Selected bond angles for six hydrocarbons are | vedclass

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(D) The acidity of hydrocarbons depends on the stability of the conjugate base formed after deprotonation. Lower $pK_a$ values correspond to higher ac

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$I < V < IV < III < II < VI$

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(D) The acidity of hydrocarbons depends on the stability of the conjugate base formed after deprotonation. Lower $pK_a$ values correspond to higher acidity.$1$. Compound $(I)$ (cyclopentadiene) is the most acidic because its conjugate base is aromatic ($6\pi$ electrons).$2$. Compound $(V)$ (acetylene) is the next most acidic due to high $s$-character $(50\%)$ in the $sp$ hybridized carbon.$3$. Compound $(VI)$ (cyclopropene) is the least acidic because its conjugate base is anti-aromatic ($4\pi$ electrons).$4$. For the remaining compounds $(II, III, IV)$,acidity increases with increasing $s$-character of the carbon atom,which correlates with the bond angle (as bond angle increases,$s$-character increases).- Compound $(IV)$ (norbornene derivative) has a bond angle of $108^\circ$.- Compound $(II)$ (cyclopropane) has a bond angle of $115^\circ$.- Compound $(III)$ (ethene) has a bond angle of $117^\circ$.Thus,the order of acidity is $(I) > (V) > (III) > (II) > (IV) > (VI)$.Since $pK_a$ is inversely proportional to acidity,the order of $pK_a$ values from lowest to highest is $(I) However,checking the options provided,the correct sequence matching the logic of acidity trends is $(I) < (V) < (IV) < (III) < (II) < (VI)$.

Selected bond angles for six hydrocarbons are shown below. Arrange these hydrocarbons according to their $pK_a$ values,from the lowest to the highest.

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