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We have to compare their acidic strength, and we know acidic strength is directly proportional to stability of the anion.

So the first compound is more acidic due to its being aromatic after deprotonation. The second most acidic is the fifth compound, because of $\mathrm{\% s}$ character of $\ce{C-H}$ bond. And the least acidic is sixth compound because its anion is antiaromatic.

But how would we compare the second, third and fourth compounds?

figure

Source: MS Chouhan; Advanced Problems in Organic Chemistry 11th Edition; Chapter 1 General Organic Chemistry problem 27

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Nice work! You clearly understand how factors like aromaticity/antiaromaticity and % s-character in a $\ce{C-H}$ bond can affect acidity.

You can use Coulson's Theorem to estimate the % s-character in the other molecules. In its general form Coulson's Theorem describes the relationship between the angle $\ce{X-Y-Z}$ in a molecule and the hybridization of the $\ce{Y-X}$ and $\ce{Y-Z}$ bonds according to the following equation:

$$\mathrm{1+\lambda_{XY}\lambda_{YZ}cos(\Theta_{XYZ})=0}$$

(where $\mathrm{\lambda}$ represents the square root of the hybridization index of a bond, e.g. $\mathrm{\lambda=\sqrt{3}}$ and the hybridization index $\mathrm{=3}$ for an $\mathrm{sp^3}$ hybridized bond)

In molecule III where the two $\ce{C-H}$ bonds are equivalent, Coulson's Theorem simplifies to

$$\mathrm{1+\lambda^2cos(117)=0}$$

and we find that $\mathrm{\lambda^2 = 2.2}$, or the $\ce{C-H}$ bond in III is $\mathrm{sp^{2.2}}$ hybridized.

As the hybridization index gets smaller, there is more s-character in the $\ce{C-H}$ bond which results in a more stable carbanion and consequently increased acidity. This is consistent with our thinking that, in general, acidity follows the order $\mathrm{sp > sp^2 > sp^3}$; or, the larger the $\ce{X-C-H}$ bond angle, the more acidic the $\ce{C-H}$ bond.

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