Question

Among the following four compounds
$\left(i\right)$ Phenol
$\left(ii\right)$ methyl phenol
$\left(iii\right)$ meta-nitrophenol
$\left(iv\right)$ para-nitrophenol
The acidity order is:

• A. $\left(iv\right)>\left(iii\right)>\left(i\right)>\left(ii\right)$
• B. $\left(iii\right)>\left(iv\right)>\left(i\right)>\left(ii\right)$
• C. $\left(i\right)>\left(iv\right)>\left(iii\right)>\left(ii\right)$
• D. $\left(ii\right)>\left(i\right)>\left(iii\right)>\left(iv\right)$

Answer 1

The correct option is A $\left(iv\right)>\left(iii\right)>\left(i\right)>\left(ii\right)$

In phenols, the presence of electron releasing groups decrease the acidity, whereas presence of electron withdrawing groups increase the acidity, compared to phenol. Among the meta and para-nitrophenols, the later is more acidic as the presence of $-N{O}_2$ group at para position stabilises the phenoxide ion to a greater extent than when it is present at meta position. Thus, correct order of acidity is : $\underset{\left(iv\right)}{\text{para-nitrophenol}}>\underset{\left(iii\right)}{\text{meta-nitrophenol}}>\underset{\left(i\right)}{\text{phenol}}>\underset{\left(ii\right)}{\text{methyl phenol}}$.