[orgoclear]

For the first question. I am confused for what Eact we need to take. for example, in the first option should I consider Eact as the barrier between the first carbocation and the corresponding TS or the rearranged carbocation and the corresponding TS

Q2. no idea

Q3. I am having some confusion deciding about strength H-bonds in phenol and methanol. (I know that EWG's increase the bond strength but am unable to decide between the two)

Q4. Can LAH be used for the purpose? (it was given in the answers) further can Zn/H+ be used ?

[orgoclear]

please help !!

[Scatter]

Well, since nobody else answered, I can at least maybe reinforce your thinking, though I do not for sure know the answers.  I'd say for #1 the Eact would be the energy needed to break the bond, not the energy needed for the rearrangement.  Which means...A) would have the lowest activation energy?

2. I'm guessing it has to do with pka.  To make phenol more acidic, you'd want a stronger base with it.

pKa:

H2SO4 = -3
Acetic Acid = 4.76
Dil. H2SO4 = 0 (water) - 3 = -3
Dil. HCl = 0 (water) - 8 = -8

Therefore, Acetic Acid is the most basic, so that would mean Phenol would be the most Acidic.  The only thing I don't know is if my method of calculating Dilute Acid pKas is correct.

3. I'd say Methanol, but only out of intuition.

4.  I don't really know...

I tried. 

[Scatter]

Well crap, looking at it again, A) and C) would be identical if it didn't have something to do with it, so I'd definately say A) because it would take more energy to get the Br in the less favorable secondary position?  Hell I don't know. haha

Hope any of this helps.  I'm just trying to use what I've learned already to intuitively come up with the answers.  I'm bored.

[orgopete]

I believe the answer to 1) is B). It is the most substituted aklene therefore the double bond will be the easiest/fastest to react. While others also give tertiary carbocations, the alkyl substituents activate the double bond more.

For number 2), I don't understand the question (but that may not help anyway).  If you protonate phenol, I would think the product of ionization, phenoxide, would be suppressed. Even if you add sodium hydroxide, I don' see how it would decrease the O-H affinity. If the question were, "in which phenolic solution would you find the lowest pH?" The it would be HCl as it is the strongest acid.

Number 3), I agree on methanol.

Number 4), these usually depend on what chemistry the o.p. knows.
A) sodium iodide.
B) sodium thiomethoxide , then Raney nickel.
C) Tributyl tin hydride and radical initiator.

[bromidewind]

1. (b) is correct, in my opinion. Like orgopete said, the more substituted alkene is the most stable. Although my instinct tells me this is a trick question of sorts..

2. Phenol is a weak acid because the formation of the phenoxide ion is stabilized by the benzene ring, although the high electrophilicity of the oxygen tends to pull hydrogen right back onto it. So to make it a stronger acid, you will want to have a solvent that makes it difficult for the oxygen to steal its hydrogen back. Sulfuric acid, when added to water (make sure you add the sulfuric acid to the water first), will be the limiting reagent, forming hydronium and hydrogen sulfate. Concentrated sulfuric acid is such a good dehydrating agent that it would most likely abstract the oxygen as well as the hydrogen.

3. I'd have to disagree on methanol having the highest intramolecular hydrogen bonding. The benzene ring is more likely to accept the electrons from the OH group within the molecule itself. I don't think that methanol will readily form hydrogen bonds with the hydrogens on CH₃. Also, if you look at the boiling points of methanol and phenol, phenol's is almost twice that of methanol. While some of this is attributed to the surface area of the phenol, I think it is mostly caused by the intramolecular hydrogen bonding.

4. 1) H₂/LiAl₅O₈
    2) I don't really understand this, but apparently it's a fairly recent method of dechlorinating chloroform using electrochemistry.
                http://sciencelinks.jp/j-east/article/200616/000020061606A0609327.php

    Just adding some reactions to orgopete's list.

I hope these answers help; I'm fairly confident that they are correct.

[Scatter]

Cool.  Well I kinda figured my take on it was wrong.  I was hoping it would make someone more apt to respond and disagree.  I just learned a bunch of new stuff from this post though.

[orgoclear]

thanks a lot everyone..

I still have the following doubts

1. The answer given is (d). This is what I think (my view). Activation energy is the energy difference between the transition state and the reactant. here attack of electrophile H+. Stabler the intermediate and more unstable is the reactant, the energy difference will be lower. So, based on this (d) can be said to be correct. as in (b) the reactant is more stable while intermediate is the same so activation energy of (b) > (d). Further although (a) and (c) are more unstable so are their intermediates ((a) and (c) should have same activation energy as the reactant and first intermediates are same). So is my view correct?

2.,3. thank you bromidewind

4. the options given were (i) LAH (ii) Zn/H+ . Can LAH be used directly without anything else?

Thanks again everyone

[bromidewind]

Your reasoning behind the first question sounds right. I never was very good with thermodynamics. I'm an organic chemist, not a physical chemist for a reason..

As for the last question, LAH will work fine by itself. I forgot that the LiAl₅O₈ method I posted was a precursor to LAH, which is a one step process. Zn/H⁺ aids in the stabilization of the cations formed when the Cl is removed. Because of the high electrophilicity of Cl, it tends to pull the CHCl₂ cation back towards it. Zn helps keep the Cl away from it. The acid helps drive the equilibrium toward the products.

[orgopete]

I agree with the activation energy reasoning, I do not agree with the result. I did not try to look thermodynamic properties, but in general the more substituted an allkene, the faster it reacts in an electrophilic reaction. This includes addition of bromine, epoxidation, and addition of HX. If B) reacts faster, then the activation energy barrier must be smaller. That is the essence of my thinking. While B) and D) share the same intermediate, that does not mean their activation energies are the same. If B) is the more stable alkene, then for D) to react faster, the must have very similar activation energies. That would argue the generality I suggested is incorrect.

The difference argument may be applied to cis and trans-2-butene in that cis would be at a higher energy level and one might reasonably argue the activation energies are similar. Therefore, the activation energy barrier must be less for the cis compound. Because I believe this argument is reflexive. If the trans isomer hypothetically reacted faster, you would find that chemists would simply argue the activation energy barrier was lower for the trans isomer.

If I am arguing with a professor, the professor is always right or that is how the exams are graded.

I agree with phenol as having the stronger hydrogen bonds. There is a constant trade between electron and proton availability. The more available the electrons, the less available the protons and the more available the protons, the less available the electrons. Thus amines and HCl do not have strong hydrogen bonds. In between are the strongest hydrogen bonds. They are amides, phenols, carboxylic acids, and HF. I erred.

For the reduction of methyl chloride, I didn't know whether LAH worked or not. I believe it works with benzylic halides. Zn/H(+) will also work with benzylic bromides (chlorides?). In general, dissolving metals will work if at a higher reduction potential, for example magnesium or lithium whether in a protic solvent or not? If not, then add water to the Grignard or methyllithium.   

[OrganicSynthesis]

Actually, I'm only in highschool so I am probably completely wrong, but I was thinking d) but since people didn't say it, I didn't want to really disagree with anyone. However, I was thinking d) opposed to b) mostly because of the steric hindering when the hydrogen is adding to the alkene. In b), the hydrogen would have to enter past the steric hinderance of two methyl groups on the alkene, whereas in d), the hydrogen would only have to enter past the steric hinderance of other hydrogens.

Once again, only a highschool student, probably a wrong answer but this is my intuition.

[bromidewind]

Actually, I'm only in highschool so I am probably completely wrong, but I was thinking d) but since people didn't say it, I didn't want to really disagree with anyone. However, I was thinking d) opposed to b) mostly because of the steric hindering when the hydrogen is adding to the alkene. In b), the hydrogen would have to enter past the steric hinderance of two methyl groups on the alkene, whereas in d), the hydrogen would only have to enter past the steric hinderance of other hydrogens.

Once again, only a highschool student, probably a wrong answer but this is my intuition.

For a highschool student, you have pretty sound thinking. I never even had the opportunity to take OChem in high school. I almost failed General Chemistry. And your thinking with steric hindrance isn't exactly right, but it is on the right track.

I feel like an idiot for not thinking of this earlier, but (B) cannot be the answer. Look at both ends of the alkene, they are both disubstituted. So when the HBr breaks the double bond, it can add to either end. In fact, it will most likely add to the top end with the two methyl groups because it is easier to navigate around two methyl groups than a methyl and an ethyl. You would get a mixture of products: 3,4-dimethyl-3-bromopentane and 2,3-dimethyl-2-bromopentane. In (D), you would only get one product. When two products are formed, I believe the activation energy is raised because the system has to spend energy forming both products.

[OrganicSynthesis]

Actually, I'm only in highschool so I am probably completely wrong, but I was thinking d) but since people didn't say it, I didn't want to really disagree with anyone. However, I was thinking d) opposed to b) mostly because of the steric hindering when the hydrogen is adding to the alkene. In b), the hydrogen would have to enter past the steric hinderance of two methyl groups on the alkene, whereas in d), the hydrogen would only have to enter past the steric hinderance of other hydrogens.

Once again, only a highschool student, probably a wrong answer but this is my intuition.

For a highschool student, you have pretty sound thinking. I never even had the opportunity to take OChem in high school. I almost failed General Chemistry. And your thinking with steric hindrance isn't exactly right, but it is on the right track.

I feel like an idiot for not thinking of this earlier, but (B) cannot be the answer. Look at both ends of the alkene, they are both disubstituted. So when the HBr breaks the double bond, it can add to either end. In fact, it will most likely add to the top end with the two methyl groups because it is easier to navigate around two methyl groups than a methyl and an ethyl. You would get a mixture of products: 3,4-dimethyl-3-bromopentane and 2,3-dimethyl-2-bromopentane. In (D), you would only get one product. When two products are formed, I believe the activation energy is raised because the system has to spend energy forming both products.

Haha, we don't actually learn ochem in my highschool, other than nomenclature. I took it upon myself to learn ochem. , and darn, I was hoping I was right XD

[orgopete]

I wanted to come back to this question, especially as it has been suggested that Q1, d) is the correct answer while I thought it should be b). I alway like to refer to actual data if possible. Unfortunately, no clear cut data is available here. The best I could do was a series of reactions in March, 3rd ed, p 670 on the bromination of some alkenes in methanol. The following relative rates are reported:

ethylene - 1
1-butene - 9.6
cis-2-pentene - 4333
2,3-dimethyl-2-butene - 9.3x10⁵

As I understand thermodynamics, a low activation energy barrier will result in a fast reaction. Consequently, 2,3-dimethyl-2-butene must have the lowest activation energy. 1-Butene and 2-pentene both give a secondary carbocation. If as has been argued that carbocation stability controls the activation energy, then a much smaller difference in rates might be expected. Unfortunately, the trans-2-pentene was not among the data reported. I don't know the relative rates of cis v trans-2-butene, but I believe it is much smaller. It is my premise that as carbon is an electron donor and thus increasing the number of carbon atoms attached to a double bond increases the ease with with a rate determining protonation of the pi-electrons occurs, then the tetra-substituted alkene should react fastest (and have the lowest activation energy).

If someone could cite better data, it would be helpful.

I do appreciate that the structures for b) should give a mixture of products. However, in the sense of the question, it should not raise the activation energy of the reaction illustrated.