Question

Organic Chemistry 1 Lab Fall 2019 Spectral Set 2 Due 10/29 - 10/31/2019, during regular lab time. For each set of spectra, label the major IR absorptions, deduce a workable formula, and calculate the DBE for that formula. Then using the splitting patterns and integrations, determine the structure of the molecule responsible for the spectra. Be sure to label each NMR signal like we did in class. S = singlet, d = doublet, t= triplet, q#quartet, p = pentet, n=nonet, and m=multiplet (cannot be resolved) DBE = Double Bond Equivalents = [(2n+2) + N] - (H +X) where n= # of C, N = nitrogens, and X = is halogens.

Answer 1

Let us analyze the spectra as below-

• From the IR spectra, it is very clear that there is a hydrogen-bonded -OH is there and C-H stretching peak. To be very frank you do not get any more information from IR.
• In the 1H-NMR, we observe 6-H's with doublet peak at 1ppm which is quite recognizable as methyl H's of  $-CH(CH_{3})_{2}$. and the multiplet is the single H of -CH group attached to TWO methyls.
• Now, obviously as you can see that 1H, singlet line at 2 ppm will be the H of -OH group attached to a -CH2 group.
• The -CH2- group is attached to both -OH and the $-CH(CH_{3})_{2}$ groups and gives doublet due to the neighboring -CH- of $-CH(CH_{3})_{2}$ .
• Up to this, you must have guessed the structure, if not, let us try to draw as described above!

Ho OH2CH S2:1pm 1H,Singlet CeH CH 2H,doublet 2 Smul tiplet TH GHdouble Molecul our Fomula L2.4+21-1 NT ie. nodoublekon) DBE 2