The number of stereoisomers possible for a compound of the molecular formula is: $C H_{3} - C H = C H - C H (O H) - M e$

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Solution

The correct option is C 4
d and l for cis form.
d and l for trans form.
So, total of 4 isomers are possible.

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The number of stereoisomers possible for a compound of the molecular formula CH₃—CH=CH—(CHCl)—Me is:

C H_{3} - C H = C H - C H_{3}

C H_{3} - C H = C H - C H (O H) - C H_{3}

C H_{3} - C H = C H - C | B r H - C H_{3}

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