Match List - I with List - II : List - I (Chemicals) (a) Alcoholic potassium hydroxide (b) Pd/ BaSO_4 (c) BHC (Benzene hexachloride) (d) Polyacetylene List - II (Use / Preparation / Constituent) (i) Electrodes in batteries (ii) Obtained by addition reaction (iii) Used for \beta - elimination reaction (iv) Lindlar's catalyst Choose the most appropriate match :

(a) - (iii), (b) - (iv), (c) - (ii), (d) - (i)

(a) - (ii), (b) - (i), (c) - (iv), (d) - (iii)

(a) - (iii), (b) - (i), (c) - (iv), (d) - (ii)

(a) - (ii), (b) - (iv), (c) - (i), (d) - (iii)

Hydrocarbons PYQ — JEE Main 2021 · 18 Mar · Shift 1 · Q8 with Solution

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Correct answer: B

Match (a): Alcoholic potassium hydroxide
Alcoholic KOH is commonly used for dehydrohalogenation, which is a $\beta$ -elimination reaction.
Hence, $(a) \to (iii)$
Match (b): $\text{Pd}/BaSO_4$
Palladium on barium sulfate is the poisoned palladium catalyst known as Lindlar's catalyst (or its base form). It is used for partial hydrogenation of alkynes to alkenes.
Hence, $(b) \to (iv)$
Match (c): BHC (Benzene hexachloride)
BHC is formed by addition of chlorine to benzene in presence of light: $C_6H_6 + 3Cl_2 \rightarrow C_6H_6Cl_6$ So it is obtained by addition reaction.
Hence, $(c) \to (ii)$
Match (d): Polyacetylene
Polyacetylene is a conducting polymer and is used in electrodes in batteries.
Hence, $(d) \to (i)$
Final matching
$(a)-(iii),\quad (b)-(iv),\quad (c)-(ii),\quad (d)-(i)$
Compare with options
This corresponds to Option B.

Therefore, the correct answer is B.