Question

(A) $Ph-OH$
(B) $-C{H}_2-OH$
(C) $Ph-C{O}_{2}H$
(D) $-C{H}_2-\stackrel{\oplus }{N}{H}_3$

Decreasing order of acid strength is:

• A. $B>A>C>D$
• B. $C>A>B>D$
• C. $C>A>D>B$
• D. $C>B>A>D$

Answer 1

The correct option is C $C>A>D>B$

$Ph-C{O}_{2}H$ is more acidic than $Ph-OH$ because $Ph-C{O}_2-$ has negative charge stabilized by resonance in benzene ring as well as between the $C-O$ bonds. In case $Ph-O-$ it is only with the ring. These both are strong acidic than $-C{H}_2-OH$ and $-C{H}_2-\stackrel{+}{N}{H}_3$ due to reason of resonance stabilization. Also, $-C{H}_2-\stackrel{\oplus }{N}{H}_3$ is more acidic than $-C{H}_2-OH$ as after the removal of $H^{+},-C{H}_2-N{H}_2$ is more stable than $-C{H}_2-O-$(owing to absence of any charged).
Hence, the order of acidic strength is: $C>A>D>B$.