text:
Energy and time considerations in chemical reactions
I'm working on hydrothermal decomposition of organic matter at high temperature and pressure (400 bar). The reactions are too complex and numerous to model individually, but proceed in a similar fashion at different temperatures.  If the reaction proceeds to the same point in 1 minute at 300degC and 1 day at 100degC, is there any way to use this information to estimate the amount of 'energy' required to do the work?  I know this is a strange question, but I want to see if the same decomposition is achieved using electrochemical energy in an ion trap. I'm not sure how I'm going to be able to compare the systems in terms of energy applied. I am quite sure that the hydrothermal decomposition depends on the force of the bonds being broken, as does the electrochemical decomposition...
Am I going down the wrong rabbit hole to calculate the specific energy of the water at this temperature?  How would time come into the equation then?

Does the Arrhenius equation not give you what you need?

Ordinarily it would - I could find an activation energy for the reaction that way... But this is a complex mixture of thousands of reactants and products, making it impossible to find something like a first-order decay constant. What I can track is the time difference to achieve the same amount of overall decay at several temperatures.

Using the Arrhenius equation will give you an effective or average activation energy that you can then use to compare with other methods.


text:
If fluorine has a lower electron affinity than chlorine, why does it have a higher ionization energy?
I have read that fluorine has a lower electron affinity than chlorine despite its lower atomic radius because its electron cloud is extremely dense. If this is the case, shouldn't the ionization energy of fluorine be lower than that of chlorine, since fluorine's dense electron cloud should eagerly repel its electrons?
Even more confusing is that electron affinity can also be defined as the ionization energy of the singly charged anion form of an atom, i.e. the ionization energy of $\ce{Cl-}$ is equal in magnitude to the electron affinity of $\ce{Cl}$. Therefore, it seems that the trends for ionization energy and electron affinity should always match, which is not the case with fluorine and chlorine.

The first ionisation energy is the energy required to remove the most loosely held electron from...

The second part is a duplicate of this question.

Notice that ionisation energy and electron affinity are two only very weakly connected physical constants.

To quote chemguide:

The first ionisation energy is the energy required to remove the most loosely held electron from one mole of gaseous atoms to produce 1 mole of gaseous ions each with a charge of 1+.

That means, the ionization energy of fluorine is the energy of the following reaction:
$$\ce{F -> F+ + e-}$$
The ionization energy of chlorine is the energy of the following reaction:
$$\ce{Cl -> Cl+ + e-}$$
Since the most loosely held electron in fluorine is closer to the nucleus than chlorine, it takes more energy to remove the electron from fluorine than in chlorine.

The ionization energy of chloride ion ($\ce{Cl-}$) is the energy of the following reaction:
$$\ce{Cl- -> Cl + e-}$$
The electron affinity of $\ce{Cl}$ is the energy of the following reaction:
$$\ce{Cl + e- -> Cl-}$$
As a result, their energies are equal in magnitude but opposite in sign.

Why does chlorine have a higher electron affinity than fluorine?
To quote this answer:

Fluorine, though higher than chlorine in the periodic table, has a very small atomic size. This makes the fluoride anion so formed unstable (highly reactive) due to a very high charge/mass ratio. Also, fluorine has no d-orbitals, which limits its atomic size. As a result, fluorine has an electron affinity less than that of chlorine.

As can be seen in the second photo, this effect seems to be true for Period 2 elements.

(source: lardbucket.org)

(source: lardbucket.org)

In that case, why is it that Cl has a lower ionization energy than F, but Cl- has a higher ionization energy than F-? (This is what I was trying to get at in the 2nd part of my question.)

@jt2000 That is answered here.

This is half an answer at best, as it doesn’t compare fluorine’s and chlorine’s electron affinity.

@Jan See the comment just above you.

@user34388 An answer should be self-containing and not rely on comments or other answers to be understood. Please edit in the relevant parts. (I’m very inclined to upvote this answer but it’s just not complete enough for me yet.)

@Jan I didn't want to include it because that part would be a duplicate. I included it notwithstanding.

@jt2000 The link I provided explains why Cl has a higher electron affinity than F, which is the same as why Cl- has a higher ionization energy than F-.

@user34388 The link you provided doesn't explain why Cl has a lower ionization energy than F, it only explains why Cl- has a higher ionization energy than F.- Also, I do understand why Cl has a higher electron affinity than F, but I'm wondering why the logic used to explain this (small atomic size) does not apply to why the ionization energy of Cl is less than that of F. It seems to me that because F has a lower electron affinity because of its small atomic size, it should also attract its own electrons less, resulting in a lower ionization energy than Cl as opposed to higher.

@jt2000 As Jan said, "Notice that ionisation energy and electron affinity are two only very weakly connected physical constants. "

@user34388 Is there a particular reason why this is true, considering that they seem closely related in physical terms, their only difference being that one relates to the singly charged anion while the other relates to the uncharged form?

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text:
Can an SN1 reaction occur in compounds with the leaving group attached to a primary carbon, which can later rearrange?
Can SN1 reactions occur in compounds where the carbocation, even though is is initially $1^\circ$, can rearrange to form a more stable carbocation?
Examples cases:

Under typical lab conditions a primary carbocation is not likely to form unless there is stabilization by an adjacent double bond or benzene ring, or unless there is a neighboring group that can provide neighboring group stabilization - but then we really don't have a simple primary carbocation in such cases.  BTW, in your first example, a H shift would produce an antiaromatic (4n electrons) cyclopentadienyl cation, so this won't occur.

Well, a benzyl cation is technically a primary one, so …

Fairly sure neopentyl iodide reacts with aqueous Ag+ to give the rearranged product t-amyl alcohol

@ron Do you mind posting an answer? Thanks!

Your thinking is correct that they can rearrange and then form stable carbocation (though your first example is wrong as it would become anti-aromatic after rearrangement) ,
but the answer to your question is NO , they would be substituted by SN2 mechanism being
1 degree . The reason behind this is that the compound will not try to form a carbocation first where it is not that stable , you can think of it in this way that at the initial stage the compound doesn't know that it can rearrange later to form stable carbocation, this is something that we can see but not the compound and thus it will go by normal SN2 mechanism.

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text:
Effect of protecting group on diastereoselectivity of LiAlH4 reduction of α-hydroxyketones

In the reactions above, taken from Tetrahedron Lett. 1982, 23 (23), 2355–2358, the choice of protecting group on the α-hydroxyl group has a large influence on the diastereoselectivity of reduction.
Why does the benzyloxymethyl protecting group lead to the formation of the anti diol (consistent wiht the Cram chelate model), whereas the tert-butyldiphenylsilyl protecting group gives the syn diol (consistent with the Felkin–Anh model)?

In order for the Cram chelate product to predominate after the addition of hydride donor to a chiral carbonyl compound, which contains a heteroatom in the alpha position, this heteroatom and part of the reagent must be able to form a five-membered ring chelate. If this is not possible, one observes Felkin-Anh proudct.
In the mentioned reaction, when protecting group is not bulky it form five-membered ring chelate easily with Li+ after addition of Hydride to carbonyl resulted in cram chelate product as major product. However, when bulky protecting group is present it doesn't five-membered ring, consequently resulted in Felkin-Anh product as major product.

I think Jan did an excellent job of answering your question, however I am writing a slightly longer answer largely with more details concerning the Felkin–Anh rule, just so that someone who is not already familiar with them can learn something too.
Assuming that there isn’t an unusually electronegative atom on the carbon next to the carbonyl, the largest group prefers a conformation where it is perpendicular to the carbonyl moiety. This gives two relatively competitive lowest energy conformations, with the medium and smallest groups differing in their proximity to the carbonyl oxygen.
For the TBDPS-protected substrate shown above, the OTBDPS group would be considered the largest group. The medium and small groups are methyl and hydrogen respectively. This placement is also consistent with a modification to the Felkin–Anh model, which places electronegative substituents (such as –NR2 or –OR) perpendicular to the carbonyl (an explanation is given at the bottom of the answer). In any case, the two conformations of interest are the two shown below:

Moreover, the nucleophile doesn't prefer to attack the carbonyl carbon at a 90° angle; rather it prefers something closer to the tetrahedral angle (Bürgi–Dunitz trajectory). Naturally, the nucleophile prefers attack from the opposite side to the large group. This leaves two possibilities: for the nucleophile to attack over the small group, or for it to attack nearest to the medium group. The former is preferred, and the major diastereomer will be the syn diol:

On the other hand, Lewis basic groups such as $\ce{-OCH2OBn}$ are also capable of forming a chelate with the metal together with the carbonyl oxygen. This feature forces the $\ce{OR}$ group to be held parallel to the carbonyl. Nucleophilic attack still occurs along the Bürgi–Dunitz trajectory, and the nucleophile attacks opposite the larger of the two remaining groups. This will invert the stereochemical outcome as compared to the Felkin–Anh model previously described, and hence leads to the formation of the anti diol.

Now, to answer your question of why does the choice of protecting group matter. The previous discussion illustrates that the reason we get the Cram Chelate product is the chelation of the metal by the $\ce{OR}$ oxygen and the carbonyl. This chelation can take place easily if the $\ce{R}$ group is small, but is greatly hindered with a bulky alkyl/silyl substituent, in which case we get the product determined by the polar Felkin–Anh relationship.
The bulk of the TBDPS group does indeed prevent chelation, and you get the observed stereochemistry in accord with the Felkin–Anh rule. However, chelation is possible when BOM is the protecting group (Jan already offered an excellent explanation), and you end up with the Cram chelate product.

When there is an electronegative atom or group α to the carbonyl, there is a slight change to the Felkin–Anh model which may be applied. This ‘slight change’ is that this electronegative group becomes the equivalent of the ‘large’ group in the Felkin–Anh model.
The reason for this that the energy of the C–X σ* antibonding orbital is rather low, and so when the C–X bond is perpendicular to the carbonyl group, this can overlap with the π* MO of the carbonyl group to make a new, lower energy LUMO.
In "human" speak, this means that this conformational arrangement is more
reactive than any other. With this change in mind, the Felkin–Anh type addition now proceeds as before. However, the stereochemical outcome will generally be different from the original Felkin–Anh model.
For more information, see: How is the Felkin-Anh model of stereoinduction correctly explained with MO theory?

The $\ce{TBDPS}$ group (tert-butyldiphenylsilyl) is a very bulky protecting group. A silicon atom is bonded to the oxygen, which already acts as a larger carbon atom. That is then substituted by three alkyl or aryl groups, one of which is the very bulky tert-butyl group. There is a lot of congestion going on next to that oxygen atom, so its ability to coordinate to Lewis acids (including via hydrogen bonds) is severely limited. Therefore, it is mainly the polar Felkin–Anh relationship that dictates the stereochemistry of the attack in the case of a $\ce{TBDPS}$ group.
On the contrary, the benzyloxymethyl (BOM) group $\ce{CH2OBn}$ used in the other example has a primary carbon attached directly to the protected oxygen. Not only is carbon a lot smaller than silicon, but it also features two hydrogen substituents ($\ce{CH2}$) and only one substituent group – an oxygen which itself only bears one other group. Therefore, the BOM-protected oxygen can still coordinate to Lewis acids easily, such as the lithium cation or aluminium as present in the example. Thus, the Cram chelate model is feasible in this case and correctly predicts the observed stereochemistry.

This is quite often the case that some protecting groups allow for chelation while others don’t. Pretty much all the silicon-based protecting groups (those that include a $\ce{O-Si}$ bond) will inhibit chelation due to their size. Typically anything that starts off as $\ce{CH2}$ will allow chelation. A non-exhaustive list of common groups of this type would be:

MOM
SEM
Bn
PMB

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text:
Why is the value of specific rotation different in different cases?

An unknown compound with a mass of $4.5~\mathrm{g}$ is dissolved in enough carbon tetrachloride to make a total volume of $250~\mathrm{cm^3}$. The observed  rotation of this solution is $+357.75^\circ$ in a $25~\mathrm{cm}$ cell using the sodium D line. But if $4.5~\mathrm{g}$ is dissolved in $125~\mathrm{cm^3}$ we observed  rotation is $+355.50^\circ$. Calculate specific rotation for this compound. (assuming length of polarimeter tube is $1~\mathrm{dm}$)

My approach is starting with the formula stated on Wikipedia.
$$\text{Specific rotation (in first case)} =
\frac{+357.75^\circ}{\left(\frac{4.5}{250}\right)(1)}=19875^\circ=75^\circ$$
$$\text{Specific rotation (in second case)} =
\frac{+355.50^\circ}{\left(\frac{4.5}{125}\right)(1)}=9875^\circ=155^\circ$$
Any idea why the two answers are different? Did I miss out something?

Note that the concentration is doubled in the seconds case. Therefore, the expected rotation should be $2\times357.75^\circ=715.50^\circ$, which is more than a full circle. Substracting one circle $(360^\circ)$ gives the indicated $715.50^\circ-360.00^\circ=355.50^\circ$.

@Loong but the length changes in the second case too!You did'nt take that into account...Initially it was 2.5 dm but finally 1 dm.

Oh.My bad....25 cm is cell length!

Yes, the remarks “in a $25\ \mathrm{cm}$ cell” and “assuming length of polarimeter tube is $1\ \mathrm{dm}$” are confusing. If $2.5\ \mathrm{dm}$ was used in the first case, and $1.0\ \mathrm{dm}$ was used in the second case, then this factor has also to be taken into account.

The "true" rotation may be the experimentally observed rotation or the experimentally observed rotation +/- (n*180°).  You must find at what multiple of 180° the experimental rotations are equal; this is the "true" rotation.

@ron You mean a rotation of $+181^{0}$ should be written as $-179^{0}$ ?

No, an experimentally observed rotation of +181° may actually be +181, +1, +361, -179, etc. That is why the experiment is run at 2 concentrations, then you can solve the equation and determine the "true" rotation.

Wait a sec!I did'nt get you @ron Which equation should I solve? (You may take the question in the post as an example if you wish )

You mean something like $\theta_1 + n\pi=\theta_2+m\pi$ ?But there can be infinite solutions of such an equation...

There is a second condition relating to the change in concentration and path length.  For example, you measure a rotation of 10°, this may really be 10, 190, -170, etc.  Now, repeat the experiment at 1/10 the concentration - is the observed rotation 1, 19, -17, etc.

@ron I'm still not sure if I understood you.Suppose in the first case we get a specific rotation $75^{0}$ and in second case we get a  specific rotation $155^{0}$ what will be the true rotation according to you?

If indeed the path lengths are equal, then rot_1=357.75 + (n180) and, due to concentration 2rot_1 = rot_2.  Solving we find that n=-2, therefore rot_1 = 357.75 +(-2*180) or -2.25.  This "true" observed value may now be converted to the specific rotation using your formula.

@Loong I took the liberty to correct that O.o

If indeed the path lengths are equal, then $rot_1=357.75 + (n*180)$ and, due to concentration $2*rot_1 = rot_2$. Solving we find that $n=-2$, therefore $rot_1 = 357.75 +(-2*180)$ or $-2.25$. This "true" observed value may now be converted to the specific rotation using your formula.
-by @ron in the comments above.

A polarimeter is typically a long cylindrical tube with flat ends. I think the 25 cm was referring to the diameter of the tube. The other wrinkle here is that the angle measurement may be + or -. So 357.75 may in fact be -2.25.
Specific Rotation (in first case)=$$\frac{-2.25^{0}}{(\frac{4.5}{250})(1)}= - 125^{0}$$
Specific Rotation (in second case)=$$\frac{-4.50^{0}}{(\frac{4.5}{125})(1)}=-125^{0}$$

Could you check my calculation in the original question please?Why is the answer not matching if I take the original values with +ve sign?

@ZOZ - How can I do the calculations wrong and get a good answer?

I never said your answer is wrong or your calculation is wrong.I'm asking why is my calculation wrong?Where did I go wrong?See the question...

Your problem was that you assumed that the reading for the solutions were positive when the are in fact negative. Think of a 360 degree compass. If you draw an angle of +357.75 degrees that is the same as drawing an angle of -2.25 degrees.

In other words think of a 360 degree compass from geometry class. They are graduated from 0 to 360. There are no negative markings. A polarimeter is setup the same way. There are no negative angles on the actual dial.

So you mean if the observed angle is >$180^{0}$ then we must take angle as negative in the calculation? I've never read something like that..could you cite a source?

Molecules can rotate light + or -. If a solution rotates light -10 degrees, then half the concentration will rotate the light -5 degrees. But my readings will be 350 degrees for the first solution and 355 degrees for the second solution since a  polarimeter doesn't have - degrees.


text:
What exactly is hydrolysis reaction in organic chemistry?What mechanism does it follow?
According to my study material:

Hydrolysis is a special type of nucleophilic substitution ($\mathrm{S_N1}$) where water acts as both nucleophile and a solvent molecule.

Is this definition correct? If so how to solve the following problem (picked up from my textbook exercise which has no solution)?

When alkyl bromides (listed here) were subjected to hydrolysis in a
  mixture of ethanol and water ($80\ \%\ \ce{C2H5OH} / 20~\%\ \ce{H2O}$)
  at $55~\mathrm{^\circ C}$ the rates of reaction showed the following order:
  $\ce{(CH3)3CBr} > \ce{CH3Br} > \ce{CH3CH2Br} > \ce{(CH3)2CHBr}$.
Provide an explaination for this order of reactivity.

But my doubt is how is this order $\ce{(CH3)3CBr} > \ce{CH3Br} > \ce{CH3CH2Br} > \ce{(CH3)2CHBr}$ possible?
I mean I know the first compound will form a tertiary carbocation. That is okay for $\mathrm{S_N1}$ but how will the second compound undergo $\mathrm{S_N1}$ (because methyl carbocation is very unstable)? Moreover if really $\mathrm{S_N1}$ is taking place the fourth compound should form a more stable carbocation compared to the third compound. These things are confusing me. Is hydrolysis really $\mathrm{S_N1}$ or something else?
According to me the order should have been $(1)>(4)>(3)>(2)$.

Do you call it a hydrolysis reaction? ... Because I just call it a substitution reaction...

@DHMO I stated what my textbook and study material wrote.I did'nt make up the definitions myself.See https://en.wikipedia.org/wiki/Solvolysis

Hydrolysis is not necessarily SN1, it can also refer to a SN2 process (especially if there is OH-). Or it can be any process really. There are many different ways of hydrolysing esters, for example. However, in this particular case, it is weird. I would expect only SN1 reactivity under the stated conditions, so I don't know why MeBr is so high up.

Your textbook is almost correct. Here is the corrected version of its statement:

Hydrolysis is a special type of nucleophilic substitution where water acts as both nucleophile and a solvent molecule.

Note me omitting the $\mathrm{S_N1}$ part. This is because a hydrolysis mechanism can be any nucleophilic substitution, whether $\mathrm{S_N1, S_N2, S_N}$ for heavy atoms or $\mathrm{S_NAr}$.
That also explains the given order of reactivity. tert-Butyl bromide is very fast at $\mathrm{S_N1}$ reactions. Bromomethane very rapidly undergoes $\mathrm{S_N2}$ reactions. The other two are in-between with bromoethane being just a tad better at $\mathrm{S_N2}$ reactions than 2-bromopropane is at $\mathrm{S_N1}$ reactions.
($\mathrm{S_N1}$ is preferred in 2-bromopropane as the solvent is polar protic and it can stabilize the intermediate carbocation well)

By the extension of your definition: In the solvolysis definition in Wikipedia too should the $\mathrm{S_N1}$ be removed? See https://en.wikipedia.org/wiki/Solvolysis

@ZOZ Yes.$%MathJax comments hurray!$

Morever,I cannot understand how can water in this case act as nucleophile in 20 percent ethanol.Water is considered to be a weak nuceophile.Can it really initiate a $\mathrm{S_N2}$ ?

@ZOZ Water isn’t that weak a nucleophile, tbh.

Well by that logic if it were 80 % ethanol and 20 % water would ethanol have acted as the nucleophile? I mean the ethyl group exerts a $+I$ effect on the O atom.So would ethanol be nucleophile in that case? @Jan

@ZOZ But ethanol is bulkier, allowing water to be a kinetically better nucleophile.

Thank you for clearing the doubt.I was really confused with that Wikipedia definition.Maybe I should edit that Wiki page :-P.


text:
Mustard gas hydrolysis (at a very high rate)

Explain mustard gas hydrolysis at a very high rate:
$\ce{Cl-CH2-CH2-S-CH2-CH2-Cl}$

I tried searching online but did not find this reaction anywhere. Can someone explain me the steps/mechanism of this reaction? Even some hints might help.
My attempt:
According to Wikipedia

Solvolysis is a special type of nucleophilic substitution ($\mathrm{S_N1}$) or
  elimination where the nucleophile is a solvent molecule. For certain
  nucleophiles, there are specific terms for the type of solvolysis
  reaction. For water, the term is hydrolysis; for alcohols, it is
  alcoholysis; for ammonia, it is ammonolysis; for glycols, it is
  glycolysis; for amines, it is aminolysis.

It clearly states that hydrolysis/solvolysis can only be $\mathrm{S_N1}$ or elimination. However I'm wondering how can $\mathrm{S_N1}$ occur on a primary carbon! Morever water doesnt seem to be a strong enough base for elimination. So what will be the preferred reaction mechanism?

The cation is somehow stabilized via S. Without S, it would be just ethyl chloride which hydrolyzes pretty slowly, as expected.

You mean SN1 occurs?What will be the end product?Can you link the mechanism?

Luckily I got the mechanism via Google Image Search.

Internal attack by sulphur atom takes place during the reaction. This is called anchimeric assistance, neighbouring group effect or $\mathrm{S_N{NGP}}$. Wikipedia describes it well.

Yes, an $S_N1$ reaction takes place. As, @Ivan pointed out, if it weren't for the it would just be like ethyl chloride. The key to this whole process is the presence of lone pair of electrons on sulfur which it uses to stabilise the primary carbocation (by forming the triangular cation depicted).

@getafix I would rather say it an extended version of $\ce{S_N1}$ i.e. $\ce{SN_{NGP}}$.It's not strictly $\ce{S_N1}$.

Of course I agree. There is clear neighbouring group participation. Sorry, my comment did not make it explicit. However, I never knew $SN_{NGP}$ was a term..learn something new everyday

@getafix Never mind.We're all here to learn :-)!

Indeed. What I meant is that the rate determine step is unimolecular, i.e $S_N1$


text:
Comparing rates of hydrolysis in cyclohexane ring (involving neighbouring group participation)

I need to compare the rates of hydrolysis of these three compounds.

MY ATTEMPT:

By definition hydrolysis is either $\ce{S_N1}$ or elimination.But water being a weak base (and weak nucleophile) $\ce{S_N1}$ is preferable.
$\ce{PhS^-}$ is a better leaving group than $\ce{Cl-}$ due to mesomeric effects.
Comparing with the middle compound the leftmost and rightmost compound will certainly have higher rate of hydrolysis due to presence of good leaving group $\ce{PhS^-}$
But how to compare between the leftmost and rightmost compound's rates?

You asked this before. http://chemistry.stackexchange.com/questions/58888/is-the-sn1-reaction-faster-with-an-axial-or-equatorial-leaving-group

In fact, no, this is slightly more complicated. I would expect there to be some degree of neighbouring group participation by the SPh (forming a sulfonium ion, followed by water attacking) This can only happen in the trans configuration, however it requires ring flip into the higher-energy trans-diaxial conformation. Complicated.

@orthocresol We don't need to worry about the ring flip I guess as it will take place automatically.

Ring flip definitely happens. The only issue is that the trans-diaxial conformation will be less populated, which will affect the rate negatively. I can't determine just from this alone which isomer is going to react faster. In the absence of NGP the cis isomer is faster, but NGP speeds up the trans isomer reaction. Is it enough to make it faster than the cis? I would guess yes, but I don't know.

@orthocresol I know it might be less populated.However internal attack mechanism rate supersedes all the external attack mechanisms.So $rate_3$ must be highest by any logic...

"When NGP is in operation it is normal for the reaction rate to be increased." -Wiki https://en.wikipedia.org/wiki/Neighbouring_group_participation

@orthocresol If you want more proof here you go http://meskvmcollege.org/upload/study_material/30275.pdf . See page 10 of the pdf. Reaction rate in trans is 670 times greater than cis (in that case).

The third reaction infact occurs via neighbouring group
  participation.

(https://en.wikipedia.org/wiki/Neighbouring_group_participation)

In the third compound $\ce{S}$ atom attacks carbon containing $\ce{Cl}$ by $\ce{SN_{NGP}}$ forming sulphonium ion. An attack by $\ce{H2O}$ on the least hindered carbon of the three member ring occurs. In this case both the carbopns in three member ring will be similar though. So the third compound has highest rate of reaction.

(I'd like to add "This can only happen in the trans configuration, however it requires ring flip into the higher-energy trans-diaxial conformation." as pointed out by - @orthocresol in the comments to the orginal question.)

Adding as a proof that trans reacts faster than cis form during NGP in cyclohexane ring.
  http://meskvmcollege.org/upload/study_material/30275.pdf (See page 10
  of the pdf)

The first compound has second highest rate. Anchimeric assistance cannot occur in cis form. Negative inductive effect by $\ce{Cl}$ enhances the rate. Reaction mechanism is $\ce{S_N1}$ and water itself acts as nucleophile.
The central compound has the lowest rate among the three. $\ce{S_N1}$ mechanism occurs and $\ce{Cl}$ is replaced by $\ce{OH}$ where water itself acts as nucleophile.

So summing up: $$\ce{rate_3>rate_1>rate_2}$$


text:
Bond energy and s-character
While studying the basics of chemical bonding, I've often come across the following statements:
1. The bond energy increases with the increase in the s-character in a hybrid orbital, i.e sp3 sp2-sp2, and so on.
I can't seem to correlate the two. If a hybrid has a higher bond energy, shouldn't it also form stronger bonds?

Let us start with the following most important part of the IUPAC definition for the bond energy.

bond energy (mean bond energy)
The average value of the gas-phase bond dissociation energies (usually
  at a temperature of 298 K) for all bonds of the same type within the
  same chemical species.

I think it should be immediately clear now that the higher a bond energy is, the stronger is the bond, so that there is no any contradiction in a bond energy increase with the increase of the $\mathrm{s}$-character in hybrid $\mathrm{sp}^n$-orbitals.
Indeed, $\mathrm{2s}$-orbitals are lower in energy that the $\mathrm{2p}$ ones, and thus, the more $\mathrm{s}$-character a hybrid orbital has, the lower is its energy: $E_{\mathrm{sp}} < E_{\mathrm{sp^2}} < E_{\mathrm{sp^3}}$. The lower an atomic or hybrid orbital energy is, the lower is the energy of a molecular orbital in formation of which it participates, $E_{\sigma_\mathrm{sp}} < E_{\sigma_\mathrm{sp^2}} < E_{\sigma_\mathrm{sp^3}}$, and consequently, the higher is the corresponding bond dissociation energy, as well as, the average of them, i.e. the bond energy. So, the bond energy indeed increases as the s-character grows, say $E_{\ce{C(\mathrm{sp})-C(\mathrm{sp})}} > E_{\ce{C(\mathrm{sp^2})-C(\mathrm{sp^2})}} > E_{\ce{C(\mathrm{sp^3})-C(\mathrm{sp^3})}}$.
Below is the MO diagram for $\sigma$ $\ce{C-C}$ bond between two $\mathrm{sp}$-, two $\mathrm{sp^2}$-, and two $\mathrm{sp^3}$-hydridized $\ce{C}$ atoms that illustrates the logic above.

@DHMO: if the sp hybrid orbital is stronger than the sp3 orbital, why is the sp-sp bond weaker than the sp3-sp3 bond?

@Kōhai, first note that orbitals aren't stronger/weaker relative to each other, but rather, they are lower/higher in energy. Now, the lower a bonding orbital is, the stronger is the corresponding bond. Finally, who  told you that $\ce{C(\mathrm{sp})-C(\mathrm{sp})}$ bond is weaker than the $\ce{C(\mathrm{sp^3})-C(\mathrm{sp^3})}$ one?

A quick Google search results in 347–356 kJ/mol for bond-dissociation energy (at 298 K) for $\sigma$ $\ce{C(\mathrm{sp})-C(\mathrm{sp})}$ bond in alkanes and 837-839 kJ/mol for the triple bond in alkynes, out of which the $\sigma$ $\ce{C(\mathrm{sp^3})-C(\mathrm{sp^3})}$ bond contributes 369 kJ/mol.

Okay, just to ensure that I've understood this correctly, lower the hybrid orbital energy, higher is the dissociation energy of the bond it forms, right? I think I messed up between the energy and the bond energy.

@Kōhai, yes, the lower a bonding orbital is, the more energy is required to break the bond. In our case the bonding orbital is formed from a hybrid orbital of C atom and some orbital of another atom X. And for a given orbital of atom X, the lower the hybrid orbital of C atom is, the lower a bonding orbital will be. I think I can try to draw this...


text:
Is the rate of an SN1 reaction really independent of the type of nucleophile?
Suppose we take two reactions
$\ce{(CH3)3CCl + H_2O \rightarrow (CH3)3COH +HCl}$
$\ce{(CH3)3CCl + CH3OH \rightarrow (CH3)3COCH3 +HCl}$
I know in $\mathrm{S_N1}$ reactions the RDS (Rate Determining Step) is the formation of a carbocation intermediate.
But I am quite certain that methanol is a more bulky and weaker nucleophile, as compared to water. Should that make the rate of the second reaction lower than the first reaction (after the RDS is crossed)?

The key concept here is that both steps of the reaction are happening simultaneously, not sequentially. The steady state theory says that there will always be a pretty small amount of carbocation in the reaction mixture that will be stable over time. The faster it is generated, the faster it will be consumed. Therefore the reaction rate is only dependent of the carbocation generation rate, as long as this step is significatly slower than the formation of the final product, enough to make the time of this second process negligible.

Nope, the rates wouldn't be affected, even if the nucleophiles used are different (though if you're dealing with highly concentrated reactants, it might be a different story...).
Judging from the way you framed the question, you probably already know a fair deal about both the Unimolecular and Bimolecular Nucleophilic Substitution (SN) reactions. So I won't start from scratch.
In the SN1 case, as the name would suggest, the rate depends on only one concentration factor, namely the substrate. You already (correctly) mentioned that a carbocation intermediate is formed. At this stage, its just a matter of whether the nucleophile performs a 'frontal attack' or a 'rear attack', and doesn't really involve the nucleophilicity or the size (at least not much in your example) of the nucleophile. This is different from the SN2 reaction, which requires species with high nucleophilicity  and appropriate size to react (rear attack) with the substrate, leading to an unstable transition state.
How about an analogy?
Suppose you've got a lot of magnets stuck on your fridge door (at least I sure do..). Now suppose you're standing at a distance from the fridge, and you want to place another magnet on the fridge by throwing it. Since you've already got a lot of magnets stuck there, its going to be really hard to make your magnet land there. The weaker your magnet, the lower the chances of you landing it on the fridge door. However, if you have this really strong neodymium magnet, this wouldn't be much of a task; one throw, and it'll land so hard, it'll send the other magnets flying. This is a lot like the SN2 scenario. The attacking group must be a stronger nucleophile that the leaving group.
Now suppose your fridge door is devoid of magnets [which is precisely what my mother requires of our fridge.... :( ]. Then clearly throwing a magnet and hoping it'll land is a very, very good possibility. In this case it doesn't really matter whether the magnet is square or circular, large or small, strong or weak (not too weak). Apart from how large your fridge door is, there's no other factor that would keep the magnet from sticking to the fridge (albeit you aim). This is a lot like your SN1 scenario. The nucleophile's characteristics don't really matter (at least not in most typical cases). As long as there's space on the door, the magnet will stick.


text:
What is the difference in the reducing power of various ionic hydrides?
By definition "Ionic or saline hydrides are composed of hydride bound to an electropositive metal, generally an alkali metal or alkaline earth metal."

What is the relative order of reactivities of all the ionic hydrides of alkali and alkaline earth metals?Is there any general trend?
Is the reactivity of alkali/alkaline-earth metal hydrides greater than $\ce{LiAlH4}$ ?
Is the reactivity of alkali/alkaline-earth metal hydrides greater than $\ce{NaBH4}$ ?

I know reducing nature of $\ce{LiAlH4}$>$\ce{NaBH4}$.

Can't NaH reduce carbonyls? @orthocresol and cant NaBH4 deprotonate compounds?

Can you explain why NaBH4 or LiAlH4 is slower than the s block hydrides ?

@orthocresol Hey,sorry for the late follow-up question.But can you clarify something?Why is breaking B-H bond tougher than breaking Na-H bond?Aren't ionic bonds stronger than covalent bonds?Or is it because NaH is ionic it already dissociates in water?

@Aaron Abraham has written a nice answer to the original question.However I would like to combine his arguments along with the points made by @orthocresol.

In case of covalent compounds larger and weaker bonds in hydrides facilitates donation of hydrogen for reduction reactions.
S-block elements too have some covalent nature.So the previous fact can be applied for them too using Fajan's rules.
In general the top most element of both the groups in s-block has maximum covalent nature.Covalent bonds are generally weaker than ionic bonds (with some exceptions).So the reducing power is maximum at top and minimum at bottom.
Alkaline earth metal hydrides tend to be more covalent than alkali metals hydrides due to higher charge density on alkaline earth metals.So Alkaline earth metal hydrides have more reducing power compared to its counterpart.
S-block hydrides are usually never used in reduction of carbonyls and similar compunds.However they are used for deprotonation.
Even $\ce{NaBH_4}$ can be used to deprotonate.However they are much weaker and slower than s-block hydrides as for them to react the $\ce{B-H}$ bond has to be broken which is difficult and slow.

Al-H bond is really weak though, LiAlH4 deprotonates stuff really fast.

But slower than s-block hydrides?Right ? @orthocresol

No idea actually.

@orthocresol Are any values available on the net to comapare the rates?Well even I have not much idea...

I don't know. I don't think that such a thing would exist, deprotonations/protonations occur at a very fast rate, and for the most part we are more concerned about the extent of de/protonation (a thermodynamic question) rather than the rate (a kinetic question).

This is a very generalized answer but I'll go for it anyhow, who knows how it'll turn out to help (if it does add to what you already know, I suspect you may already know about what I'm going to type....)
The metrics you're considering are 'Reactivity' and (since you brought up LiAlH) 'Reducing character'. Clump these together, and it turns out what you're looking for is the ability for these hydrides to donate/release their hydrogen.
Now I've reasons to believe you already know the 'reactivity' trend for p-block hydrides (correct me if I'm wrong). Figuring out the reducing character of the s-block ionic hydrides involves similar logic.
In the p-block hydrides, you've probably learn't that as you move down the group, the E-H bond length increases (I use 'E' to denote p-block elements). If the bond length increases, the easier it is to cleave, therefore the hydrogen is released more readily. Here's an analogy if that bit seems a bit hard to grasp. Think of the E-H bond as a pencil. The longer it is, the easier it is to break; the shorter it gets the more difficult it is to break (try it......P.S-this credits for this analogy goes to my Chem. Teacher).
In a nutshell, for the p-block hydrides, as you move down the group the tendency to donate hydrogen increases.
Now back to the s-block. You can't apply the logic you used for p-block hydrides here (at least not directly ;) ).
At first glance, the main impediment you'd see is "s-block hydrides are ionic but p-block hydrides are covalent, so I can't figure out the trend the same way...". You'd be correct, but if you trying extrapolating (in reverse) the logic you use for p-block hydrides, see what happens.....
Since they're all ionic hydrides, and since you want the reactivity trend, what you'd want to do is compare the covalent nature of each s-block hydride.
As you'd already know, an ionic bond isn't 100% ionic, and the same can be said for covalent bonds. You call them ionic or covalent, if the ionic or covalent nature of a bond is relatively significant.
So by using Fajan's rules (assuming you know those as well) you can figure out that LiH is the most 'covalent' of the ionic s-block hydrides, and therefore more prone to donating hydrogen thereby acting as a reducing agent. This Wikipedia page just confirms that...
As you move down Groups 1 and 2, the reducing character of the s-block hydrides decreases.
Also LiAlH and NaBH are much, much stronger reducing agents than plain ole' s-block hydrides.

Can you give reasons for your last sentence?

$\ce{BeH2}$ is the most covalent hydride among s-block hydrides.


text:
What exactly is the use of photographic films in cameras? How are the pictures generated?
By definition:

Photographic film is a strip or sheet of transparent plastic film
  base coated on one side with a gelatin emulsion containing microscopically small light-sensitive silver halide crystals. The
  sizes and other characteristics of the crystals determine the
  sensitivity, contrast and resolution of the film.

But how exactly is a photographic film utilized in a camera?
I know that it has a sort of a silver coating. But how does it work? I read the Wikipedia article but found it a bit difficult to understand.

While it is common to find claims on the Internet that cameras once used a thin layer of matter (or 'film') nanoengineered to capture and record images completely passively and without the use of electronics, the technology to do this is lost, like Atlantis, in the mists of prehistory, if it ever existed. A camera, like the one in your cellphone, uses an array of photosensors to capture images.

A photographic film consists of a light sensitive emulsion of fine particles (grains) of silver salts in gelatine
spread on a clear celluloid strip or a glass plate.
The grain size is very important to photographers, as this
affects the quality of the pictures produced.
AgBr is mainly used as the light sensitive material.
Some AgI is used in "fast" emulsions. The film is placed in a camera. When the photograph is exposed, light from the subject enters the camera
and is focussed by the lens to give a sharp image on the film.
The light starts a photochemical reaction by
exciting a halide ion, which loses an electron.
The electron moves in a conduction band to the surface of the
grain, where it reduces a $\ce{Ag+}$
ion to metallic silver.

$\ce{2AgBr + h\nu \rightarrow 2Ag + Br2}$

In modern photography only a short exposure of perhaps 1/100th of a second is used.
In this short time, only a few atoms of silver (perhaps 10-50) are produced in each grain exposed to light.
Parts of the film which
have been exposed to the bright parts of the subject contain a lot of grains with some silver.
Parts exposed to paler parts of the subject contains a few grains with some silver, whilst parts not exposed contain none.
Thus the film contains a latent image of the subject.
However, the number of silver atoms produced is so small that the image is not visible to the eye.
Next the film is placed a developer solution. This is a mild reducing agent, usually containing quinol.
Its purpose is to reduce more silver halide to $\ce{Ag}$ metal.$\ce{Ag}$ is deposited mainly where there are already some $\ce{Ag}$ atoms. Thus the developing process intensifies the latent image on the film so it becomes visible. The
correct conditions for processing must be used to obtain an image of the required blackness.

If the film was brought out into daylight at this stage, the unexposed parts of the emulsion would turn black
and thus destroy the picture.
To prevent this happening any unchanged silver halides are removed by placing
the film in a fixer solution.
A solution of sodium thiosulphate is used as fixer. It forms a soluble complex with silver halides.

$\ce{AgBr + 2Na2S2O3 \rightarrow Na3 [Ag(S2O3)2] + NaBr}$

After fixing, the film can safely be brought out into daylight. Parts blackened by silver represent the light parts of the original picture. This is therefore a negative.
To obtain an image with light and dark the right way round, a print must be made.
Light is passed through the negative onto a piece of paper coated with $\ce{AgBr}$ emulsion. This is then developed and fixed in the same way
as before.
EDIT: Wikipedia writes:

In black-and-white photographic film there is usually one layer of
  silver halide crystals. When the exposed silver halide grains are
  developed, the silver halide crystals are converted to metallic
  silver, which blocks light and appears as the black part of the film
  negative. Color film has at least three sensitive layers,
  incorporating different combinations of sensitizing dyes. Typically
  the blue-sensitive layer is on top, followed by a yellow filter layer
  to stop any remaining blue light from affecting the layers below. Next
  come a green-and-blue sensitive layer, and a red-and-blue sensitive
  layer, which record the green and red images respectively. During
  development, the exposed silver halide crystals are converted to
  metallic silver, just as with black-and-white film. But in a color
  film, the by-products of the development reaction simultaneously
  combine with chemicals known as color couplers that are included
  either in the film itself or in the developer solution to form colored
  dyes. Because the by-products are created in direct proportion to the
  amount of exposure and development, the dye clouds formed are also in
  proportion to the exposure and development. Following development, the
  silver is converted back to silver halide crystals in the bleach step.
  It is removed from the film during the process of fixing the image on
  the film with a solution of ammonium thiosulfate or sodium thiosulfate
  (hypo or fixer).Fixing leaves behind only the formed color dyes, which
  combine to make up the colored visible image. Later color films, like
  Kodacolor II, have as many as 12 emulsion layers,with upwards of 20
  different chemicals in each layer.

Well you describe B/W film well. Color film is the same sort of thing but has different layers that respond to red/green/blue light like our eyes.

@MaxW Well the answer is about B/W films only...BTW I added about color films too..

@orthocresol Google Images :-P...I lost the links.The third image is from my texbook.The last image is from Wikipedia...

First image from https://tutsplus-media.s3.amazonaws.com/photo.tutsplus.com/uploads/2013/02/iso_05.jpg


text:
Anomalous oxidation states of Transition Metals

I have noted down the available positive oxidation states of the first
  row of transition elements (on the Periodic Table) from the respective Wikipedia articles of
  the elements.
$\ce{Sc} - 3, 2,1$
$\ce{Ti} - 4, 3, 2, 1 $
$\ce{V} - 5, 4, 3, 2, 1$
$\ce{Cr} - 6, 5, 4, 3, 2, 1$
$\ce{Mn} - 7, 6, 5, 4, 3, 2, 1$
$\ce{Fe} - 6,5,4,3,2,1 $
$\ce{Co} - 5,4,3,2,1 $
$\ce{Ni}-  4,3,2,1$
$\ce{Cu} - 4,3,2,1$
$\ce{Zn} - 2,1$

Why does $\ce{Fe}$ show not show $+7$ or $+8$ oxidation state?
After $\ce{Fe}$ from $\ce{Co}$ why is there a decrease in number of possible    oxidation states?
Why don't $\ce{Fe,Co,Ni,Cu,Zn}$ show higher oxidation states though they have valence electrons left in $3d$ subshell?

For a full picture, you need to add the possible negative oxidation states.

Regarding the maximum oxidation state achievable, there are two competing trends. Both have to do with ionisation energies.
In going to a higher oxidation state, the additional IEs are always balanced against the release in energy in the formation of stronger ionic/covalent bonds. However, if the additional IE is too large, then the favourable bonding will not be sufficient to bring out the higher oxidation state. A simple example is sodium, which cannot adopt the +2 oxidation state because its second IE is simply too large, even though the lattice energy of a hypothetical $\ce{NaCl2}$ crystal is larger than that of $\ce{NaCl}$.
Firstly, the maximum oxidation state is limited by the number of valence electrons available. This is exactly analogous to the case of sodium; therefore, manganese does not exhibit the +8 oxidation state, because its eighth IE involves ionisation from the 3p subshell, much lower in energy than either the 3d or 4s. This therefore accounts for the increase in the maximum oxidation state going from Sc to Mn.
Secondly, the ionisation energies generally increase going across the 3d block, because of an increase in effective nuclear charge. This is the factor which leads to the decrease in the maximum oxidation state after Mn. For example, the seventh IE of Fe is larger than that of Mn; therefore, Fe does not exhibit the +7 oxidation state.
You can look up the IEs in any relevant textbook. However, note that since the +2 oxidation state is very common in the later half of the 3d block, it is usually the third and higher IEs that are more important in this discussion. So, looking at the trend of first/second IEs won't get you anywhere.

But that's not a good enough reason.You are justifying using IE.But what if you did'nt have access to I.E. values.Then how would you justify the sudden decrease in available oxidation states after $\ce{Fe}$ just by using the electronic configurations ?

You don't need to look up the IE values to know that there is an increasing trend.

You did not understand my question.Your answer does not explain why is there a decrease in available oxidation states.Say,why doesnt Fe show +7 oxidation state,why doesnt Co show +6 oxidation state though electrons are left in 3d? I know you can just look up I.E. values but that is not a reason.

Because to get cobalt to the +6 oxidation state, you need to remove six electrons. Six electrons are too hard to remove in cobalt, because there is a large effective nuclear charge, and the ionisation energies are large, despite the electrons still being in the "valence" 3d orbital. In fact, that's exactly what I wrote in my answer.

@ZOZ please use correct spacing before and after punctuation marks. No space before one, but one space when you're beginning a new sentence.

@Rubisco I never knew of that.I thought that was MS-Word style :-P.Anyway I will follow it in future if you say so...

"Iron does not exhibit the +7 oxidation state". Lu et al., https://doi.org/10.1039%2FC6CP06753K. But yes, we would not see iron(VII) in normally encountered laboratory settings.

In some cases what's anomalous is not only the number but the environment in which the oxidation state is realized. We might expect high oxidation states to be realized by combining the element with oxide or fluoride ligands, but for instance one of the most widely used lead(IV) compounds for many years was $\ce{Pb(C2H5)4}$.
Palladium(V) offers an even more extreme example. Silicon as such has low electronegativity, but the molecular orbital structure of silyl ligands makes them more electronegative and metals bonded to such ligands are considered to be at the positive end of such bonds. Palladium is combined with five silyl ligands to give the $\ce{Pd(V)}$ species identified in Ref. 1.
Reference

Shimada, Shigeru; Li, Yong-Hua; Choe, Yoong-Kee; Tanaka, Masato; Bao, Ming; Uchimaru, Tadafumi (2007). "Multinuclear palladium compounds containing palladium centers ligated by five silicon atoms". Proceedings of the National Academy of Sciences. 104 (19): 7758–7763. https://doi.org/10.1073/pnas.0700450104.


text:
Why can't primary alcohols react with sodium bromide via an SN2 mechanism?
I was told that a primary alcohol reacts with $\ce{HBr}$ for example, but not with $\ce{NaBr}$ through the $\mathrm{S_N2}$ mechanism in the same reaction environment. Why?
This reaction can occur:
$$\ce{R-OH + HBr -> R-OH2+ + Br- -> R-Br + H2O}$$
But this reaction does not occur:
$$\ce{R-OH + NaBr -> R-OHNa+ + Br- -> R-Br + NaOH}$$

For $\mathrm{S_N2}$ to occur, the $\ce{-OH}$ has to be "helped along" to leave, by converting it to a good leaving group (good leaving groups = weak bases). If you protonate it with a strong acid, $\ce{Br-}$ can attack, and a molecule of water will be kicked out as the leaving group.
So why doesn't it work with $\ce{NaBr}$? A proton is a strong lewis acid, meaning $\ce{HBr}$ can protonate a small fraction of the $\ce{-OH}$ groups. $\ce{Na+}$ is not, on the other hand; it is too electropositive (just as $\ce{Na}$ tends to give up an electron and become $\ce{Na+}$). It simply cannot hold on to the electrons from a formally neutral oxygen-atom at all. Hence, it's like you are still trying to get $\ce{HO-}$ to leave.

Extra - when oxygen has a formal negative charge

If you go on and study carbonyl chemistry you will find places where $\ce{Li+}$ (which has a better size to interact with $\ce{O-}$ than $\ce{Na+}$) can stabilise the so-called tetrahedral intermediate, and help keep it from collapsing.
This is useful to stop the reduction of, for example, esters at the aldehyde, instead of continuing the reduction to the primary alcohol.
To get the lithium-ion to dissociate, you just add water. As DHMO explained, as soon as you have water, the positively charged group I metal ions are simply solvated so well by the water, that they are virtually inert.

This can be reasoned simply by looking at the acid/base strengths of the species present. Equilibrium will favor the side of the reaction with the weaker acid/base character.
$$\mathrm{p}K_\mathrm{a}(\ce{HBr})=-9 \hspace{1cm} \mathrm{p}K_\mathrm{a}(\ce{H2O})=15.7$$
$$K_\mathrm{eq}=\frac{[\ce{H2O}]}{[\ce{HBr}]}=10^{15.7-(-9)}=10^{24.7}$$
You don't need me to tell you that that's a really, really big number. Let's take a look at the other reaction, though. This will be the same type of comparison, except now we will be comparing basicities.
$$\mathrm{p}K_\mathrm{b}(\ce{Br-})=23 \hspace{1cm} \mathrm{p}K_\mathrm{b}(\ce{OH-})=-1.7$$
$$K_\mathrm{eq}=\frac{[\ce{OH-}]}{[\ce{Br-}]}=10^{-1.7-23}=10^{-24.7}$$
This equilibrium constant is as small as the other was large. So while the reaction you proposed may actually happen (though $\ce{Na+}$ will coordinate rather than form a bond as you show), the formation of the reactants is so favored the products won't be seen.

An $\mathrm{S_N2}$ reaction is a bimolecular reaction. As such, it requires both partners to be sufficiently reactive; that is we need both a good-enough nucleophile to attack and a good-enough leaving group. If we have a slightly better leaving group, we can make do with a worse nucleophile and vice-versa, but overall they have to be strong enough in combination.
Leaving groups in the $\mathrm{S_N2}$ reaction are good ones if they have a good charge distribution (charge per volume), which correlates well with being the conjugate base of a strong acid. If your leaving group is cationic (i.e. becomes neutral after leaving), it is a very good leaving group as there is no charge that needs to be distributed. Hence why protonating a hydroxy group to give $\ce{-OH2+}$ turns it into a good leaving group.
This is also why bromide is a rather good leaving group: Not only is it the conjugate base of a very strong acid $\ce{HBr}$, it is also a very large anion with a small charge and thus a rather good charge distribution. On the other hand, a hydroxide, which would be the leaving group in an $\mathrm{S_N2}$ attack onto an alcohol, is much smaller but still has the same charge. Thus, it is an absolutely terrible leaving group and nucleophilic attacks that liberate hydroxide effectively don’t happen.
Thankfully for synthetic organic chemistry, there are many ways of turning hydroxy groups into good leaving groups; most of these involve adding an acyl residue of some sort that creates a stabilised anion upon leaving with a much better charge distribution. The most popular ones (given as the actual leaving group) are probably mesylate $\ce{MsO-}$, tosylate $\ce{TsO-}$ and triflate $\ce{TfO-}$ — all sulphonic acid derivatives.

As a side note: You wrote an addition of $\ce{Na+}$ to the hydroxy group to generate a charged species. But unlike a proton — which is such a strong Lewis acid that it will never walk alone — a sodium cation will never form that strong a bond with a hydroxy group that they would leave as $\ce{NaOH}$. In fact, $\ce{NaOH}$ if it doesn’t precipitate due to low solubility will always dissociate into $\ce{Na+}$ and $\ce{OH-}$ while water molecules will not in typical organic solvents.


text:
What is the hybridization of diatomic nitrogen, diatomic anything like oyxgen, bromine; hydrogen and chlorine in HCl?
I'm guessing sp hybridization for nitrogen in diatomic nitrogen based on the lone pair on nitrogen and the triple bond but there is no mixing of s and p orbitals so I want to say that there is no hybridization. Are there any chemicals without hybridization? What are some examples of chemicals without hybridization
Why are lone pairs counted in hybridization if they do not contribute to the mixing of the orbitals, a.k.a. hybridization?

Hybridisation is something you typically deduce from the geometry of the compound you are observing. It is primarily a mathematical concept.
That said, to form the diatomic molecules you asked about, no hybridisation is needed. In fact, you get much better results from assuming unhybridised atoms and then combining their atomic orbitals to molecular ones. Except for hydrogen (which only has 1σ and 2σ), the molecular orbitals of all these diatomic compounds ($\ce{N2, O2, F2, Cl2, Br2, I2}$) are basically identical save the energy levels of the MOs and their populations. Thus, these would all count as unhybridised compounds.
Also, there is no nead to invoke hybridisation to explain the bonding situation in $\ce{HCl}$ either. Instead agan consider a σ bond formed between hydrogen’s 1s orbital and chlorine’s 3p. All of chlorine’s other orbitals will sit there happily filled and unhybridised.


text:
Configuration of endocyclic double bond in Vitamin A

I think the double bond inside the ring of vitamin A should have an (E) configuration. Am I wrong? I don't think I'm assigning the priority correctly.

The endocyclic double bond of vitamin A has a (Z) configuration. Here’s why:
First, let’s determine priorities of the atoms connected on the top side of it:

C

C (ring)
C (methyl 1)
C (methyl 2)

C

C
(C) (ghost atom; due to the double bond)
H

So the ‘ring side’ — the atom on the left in the way it is drawn at the moment — has a higher priority than the side chain (C at (1.3) beats H at (2.3)).
What about the bottom side?

C (ring)

C
H
H

C (methyl group)

H
H
H

Here again the ‘ring side’ has a higher priority than the methyl group: (1.1) carbon beats (2.1) hydrogen.
Both left-hand sides have a higher priority than their corresponding right-hand side. Thus, the double bond is (Z) configured.

There are two groups of compounds with Vitamin A, first is retinoids, also know as vitamin A and provitamin A, known as carotenoids which is a vitamin A precursors( which contain at least one molecule of retinol).
Naturally retinoids exist in the form of retinol, retinal and retionic acid and they are derived from animal sources. There are hundered of Carotenioids which are mainly synthesized by plants and only 50 of them can act as vitamin A precursors.
Technically speaking, large number of geometrical isomers are theoretically possible for retinoids and carotenoids but due to steric hinderence many mono-cis and poli-cis isomer are rarely occurred in nature.
All-trans forms are predominant in nature and more stable, but only those cis isomers occur which offer less steric hinderence and thermodynamically stable.

5-cis-retinol geometrical isomer  is rather unstable and do not occur in that configuration in nature.

Two of your molecules lost a double bond. 2) There is no cis-configured extracyclic double bound in your ‘13-cis-retinol’. 3) This does not answer the question.

edited and notified correction has been made. This picture only posted to show that 13-cis isomer is possible and stable,

The pictures still does not show 13-cis in spite of it claiming to do so. The question is still not answered.


text:
Does trans-decalin really have a plane of symmetry?
My teacher told me that trans-decalin (see below) is achiral due to the presence of both, a centre of symmetry and a plane of symmetry. But I could not spot the plane of symmetry untill now. Can someone point out the plane of symmetry in a diagram?

It passes through the two central carbons, perpendicular to the molecule.

It's not easy to see from a diagram, because it distorts bonds and angles. I recommend building it with a balls-and-sticks model set. You can also use a molecular viewer to model it; there are a couple of open-source (or at least free) ones out there.
I have calculated the molecule on the DF-BP86/def2-SVP level of theory. The point group of the molecule is C2h. In the following model I have highlighted the plane of symmetry, and the rotational C2-axis. At their intersection is an inversion centre Ci.

(I needed to downscale this a lot, click on the image to get to a high resolution still. Images created with ChemCraft, assembled with GIMP.)

Is it possible for one ring to have a boat conformation at room temperature? In this case the plane disappears.

@porphyrin I found another structure, where one ring is in twist-boat conformation. On the GFN2-xTB level of theory (semi-empirics) it is about 23 kJ/mol higher in energy. This is at room temperature negligible.

The question has already been answered nicely by Martin - マーチン♦ who pointed out that there is a mirror plane (i.e. a $\sigma = S_1$ element) and a center of inversion (i.e. an $S_2$ element), which both are already enough for a molecule to become achiral (existence of any $S_n$ element suffices for that). There is one important caveat though that often bugged me during stereochemistry lectures:  Even if the ground state of a substance might not be achiral (for example if you placed some substituents on your decalin), then it can still be possible for the chemical to be achiral due to quick racemization when an achiral transition state with low energy is available (e.g. inversion of $\mathrm{NR_3},\mathrm{PR_3}$ systems or the $^1C_4\mapsto ^4C_1$ inversion of cyclohexanes). I think that is a little trap that is often missed in stereochemistry lectures.


text:
Polyatomic Cations
While learning about poly-atomic ions, our instructor stated that ammonium is one of the few poly-atomic cations. Are there other poly-atomic cations? And why are positively-charged poly-atomic ions "rarer" than negatively-charged poly-atomic ions?

https://en.wikipedia.org/wiki/Polyatomic_ion#Examples_of_common_polyatomic_ions

There is an infinite number of polyatomic cations, much like there is an infinite number of polyatomic anions. So your instructor is wrong there. But there is still some reason in the statement. Why? Polyatomic simple cations are rare (if you discount metal-ligand complexes).
Generating a simple polyatomic anion is easy: Take an element that is not fluorine, bond it to as many hydroxy groups as its valency is and deprotonate the hydrogens — voilà. You can then play around with oxidation states (i.e. add more oxygen) to generate even more anions, e.g. with $\ce{PO3^3-}$ and $\ce{PO4^3-}$.
This works because oxygen is a very electronegative element that, when bonded to a hydrogen, is strong enough to displace said hydrogen. For a polyatomic cation to form, we would need to reverse the idea: Take a stable, neutral molecule that has a lone pair and then add a hydrogen (one of the few rather electropositive elements that still has a tendency to form bonds).
And here are the problems. Most elements simply aren’t basic enough to cope with the additional hydrogen. In fact, basically any element that isn’t nitrogen will be too acidic in its protonated state. Also, you can’t really play with additional oxygens to create further oxidation states since increasing the oxidation state would destabilise the cation. Thus, simple polyatomic cations are few while simple polyatomic anions are plenty.

That was a great exploration, I didn't know that electropositivity was a thing. Thanks!

I think it could phrased better as inorganic polyatomic cations are rare. Every organic amine could form a cation of course.

@MaxW Hence simple. Taking my arbitrary definition of simple, it would only include two elements, but any organic amine requires at least three elements (carbon, nitrogen and hydrogen).

Jan - understand. My point was just that "inorganic" is a fairly well defined term but "simple" is more fuzzy.


text:
Why does aqueous NaCl conduct electricity?
Could someone please explain why an aqueous solution of NaCl conducts electricity? I've Googled this up but couldn't get a satisfactory answer. This has an answer on Physics.SE but that's too complicated. My textbook doesn't answer my question.

Essentially you understand that it is because it dissociates, and your follow-up question is why it dissociates, which, as someone else pointed out, is another question.

See the answer, and linked reference therein, by @M.Farooq: https://chemistry.stackexchange.com/a/118441/79678.

Original Question:
Why an aqueous solution of $\ce{NaCl}$ conducts electricity
Because $\ce{NaCl}$ is an electrolyte. This means it yields ions in solution.
Simply put, solid $\ce{NaCl}$ consists of $\ce{Na+}$ cations and $\ce{Cl-}$  anions bound together in a rigid crystal lattice. When it melts or is dissolved in water, the crystal lattice breaks. The ions are now able to move around. Similar to charged particles in a metal conductor (in this case electrons), in liquid form or aqueous solution the ions are the charged particles that can move, allowing the solution to conduct electricity.
Follow-up question:
Why $\ce{NaCl}$ dissociates in water
The (rather oversimplified) answer is that dissociation of an ionic compound is facilitated by ion-dipole attractions between the ions of the compound and the polar water molecules.
The water molecules (dipoles) are attracted to the ions  and cause the crystal lattice to destabilize and ions to dissociate. The charged ions in solution are surrounded and stabilized by the water molecules (dipoles).
Note: Some ions migrate as ion pairs, but for a strong electrolyte most or all the ions will be dissociated and surrounded by water molecules.

As you seem dissatisfied with the answers received so far, I assume you want to fully understand the mechanism involved in dissociation. Unfortunately, this is rather complex.
There is an excellent  paper by Ballard & Dellago [1] that explains their work on the subject, but you will probably need a bit more physical chemistry / thermodynamics knowledge to understand it fully.
Reference

Ballard, AJ & Dellago, C "Toward the Mechanism of Ionic Dissociation in Water," *J. Phys. Chem. *B 2012, 116, 45, 13490–13497
Publication Date:October 19, 2012
https://doi.org/10.1021/jp309300b

NaCl is an electrolyte. When in solution it dissociates into Na+ and Cl-. When you put electrodes in the solution, the cations are drawn to the cathode and the anions to the anode. This movement produces a current and that is why NaCl solutions can conduct electricity.

Why does it dissociate?

I think this answer needs some elaboration. Does this only work with AC? If not, what happens with DC when the bulk of  the ions have already migrated?

With a DC current, you'll probably get some electrolysis of water and possible oxidation of the chloride (which helps by removing negative charge from the positive electrode).

@Zhe Agreed. I was just trying to prompt the psoter into fleshing out his answer for the OP.

$\ce{NaCl}$ dissociates into $\ce{Na+}$ and $\ce{Cl-}$ when it dissolves into water. Hydration stabilizes the ions formed. There is $\delta+$ charge on hydrogens of water and $\delta-$ charge on the atoms of oxygen. When $\ce{NaCl}$ is dissolved into water, the hydrogens are attracted to the $\ce{Cl-}$ ions and oxygen atoms to the $\ce{Na+}$ ions. Thus water molecules surround and separate the ions.

Hence, in an aqueous solution of $\ce{NaCl}$, there are positive and negative ions available to move freely (of course with some resistance due to other molecules in vicinity). When a potential difference is applied, the positive ions are attracted to the negative terminal and vice versa. Thus, a current is established.

@Abcd See this video on YouTube. link

I do not think that actual $\ce{OH-}$ ions are attracted to $\ce{Na+}$ because  the dissociation constant of water is quite low, $pK_w = 14$. So, I don't think that free $H+$ and $OH-$ ions are available. A water molecule as a whole orients itself in a particular way and surrounds the ions.

Sorry, I didn't mean that $OH-$ ions are not attracted to $Na+$ but I think that the effect wouldn't be as much.

Thankyou. So can we say that dissociation of NaCl takes place because of orientation of water molecules?

And where did u get that awesome picture from ?

Yes. I think so.

I googled it(images).

The simple answer is that water molecules have a dipole. The hydrogen atoms are slightly positively charged and the oxygen atoms slightly negatively charged. So, as you can see in the diagram above, water molecules around positively charged sodium orient themselves such that the oxygen is closer to sodium than the hydrogens. This is known as an ion-dipole interaction. The opposite is true for chlorine. @Abcd

@orthocresol and therfore they dissociate ?

@Abcd yes. In the absence of water, NaCl does not dissociate, because it is held together by ionic bonds. The dissociation of NaCl in water also requires the breakage of ionic bonds, but that is compensated for the formation of such ion-dipole interactions, which is the driving force for the dissociation.

@orthocresol sorry if i am troubling you by questioning too much .. what do u mean by driven by and how does the bond break?

@Abcd Ionic bonds are only electrostatic forces of attraction. From physics, you will know that the strength of such forces depend on the distance between the two charged species. In solid NaCl, the ions are close to each other, so a bond is said to exist. In aqueous NaCl, the ions are separated from each other, so the bond is said to be broken. A driving force simply refers to something that favours a process. So, the dissociation process is favoured by the fact that ion-dipole interactions are formed.

$\ce{NaCl}$ is strongly ionic compound. It gets completely ionized and dissociates in $\ce{Na+}$ and $\ce{Cl-}$. Both $\ce{Na+}$ and $\ce{Cl-}$ are surrounded by water molecules, $\ce{Na+}$ is surrounded with $\ce{O}$ of $\ce{H2O}$ facing towards $\ce{Na+}$, similarly $\ce{Cl-}$ is surrounded by $\ce{H2O}$ molecules with $\ce{H}$ facing towards $\ce{Cl-}$. Presence of positive and negative ions helps in conduction of electricity.
When electrodes are inserted and current is passed through these electrodes, movement of ions in opposite direction creates current.
After passing current for a long time (especially DC current), the concentration of $\ce{Cl-}$ ions will decrease. The concentration of $\ce{OH-}$ will increase gradually and there will be competition between $\ce{Cl-}$ and $\ce{OH-}$.
Electrolysis of $\ce{NaCl}$
Anode: $\ce{2 Cl- -> Cl2 + 2 e-}\quad E=-1.36\ \mathrm V$
Cathode: $\ce{H2O + 2 e- -> H2 + 2 OH-}\quad E=-0.83\ \mathrm V$
Electrolysis of $\ce{NaOH}$
Anode: $\ce{4 OH- -> O2 + 2 H2O + 2 e-}\quad E=-0.40\ \mathrm V$
Cathode: $\ce{H2O + 2 e- -> H2 + 2 OH-}\quad E=-0.83\ \mathrm V$
After Sufficient long time there will be competition between $\ce{Cl-}$ and $\ce{OH-}$ to get oxidized on Anode, whereas there is only one possible reaction on cathode. Reaction on Anode depends upon the concentration and reduction potential of $\ce{Cl-}$ and $\ce{OH-}$. Actually on cathode there is also possibility of reduction of $\ce{Na+}$ but reduction potential is $-2.7\ \mathrm V$ which is difficult than $\ce{H2O}$ so only one reaction takes place on cathode.


text:
How does sulfuric acid react to heating?
How would $\ce{H2SO4}$ react to heat? I assume that it would form toxic sulfur (di/tri)oxide, but I can't find any information on this.
I'm concerned about fire hazards specifically, so assume very intense heating and very low concentration.

"very low concentration"? Then that's slightly acidic water and there's  not much difference what acid is there.

Sulfuric acid, $\ce{H2SO4}$, having an enthalpy of formation of $-814\ \mathrm{kJ/mol}$, is quite stable and won't decompose easily.
According to A Kinetic Study of the Decomposition of Spent Sulfuric Acids at High Temperature, Dominique Schwartz, Roger Gadiou, Jean-François Brilhac, Gilles Prado, and Ginès Martinez:

The decomposition of $\ce{H2SO4}$ to $\ce{H2O}$ and $\ce{SO3}$ is predominant between $400$ and $700\ \mathrm K$. The formation of a small amount of gaseous sulfuric acid can be observed. Above $673\ \mathrm K$, the equilibrium constant of the reaction R1 becomes higher than 1 and increases rapidly.
$$\ce{H2SO4 <=> H2O + SO3}\tag{R1}$$
The second process is the reduction of sulfur trioxide to $\ce{SO2}$. This endothermic reaction needs a high temperature to take place, the equilibrium constant of the reaction R2 being higher than 1 above $1050\ \mathrm K$.
$$\ce{SO3 <=> SO2 + \dfrac12O2}\tag{R2}$$

So, you only have to worry when the temperature reaches $400\ \mathrm K$, or $127\ \mathrm{^\circ C}$ (false precision).

@YdobEmos Usually available sulfuric acid is dissolved in water (except sulfuric acid in which case there is little water), so the temperature would be limited by the water.

In aqueous solution, you also have to worry about the water boiling and spitting acid at you. A speck of hot sulphuric acid on a mucus membrane is not a pleasant experience.

First, you don't need to worry about fire: sulphuric acid is non-flammable.
Around $1000$ kelvin, this reaction would occur:
$$\ce{2H2SO4 <=> 2SO2 + 2H2O + O2}$$
This is industrially used for the production of hydrogen gas, and the total reaction cycle is called the sulphur-iodine cycle.


text:
Is cis-1,2-dimethylcyclohexane a meso compound?
I was watching a video lecture today on hydrocarbons and came across this. The instructor says that, as there is a plane of symmetry, cis-1,2-dimethylcyclohexane is a meso-compound:

However, the cyclohexane ring is never planar as it adopts a chair conformation. I cannot find any plane of symmetry in chair form. Like this:

Is the compound meso or not?

cis-1,2-Dimethylcyclohexane is achiral, not because there is a plane of symmetry, but because it consists of two enantiomeric conformations which interconvert rapidly via ring flipping at normal temperatures.
This is exactly the same case as amine inversion. "Chiral nitrogens" such as that in $\ce{NHMeEt}$ do not lead to chirality or optical activity because of rapid inversion of configuration at the nitrogen atom, leading to interconversion of the two enantiomeric forms 1a and 1b.

Likewise, the ring flip in cis-1,2-dimethylcyclohexane leads to two different conformers. I have deliberately chosen to depict the ring flip in the following fashion, to make the mirror image relationship more obvious. The green methyl group, equatorial in conformer 2a, is changed into an axial methyl group in conformer 2b. Likewise, the blue methyl group goes from axial to equatorial. The 1,2-cis relationship between the two methyl groups is retained in both conformers.

Each individual conformer can be said to be chiral, but just like how the amine is considered achiral, cis-1,2-dimethylcyclohexane as a whole is considered achiral.

Is the compound meso? According to the IUPAC Gold Book, a meso-compound is defined as:

A term for the achiral member(s) of a set of diastereoisomers which also includes one or more chiral members.

1,2-Dimethylcyclohexane possesses two diastereomers, one cis and one trans form. The trans form is chiral, but the cis form is achiral, as explained above. Therefore, the cis form satisfies the above definition and is considered a meso-compound.

Can we get it cold enough to freeze the enantiomers long enough to detect them?   Can we see evidence of non-racemization in an optically active environment?

@OscarLanzi I don't know. I presume it is possible, but for all intents and purposes it's achiral.

Sir, I'd like to ask one thing, suppose we have the compound 1-(chloromethyl)-3,5-dimethylcyclohexane in which all three substituents (the 2 methyl and one chloromethyl) are coming out of the plane, is this compound meso? Due to the same reason as explained in this answer (flipping of ring)? It does have two stereogenic centres (the 3,5 positions)...

@LightYagami Yes, it is the same thing.

Is it wrong then to see "plane of symmetry" at all in the rings? Because that is not how in reality it exists, but is it a conicidence that if we answer according to the 2D picture (using POS idea), the answer still comes out to be the same. I am also not able to see why the same argument for the trans form cannot be applied in this case.

Is the answer to my second question that, the same argument cannot be applied to the trans form because in cis,due to ring flip we were getting both $(R,S)$ and $(S,R)$ but in trans, if the isomer is $(R,R)$, then it will remain $(R,R)$ only even after flipping and hence the other conformer is not obtained, making it chiral?

@LightYagami I think you should ask a new question, this is getting too substantial for the comments. To be honest, I also don't understand what you mean, so you will need to explain it a bit more clearly. Your compound has three different groups on the cyclohexane, you can't call it cis or trans. (R) and (S) descriptors are to do with configuration and not conformation. A ring flip doesn't interconvert (R) and (S) under any circumstances whatsoever.

@orthocresol cis-1,2-dimethyl cyclohexane at room temperature is optically inactive due ring flipping right? It means in mixture we have equal percentages of enantiomers. So why should we call it as a meso reather than a racemic mixture.

And also I have a doubt, if we write absolute configurations at two chiral carbons in compound 2a of your post we R and S . This means that the compounds 2a ,2b possess internal compensation. So none of the compounds 2a ,2b are optical active. Then how both are enatiomers?

@Infinite (1) It's a bit of a grey area, but the key point is that the enantiomers are not isolable at RT. "Racemic" implies that you can separate and isolate the two enantiomers. (2) It being (R,S) means that the compound is optically inactive, which completely agrees with what I wrote. This, however, does not rule out that a specific conformation may be chiral. On balance, when all conformations are considered, it will be optically inactive. Note that the same is true of, say, meso-tartaric acid, for which one can easily construct a chiral conformation.

@Infinite Note, further, that if you are describing one particular species or conformation, then it is better to use the term "chiral" or "achiral". On the other hand, "optically active" or "optically inactive" should only be used to refer to a bulk sample of such a compound, because optical activity is something you can only measure on a bulk sample. You can't go into the lab and determine how a single molecule rotates plane-polarised light. You can, however, use symmetry arguments etc. to show that the molecule is or isn't superimposable on its mirror image: hence chiral or achiral.

I'd like to add a few points definition wise from the IUPAC goldbook.
(Because recently the original website seems to be broken I am using links via the internet archive.)

meso
  A prefix to indicate the presence of a symmetry element of the second kind (see chirality) in a chemical species which is usually one of a set of diastereoisomers that also includes a chiral member.

However, there is a more general definition, too:

meso-compound
  A term for the achiral member(s) of a set of diastereoisomers which also includes one or more chiral members.

The first part clearly does not apply to the cis-isomer, (1R,2S)-1,2-dimethylcyclohexane, the second does because of the definition of chirality.

chirality
  The geometric property of a rigid object (or spatial arrangement of points or atoms) of being non-superposable on its mirror image; such an object has no symmetry elements of the second kind (a mirror plane,  σ = S1, a centre of inversion,  i = S2, a rotation-reflection axis, S2n). If the object is superposable on its mirror image the object is described as being achiral.

I think the key point of this definition is in this case, that the object needs to be rigid, which (1R,2S)-1,2-dimethylcyclohexane is not at room temperature. If you cool down the solution, inhibiting the ring flip, you would probably be able to find both configurations; that is a racemic mixture of them.
Since (1R,2R)-1,2-dimethylcyclohexane and (1S,2S)-1,2-dimethylcyclohexane, the trans-conformers, are both chiral (and enantiomers), the remaining member of the set (1R,2S)-1,2-dimethylcyclohexane can be considered a meso-compound, since it is achiral.


text:
How does boiling water extract dyes from vegetables?
Why does boiling water always seem to release and become the colors of whatever vegetables I am boiling?
For example, beet root and red cabbage both vividly color hot water.
I'd assume it's something to do with collisions with water particles breaking down cell walls and releasing a pigment, but how would this work on a physical level?

I believe this is a Chemistry question.

@QuantumBrick I'm very interested in analysing it from a physics perspective, specifically that of how the kinetic energies of the particles in the water succeed in breaking down the solid cell wall.

cross post at: http://biology.stackexchange.com/questions/52533/how-does-boiling-water-release-dyes-from-vegetables/52542?noredirect=1#comment92063_52542

These dyes (NOT pigments$^*$) are highly water soluble. As you pointed out, boiling causes cell walls to rupture, releasing the cell plasma into the water. Boiling of the plasma fluid causes pressure build up, something the cell walls aren't designed to withstand.
You can also extract them with cold water by mashing up the raw material in a food blender or grinding it in a mortar and pestle. This has the same effect of rupturing the cell walls.
An interesting fact is that many of these dyes are pH sensitive: their colour is different in acid or alkaline conditions. Notably red cabbage extract changes from wine red to greenish blue when you add some sodium bicarbonate (baking powder) to it, because 'Bicar' is a (weak) alkali. Then adding a lot of vinegar (a weak acid) will change it back to its original colour.
$^*$ pigments are inorganic substances (many are refined minerals) and mostly insoluble in water.

But how does the water actually cause the cell walls to rupture?

Just edited it: it's due to pressure.


text:
Can the following molecules be called meso compounds (cyclic structures which usually exist in non-planar forms)?

In the planar form the following molecules have one plane of symmetry.However in
  the chair-form (the most stable form) for the first molecule and the non-planar form for the second molecule there is no plane of
  symmetry.Will the molecules still be called meso?

BTW I do not think the molecules ever attain completely planar form.

Just for quick reference I'm adding the non-planar forms of the two molecules.

These molecules can never be planar. There are no internal symmetry planes and therefore they are not meso.

I've just tried building 3D models of the two enantiomers of the 1,2,4,5-tetramethylcyclohexane with my molymod kit and they are not superimposable - I tried chair, boat, half-chair and twist-boat conformations. I haven't tried the cyclooctene but I'm pretty sure it's not meso and it can't be planar, neither in cis nor trans.


text:
Can the following compound having a centre of symmetry be called both optically inactive and meso?

The Centre Of Symmetry (COS) has been shown by a blue dot.

My questions are:

Can this molecule be called optically inactive? I have this doubt because single bonds permit free rotation and the molecule hardly
remains in the given form for long.
Can this molecule be called a meso-compound as it optically inactive?
Slightly unrelated: I'm actually a bit confused with the definition of meso. I learnt that meso compounds must have a plane of symmetry. Am
I correct?

According to Wikipedia a meso compound or meso isomer is a non-optically active member of a set of stereoisomers, at least two of which are optically active.
Update: I realized my mistake in thought procedure after asking the question.(I fell into the "Meso Trap" :-P )
I would like to add this link as it might help future viewers.
http://www.masterorganicchemistry.com/2011/01/12/the-meso-trap/

You are correct that the molecule is not chiral. It is indeed a meso compound due to the internal symmetry plane. Also, don't worry about free-rotation of single bonds.

Ok my doubt is that if it was not meso due to plane of symmetry would it still be optically inactive due to centre of symmetry?

Also can you say why "If you actually draw a Newman projection, the most stable form of the molecule is symmetrical on both sides of the plane." ?

Sorry,  ignore the part about the Newman projection. With the centre of symmetry or also called centre of inversion is a symmetry element,  you should end up with the same molecule after the operation. However,  I am not sure if that's enough on its own to prove thay it is a meso compound.

By the Wiki definition "A meso compound or meso isomer is a non-optically active member of a set of stereoisomers, at least two of which are optically active. This means that despite containing two or more stereogenic centers, the molecule is not chiral." ---- Can we conclude something from this?I think a COS should be enough to prove that it is meso...

I have build a model again and determined the point group which is Cs. Therefore, there is no centre of inversion, there is only internal symmetry plane.

Could you upload a picture of your model? According to the picture a COS exists...

I apologize, it does exist. The point group is Ci. It it obvious that the molecule is a meso compound as it's superimposable on its mirror image and therefore, it seems the centre of inversion should be enough to confirm it as it's the only symmetry element present apart from the identity.

Thank you!Can you answer this question ? http://chemistry.stackexchange.com/questions/61020/stereochemistry-of-cycloalkanes-having-substituents


text:
Stereochemistry of cycloalkanes having substituents

My teacher said that from $3$ to $6$ member cycloalkanes having any number
  and type of substituents we can always draw the planar structure and
  decide the specifics of stereochemistry like:

Are they optically active or not?
Are they meso compunds or not?
Do they have a plane of symmetry/centre of symmetry/axis of symmetry/alternating axis of symmetry or not?

But I feel that the molecules are hardly in planar form. It would be incorrect to assume they are planar and find the specifics of stereochemistry. Only for cyclopropane the points should be correct as it is always planar. Maybe my teacher said that mistakenly in a hurry.
So regarding points $(1),(2)$ and $(3)$ who is correct? Why? Or are we both partially correct? I need a slightly detailed answer.

For a compound to be optically active it must not be superimposable with its mirror image. It's as simple as that. Also note that stereogenic centres are not always required for chirality. There are other types of chirality such as axial and planar chirality where a compound possesses either an axis or plane of chirality. Sometimes it can be the case that just the sterics of a molecule does not allow for free rotation of bonds and the molecule is locked one configuration. I think in the simplest cases of open chain molecules, you could indeed use the structural formula, but with rings which adopt specific 3D configurations, you must refer to the 3D structure.
Again, as said in 1., to determine if a molecule is chiral, it's better to refer to its 3D structure, especially when with meso, when you are dealing with more than one stereogenic  centre. For a compound to be meso, it must not be chiral, but posses stereogenic centeres (look up the definition).
No, you cannot use the simple structure to determine symmetry elements present if that structure is not really planar. It's not necessary to build a model, but you must use wedged/dashed bonds to show stereochemistry.

If someone thinks differently, please comment and correct me.

Axially and planar-ly chiral molecules do have "stereogenic units", even if they don't have "stereogenic centers".

Yes, I have referred to those as axis and plane of chirality and they are stereogenic indeed as two enantiomers are observed.


text:
Does bicyclo[3.2.0]heptane have a plane of symmetry?

Most stable form of cyclobutane is the "envelope form" and most stable
  conformation of cyclopentane is either the "envelope form" or the "twist"
  form.

In the above compound do the individual rings exist in their most stable conformation? Or is the actual situation different?
What is the most stable conformation of the above compound? And does it have a plane of symmetry?

I figure that the question "why is the cyclobutane ring planar" is a good question in itself for main site. But you'll probably have to specify that you are looking for an org chem answer and not simply some calculations that say "because it is the most stable conformation".

This question is a prime example of overthinking!

It turns out that bicyclo[3.2.0]heptane does have a plane of symmetry (courtesy PubChem):

According to The Molecular Structures of Bicyclo[3.2.0]heptane and $\Delta^6$-Bicyclo[3.2.0]heptene. An Electron-diffraction Study of Gaseous $\ce{C7H12}$ and Molecular Mechanics Calculations on $\ce{C7H12}$ and $\ce{C7H10}$. Robert Glen, Grete Gundersen, Peter Murray-Rust, and David W. H. Rankin, here are the bond angles in degrees:
The link explains the experimental procedure in deduction of the actual structure.There are two possible stable structures/forms for the molecule namely endo and exo.The endo form is supposed to be more favourable than the exo form.

From here, we can conclude Cs symmetry.

I'm sorry to say.But you have given a very vague answer with no reason at all.Why doesnt the cyclopentane take up an envelope or twist form here?Why doesnt the cyclobutane take up an envelope form?Really vague answer :-(

@ZOZ Satisfied?

No :-(!You gave the bond angles.Ok good.You gave the predicted structure.Good.But you still did not give the reason for the structure's occurence and stability..and you did'nt answer my previous questions..in the comments as well as the original question..

@ZOZ Reason? This is experimentally determined. You could solve the Schroedinger's equation if that counts for a reason. There's no reason. Everything is just rationalization.

@ZOZ Even VSEPR is a rationalization.

It would not be correct to state that there is no reason.Rationalization is a pathway to reasoning.BTW the document you linked was very helpful and tries to explain the experimental procedure along with an attempt to explain why the endo form is more favourable.Thanks for that link.

Just a thought - I suppose more diaxial interactions in exo form could account for its lower stability...

@ZOZ Rationalization is not a reason. The reason is that this conformation has the lowest energy. The reason that this conformation has the lowest energy is because of, you know, too many parameters.

I never said rationalization is a reason.But it is a pathway to reason.Atleast it gives a head start and that's all I wanted in an answer.Anyway this is not the place to discuss which definition is correct.That belongs to English SE. Anyway +1.

If the solution of the Schrodinger equation is the only admissible "reason", then chemistry is in a lot of trouble. @ZOZ The cyclopentane ring is already in an envelope conformation, as pH13 notes. Remember that any distortion of geometry in the cyclobutane ring is transferred to the cyclopentane ring via the bridging carbons. Presumably, distortion of the cyclobutane ring towards  a puckered geometry simply creates more strain (bond/angle) in the rest of the molecule than is relieved in the cyclobutane ring.

Most of the discussion about the rationalisation of the most favourable structure is irrelevant to the original question especially since crystals have significant molecule-molecule interactions that might matter (though in this case the molecule has a plane of symmetry within error bars anyway). In solution several conformations most likely equilibrate to an average which has a plane of symmetry unless there are strong barriers to conformational shifts.

The most stable conformation is the endo-form which obeys $C_\mathrm s$ symmetry. The cyclobutane-part is in envelope form and the cyclopentane-part is also in envelope form. The endo version with the methylene group bent towards the cyclobutane moiety is the preferred one.
The next most stable form would then be the exo form, where neither the cyclobutane- nor the cyclopentane-part is in its favorite form, both are slightly distorted. At least after my optimization, this molecule does not have a symmetry, except $C_1$. Though, I did not try a symmetric version.
With a quite high energy gap of ~90 kJ/mol the most unstable form is where both parts are in their twisted forms. Though, this molecule is $C_2$ symmetric.

The structures were optimized with RI-TPSS-D3BJ/def2-TZVP and a following frequency calculation showed that the results are local minima. Those geometries where then used for a single point energy calculation, using the Tight-PNO DLPNO-CCSD(T)/def2-TZVP scheme. All calculations have been made with ORCA 3.0.3.

I see my answer here not really as an answer but more as an additional help for some qualified people to discuss.