ORGANIC CHEMISTRY LL BUNDLE
ORGANIC CHEMISTRY LL BUNDLE
4th Edition
ISBN: 9781119761112
Author: Klein
Publisher: WILEY
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Chapter 22, Problem 87IP
Interpretation Introduction

Interpretation: The possible structure for the compound C6H15N that exhibits two types of signals in its proton NMR spectrum and two signals in 13C NMR spectrum has to be drawn.

Concept introduction:

The arrangement of atoms that are bonded together determines its constitution and molecular formula of that particular compound.  This concept is referred as structural isomers or in more modern term constitutional isomers.  Each atom has a typical valency or valence which is defined as the ability of an atom to form a chemical bond with other atoms.  For example, carbon has four valence or tetravalent that means carbon has the capacity to form four bonds with other elements or other atoms.  Nitrogen atoms are trivalent whereas hydrogen atoms are monovalent in nature.

In proton NMR, the signals are based on the arrangement of hydrogen atoms that are connected to carbon atoms.  In 13C NMR, the signals are based on the arrangement of carbon atoms.

To find: Draw the possible structure for the compound C6H15N that exhibits two types of signals in its proton NMR spectrum and two signals in 13C NMR spectrum

Find the valency for carbon (C), hydrogen (H) and nitrogen (N) in C6H15N

ORGANIC CHEMISTRY LL BUNDLE, Chapter 22, Problem 87IP

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Predict the organic products that form in the reaction below: OH H+ H+ + ☑ Y Note: You may assume you have an excess of either reactant if the reaction requires more than one of those molecules to form the products. In the drawing area below, draw the skeletal ("line") structures of the missing organic products X and Y. You may draw the structures in any arrangement that you like, so long as they aren't touching. Click and drag to start drawing a structure. ✓ m
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Please help, this is all the calculations i got!!! I will rate!!!Approx mass of KMnO in vial: 3.464 4 Moss of beaker 3×~0. z Nax200: = 29.9219 Massof weacerv after remosimgain N2C2O4. Need to fill in all the missing blanks. ง ง Approx mass of KMnO4 in vials 3.464 Mass of beaker + 3x ~0-304: 29.9219 2~0.20 Miss of beaker + 2x- 29.7239 Mass of beaker + 1x~0.2g Naz (204 29-5249 Mass of beaver after removing as qa Na₂ C₂O T1 T2 T3 Final Buiet reading Initial butet reading (int)) Hass of NaOr used for Titration -reading (mL) calculation Results: 8.5ml 17mL 27.4mL Oml Om Oml T1 T2 T3 Moles of No CO Moles of KMO used LOF KM. O used Molenty of KMNO Averagem Of KMOWL

Chapter 22 Solutions

ORGANIC CHEMISTRY LL BUNDLE

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