The number of isomeric tetraenes (NOT containing sp-hybridized carbon atoms) that can be formed from the following reaction sequence is_____.

The correct answer is 2.

To determine the number of isomeric tetraenes formed, let us break down the reaction sequence step-by-step:

Step 1: Reduction of Alkyne using Na / liquid NH₃

The starting material shown in the image is 3-(but-2-yn-1-yl)cyclohex-1-ene, which contains both a ring double bond and an internal alkyne group in the side chain:

Reagent 1, sodium in liquid ammonia ($\text{Na/liquid}{\text{NH}}_3$), selectively reduces internal alkynes into trans-alkenes via a free-radical mechanism. The isolated double bond present in the cyclohexene ring remains unreacted. Thus, the alkyne group $-\text{C}\equiv \text{C}-$ is converted into a trans-alkene $-\text{CH}=\text{CH}-$, resulting in a diene intermediate.

Step 2: Halogenation with excess Br₂

Adding excess bromine (${\text{Br}}_2$) results in electrophilic addition across both double bonds (the ring double bond and the side-chain double bond). This adds a total of 4 bromine atoms to the molecule, yielding a tetrabrominated saturated intermediate.

Step 3: Dehydrohalogenation with alcoholic KOH

Treatment with alcoholic $\text{KOH}$ induces elimination of 4 molecules of hydrogen bromide ($\text{HBr}$). Elimination of 4 $\text{HBr}$ units from the 10-carbon tetrabromo derivative generates 4 double bonds (a tetraene system).

Since the problem specifies that the tetraene must NOT contain sp-hybridized carbon atoms (meaning no cumulated double bonds/allenes or alkynes are allowed, only sp² and sp³ carbons), double bonds must be conjugated across the cyclic ring and side chain.

Considering the geometric isomers and possible conjugated double-bond positional arrangements around the 10-carbon framework without forming sp-carbons, exactly 2 isomeric tetraenes are formed.