[abhi]

these are not the ques from any book but are my basic doubts based on 1 reaction which have just not let me sleep over nights i am sure you guys would certainly get it removed

>C=C< + H2O ( acidic media) >CH-C(OH)<

(a) why here acidic media is required when H2O can itself act as a source of proton,(wait a sec...) i hope your probable answer is that H2O is very weak acid but my question is what makes H2O so much weak acidic that it cannot even initiate this reaction

(b)the basic purpose of taking acidic media is that it acts as a source of proton as given in example above surely it acts as a source of proton since water is given

but my ques is that if in any ques suppose H2Ois not given
then can we suppose by ourself that acidic media would acts as a source of proton

could you explain me this text which was written in my textbook that "for dehydrating alcohols ,a concentrated dehydrating acid (such as H2(SO)4 and H3(PO)4) is used to favour the equilibrium to the formation of alkenes
hydration equilibrium shifts to the formation of an alcohol by addition of excess water"

[Schrödinger]

I advise u to get to know the mechanism and the concepts thoroughly... That is really very helpful...

(a)First of all, the reaction :

 >C=C< + H₂O  (acidic medium)   >CH-C(OH)<

is initiated by the attack of a proton(electrophile) on the nucleophile (one of the double bonded carbons). This is the slow step (Rate Determining Step). The intermediate thus formed is a carbocation....i hope u can visualise, as i am not good at using latex...

This Carbocation is obviously an electrophile due the positive charge on the carbon. The second step is due to the attraction of the lone pair of electrons on water(nucleophile) towards the carbocation. The will give u the second intermediate.

The second intermediate will have H₂O+ , i.e, positively charged water molecule attached to the carbon which initially had the positive charge...something like coordination of the lone pairs of water to the carbocation.

The third step is deprotonation of the charged water species to yield the alcohol. This is initiated by the conjugate base of the acid we have used or water itself, as it is also a base.

For strong acids, pH<7....and you know that pH = -log(H₃O+).

HA + H₂O  H₃O+  +  A- 
H₂O  +  H₂O     H₃O+  +   OH-   

This way, at equilibrium, the concentration of H₃O+ ions for other acids will be greater than for water. So, they are stronger acids....Water on the other hand can behave as a base as well as acid....It is not a strong acid....

(b) There can't be any acidic property exhibited by an acid without water, because the dissociation of acid into H₃O+ and the conjugate base is due to the presence of water....we can't do away with it....

I think you can understand the last part of your question by now.....More the water, more is the formation of alcohol....

[orgopete]

>C=C< + H2O ( acidic media) >CH-C(OH)<

(a) Why is acidic media required when H2O can act as a source of proton? Your probable answer is that H2O is very weak acid but my question is why is H2O too weak to initiate this reaction?

(b) The purpose of an acidic media is as a proton source, example above. Then it acts as a source of proton since water is used. My question is, "Suppose H2O is not present, can we suppose that acidic media would also act as a proton source?"

Explain why "for dehydrating alcohols, a concentrated dehydrating acid (such as H2(SO)4 and H3(PO)4) is used for the formation of an alkene while an acid can give an alcohol by addition of excess water to an alkene."

Even though this question has been answered, I wish to provide a slightly different answer.

In one sense, this is basically a question of acid-base equilibria. Let me answer (b) first. In the absence of water, one can indeed had a proton source from the media. For example, if gaseous HCl were added to propene, HCl would be the proton source and the final product would be 2-chloropropane. In this case, we can imagine that an equilibrium exists between a proton-chloride ion and a proton-propene. The equilibrium will be determined by the affinity of the chloride and the propene electrons to a proton. The equilibrium need not be entirely to a proton-propene complex as that complex can react with a chloride to give 2-chloropropane. In 2-chloropropane, a proton-chlorine complex is probably too weak to effect a reverse reaction. Therefore the reaction can still proceed to product.

Now, let us return to (a). If water were used, the products of its ionization would be HO^- and H⁺. The affinity of HO^- would be much greater than an alkene for a proton that an insufficient amount of protonation of the alkene would exist to cause a further reaction to occur. If an acid such as HCl is added to water, water is still more basic than an alkene and protonation of water occurs. If an unactivated alkene is used, the equilibrium is not sufficient to result in a measurable reaction. Increasing the protonation of an alkene by adding electron donating groups (carbons) enhance the acid catalyzed addition of water.

If an alcohol is treated with an acid, the hydroxyl group can be readily protonated. That product can lose water to give an carbocation. Deprotonation of the carbocation will result in an alkene. (It could also be an E2 elimination with loss of a proton to be concerted with a loss of water.) This is the reverse of the addition reaction. Overall, these two reactions are in equilibria. What would happen to the product if water were removed from the equilibrium? The reaction would form an alkene. This is an application of Le Chatelier's principle. Thus, dehydrating acids are used. (While not mentioned in the reaction, the alkene is usually removed from the sulfuric acid by distillation to prevent a reverse reaction.)

That should logically limit the effectiveness of an addition of water to an alkene. If the acid catalyzes the loss of water as well as its addition, then it can be difficult to add water to unactivated alkenes. That is because protonation of the alcohol catalyzes the reverse reaction more strongly than a forward reaction. Remedies to this are to use another acid which might kinetically complex more strongly with an alkene or to have another nucleophile intercept the carbocation intermediate than water. With ethylene or propene, sulfuric acid is used and the product is the sulfate ester. That is, bisulfate anion reacts with the carbocation intermediate. Because bisulfate is a much weaker base than ethanol, the reverse reaction does not take place. This is similar to the addition of HCl to propene discussed above.

The more practical solution to the addition is to use mercuric sulfate in water. Mercury complexes with an alkene to catalyze the addition of water similar to an acid. The mercury must be removed by a second reduction step. An alternate method is to use acetic acid as the protic medium with a strong acid, such as sulfuric acid. Now, acetic acid will react with the carbocation intermediate. Because an acetate is a much weaker base, the reversion back to an alkene is also suppressed. If an alcohol is desired, a second hydrolysis step is needed.

If a trisubstituted or a 1,1-disubstituted alkene is used, water can be readily added to it with acid catalysis. With other alkenes, more vigorous conditions may be needed and thus one of the methods above will work better.