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Chapter 14: Q.21558-14-6P. (page 535)

Label the protons in each highlighted CH2 group as enantiotopic, diastereotopic, or homotopic.

a.

b.

c.

Short Answer

Expert verified

a. Enantiotopic

b. Neither

c. Diastereotopic

Step by step solution

01

Homotopic, enantiotopic, and diastereotopic protons

鈥� Homotopic protons are formed when the same product is formed by the substitution of two H by Z.

鈥� Enantiotopic protons are formed when two H are equivalent and occur at the same NMR signal.

鈥� Diastereotopic protons are formed when two H are not equivalent and occur at two different signals.

02

Explanation for highlighted proton

a.

The replacement of H with X is shown below.

Enantiotopic proton

b.

The replacement of H with X is shown below.

Neither enantiotopic nor diastereotopic

c.

The replacement of H with X is shown below.

Diastereotopic proton

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Most popular questions from this chapter

Label the signals due to \({{\bf{H}}_{\bf{a}}}\), \({{\bf{H}}_{\bf{b}}}\), and \({{\bf{H}}_{\bf{c}}}\) in the 1 H NMR spectrum of acrylonitrile \(\left( {{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{ = CHCN}}} \right)\). Draw a splitting diagram for the absorption due to the \({{\bf{H}}_{\bf{c}}}\) proton.

Question: How many 13C NMR signals do each compound exhibit?

The 1H NMR spectrum 1,2-dimethoxyethane (CH3OCH2CH2OCH3) recorded on a 300 MHz NMR spectrometer consists of signals at 1017 Hz and 1065 Hz downfield from TMS. (a) Calculate the chemical shift of each absorption. (b) At what frequency would each absorption occur if the spectrum were recorded on a 500 MHz NMR spectrometer?

Question:Propose a structure consistent with each set of spectral data:

a. C4H8Br2: IR peak at 3000鈥�2850cm-1 ;

NMR (ppm):

1.87 (singlet, 6 H)

3.86 (singlet, 2 H)

b. C3H6Br2: IR peak at 3000鈥�2850cm-1 ;

NMR (ppm):

2.4 (quintet)

3.5 (triplet)

c. C5H10O2: IR peak at 1740cm-1 ;

NMR (ppm):

1.15 (triplet, 3 H) 2.30 (quartet, 2 H)

1.25 (triplet, 3 H) 4.72 (quartet, 2 H)

d . C6H14O: IR peak at 3600-3200 cm-1 ;

NMR (ppm):

0.8 (triplet, 6 H) 1.5 (quartet, 4 H)

1.0 (singlet, 3 H) 1.6 (singlet, 1 H)

e. C6H14O: IR peak at 3000-2850cm-1 ;

NMR (ppm):

1.10 (doublet, relative area = 6)

3.60 (septet, relative area = 1)

f . C3H6O: IR peak at 1730cm-1 ;

NMR (ppm):

1.11 (triplet)

2.46 (multiplet)

9.79 (triplet)

For each compound, which of the protons on the highlighted carbons absorbs farther downfield?

a.

b.

c.

See all solutions

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