Chemistry MCQs for NEET — Practice Questions with Answers

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.

(b) Ketones can be prepared by the oxidation of secondary alcohols by using oxidising agent such as K₂Cr₂O₇/H₂SO_4.

     ${}_{{\mathrm{H}}_3\mathrm{C}}{}^{{\mathrm{H}}_3\mathrm{C}}>\mathrm{CHOH}$ + [O] $\to$  ${}_{{\mathrm{H}}_3\mathrm{C}}{}^{{\mathrm{H}}_3\mathrm{C}}>$C=O + H₂O

        2$°$ alcohol               Dimethyl ketone

In the aldol condensation reaction, two aldehydes or a ketone and an aldehyde undergo nucleophilic addition followed by dehydration to form a β-hydroxy aldehyde or a β-hydroxy ketone respectively. Therefore, the product formed is a β-hydroxy aldehyde or a β-hydroxy ketone.

The given reaction sequence involves the dehydroxylation of phenol to benzene using Zn dust, followed by chlorination of benzene with anhydrous AlCl3 to form chlorobenzene, and finally oxidation of chlorobenzene with alkaline KMnO4 to give benzoic acid.

(a) Clemmensen reduction Aldehydes and ketones are reduced to the corresponding alkanes by means of amalgamated zinc and HCl.

>C=O + 4[H]$\stackrel{\mathrm{Zn}-\mathrm{Hg}/\mathrm{HCl}}{\to }$>CH₂ + H₂O

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

(a) The -OH group of alcohol or the -COOH group of a carboxylic acid is replaced by -Cl using phosphorus pentachloride (i.e. PCl₅)

  ROH + PCl₅ $\to$RCl + POCl₃ + HCl

Alcohol

  RCOOH + PCl₅$\to$RCOCl + POCl₃ + HCl

   Acid

(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)