Carboxylic Acid MCQ Questions & Answers in Organic Chemistry | Chemistry

Acid chlorides, anhydrides, acids give ester with alcohol
$$\eqalign{ & ROH + R'COCl \to ROCOR' + HCl \cr & ROH + {\left( {R'CO} \right)_2}O \to ROCOR' + RCOOH \cr & ROH + R'COOH \to ROCOR' + {H_2}O \cr} $$

Acidic strength depends upon the stability of anion formed of stability of anion formed is high, its acidic strength will also be high. Order of stability of formed anion :
$$HC \equiv CO{O^ - } > {H_2}C = CHCO{O^ - } > {C_6}{H_4}\left( {OMe} \right)CO{O^ - } > {H_3}CC{H_2}CO{O^ - }$$

The electron withdrawing $$\left( { - I} \right)$$  group $$ - Cl$$  withdraws electrons from $$O - H$$  bond and thus helps the cleavage of the $$O - H$$  bond releasing hydrogen as $${H^ + }.$$

Claisen condensation

Carboxylic acid is stronger than alcohol and water because after removal of proton, carboxylate ion is stabilised by resonance. Hence, correct order of acid strength is
$$RCOOH > HOH > ROH > HC \equiv CH$$
Which is based upon the rate of donation of proton or strength of base, thus order of basic strength is $$\mathop C\limits^ - \equiv CH > R - \mathop O\limits^ - > O{H^ - } > RCO{O^ - }$$

Due to $$ortho$$  - effect, (I) and (II) are stronger acids than (III) and (IV). Due to two $$ortho$$  hydroxyl groups in (I), it is stronger acid than (II). (III) is a stronger acid than (IV) because at $$m$$ - position, $$-OH$$  group cannot exert $$+R$$  effect but can only exert $$-I$$  effect while at $$p$$ - position, $$-OH$$  group exerts its strong $$+R$$  effect. Thus, the correct order of acidity is : I > II > III > IV.

Electron withdrawing group attach to the benzene ring increases the reactivity towards nucleophilic sustitution reaction. Since $$ - N{O_2}$$  group is strong electron withdrawing group. Hence in basic medium ester containing $$ - N{O_2}$$  group Will hydrolysed most easily.

According to the above equation, it is clear that reduction of hydrolysed product of ester by $$LiAl{H_4}$$  gives two alcohols.

\[\underset{\text{Acetic acid}}{\mathop{C{{H}_{3}}COOH}}\,+\underset{\begin{smallmatrix} \text{Sodium} \\ \text{carbonate} \end{smallmatrix}}{\mathop{NaHC{{O}_{3}}}}\,\to \]       $$C{H_3}COON\mathop a\limits^ + \, + {H_2}O + C{O_2} \uparrow $$