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

1 Answer
Jul 2, 2017

Answer:

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,

paint

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:

paint

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:

paint

Lastly, looking at #color(orange)"4-ethyl"#, we place an ethyl group at carbon number #4#. Doing so, our finished structure looks like this:

enter image source here
#color(white)(aaaaa)"4-ethyl-2-isopropyl-1-methylcyclopentane"#