Q.Write the mechanism involved in the esterification of a carboxylic acid with alcohol.

Fischer esterification is acid-catalysed nucleophilic acyl substitution: the acid catalyst first activates the carbonyl by protonation, the alcohol then adds to give a tetrahedral intermediate, an internal proton shift converts one hydroxyl into a good leaving group ($H_2O$), and its loss followed by deprotonation gives the neutral ester.

Overall reaction

$RCOOH + R'OH \underset{H^+,\ -H_2O}{\rightleftharpoons} RCOOR'$

(reversible; catalysed by a strong acid such as conc. $H_2SO_4$ or dry HCl gas)

Step-by-step mechanism

Step 1 — Protonation of the carbonyl oxygen. The acid catalyst ($H^+$) protonates the carbonyl oxygen of the carboxylic acid, which increases the electrophilicity of the carbonyl carbon (the positive charge is delocalised over both oxygens by resonance):

$RCOOH + H^+ \rightarrow R-C(OH)=OH^+ \ \ (\text{resonance-stabilised oxocarbenium ion})$

Step 2 — Nucleophilic addition of the alcohol. The lone pair on the alcohol's oxygen attacks the now highly electrophilic carbonyl carbon, forming a tetrahedral intermediate carrying a positively charged oxonium centre on the incoming alkoxy group:

$R-C(OH)=OH^+ + R'OH \rightarrow R-C(OH)_2(\overset{+}{O}HR')$

Step 3 — Proton transfer (tautomerisation). A proton migrates (intermolecularly, via the medium) from the newly added $-\overset{+}{O}HR'$ group to one of the two $-OH$ groups already on the tetrahedral carbon, converting that $-OH$ into the excellent leaving group $-OH_2^+$:

$R-C(OH)_2(\overset{+}{O}HR') \rightarrow R-C(OH)(OR')(OH_2^+)$

Step 4 — Loss of water. The protonated hydroxyl leaves as a neutral water molecule, regenerating a resonance-stabilised, protonated carbonyl (now bearing the ester's $-OR'$ group):

$R-C(OH)(OR')(OH_2^+) \rightarrow R-C(OR')=OH^+ + H_2O$

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