NEET Practice Questions (MCQs) with Answers & Solutions

(d) ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COOC}}_2{\mathrm{H}}_5\underset{\mathrm{Boiling}}{\overset{\mathrm{KOH}}{\to }}$   ${\mathrm{C}}_6{\mathrm{H}}_6\mathrm{COOK} + {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$

    Ethyl benzoate

   ${\mathrm{C}}_6{\mathrm{H}}_6\mathrm{COOK}\stackrel{{\mathrm{H}}^{+}/{\mathrm{H}}_2\mathrm{O}}{\to }$  C₆H₅COOH
                                     (Benzoid Acid)

When malonic acid is heated with urea, it forms barbituric acid. This reaction is known as the Grimaux condensation reaction, where two molecules of malonic acid condense with one molecule of urea to form barbituric acid.

When a primary alcohol undergoes catalytic dehydrogenation, it forms an aldehyde. This reaction involves the removal of two hydrogen atoms from the alcohol group, resulting in the formation of a carbonyl group, which is characteristic of aldehydes.

 CH₃COOCH₃ $\stackrel{{\mathrm{LiAlH}}_4}{\to }$CH₃CH₂OH + CH₃OH

 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]^-