NEET Practice Questions (MCQs) with Answers & Solutions

ROH + CH₃COCl ------- ROCOCH₃ + HCl 

    M                                   (M + 42)

 Acylation of each OH group results in an increase of molecular mass of 42. In the given problem molecular mass of the compound increases by 84 on acylation. Therefore, the compound has two OH groups.

Na reacts with alcohol and acid .. but here all options r alcohol

thy form di ethyl ether , it means alcohol has 2 carbon , so its ethanol

This is Reimer-Tiemann reaction.

When ethylene oxide is treated with a Grignard reagent (RMgX), it undergoes nucleophilic addition to form a primary alcohol. The Grignard reagent attacks the electrophilic carbon of the epoxide ring, and the magnesium salt is then hydrolyzed to yield the primary alcohol.

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.