Q11. Give the IUPAC name of CH3CH2C≡CCH(CH3)2.
2-Methylhex-3-yne. The chain has six carbons and the triple bond starts at carbon 3 from either end; the tie is broken by giving the methyl group the lower number, 2 rather than 5.
Q12. Write the equation for the reaction of propyne with sodamide and name the product.
CH3C≡CH + NaNH2 → CH3C≡C−Na+ + NH3. The product is sodium propynide (NCERT equation 9.61).
Q13. Why do alkynes not show geometrical isomerism?
The sp carbons make the C–C≡C–C unit linear, and each triply bonded carbon carries only one other group. There is no pair of groups on each carbon that could lie on the same or opposite sides.
Q14. How many moles of H2 and of Br2 can one mole of but-2-yne add? Name the final products.
Two moles of each. With H2 (Pt, Pd or Ni) the final product is butane, C4H10; with Br2 it is 2,2,3,3-tetrabromobutane, CH3CBr2CBr2CH3.
Q15. Arrange HC≡CH, CH3C≡CH and CH3C≡CCH3 in decreasing order of acidity, and explain.
HC≡CH > CH3C≡CH ≫ CH3C≡CCH3 (NCERT page 316). The +I methyl group of propyne destabilises its anion, so propyne is weaker than ethyne; but-2-yne has no hydrogen on an sp carbon.
Q16. What is formed when but-2-yne is warmed with water, HgSO4 and dilute H2SO4?
Butanone, CH3COCH2CH3. Water adds across the triple bond to give the enol CH3C(OH)=CHCH3, which rearranges to the ketone.
Q17. Write the IUPAC name of CH2=CH–C≡C–CH3 (NCERT Exercise 9.2 (b)).
Pent-1-en-3-yne. The five-carbon chain is numbered from the end nearer a multiple bond, which puts the double bond at 1 and the triple bond at 3; -ene comes before -yne in the name.
Q18. Write the equation for the complete combustion of hexyne (NCERT Exercise 9.8 (iii)).
C6H10 + 17/2 O2 → 6CO2 + 5H2O, or 2C6H10 + 17O2 → 12CO2 + 10H2O. A hydrocarbon CxHy needs x + y/4 mol O2; here 6 + 10/4 = 8.5.