The major product [B] in the following sequence of reactions is:

• Step 1: Hydroboration-Oxidation (Formation of Alcohol)

  The starting alkene undergoes hydroboration-oxidation using $$\text{B}_2\text{H}_6$$ followed by alkaline $$\text{H}_2\text{O}_2$$. This reaction results in the net anti-Markovnikov addition of water across the double bond with syn-stereochemistry, yielding the secondary alcohol:

  $$\text{2,4-dimethylhexan-3-ol} \quad \left[\text{H}_3\text{C--CH}\left(\text{CH}(\text{CH}_3)_2\right)\text{--CH(OH)--CH}_2\text{--CH}_3\right]$$



• Step 2: Carbocation Formation via Dehydration

  Treating this alcohol with dilute sulfuric acid and heat ($$\text{dil. H}_2\text{SO}_4/\Delta$$) leads to protonation of the hydroxyl group followed by the loss of a water molecule ($$-\text{H}_2\text{O}$$). This generates a secondary carbocation:

  $$\text{H}_3\text{C--CH}\left(\text{CH}(\text{CH}_3)_2\right)\text{--}\text{CH}^+\text{--CH}_2\text{--CH}_3$$



• Step 3: Carbocation Rearrangement (1,2-Hydride Shift)

  To achieve greater stability, a 1,2-hydride shift ($$\text{H}^-$$ shift) occurs from the adjacent tertiary carbon to the positively charged secondary carbon atom. This rearranges the intermediate into a highly stable tertiary carbocation.




• Step 4: Deprotonation to the Saytzeff Product (Formation of B)

  Finally, elimination of a proton ($$-\text{H}^+$$) occurs from the adjacent carbon following Saytzeff's rule to form the most highly substituted, conjugated, and thermodynamically stable alkene as the major product:

  $$\text{Product B} = \text{2,3-dimethylhex-2-ene} \quad \left[(\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)\text{CH}_2\text{CH}_2\text{CH}_3\right]$$

Conclusion:

The reaction sequence goes through an anti-Markovnikov alcohol formation, followed by a carbocation generation that undergoes a thermodynamic hydride shift to yield the most substituted alkene (2,3-dimethylhex-2-ene).

Answer: Option B — 2,3-dimethylhex-2-ene