Question

Consider the following compound (X)

$\mathrm{H}-\stackrel{\mathrm{I}}{\mathrm{C}}=\stackrel{\mathrm{II}}{\mathrm{C}}-\stackrel{\mathrm{III}}{\mathrm{CH}_{2}}-\stackrel{\mathrm{CH}}{\mathrm{CH}}-\stackrel{\mathrm{IV}}{\mathrm{CH}_{3}}$

Answer 1

There are two stereoisomers of (X): the (E)- and (Z)-isomers of 1,4‑pentadiene.

Answer 2

Solution:

We analyze the structure given by

$$\mathrm{H}-\overset{I}{C}=\overset{II}{C}-\overset{III}{CH_2}-\overset{\,}{CH}=\overset{IV}{CH_3}$$

Rewriting this in a more conventional form, we have

$$CH_2=CH-CH_2-CH=CHCH_3.$$

Step 1. Identify the double bonds:

• The first double bond is at $CH_2=CH-$. Here the left carbon is $CH_2$ (two hydrogens) so it cannot exhibit cis–trans isomerism.
• The second double bond is at $-CH=CHCH_3$. On the left carbon of this double bond, one group is the $CH_2$ (from the adjoining $-CH_2-$) and the other is a hydrogen. On the right carbon, one substituent is a methyl group ($CH_3$) and the other is a hydrogen. Since both carbons have two different groups, this double bond can exist in two geometric forms (E and Z).

Step 2. Conclusion:

Thus, the molecule is 1,4‑pentadiene. Only the second double bond is stereogenic, giving rise to two geometric isomers: the (E)- and (Z)-1,4-pentadiene.

For a visual representation in mermaid:

Minimal Explanation of the Core Steps:

1. Rewrite the structure as $CH_2=CH-CH_2-CH=CHCH_3$.
2. Note that the terminal double bond (C₁=C₂) has a CH₂ group (no isomerism).
3. Recognize that the internal double bond (C₄=C₅) has two different substituents on each sp² carbon.
4. Conclude that two stereoisomers ((E) and (Z)) are possible.