/*! 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} Q62P The reaction of \({\left( {{\bf{... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The reaction of \({\left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}} \right)_{\bf{3}}}{\bf{CCHO}}\) with \({\left( {{{\bf{C}}_{\bf{6}}}{{\bf{H}}_{\bf{5}}}} \right)_{\bf{3}}}{\bf{P = C}}\left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}} \right){\bf{OC}}{{\bf{H}}_{\bf{3}}}\), followed by treatment with aqueous acid, afford R \(\left( {{{\bf{C}}_{\bf{7}}}{{\bf{H}}_{{\bf{14}}}}{\bf{O}}} \right)\). R has strong absorption in its IR spectrum at 1717 \({\bf{c}}{{\bf{m}}^{{\bf{ - 1}}}}\) and three singlets in its \({}^{\bf{1}}{\bf{H}}\)-NMR spectrum at 1.02 (9 H), 2.13 (3 H), and 2.33 (2 H) ppm. What is the structure of R? We will learn about this reaction in Chapter 21.

Short Answer

Expert verified

The olefination reaction involving the conversion of a carbonyl compound into an alkene using the triphenylphosphonium ylide is known as the Wittig reaction.

Step by step solution

01

Wittig reaction

The olefination reaction involving the conversion of a carbonyl compound into an alkene using the triphenylphosphonium ylide is known as the Wittig reaction.

02

Analysing the data

Given data:

Chemical formula =\({{\rm{C}}_7}{{\rm{H}}_{{\rm{14}}}}{\rm{O}}\)

IR absorption at = 1717 \({\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}\)

\({}^{\rm{1}}{\rm{H}}\)-NMR spectra = 1.02 ppm (9H), 2.13 ppm (3H), 2.33 ppm (2H)

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

where

n= number of carbon atoms

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

On substituting the values,

the degree of unsaturation (IHD) \(\begin{aligned}{c} = \frac{{\left( {\left( {{\rm{2}} \times 7{\rm{ + 2}}} \right){\rm{ - 14}}} \right)}}{{\rm{2}}}\\ = 1\end{aligned}\).

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

The singlet at 1.02 ppm attributes to 9H having a neighboring carbon with no hydrogen atoms.

The singlet at 2.13 ppm attributes to 3H having a neighboring carbon with no hydrogen atoms.

The singlet at 2.33 ppm attributes to 2H having a neighboring carbon with no hydrogen atoms.

The IR stretching at 1717 \({\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}\) indicates the presence of a ketone group.

03

Molecular structure

The required molecular structure of R is as follows:

Molecular structure of R

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

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: The 1 H NMR spectrum of N,N-dimethylformamide shows three singlets at 2.9, 3.0, and 8.0 ppm. Explain why the two groups are not equivalent to each other, thus giving rise to two NMR signals.

Question: How many \(^{\bf{1}}{\bf{H}}\) NMR signals does each compound show?

a.

b.

c.

d.

e.

f.

g.

h.

Question: The \(^{\bf{1}}{\bf{H}}\) NMR spectrum of \({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{OH}}\) recorded on a 500 MHz NMR spectrometer consists of two signals, one due to the \({\bf{C}}{{\bf{H}}_{\bf{3}}}\) protons at 1715 Hz and one due to the OH proton at 1830 Hz, both measured downfield from TMS. (a) Calculate the chemical shift of each absorption. (b) Do the \({\bf{C}}{{\bf{H}}_{\bf{3}}}\) protons absorb upfield or downfield from the OH proton?

Question:Reaction of unknown A with HCl forms chlorohydrin B as the major product. A shows no absorptions in its IR spectrum at 1700 cm-1 or 3600-3200 cm-1 , and gives the following 1H NMR data: 1.4 (doublet, 3 H), 3.0 (quartet of doublets, 1 H), 3.5 (doublet, 1 H), 3.8 (singlet, 3 H), 6.9 (doublet, 2 H), and 7.2 (doublet, 2 H) ppm.

(a) Propose a structure for A, including stereochemistry.

(b) Explain why B is the major product in this reaction.

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.