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Chapter 14: Q.21558-14-71P (page 568)

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

a. A compound X (molecular formula C6H12O2) gives a strong peak in its IR spectrum at 1740 cm-1. The 1 H NMR spectrum of X shows only two singlets, including one at 3.5 ppm. The 13C NMR spectrum is given below Propose a structure for X.

b. A compound Y (molecular formula C6H10 ) gives four lines in its 13C NMR spectrum (27, 30, 67, and 93 ppm), and the IR spectrum given here. Propose a structure for Y.

Short Answer

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Answer

a.

b.

Step by step solution

01

Structure determination

If the 1740 peak is present in given data, it means the ester group is present.13C spectra indicate the types of carbon present in the structure.

1 degree of unsaturation IR peaks at 1740cm-1 : C=O 1H NMR, 2 signals: 2 types of H's 13C NMR, and 4 signals: 4 kinds of C's, including one at approx 170 ppm due to a C=O.

Structure of C6H12O2

02

Structure determination

In C6H10 structure, two degrees of unsaturation are present, which means it has either two double bonds or one triple bond or two rings present in the molecule.

In the given IR spectra, the sharp peak is present at 3500-3200 that indicates Csp3-H carbon is present in the molecule.

Two degrees of unsaturation IR peak at 3000 cm-1 : Csp3–H bonds peak at 3300cm-1 : Csp–H bond peak at 2150 cm-1 : C≡C bond 13C NMR: 4 signals: 4 kinds of C's

structure of C6H10

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Most popular questions from this chapter

Draw all constitutional isomers of molecular formula C3H6Cl2.

a. How many signals does each isomer exhibit in its 1HNMR spectrum?

b.How many lines does each isomer exhibit in its 13CNMR spectrum?

c. When only the number of signals in both 1Hand 13C NMR spectroscopy is considered, is it possible to distinguish all of these constitutional isomers?

Identify the carbon atoms that give rise to the signals in the \)^{{\bf{13}}}{\bf{C}}\) NMR spectrum of each compound.

a. \){\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{OH}}\); \)^{{\bf{13}}}{\bf{C}}\) NMR: 14, 19, 35, and 62 ppm

b. \){\left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}} \right)_{\bf{2}}}{\bf{CHCHO}}\); \)^{{\bf{13}}}{\bf{C}}\) NMR: 16, 41, and 205 ppm

c. \){\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{ = CHCH}}\left( {{\bf{OH}}} \right){\bf{C}}{{\bf{H}}_{\bf{3}}}\); \)^{{\bf{13}}}{\bf{C}}\) NMR: 23, 69, 113, and 143 ppm

Identify products A and B from the given 1HNMR data.

a. Treatment of CH2=CHCOCH3 with one equivalent of HCl forms compound A. A exhibits the following absorptions in its 1HNMR spectrum: 2.2 (singlet, 3 H), 3.05 (triplet, 2 H), and 3.6 (triplet, 2H) ppm. What is the structure of A?

b.Treatment of acetone [(CH3)2C=O] with dilute aqueous base forms B. Compound B exhibits four singlets in its 1HNMR spectrum at 1.3 (6 H), 2.2 (3 H), 2.5 (2 H), and 3.8 (1 H) ppm. What is the structure of B?

Question: Explain why the carbonyl carbon of an aldehyde or ketone absorbs farther downfield than the carbonyl carbon of an ester in a 13C NMR spectrum

Identify products A and B from the given \({}^{\bf{1}}{\bf{H}}\) NMR data.

(A) Treatment of \({\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{ = CHCOC}}{{\bf{H}}_{\bf{3}}}\) with one equivalent of HCl forms compound A. A exhibits the following absorptions in its \({}^{\bf{1}}{\bf{H}}\)NMR spectrum: 2.2 (singlet, 3 H), 3.05 (triplet, 2 H), and 3.6 (triplet, 2H) ppm. What is the structure of A?

(B) Treatment of acetone \(\left( {{{\left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}} \right)}_{\bf{2}}}{\bf{C = O}}} \right(\)with dilute aqueous base forms B. Compound B exhibits four singlets in its \({}^{\bf{1}}{\bf{H}}\) NMR spectrum at 1.3 (6 H), 2.2 (3 H), 2.5 (2 H), and 3.8 (1 H) ppm. What is the structure of B?
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