Question

Mark the correct order of decreasing acid strength of the following compounds.

a.
b.
c.
d.
e.

• A. $b>d>a>c>e$
• B. $e>d>b>a>c$
• C. $b>d>c>a>e$
• D. $e>d>c>b>a$
• E. $d>e>c>b>a$

Answer 1

The correct option is C $b>d>c>a>e$

$\text{Comparing the acidic strengths of the given compounds}$

Presence of electron withdrawing groups in phenol increases the acidic strength whereas presence of electron donating groups generally decreases the acidic strength.

$\text{Analysis of phenol and p-methoxyphenol}:$

• In $p$-methoxyphenol, electron density on $O$ atom of $-OH$ group increases due to $+R$ effect of $-OC{H}_3$ group. This makes release of $H^{+}$ difficult and decreases acidity.
• Thus, phenol is more acidic than $p$-methoxyphenol.

$\text{Analysis of}m\text{-nitrophenol and phenol}:$

• In $m$-nitrophenol, $N{O}_2$ shows $-I$ effect due to which electron density on $O$ atom of $-OH$ group decreases. This makes release of $H^{+}$ easier and increases acidity.
• Thus $m$-nitrophenol is more acidic than both phenol and $p$-methoxyphenol.

$\text{Analysis of}p\text{-nitrophenol and}m\text{-nitrophenol}$

• In $p$-nitrophenol, electron density on $O$ atom of $-OH$ group decreases due to $-R$ effect of $-N{O}_2$ group. This makes release of $H^{+}$ easier and increases acidity. Moreover, $p$-nitrophenol is more acidic than $m$-nitrophenol since in $p$-nitrophenol both -$R$ and -$I$ effects of $-N{O}_2$ tend to increase acidity but in case of $m$-nitrophenol only –$I$ effect contribute significantly towards increasing acidity not -$R$ effect.
• Thus $p$-nitrophenol is more acidic than $m$-nitrophenol

Resonating Structures of $p$-nitro phenoxide ion


$\text{Analysis of phenol and}m\text{- methoxyphenol}:$

• In $m$-methoxyphenol, $-OC{H}_3$ shows –$I$ effect due to which electron density on $O$ atom of $-OH$ group decreases. This makes release of $H^{+}$ easier and increases acidity, Thus $m$-methoxyphenol is more acidic than phenol.

$\text{Analysis of m-nitrophenol and}m\text{-methoxyphenol}:$

• Both $m$-nitrophenol and $m$-methoxyphenol have electron withdrawing groups with respect to the $-OH$ group, and increase acidity. But $-I$ effect of $-N{O}_2$ group is greater than the $-I$ effect of $-OC{H}_3$ group i.e., $m$-nitrophenol will be more acidic than $m$-methoxyphenol.

Thus, the correct order of decreasing strengths of the acids will be:

$p$-nitrophenol > $m$-nitrophenol >
$m$-methoxyphenol > phenol >
$p$-methoxyphenol or

$b>d>c>a>e$

So, the correct answer is option $\left(E\right)$.