/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q.21558-14-63P Question. Reaction of aldehyde D... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Rank each group of protons in order of increasing chemical shift.

a.

b.

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 can you use 1H NMR spectroscopy to distinguish between CH2=C(Br)CO2CH3 and methyl (E)-3-bromopropenoate, BrCH=CHCO2CH3?

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?

Label each statement as True or False.

a. When a nucleus is strongly shielded, the effective field is larger than the applied field and the absorption shifts downfield.

b. When a nucleus is strongly shielded, the effective field is smaller than the applied field and the absorption is shifted upfield.

c. A nucleus that is strongly deshielded requires a lower field strength for resonance.

d. A nucleus that is strongly shielded absorbs at a larger δ value.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.