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

(c) Glycerol decolourises pink colour of phenolphthalein in borax solution in cold, which however regenerates on heating.

(a) Pyroligneous acid obtained during destructive distillation of wood contains mainly acetic acid (9 10%), methyl alcohol (2-25%) and acetone about 0.5%; the other distillation products are wood gas, wood charcoal, wood tar.

Acetone reacts with iodine in the presence of sodium hydroxide (NaOH) to form iodoform (CHI₃) and sodium acetate. This reaction is called the iodoform test and is used to distinguish methyl ketones from other types of ketones and aldehydes.

(a) 2-methyl-2-butene (molecular formula C₅H₁₀) yields acetone on ozonolysis.

CH₃COOH + NaHCO₃ $\to$CH₃COONa + H₂O + CO₂

  Acetic acid       Sodium 

                       Carbonate

NaHSO₃ gives the addition reaction with aldehyde and only aliphatic ketone. Acetophenone is the aromatic ketone so it does not give the addition product with NaHSO₃ aldehyde from the addition product with NaHSO₃ which on treatment with acid or base give again aldehyde.

$$\begin{gathered} \mathrm{RCHO}+{\mathrm{HSO}}_3\mathrm{Na}\to \mathrm{R}-\underset{{\mathrm{SO}}_3\mathrm{nA}}{\overset{\mathrm{OH}}{\underset{\left|}{\overset{\right|}{\mathrm{C}}}}}-\mathrm{H}\underset{{\mathrm{OH}}^{-}}{\overset{{\mathrm{H}}^{+} \mathrm{or}}{\to }}\mathrm{R}-\mathrm{CHO} \\ {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COCH}}_3+{\mathrm{NaHSO}}_3\to \mathrm{No} \mathrm{reaction} \end{gathered}$$

Both Iodoform reaction and reaction with NaHSO₃ can distinguish between 3-pentanone and 2-pentanone.

${\mathrm{CH}}_3\mathrm{CHO}+{\mathrm{Na}}^{+}{\mathrm{HSO}}_3^{-}\to {\mathrm{CH}}_3-\underset{\mathrm{H}}{\overset{\mathrm{OH}}{\underset{\left|}{\overset{\right|}{\mathrm{C}}}}}-{\mathrm{SO}}_3^{-}{\mathrm{Na}}^{+}$

NaHSO₃ reacts with aldehyde or methyl ketone to give above comp.

The given reaction is the ozonolysis of 2-butyne, followed by hydrolysis of the ozonide intermediate. The products of this reaction are acetic acid (CH3COOH) and propanoic acid (HOOC.CH2CH3). The ozonide first forms, which then undergoes hydrolysis to give these two carboxylic acids.

The alkene that will give acetaldehyde (CH3CHO) and acetone (CH3COCH3) on ozonolysis is 2-methylpropene (CH3CH=C(CH3)2). Ozonolysis cleaves the carbon-carbon double bond, and the resulting ozonide undergoes hydrolysis to give the two carbonyl compounds mentioned.