[silver]

Hi there. My question is:

Predict the final product likely to be obtained when (S)-2-methyl-3-pentanol is subjected to the  following sequence of reactions: i) Tosyl chloride; ii) NaOH (aq), heat
A) (R,S)-2-methyl-3-pentanol and 2-methyl-2-pentene in more or less equal amounts
B) (R,S)-2-methyl-3-pentanol
C) 2-methyl-2-pentene
D) (R)-2-methyl-3-pentanol
E) (S)-2-methyl-3-pentanol

So far I figured that the Tosyl chloride would act as a nucleophile and the Ts will replace the Hydrogen of the alcohol group then the OH- from the NaOH will come in and attack. The problem is I can't figure out whether it would be elimination or another substitution. Pretty much I am debating between answers D and C.
I never really understood how to determine between substitution and elimination.

I also have this question:

Alkaline hydrolysis of an ester involves initial attack by hydroxide ion on the carbonyl carbon. The  presence of substituents on the aromatic ring of ethyl benzoate may be expected to increase/decrease  the rate of hydrolysis of this ester. In what order should the five substituents below be arranged to  represent the decreasing order of the rates of hydrolysis when these substituents are present in the  para- position of the aromatic ring in ethyl benzoate?
-NO2, -Cl, -H, -CH3, -OCH3

So far I think it could have something to do with stability of the compounds but I am more lost on this one.

Any help would be greatly appreciated. Thanks in advance.

[orgopete]

Tosyl chloride will be an electrophile and the alcohol, after deprotonation will be the nucleophile.

I too find it difficult to accurately predict wether a substitution or elimination should should occur in some reactions. However, one can usually expect a primary halide will give substitution and tertiary elimination. For secondary halides, a rule of thumb from Brown, Foote, and Iverson that I like is that a nucleophile with a pKa greater then 11 or 12 (?) will give elimination. Hydroxide fits (pKa 16), so I predict elimination.

For the second question, it is asking you what the rate limiting step must be. While the stability of the compounds (in hydroxide) is related, I doubt the rate limiting step is the deprotonation of the acid. Think of an earlier step.

[zengjia1029]

1, D) (R)-2-methyl-3-pentanol
2 NO2->Cl->H->CH3->CH3O-.

PS:

• TSO- group always prefers to replace by other nucleophilic group over elimination.
  
  The carbon in ester group will become more positive with a EWG at the p position 

[silver]

Thanks guys!!!

[orgopete]

See elimination reactions of butyl and pentyl tosylates with ethoxides, http://pubs.acs.org/doi/abs/10.1021/ja00951a067

[zengjia1029]

considering the answer C, you can see there must be running a E2 reaction with attacking by alkali at the tertiary carbon, but it is not easy due to the stereo hindrance.

[orgopete]

Although my discussion of points noted by zengjia shall go beyond the scope of the original posters question, zengjia makes a good point. One could arguably suggest that all E2 elimination reactions should prefer Hoffmann elimination products due to the steric hindrance argument, yet Zaitsev products predominate. I find myself compelled to envision a wider range of reaction timing than the simple models suggested by concerted reactions. A real example that are noted in some textbooks are E2 and E1cb eliminations. Such differences are inferred by an application of the Hammond Postulate or more explicitly stated by Arigoni in analyses of enzyme mechanisms, namely that there are not concerted reactions. That is, if time could be divided into infinitely small portions, then three billiard balls cannot collide at the same time. One ball will always make contact before the other.

If you apply that model, then Zaitsev products can be thought to be E1-like products. That is, if R-X bond cleavage were to begin before proton abstraction, then one might find those electrons best donated to the developing charge should predominate. These will lead to Zaitsev products. If proton abstraction precedes R-X bond cleavage, then Hoffmann-like products should predominate. Because carbons can be thought of as electron donors, then less substitution will be the most acidic, and give Hoffmann products.

In my opinion, most elimination reactions leading to Zaitsev products are in polar solvents, such as ethanol. Note, I am not suggesting a change in mechanism. The rate limiting step is still proton removal. I am merely suggesting that the C-H electrons begin migrating to the back of the R-X bond in a Zaitsev elimination.

[zengjia1029]

Thank you,orgopete! Here you have given us a very detailed and convincible explanation, and I learned many new chemistry knowledge. As you said, there are no pure E1, E2 and SN1, SN2 mechanism, only similar to them can be used. so considering this title, An E1-like mechanism will result to forming the Saitzev product which also is the predominant product. Thank you again to clarify some ambiguous arguments.