If 3 – bromo 4 – methyl hexane is treated with alc. $KOH$  solution, how many different alkenes would be formed. (Including stereo isomers)

When 3-bromo-4-methylhexane is treated with alcoholic KOH, elimination (E2) reaction occurs to form alkenes by removal of a β-hydrogen from the adjacent carbon atoms.

Stepwise analysis:

• The bromo substituent is at carbon-3.
• β-hydrogens available for elimination are from C-2 and C-4.
• From C-2 elimination → double bond between C2 and C3.
• From C-4 elimination → double bond between C3 and C4.

Counting alkenes:

1. Alkene with double bond between C2=C3:
   This alkene does not have any stereocenter about the double bond (no cis/trans possible due to one side being a methyl and the other hydrogen).
2. Alkene with double bond between C3=C4:
   This alkene has substituents on both carbons of the double bond, so cis (Z) and trans (E) isomers are possible, giving 2 stereoisomers.

Total different alkenes (including stereoisomers):

• 1 (C2=C3 alkene) + 2 (C3=C4 cis/trans) = 3 alkenes

Final answer: Three different alkenes can be formed upon treatment of 3-bromo-4-methylhexane with alcoholic KOH, including stereoisomers.