Aldehyde ketones & carboxylic Acid MCQs for NEET — Chemistry Questions with Answers

 HCOOH reacts with NaHCO₃ giving out effervescences of CO₂. Note that HCOOH is also strong reducing agent.

 C₆H₅COCH₃ $\stackrel{{\mathrm{Cl}}_2}{\to }$C₆H₅COCH₂Cl

                             Tear gas

(b) CH₃COCH₃ + 3I₂ + 4NaOH $\to$CHI₃ + CH₃COONa + 3NaI + 3H₂O

      Acetone                               Iodoform

(b) All aldehydes are easily oxidised to respective acids and thus, acts as reducing agents.

The compound is pentan-3-one.

phenyl hydrazone indicate, presence of C=0

-ve tollen, means, not aldehyde, -ve iodoform mean no methyl gr at any terminal side

As the given compound does not answer Tollen's test, it is a ketone. The ketone is not a methyl ketone, as it fails to answer iodo form reaction.  Ketones on reaction with phenyl hydrazine gives phenyl hydrazone.

The ketone is 3-pentanone. That on reduction using Zn/Hg and HCl gives hydrocarbon pentane. Thus the formula of the compound A is ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{COCH}}_2{\mathrm{CH}}_3$

The possible reactions 

$$\begin{gathered} \underset{3-\mathrm{pentanone}}{{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{COCH}}_2{\mathrm{CH}}_3}\stackrel{{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NHNH}}_2}{\to }{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{C}\left({\mathrm{CH}}_2{\mathrm{CH}}_3\right)=\underset{\mathrm{Phenyl} \mathrm{hydrozone}}{{\mathrm{NNHC}}_6{\mathrm{H}}_5} \\ {\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{COCH}}_2{\mathrm{CH}}_3\stackrel{\mathrm{Zn}/\mathrm{Hg} \mathrm{and} \mathrm{HCl}}{\to }\underset{\mathrm{Pentane}}{{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_3} \end{gathered}$$

Thus the compound A is 3- pentanone

(c) RCOCl + AlCl₃ $\to$R$\stackrel{+}{\mathrm{C}}$O + [AlCl₄]^-

(a) Aldol condensation in aldehydes is due to presence of $\mathrm{\alpha }$-hydrogen atoms. These aldehydes which does not have $\mathrm{\alpha }$-hydrogen atom like HCHO, does not give aldol condensation.

LiAlH₄ is used for converting $-$COOH to $-$CH₂OH

(d) Three electron-withdrawing groups attached on -CH.   

Among the given options, only 2-hydroxy propane (isopropyl alcohol) can be oxidized to the corresponding carbonyl compound, which is acetone. The other options, such as ortho-nitro phenol, phenol, and 2-methyl-2-hydroxy propane, cannot be easily oxidized to carbonyl compounds.