Isomeric Tetraenes Formed from Alkyne Reaction Sequence

To determine the total number of isomeric tetraenes (containing no $sp$-hybridized carbon atoms) formed from the given reaction sequence, we analyze each step sequentially:

Step 1: Reduction with $\text{Na / liquid NH}_3$

• Starting Material: 3-(but-2-yn-1-yl)cyclohex-1-ene.
• Reaction: Dissolving metal reduction ($\text{Na / liquid NH}_3$) selectively reduces alkynes to $(E)$-alkenes (trans-alkenes) while leaving non-conjugated ring double bonds unreacted.
• Product 1: $(E)$-3-(but-2-en-1-yl)cyclohex-1-ene.

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Step 2: Bromination with $\text{Br}_2$ (excess)

• Reaction: Excess $\text{Br}_2$ undergoes electrophilic addition across both double bonds (the ring double bond at $C_1=C_2$ and the side-chain double bond at $C_8=C_9$).
• Product 2: Tetrabromo intermediate, 1,2-dibromo-3-(2,3-dibromobutyl)cyclohexane.
  • Bromine atoms are attached at carbons $C_1, C_2, C_8,$ and $C_9$.

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Step 3: Elimination with Alcoholic $\text{KOH}$

• Reaction: Dehydrohalogenation ($\text{E2}$ elimination) removes four molecules of $\text{HBr}$ from the tetrabromo intermediate to form a tetraene.
• Constraint: The resulting tetraene must NOT contain any $sp$-hybridized carbon atoms (i.e., no allenes or alkynes).

1. Ring Double Bonds:

   • To eliminate $\text{HBr}$ from $C_1(\text{Br})$ and $C_2(\text{Br})$ without forming an allene ($C_1=C_2=C_3$), double bonds must be formed at $C_6=C_1$ and $C_2=C_3$. This yields a stable, conjugated diene within the 6-membered ring.

2. Side-Chain Double Bonds:

   • To eliminate $\text{HBr}$ from $C_8(\text{Br})$ and $C_9(\text{Br})$ without forming an allene ($C_7=C_8=C_9$ or $C_8=C_9=C_{10}$), double bonds must be formed at $C_7=C_8$ and $C_9=C_{10}$.

Thus, the structural formula of the tetraene is: $\text{cyclo-}(C_6=C_1-C_2=C_3)-C_7\text{H}=C_8\text{H}-C_9\text{H}=C_{10}\text{H}_2$

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Counting the Isomers

• Ring Double Bonds ($C_6=C_1$ and $C_2=C_3$): Due to ring strain in a 6-membered ring, both double bonds are restricted strictly to the $(Z)$-configuration.
• Terminal Double Bond ($C_9=C_{10}$): Since carbon $C_{10}$ is bonded to two identical hydrogen atoms ($=\text{CH}_2$), it cannot show geometrical isomerism.
• Acyclic Double Bond ($C_7=C_8$): Carbon $C_7$ is bonded to $-\text{H}$ and the cyclohexadienyl ring, while carbon $C_8$ is bonded to $-\text{H}$ and $-\text{CH}=\text{CH}_2$. Therefore, the $C_7=C_8$ double bond exhibits geometrical isomerism and can exist in two forms:
  1. $(E)$-isomer
  2. $(Z)$-isomer

Thus, the total count of isomeric tetraenes formed is 2.