How Do You Draw (E)-1-Phenyl-1,3-butadiene?

The key is to identify which double bond carries the stereochemical descriptor. In (E)-1-phenyl-1,3-butadiene, the relevant bond is between carbons 1 and 2. Draw the phenyl group and the rest of the diene chain on opposite sides of that double bond.

Identify the structure and the stereochemical bond

Number the parent chain to give the double bonds the lowest locants. Here, either direction gives locants 1,3, so number from the phenyl-substituted end to give the phenyl group locant 1. The parent chain is buta-1,3-diene, and a phenyl group is attached to carbon 1. The connectivity is therefore:

$\mathrm{Ph-CH=CH-CH=CH_2}$

Here, “Ph” means a phenyl group, $\mathrm{C_6H_5}$. The first double bond is $\mathrm{C1=C2}$; the second is $\mathrm{C3=C4}$. The E descriptor applies to the first bond, not to the diene as a whole without regard to its numbering.

Decide what E means here

For each atom of a double bond, compare the two groups attached to it using the Cahn–Ingold–Prelog priority rules. At C1, phenyl has higher priority than hydrogen. At C2, the carbon chain continuing to C3 has higher priority than hydrogen. The two higher-priority groups must lie on opposite sides of the C1=C2 bond for the configuration to be E.

The chain continuing from C2 is $\mathrm{-CH=CH_2}$; it is not a hydrogen or a separate substituent to ignore. It is the higher-priority group on C2, so its position relative to phenyl determines the E/Z assignment.

Draw a worked sketch

One simple layout is to draw the first double bond horizontally and put phenyl above it at C1. Put the chain continuing from C2 below it:

          Ph       H
           \       /
            C1 = C2
           /       \
          H         C3 = C4H2

The slanted strokes indicate the bonds from C1 to phenyl and from C2 to C3; the hydrogens are shown to make the geometry clear. Phenyl and the C3 chain are on opposite sides of C1=C2, so this is the E arrangement. The drawing can be rotated or reflected as a whole and still represent the same stereoisomer, provided those two groups remain opposite across the double bond.

In a conventional skeletal drawing, you may omit carbon labels and most hydrogens. Keep the connectivity and relative positions clear: draw the phenyl-bearing alkene carbon, the other alkene carbon, and the continuation of the conjugated chain. A condensed version is useful for checking connectivity, but by itself, $\mathrm{Ph-CH=CH-CH=CH_2}$ does not show whether the first double bond is E or Z. The arrangement in the drawing must show it.

Check the other double bond

The second double bond is between C3 and C4. C4 is the terminal $\mathrm{CH_2}$ carbon, so it has two identical hydrogen substituents. E/Z notation requires each alkene carbon to have two different substituents; consequently, C3=C4 cannot be assigned E or Z. You do not need to impose a separate E/Z arrangement on that end of the molecule.

The molecule is a conjugated diene because its double bonds are separated by one single bond. Rotation around the central C2–C3 single bond can produce different s-cis and s-trans conformations. Those conformations are a separate issue from the E geometry at C1=C2: do not use an s-cis/s-trans label as a substitute for the E descriptor.

Common drawing mistakes

• Putting phenyl and the C3 chain on the same side. That gives the Z configuration at C1=C2, not E.
• Assigning E/Z to the terminal double bond. C4 has two identical hydrogens, so that bond has no E/Z descriptor.
• Treating the phenyl group as the entire molecule. The group attached to C2 is the continuation of the diene chain, and it must be included when comparing the two sides of C1=C2.
• Confusing alkene geometry with diene conformation. E/Z describes the arrangement at a double bond; s-cis and s-trans describe conformations around the central single bond.

A final check is to point to C1=C2 and identify its higher-priority groups: phenyl at C1 and the C3 chain at C2. If they appear on opposite sides, the drawing has the requested E geometry.