you mentioned it contains an extensive review of the Wittig )


text:
Is carbon dioxide IR inactive?
Wouldn't CO2 be IR inactive because of its non-polar bonds?

The best way to consider IR activity of vibrations is to study symmetry and molecular point groups. It takes perhaps a few lectures, but with practice, anyone can identify vibrations and their IR or Raman activity with nothing but a pencil and paper and the symmetry character tables.

Incidentally, $\ce{CO2}$ is not only IR active, but a known and major greenhouse gas because of the IR absorption in the atmosphere.

I am told that carbon dioxide is IR inactive.

You're right, that's not true.  Since carbon dioxide is linear it has $3n-5 = 4$ vibrations and they are pictured below.

The symmetric stretch does not result in a change (of the initially
zero dipole moment), so it is ir-inactive.
The asymmetric stretch does result in a change in dipole moment so it is ir-active.
The bend also results in a change in dipole moment so it too is ir-active.

We expected 4 vibrations and I've only listed 3. This is because the "bend" (let's start by placing the molecule along the x-axis) can occur in the y direction and the z direction.  But these two motions are the same, just deforming in different directions, the bend is said to be degenerate, accounting for the "fourth" vibration.
To sum up, carbon dioxide has 2 ir-active vibrations.
Edit - response to example added (question d) by OP
Question d is incorrect.  Either the author 1) inadvertently switched the column headings (IR active, IR inactive) or 2) meant to use some molecule other than carbon dioxide.
The first 3 rules you learn for interpreting IR and Raman spectra are

The number of molecular vibrational modes equals 3n-6 (3n-5 for linear molecules), where n is the number of atoms.
An ir active band will be observed if a vibration results in a change of the dipole moment. The initial dipole moment in the molecule's equilibrium geometry can be zero; all you need is a change.  The terms "polar" and "non-polar" can be confusing, they often mean different things to different people.  Leave "polar" out of the criteria for ir activity and stick with dipole moment, it is a much better understood term.
the rule of mutual exclusion, it states that, for centrosymmetric molecules (molecules with a center of symmetry, like carbon dioxide), vibrations that are IR active are Raman inactive, and vice versa.  So for carbon dioxide there is 1 Raman band and two IR bands.

Here's a link to a recent SE Chem question: How can I deduce the linearity of XeF2 from the IR spectrum? where these rules were used to determine the structure of a molecule.

@Dissenter to be IR active, a vibration must change the dipole moment. A polar bond can give IR active vibrations, but as the CO2 example shows, polar bonds can also give symmetric, IR inactive vibrations too.

And nonpolar bonds can have vibrations that change the dipole moment (for example $\ce{C=C}$ and $\ce{C#C}$). If you need proof that $\ce{CO2}$ is IR active, breath into an IR spectrometer while it is running.

It's just plain wrong.  Look here, they analyze carbon dioxide and get the same result as posted here.  They also analyze acetylene, a non-polar molecule, and again predict and observe ir-active vibrations.

Not true at all.  The $\ce{C=O}$ bond is one of the most strongly IR active bonds there is (and the IR activity of $\ce{CO2}$ is the reason it's a greenhouse gas).
I suspect the person who told you this was thinking that because $\ce{CO2}$ doesn't have a static dipole, it can't be IR active.  However, IR activity is the result of dynamic dipoles (meaning the dipole changes with some type of deformation motion; in the case of $\ce{CO2}$, this occurs with bending motion and asymmetric stretching, as another answerer described), not static dipoles.
In some symmetric molecules, like $\ce{N2}$ or $\ce{O2}$, the only vibrational modes that can exist are stretching of the only bond, which because it's symmetric, doesn't lead to a dipole change.  Thus, those species are not IR active.


text:
Calcium chloride + starch desiccant packs inside PPE for ebola workers?
(First question on stackexchange, apologies for any faux pas)
Ebola workers get drenched in sweat while working inside their personal protective equipment.  There's a discussion happening over on OpenIdeo about whether the desiccant packs used in shipping containers might help reduce humidity:
https://openideo.com/challenge/fighting-ebola/ideas/desiccant-embedded-ppe-suit-to-keep-ppe-dry-and-bearable
Questions for the chemists:

What are the reaction products of a calcium chloride + starch desiccant pack with humid air?  In particular, would it generate any gases?  (That'd obviously be a problem inside an impermeable suit)
The worker sweats to carry away heat evaporatively.  Presumably most of that body heat would need to pass through the desiccant pack as the sweat gets absorbed, and would need to be transmitted outside the suit in order to be worthwhile.  It looks like the reaction is also exothermic.  How much total heat would need to be removed from the pack, and is this going to be too much to dissipate?
Any other risks to be aware of with this idea?

If the pack contains nothing but calcium chloride and starch, there would be no gaseous products at all.  The products would be hydrated calcium chloride and soggy starch (library paste if it gets wet enough.)  The assumption that the pack contains nothing but CaCl2 and starch may not be a good one though.  I'm not so much concerned with a gaseous product as with a dissolved product after the packet comes in contact with a large amount of sweat.


text:
How to use gas laws to calculate the partial pressure of a gas that has partly condensed from a mixture?

A mixture of three gases N2, CO2 and O2 has a total pressure of 106000 Pa. The mixture is cooled to -100°C and the resulting pressure is found to have decreased to 82500 Pa. What gas is condensed and what is its partial pressure?
  (BP N2 = -196°C, BP O2 = -183°C and BP CO2 = -78°C)

I've had a look at all the formulas and laws that have been covered in my Chemistry textbook like Dalton's law of partial pressure and the combined gas equation but it appears that there is not enough information to solve it. I mean, if it asked to find the initial temperature, I could do it, but I'm not sure how to find the partial pressure of the gas(es) from the information provided.

Homework questions should show some attempt to answer the question and ask about a specific concept with which you are struggling.  How to ask a good homework question on Chemistry Exchange. Is it the terms "condensed" and "partial pressure?" The particular question you've put in block quotations is unanswerable as written as well.  We would need to know the initial temperature of the gas mixture (when P = 106kPa.)

That's what confused me too, it appears to be missing information which is why I haven't been able to attempt it. If that's the case I'll ask my teacher to clarify as he made the question. I've made amendments to the question which elaborates abit more.

You would need the initial temperature to be able to solve this.
You could use the combined gas law (or Gay-Lussac's Law) to determine the total pressure of the gas at -100°C, assuming none had condensed.  This ideal gas pressure would differ from the 82.5kPa by the partial pressure of the condensed gas at -100°C.
The question doesn't specify at which conditions it wants to know the partial pressure of the missing gas.  If it's -100°C, you'd be done at that point.  If it's at the original temperature, you would have to apply the same gas law in reverse to get back to your original conditions.
The boiling point information you added should allow you to determine which of the three gases condensed.  Only one would be a liquid or solid at -100°C.


text:
Difference between the absolute thermodynamic scale and ITS-90
What is the difference between the absolute thermodynamic temperature scale and the practical temperature scale: ITS-90?

The absolute thermodynamic temperature scale is a scale fixed by absolute zero and (at least for the kelvin, I'm not sure about the Rankine scale) the triple point of water, where absolute zero is defined as 0 K and the triple point of water is defined as 273.16 K. The rest of the scale is just a straight line from zero through 273.16 K off to infinity. However, this scale relies on only two defined points (only one of which can actually be measured, absolute zero being established from thermodynamic relations).
In practical use, for the purposes of calibrating thermometres and the like, this is not a great scale as the two calibration points may be far away from the temperature region one is interested in actually measuring. The usual way to deal with this is to use other reference points outside the actual definition. So that different people do this as consistently as possible, ITS-90 was created as a practical implementation of the absolute scale, setting additional reference points (all phase transitions) to defined values, interpolating between them to get other values. It's technically not a scale in itself, just a standard for calibrating temperatures that are as close to the absolute scale as possible. By making sure everyone uses the same reference points, measurements can be more consistent than if reference points are chosen arbitrarily and possibly assigned different values.
For example, if one wanted to make a thermometre for general lab use, one might pick the triple point of mercury, the triple point of water, and the melting point of gallium from the ITS-90 standard for reference points, knowing that other thermometres in this range conforming to the standard would also be calibrated this way, rather than choosing say the melting and boiling points of water, increasing the likelihood of consistent measurements with other thermometres.


text:
Optical Density numbers
If it's OD of 600nm I know that's the wavelength used, but what are the other numbers in something like  "OD 600nm of 0.6-0.7". The 0.6-0.7?

That is the optical density (absorbance). It's just a range of between 0.6 and 0.7. Without the context, it's hard to say why they don't have a precise value, but in something like a protocol, it's common to see that kind of thing when there is some natural variation or the exact number isn't important.

It's a (sloppy) spectroscopists' way to denote a concentration range without actually knowing it.
It happens when you don't start from a properly prepared stock solution and use an UV-VIS spectrometer instead of a balance. It typically goes like this:

Add some compound (tip of a spatulum) to a cuvette and dissolve it.
Measure the absorption. Typically it's much too high! ;)
Remove a part of the solution from the cuvette and add some solvent
Record the absorption spectrum again.

Repeat 3. and 4. until your absorption is in a medium range: far from saturation but still high enough to observe smaller signals (as the result of a possible reaction).


text:
Gilmans and carbon dioxide
I know that Grignards react with carbon dioxide to give carboxylate anions.
But will Gilmans do the same? Will Gilmans react with carbon dioxide in a productive fashion?

Can you be a bit more precise? Will Gilman reagents react with carbon dioxide?

Yes, cuprates do react with carbon dioxide to give access to carboxylates / carboxylic acids.
Running a SciFinder search, I do not see any examples of traditional Gilman reagents (alkyl cuprates) reacting directly with carbon dioxide. This shouldn't be surprising though, since the alkyl cuprates are generated from the corresponding alkyl lithium compound. If you have the alkyl lithium compound, then you can just react that with carbon dioxide directly.
A more synthetically useful situation is when the alkyl cuprate is reacted with an alkyne, resulting in a syn-addition of the alkyl group and copper across the alkyne. The alkenyl (vinyl) cuprate can then be trapped with carbon dioxide to give a carboxylate.

A couple examples from the literature:
Z,Z-Dienes via acetylene carbocupration:  synthesis of navel orangeworm pheromone
The silylcupration of acetylenes: a synthesis of vinylsilanes
Also check the write up in Organic Reactions by Lipshutz (the world's expert in organocopper chemistry).
More complicated (non-Gilman) organocopper compounds have been reacted with carbon dioxide to give copper carboxylates:
Carbon-carbon coupling of activated alkanes with carbon dioxide by [(phen)(Ph3P)Cu(O2COH)]
Synthesis, Structure, and CO2 Reactivity of a Two-Coordinate (Carbene)copper(I) Methyl Complex

Informative, as usual.  But I still wonder about Gilman reagents adding to carbon dioxide.  Gilman reagents behave like a weaker base than Grignards or alkyl lithium reagents.  Gilman reagents don't typically have the "oomph" to add to ketones, while the stronger Grignards and alkyl lithiums do.  Further evidence that Gilman reagents are weaker, they typically add in conjugate fashion to alpha-beta unsaturated enones, while the other reagents can add directly to the carbonyl.  Like I said, just wondering.

@ron I bet that reaction works... The trend in cuprate reactivity is that spC-Cu is essentially unreactive, sp2C-Cu is more reactive, and sp3C-Cu is the most reactive. In the examples I listed, the vinyl cuprates react with CO2. Also, although I've never seen it directly compared to other carbonyls, I think CO2 is a pretty reactive electrophile with the two oxygens pulling electrons away from carbon through resonance. Toss in the fact that it's really small with two equivalent LUMO's so there are many possible modes that a nucleophile can approach.


text:
Factors to consider to predict EPR
What are the factors to consider when predicting the EPR spectrum of $\ce{Cu(Ox)P(CH3)3}$ where $\ce{OX}$ is the oxalate ion?
I understand that there are several factors to consider for EPR (Electron Paramagnetic Resonance) when we have transition metal complexes like: coupling, anisotropy, high spin/low spin, Jahn Teller distortion, Zero Field Splitting, or Kramer's rule. My question is which of these are important for this special copper complex and why?

The EPR experiment measures the interaction of unpaired spin density ($\rho(\alpha)$) at the nuclei.  This is quantified by an equation called the Fermi contact interaction.
If you have some prior knowledge of "where to put the dot" in your structure (highly localized), then generally you can assume that the EPR pattern is split most by that nucleus.  Splitting patterns are determined by standard rules in accord with the Zeeman effect.
The nucleus must have a non-zero nuclear spin to be observable. Certain nuclei such as hydrogen-1 and many transition metals have spin. It is very common that isotopic impurities (e.g. carbon-13) appear though the most prevalent isotope is EPR-silent.
(There are other complications, such as nuclear quadrupole moment and broadening due to exchange of chemical sites.)
An additional issue is that the spin density is generally spread over the molecule, and interacts with multiple nuclei.  It turns out to be easier to measure the spectrum, and then rationalize it given the particular structure and nuclear composition.
Accurate prediction of EPR spectra is a rather difficult problem, as quantum chemical calculations generally don't emphasize basis function placement at the nucleus, as they tend to focus on the "valence" and not the cusp-like nature of the radial distribution.  (Here, Slater-type orbitals would be better in principle than Gaussian-type, but STOs of sufficient number and of high-enough quality of Hamiltonian are virtually impossible).  Finally, solvent and powder effects can render the gas-phase ab initio calculations virtually irrelevant.
EPR simulation/interpretation of fine structure is much more difficult than NMR spectra, because in NMR spectra, one has the chemical shift to conveniently move splitting patterns out of the way.  In EPR, the splittings are often "piled" on top of each other, and make it necessary to use computer simulation to obtain precise fits.
In your specific example, I would identify the nuclei with spin (probably Cu, certainly phosphorus-31, and then hydrogen-1).  Then, identify where you think the unpaired density will be, and then take the nuclear spin + 1 and that will be the "guesstimated" isotropic spectrum.  The pattern will follow some variant Pascal's triangle if I=1/2, otherwise, you have to figure it out by writing out the number of possible configurations.
Edit: I have not even started on the characteristics of triplet spectra, as here the electrons interact.  One thing you can hope for is that all of your electrons are paired (singlet), and so it's EPR silent.


text:
How to predict the type of intrinsic defect?
Intrinsic defects occur when a pure material crystal shows either vacancy or interstitialcy. When an ionic compound has vacancy, the imbalance in charge has to be compensated with another defect. When a vacancy is balanced with another vacancy it is called Schottky defect , when it is compensated by an interstitial defect, it is called Frenkel defect. When species are exchanged on the sites, the defect is called anti-site defect.
Knowing all of that, how do we predict the type of intrinsic defects in a given material such as $\ce{HgSe}$ or $\ce{Ba3N2}$ ?

first google landed me here,is it u? http://lfhck.com/questions/how-to-predict-the-type-of-intrinsic-defect

I posted this question 2 years ago, that lfhck is a mirror of the original post. I am not sure whether direct googling would help. One needs to really know inorganic chemistry.

seemingly why i kind of gave up, all i know is organic, sorry

In general, there are not good heuristics to predict what defects in a crystal will have the lowest formation energies, much less diffusion energies. Even in the alkali halides there are anomalies in which defects predominate.

Don't know about anti site defects but frenkel defects occur when there is significant covalency (like in Mercury selenide) but you get schottky defects with more ionic materials (like barium nitride)

@RobChem, interesting! Could you explain a bit why is that the case? Why do we get vacancy balancing defects when there is ionic material? Why do we get interstitials when we have covalency? If you put it in answer format it would be even better.


text:
What is a chemical daily life experience that can be modelled by superior mathematics?
I'm searching for a chemical daily life experience that can be modelled by superior mathematics so that the origin of the equations involved can be easily explained to a beginner in chemistry.
By superior mathematics I mean things like partial and ordinary differential equation, Integral Calculus, group theory, differential geometry. But I want to understand how are the equations derived.
For example, in physics, I could model the trajectory of the rear wheel of a bicycle (a daily life experience) and I fully understand all the equations involved, without being a physicist. I want something similar to find in chemistry.

Diffusion is one example from the daily life. For example, the absorption of ethanol from the small intestine to blood can be modeled using differential equations. The extraction of caffeine from grounded coffee is also a diffusion phenomenon.


text:
Is (A->B<-C) a multi-step reaction?
I want to know if:
A -> B <- C

a multi-step reaction OR are multiple reactions:
A -> B, C -> B

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site. Finally, we have an important policy: your questions (especially homework questions), should show your own work or thinking that you have already done in an initial attempt to answer the question.

A reaction is generally something along the lines of
$$\ce{A->Z}$$
where A is your starting material and Z the product.
This reaction can also proceed stepwise, here shown via an intermediate Y:
$$\ce{A->Y->Z}$$
You could also add intermediate "steps" in the reaction, where you introduce more reactants (here B) to give the product:
$$\ce{A->Y ->[\ce{B}] Z}$$
One commonality that all these reactions share is that the arrow is always pointing one way (and usually to the right). This allows us to reorder your reaction as follows:
$$\ce{A->B<-C} \begin{cases} \ce{A->B}\\ \ce{C->B} \end{cases} $$
You should be able to figure it out from here.

+1: Thanks. This means that for a valid single-step and multi-step reaction, (TotalIntermediates = TotalSteps - 1) will hold true.

@A9S6 No, not really. It depends on the type of reaction (uni-, bi- or trimolecular). Only then do you know how many intermediate species you have.


text:
MO theory methods other than LCAO
What methods other than linear combination of atomic orbitals does MO theory use? Which one is the most accurate and comprehensive?

Well, this gets into the concept of basis sets.
LCAO implies that the number of molecular orbitals will be the same as the number of atomic orbitals involved.
In fact, we usually use more basis functions than just the simple number of atomic orbitals. We might add diffuse or polarization functions to better describe electron density away from the nucleus.
Moreover, rather than the physically correct Slater-type orbitals:
$$R_l(r) = A(l,\alpha) r^l e^{-\alpha r}$$
We usually use Gaussian-type orbitals to express the radial component.
$$R_l(r) = B(l,\alpha) r^l e^{-\alpha r^2}$$
Now GTOs don't express the radial part of the wave function well -- they decay too quickly as you get far from the nucleus. So we use a linear combination of multiple Gaussian orbitals to approximate the physically-correct Slater terms. Still, it's much faster to use a set of multiple GTOs.
Beyond STOs and GTOs, people do use other types of orbitals. For example:

Wavelets
Plane waves
Muffin-tin


text:
Reaction of Grignard reagents with esters
Suppose I have a phenyl Grignard and methyl benzoate.
If I react the two, can I expect benzophenone and a triphenylmethoxide ion as my products?
I ask because I know that Grignards react with ketones and aldehydes. So as the benzophenone is generated, couldn't more Grignard attack the benzophenone as well as the original reactant?
Expected products and mechanism shown below:

Reaction of a Grignard reagent with an ester is a standard method for producing tertiary alcohols where at least two of the substituents (the "$\small\ce{R_2}$" group attached to the Grignard) are the same.  Judicious choice of the starting ester allows for the preparation of a tertiary alcohol where all 3 substituents are the same. Alternately, you could react 3 equivalents of the Grignard with a cheap dialkyl carbonate (a "double" ester, if you will) to obtain a tertiary alcohol where all 3 groups are the same.

Typically a slight excess of Grignard reagent is used and the tertiary alcohol is produced after acid work-up.  If an excess of Grignard is not used then a mixture (ugh!) of ester, intermediate ketone and alcohol may result.

So I world need an acid workup after creating the anion?

If you mean the anion of the tertiary alcohol, then yes.

Wait @Ron I'm confused, I'm not showing a grignard directly reacting with the anion

Oh, I thought the reactant in the top line of your drawing was a carboxylate anion.  I guess that's not a minus sign but rather a bond where you just didn't draw the alkyl group on the end.

yes, that's a methyl group on that oxygen.

Yes, that is exactly what happens. And since ketones are typically more reactive than esters, you should expect that the tertiary alcohol (after workup) will be the predominant product.
In order to access the ketone, add the Grignard to a nitrile, which will give an imine (after protonation) that can be hydrolyzed to a ketone. Alternatively, add the Grignard to an aldehyde and oxidize the resulting alcohol to the ketone.


text:
Conditions for spontaneity
So the two conditions for a reaction to be spontaneous are:

tendency to achieve minimum energy
tendency to achieve maximum randomness.

Both these things are contrasting and the only way I see them working are when they work independent of each other. Do these work simultaneously?
I am really having a hard time in figuring out a reaction where energy decreases and entropy increases. Can this happen?

Luckily for us, a really smart guy named Josiah Willard Gibbs figured out how these two tendencies work together back in the late 1800's. As a result, in chemical thermodynamics we now have a state function called the Gibbs Free Energy that describes what is called the thermodynamic potential of a system at constant pressure, temperature, and number of molecules.
Since most of the systems we work with under lab conditions are at constant pressure and temperature and are typically closed to mass transfer (abbreviated as NPT), Gibbs free energy is a very convenient way of determining what will happen to a given chemical system under normal lab conditions.
The reason it works is that it is a potential energy function - it describes how a system's potential energy will change as different state variables change, and therefore lets us predict where the system will "try" to go. An analogy that works pretty well is to imagine it like the surface of a planet, where gravity is the potential energy. If the derivative or slope is negative in a given direction, then things will tend to "roll downhill" at that point. On the other hand, to make things go uphill, you need to add energy.
The other thing we can get from this analogy is that given the opportunity, things will eventually move from a state of higher potential energy to a state of lower potential energy. So we can compare the energy at two different states (the reactants and products, for example), and decide based on the difference in potential energy whether the reaction would be spontaneous under those conditions.
The equation for Gibbs free energy is:
$G = H - TS$
Where $H$ is enthalpy (a.k.a heat of reaction), $T$ is absolute temperature (usually measured in Kelvin) and $S$ is entropy.
If you take the derivative at constant temperature and pressure, you get:
$\Delta G = \Delta H - T\Delta S$
(This is for large changes. for small changes replace $\Delta$ with $\delta$). What this tells us is that change in Gibbs Free Energy is a function of change in enthalpy, change in entropy, and the absolute temperature.
To illustrate how this works, take a look at the following diagram.

Here the line represents the Gibbs Free energy "surface" - the thermodynamic potential as a function of some variable (a reaction coordinate, for example). The ball represents the system at some point along that coordinate axis. If this were regular potential energy, it would be easy to see what happens - the ball will roll downhill if it gets the chance, and it will stop at the lower energy state. To get back up the hill, someone would have to put in that amount of energy. For thermodynamic potentials, it works the same way - the system will move along the coordinate axis in the direction of decreasing thermodynamic potential energy ($G$ in this case).
This means that under constant NPT conditions, any process that involves an overall decrease in Gibbs Free energy will be spontaneous. It will also tend to move in the "direction" that has the most negative slope in G at any given time. In mathematical terms,
$\Delta G < 0$ - overall process is spontaneous
$dG < 0$ - process will be moving in that direction
$dG = 0$ - process is at equilibrium
So how does this all fit in with what you described about systems trying to reach minimum energy, and maximum entropy? The answer is: those are both thermodynamic potentials for different types of systems. For a system with constant entropy, volume, and number of particles, (NSV) the internal energy (or total energy) is the thermodynamic potential. For a system at constant entropy and pressure (NSP), enthalpy is the thermodynamic potential. And for a system at constant volume and internal energy (NVE), the negative of entropy is the thermodynamic potential.
In other words, all of these thermodynamic variables are interconnected, and a change in one affects all of the others. You can hold three constant at any given time and still allow the system to "move" through phase-space. Which thermodynamic potential you need to describe how the system will move depends on which variables you choose.
Let's look at the equation for $\Delta G$ and see how changes in enthalpy and entropy affect it.
$\downarrow \Delta G = \space \downarrow \Delta H - T \Delta S$
If $\Delta H$ decreases, it will make $\Delta G$ decrease as well. This makes sense, since we know that exothermic processes tend to be spontaneous, because they are releasing energy and therefore the final system energy is lower than the initial.
$\downarrow \Delta G = \Delta H - T \space \uparrow \Delta S$
On the other hand, $\Delta G$ tends to decrease as $\Delta S$ increases - this is because the change in entropy is subtracted in the equation. This also matches up with what you know - an increase in entropy indicates a spontaneous process.
For your last question:

I am really having a hard time in figuring out a reaction where energy decreases and entropy increases. Can this happen?

Yes, it can! In fact, under these conditions, the process is guaranteed to be spontaneous - if $\Delta H$ is negative, and $\Delta S$ is positive, then $\Delta G$ has to be negative - the reaction would be spontaneous at any temperature under these conditions.
Let's look at the other possibilities:
$\Delta H > 0; \Delta S < 0$
In this case, since $\Delta H$ is always positive, and we are subtracting a negative $\Delta S$, the reaction can never be spontaneous.
$\Delta H > 0; \Delta S > 0$
$\Delta H < 0; \Delta S < 0$
In these two cases, the reaction could be spontaneous or it could be non-spontaneous - it depends on the relative magnitudes of the enthalpy and energy terms as well as the temperature at which it occurs.
To summarize - all thermodynamic variables are related to each other in fairly complicated ways. When you hold three of them constant (two if you don't count number of molecules), you can derive thermodynamic potential energy functions that describe the behavior of the system in terms of a single quantity. For most cases in chemistry, Gibbs Free Energy is the thermodynamic potential that we use. It gives us the relationship between enthalpy, entropy and temperature under constant pressure conditions. Since it's the thermodynamic potential, it also lets us predict how the system will behave - how it will move through phase space.

I feel that the Original Poster actually needed examples of :
$\Delta H < 0 ;  \Delta S > 0$
since majority of the common reactions do not fall under this criteria . If not then Thomij's answer is perfect for the understanding of Gibbs Free Energy .
So here are a few examples :

Hydrogen peroxide is thermodynamically unstable and decomposes to form water and oxygen
$\ce{2 H_2O_2 → 2 H_2O + O_2}$
$ \Delta H = -98.2~\mathrm{kJ\, mol^{-1}}$
$ \Delta S = 70.5~\mathrm{kJ\, mol^{-1}K^{-1}}$
Decomposition of Ammonium dichromate
$\ce{(NH_4)_2Cr_2O_7 → Cr_2O_3 + N_2 + 4 H_2O}$
$ \Delta H = -1794.9.2~\mathrm{kJ\, mol^{-1}}$
$ \Delta S > 0 $ due to formation of gases from the solid


text:
What should the Rehm-Weller equation look like?
The Rehm-Weller equation, which defines the energy change for photoinduced electron transfer according to Lakowicz "Principles of Fluorescence Spectroscopy" [1, p. 337] should look like:
$$\Delta G=E^\mathrm{red}(D^+/D)-E^\mathrm{red}(A/A^-)-\Delta G_{00} - \frac{e^2}{\epsilon d} \tag{1}$$
So you can see the first two components are reduction potentials of donor $D^+$ and acceptor $A$. But sometimes you can find another definition:
$$\Delta G=E^\mathrm{ox}(D/D^+)-E^\mathrm{red}(A/A^-)-\Delta G_{00} - \dfrac{e^2}{\epsilon d} \tag{2}$$
Which one is correct? I'm totally confused.
References

Lakowicz, J. R. Principles of Fluorescence Spectroscopy, 3rd ed.; Springer: New York, 2006. ISBN 978-0-387-31278-1.

Both expressions are identical. It's just juggling of the signs, since
$$E^{\mathrm{red}}(D^{\cdot+}/D) = - E^{\mathrm{ox}}(D/D^{\cdot+})$$
Relevant primary references would be:

Rehm, D.; Weller, A. H. Isr. J. Chem. 1970, 8, 259-271.
Rehm, D.; Weller, A. H. Ber. Bunsen-Ges. Phys. Chem. 1969, 73,
834-839.
Weller, A. H. Z. Phys. Chem. NF 1982, 133, 93-98.

A normative glossary of terms in photochemistry is available from the IUPAC.

Actually, both expressions are then not identical. There is a minus sign missing in the second one.

@tschoppi exactly. I was surprised when I saw that. It looks that some researchers does not understand something or what it is more likely, that I do not understand something.

By convention both donor (D) and acceptor (A) potentials are listed as reduction potentials. For a nice description, see Turro's explanation of the Rehm-Weller equation in his photochemical treatise "Modern Molecular Photochemistry of Organic Molecules"
To clarify, the equation that is being referenced is not actually the Rehm-Weller equation but is the "Gibbs free energy of photoinduced electron transfer".
Please refer to the IUPAC definition of Gibb's energy of photoinduced electron transfer.
The Gibbs energy of photoinduced electron transfer is part of the Rehm-Weller equation. The Rehm-Weller equation is the correlation between second order rate constants and Gibbs energy of photoinduced electron transfer.
See the IUPAC definition of the Rehm-Weller equation


text:
What is this glassware for a distillation apparatus called?
I'm messing around with building a solar distillation apparatus, and one of the parts I think I need looks kind of like the following, except I have no idea what it is called (and thus can't look it up to see if that site even sells them).

So, can anyone tell me what this thing is commonly called?

We called them kugelrohr bulbs.

It's part of a Kugelrohr distillation apparatus. It is more sophisticated and more expensive but more efficient for fractional distillation especially for higher boiling point compounds and probably easier to use when you know what you are doing. This is a link with a video  @ron you are so fast! I initially thought it looked like a glass column head/reservoir...

The "A" on the glassware gives it away as an Aldrich glassware component;
and Ron is correct calling it a Kugelrohr bulb. It is actually a single bulb ball tube.
This specific item can be found in the catalogue at Aldrich's website


text:
Formation of methyl esters
Can't we perform Fischer esterification using methanol to make a methyl ester from a carboxylic acid?
If so, why might we use diazomethane? Would the release of nitrogen gas confer an entropic advantage to the reaction? As I understand, the Fischer esterification process is an equilibrium process and at equilibrium there is often significant amounts of both product and reactant. So we might have to add excess alcohol to drive the reaction to the right.
I ask because my book just presents the use of diazomethane as a way of making methyl esters without any discussion of why might someone do this rather than just use a few gallons of methanol.
On the other hand I know from working in a research lab that methanol is poisonous and protocol calls for its use in a fume hood ... could that be another reason? On the other hand diazomethane is also toxic (and explosive to boot) ... so I would definitely be inclined to use that in a fume hood (perhaps more so than when using methanol).

Both methods are used in practice (and a a lot more exist). Diazomethane is a last resort method when everything else fails. For example you might have a molecule that is acid sensitive or heat sensitive and then $\ce {MeOH/H^+}$ is out of the question. Although diazomethane is an excellent reagent to do the job, as it works selectively with almost every substrate, to give good yields under mild conditions, generally no extensive work-up or purification are required, it has some major disadvantages: It is not commercially available, it has to be prepared fresh before the reaction using a (somewhat) specialised equipment and a toxic starting material (Diazald). It is explosive and very toxic and obviously can not be stored. An alternative to using diazomethane is TMS diazomethane that is stable enough to be stored as a solution and it is commercially available however it is not innocent at all as this article  shows.


text:
Electrophilic Addition Reaction
When HBr reacts with ethene it proceeds by the attack of proton on Ethene to give Ethyl carbocation and then attack of nucleophilic bromide ion to give Bromoethane.
My question is why does the proton react with Ethene earlier than the bromide ion? Is a proton more reactive than bromide ion or is there some other reason?

The mechanism of the addition of hydrogen bromide to ethene is called electrophilic addition. The region of highest electron density in ethene is the double bond, where the π electrons are. Hydrogen bromide acts as an electrophile toward the π electrons of the double bond. The proton of hydrogen bromide is positively polarized and thus electrophilic. (Obviously, the negatively charged bromide ion is not electrophilic.) Hence, the first step is an acid-base reaction in which hydrogen bromide donates a proton to ethene, forming a carbocation.

Also note that ethene is not usually reacted with $\ce{HBr(aq)}$. Ethene is usually reacted with $\ce{HBr(g)}$, in which case the acid is not ionized. In this case, the electron-rich (nucleophilic) part of ethene (the $\pi$ bond) reacts with the electron poor (electrophilic) part of $\ce{HBr}$ (the hydrogen atom).


text:
what is the acceptable percentage for food grade ethanol?Is 95% ethanol plus 5% water consider as food grade ethanol?
Is 95% ethanol plus 5% water consider as food grade ethanol?

Welcome to the site! Please take a moment to acquaint yourself with the customs here, primarily by sharing your thoughts on the matter you're asking questions about.

The grade of ethanol does not depend on the water and ethanol percentages, but rather on what else is in there. If your mixture is truly 95-5 ethanol-water, with nothing else in it, than that would be safe to consume (accounting for the health hazards of consuming too much ethanol of course).
However, there might be traces of e.g. methanol or benzene in the ethanol, which are both highly toxic to humans and therefore cannot be present in food grade ethanol mixtures. According to this website the FCC grade and above are food grade, in particular due to some heavy metal specifications, so apparently there might also be trace heavy metals in the ethanol.

Thanks Michiel for the explanation. If methanol considered as highly toxic if consumed by human, how about the food grade methanol? Can we used it in manufacturing of food?

In manufacturing yes, because you won't be `serving' the methanol. But food grade methanol is by no means meant for direct consumption


text:
Galvanic cell using NaCl, Cu and Zn, why does it work?
Got a paper due tomorrow in chemistry, and my friends and I are totally stuck on this:
Basically we mixed some regular salt (NaCl) into a glass of water and stuck a zinc nail and a copper nail into the solution. We connected the nails and measured a voltage of .6V. Why does this happen?
I understand galvanic cells of other types and electrolysis to a pretty OK extent, but I can't for the life of me understand what's happening at a chemical level in this reaction. I also need chemical formulas and the likes. Thanks in advance!

I want you to try and work out the details yourself, but there's one aspect of this kind of cell that's not straightforward. At one electrode, you have a reduction reaction and at the other, an oxidation, just like any other galvanic cell. In the oxidation reaction, one of the metal electrodes is oxidized to put cations in solution. (think about which metal is more easily oxidized) In the type of galvanic cell that you may be more familiar with, a metal cation in solution is reduced onto the other electrode, but in this case, the only metal cation is $\ce{Na+}$, which is extremely difficult to reduce. Therefore, some other reduction reaction must be happening. The only other reducible substance present is water.
The two reactions occurring can be found on this table, and using the potentials in the table, you should be able to work out which reactions are occurring and at which electrodes.

Good answer, thanks, but I believe I'm not quite yet informed enough on this subject to quite understand what's happening. This is my first semester of science (meaning only a part of it is chemistry) at school, and I think my teacher made a mistake in assigning this task to us. I found a site that helped me at least get some kind of a solution to this problem, but I'm not 100% sure it's correct: http://sci-toys.com/scitoys/scitoys/echem/batteries/batteries.html

That is almost the same cell you're thinking of. Unfortunately, their explanation is wrong with only NaCl electrolyte. It's the same as the classic lemon battery, though. Take a look at this: http://surface.syr.edu/cgi/viewcontent.cgi?article=1001&context=che


text:
Why doesn't rubidium react with nitrogen gas?
I ran across a problem in which I was asked to predict the result of a reaction between $\ce{Rb}$ and $\ce{N2}$. I was frustrated until I found a note in my textbook which stated that out of all the alkali metals, only $\ce{Li}$ reacts with $\ce{N2}$. I am curious to know the reason for this.

Lithium is the only group 1 element that can form a stable alkali metal nitride (at standard conditions), the other group 1 elements (such as sodium, potassium, rubidium,...) will react (at different reaction conditions) to form their respective nitrides (in the case of rubidium, $\ce{Rb3N}$ (rubidium nitride)), but the compound will be unstable and decompose quickly.  The reactions of the alkali metals (other than lithium) with nitrogen would not release enough lattice energy and would thus be endothermic, so they do not form nitrides at standard conditions
Read more here : http://en.wikipedia.org/wiki/Alkali_metal#Nitrides_and_pnictides


text:
Enolizablity of pyruvate and lactate
I understand that to create an enol, you simply eliminate a proton alpha to the relevant $\ce{C=O}$ and then tack the proton onto the $\ce{O}$ and there you have it — an alkene and an alcohol — or enol.
Given this, between pyruvate and lactate, which is more enolizable?
I'm thinking that pyruvate doesn't even have an enol tautomer … there are no protons alpha to a carbonyl group. Sure, there is the carboxylate group … but that doesn't have anything to do with keto-enol tautomerism (does it?)

You're right and wrong.  Your right that pyruvate anion has a higher enol content than lactate anion, at least that would be my guess (pyruvate anion has 3 enolizable hydrogens, while lactate anion has one and needs to destroy the carboxylate resonance in order to create the carbonyl required for enolization...read on).
But I think that lactic acid could enolize, after all esters can enolize.  I suspect that the enol content in the lactate carboxylate anion would be non-measurable - you would need to disrupt the carboxylate anion resonance in order to generate the carbonyl required for enolization.  If we use 36 kcal/mol as an estimate of the carboxylate resonance energy and say that for every 1.4 kcal/mol we can shift an equilibrium by a power of 10, then we might expect 1 enol molecule in every $10^{26}$ lactate carboxylate molecules.  In the lactic acid itself, we have no carboxylate resonance to disrupt, the enol content would be higher.


text:
Potassium permanganate oxidation of sugars
I am told that all three sugars below can be oxidized to their corresponding carboxylic acids using potassium permanganate.
This strikes me as bloody wrong because only hemiacetal sugars interconvert in solution with their open chain forms, and we need that open chain form to access the aldehyde/ketone for oxidation. Potassium permanganate as far as we know creates C-O bonds while not breaking any C-C bonds ... hence ... the general rule in our class is that ketones cannot be further oxidized (further oxidation would rupture C-C bonds).

Each of your molecules also contains at least one primary alcohol appendage.  Primary alcohols are readily oxidized by potassium permanganate to the corresponding carboxylic acid (see this link for more information).


text:
Why hybridization occurs in diatomic molecule
In diatomic nitrogen
There are two electron configurations I'd like to compare:

1s2 2s2(lone pair) 2p3(the bonding electrons sigma,pi,pi)
1s2 2sp(lone pair) 2sp(sigma bonding) 2p2(pi,pi bonding)

Why is option 2 favored over option 1?
I thought having the lone pair in a 2s orbital would be favored over 2sp.

I thought having the lone pair in a 2s orbital would be favored over
  2sp

That's true, placing the lone pair in a 2s orbital would be put the lone pair electrons at lower energy then if they were in a 2sp orbital.  However you're only looking at a part of the puzzle.  As you note, in this case the sigma bonding electrons would be in a pure p orbital and they would then be higher in energy then if they were in an sp orbital. You need to optimize the total energy of the molecule, so you need to consider all electrons and find the hybridization that gives the lowest possible energy to both the bonding and non-bonding electrons.
Which hybridization scheme a molecule adopts is the result of a delicate balance between electron stabilization and other factors such as bond strengths, steric and electrostatic effects.  The balance of these factors can shift and that is why some molecules hybridize one way and other molecules hybridize differently.


text:
Imine stability in acids and bases
I understand that imines can be hydrolyzed in acidic medium. But what about in basic medium? Can imines still be hydrolyzed?
I have a plausible mechanism ... nitrogen's withdrawal of electron density makes the alpha proton relatively acidic; base abstracts the proton in an E2 reaction; nitrogen then takes back a proton to regenerate the base catalyst.
Is this correct? Does this actually happen?

In analogy to a carbonyl, why not attack directly at the imine's electrophilic carbon with your basic nucleophile to create a tetrahedral intermediate?  Your proposed removal of an alpha-proton does occur in the enamine formation process.


text:
Osmium tetroxide and meta-chloroperoxybenzoic acid
Can these be used to induce acetal ring closures? For example, when dealing with a molecule that has both a C=O functional group and an alkene as part of its structure and there is the possibility of creating a 5 or 6 membered ring through intramolecular -OH attack on the C=O ... can we just use the $\ce{OsO4}$ to create the diol and expect it to provide us sufficient acidity to catalyze ring closure?
Same question goes for meta-chloroperoxybenzoic acid.

It's not unusual for hydroxy-esters to cyclize and form a lactone especially if a 5- or 6-membered ring can be formed.  With a diol (from osmium tetroxide) you might get a thermodynamically controlled mixture of products. M-chloroperoxybenzoic will produce an epoxide. Acid catalyzed opening of the epoxide could yield stereoisomers which could also lead to a mixture of possible lactones.

@ron but will the osmium tetroxide be acidic enough to form an lactone?

Just a guess, but I would think so. If the lactone is thermodynamically more stable than the hydroxyl-ester, then closure should occur under neutral conditions.

With osmium tetroxide, no it does not appear that dihydroxylation and acetalization can be done simultaneously. Usually dihydroxylation is carried out with catalytic osmium tetroxide with a stoichiometric amount of N-methylmorpholine N-oxide (NMO). The latter regenerates the Os(VIII) species necessary to carry out the dihydroxylation and results in the formation of N-methylmorpholine, which is a weak base. Without an acid available for catalysis, the acetalization does not occur. The sequence can be accomplished in "one-pot" however. In practice, you can monitor the progress of the dihydroxylation reaction. When it is deemed complete, add catalytic acid directly to the same reaction mixture, which allows the acetal formation.
Reference: 10.1016/j.tetlet.2012.02.045
Similarly, the overall transformation has been accomplished with mCPBA, but it's not clear if happens without any further intervention. In the referenced paper below, it's not entirely clear to me when the epoxide opens, because the procedure specifies that the epoxidation step is carried out with exclusion of water. It's likely that the epoxide is stable until addition of water, which opens the epoxide to the diol and then the acetalization can occur, catalyzed by the meta-chlorobenzoic acid byproduct from the epoxidation. In some cases, they simply isolate the epoxyketone as well.
Reference: 10.1002/ejoc.201101087

Can dihydroxylation also be carried out with OsO4 and HOOH? Prof never heard of NMO.

I think so. One advantage of NMO is that you can buy it as the pure compound, while peroxide has to be purchased as a solution. In my experience, the concentration of peroxide is not always reliable.