Question

If someone explain how to find these unknowns..And what they are and how you got the answer. I am having a hard time really understanding NMR and IR.. Thanks in advance

Unknowns for 1H NMR Final Report 4) Chemical Formula: CH360 IR: strong broad peak at 3300cm PPM 5) Chemical Formula: CsHN IR: Two weak peaks at 3350cm Doublet, 1H Triplet, IH Triplet, 1H Doublet, IH PPM 6) Chemical Formula:CH,202 IR: strong peak 1750cm! 3H SH PPM

Answer 1

4) The formula is C₄H₁₀O

• Degree of Unsaturation = (2*C+2-H)/2= (2*4+2-10)/2 = 0
• This means that no double bonds are present in the molecule.
• A compound with Oxygen having no double bond can either be ether or alcohol.
• Further, the IR peak is at 3300 cm^-1 . This suggests an alcohol group.
• Further, in nmr spectroscopy too, the peak lying aroung 4.25 ppm suggests an alcohol group (-OH)
• Further, a peak containing 6H at around 0.9 ppm suggests two methyl groups (-CH₃).
• A peak containing 2H at around 3.4 ppm suggests a -CH₂- group. Further, the chemical shift is high which suggests an electronegative atom attached to the carbon (Oxygen). This means that -CH₂OH must be present in the compound.
• A peak containing 1H at around 1.4 ppm suggests a CH group. This peak is a multiplet which suggests adjoining carbons have multiple Hydrogens.
• Putting together everything, we get the following structure-

I I-OOMI I OH I– All Hydrogens shown Line notation

The compound is 2-Methyl-1-propanol as shown in the figure. The two methyl group has NMR of 0.9 ppm at 6H. The CH group has 1H at 1.4 ppm. The adjoining CH₂ group has 2H at 3.4 ppm and the Hydrogen of the alcohol group has nmr of 4.25 ppm