#1 quiz 3 preparation

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Angela_Kim_1N
Posts: 21
Joined: Fri Jul 22, 2016 3:00 am

#1 quiz 3 preparation

Postby Angela_Kim_1N » Tue Mar 07, 2017 10:58 am

Name the compound shown below.

..........................................CH(CH3)2
.............................................I
CH3 CH2 CH2 CH2 CH2 CH2 CH CH2 CH2 CH3

(ignore the periods)

The answer for this problem is 4-isopropyldecane or 4-(1-methylethyl) decane but I don't understand how to come up with both of these names for this structure.

Thanks!

Paul Sedaros 1B
Posts: 23
Joined: Fri Sep 25, 2015 3:00 am

Re: #1 quiz 3 preparation

Postby Paul Sedaros 1B » Tue Mar 07, 2017 12:13 pm

Since there is a carbon chain with 10 carbons, we know the base of the chain is a decane. We also see an isopropyl substituent (CH(CH3)2) attached to the 4th carbon in the decane chain. This is how we get 4-isopropyldecane. For the other answer, if we draw out the CH(CH3)2, we can see a methyl and an ethyl group as substituents. The methyl group comes off the 1st carbon, and since both the methyl and ethyl still are on the 4th carbon, we can name it as 4-(1-methylethyl decane).

Manpreet Singh 1N
Posts: 41
Joined: Wed Sep 21, 2016 2:59 pm
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Re: #1 quiz 3 preparation

Postby Manpreet Singh 1N » Tue Mar 07, 2017 12:25 pm

Aadding on, we use the carbon count from right to left because it gives us a small number. From left to right the isoporpyl would have been on the 7th carbon, but from right to left it is attached to the 4th carbon. Professor Lavelle said we want the smaller number when writing the name of the compound.

Hope that helps :)


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