MCQ
Order of acidic nature
- A$a > c > d > b$
- ✓$a > b > d > c $
- C$a > b > c > d$
- D$d > c > b > a$
✓
Answer
Correct option: B.
$a > b > d > c $
b
In compound $(a)$, $NO _{2}$ possesses the $- M$ effect at para position due to which it is highly acidic than other three compounds. The compound $(b)$ has intermolecular hydrogen bonding and it is less acidic than $(a)$. In compound $(d)$, there is absence of an electron donating group, due to which it is more acidic than compound $(c)$, which has electron donating methyl group. Therefore, the correct order of acidic nature is $a > b > d > c$
In compound $(a)$, $NO _{2}$ possesses the $- M$ effect at para position due to which it is highly acidic than other three compounds. The compound $(b)$ has intermolecular hydrogen bonding and it is less acidic than $(a)$. In compound $(d)$, there is absence of an electron donating group, due to which it is more acidic than compound $(c)$, which has electron donating methyl group. Therefore, the correct order of acidic nature is $a > b > d > c$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Formic acid is a weak acid and hydrochloric acid is a strong acid. It follow that the
Steric inhibition of resonance take place in
The major product of the following reaction is
→$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,} \\
{|\,\,\,\,\,\,\,\,} \\
{C{H_3} - C - CHC{H_3}} \\
{\,|\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,H\,\,\,\,\,\,\,Br}
\end{array}$ $\xrightarrow{{C{H_3}OH}}$
$+\, CH_3SNa$ $\xrightarrow{{{C_2}{H_5}OH}}$ ?
→The product formed is
The major product of the given reaction is
→ $[Figure]$ $\xrightarrow[{(ii)\,{H_2}S{O_4}\,,\,heat}]{{(i)\,OHCC{H_2}COCl}}$
Which of the following is an uncommon hydrolysis product of $XeF_2$ and $XeF_4$?
→During titration of acetic acid with aq. $NaOH$ solution, the neutralisation graph has a vertical line. This line indicates
→Oxidation number of $N$ in $HN{O_3}$ is
→The oxidation states of nitrogen in $NO , NO _{2}, N _{2} O$ and $NO _{3}^{-}$ are in the order of :
→Zirconium phosphate $[Zr_3 (PO_ 4)_4]$ dissociates into three zirconium cations of charge $+ 4$ and four phosphate anions of charge $- 3$. If molar solubility of zirconium phosphate is denoted by $S$ and its solubility product by $K_{sp}$ then which of the following relationship between $S$ and $K_{sp}$ is correct?
→