Question

\[1,2\;\]di-bromocyclohexane on dehydrohalogenation gives
A.
B.
C.
D.

Answer 1

Hint: \[1,2\;\]di-bromocyclohexane has the following structure


Dehydrohalogenation is a chemical reaction that undergoes removal of halogen and hydrogen to form alkene. ‘de’ means removal, ‘hydro’ means hydrogen, ‘halogen’ means halogen. The halogen and hydrogen must be present on adjacent positions.

Complete step-by-step answer:Dehydrohalogenation takes place in presence of a strong base like $KOH$. The $O{H^ - }$ accepts a hydrogen from the adjacent position of a good leaving group resulting in formation of positive charge which converts into a double bond by removal of the leaving group, namely halide for this particular reaction. In reality, the positive charge is actually not formed because the entire process occurs in a concerted manner. It is only to understand the basic mechanism. The elimination follows $E2$ mechanism which is a one step process with no intermediate (or positive charge) but only proceeds through the transition state. In \[1,2\;\] di-bromocyclohexane , the two bromides are present at \[1,2\;\] positions on cyclohexane. Each bromide will leave along with adjacent hydrogen to form a product similar to option A.
Hence A is the correct answer.

Option ‘A’ is correct

Additional Information: Alkene formation can follow Saytzeff elimination (more substituted alkene) or Hofmann elimination (less substituted alkene). Saytzeff elimination is followed in case of neutral molecules and Hofmann in case of charged species.

Note: Dehydrohalogenation requires proper stereo or proper orientation of the atoms. To undergo elimination, halogen and hydrogen must be anti-planner to each other. Hence, dehydrohalogenation is a stereospecific reaction. Dehalogenation reaction is also a type of reaction which eliminates two halogens present at \[1,2\;\]. But it is carried out only in presence of reagents like zinc dust,$NaI$, $MeOH$OR $EtOH$.