Animated Solution for Chemistry - Organic Chemistry: The number of isomeric tetraenes (NOT containing sp-hybridized carbon atoms) that can be formed from the following reaction sequence is ________.

Look closely at the starting material provided in the problem. We are dealing with a molecule that contains two distinct functional groups: an isolated cyclohexene ring and an internal alkyne side chain. Specifically, the molecule is $3\text{-(but-2-yn-1-yl)cyclohex-1-ene}$.

Having two different reactive sites means we must carefully evaluate the chemoselectivity of each reagent in the sequence. The question asks us to determine the total number of isomeric tetraenes that can be formed, with a strict condition: the final products must NOT contain any sp-hybridized carbon atoms. This constraint immediately rules out the formation of alkynes or allenes (cumulenes) in the final step.

The first reagent in our sequence is sodium metal dissolved in liquid ammonia (${\text{Na, liquid NH}}_3$). This is the classic setup for a Birch reduction.

The Birch reduction is highly specific; it provides solvated electrons that reduce internal alkynes to trans-alkenes. Crucially, isolated alkenes (like the one in our cyclohexene ring) are generally unreactive under these conditions because their Lowest Unoccupied Molecular Orbital (LUMO) is too high in energy to readily accept an electron.

Therefore, after the first step, the alkyne $-\text{C}\equiv \text{C}-$ group is cleanly converted into a trans $-\text{CH}=\text{CH}-$ group, leaving the ring's double bond completely untouched. Our first intermediate is $3\text{-(but-2-en-1-yl)cyclohex-1-ene}$.

Next, we introduce bromine in excess (${\text{Br}}_2$). Bromine is a classic reagent for the electrophilic addition across carbon-carbon double bonds.

Because the bromine is in excess, it will attack both available double bonds in our intermediate molecule. Two bromine atoms will add across the double bond in the cyclohexene ring, and two more will add across the newly formed trans double bond in the side chain.

This exhaustive bromination yields a tetrabromide intermediate. We now have a cyclohexane ring with two bromine atoms at adjacent positions, and a side chain that has been converted into a $2,3\text{-dibromobutyl}$ group.

The third and final reagent is alcoholic potassium hydroxide ($\text{alc. KOH}$). This is a strong base and a classic reagent for promoting E2 elimination (dehydrohalogenation).

Since our intermediate has four bromine atoms, the strong base will drive the elimination of four equivalents of $\text{HBr}$ to form four new double bonds, resulting in a tetraene. Let's break this down by region:

1. The Ring Elimination: The ring contains two bromine atoms at adjacent carbons (let's call them C1 and C2). To eliminate $\text{HBr}$, the base must remove a proton from a carbon adjacent to the bromine-bearing carbon. Removing a proton from C6 and the bromine from C1 forms a double bond between C1 and C6. Similarly, removing a proton from C3 and the bromine from C2 forms a double bond between C2 and C3. This specific elimination pathway yields a highly stable, conjugated $1,3\text{-cyclohexadiene}$ system. Forming an allene in a six-membered ring is sterically prohibited, so this is the only viable diene structure for the ring.

2. The Side Chain Elimination: The side chain is a $-{\text{CH}}_2-\text{CH}(\text{Br})-\text{CH}(\text{Br})-{\text{CH}}_3$ group. Double elimination of $\text{HBr}$ here will also favor the formation of a stable conjugated system. The base removes a proton from the terminal $-{\text{CH}}_3$ group and the adjacent bromine to form a terminal double bond ($={\text{CH}}_2$). It also removes a proton from the $-{\text{CH}}_2-$ group and the remaining bromine to form an internal double bond ($-\text{CH}=$). The resulting structure is a conjugated $1,3\text{-butadiene}$ side chain: $-\text{CH}=\text{CH}-\text{CH}={\text{CH}}_2$.

Now that we have the complete structure of our tetraene, we must count the possible stereoisomers.

First, look at the $1,3\text{-cyclohexadiene}$ ring. It contains no chiral centers, and because it is a six-membered ring, the double bonds are geometrically constrained to be strictly cis. Therefore, the ring contributes only $1$ structural variation.

Next, examine the conjugated side chain: $-\text{CH}=\text{CH}-\text{CH}={\text{CH}}_2$. The terminal double bond ($={\text{CH}}_2$) has two identical hydrogen atoms, meaning it cannot exhibit geometric isomerism. However, the internal double bond ($-\text{CH}=\text{CH}-$) can exist in either the E (trans) or Z (cis) configuration.

Since the ring has $1$ isomer and the side chain has $2$ isomers, the total number of isomeric tetraenes formed is $1\times 2=2$.