Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Write structures for compounds with the following descriptions (There may be more than one correct answer, but only one answer is required.) a. \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}\) with one proton \(\mathrm{nmr}\) shift b. \(\mathrm{C}_{6} \mathrm{H}_{12}\) with one proton nmr shift c. \(\mathrm{C}_{5} \mathrm{H}_{12}\) with one proton \(\mathrm{nmr}\) shift d. \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}\) with two different proton \(\mathrm{nmr}\) shifts e. \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}\) with three different proton \(\mathrm{nmr}\) shifts f. \(\mathrm{C}_{4} \mathrm{Cl}_{8}\) with two different \({ }^{13} \mathrm{C} \mathrm{nmr}\) shifts

Short Answer

Expert verified
a) Ethanol, b) Cyclohexane, c) Neopentane, d) Butanone, e) Ethyl acetate, f) Tetrachloromethane.

Step by step solution

01

Identify Molecular Formulas

To start solving the exercise, identify each molecular formula in the given descriptions, such as \( \text{C}_2 \text{H}_6 \text{O} \), \( \text{C}_6 \text{H}_{12} \), \( \text{C}_5 \text{H}_{12} \), \( \text{C}_4 \text{H}_8 \text{O} \), \( \text{C}_4 \text{H}_8 \text{O}_2 \), and \( \text{C}_4 \text{Cl}_8 \). This will guide us in determining possible structures for each compound.
02

Determine Functional Groups and Structures

For each compound, based on the molecular formula, hypothesize potential functional groups or structural isomers: a. For \( \text{C}_2 \text{H}_6 \text{O} \), consider ethyl alcohol.b. For \( \text{C}_6 \text{H}_{12} \), consider a symmetrical alkene like cyclohexane which will show one proton NMR shift.c. For \( \text{C}_5 \text{H}_{12} \), consider neopentane with a symmetrical structure to show one proton NMR shift.d. For \( \text{C}_4 \text{H}_8 \text{O} \), consider butanone which would yield two proton NMR shifts due to its structure.e. For \( \text{C}_4 \text{H}_8 \text{O}_2 \), consider ethyl acetate, which will give three distinct proton NMR shifts.f. For \( \text{C}_4 \text{Cl}_8 \), a symmetrical structure like carbon tetrachloride will fit, but to meet the requirement of 2 different \(^{13}C\) NMR, consider tetrachloromethane with two unique carbon environments.
03

Verify NMR Shifts

Ensure that the structures align with the given NMR shift requirements: a. Ethanol has symmetrical hydrogens in its methyl group and hydroxyl group, leading to one type of NMR shift. b. Cyclohexane's symmetry gives just one proton type, resulting in one shift. c. Neopentane's symmetry also provides a single type of proton for one shift. d. Butanone has two main environments for protons, giving two NMR shifts. e. Ethyl acetate's protons are in three different environments, resulting in three shifts. f. Tetrachloromethane with a periodical substitution shows two types of carbon environments for two NMR shifts.
04

Draw Chemical Structures

Translate molecular formula and symmetry considerations into chemical structures: a. Ethyl alcohol can be drawn as CH3-CH2-OH. b. Cyclohexane is drawn by a hexagonal ring of carbon atoms, with two hydrogens attached to each. c. Neopentane is illustrated as a central carbon with four CH3 groups branching out. d. Butanone is drawn as CH3-CO-CH3-CH3, with a carbonyl group in the middle. e. Ethyl acetate is depicted as CH3-COO-CH2-CH3. f. Draw a structure of tetrachloromethane with variations leading to two environments.

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 Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular formulas
Molecular formulas are a blueprint that details the types and numbers of atoms within a molecule. They act like coded messages indicating the constituent elements of a compound. For example, the formula \( \text{C}_2 \text{H}_6 \text{O} \) suggests a molecule consisting of two carbon atoms, six hydrogen atoms, and one oxygen atom. This could correspond to familiar compounds such as ethanol or dimethyl ether.
  • These formulas help in predicting possible structures and thereby possible chemical and physical properties of a substance.
  • They are integral for calculating molar masses, determining empirical formulas, and understanding stoichiometry in chemical reactions.
However, they don't provide information about the molecule's structure, such as how the atoms are connected or spatially arranged. That's where structural formulas and corresponding NMR shifts come into play.
Functional groups
Functional groups are specific clusters of atoms within molecules that are responsible for the characteristic chemical reactions of those molecules. Identifying these groups is crucial for understanding the behavior of the molecules under various conditions. For instance, in the formula \( \text{C}_2 \text{H}_6 \text{O} \), the compound could potentially have an alcoholic (OH group) or an ether functional group.
  • Alcoholic functional groups, such as in ethanol, tend to participate in hydrogen bonding, affecting boiling points and solubility.
  • Ethers are generally less reactive but can serve as good solvents due to their polar nature.
By recognizing these key structures, chemists can better predict reactions and interactions, assisting in tasks from synthesis to pharmaceutical development.
Chemical structures
Chemical structures visually represent the arrangements of atoms within a molecule. Different structures can exist for a given molecular formula, known as isomers. This aspect can significantly impact chemical properties and reactivity. For example, \( \text{C}_5 \text{H}_{12} \) could represent pentane, isopentane, or neopentane – all unique in the spatial arrangement of their atoms.
  • Pentane is a straight-chain alkane, leading to different physical properties compared to its branched counterparts.
  • Isomers can exhibit variations in boiling points, solubility, and chemical reactivity.
Determining the chemical structure provides insight beyond the mere count of atoms, helping elucidate how molecules will interact in different conditions or reactions.
Proton NMR shifts
Proton NMR shifts are key indicators used to determine the chemical environment of hydrogen atoms in a molecule. They arise because hydrogen nuclei (protons) resonate at different frequencies in a magnetic field, affected by nearby electron clouds. Each unique hydrogen environment will produce a separate signal or shift. This property allows chemists to ascertain molecular structure details.
  • For molecules like \( \text{C}_6 \text{H}_{12} \), a compound such as cyclohexane will exhibit only one proton NMR shift due to its symmetry, meaning all hydrogens are equivalent.
  • In contrast, a molecule like butanone (\( \text{C}_4 \text{H}_8 \text{O} \)) will have two distinct shifts due to differing local hydrogen environments.
These shifts are crucial for identifying compounds and understanding molecular architectures, serving as a molecular fingerprint in chemistry.
Carbon-13 NMR shifts
Carbon-13 NMR shifts operate on principles similar to those for protons but provide insights into the environments of carbon atoms within a molecule. Since carbon-13 is a naturally occurring isotope with a low natural abundance, these shifts are slightly less sensitive but immensely informative.
  • Distinct carbon environments, as seen in compounds like \( \text{C}_4 \text{Cl}_8 \), will generate unique shifts, allowing chemists to identify and distinguish between different carbon atoms within a molecule.
  • Symmetrically equivalent carbons will lead to fewer distinct shifts in the NMR spectrum.
Carbon-13 NMR spectroscopy is invaluable for constructing a detailed understanding of molecular frameworks, playing a pivotal role in modern analytical chemistry.

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

a. Calculate the relative intensities of the \((\mathrm{M}+1)^{+}\) and \((\mathrm{M}+2)^{+}\) ions for a molecule of elemental composition \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NO}_{2}\) b. The \(\mathrm{M}^{+},(\mathrm{M}+1)^{+}\), and \((\mathrm{M}+2)^{+}\) ion intensities were measured as \(100,8.84\), and \(0.54\) respectively, and the molecular weight as 120 . What is the molecular formula of the compound? c. In our example of how natural \({ }^{13} \mathrm{C}\) can be used to determine the number of carbon atoms in a compound with \(\mathrm{M}^{+}=86\) and a \((\mathrm{M}+1)^{+} / \mathrm{M}^{+}\) ratio of \(6.6 / 100\), we neglected the possible contribution to the \((\mathrm{M}+1)^{+}\) peak of the hydrogen isotope of mass 2 (deuterium). The natural abundance of deuterium is \(0.015 \%\) For a compound of composition \(\mathrm{C}_{6} \mathrm{H}_{14}\), how much do you expect the deuterium to contribute to the intensity of the \((\mathrm{M}+1)^{+}\) peak relative to the \(\mathrm{M}^{+}\) peak?

a. Show how the assignment of \(J_{A B}=J_{B C}=2 J_{A C}\) leads to the prediction of four equally spaced and equally intense lines for the methyl resonance of 2 -phenylpropene. b. What would the splittings of the alkenic and methyl protons look like for trans-1-phenylpropene if \(J_{A B}=16 \mathrm{~Hz}\), \(J_{A C}=4 \mathrm{~Hz}\), and \(J_{B C}=0 \mathrm{~Hz} ?\)

Which compound in each group would have the most intense infrared absorption band corresponding to stretching vibrations of the bonds indicated? Give your reasoning. a. \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{O},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CH}_{2}\) (multiple bond) b. \(\mathrm{CH}_{3}-\mathrm{CH}_{3}, \mathrm{CH}_{3}-\mathrm{O}-\mathrm{CH}_{3}(\mathrm{C}-\mathrm{C}\) vs. \(\mathrm{C}-\mathrm{O})\) c. \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH}, \mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{3}\) (multiple bond) d. \(\mathrm{H}-\mathrm{Cl}, \mathrm{Cl}-\mathrm{Cl}\)

Sketch the proton chemical shifts in ppm and \(\mathrm{Hz}\) as well as the integral you would expect for each of the following substances at \(60 \mathrm{MHz}\). (The spin-spin splitting of the resonance lines evident in Figures 9-23 and 9-27, but not seen in Figure \(9-31\), can be safely neglected with all of the compounds listed.) a. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{OCH}_{3}\) b. \(\mathrm{CH}_{2} \mathrm{COC}\left(\mathrm{CH}_{3}\right)_{3}\) c. \(\mathrm{HCOC}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHO}\) d. e. \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CCl}_{2}\) f. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COC} \equiv \mathrm{CH}\) g. \(\left(\mathrm{CH}_{2} \mathrm{Cl}\right)_{3} \mathrm{CCO}_{2} \mathrm{H}\) h.* cis-1-methyl-4-tert-butyl-1,2,2,3,3,4,5,5,6,6-decachlorocyclohexane

Sketch the proton \(\mathrm{nmr}\) spectrum and integral expected at \(60 \mathrm{MHz}\), with TMS as standard, for the following substances. Show the line positions in \(\mathrm{Hz}\); neglect spin-spin couplings smaller than 1 to \(2 \mathrm{~Hz}\) and all second-order effects. Remember that chlorine, bromine, and iodine (but not fluorine) act as nonmagnetic nuclei. a. \(\mathrm{CH}_{3} \mathrm{Cl}\) b. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\) c. \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}\) d. \(\mathrm{CH}_{3} \mathrm{CCl}_{2} \mathrm{CH}_{2} \mathrm{Cl}\) e. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) f. \(\mathrm{CHCl}_{2} \mathrm{CHBr}_{2}\) g. \(\mathrm{CH}_{3} \mathrm{CHClCOCH}_{3}\) h. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) i. \(\mathrm{ClCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{I}\) j. \(\left(\mathrm{ClCH}_{2}\right)_{3} \mathrm{CH}\)

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.

Sign-up for free