Given a compound with the constitution $C_4{H}_8$ that contains two double bonds (i.e., a diene), how many stereoisomers are possible for this compound?

• A. $4$
• B. $1$
• C. $3$
• D. $2$

Answer: (A)

Verified step by step guidance

1. Identify the molecular formula and the type of compound: C\_4H\_8 with two double bonds means the compound is a diene with four carbons and two double bonds.
2. Determine the possible structures: For C\_4H\_8 with two double bonds, the compound is likely a conjugated or isolated diene, such as 1,3-butadiene or 1,2-butadiene. Consider the positions of the double bonds and the possibility of cis/trans (E/Z) isomerism around each double bond.
3. Analyze each double bond for stereoisomerism: Each double bond can have two configurations (E or Z) if the substituents on the double-bonded carbons are different. Check if both double bonds can exhibit E/Z isomerism independently.
4. Calculate the total number of stereoisomers: If both double bonds can independently be E or Z, the total number of stereoisomers is given by \$2^n\(, where \)n$ is the number of double bonds capable of E/Z isomerism.
5. Consider any symmetry or identical substituents that might reduce the number of stereoisomers: If the molecule has symmetry or identical groups that make some stereoisomers identical, adjust the count accordingly to find the final number of unique stereoisomers.