NEET Practice Questions (MCQs) with Answers & Solutions

$ CH_2 = CH_2 + Br_2 \rightarrow BrCH_2 - CH_2 + Br $ (Brown) (Colour less) Disappearance of brown colour is the test for unsaturation

$ 2CHCI_3 + 6 Ag \rightarrow HC \equiv CH + 6 AgCl $

$ CH_3 - CHCl_2 + 2 alc. KOH \rightarrow HC \equiv CH + 2KCl + 2 H_2O $

Ethylene dibromide (1,2-dibromoethane) on heating with alcoholic KOH undergoes dehydrohalogenation to form acetylene (C2H2). The reaction involves the elimination of two molecules of HBr, resulting in the formation of a triple bond.

Propyne ($CH_3C≡CH$) is formed by the reaction of $CH_3Br$ (methyl bromide) with sodium acetylide ($NaC≡CH$). The sodium acetylide acts as a nucleophile and displaces the bromine atom from the methyl bromide, forming propyne.

When acetylene ($C_2H_2$) is passed through a red hot iron tube, it undergoes a trimerization reaction to form benzene ($C_6H_6$). This is a well-known industrial process for producing benzene.

The reagent $Hg^{2+} / H_2SO_4$ is used to convert alkynes to ketones. In this case, But-1-yne would be converted to Butane-2-one through a hydration reaction in the presence of mercury(II) sulfate as a catalyst. This reaction involves the addition of water (H-OH) across the triple bond, following Markovnikov's rule, resulting in the formation of a ketone.