# How can I draw 3–methylpent–2–ene, (CH_3CH = C(CH_3) CH_2CH_3), and know if it shows E-Z isomerism?

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#### Explanation

Explain in detail...

#### Explanation:

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3
Aug 24, 2015

First you expand the condensed structural formula to a bond-line view, and then you check whether there are two different groups on each end of the $\text{C=C}$ bond.

#### Explanation:

Step 1. Draw the bond-line structure.

Step 2. Check for two different substituents on each end.

In the structure above, the left hand $\text{C}$ atom has an $\text{H}$ and a ${\text{CH}}_{3}$ (two different substituents).

The right hand end has a ${\text{CH}}_{3}$ and a ${\text{CH"_2"CH}}_{3}$ (two different substituents).

The compound shows $E \text{/} Z$ isomerism.

In the above structure, the ${\text{CH}}_{3}$ group on the left hand end has priority 1, while the ${\text{CH"_2"CH}}_{3}$ group on the right hand end has priority 1.

Since the two high-priority groups are on opposite sides of the double bond, the compound is ($E$)-3-methylpent-2-ene.

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