Why concentrated sulphuric acid is used as a catalyst instead of dilute sulphuric acid or a weak acid in esterification?

Or in symbols.... $"H"_3"CC(=O)OH + HOCH"_2"CH"_3 stackrel(Delta, "H"^+)rarr$ $"H"_3"CC(=O)OCH"_2"CH"_3 +"H"_2"O"$ Alternatively, the acid may be heated at reflux to give the acid halide... $"H"_3"CC(=O)OH + SOCl"_2 rarr"H"_3"CC(=O)Cl+HCl(g)"uarr+"SO"_2uarr$ And the acid...

The Fischer esterification is an acid-catalyzed equilibrium reaction. $underbrace("RCO-OH")_color(red)("acid") + underbrace("H-OR'")_color(red)("alcohol") ⇌ underbrace("RCO-OR'")_color(red)("ester") + "H"_2"O"$ You need the acid to catalyze the reaction and make it run at a reasonable...

Remember that esterification is an equilibrium process. I.e.: $Alcohol+Carbo xylic acid rightleftharpoons Ester$ This means that your product will still contain unreacted reactants at a dynamic equilibrium

Fischer esterification is the formation of esters by nucleophilic acyl substitution, reacting a carboxylic acid with an alcohol using acid catalysis (typically sulphuric). There are other types of esterification...

The leaving group is water (i.e. the carboxyl hydroxide that is protonated off as water...) $"H"_3"CCO"_2"H" + "HOCH"_2"CH"_3 stackrel("H"^+)rarr"H"_3"CCO"_2"CH"_2"CH"_3 + "H"_2"O"$ And this will be explained in much greater detail,...

Ph stands for a phenyl group. Here the carboxylic acid is first protonated by sulfuric acid, making the carbonyl carbon more electrophilic. The Oxygen of the alcohol then attacks this...

......so we would predict the reaction to be exothermic: $RCO_2H + HOR' rarr RCO_2R' + H_2O$ Of course, most of the time, esterifications are conducted under reflux conditions. The activation...

Esterification is a very slow reaction, so in order to make it fast we use the sulfuric acid(H2SO4) as a catalyseur

You have performed the condensation... $R'OH + HO(O=)CRstackrel(H_3O^+)rightleftharpoonsR'O(O=)CR+H_2O$ Extraction with sodium carbonate solution will remove ANY unreacted acid from the organic phase... $RC(=O)OH(aq) + Na_2CO_3(aq) rarrunderbrace(RC(=O)O^(-)""^(+)Na)_"water soluble"+Na^(+)""^(-)HCO_3$

$(H_3C)_2CHOH+HC(=O)OHstackrel(H^+, Delta)rarr(H_3C)_2CHO(O=)CH$ The product has a normal boiling point of $68.2$ $""^@C$, whereas formic acid has a normal boiling point of $100.2$ $""^@C$. The other starting material, isopropyl alcohol, has...