The number of stereoisomers possible for a compound of the molecular formula $$CH_3 - CH = CH - CH(OH) - Me$$ is :

The molecule may be written in condensed form as $$CH_3-CH = CH-CH(OH)-CH_3$$, i.e. 3-pent-2-en-4-ol.

Two different types of stereogenic elements are present:

1. Alkene double bond $$C_2 = C_3$$
   • At $$C_2$$ the two substituents are $$H$$ and $$CH_3$$ (different).
   • At $$C_3$$ the two substituents are $$H$$ and $$CH(OH)CH_3$$ (different).
   Hence the double bond can show geometrical (E/Z or cis/trans) isomerism.
   Number of possible configurations here = $$2$$.

2. Chiral centre at $$C_4$$ [the carbon bearing $$OH$$]
   The four different groups attached are: $$OH$$, $$H$$, $$CH_3$$, and the allylic fragment $$CH=CHCH_3$$.    Therefore $$C_4$$ is an asymmetric carbon and can exist in both $$R$$ and $$S$$ configurations.
   Number of possible configurations here = $$2$$.

Because the alkene configuration and the chiral configuration are independent, the total number of stereoisomers is obtained by the product rule:

Total stereoisomers $$= 2 \times 2 = 4$$.

There is no meso form, since only one chiral centre is present; mirror images are non-superposable.

Hence the number of stereoisomers possible is $$4$$.

Option C which is: $$4$$