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Chapter 34: Problem 769

(a) Write out all of the possible carbon skeletons for acyclic terpene and sesquiterpene hydrocarbons that follow the isoprene rule. Do not consider double-bond position isomers. (b) Do the same for monocyclic terpenes with a six-membered ring.

Short Answer

Expert verified
(a) The possible carbon skeletons for acyclic terpene hydrocarbons are: 1. Linear arrangement: \(CH_2=C(CH_3)CH_2CH_2C(CH_3)=CH_2\) 2. Branched arrangement: \(CH_2=C(CH_3)CH_2C(CH_3)(CH_2CH_2C(CH_3)=CH_2)\) For acyclic sesquiterpene hydrocarbons, there are three possible arrangements: 1. Linear arrangement: \(CH_2=C(CH_3)CH_2CH_2C(CH_3)=CH_2CH_2C(CH_3)CH_2CH_2C(CH_3)=CH_2\) 2. Middle isoprene branching: \(CH_2=C(CH_3)CH_2C(CH_3)(CH_2CH_2C(CH_3)=CH_2)CH_2CH_2C(CH_3)=CH_2\) 3. Terminal isoprene branching: \(CH_2=C(CH_3)CH_2CH_2C(CH_3)=CH_2C(CH_3)(CH_2CH_2C(CH_3)=CH_2)CH_2CH_2C(CH_3)=CH_2\) (b) For monocyclic terpenes with a six-membered ring, the carbon skeleton is: \(C_6H_{10}-CH_2CH_2C(CH_3)=CH_2\)

Step by step solution

01

Understanding the isoprene rule

The isoprene rule states that terpenes and sesquiterpenes are composed of an integer number of isoprene units (C5H8) joined together. Terpenes have 10 carbons (2 isoprene units) and sesquiterpenes have 15 carbons (3 isoprene units).
02

Acyclic terpene hydrocarbons

For acyclic terpenes, we need to draw carbon skeletons containing 10 carbon atoms, with 2 isoprene units joined together. There are two possible ways to do this: 1. Linear arrangement: The two isoprene units can join in a straight chain, without any branching. In this case, the carbon skeleton will be: \(CH_2=C(CH_3)CH_2CH_2C(CH_3)=CH_2\) 2. Branched arrangement: One of the isoprene units forms a branch on the main chain. In this case, the carbon skeleton will be: \(CH_2=C(CH_3)CH_2C(CH_3)(CH_2CH_2C(CH_3)=CH_2)\)
03

Acyclic sesquiterpene hydrocarbons

For acyclic sesquiterpenes, we need to draw carbon skeletons containing 15 carbon atoms, with 3 isoprene units joined together. There are three possible ways to do this: 1. Linear arrangement: The three isoprene units join in a straight chain, without any branching. In this case, the carbon skeleton will be: \(CH_2=C(CH_3)CH_2CH_2C(CH_3)=CH_2CH_2C(CH_3)CH_2CH_2C(CH_3)=CH_2\) 2. Branched arrangement: One of the isoprene units forms a branch on the main chain. There are two possibilities for this: a) Middle isoprene branching: \(CH_2=C(CH_3)CH_2C(CH_3)(CH_2CH_2C(CH_3)=CH_2)CH_2CH_2C(CH_3)=CH_2\) b) Terminal isoprene branching: \(CH_2=C(CH_3)CH_2CH_2C(CH_3)=CH_2C(CH_3)(CH_2CH_2C(CH_3)=CH_2)CH_2CH_2C(CH_3)=CH_2\)
04

Monocyclic terpenes with a six-membered ring

For monocyclic terpenes, we need to draw carbon skeletons containing a six-membered ring and only 10 carbon atoms. Since there are only two isoprene units in terpenes, this means one of the isoprene units will form a six-membered ring. This results in only one possible carbon skeleton: \(C_6H_{10}-CH_2CH_2C(CH_3)=CH_2\)

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