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Chapter 14: Q.21558-14-63P (page 566)

Question. Reaction of aldehyde D with amino alcohol E in the presence of NaH forms F (molecular formula C11H15NO2). F absorbs at 1730 cm-1in its IR spectrum. F also shows eight lines in its 13C-NMR spectrum, and gives the following -NMR spectrum: 2.32 (singlet, 6 H), 3.05 (triplet, 2 H), 4.20 (triplet, 2 H), 6.97 (doublet, 2 H), 7.82 (doublet, 2 H), and 9.97 (singlet, 1 H) ppm. Propose a structure for F. We will learn about this reaction in Chapter 18.

Short Answer

Expert verified

Answer

The structure of F is as follows:

Step by step solution

01

Reaction of tertiary amines

Due to lone pair of electrons, amines would act as nucleophiles and bond at an electrophilic center.

The benzaldehyde reacts with amino alcohol in the presence of the strong base sodium amide to form a tertiary ammonium salt.

Formation of F

02

Analyzing data

Given data

Chemical formula =C11H15NO2

IR absorption= 1730cm-1

TheDegreeofunsaturation(1HD)=((2n+2)-x)2

Where,

  • n= Number of carbon atoms
  • x= (Number of hydrogen atoms) + (Number of halogen atoms) - (Number of nitrogen atoms)

On substituting values,

Thedegreeofunsaturation(1HD)=((2×11+2)-(15-1))2

=5

The value of IHD suggests the presence of a benzene ring (3 pi-bonds and the ring) and another pi-bond. The eight signals in 13C-NMR suggests the presence of eight types of carbon atoms.

From the NMR spectra,

  • The singlet at 2.32 ppm attributes to the 6H atoms corresponding to the two methyl groups on the nitrogen atom.
  • The triplet at 3.05 ppm attributes to the 2H atoms adjacent to the nitrogen atom
  • The triplet at 4.20 ppm attributes to the 2H atoms adjacent to the hydroxyl group.
  • The doublet at 6.97 ppm attributes to the 2H atoms on the benzene ring that are equivalent in nature.
  • The doublet at 7.87 ppm attributes to the 2H atoms on the benzene ring that are equivalent in nature and are present next to the alpha-carbon atom.
  • The singlet at 9.97 ppm attributes to the 1H atom of the 1H located para to the alpha- carbon atom.

The IR vibrational stretching at 1730 cm-1 indicates the presence of carbonyl stretching of a ketone group.

03

Proposing the structure

Structure of F

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Most popular questions from this chapter

Question:Answer the following questions about each of the hydroxy ketones: 1-hydroxybutan-2-one (A) and 4-hydroxybutan-2-one (B).

a. What is the molecular ion in the mass spectrum?

b. What IR absorptions are present in the functional group region?

c. How many lines are observed in the 13C NMR spectrum?

d. How many signals are observed in the 1H NMR spectrum?

e. Give the splitting observed for each type of proton as well as its approximate chemical shift.

How can you use \(^{\bf{1}}{\bf{H}}\) NMR spectroscopy to distinguish between \({\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{ = C}}\left( {{\bf{Br}}} \right){\bf{C}}{{\bf{O}}_{\bf{2}}}{\bf{C}}{{\bf{H}}_{\bf{3}}}\)and methyl (E)-3-bromopropenoate, \({\bf{BrCH = CHC}}{{\bf{O}}_{\bf{2}}}{\bf{C}}{{\bf{H}}_{\bf{3}}}\)?

Question. Treatment of butan-2-one (CH3COCH2CH3) with a strong base followed by CH3l forms a compound Q, which gives a molecular ion in its mass spectrum at 86. The IR (> 1500 only) and 1H-NMR spectrum of Q is given below. What is the structure of Q?

Question: Into how many peaks will the signal for each of the labeled protons be split?

Question: Treatment of 2-methylpropanenitrile [(CH3)2CHCN] with CH3CH2CH2MgBr, followed by aqueous acid, affords compound V, which has molecular formula C7H14O. V has a strong absorption in its IR spectrum at 1713 cm-1, and gives the following 1 H NMR data: 0.91 (triplet, 3 H), 1.09 (doublet, 6 H), 1.6 (multiplet, 2 H), 2.43 (triplet, 2 H), and 2.60 (septet, 1 H) ppm. What is the structure of V? We will learn about this reaction in Chapter 22.

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