Chapter 2 : Alkanes

Ans 1 : The molecular formula of C₆H₁₄ contains the maximum number of H atoms for 6C (C_nH_2n+2) so it can cannot contain double bonds or rings. Therefore the only functional group that can be present is an alkane.  When you draw them it is a good idea to have a system whereby you start with the longest chain, then move to shorter chains with an increasing number of substitutents, moving them to different positions on the chain as you go. Watch that you don't repeat structures.

Ans 2 : The molecular formula of C₅H₁₀ contains 2 less H atoms than pentane (C₅H₁₂) so that it must contain either a double bond or one ring. Therefore in terms of functional groups we need to consider alkenes and cycloalkanes.

Ans 4 : In order to answer the question, it is a very good idea to draw a diagram that relates the following three equations that can be use to create the Hess's Law  "triangle" that relates the alkanes, their component elements and their combustion products.

From Hess's Law we get that ΔH_f (alkane) = ΔH_c(e) - ΔH_c(alkane)

ΔH_c(e)= 8 x ΔH_c(C) + 9 x ΔH_c(H₂) = 8 x -94.05 + 9 x -57.8 = -1272.6 kcal/mol.

So for 2,2-dimethylhexane, I, ΔH_f = -1272.6 - - 1304.6 = 32 kcal/mol

and for 2,2,3,3-tetramethylbutane, II,  ΔH_f = -1272.6 - -1303.0 = 30.4 kcal/mol

Thus 2,2,3,3-tetramethylbutane is the more stable since it has the more exothermic ΔH_f. (care ! both are calculated to be endothermic).  This is also evident in that II
  has the less exothermic ΔHc, and is the most branched of the two hydrocarbons. The relative stability is most clearly demonstrated and appreciated by the diagram, otherwise be very cautious with the signs of your calculations.

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