$$\text{The possible number of stereoisomers for 5-phenylpent -4-en-2-ol is } \underline{\hspace{2cm}}.$$

We are asked to find the number of possible stereoisomers for 5-phenylpent-4-en-2-ol.

The structure of 5-phenylpent-4-en-2-ol can be represented as $$C_6H_5-CH=CH-CH_2-CH(OH)-CH_3$$; numbering from the OH end gives C1 as $$CH_3$$, C2 as $$CH(OH)$$, C3 as $$CH_2$$, while the C4=C5 double bond connects to a phenyl group at C5.

At C2 there is a chiral center bearing four different groups ($$-CH_3$$, $$-OH$$, $$-H$$, and $$-CH_2CH=CHC_6H_5$$), which results in 2 optical isomers (R and S).

Moreover, the C4=C5 double bond allows geometric isomerism because C5 carries phenyl ($$C_6H_5$$) and H, while C4 carries H and $$-CH_2CH(OH)CH_3$$, so E/Z isomerism is possible, giving 2 geometric isomers.

Combining these possibilities yields a total of 2 optical isomers times 2 geometric isomers, giving $$2 \times 2 = 4$$ stereoisomers.

The number of possible stereoisomers for 5-phenylpent-4-en-2-ol is 4.