JEE Main 2022 · ChemistryMultiple choiceDiagram basedMediumApplication
JEE Main 25 July 2022, Shift 2, Chemistry Q73
Question 73 of 90 in this shift, Chemistry question 13 of 30, Section A.
Major product of the following reaction is
[figure: skeletal drawing, left to right: a C=C double bond at the left end whose inner carbon is bonded directly to the carbonyl carbon (C=O, O drawn below); the carbonyl carbon is also bonded to a chain oxygen; that O is joined to a chain ending in a C=C double bond at the right end] --2 HBr-->
The figure, in words
Reactant drawn as skeletal formula: C=C at left end bonded directly to the carbonyl carbon C(=O) (O below), which is bonded to O, which is attached to a chain ending in C=C at right. Reagent 2 HBr over the arrow. Options are skeletal structures with Br atoms placed as described.- (1)[figure: skeletal structure; on the left a carbon bearing Br is bonded to the carbonyl carbon (C=O, O below) and to a one-bond branch; the carbonyl carbon is bonded to O, which is joined to a zigzag chain whose terminal carbon carries Br]
- (2)[figure: skeletal structure; on the left a carbon bearing Br is bonded to the carbonyl carbon (C=O, O below) and to a one-bond branch; the carbonyl carbon is bonded to O, which is joined to a carbon bearing Br and a one-bond branch]
- (3)[figure: skeletal structure; a zigzag chain on the left, terminal carbon carrying Br, joined to the carbonyl carbon (C=O, O below); the carbonyl carbon is bonded to O, which is joined to a zigzag chain whose terminal carbon carries Br]
- (4)[figure: skeletal structure; a zigzag chain on the left, terminal carbon carrying Br, joined to the carbonyl carbon (C=O, O below); the carbonyl carbon is bonded to O, which is joined to a carbon bearing Br and a one-bond branch]Official answer
Official answer
Option 4
NTA final key.
- Chapter
- Hydrocarbons
- Topic
- Alkenes and Alkynes
- Idea tested
- Markovnikov's Rule Application