Concept explainers
a)
Interpretation:
The structure of an given molecular formula C8H16 to be predicted using 13CNMR spectra.
Concept introduction:
The 13CNMR spectrum gives information on the different electronic environments of carbon. As like 1HNMR, the number of signals generated in 13CNMR are predicted by performing symmetry operations (rotation or reflection symmetry). Only chemical shift values are reported in the spectrum but not the multiplicity and integration values because the coupling between two neighboring 13C - 13C nuclei are weakly involved due to the low abundance of 13C isotopes of carbon atom.
Broadband-decoupled 13CNMR spectrum: The spectra provide information regarding the total number of carbon environments.
DEPT (Distortionless enhancement by polarization transfer):
i) DEPT -90: The spectrum exhibits signal only from CH group and no signals from CH3, CH2, CH and quaternary carbon (carbon with no protons).
ii) DEPT -135: The spectrum exhibits CH3 groups and CH groups as positive signals (pointing up); CH2 groups appear as negative signals (pointing down) and quaternary carbon does not appear.
The signals appear in each type of spectrum:
To identify:
The structure of given molecular formula C8H16.
Answer to Problem 48AP
Explanation of Solution
Calculate HDI:
The HDI calculation led to confirm about the presence of a double bond or a ring.
Interpret the given 13CNMR spectrum:
Broadband - decoupled spectrum:
The spectrum shows five signals whereas the given molecular formula also has eight carbon atoms. Thus all the eight carbons have chemically different electronic environments showing signals.
i) The five signals in the region of 0-50ppm indicate the sp3 hybridized carbon atoms which can be methyl / methylene or Methine groups.
DEPT -90: Spectrum has no signals led to confirmation of no CH group in the structure.
DEPT -135 (gives signals of CH2, CH3 and C groups):
ii) The two positive signals indicate the presence of two methyl groups as no signals appear in the DEPT -90 spectrum.
iii) The three negative signals indicate the presence of four methylene groups; only methylene groups appear negative in the spectrum.
iv) So far the group of fragments obtained is
Two -CH3, -C-, four -CH2
From the fragments (CH3 + CH3 + (CH2)5 + C = 16 protons) sixteen protons are obtained whereas the total number of protons from the molecular formula is also sixteen.
The possible structures are:
The first structure possesses rotational symmetry and exhibits fewer signals in the spectrum and gets cancelled out.
The structure of a given molecular formula C8H16 is predicted using 13CNMR spectra.
b)
To identify:
The structure of given molecular formula C3H8O.
Answer to Problem 48AP
Explanation of Solution
Calculate HDI:
The HDI calculation led to confirm about the absence of a double bond or a ring.
Interpret the given 13CNMR spectrum:
Broadband - decoupled spectrum:
The spectrum shows three signals whereas the given molecular formula also has three carbon atoms. Thus all the three carbons have chemically different electronic environments showing signals.
i) The two signals in the region of 0-50ppm indicate the sp3 hybridized carbon atoms which can be methyl / methylene groups.
DEPT -90: Spectrum has no signals led to confirmation of no CH group in the structure.
DEPT -135 (gives signals of CH2, CH3 groups):
ii) The two positive signals indicate the presence of two methyl groups as no signals appear in the DEPT -90 spectrum.
iii) The one negative signals indicate the presence of methylene groups; only methylene groups appear negative in the spectrum.
iv) So far the group of fragments obtained is
Two -CH3, -CH2
From the fragments (CH3 + CH3 + CH2 = 8 protons) eight protons are obtained whereas the total number of protons from the molecular formula is also eight.
The structure of a given molecular formula C3H8O is predicted using 13CNMR spectra.
c)
To identify:
The structure of given molecular formula C10H20.
Answer to Problem 48AP
Explanation of Solution
Calculate HDI:
The HDI calculation led to confirm about the presence of a double bond or a ring.
Interpret the given 13CNMR spectrum:
Broadband - decoupled spectrum:
The spectrum shows six signals whereas the given molecular formula also has ten carbon atoms. Thus all the ten carbons have chemically different electronic environments showing signals.
i) The two signals in the region of 0-50ppm indicate the sp3 hybridized carbon atoms which can be methyl / methylene groups.
DEPT -90: Spectrum has signals led to confirmation of CH group in the structure.
DEPT -135 (gives signals of CH2, C, CH3 groups):
ii) The two positive signals indicate the presence of three methyl groups as no signals appear in the DEPT -90 spectrum.
iii) The three negative signals indicate the presence of five methylene groups; only methylene groups appear negative in the spectrum.
iv) So far the group of fragments obtained is
Two -CH3, -CH2
From the fragments (CH3)3 + (CH2)5 + CH + C = 20 protons) twenty protons are obtained whereas the total number of protons from the molecular formula is also twenty.
The structure of a given molecular formula C10H20 is predicted using 13CNMR spectra.
d)
To identify:
The structure of given molecular formula C6H10.
Answer to Problem 48AP
Explanation of Solution
Calculate HDI:
The HDI calculation led to confirm about the presence of a double bond or a ring. (two unsaturated degree).
Interpret the given 13CNMR spectrum:
Broadband - decoupled spectrum:
The spectrum shows six signals whereas the given molecular formula also has six carbon atoms. Thus all the six carbons have chemically different electronic environments showing signals.
i) The two signals in the region of 0-50ppm indicate the sp3 hybridized carbon atoms which can be methyl / methylene groups.
DEPT -90: Spectrum has signals led to confirmation of CH group in the structure.
DEPT -135 (gives signals of CH2, C, CH3 groups):
ii) The four positive signals indicate the presence of two methyl groups as no signals appear in the DEPT -90 spectrum.
iii) The negative signals indicate the presence of methylene groups; only methylene groups appear negative in the spectrum.
iv) So far the group of fragments obtained is
Two -CH3, -CH2
From the fragments (CH3)2 + CH2 + (CH)2 + C = 10 protons) ten protons are obtained whereas the total number of protons from the molecular formula is also ten.
The structure of a given molecular formula C6H10 is predicted using 13CNMR spectra.
e)
To identify:
The structure of given molecular formula C8H16.
Answer to Problem 48AP
Explanation of Solution
Calculate HDI:
The HDI calculation led to confirm about the presence of a double bond or a ring.
Interpret the given 13CNMR spectrum:
Broadband - decoupled spectrum:
The spectrum shows four signals whereas the given molecular formula also has eight carbon atoms. Thus all the eight carbons have chemically different electronic environments showing signals.
i) The four signals in the region of 0-50ppm indicate the sp3 hybridized carbon atoms which can be methyl / methylene or Methine groups.
DEPT -90: Spectrum has no signals led to confirmation of CH group in the structure.
DEPT -135 (gives signals of CH2, CH3 and CH groups):
ii) The two positive signals indicate the presence of two methyl groups as no signals appear in the DEPT -90 spectrum.
iii) The two negative signals indicate the presence of four methylene groups; only methylene groups appear negative in the spectrum.
iv) So far the group of fragments obtained is
Two -CH3, -CH-, four -CH2
From the fragments (CH3 + CH3 + (CH2)4 + (CH)2 = 16 protons) sixteen protons are obtained whereas the total number of protons from the molecular formula is also sixteen.
The structure of a given molecular formula C8H16 is predicted using 13CNMR spectra.
f)
To identify:
The structure of given molecular formula C6H10O.
Answer to Problem 48AP
Explanation of Solution
Calculate HDI:
The HDI calculation led to confirm about the presence of a double bond or a ring.
Interpret the given 13CNMR spectrum:
Broadband - decoupled spectrum:
The spectrum shows four signals whereas the given molecular formula also has six carbon atoms. Thus all the six carbons have chemically different electronic environments showing signals.
i) The four signals in the region of 0-50ppm indicate the sp3 hybridized carbon atoms which can be methyl / methylene or Methine groups.
DEPT -135 (gives signals of CH2, C groups):
ii) The two negative signals indicate the presence of four methylene groups; only methylene groups appear negative in the spectrum.
iii) So far the group of fragments obtained is
C=O, -C-, four -ch2
From the fragments (CH2)5 + C = 10 protons) ten protons are obtained whereas the total number of protons from the molecular formula is also ten.
The structure of a given molecular formula C6H10O is predicted using 13CNMR spectra.
Want to see more full solutions like this?
Chapter 13 Solutions
Student Value Bundle: Organic Chemistry, + OWLv2 with Student Solutions Manual eBook, 4 terms (24 months) Printed Access Card (NEW!!)
- Nonearrow_forwardPart II. Given below are the 'H-NMR spectrum at 300 MHz in CDC13 and mass spectrum using electron ionization of compound Brian. The FTIR of the said compound showed a strong peak at 1710 cm"). Determine the following: (a) molecular Formula and Degree of unsaturation of compound Brian (b) Basing on the given H-NMR spectrum tabulate the following (i) chemical shifts (ii) integration, ciii) multiplicity and (iv) interferences made for each signal (c) Draw the structure of compound Brian. ) ΕΙ 43 41 27 71 114 (M+) Hmmm 20 30 40 50 60 70 80 90 100 110 120 1H NMR spectrum 300 MHz in CDCl3 2.0 alle 1.0arrow_forwardThe iron-iron carbide phase diagram. In the diagram, the letter L indicates that it is a liquid. Indicate what its components are. Temperature (°C) 1600 10 Composition (at% C) 15 25 1538°C -1493°C 8 1400 1200 1394°C y+L L 2500 1147°C y. Austenite 2.14 4.30 2000 1000 912°C y + Fe3C 800ㅏ 0.76 0.022 600 a, Ferrite a + Fe3C 400 0 (Fe) Composition (wt% C) 727°C 1500 Cementite (Fe3C) 1000 6 6.70 Temperature (°F)arrow_forward
- Part V. Choose which isomer would give the 1H-NMR spectrum below. Justify your reasoning by assigning important signals to the Corresponding protons of the correct molecule. A D on of of of H H 88 2 90 7.8 7.6 7.4 80 5 6 [ppm] 7.2 6.8 6.6 6.4 ō [ppm]arrow_forwardShow work with explanation. don't give Ai generated solutionarrow_forwardQ7. a. Draw the line-bond structure of the major product for the following reaction, if a reaction occurs, assume monohalogenation. b. Calculate the product ratios using the following information (hint: use the number of hydrogens in each category present to calculate the ratios). Chlorination: 1° Reactivity=1 2° Reactivity=4 Heat + Cl2 3° Reactivity=5arrow_forward
- Please correct answer and don't use hand rating and don't use Ai solutionarrow_forwardQ10: Alkane halogenation a. Give the name and structures of the five isomeric hexanes. Page 4 of 5 Chem 0310 Organic Chemistry 1 Recitations b. For each isomer, give all the free radical monochlorination and monobromination products that are structurally isomeric.arrow_forwardQ9. The insecticide DDT (in the box below) is useful in controlling mosquito populations and has low toxicity to humans, but is dangerous to birds and fish. Hoping to alleviate the dangers, little Johnny Whizbang, an aspiring chemist, proposes a new version of DDT ("Bromo-DDT") and shows his synthesis to his boss. Will Johnny Whizbang's synthesis work? Or will he be fired? Assume there is an excess of bromine and polybrominated products can be separated. Explain why. CH3 Br2, light CBR3 ok-ok Br Br Br Br CI "Bromo-DDT" CCl 3 DDT (dichlorodiphenyltrichloroethane) CIarrow_forward
- Differentiate the terms Monotectic, Eutectic, Eutectoid, Peritectic, Peritectoid.arrow_forwardQ5. Predict the organic product(s) for the following transformations. If no reaction will take place (or the reaction is not synthetically useful), write "N.R.". Determine what type of transition state is present for each reaction (think Hammond Postulate). I Br₂ CH3 F2, light CH3 Heat CH3 F₂ Heat Br2, light 12, light CH3 Cl2, lightarrow_forwarda. For the following indicated bonds, rank them in order of decreasing AH° for homolytic cleavage. Based on your answer, which bond would be most likely to break homolytically? (a) (c) H3C CH3 .CH3 CH3 CH3 (b) Page 1 of 5 Chem 0310 Organic Chemistry 1 Recitations b. Draw all the possible radical products for 2-methylbutane, and determine which bond is most likely to be broken.arrow_forward
- Principles of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning