NEET Practice Questions (MCQs) with Answers & Solutions

The given reaction sequence phenol to X (by Zn dust), then X to Y (by anhydrous AlCl3 and CH3Cl), and finally Y to Z (by alkaline KMnO4) is the oxidation of a side chain. It converts a methyl group to a carboxyl group, so the final product Z is benzoic acid.

${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\stackrel{\mathrm{P}+{\mathrm{I}}_2}{\to }{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{I}\stackrel{\mathrm{Mg}}{\to }{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{MgI}\stackrel{\mathrm{HCHO}}{\to }{\mathrm{C}}_2{\mathrm{H}}_5-{\mathrm{CH}}_2-\mathrm{OMgI}\stackrel{\mathrm{HOH}}{\to }{\mathrm{C}}_2{\mathrm{H}}_5-{\mathrm{CH}}_2-\mathrm{OH}$

(d) In Lucas test when Lucas reagent is treated 1$°$,2$°$ and 3$°$ alcohols, then turbidity appears, if turbidity is appeared immediately, then alcohol is tertiary. 2-methyl propan-2-ol is a tertiary alcohol. Hence, it reacts fastest with Lucas reagent

CH₃CH(OH)CH₂.CH₂.CH₃$\stackrel{\left[\mathrm{O}\right]}{\to }$CH₃CO.CH₂CH₂.CH₃$\underset{\mathrm{Poff}\text{'}\mathrm{s} \mathrm{rule}}{\overset{\left[\mathrm{O}\right]}{\to }}$CH₃COOH + CH₃CH₂COOH

(d) Reduction by H₂ is favoured by catalyst

(c) CH₃OH$\stackrel{\left[\mathrm{O}\right]}{\to }$HCHO$\stackrel{\left[\mathrm{O}\right]}{\to }$HCOOH

                     Aldehyde

(c) 3CH₃CH=CH₂ $\stackrel{{\mathrm{B}}_2{\mathrm{H}}_6}{\to }$(CH₃-CH₂-CH₂)₃B $\stackrel{{\mathrm{H}}_2{\mathrm{O}}_2/{\mathrm{OH}}^{-}}{\to }$CH₃-CH₂-CH₂-OH

                                                                            Propan-1-ol

Here, Half mol of (B2H6) diborane react with propane by Markownikoff's addition it gives tripropyl borane called hydroboration. In presence of H₂O₂  in basic medium tripropyl borane  gives alcohol. Remember that product is Anti-Markownikoff's rule that is 1-propanol. Reaction is called hydroboration oxidation.

(c)  p-nitrophenols are more acidic.

 CH₃CHOHCH₃ $\stackrel{{\mathrm{PCl}}_5}{\to }$CH₃CHClCH₃$\stackrel{\mathrm{Alc}.\mathrm{KOH}}{\to }$CH₃CH=CH₂$\stackrel{{\mathrm{H}}_2\mathrm{O}}{\to }$CH₃CHOHCH₃

              (Z)                    (X)                          (Y)                       (Z)