Return to Contents Chapter 2 : Alkanes  Ch 2 contents
Isomers Answers

Ans 1 : The molecular formula of C6H14 contains the maximum number of H atoms for 6C (CnH2n+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.

Constitutional isomers of C6H14

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.

Constitutional isomers of C5H10

Ans 3 : The molecular formula of C3H6O contains 2 less H atoms than propane (C3H8) so that it must contain either a double bond or one ring.
The combination of the double bond, ring and oxygen atoms means that the following functional groups are possible: aldehyde, alcohol, alkene, cycloalkane, epoxide, ether, ketone.

Constitutional isomers of C3H8O

Thermodynamics Answers

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.

Hess's Law cycle for C8H18 combustion

Reaction ? Eqn
C8H18 ----------> 8 CO2 + 9 H2O Combustion (known), ΔHc(alkane) 1
8 C + 9 H2 -----> 8 CO2 + 9 H2O Combustion (calculate), ΔHc(elements) 2
8 C + 9 H2 ----------> C8H18 Formation (required) ΔHf(alkane) 3

From Hess's Law we get that ΔHf (alkane) = ΔHc(e) - ΔHc(alkane)

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

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

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

Thus 2,2,3,3-tetramethylbutane is the more stable since it has the more exothermic ΔHf. (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.

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