Discussion
An organic compound having molecular formula C6H11BrO2 exhibits the following peaks in 1H-NMR spectrum : δ 4.1(2H, q, J = 7.5 Hz), 4.0(2H, t, J = 7.5 Hz), 1.5-2.2(4H, m), 1.25(3H, t, J = 7.5 Hz). The structure of the compound is :
- A.
- B.
- C.
- D.
Answer: C
The structure is (C), ethyl 4-bromobutanoate, Br-CH2CH2CH2-CO-OCH2CH3.
The quartet at 4.1 (2H) and triplet at 1.25 (3H) with the same J show an ethyl ester, OCH2CH3. The remaining 2H triplet and the 4H multiplet need a chain of three CH2 groups: the triplet is the CH2 at one end of the chain (the key assigns it to CH2Br), and the multiplet is the two inner CH2 groups.
(A) and (B) contain a CH bearing Br and would show a methine signal and, for (A), a methyl doublet. (D) has no ethyl ester: it would show no quartet and its CH3 would be part of a propyl group.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development