Learning Activity #2 – Corrections

Some corrections and clarifications for Learning Activity #2 follow. Let me know if you have other suggestions.

  • #DU. A ring and a double-bond both count as 1 DU. Consider some three-carbon compounds. CH3CH2CH3 is propane, the saturated molecule. CH2=CHCH3 is C3H6. This molecule is one H2 short of saturation (1 DU). Cyclopropane, (CH2)3, contains only single bonds in a three-atom ring. Because its molecular formula is C3H6, it is also one H2 short of saturation (1 DU).
  • Model 4. The definition of alkane might be reworded to make it clearer. Let's try alkane = saturated hydrocarbon.
  • Fact 2.3 gives an incomplete definition of branched alkane. A branched alkane must contain at least one methine (CH) group OR one quaternary C.
  • The right-hand structure in Fig. 2.5 is incorrect. Add another CH3 group to the left end of the molecule so that there are 8 C's in the chain.
  • Re-word Q#19 to read, "Circle each group of C in Fig. 2.5 …". The idea is to draw a single circle around each alkyl substituent.
  • Q#20-22 might lead to some confusion (Sorrell will straighten you out). An alkane is a molecule in which all of the C and H bonding patterns are obeyed. An alkyl group contains one C with only three bonds, a rule violation. Thus, methyl is derived from methane. Q#20-21: the suffix that methyl and ethyl share is 'yl', not 'ethyl'. Applying this in Q#22 leads to CH3CH2CH2 = propyl and so on.

Friday, Aug 31 update. Several students called my attention to an apparent contradiction in Model 6 and Fig. 2.6. The molecule on the left is named 2-ethyl-1,4-dimethylcyclohexane. This name appears to contradict the rule given in Model 6 that states, "C#1 is the ring carbon bonded to the substituent that comes first in the alphabet (prefixes like “di” and “sec-” are ignored)". The rule suggests 'ethyl' is located at C#1, but the name that was provided clearly assigns a methyl to C#1.

I have checked various sources to see where I found this molecule and this rule. I believe the molecule is named correctly so I must have given you the wrong rule. I have found a book that says, "If there are exactly two alkyl groups, decide which C gets "1" based on alphabetical order". This rule doesn't apply here, however, because there are three alkyl groups. The same book also says, "[groups] are listed in a name alphabetically (not including prefixes di, tri, sec, or tert)", but this rule refers to the position of groups in the name and not the numbering of the ring. So I can't find any basis for the rule I gave you (except when there are exactly two alkyl groups) and I think my error was to combine two statements from my reference book into one. Yikes.

Fortunately, you are not being asked to learn how to assign names to formulas so I hope this error won't cause too much confusion.

This entry was posted in Post-lecture. Bookmark the permalink.