# How do you draw this cyclic pentane? 4-ethyl-2-isopropyl-1-methylcyclopentane

Jul 2, 2017

See below

#### Explanation:

Let's break this down a bit.

First, your job is to identify the parent chain, which is the longest carbon chain.

That's easy as the last part of the structure's name gives it away: $\text{cyclopentane}$. Pentane means it's a 5 carbon alkane and the prefix cyclo- means the parent chain is cyclic. It looks like this without any alkyl chains attached to it.
$\textcolor{w h i t e}{a a a a a a a a a a a}$

Next, you have to identify the substituents. Well, knowing that cyclopentane is the parent chain, whatever else that we have left are treated as the substituents. Let's write them down.

• $\textcolor{\mathmr{and} a n \ge}{\text{4-ethyl}}$
• $\textcolor{b l u e}{\text{2-isopropyl}}$
• $\textcolor{m a \ge n t a}{\text{1-methyl}}$

If we take our cyclopentane and first number all the carbons,

we can then, starting in numerical order, attach the substituents one-by-one.

$\textcolor{w h i t e}{a a a a}$

$\underline{\text{Adding Substituents}}$

Looking at $\textcolor{m a \ge n t a}{\text{1-methyl}}$, we can see we have a methyl group attached at carbon number $1$. Placing it at the appropriate position, we get the following structure:

Next, we take a look at $\textcolor{b l u e}{\text{2-isopropyl}}$. We have an isopropyl group attached at carbon number $2$. Placing this alkyl group at its appropriate position we get the following structure thus far:

Lastly, looking at $\textcolor{\mathmr{and} a n \ge}{\text{4-ethyl}}$, we place an ethyl group at carbon number $4$. Doing so, our finished structure looks like this:

$\textcolor{w h i t e}{a a a a a} \text{4-ethyl-2-isopropyl-1-methylcyclopentane}$