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Answer 1 · 2016-07-02T14:28:32

Here's how to generate the IUPAC name.

The prime rule is that you choose the longest continuous chain that includes both the carboxylic acid and the alkene double bond.

Start from the $COOH$ $"COOH"$ and count to the left hand end. That gives you a 5-carbon chain:

$C-C=C-C-COOH$ $"C-C=C-C-COOH"$

The base name for this chain becomes "pentenoic acid".

The carboxyl carbon is number 1, so the double bond is between carbons 3 and 4.

The name becomes "pent-3-enoic acid", with the "3" coming as close as possible before the "en".

Now, we have two methyl groups on carbon 2:

${CH}_{3} CH=CHC {({CH}_{3})}_{2} COOH$ $"CH"_3"CH=CHC"("CH"_3)_2"COOH"$

The IUPAC name is 2,2-dimethylpent-3-enoic acid.