Arrange benzene,$n-$hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour.

(D) The acidic character of a hydrocarbon depends on the ease with which it can lose a proton $(H^+)$. This is directly related to the hybridization state of the carbon atom to which the hydrogen is attached.

Compound	Hybridization of Carbon	$s-$character
$n-$Hexane $(CH_3CH_2CH_2CH_2CH_2CH_3)$	$sp^3$	$25\%$
Benzene $(C_6H_6)$	$sp^2$	$33\%$
Ethyne $(HC \equiv CH)$	$sp$	$50\%$

As the $s-$character of the hybrid orbital increases,the electronegativity of the carbon atom increases. This causes the electron pair of the $C-H$ bond to be held more strongly by the carbon atom,making the hydrogen atom more acidic (easier to release as $H^+$).
The $s-$character increases in the order: $sp^3 < sp^2 < sp$.
Therefore,the decreasing order of acidic behavior is: Ethyne $>$ Benzene $>$ $n-$Hexane.