Organic Chemistry Write the products of the following acid-base reactions: (a) C

Question

Organic Chemistry

Write the products of the following acid-base reactions:

(a) CH3OH + H2SO4 <--> ?

(b) CH3OH + NaNH2 <--> ?

(c) CH3NH3 + Cl- + NaOH <--> ?

- I do not necessarily need the solutions but what I REALLY need help in is understanding why one reactant is more acidic/basic than the other or even how I would calculate this. I actually have the textbook solutions already BUT I don't undeerstand how the book got to these answers.

Example:

(a) The solutions manual provided this to me:

CH3OH (Stronger Base) + H2SO4 (Stronger Acid) <--> CH3OH2+ (Weaker Acid) + HSO4- (Weaker Base)

My first questions from come from seeing this solution:

1. How do I determine which one is base/acid between the reactants? Both Methanol and Sulfuric acid has lone pairs and Hydrogen atoms, so why is Sulfuric acid the one to give up the H+. How did you determine, comparing to Methanol, that Sulfuric Acid was the one to give up the H+ vs the Methanol?

2. When the H+ moved from Sulfuric Acid to Methanol, why would it attach to Methanol vs just being a floating H+ ion in the products side?

Explanation / Answer

The products of the reaction:

a)CH3OH + H2SO4   ----------> CH3OSO3H   + H2O : Esterification reaction

b)CH3OH   + NaNH2 ----------> CH3ONa    + NH3      : Abstraction of proton from methanol by strong base NaNH2

c)CH3NH3    + Cl - + NaOH -------> Check formula of srarting material (CH3NH3) & reaction as well.

Explanation to the reaction:

The following reaction is an esterification reaction. i.e. reaction of alcohol with inorganic acid to form sulphate ester

CH3OH + H2SO4   ----------> CH3OSO3H   + H2O

As per definition an acid is the substance which gives H+ ions in the solution. Methanol is a neutral compound as it contains hydroxy oxygen bearing a lone pair of electrons . It accepts the the proton fron acid as per Lewis concept of acids & bases. But it is not a strong base . After protonation OH it is removed in the form of water & + CH3   attacks HSO4- as conjugate base of acid H2SO4 to form an ester.   

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