Q17 / 60 · JEE Advanced 2013 (Paper 2) · Chemistry
Match the chemical conversions in List I with the appropriate reagents in List II and select the correct answer using the code given below the lists :
List - I List - II (A) 
(P) (B) 
(Q) (C) 
(R) (D) 
(S)
| List - I | List - II | ||
| (A) | ![]() | (P) | |
| (B) | ![]() | (Q) | |
| (C) | ![]() | (R) | |
| (D) | ![]() | (S) | |
- Aa-p;b-r;c-q;d-s;
- Ba-q;b-r;c-s;d-p;
- Correct answerCa-q;b-r;c-p;d-s;
- Da-q;b-p;c-r;d-s;
Solution
A tertiary alkyl halide with strong base undergoes elimination reaction, while a primary alkyl halide with butoxide gives ether by a substitution reaction. Alkenes undergo hydration (addition of water) by Hydroboration-Oxidation, giving anti-Markovnikov's products and by Oxymercuration-Demercuration, giving Markovnikov's products.
Hence, the correct answer is option: 'C'.