1. Molecular formula: C3H6Br2 IR: 3000-2850 cm-1 1H NMR: 2.4 (quintet, 2H); 3.5

Question

1. Molecular formula: C3H6Br2 IR: 3000-2850 cm-1 1H NMR: 2.4 (quintet, 2H); 3.5 (triplet, 4H) 2. Molecular formula: C5H10O2 IR: 1740 cm-1 1H NMR: 1.15 (triplet, 3H); 1.25 (triplet, 3H); 2.30 (quartet, 2H); 4.72 (quartet, 2H) 3. Compound J Molecular ion: m/z=72 IR: 1710 cm-1 1H NMR: 1.0 (triplet, 3H); 2.1 (singlet, 3H), 2.4 (quartet, 2H) PLEASE don't just give answer. I really need to understand this! I think the first one is 1,3 dibromopropane, but I am not sure if is cis or trans. The IR for 2 and 3 seems to indicate C=O to me, but I am having trouble seeing the whole picture.

Explanation / Answer

1) 1,3 dibromo propane ( cis , trans cannot arise since it is alkane not alkene), 2) degree of unsaturation = (5x2+2 -10)/2 = 1 , so 1 double bond, IR at 1740 indicates C=O , NMR 1.15 (3H) triplet indicates CH adjacent to CH2, same with 1.25(3H) triplet , CH3-CH2, 2.3 quatret(2H) CH2 adjacent to CH3 and C=O , 4.72 quatret (2H) means Ch2 adjacent to CH3 and O. Joining above four fragments we get CH3-CH2-C(=O)-O-CH2-CH3 ( ethyl propanoate) , 3) IR at 1710 cm- shows C=O of ketone , hence compound has C=O ( ketone), m/z = 72 = 28 (fromCO) + CnH2n+2 , n = 3 , compound is 2butanone ( only 1 ketone possible with 4 carbon chain) , NMR 1triplet( 3H) shows CH3 adjacent to CH2. 2.1 singlet(3H) shows CH3 adjacent to C=O, 2.4 quatret (2H) shows CH2 adjacent to C=O and CH3, joining above we get CH3-C(=O)-CH2-CH3 i.e 2 butanone

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.