Chemistry NCERT Exemplar Solutions Class 11th Chapter Thirteen

The intermediate carbocation formed in the reactions of HI, HBr and HCl with propene is the same and the bond energy of HCl, HBr and HI is 430.5 kJ mol^–1, 363.7 kJ mol^–1 and 296.8 kJ mol^–1 respectively. What will be the order of reactivity of these halogen acids?

Rotation around carbon-carbon single bond of ethane is not completely free. Justify the statement.

Alkynes on reduction with sodium in liquid ammonia form trans alkenes. Will the butene thus formed on reduction of 2-butyne show the geometrical isomerism?

Alkynes on reduction with sodium in liquid ammonia form trans alkenes. Will the butene thus formed on reduction of 2-butyne show the geometrical isomerism?

Why do alkenes prefer to undergo electrophilic addition reaction while arenes prefer electrophilic substitution reactions? Explain.

In the presence of peroxide addition of HBr to propene takes place according to anti Markovnikov's rule but peroxide effect is not seen in the case of HCl and HI. Explain.

An unsaturated hydrocarbon 'A' adds two molecules of H₂ and on reductive ozonolysis gives butane-1,4-dial, ethanal and propanone. Give the structure of 'A', write its IUPAC name and explain the reactions involved.

896 mL vapour of a hydrocarbon 'A' having carbon 87.80% and hydrogen 12.19% weighs 3.28g at STP. Hydrogenation of 'A' gives 2-methylpentane. Also 'A' on hydration in the presence of H₂SO₄ and HgSO₄ gives a ketone 'B' having molecular formula C₆H₁₂O. The ketone 'B' gives a positive iodoform test. Find the structure of 'A' and give the reactions involved.

An alkyl halide C₅H₁₁Br (A) reacts with ethanolic KOH to give an alkene 'B', which reacts with Br₂ to give a compound 'C', which on dehydrobromination gives an alkyne 'D'. On treatment with sodium metal in liquid ammonia one mole of 'D' gives one mole of the sodium salt of 'D' and half a mole of hydrogen gas. Complete hydrogenation of 'D' yields a straight chain alkane. Identify A,B, C and D. Give the reactions invovled.