Double E2 elimination yields isomeric hexadienes Steps:

• 2,5-Dibromohexane undergoes two E2 eliminations with ethanolic KOH, forming C6H10 dienes by removing HBr from C1-C2/C5-C6 (terminal) or C2-C3/C4-C5 (internal).
• Both terminal eliminations produce hexa-1,5-diene (CH2=CHCH2CH2CH=CH2), 1 achiral isomer.
• One terminal and one internal produce hexa-1,4-diene (CH2=CHCH2CH=CHCH3), with E/Z stereoisomerism at the internal double bond (2 isomers).
• Both internal eliminations produce hexa-2,4-diene (CH3CH=CHCH=CHCH3), with (2E,4E), (2Z,4Z), and (2E,4Z) stereoisomers (3 isomers).
• Total: 1 + 2 + 3 = 6 isomers.

Why C is correct:

• Six isomers follow from structural diene types and E/Z stereoisomerism rule, where each trisubstituted double bond allows cis/trans configurations.

Why the others are wrong:

• A: Counts only structural isomers, ignoring E/Z stereoisomers.
• B: Omits one 2,4-hexadiene stereoisomer, such as (2E,4Z).
• D: Includes nonexistent products like 1,3-hexadiene, impossible without Br at C3/C4.

Final answer: C