Question

Give a structure or structure consistent with each of the following sets of nmr data. (a) \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{Cl}_{5}\) (d) \(\quad \mathrm{C}_{10} \mathrm{H}_{14}\) a triplet, \(64.52,1 \mathrm{H}\) a doublet, \(\delta 0.88,6 \mathrm{H}\) b doublet, \(\delta 6.07,2 \mathrm{H}\) b multiplet, \(\delta 1.86,1 \mathrm{H}\) c doublet, \(\delta 2.45,2 \mathrm{H}\) d singlet, \(\delta 7.12,5 \mathrm{H}\) (b) \(\quad \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{Cl}_{3} \quad\) (e) \(\mathrm{C}_{10} \mathrm{H}_{13} \mathrm{Cl}\) a singlet, \(\delta 2.20,3 \mathrm{H}\) a singlet, \(61.57,6 \mathrm{H}\) c singlet, \(\delta 7.27,5 \mathrm{H}\) (c) \(\mathrm{C}_{10} \mathrm{H}_{14}\) a singlet, \(\delta 1.30,9 \mathrm{H}\) a singlet, \(\delta 7.28,5 \mathrm{H}\)

Short answer

The structures consistent with the given NMR data are as follows: (a) \(\mathrm{CCl}_{2} \mathrm{C}( \mathrm{Cl}_{2} ) \mathrm{CHCl}_{2}\) (b) \(\mathrm{CH}_{3} \mathrm{CCl}_{2} \mathrm{CHCl}_{2}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C}(\mathrm{CH}_{3})_{3}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (e) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C}(\mathrm{CH}_{3})_{2} \mathrm{CH}_{2} \mathrm{Cl}\)

Step-by-step solution

(a) Analyzing NMR data of C3H3Cl5

In this case, we have no given data. It's a simple compound consisting of 3 carbon atoms, 3 hydrogen atoms, and 5 chlorine atoms. The only possible structure for this compound is \(\mathrm{CCl}_{2} \mathrm{C}( \mathrm{Cl}_{2} ) \mathrm{CHCl}_{2} \).

(b) Analyzing NMR data of C3H5Cl3

We are given the following NMR peaks for this compound: 1. a singlet, \(\delta 2.20, 3 \mathrm{H}\) We notice that there are only 3 hydrogen atoms in this compound, all of which show up as a singlet, indicating they are all equivalent. Thus, the only possible structure with 3 carbon atoms, 5 hydrogen atoms, and 3 chlorine atoms is \(\mathrm{CH}_{3} \mathrm{CCl}_{2} \mathrm{CHCl}_{2} \).

(c) Analyzing NMR data of C10H14

We are given the following NMR peaks for this compound: 1. a singlet, \(\delta 1.30, 9 \mathrm{H}\) 2. a singlet, \(\delta 7.28, 5 \mathrm{H}\) Based on the data, there is no indication of any splitting, which means all the hydrogen atoms are equivalent in each group. A possible structure that fits this description is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C}(\mathrm{CH}_{3})_{3}\) (a methyl-substituted phenyl ring).

(d) Analyzing NMR data of C10H14

We are given the following NMR peaks for this compound: 1. a triplet, \(\delta 0.88, 1 \mathrm{H}\) 2. a doublet, \(\delta 6.07, 2 \mathrm{H}\) 3. a multiplet, \(\delta 1.86, 1 \mathrm{H}\) 4. a doublet, \(\delta 2.45, 2 \mathrm{H}\) 5. a singlet, \(\delta 7.12, 5 \mathrm{H}\) From the analysis, we have the following information: - One hydrogen with a triplet (connected to a carbon bonded to two other hydrogens) - Two hydrogens with a doublet (connected to a carbon bonded to one other hydrogen) - One hydrogen with a multiplet - Two hydrogens with a doublet (connected to a carbon bonded to one other hydrogen) - A group of 5 hydrogens showing up as a singlet, which suggests they are all equivalent and possibly in a phenyl ring. A possible structure fitting this description is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (a n-propyl-substituted phenyl ring).

(e) Analyzing NMR data of C10H13Cl

We are given the following NMR peaks for this compound: 1. a singlet, \(\delta 1.57, 6 \mathrm{H}\) 2. a singlet, \(\delta 7.27, 5 \mathrm{H}\) Again, all hydrogens in each group are equivalent, which points towards a simple substitution of the phenyl ring. A possible structure fitting this description is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C}(\mathrm{CH}_{3})_{2} \mathrm{CH}_{2} \mathrm{Cl}\) (a Cl-substituted isopropyl phenyl ring).

Key concepts

Chemical Shift

The chemical shift is an essential term in NMR spectroscopy analysis, representing the resonance frequency of nuclei relative to a standard reference. In simple terms, it tells us about the electronic environment surrounding a nucleus—typically a hydrogen atom in proton NMR—by indicating how much the magnetic field at the nucleus has been affected by electron distribution.

In the exercise given, when you see a chemical shift like \(\delta 0.88, 6 \mathrm{H}\), it implies that the 6 hydrogens are in a particular electronic environment, likely near electronegative atoms or within a certain molecular framework. It's important to remember that the values of chemical shift are measured in parts per million (ppm) and they can help us determine the type of chemical group present. For instance, hydrogens attached to carbons bonded to electronegative atoms often have higher chemical shift values, whereas hydrogens on carbons without such bonds typically resonate at lower ppm values.

Multiplicity in NMR

Multiplicity refers to the splitting pattern of NMR signals and is determined by the number of neighboring hydrogen atoms that a given hydrogen atom (or group of equivalent hydrogens) has. It arises from a phenomenon known as spin-spin coupling.

For example, in the NMR data of \(\mathrm{C}_{10} \mathrm{H}_{14}\), we see a 'triplet' at \(\delta 0.88, 1 \mathrm{H}\), suggesting that the hydrogen is adjacent to two hydrogens (n+1 rule, where n equals the number of neighboring hydrogens). Similarly, a 'doublet' could indicate the presence of one neighboring hydrogen. Learning to read these patterns helps us understand the structure of a molecule by revealing connectivity between atoms.

Hydrogen Equivalence

Hydrogen equivalence is a notion in NMR that posits when two or more hydrogens are in identical chemical environments, they are considered equivalent and will produce a single signal. This can be pivotal when deducing a compound's structure.

In the structure \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{Cl}_{3}\), we observe a 'singlet' indicating that all three hydrogen atoms are equivalent, leading to the conclusion that they are likely part of a methyl group (\(\mathrm{CH}_{3}\)) since they exhibit no splitting. This principle aids in simplifying complex NMR spectra by reducing the number of signals to interpret.

NMR Spectroscopy Analysis

NMR spectroscopy analysis involves interpreting the data from NMR spectra to deduce the structure of organic compounds. The process often begins with identifying the molecular formula of the compound, followed by analysis of the chemical shift, multiplicity, and integration of the signals reported.

For instance, with the given compound \(\mathrm{C}_{10} \mathrm{H}_{13} \mathrm{Cl}\), the analysis includes identifying that the six hydrogens with a singlet at \(\delta 1.57\) are likely part of two identical groups (suggested by their equivalence), while the five hydrogens showing a singlet at \(\delta 7.27\) are part of a different chemical environment. By piecing together such clues, chemists can construct the most probable molecular structure that matches the NMR data.