1. Write the structure clearly

The compound is 5-phenylpent-4-en-2-ol.

Main chain: pent-4-en-2-ol

• 5 carbons in chain
• double bond between C4 and C5
• OH group at C2
• phenyl group at C5

So the structure is:

$\text{CH}_3-\text{CH(OH)}-\text{CH}_2-\text{CH}=\text{C(H)(Ph)}$

Equivalently,

$\text{CH}_3-\text{CH(OH)}-\text{CH}_2-\text{CH}=\text{CH-Ph}$

2. Find chiral centers

Check carbon-2:

$\text{CH(OH)}$

It is attached to four different groups:

• $-OH$
• $-H$
• $-CH_3$
• $-CH_2-CH=CH-Ph$

Hence, C2 is a chiral center.

So this gives:

$2 \text{ optical isomers}$

3. Check geometrical isomerism of the double bond

Double bond is between C4 and C5.

• At C4, the two groups are:

  • $H$
  • $-CH_2-CH(OH)-CH_3$

• At C5, the two groups are:

  • $H$
  • phenyl

Since each double-bonded carbon has two different substituents, the alkene shows E/Z isomerism.

So this gives:

$2 \text{ geometrical isomers}$

4. Total possible stereoisomers

One chiral center and one $E/Z$ double bond are independent stereogenic elements.

Therefore total stereoisomers:

$2 \times 2 = 4$

5. Check for meso possibility

A meso form is not possible because there is only one chiral center and the molecule has no internal plane of symmetry that cancels chirality.

Hence the total remains:

$4$

Therefore, the possible number of stereoisomers is:

$\boxed{4}$