Question

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

Answer 1

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: `"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.
#color(white)(aaaaaaaaaaa)#![http://www.sigmaaldrich.com/catalog/product/aldrich/c111805?lang=en&region=US]()

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.

• `color(orange)"4-ethyl"`
• `color(blue)"2-isopropyl"`
• `color(magenta)"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.

`color(white)(aaaa)`

`ul"Adding Substituents"`

Looking at `color(magenta)"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 `color(blue)"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 `color(orange)"4-ethyl"`, we place an ethyl group at carbon number `4`. Doing so, our finished structure looks like this:

`color(white)(aaaaa)"4-ethyl-2-isopropyl-1-methylcyclopentane"`