Question

Predict the major product for the following reactions considering competing substitution and elimination pathways. CI NaOEt NaSH NaOH Explain why a completely nonpolar bond will not give a stretching signal in the IR spectra. Would you expect to see a signal for C-H stretching for a nonpolar molecule? Why or why not?

Answer 1

Naofty Elemination Reuction NasH SH Substitution Reaction NaoH Cl Substitution Reach on

. The normal range of infrared frequency used is 4000 cm-1 to 400 cm-1. Completely nonpolar bonds do not absorb IR radiation. For example, the C-C bond in ethane does not absorb radiation, whereas the CH bond does. The intensity of the absorption increases with the polarity of the bond so a C-O stretching band is more intense than a C-H band. Also, intensity increases with the number of the bands. In general, stretching bands are more intense than bending bands.

Alkanes: In alkanes there is always a C-H stretch at 2970-2850 cm-1, just below 3000 cm-1. Don't forget that every organic molecule has C-H in it.

Alkenes: In alkenes there is sp2 C-H stretching at 3100-3010 cm-1, just above 3000 cm-1. Also there is C=C stretching at about 1650 cm-1. You can also tell the substitution pattern of your alkene and determine whether it is trisubstituted, disubstituted, cis, trans, etc.

Alkynes: There is stretching for sp-hybridized C-H at 3300 cm-1. There is also a C-C triple bond stretch for terminal alkynes at 2100-2140 cm-1.

Alcohols: Alcohols have a very broad O-H stretch at 3300-3350 cm-1. You can also tell whether it is a primary, secondary or tertiary alcohol from the C-O stretch