Equivalent Group Exercise

### First Help Section

A, B, C & D are C5H12O 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, F & G are C5H10 isomers.
| 3. How many rings and/or double bonds are suggested by this formula? enter a number.
| 4. What functional group is suggested by the hydrogen addition to give H? enter a formula or name.

Consider all possible structural formulas for the isomers of H, 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

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### Second Help Section

You should have concluded that H is (CH3)2CHCH2CH3, and that compounds A, B, C & D are alcohols having this carbon skeleton.
| 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 H is the key to solving this problem. From its formula H is clearly an alkane. There are three possible constitutional isomers having this formula. They are pentane, CH3(CH2)3CH3, 2-methylbutane, (CH3)2CHCH2CH3, and 2,2-dimethylpropane, (CH3)4C. For each of these carbon frameworks (skeletons) the number of possible alcohol derivatives is different, as noted in the following table.

Parent Alkane | Derived Alcohols
--- | ---
pentane | three (chain locations 1, 2 & 3)
2-methylbutane | 4 (chain locations 1, 2, 3 & 4)
2,2-dimethylpropane | one (chain location 1)

The assignment of structural formulas to B and C is straightforward, since B 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 D, both of which react with PCC to give isomeric aldehydes (M & N), it is necessary to consider the alkenes produced by dehydration with hot phosphoric acid. A gives an alkene E, which is also one of the alkenes produced by dehydration of B. Likewise D gives an alkene G, which is one of the alkenes produced by dehydration of C. This relationship permits an unambiguous assignment of structure to these alcohols.