To draw the Kekulé structure of trans 9-(2-phenylethenyl)anthracene, we start by recognizing that anthracene is a polycyclic aromatic hydrocarbon consisting of three fused benzene rings. The structure can be represented with alternating single and double bonds.

The 2-phenylethenyl group is attached at the 9-position of the anthracene. The trans configuration indicates that the substituents on either end of the double bond are positioned opposite each other, which affects the overall structure.

While I can’t draw here, you can visualize the structure as three linearly fused benzene rings with one of the terminal carbons of the anthracene bonded to a vinyl group (C=C) that connects to a phenyl group (C6H5) at the end.

Now, let’s calculate the molecular formula:

• Anthracene has the formula C14H10.
• The 2-phenylethenyl group adds C8H8 due to its structure (C6H5-CH=CH2).

When we combine these, we have:

• Total carbons (C): 14 (from anthracene) + 8 (from 2-phenylethenyl) = 22
• Total hydrogens (H): 10 (from anthracene) + 8 (from 2-phenylethenyl) = 18

Thus, the molecular formula in CxHy format is: C22H18.

Next, we calculate the degree of unsaturation (DU). The formula for calculating the degree of unsaturation is:

DU = (2C + 2 + N – H – X) / 2

Where C is the number of carbons, H is the number of hydrogens, N is the number of nitrogens (if any), and X is the number of halogens (if any). For our compound:

• C = 22
• H = 18
• N = 0
• X = 0

Plugging these values into the equation gives:

DU = (2(22) + 2 – 18) / 2 = (44 + 2 – 18) / 2 = 28 / 2 = 14.

Therefore, the degree of unsaturation for trans 9-(2-phenylethenyl)anthracene is 14.