Q.For the following reactions (A) CH₃CH₂CH₂Br + KOH → CH₂=CH₂ + KBr + H₂O (B) (CH₃)₃CBr + KOH → (CH₃)₃COH + KBr (C) Cyclohexene + Br₂ → 1,2-dibromocyclohexane. Which of the following statement is correct?

Step 1. Reaction (A): a propyl bromide reacts with KOH and the product is an alkene plus KBr and water -- hydrogen and bromine are removed from adjacent carbons and a new C=C double bond forms in their place. This is the defining pattern of a $\beta$-elimination reaction.

Step 2. Reaction (B): tert-butyl bromide, $(CH_3)_3C$-Br, reacts with KOH to give tert-butanol, $(CH_3)_3C$-OH, plus KBr -- the bromine atom on carbon has simply been replaced by a hydroxyl group, with no change in the degree of saturation. This is a substitution (displacement) reaction (proceeding, for this hindered tertiary substrate, via the $S_N1$ pathway through the tert-butyl carbocation).

Step 3. Reaction (C): cyclohexene reacts with $Br_2$ to give 1,2-dibromocyclohexane -- the ring's C=C double bond is consumed as two new C-Br sigma bonds form, exactly the two-new-bonds-for-one-broken-$\pi$-bond pattern of an addition reaction (electrophilic addition of $Br_2$ across the alkene).

Step 4. So (A) is elimination, (B) is substitution, and (C) is addition -- matching option (d).

Classify each reaction by what structurally happens to the substrate: a leaving group plus an adjacent H forming a new double bond is elimination; one group swapped for another with saturation unchanged is substitution; two new bonds replacing one broken $\pi$ bond is addition.

• Assuming KOH always causes elimination -- with a hindered tertiary substrate like (B), KOH instead substitutes, since a stable carbocation intermediate is readily accessible ($S_N1$).
• Overlooking that (C) has no leaving group and no base at all -- $Br_2$ simply adds across the ring's double bond, which can only be an addition, never a substitution or elimination.