The total number of hydrogen atoms in product A and product B is __________.

Formation of Product A (Elimination)

• Reaction: Bromoethane ($$\text{CH}_3\text{CH}_2\text{Br}$$) reacting with $$\text{NaOH}$$ in alcoholic medium ($$\text{C}_2\text{H}_5\text{OH}$$).
• Mechanism: Alcoholic $$\text{NaOH}$$ acts as a strong base, favoring $$\text{E}2$$ dehydrohalogenation.
• Product A: Ethene ($$\text{CH}_2=\text{CH}_2$$)
• Hydrogen Count in A: 4

2. Formation of Product B (Substitution)

• Reaction: Bromoethane ($$\text{CH}_3\text{CH}_2\text{Br}$$) reacting with $$\text{NaOH}$$ in aqueous medium ($$\text{H}_2\text{O}$$).
• Mechanism: Aqueous $$\text{NaOH}$$ provides strong nucleophilic hydroxyl ions ($$\text{OH}^-$$), favoring $$\text{S}_\text{N}2$$ substitution.
• Product B: Ethanol ($$\text{CH}_3\text{CH}_2\text{OH}$$)
• Hydrogen Count in B: 6