Question

check the box under each compound that exists as a pair of cis/trans isomers. if none of them do, check the none of the above box under the table. table with four compounds:

1. ( ce{ho-c(ch2oh)=c-ch3} ) (structure with triple bond? or double bond, typo)
2. ( ce{cl-c=c-cl} )
3. ( ce{ho-ch=ch-oh} )
4. ( ce{cl-c=ch-cl} )

plus a none of the above checkbox.

Explanation:

To determine if a compound has cis - trans isomers, we need to check two conditions: (1) The compound has a carbon - carbon double bond ($\ce{C = C}$); (2) Each carbon atom in the double bond has two different groups attached to it.

Step 1: Analyze the first compound $\ce{HO - C\equiv C - CH3}$ (with an additional $\ce{CH2OH}$? Wait, the structure seems to have a triple bond $\ce{C\equiv C}$. Compounds with triple bonds do not have cis - trans isomers because the carbon atoms in a triple bond are sp - hybridized and the bond is linear, so there is no possibility of different spatial arrangements around the bond. So this compound (with a triple bond) does not have cis - trans isomers.

Step 2: Analyze the second compound $\ce{Cl - C = C - Cl}$ (with two $\ce{Cl}$ on each carbon? Wait, the structure is $\ce{Cl - C = C - Cl}$? Wait, no, the given structure is $\ce{Cl - C = C - Cl}$ with two $\ce{Cl}$ on each carbon? Wait, no, the left carbon has two $\ce{Cl}$? Wait, no, the structure is $\ce{Cl - C = C - Cl}$? Wait, no, the original structure: first row second column: $\ce{Cl - C = C - Cl}$ (each carbon has two $\ce{Cl}$? No, that can't be. Wait, no, the correct way: for a double bond $\ce{C = C}$, each carbon must have two different groups. In $\ce{Cl - C = C - Cl}$, each carbon is attached to two $\ce{Cl}$ atoms (assuming the structure is $\ce{Cl2C = CCl2}$? No, the drawing shows $\ce{Cl - C = C - Cl}$ with two $\ce{Cl}$ on each carbon? Wait, no, maybe it's $\ce{Cl - C = C - Cl}$ where each carbon has one $\ce{Cl}$ and one... Wait, no, if both carbons in the double bond have the same groups (both have $\ce{Cl}$ and the other group is also $\ce{Cl}$? No, that would be the same. Wait, no, let's re - examine.

Wait, the third compound: $\ce{HO - CH = CH - OH}$. Let's check this. The double bond is $\ce{C = C}$. Each carbon in the double bond: the left carbon is attached to $\ce{HO -}$ and $\ce{H -}$ (wait, no, the formula is $\ce{HO - CH = CH - OH}$, so each carbon in the double bond is attached to $\ce{OH}$ and $\ce{H}$? Wait, no, $\ce{HO - CH = CH - OH}$: the first carbon (left) is bonded to $\ce{HO -}$ and $\ce{H -}$ (since it's $\ce{CH}$), and the second carbon (right) is bonded to $\ce{OH -}$ and $\ce{H -}$? Wait, no, $\ce{HO - CH = CH - OH}$: each carbon in the double bond has two different groups? Wait, no, the left carbon: $\ce{HO -}$ and $\ce{H -}$ (from $\ce{CH}$), and the right carbon: $\ce{OH -}$ and $\ce{H -}$ (from $\ce{CH}$). Wait, but actually, in $\ce{HO - CH = CH - OH}$, the two carbons in the double bond: each is attached to $\ce{OH}$ and $\ce{H}$? Wait, no, the formula is $\ce{HO - CH = CH - OH}$, so the left $\ce{CH}$ is bonded to $\ce{HO -}$ and the right $\ce{CH}$ is bonded to $\ce{OH -}$. Wait, no, let's draw it: $\ce{H - O - C = C - O - H}$. Wait, no, the correct structure for $\ce{HO - CH = CH - OH}$ is $\ce{HOCH = CHOH}$, so the double bond is between the two $\ce{CH}$ groups. Each $\ce{CH}$ group: one is bonded to $\ce{OH}$ and $\ce{H}$, and the other is also bonded to $\ce{OH}$ and $\ce{H}$? No, that would be the same. Wait, no, maybe I made a mistake.

Wait, the fourth compound: $\ce{Cl - C = CH - Cl}$ (with a $\ce{Cl}$ on the top of the left carbon). So the left carbon in the double bond is attached to two $\ce{Cl}$ atoms (one on the top, one on the left) and the right carbon is attached to $\ce{CH -}$ and $\ce{Cl -}$. Wait, no, the left carbon: $\ce{Cl}$ (top), $\ce{Cl}$ (left), and the double bond to the right carbon. The right carbon: $\ce{CH -}$ and $\ce{Cl -}$ (right). So the left carbon has two $\ce{Cl}$ atoms, so it does not have two different groups.

Wait, let's recall the cis - trans isomerism rules. For a compound to have cis - trans isomers, it must have a carbon - carbon double bond, and each carbon in the double bond must have two different substituents.

First compound: has a triple bond ($\ce{C\equiv C}$), so no cis - trans isomers.

Second compound: $\ce{Cl - C = C - Cl}$ (assuming both carbons have two $\ce{Cl}$ atoms, so same substituents on each carbon, so no cis - trans isomers).

Third compound: $\ce{HO - CH = CH - OH}$. Let's look at the two carbons in the double bond. The left carbon: bonded to $\ce{HO -}$ and $\ce{H -}$ (from $\ce{CH}$), the right carbon: bonded to $\ce{OH -}$ and $\ce{H -}$ (from $\ce{CH}$). Wait, but $\ce{HO -}$ and $\ce{OH -}$ are different? No, $\ce{HO -}$ is the same as $\ce{OH -}$ (just the direction of bonding, but in terms of substituents, $\ce{OH}$ is the same). Wait, no, the left carbon is $\ce{CH}$ bonded to $\ce{OH}$ (as $\ce{HO -}$) and the right carbon is $\ce{CH}$ bonded to $\ce{OH}$ (as $\ce{OH -}$). Wait, actually, in $\ce{HOCH = CHOH}$, the two carbons in the double bond: each has $\ce{OH}$ and $\ce{H}$ as substituents. So we have a double bond, and each carbon has two different substituents ($\ce{OH}$ and $\ce{H}$). So this compound can have cis - trans isomers. Wait, but let's check again.

Wait, the fourth compound: $\ce{Cl - C = CH - Cl}$ (left carbon has two $\ce{Cl}$ atoms, so it has two identical substituents, so no cis - trans isomers).

The second compound: $\ce{Cl - C = C - Cl}$ (if both carbons have two $\ce{Cl}$ atoms, then each carbon has two identical substituents, so no cis - trans isomers).

The first compound: triple bond, no cis - trans isomers.

Wait, maybe I made a mistake with the third compound. Let's re - express the third compound: $\ce{HO - CH = CH - OH}$. The double bond is between the two $\ce{CH}$ groups. Each $\ce{CH}$ group: one is bonded to $\ce{OH}$ and $\ce{H}$, and the other is also bonded to $\ce{OH}$ and $\ce{H}$. Wait, but that would mean that the two substituents on each carbon are $\ce{OH}$ and $\ce{H}$, so they are different. So the compound $\ce{HO - CH = CH - OH}$ (i.e., ethylene glycol with a double bond? Wait, no, ethylene glycol is $\ce{HOCH2CH2OH}$, but this is $\ce{HOCH = CHOH}$, which is a different compound, called 1,2 - dihydroxyethene. In this compound, the two carbons in the double bond each have $\ce{OH}$ and $\ce{H}$ as substituents, so it can exist as cis - trans isomers.

Wait, but let's check the other compounds again.

First compound: $\ce{HO - C\equiv C - CH3}$ (with an additional $\ce{CH2OH}$? The structure shows a triple bond, so no cis - trans isomers.

Second compound: $\ce{Cl - C = C - Cl}$ (if the structure is $\ce{Cl2C = CCl2}$, then each carbon has two $\ce{Cl}$ atoms, so same substituents, no cis - trans isomers.

Fourth compound: $\ce{Cl - C = CH - Cl}$ (left carbon has two $\ce{Cl}$ atoms, so same substituents, no cis - trans isomers.

So the only compound that can have cis - trans isomers is $\ce{HO - CH = CH - OH}$, so we check the box under $\ce{HO - CH = CH - OH}$.

Wait, but maybe I made a mistake. Let's recall the exact rule: for cis - trans isomerism, the double bond must be non - rotatable (which it is), and each carbon in the double bond must have two different groups. So in $\ce{HO - CH = CH - OH}$, each carbon has $\ce{OH}$ and $\ce{H}$, so two different groups. So this compound has cis - trans isomers.

Answer:

We check the box under the compound $\boldsymbol{\ce{HO - CH = CH - OH}}$. If none of the compounds had met the criteria, we would check "none of the above", but in this case, $\ce{HO - CH = CH - OH}$ meets the criteria for cis - trans isomerism.