Read each statement A-F carefully (see attached image). Then, select all of the

Question

Read each statement A-F carefully (see attached image). Then, select all of the letters below that represent false statements.

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MeBr will have an IR absorbance signal show up in Zone 3 B. Knowing that a typical O-H stretch appears at 3400 cm in the IR, the relative change in vibrational frequency wil be less for an 1"O-H stretch than it would be an 16O-D stretch. C. Carbonyl (C-O) stretches typically result in strong vibrational frequencies in the H NMR due to the large change in dipole moment associated with this bond. D. Ring current is a phenomenon associated with a current caused by -electrons. When the current induces a magnetic field which reinforces the external magnetic field, the result will be a deshielding effecet on the nuclei. E. tert-Butyl phenyl acetylene when treated with Na Golid), NHsdiqui) in methanol will 3(liquid) n methanol W1 produce an alkene where each vinyl proton will be a doublet with an expected coupling constant (J) somewhere between the range of 12-18 Hz. F. The product formed after the addition of cyclopentanol to PBr should, in theory, produce a 'H NMR spectrum having 3 peaks

Explanation / Answer

A) The product will have an absortion between 2100-2400 cm-1 because the triple bond remains with the substitution reaction. It is true

b) it is true because a reduced mass for 17O-H and 16O-H as almost the same. So the change is tiny.

C) False. NMR has nothing to do with dipole changes

d) True. The ring current reinforces the magnetic field in protons of the ring.

e) alkynes reduced by Na/NH3 produce trans alkene. The protons in a trans alkene has a large coupling constant of about 12-18 Hz, So this is true.

F) it is true because you have three kinds of protons. But of you consider the spacial structure you could have until four peaks.

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