Question

ii) (4 marks) Briefly explain how 1H-NMR spectroscopy would allow you to distinguish between the 3 isomers of butyl alcohol. MOH LOH

Answer 1

O 1-utanol primary alcohol 2-methy 1. propanol Secondary alcohol 2-Methyl 2-propanol testary alchol OH - ь The 8ppm range of alcohas 0.5 to 5:5ppm at The sppm value of -of Hydrogen is 2.25s singlet- -Chy - a) Hydrogens two Hydrogen 5 triplet at S = 3.65 ppm ce) - CH₂ - (b) Hydrogens two Hydrogens pentet at s=1.9 ppm -CH, - ©) Hydrog ens two Hydrogens Sesker at Saloy ppm CH₂ - (d) Hydrogens three Hydrogens trepler at S=0.9 ppm ins Scat a torpher CH2 1.98 CH₂ 2.256 tolpec CH2 3.655 0-H 0.95 Singlet persen penter Is Scanned with CamScanner.

OH NOH Isobutanol b a CH-CH-OHd The Molecule gives 4. H-NMR signals -OH (1H) (d) Hydrogen -one Hydrogen Singler at s= 3.45 -CH2- Q Hydrogens - to o Hydrogens doublet al S = 3.7 $ (24) 5 -CH - (IH) s -one Hydrogen Multiplet at s= 1. 25 Hydrogens Spa - Ste hydrogens doublet at sao.98. He ad 3 CH-CH2-OH 3:45 0.95 Singler - 1.28 3.75 maltsplet doublet- - The molecule giver two signals -Oh eas thydrogen -one Hydrogen singlet-at- &= 3.25 i ch Hydrogens - nine Hydrogens Hz OH - Single- Singlet at g = 0.9 194 cs Scanned this I c Songlet og 0.98,