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Order of esterification of alcohol is
A. $3^\circ > 2^\circ > 1^\circ $
B. \[2^\circ > 3^\circ > 1^\circ \]
C. \[1^\circ > 2^\circ > 3^\circ \]
D. none of these

Answer
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Hint: Formation of ester from alcohol and carboxylic acid in presence of concertation ${H_2}S{O_4}$ mirror is called esterification.
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Step by step answer:
Let us explain the general reaction of esterification.
Alcohol and phenol react carboxylic acid, acid chlorides and acid anhydride to form esters.
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The reaction is reversible therefore water is removed by absorbing it by concentration ${H_2}S{O_4}$.
 With acyl chloride,${H_4}$ is removed by pyridine.
So that reaction takes place in a forward direction.
In the esterification reaction H-atoms come from alcohol and $ - OH$ comes from carboxylic acid.
Acidic character of alcohol due to plan nature of $O - H$ bond.
An electron releasing group like $ - C{H_3}, - {C_2}{H_5}$ increases electron density on oxygen.
It decreases polarity of$O - H$bond. This decreases acid strength.
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Steric hindrance (far bulkiness) increases from primary to secondary to tertiary alcohol, the order of esterification decreases. Thus the relative order of esterification of alcohol is \[1^\circ > 2^\circ > 3^\circ \].
Therefore, from the above explanation the correct option is (C) \[1^\circ > 2^\circ > 3^\circ \].


Note: In esterification carbonyl oxygen of carboxylic acid is protonated.
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This increases the positive character of C-atoms.
The nucleophile alcohol attacks the carbonyl carbon.
Then followed by intramolecular ${H^ + }$ transfer and loss of water molecules.
In the last step deprotonation gives ester.

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