Learn Carboxylic Acid MCQ questions & answers in Organic Chemistry are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
51.
The compound $$(X)$$ is
A
$$C{H_3}COOH$$
B
$$BrC{H_2} - COOH$$
C
$${\left( {C{H_3}CO} \right)_2}O$$
D
$$CHO - COOH$$
Answer :
$${\left( {C{H_3}CO} \right)_2}O$$
This reaction is an example of “Perkin reaction”. The compound X should be $${\left( {C{H_3}CO} \right)_2}O.$$
In this step the carbanion is obtained by removal of an $$\alpha - H$$ atom from a molecule of an acid anhydride, the anion of the corresponding acid acting as a necessary base.
52.
A compound $$(X)$$ with a molecular formula $${C_5}{H_{10}}O$$ gives a positive 2, 4-$$DNP$$ test but a negative Tollens' test. On oxidation it gives a carboxylic acid $$(Y)$$ with a molecular formula $${C_3}{H_6}{O_2}.$$ Potassium salt of $$(Y)$$ undergoes Kolbe's reaction and gives a hydrocarbon $$(Z).$$ $$(X), (Y)$$ and $$(Z)$$ respectively are
A
pentan-3-one, propanoic acid, butane
B
pentanal, pentanoic acid, octane
C
2-methylbutanone, butanoic acid, hexane
D
2, 2-dimethylpropanone, propanoic acid, hexane
Answer :
pentan-3-one, propanoic acid, butane
Since the compound gives positive 2, 4-$$DNP$$ test and negative Tollens' test, it is a ketone.
53.
The compound formed when malonic acid is heated with urea, is
A
cinnamic acid
B
butyric acid
C
barbituric acid
D
crotonic acid
Answer :
barbituric acid
54.
Which of the following will not undergo $$HVZ$$ reaction?
A
Propanoic acid
B
Ethanoic acid
C
2-Methylpropanoic acid
D
2, 2-Dimethylpropanoic acid
Answer :
2, 2-Dimethylpropanoic acid
2, 2-Dimethylpropanoic acid will not undergo $$HVZ$$ reaction due to absence of $$\alpha {\text{ - }}H$$ atom.
55.
In the following sequence of reactions, what is $$D?$$ \[\xrightarrow{\left[ O \right]}A\]
\[A\xrightarrow{SOC{{l}_{2}}}B\xrightarrow{Na{{N}_{3}}}C\xrightarrow{\text{heat}}D\]
A
Primary amine
B
An amide
C
Phenyl isocyanate
D
A higher hydrocarbon
Answer :
Phenyl isocyanate
\[{{C}_{6}}{{H}_{5}}C{{H}_{3}}\xrightarrow{\left[ O \right]}\underset{\left[ A \right]}{\mathop{{{C}_{6}}{{H}_{5}}COOH}}\,\] \[\xrightarrow{SOC{{l}_{2}}}\underset{\left[ B \right]}{\mathop{{{C}_{6}}{{H}_{5}}COCl}}\,\xrightarrow{Na{{N}_{3}}}\] \[\underset{\left[ C \right]}{\mathop{{{C}_{6}}{{H}_{5}}CO{{N}_{3}}}}\,\xrightarrow{\text{heat}}\] \[\underset{\text{Phenyl isocyanate,}\left[ D \right]}{\mathop{\to {{C}_{6}}{{H}_{5}}NCO}}\,\]
56.
The major product $$H$$ of the given reaction sequence is
\[C{{H}_{3}}-C{{H}_{2}}-CO-C{{H}_{3}}\xrightarrow{\overset{\Theta }{\mathop{C}}\,N}\] \[G\xrightarrow[{{\text{Heat}}}]{{95\% \,\,{H_2}S{O_4}}}H\]
A
B
C
D
Answer :
57.
When acetaldehyde is heated with Fehling’s solution it gives a precipitate of
58.
Which of the following represent the correct decreasing order of acidic strength of
following?
(i) Methanoic acid
(ii) Ethanoic acid
(ii) Propanoic acid
(iv) Butanoic acid
A
(i) > (ii) > (iii) > (iv)
B
(ii) > (iii) > (iv) > (i)
C
(i) > (iv) > (iii) > (i)
D
(iv) > (i) > (iii) > (ii)
Answer :
(i) > (ii) > (iii) > (iv)
The correct order of acidic strength is methanoic acid > ethanoic acid > propanoic acid > butanoic acid because the $$+I$$ - effect of alkyl group increases in the order.
$$C{H_3} < {C_2}{H_5} < {C_3}{H_7} < {C_4}{H_9}$$
$${\text{Acidic Nature}} \propto \frac{{ - I{\text{ - effect}}\left( {EWG} \right)}}{{ + I{\text{ - effect}}\left( {ERG} \right)}}$$
$$-I$$ - effect increases hence, acidic nature increases.
59.
\[C{{H}_{3}}C{{H}_{2}}COOH\xrightarrow[\text{Red}\,P]{B{{r}_{2}}}X\xrightarrow{N{{H}_{3}}\left( alc. \right)}Y\]
$$Y$$ in the above reactions is
60.
Which of the following compounds will react with \[NaHC{{O}_{3}}\] solution to give sodium salt and carbon dioxide ?
A
Acetic acid
B
$$n$$ - Hexanol
C
Phenol
D
Both (B) and (C)
Answer :
Acetic acid
Only acetic acid is more acidic than carbonic acid ( conjugate acid of \[NaHC{{O}_{3}}\] ) hence it dissolves in \[NaHC{{O}_{3}},\]
\[\underset{\text{stronger acid}}{\mathop{C{{H}_{3}}COOH}}\,+NaHC{{O}_{3}}\to \] \[C{{H}_{3}}CO{{O}^{-}}N{{a}^{+}}+\underset{\text{weaker acid}}{\mathop{{{H}_{2}}C{{O}_{3}}}}\,\left( {{H}_{2}}O+C{{O}_{2}} \right)\]
while phenol and $$n$$ - hexanol are less acidic than carbonic acid and hence do not dissolve in \[NaHC{{O}_{3}}.\]