Question

several alkene compounds are shown. indicate if each compound can exist as stereoisomers.

can $\ce{ch3ch=chch3}$ exist as stereoisomers?
$\bigcirc$ yes
$\bigcirc$ no

can $\ce{ch3ch=cbrch2ch3}$ exist as stereoisomers?
$\bigcirc$ yes
$\bigcirc$ no

can
\

$$\begin{matrix} \\ce{h} & & & \\\\ & \\ce{c}=\\ce{c} & & \\\\ & & & \\\\ \\end{matrix}$$

(with appropriate substituents as shown in the image) exist as stereoisomers?
$\bigcirc$ no
$\bigcirc$ yes

can
\

$$\begin{matrix} \\ce{h3c} & & \\ce{ch3} \\\\ & \\ce{c}=\\ce{c} & \\\\ \\ce{cl} & & \\ce{cl} \\end{matrix}$$

exist as stereoisomers?
$\bigcirc$ no
$\bigcirc$ yes

Explanation:

For an alkene to exist as stereoisomers (cis-trans isomers), each carbon in the double bond must be bonded to two distinct groups.

1. For $\text{CH}_3\text{CH=CHCH}_3$: Each double-bonded C has $\text{CH}_3$ and H. Both carbons have two unique groups, so stereoisomers exist.
2. For $\text{CH}_3\text{CH=CBrCH}_2\text{CH}_3$: The first double-bonded C has $\text{CH}_3$ and H; the second has $\text{Br}$ and $\text{CH}_2\text{CH}_3$. All groups are distinct per C, so stereoisomers exist.
3. For the drawn alkene: One double-bonded C has two ethyl groups, so it does not have two distinct groups. No stereoisomers exist.
4. For $\text{(CH}_3\text{)CCl=CCl(CH}_3\text{)}$: Each double-bonded C has $\text{CH}_3$ and $\text{Cl}$. Both carbons have two unique groups, so stereoisomers exist.

Answer:

1. Can $\text{CH}_3\text{CH=CHCH}_3$ exist as stereoisomers? yes
2. Can $\text{CH}_3\text{CH=CBrCH}_2\text{CH}_3$ exist as stereoisomers? yes
3. Can the drawn branched alkene exist as stereoisomers? no
4. Can $\text{(CH}_3\text{)CCl=CCl(CH}_3\text{)}$ exist as stereoisomers? yes