JEE PYQ: Aldehydes And Ketones Question 19
Question 19 - 2020 (06 Sep Shift 1)
‘A’ ($\text{C}7\text{H}{14}$): Ozonolysis gives B (positive iodoform + silver mirror) and C (no iodoform, Lucas test positive after LiAlH$_4$ reduction). A is:
(1). (2-methylhex-3-ene)
(2). (3-methylhex-3-ene)
(3). (2-methylhex-2-ene)
(4). (3-methylhex-2-ene)
Show Answer
Answer: (4)
Solution
B = CH$_3$CHO (iodoform + Tollens positive). C = ketone (no iodoform). LiAlH$_4$ reduces C to secondary alcohol (Lucas test in 5 min). A = 3-methylhex-2-ene gives CH$_3$CHO + pentan-3-one on ozonolysis.
BETA
