Compound Q has molecular formula \({{\bf{C}}_{\bf{5}}}{{\bf{H}}_{\bf{9}}}{\bf{Cl}}{{\bf{O}}_{\bf{2}}}\). Deduce the structure of P from its \({}^{\bf{1}}{\bf{H}}\) and \({}^{{\bf{13}}}{\bf{C}}\)-NMR spectra.

Short Answer

Expert verified

The structure of Q is as follows:

Step by step solution

01

Molecular structure

The most accepted arrangement of atoms and elements within a compound to satisfy the barriers of energy and steric effects is the molecular structure of a compound.

02

Analyzing data

Given data

Chemical formula =\({{\rm{C}}_5}{{\rm{H}}_9}{\rm{Cl}}{{\rm{O}}_{\rm{2}}}\)

\({\rm{The degree of unsaturation }}\left( {{\rm{IHD}}} \right){\rm{ = }}\frac{{\left( {\left( {{\rm{2n + 2}}} \right){\rm{ - x}}} \right)}}{{\rm{2}}}\)

where,

n = number of carbon atoms

x = (number of hydrogen atoms) + (number of halogen atoms) - (number of nitrogen atoms)

On substituting values,

\(\begin{array}{c}{\rm{The degree of unsaturation }}\left( {{\rm{IHD}}} \right){\rm{ = }}\frac{{\left( {\left( {{\rm{2 \times 5 + 2}}} \right){\rm{ - }}\left( {{\rm{9 + 1}}} \right)} \right)}}{{\rm{2}}}\\{\rm{ = 1}}\end{array}\)

The value of IHD suggests the presence of a pi-bond.

The five signals in \({}^{{\rm{13}}}{\rm{C}}\)-NMR suggests the presence of five different types of carbon atoms.

From the \({}^{\rm{1}}{\rm{H}}\)-NMR spectra,

  • The quartet at 4.2 ppm attributes to 2H atoms corresponding to split of two methyl groups,
  • The triplet at 1.3 ppm attributes to 3H atoms due to split of 2H’s atoms,
  • The triplet at 2.8 ppm attributes to 2H atoms due to split of 2H’s atoms,
  • The triplet at 3.7 ppm attributes to 2H atoms due to split of 2H’s atoms,
03

Proposing the structure

Structure of Q.

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Most popular questions from this chapter

Which of the highlighted carbon atoms in each molecule absorbs farther downfield?

a.

b.

c.

d.

Question: Cyclohex-2-enone has two protons on its carbon-carbon double bond (labeled Ha and Hb ) and two protons on the carbon adjacent to the double bond (labeled Hc ). (a) If Jab = 11 Hz and Jbc = 4 Hz, sketch the splitting pattern observed for each proton on the hybridized carbons. (b) Despite the fact that Ha is located adjacent to an electron-withdrawing C=O, its absorption occurs up-field from the signal due to Hb(6.0 vs. 7.0 ppm). Offer an explanation.

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

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

NMR (ppm):

1.87 (singlet, 6 H)

3.86 (singlet, 2 H)

b. C3H6Br2: IR peak at 3000–2850cm-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)

Question: Reaction of C6H5CH2CH2OH with CH3COCl affords compound W, which has molecular formula C10H12O2. W shows prominent IR absorptions at 3088–2897, 1740, and 1606cm-1 . W exhibits the following signals in its 1 H NMR spectrum: 2.02 (singlet), 2.91 (triplet), 4.25 (triplet), and 7.20–7.35 (multiplet) ppm. What is the structure of W? We will learn about this reaction in Chapter 22.

Question: Identify the structures of isomers H and I (molecular form C8H11N)

a.

b.

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