Learn Carboxylic Acid MCQ questions & answers in Organic Chemistry are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
31.
The compound that does NOT liberate $$C{O_2}$$ , on treatment with aqueous sodium bicarbonate solution, is
A
Benzoic acid
B
Benzenesulphonic acid
C
Salicylic acid
D
Carbolic acid (Phenol)
Answer :
Carbolic acid (Phenol)
Carbolic acid (Phenol) is weaker acid than carbonic acid and hence does not liberate $$C{O_2}$$ on treatment with aq. $$NaHC{O_3}$$ solution.
32.
Acetic acid can be halogenated in presence of phosphorus and chlorine. Formic acid cannot be halogenated with same way because of
A
presence of $$\alpha {\text{ - }}H$$ atom in formic acid
B
presence of $$\alpha {\text{ - }}H$$ atom in acetic acid
C
absence of $$\alpha {\text{ - }}H$$ atom in $$C{H_3}COOH$$
D
higher acidic strength of acetic acid than formic acid
Answer :
presence of $$\alpha {\text{ - }}H$$ atom in acetic acid
No explanation is given for this question. Let's discuss the answer together.
33. \[\xrightarrow[\left( ii \right)\,{{H}^{+}}/{{H}_{2}}O]{\left( i \right)\,NaOH/{{100}^{\circ }}C}?\]
The major product in the above reaction is
A
B
C
D
Answer :
This is intramolecular Cannizzaro’s reaction.
34.
An ester $$(A)$$ with molecular formula $${C_9}{H_{10}}{O_2}$$ was treated with excess of $$C{H_3}MgBr$$ and the complex so formed was treated with $${H_2}S{O_4}$$ to give an olefin $$(B).$$ Ozonolysis of $$(B)$$ gave a ketone with molecular formula $${C_8}{H_8}O$$ which shows $$+ve$$ iodoform test. The structure of $$(A)$$ is
A
$${C_6}{H_5}COO{C_2}{H_5}$$
B
$$C{H_3}COC{H_2}CO{C_6}{H_5}$$
C
$$p{\text{ - }}C{H_3}O{\text{ - }}{C_6}{H_4}{\text{ - }}COC{H_3}$$
D
$${C_6}{H_5}COO{C_6}{H_5}$$
Answer :
$${C_6}{H_5}COO{C_2}{H_5}$$
Since ketone with molecular formula $${C_8}{H_8}O$$ shows $$+ve$$ iodoform test, therefore, it must be a methyl ketone, i.e., $${C_6}{H_5}COC{H_3}.$$ Since this ketone is obtained by the ozonolysis of an olefin $$(B)$$ which is obtained by the addition of excess of $$C{H_3}MgBr$$ to an ester $$(A)$$ with molecular formula $${C_9}{H_{10}}{O_2},$$ therefore, ester $$(A)$$ is $${C_6}{H_5}COO{C_2}{H_5}$$ and the olefin $$(B)$$ is
\[\overset{\begin{smallmatrix}
C{{H}_{3}} \\
|\,\,\,\,\,
\end{smallmatrix}}{\mathop{{{C}_{6}}{{H}_{5}}C=C{{H}_{2}}}}\,\] as explained below :
\[\underset{A\left( M.F.\,{{C}_{9}}{{H}_{10}}{{O}_{2}} \right)}{\mathop{{{C}_{6}}{{H}_{5}}COO{{C}_{2}}{{H}_{5}}}}\,\xrightarrow[\left( \text{ii} \right)\,\frac{{{H}^{+}}}{{{H}_{2}}O}]{\left( \text{i} \right)\,2C{{H}_{3}}MgBr}\] \[{{C}_{6}}{{H}_{5}}\underset{\begin{smallmatrix}
| \\
\,\,\,C{{H}_{3}}
\end{smallmatrix}}{\overset{\begin{smallmatrix}
\,\,\,\,\,C{{H}_{3}} \\
|
\end{smallmatrix}}{\mathop{-C-}}}\,OH\xrightarrow[-{{H}_{2}}O]{{{H}_{2}}S{{O}_{4}}}\]
\[{{C}_{6}}{{H}_{5}}\underset{\left( B \right)}{\overset{\begin{smallmatrix}
\,\,\,\,\,C{{H}_{3}} \\
|
\end{smallmatrix}}{\mathop{-C=}}}\,C{{H}_{2}}\xrightarrow[Zn]{{{O}_{3}}}\underset{\text{M}\text{.F}\text{.}\,\,{{C}_{8}}{{H}_{8}}O}{\mathop{{{C}_{6}}{{H}_{5}}COC{{H}_{3}}}}\,+HCHO\]
35.
Which of the following reactions does not occur?
A
B
C
D
Answer :
No explanation is given for this question. Let's discuss the answer together.
36.
\[C{{H}_{3}}C{{H}_{2}}COOH\xrightarrow[\text{red}\,P]{C{{l}_{2}}}A\] \[\xrightarrow{\text{alc}\text{.}\,KOH}B.\] What is $$B$$ ?
37.
In a set of the given reactions, acetic acid yielded a product $$C.$$
$$C{H_3}COOH + PC{l_5} \to $$ \[A\xrightarrow[\text{ Anhy}.\,AlC{{l}_{3}}]{{{C}_{6}}{{H}_{6}}}B\xrightarrow[\text{ether}]{{{C}_{2}}{{H}_{5}}MgBr}C\]
Product $$C$$ would be
A
$$C{H_3}CH\left( {OH} \right){C_2}{H_5}$$
B
$$C{H_3}CO{C_6}{H_5}$$
C
$$C{H_3}CH\left( {OH} \right){C_6}{H_5}$$
D
\[\overset{\begin{smallmatrix}
{{C}_{2}}{{H}_{5}}\, \\
|\,\,\,\,\,\,\,\,\,
\end{smallmatrix}}{\mathop{C{{H}_{3}}-C\left( OH \right){{C}_{6}}{{H}_{5}}}}\,\]
The correct order of acidity is $$CC{l_3}COOH > CHC{l_2}COOH$$ $$ > C{H_2}ClCOOH > C{H_3}COOH$$
39.
Which one of the following orders of acid strength is correct ?
A
\[RCOOH>HOH>ROH>HC\equiv CH\]
B
\[RCOOH>HOH>HC\equiv CH>ROH\]
C
\[RCOOH>HC\equiv CH>HOH>ROH\]
D
\[RCOOH>ROH>HOH>HC\equiv CH\]
Answer :
\[RCOOH>HOH>ROH>HC\equiv CH\]
\[RCOOH>ROH>HC\equiv CH.\] Water is more acidic then alcohol.
40.
The end product $$(Z)$$ in the given sequence of reactions is
\[C{{H}_{3}}CH=CHCHO\xrightarrow{NaB{{H}_{4}}}\] \[X\xrightarrow[ZnC{{l}_{2}}]{HCl}Y\xrightarrow[\left( \text{ii} \right)\,{{H}^{+}}]{\left( \text{i} \right)\,KCN}Z\]