Return to Contents Chapter 9 : Alkynes Ch 9 contents
Alkyne Answers
Qu 1:  The π bond in an alkene is given by : C=C - C-C 

611 kJ/mol (146 kcal/mol) - 368 kJ/mol (88 kcal/mol) = 243 kJ/mol (58 kcal/mol)

The second π bond in an alkyne is given by : C≡C - C=C
  820 kJ/mol (196 kcal/mol) - 611 kJ/mol (146 kcal/mol) = 209 kJ/mol (50 kcal/mol)

This indicates that it should be recognised that the extra π bond in an alkyne is a weaker bond than the p bond in an alkene. 

Qu 2: Consider the following points about the reaction of 2-butyne with HBr

(a) Since the reaction gives predominantly the 2-bromo-2-butene, the product must be less reactive than the alkyne starting material otherwise a mixture the mono-bromide and the dibromide would be formed.

(b) carbocations from the protonation of 2-bromo-2-butene
Protonation of 2-bromo-2-butene gives a secondary carbocation regardless of which of the two C get protonated. 
However, since the only product is 2,2-dibromobutane, the intermediate must be that from the carbocation with the +ve charge formed adjacent to the bromine. 
This can be justified by considering the resonance stabilising influence the bromine can provide.

Qu 3: The final product is 2-butyne.

 The steps are: (i) propene to 1,2-dibromopropane (addition) then (ii) elimination to propyne, a terminal alkyne, and (iii) removal of the terminal H gives the carbanion which undergoes an SN2 with the methyl iodide forming a new CC bond.

Qu 4: The major products are:
        (a) 1-pentyne

(b) 2-pentyne


organic chemistry © Dr. Ian Hunt, Department of Chemistry University of Calgary