Equivalent Group Exercise

### First Help Section

A, B, C & D are C6H14O constitutional isomers.
| 1. How many rings and/or double bonds are suggested by this formula? enter a number.
| 2. What functional group is suggested by the reaction with sodium? enter a formula or name.
E is a C6H12 compound.
| 3. How many rings and/or double bonds are suggested by this formula? enter a number.
| 4. What functional group is suggested by the reactions with bromine and ozone? enter a formula or name.

Consider all possible structural formulas for the formula of E, and select the one you believe agrees best with the evidence presented in this problem. Check your selection by drawing the structure.

Go to Second Help Section

This is the end of the designated help screen>

### Second Help Section

You should have concluded that E is 3-methyl-2-pentene (cis & trans isomers), and that compounds A, B, C & D are alcohols having this carbon skeleton (Wolff-Kishner reduction of each gives 3-methylpentane).
| 1. What class of alcohols is not oxidized by PCC? enter a number or a name.
| 2. What class of alcohols gives aldehydes on oxidation with PCC? enter a number or a name.

After completing this section, you should be able to write structural formulas for all the unknown compounds described in this problem. Try to do so before viewing the answers.

View Puzzle Answers

Deducing the structure of E is the key to solving this problem. From its formula E is clearly an alkene or cycloalkane. Qzonolysis shows it to be an alkene, and also locates the double bond. The compound is 3-methyl-2-pentene (cis and/or trans isomers), and the corresponding alkane is 3-methylpentane. For this carbon framework (skeleton) there are three possible alcohol derivatives.
The assignment of structural formulas to C and D is straightforward, since D does not react with PCC and is therefore a 3¡-alcohol (only one is possible) and C reacts to give a ketone (K) and is therefore a 2¡-alcohol. To distinguish the 1¡-alcohols A and B, both of which react with PCC to give isomeric aldehydes (M & N), it is necessary to consider their chirality. A is chiral and its derived aldehyde M is also chiral. B is achiral, as is its derived aldehyde N.