Question

Short answer

b. A-1740\({\rm{c}}{{\rm{m}}^{ - 1}}\)for CO and 3600\({\rm{c}}{{\rm{m}}^{ - 1}}\)for OH

B-1730\({\rm{c}}{{\rm{m}}^{ - 1}}\)for CO and 3600\({\rm{c}}{{\rm{m}}^{ - 1}}\)for OH

c. 4 for A and B both.

d. 4 for A and B both.

e. A-(t)\({\rm{C}}{{\rm{H}}_{\rm{3}}}\), (q)\({\rm{C}}{{\rm{H}}_2}\,\), (s) 1H, (s)\({\rm{C}}{{\rm{H}}_2}\,\)

B--(s)\({\rm{C}}{{\rm{H}}_{\rm{3}}}\), (t)\({\rm{C}}{{\rm{H}}_2}\,\), (s) 1H, (t)\({\rm{C}}{{\rm{H}}_2}\,\)

Step-by-step solution

Step 1: \(^{{\bf{13}}}{\bf{C}}\)NMR spectrum

\(^{{\rm{13}}}{\rm{C}}\)NMR spectrum gives the location of all non-equivalent carbons. The equivalent carbon has the same spectrum. The substituents affect the most in the\(^{{\rm{13}}}{\rm{C}}\)spectrum signals.

• If the electron-withdrawing group is adjacent to the corresponding\(^{{\bf{13}}}{\bf{C}}\),the spectra would be downfield.
• If the electron-donating group is adjacent to the corresponding\(^{{\bf{13}}}{\bf{C}}\),the spectra would be upfield.

Explanation for a and b

a. Molecular ion in the mass spectrometry is obtained by ionizing the molecule.

The molecular ions are shown as:

b.

IR absorption for the ketone group in A is 1740\({\rm{c}}{{\rm{m}}^{ - 1}}\), which is due to the presence of the OH group. The OH group attracts the electron towards itself and increases the stretching frequency of the ketone group.

The IR absorption for OH group in A and B is 3600\({\rm{c}}{{\rm{m}}^{ - 1}}\).

IR absorption for the ketone group is 1730\({\rm{c}}{{\rm{m}}^{ - 1}}\) due to the OH group being away from the CO. The OH group attracts the electron towards itself and increases the stretching frequency of the ketone group slightly less than A.

Explanation for c, d, and e

c.

The number of non-equivalent\(^{{\rm{13}}}{\rm{C}}\)signals for both A and B are 4 – two signals for two\({\rm{C}}{{\rm{H}}_2}\,\), one\({\rm{C}}{{\rm{H}}_{\rm{3}}}\)and one carbon for CO.

d.

The number of non-equivalent\(^{\rm{1}}{\rm{H}}\)signals for both A and B are 4 – four signals for two\({\rm{C}}{{\rm{H}}_2}\,\), one\({\rm{C}}{{\rm{H}}_{\rm{3}}}\)and one carbon for OH.

e.

The splitting signals for A and B are given as:

A-(t)\({\rm{C}}{{\rm{H}}_{\rm{3}}}\), (q)\({\rm{C}}{{\rm{H}}_2}\,\), (s) 1H, (s)\({\rm{C}}{{\rm{H}}_2}\,\)

B--(s)\({\rm{C}}{{\rm{H}}_{\rm{3}}}\), (t)\({\rm{C}}{{\rm{H}}_2}\,\), (s) 1H, (t)\({\rm{C}}{{\rm{H}}_2}\,\)