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Schaum's Outline of Organic Chemistry

George Hademenos, George Hademenos

Chapter 12

SPECTROSCOPY AND STRUCTURE - all with Video Answers

Educators

Chapter Questions

04:56

Problem 1

(a) The wavelengths are substituted into the equation $v=c / \lambda$, where $c=$ speed of light $=3.0 \times 10^8 \mathrm{~m} / \mathrm{s}$. Thus
Violet: $\quad v=\frac{3.0 \times 10^8 \mathrm{~m} / \mathrm{s}}{400 \times 10^{-9} \mathrm{~m}}=7.5 \times 10^{14} \mathrm{~s}^{-1}=750 \mathrm{THz}$
Red: $\quad v=\frac{3.0 \times 10^8 \mathrm{~m} / \mathrm{s}}{750 \times 10^{-9} \mathrm{~m}}=4.0 \times 10^{14} \mathrm{~s}^{-1}=400 \mathrm{THz}$
where $1 \mathrm{THz}=10^{12} \mathrm{~Hz}=10^{12} \mathrm{~s}^{-1}$. Violet light has the shorter wavelength and higher frequency.
(b) The frequencies from part (a) are substituted into the equation $E=h v$, where $h=6.624 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}$ (Planck's constant). Thus
$$
\begin{aligned}
\text { Violet: } & E=\left(6.624 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(7.5 \times 10^{14} \mathrm{~s}^{-1}\right)=5.0 \times 10^{-19} \mathrm{~J} \\
\text { Red: } & E=\left(6.624 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(4.0 \times 10^{14} \mathrm{~s}^{-1}\right)=2.7 \times 10^{-19} \mathrm{~J}
\end{aligned}
$$

Photons of violet light have more energy than those of red light.

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
08:10

Problem 1

(a) Suggest a structure for a compound $\mathrm{C}_9 \mathrm{H}_{12}$ showing low-resolution nmr signals at $\delta$ values of $7.1,2.2,1.5$, and $0.9 \mathrm{ppm}$. (b) Give the relative signal areas for the compound.
(a) The value $7.1 \mathrm{ppm}$ indicates $\mathrm{H}$ 's on a benzene ring. The formula shows three more $\mathrm{C}$ 's, which might be attached to the ring as shown below (assuming that, since this is an alkylbenzene, all aromatic $\mathrm{H}$ 's are equivalent):
(1) $3 \mathrm{CH}_3$ 's in trimethylbenzene, $\left(\mathrm{CH}_3^a\right)_3 \mathrm{C}_6 \mathrm{H}_3^b$
(2) a $\mathrm{CH}_3$ and a $\mathrm{CH}_2 \mathrm{CH}_3$ in $\mathrm{CH}_3^a \mathrm{C}_6 \mathrm{H}_4^b \mathrm{CH}_2^c \mathrm{CH}_3^d$
(3) a $\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ in $\mathrm{C}_6 \mathrm{H}_5^a \mathrm{CH}_2^b \mathrm{CH}_2^c \mathrm{CH}_3^d$
(4) a $\mathrm{CH}\left(\mathrm{CH}_3\right)_2$ in $\mathrm{C}_6 \mathrm{H}_5^a \mathrm{CH}^h\left(\mathrm{CH}_3^c\right)_2$
Compounds (1) and (4) can be eliminated because they would give two and three signals, respectively, rather than the four observed signals. Although (2) has four signals, $\mathrm{H}^a$ and $\mathrm{H}^c$ are different benzylic $\mathrm{H}$ 's and the compound should have two signals in the region $3.0-2.2 \mathrm{ppm}$ rather than the single observed signal. Hence (2) can be eliminated. Only (3) can give the four observed signals with the proper chemical shifts.
(b) $5: 2: 2: 3$.

Dr. Satish Ingale
Dr. Satish Ingale
Numerade Educator
05:04

Problem 2

Express 10 micrometers $(a)$ in centimeters, $(b)$ in angstroms $\left(1 \AA=10^{-10} \mathrm{~m}\right),(c)$ in nanometers, (d) as a wave number.
(a) $10 \mu \mathrm{m}=\left(10 \times 10^{-6} \mathrm{~m}\right)\left(\frac{100 \mathrm{~cm}}{1 \mathrm{~m}}\right)=10^{-3} \mathrm{~cm}$
(b) $10 \mu \mathrm{m}=\left(10 \times 10^{-6} \mathrm{~m}\right)\left(\frac{1 \AA}{10^{-10} \mathrm{~m}}\right)=10^5 \mathrm{~A}$
(c) $10 \mu \mathrm{m}=\left(10 \times 10^{-6} \mathrm{~m}\right)\left(\frac{10^9 \mathrm{~nm}}{1 \mathrm{~m}}\right)=10^4 \mathrm{~nm}$
(d) $\tilde{v}=\frac{1}{10 \times 10^{-6} \mathrm{~m}}=10^5 \mathrm{~m}^{-1}=\left(10^5 \mathrm{~m}^{-1}\right)\left(\frac{1 \mathrm{~cm}^{-1}}{100 \mathrm{~m}^{-1}}\right)=10^3 \mathrm{~cm}^{-1}$

In a typical spectrophotometer, a dissolved compound is exposed to electromagnetic radiation with a continuous spread in wavelength. The radiation passing through or absorbed is recorded on a chart against the wavelength or wave number. Absorption peaks are plotted as minima in infrared, and usually as maxima in ultraviolet spectroscopy.
At a given wavelength, absorption follows an exponential law of the form
$$
A=\varepsilon C l
$$
where $A \equiv$ absorbance $\equiv-\log _{10}$ (fraction of incident radiation transmitted)
$\varepsilon \equiv$ molar extinction coefficient, $\mathrm{cm}^2 / \mathrm{mol}$
$C \equiv$ concentration of solution, $\mathrm{mol} / \mathrm{L}\left(=\mathrm{mol} / \mathrm{cm}^3\right)$
$l \equiv$ thickness of solution presented to radiation, $\mathrm{cm}$

The wavelength of maximum absorption, $i_{\max }$, and the corresponding $\varepsilon_{\max }$ are identifying properties of a compound. Units are normally omitted from specifications of $\varepsilon$.

Suzanne W.
Suzanne W.
Numerade Educator
02:01

Problem 3

The relative energy for various electronic states (MO's) is:

List the three electronic transitions detectable by uv spectrophotometers in order of increasing $\Delta E$.
$$
\mathrm{n} \longrightarrow \pi^*<\pi \longrightarrow \pi^*<\mathrm{n} \longrightarrow \sigma^*
$$

Nicole Smina
Nicole Smina
Numerade Educator
08:24

Problem 4

Methanol is a good solvent for uv but not for ir determinations. Why?
Methanol absorbs in the uv in $183 \mathrm{~nm}$, which is below $190 \mathrm{~nm}$, the cutoff for most spectrophotometers, and therefore it doesn't interfere. Its ir spectrum has bands in most regions and therefore it cannot be used. Solvents such as $\mathrm{CCl}_4$ and $\mathrm{CS}_2$ have few interfering bands are are preferred for ir determinations.

Temi Ajayi
Temi Ajayi
Numerade Educator
01:37

Problem 4

List all the electronic transitions possible for (a) $\mathrm{CH}_4$, (b) $\mathrm{CH}_3 \mathrm{Cl},(c) \mathrm{H}_2 \mathrm{C}=\mathrm{O}$.
(a) $\sigma \rightarrow \sigma^*$. (b) $\sigma \rightarrow \sigma^*$ and $\mathrm{n} \rightarrow \sigma^*$ (there are no $\pi$ or $\pi^*$ MO's). (c) $\sigma \rightarrow \sigma^*, \sigma \rightarrow \pi^*, \pi \rightarrow \sigma^*, \mathrm{n} \rightarrow \sigma^*$, $\pi \rightarrow \pi^*$ and $\mathrm{n} \rightarrow \pi^*$.

Adriano Chikande
Adriano Chikande
Numerade Educator
01:22

Problem 5

The uv spectrum of acetone shows two peaks of $i_{\max }=280 \mathrm{~nm}, \varepsilon_{\max }=15$ and $i_{\max }=190$, $\varepsilon_{\max }=100$. (a) Identify the electronic transition for each. $(b)$ Which is more intense?
(a) The longer wavelength $(280 \mathrm{~nm})$ is associated with the smaller-energy $\left(\mathrm{n} \rightarrow \pi^*\right)$ transition. $\pi \rightarrow \pi^*$ occurs at $190 \mathrm{~nm}$.
(b) $\pi \rightarrow \pi^*$ has the larger $\varepsilon_{\max }$ and is the more intense peak.

Lottie Adams
Lottie Adams
Numerade Educator
02:27

Problem 6

Draw conclusions about the relationship of $\lambda_{\max }$ to the structure of the absorbing molecule from the following $i_{\max }$ values (in nm): ethylene (170), 1,3-butadiene (217), 2.3-dimethyl-1,3-butadiene (226), 1,3-cyclohexadiene (256), and 1,3,5-hexatriene (274).
1. Conjugation of $\pi$ bonds causes molecules to absorb at longer wavelengths.
2. As the number of conjugated $\pi$ bonds increases, $i_{\max }$ increases.
3. Cyclic polyenes absorb at higher wavelengths than do acyclic polyenes.
4. Substitution of alkyl groups on $\mathrm{C}=\mathrm{C}$ causes a shift to longer wavelength (red shift).

Tianyu Li
Tianyu Li
Numerade Educator
02:34

Problem 7

Account for the following variations in $i_{\max }(\mathrm{nm})$ of $\mathrm{CH}_3 \mathrm{X}: \mathrm{X}=\mathrm{Cl}(173), \mathrm{Br}(204)$, and I(258).

The transition must be $\mathrm{n} \rightarrow \sigma^*$ [Problem 12.4(b)]. On going from $\mathrm{Cl}$ to $\mathrm{Br}$ to $\mathrm{I}$ the $\mathrm{n}$ electrons $(a)$ are found in higher principal energy levels (the principal quantum numbers are $3,4,5$, respectively), $(b)$ are futher away from the attractive force of the nucleus, and (c) are more easily excited. Hence absorption occurs at progressively higher $\dot{\lambda}_{\max }$ since less energy is required.

Pronoy Sinha
Pronoy Sinha
Numerade Educator
02:09

Problem 8

Identify the two geometric isomers of stilbene, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}=\mathrm{CHC}_6 \mathrm{H}_5$, from their $i_{\max }$ values, $294 \mathrm{~nm}$ and $278 \mathrm{~nm}$.

The higher-energy cis isomer has the shorter wavelength. Steric strain prevents full coplanarity of the cis phenyl groups, and the conjugative effect is attenuated.

Freddie Montague
Freddie Montague
Numerade Educator
01:28

Problem 9

The complementary color pairs are: violet-yellow, blue-orange, and green-red. Given a red, an orange and a yellow polyene, which is most and which is least conjugated?

The orange polyene absorbs blue, the red absorbs green and the yellow absorbs violet. The most conjugated polyene, in this case the red one, absorbs the color of longest wavelength, in this case green. Violet has the shortest wavelength and therefore the yellow polyene is the least conjugated.

Barbara Kipreos
Barbara Kipreos
Numerade Educator
04:03

Problem 10

How do the following factors affect absorption frequencies? Use data in Tables 12-1 and 12-2. (a) For $\mathrm{C}-\mathrm{H}$ stretch, the hybrid orbitals used by $\mathrm{C}$. (b) Bond strength; i.e., change in bond multiplicity. (c) Change in mass of one of the bonded atoms; e.g., $\mathrm{O}-\mathrm{H}$ versus $\mathrm{O}-\mathrm{D}$. (d) Stretching versus bending. (e) $\mathrm{H}$-bonding of $\mathrm{OH}$.
(a) The more $s$ character in the $\mathrm{C}-\mathrm{H}$ bond, the stiffer the bond and the higher the frequency:
(b) Stretching frequencies parallel bond strengths. Because bond strength increases with the number of bonds between two given atoms, absorption frequencies increase with bond multiplicity:
(c) Frequencies are inversely related to the masses of the bonded atoms. Therefore, changing the lighter $\mathrm{H}$ to the heavier D causes a decrease in the stretching frequency.
(d) Most of the stretching frequencies in Table 12-1 are higher than the bending frequencies in Table 12-2.
(e) $\mathrm{H}$-bonding causes a shift to lower frequencies $\left(3600 \mathrm{~cm}^{-1} \rightarrow 3300 \mathrm{~cm}^{-1}\right)$. The band also becomes broader and less intense.

Tianyu Li
Tianyu Li
Numerade Educator
06:39

Problem 11

Identify the peaks marked by Roman numerals in Fig. 12-2, the ir spectrum of ethyl acetate,
<smiles>CCC(=O)CC</smiles>
The deep valleys represent transmission minima and are therefore absorption "peaks" or bands. At about $2800 \mathrm{~cm}^{-1}$, peak I is due to $\mathrm{H}-\mathrm{C}_{s p}$, stretching. The peak at $1700 \mathrm{~cm}^{-1}$, II, is due to stretching of thegroup. The two bands at $1400-1500 \mathrm{~cm}^{-1}$, IIl, are again due to $\mathrm{C}-\mathrm{H}$ bonds. The one at $1250 \mathrm{~cm}^{-1}$, IV, is due to the $\mathrm{C}-\mathrm{O}$ stretch. (It is extremely difficult and impractical to attempt an interpretation of each band in the spectrum.)

Zubair Abdulla
Zubair Abdulla
Numerade Educator
06:47

Problem 12

Which of the following vibrational modes show no ir absorption bands? (a) Symmetrical $\mathrm{CO}_2$ stretch, (b) antisymmetrical $\mathrm{CO}_2$ stretch, (c) symmetrical $\mathrm{O}=\mathrm{C}=\mathrm{S}$ stretch. $(d) \mathrm{C}=\mathrm{C}$ stretch in $o-x y l e n e,(e) \mathrm{C}=\mathrm{C}$ stretch in $p$-xylene and $(f) \mathrm{C}=\mathrm{C}$ stretch in $p$-bromotoluene.

Those vibrations which do not result in a change in dipole moment show no band. These are $(a)$ and $(e)$, which are symmetrical about the axis of the stretched bonds.

Tianyu Li
Tianyu Li
Numerade Educator
05:57

Problem 13

Which of the following atoms do not exhibit nuclear magnetic resonance? ${ }^{12} \mathrm{C},{ }^{16} \mathrm{O},{ }^{14} \mathrm{~N},{ }^{15} \mathrm{~N},{ }^2 \mathrm{H}$, ${ }^{19} \mathrm{~F},{ }^{31} \mathrm{P},{ }^{13} \mathrm{C}$, and ${ }^{32} \mathrm{~S}$.

Atoms with odd numbers of protons and/or neutrons are nmr-active. The inactive atoms are: ${ }^{12} \mathrm{C}(6 p, 6 n)$, ${ }^{16} \mathrm{O}(8 p, 8 n)$, and ${ }^{32} \mathrm{~S}(16 p, 16 n)$. To detect the $n m r$ activity of atoms other than ${ }^1 \mathrm{H}$ requires alteration of the nmr spectrometer. The ordinary spectrometer selects the range of radiowave frequency that excites only ${ }^1 \mathrm{H}$.

Zubair Abdulla
Zubair Abdulla
Numerade Educator
06:07

Problem 14

Use Table $12-3$ to assign approximate $\delta$ values for the chemical shift of the one type of $\mathrm{H}$ in (a) $\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_3\right)_2,($ b $)\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{O},($ c $)$ benzene, $($ d $) \mathrm{O}=\mathrm{CH}-\mathrm{CH}=\mathrm{O}$.
(a) $1.7 \mathrm{ppm}$,
(b) $2.3 \mathrm{ppm}$,
(c) $7.2 \mathrm{ppm}$,
(d) $9.5 \mathrm{ppm}$.

Banhishikha Sinha
Banhishikha Sinha
Numerade Educator

Problem 15

Give the numbers of kinds of $\mathrm{H}$ 's present in (a) $\mathrm{CH}_3 \mathrm{CH}_3$, (b) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3$, (c) $\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}_2 \mathrm{CH}_3,($ d $) \mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2,(e) \mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2,(f) \mathrm{C}_6 \mathrm{H}_5 \mathrm{NO}_2,(g) \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3$.
(a) One (all equivalent).
(b) Two: $\mathrm{CH}_3^a \mathrm{CH}_2^b \mathrm{CH}_3^a$.
(c) Four: $\left(\mathrm{CH}_3^a\right)_2 \mathrm{CH}^b \mathrm{CH}_2^c \mathrm{CH}_3^d$.
(d) One (all equivalent).
(e) Four:
The $=\mathrm{CH}_2$ H's are not equivalent since one is cis to the $\mathrm{CH}_3$ and the other is trans. Replacement of $\mathrm{H}^c$ by $\mathrm{X}$ gives the cis-diastereomer. Replacement of $\mathrm{H}^d$ gives the trans-diastereomer.
(f) Three: Two ortho, two meta, and one para.
(g) Theoretically there are three kinds of aromatic H's, as in $(f)$. Actually the ring H's are little affected by alkyl groups and are equivalent. There are two kinds: $\mathrm{C}_6 \mathrm{H}_5^a \mathrm{CH}_3^b$.

Check back soon!
06:14

Problem 16

How many kinds of equivalent $\mathrm{H}$ 's are there in the following?
(a) $\mathrm{CH}_3 \mathrm{CHClCH}_2 \mathrm{CH}_3$
(b) $p-\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{C}_6 \mathrm{H}_4-\mathrm{CH}_2 \mathrm{CH}_3$
(c) $\mathrm{Br}_2 \mathrm{CHCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}$
(a) Five as shown:
The two $\mathrm{H}^{\prime}$ s of $\mathrm{CH}_2$ are not equivalent, because of the presence of a chiral $\mathrm{C}$. Replacing $\mathrm{H}^c$ and $\mathrm{H}^d$ separately by $\mathrm{X}$ gives two diastereomers; $\mathrm{H}^c$ and $\mathrm{H}^d$ are diasteriomeric $\mathrm{H}$ s.
(b) Three: All four aromatic H's are equivalent, as are the six in the two $\mathrm{CH}_3$ and the four in the $\mathrm{CH}_2$ 's.
(c) Four: $\mathrm{Br}_2 \mathrm{CH}^a \mathrm{CH}_2^b \mathrm{CH}_2^c \mathrm{CH}_2^d \mathrm{Br}$.

Kevin Barayuga
Kevin Barayuga
Numerade Educator
06:24

Problem 17

How many kinds of H's are there in the isomers of dimethylcyclopropane?

Dimethylcyclopropane has three isomers, shown with labeled H's to indicate differences and equivalencies.
<smiles>C[C@H]1[C@H](C)[C@@H](C)[C@H]1C</smiles>
1,1 -
Dimethylcyclopropane
(I)
<smiles>CC1C(C)(C)C1(C)C</smiles>
cis-1,2-
Dimethylcyclopropane
(II)
<smiles>CC1CC[C@H]1C</smiles>
trans-1,2-
Dimethylcyclopropane
(III)

In II, $\mathrm{H}^c$ and $\mathrm{H}^d$ are different since $\mathrm{H}^c$ is cis to the $\mathrm{CH}_3$ 's and $\mathrm{H}^d$ is trans. In III the $\mathrm{CH}_2 \mathrm{H}$ 's are equivalent; they are each cis to a $\mathrm{CH}_3$ and trans to a $\mathrm{CH}_3$.

Sandra Lundell
Sandra Lundell
Numerade Educator
08:10

Problem 19

What compound $\mathrm{C}_7 \mathrm{H}_8 \mathrm{O}$ has $\mathrm{nmr}$ signals at $\delta=7.3,4.4$, and $3.7 \mathrm{ppm}$, with relative areas $7: 2.9: 1.4$, respectively?

The relative areas of the three different kinds of $\mathrm{H}$ become $5: 2: 1$ on dividing by 1.4 . That is, five H's contribute to the $\delta=7.3,2 \mathrm{H}$ 's to $\delta=4.4$ and $1 \mathrm{H}$ to $\delta=3.7$, for a total of eight H's, which is consistent with the formula. The five $\mathrm{H}$ 's at $\delta=7.2$ are aromatic, indicating a $\mathrm{C}_6 \mathrm{H}_5$ compound. The remaining portion of the formula comprises the $\mathrm{CH}_2 \mathrm{OH}$ group. The $\mathrm{H}$ at $\delta=3.7$ is part of $\mathrm{OH}$. The two H's at $\delta=4.4$ are benzylic and alpha to $\mathrm{OH}$. The compound is $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OH}$, benzyl alcohol.

Dr. Satish Ingale
Dr. Satish Ingale
Numerade Educator
View

Problem 20

The pmr (proton magnetic resonance) spectrum of $\mathrm{CH}_3 \mathrm{OCH}_2 \mathrm{CH}_2 \mathrm{OCH}_3$ shows chemical shifts of 3.4 and $3.2 \mathrm{ppm}$, with corresponding peak areas in the ratio $2: 3$. Are these numbers consistent with the given structure?
Yes. Since the $\mathrm{CH}_3$ 's and $\mathrm{CH}_2$ 's are each equivalent, only two signals appear. Both are shifted downfield by the $\mathrm{O}$, the $\mathrm{CH}_2$-signal more than the $\mathrm{CH}_3$-signal. Integration provides the correct ratio, $2: 3=4: 6$, the actual numbers of $\mathrm{H}$ 's engendering the signals.

Susan Hallstrom
Susan Hallstrom
Numerade Educator
02:08

Problem 21

In which of the following molecules does spin-spin coupling occur? If splitting is observed give the multiplicity of each kind of $\mathrm{H}$.
Splitting is not observed for $(a)$ or $(d)$, which each have only equivalent H's, or for $(c)$, which has no nonequivalent H's on adjacent $\mathrm{C}$ 's. The $\mathrm{H}$ 's of $\mathrm{CH}_2$ in $(b)$ are nonequivalent and each signal is split into a triplet $(n=2 ; 2+1=3)$. In $(e)$ the two H's are not equivalent and each generates a doublet. The vinyl H's in $(f)$ are nonequivalent since one is cis to $\mathrm{Cl}$ and the other is cis to $\mathrm{I}$; each gives rise to a doublet. In this case the interacting $\mathrm{H}$ 's are on the same $\mathrm{C}$. Compound $(\mathrm{g})$ gives a singlet for the equivalent uncoupled aromatic $\mathrm{H}$ 's, quartet for the $\mathrm{H}$ 's of the two equivalent $\mathrm{CH}_2$ groups coupled with $\mathrm{CH}_3$, and a triplet for the two equivalent $\mathrm{CH}_3$ groups coupled with $\mathrm{CH}_2$.

Dan Ni
Dan Ni
Numerade Educator
04:03

Problem 22

Why is splitting observed in 2-methylpropene but not in 1-chloro-2,2-dimethylpropane?
See Fig. 12-5. In $(d), \mathrm{H}^a$ is more downfield than $\mathrm{H}^b$ because $\mathrm{Cl}$ is more electron-withdrawing than $\mathrm{Br}$. In $\left(\mathrm{CH}_3^a\right)_3 \mathrm{C}-\mathrm{CH}_2^b \mathrm{Cl}, \mathrm{H}^a$ and $\mathrm{H}^b$ are not on adjacent $\mathrm{C}$ 's and are too far away from one another to couple. In
<smiles>C=C(C)C</smiles>
although $\mathrm{H}^a$ and $\mathrm{H}^b$ are not on adjacent C's, they are close enough to couple because of the shorter $\mathrm{C}=\mathrm{C}$ bond.

Himanshu Kushwaha
Himanshu Kushwaha
Numerade Educator

Problem 23

Sketch the nmr spectra of (a) 1,1-dichloroethane, (b) 1,1,2-trichloroethane, (c) 1,1,2,2-tetrachloroethane and $(d)$ 1-bromo-2-chloroethane. In each case indicate the "staircase" curve of relative areas.

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07:28

Problem 24

F's couple H's in the same way as do other H's. Predict the splitting in the nmr spectrum of 2,2difluoropropane.

In $\mathrm{CH}_3^a \mathrm{CF}_2 \mathrm{CH}_3^a$ the two F's split the $\mathrm{H}^a$-signal into a $1: 2: 1$ triplet. The F-signal, when detected by a special probe, would be a septet.

Nima Gharibi
Nima Gharibi
Numerade Educator
01:52

Problem 25

Deuterium does not give a signal in the proton nmr spectrum nor does it split signals of nearby protons. Thus D's might just as well not be there. What is the difference between nmr spectra of $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Cl}$ and $\mathrm{CH}_3 \mathrm{CHDCl}$ ?

Alejandro Hernandez
Alejandro Hernandez
Numerade Educator
01:52

Problem 26

The stable anti conformer of $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Cl}$ shows a nonequivalency of the $\mathrm{CH}_3 \mathrm{H}$ 's:
<smiles>ClC1CC2CC1C2Cl</smiles>
$\mathrm{H}^*$ is anti to the $\mathrm{Cl}$ while the $\mathrm{H}_{\Lambda}$ 's are gauche. Why does $\mathrm{H}^*$ not give a signal different from the $\mathrm{H}_{\Delta}$ 's? (Instead, the three H's produce an equivalent triplet,)

Rotation around the $\mathrm{C}-\mathrm{C}$ bond is rapid. Detection by the nmr spectrometer is slower. The spectrometer therefore detects the average condition, which is the same for each $\mathrm{H} ; 1 / 3$ anti and $2 / 3$ gauche.

Alejandro Hernandez
Alejandro Hernandez
Numerade Educator
08:10

Problem 27

What information can you deduce from the fact that one signal in the nmr spectrum of $2,2,6,6$ tetradeuterobromocyclohexane changes to two smaller signals when the spectrum is taken at low temperatures?

As the ring changes its conformation from one chair form to another, the $\mathrm{Br}-\mathrm{C}-\mathrm{H}$ proton changes its position from axial to equatorial (Fig. 12-6). An axial $\mathrm{H}$ and an equatorial $\mathrm{H}$ have different chemical shifts. But at room temperature the ring "flips" too fast for the instrument to detect the difference; it senses the average condition. At low temperatures this process becomes slow enough so that the instrument can pick up the two different $\mathrm{H}_{\mathrm{ax}}$ and $\mathrm{H}_{\mathrm{eq}}$ signals. D's are used to ensure that the $\mathrm{H}$ under study is a singlet.

Dr. Satish Ingale
Dr. Satish Ingale
Numerade Educator
02:20

Problem 28

A compound, $\mathrm{C}_2 \mathrm{H}_2 \mathrm{BrCl}$, has two doublets, $J=16 \mathrm{~Hz}$. Use Table $12-4$ to suggest a structure.
The three possibilities showing $J$ values for two doublets are:
<smiles>ClC=C(Cl)Br</smiles>
<smiles>ClC=CBr</smiles>
<smiles>ClC=CBr</smiles>
gem-vinyl $\mathrm{H}$ 's
cis $\mathrm{H}$ 's
trans $\mathrm{H}$ 's
$I=0-3 \mathrm{~Hz}$
$J=7-12 \mathrm{~Hz}$
$J=13-18 \mathrm{~Hz}$

The trans isomer fits the data.

Zubair Abdulla
Zubair Abdulla
Numerade Educator
01:29

Problem 29

Why is spin-spin coupling between adjacent ${ }^{13} \mathrm{C}$ 's not observed?
Since the natural abundance of this isotope is so low, the chance of finding two ${ }^{13} \mathrm{C}$ 's next to each other is practically nil. However, if a compound were synthesized with only ${ }^{13} \mathrm{C}$ 's, then coupling would be observed.

Sharfa Farzandh
Sharfa Farzandh
Numerade Educator
04:06

Problem 30

How many peaks would be evidenced in the decoupled spectrum of $(a)$ methylcyclohexane? $(b)$ cyclohexene? (c) 1-methylcyclohexene?

Zubair Abdulla
Zubair Abdulla
Numerade Educator
03:34

Problem 31

(a) What molecular formulas containing only $\mathrm{C}$ and $\mathrm{H}$ can be assigned to a cation with $m / e$ equal to (i) 43, (ii) 65 , (iii) 91 ? (Assume that $e=+1$.) (b) What combination of $\mathrm{C}, \mathrm{H}$ and $\mathrm{N}$ can account for an $m / e$ of (i) 43 , (ii) 57 ? (Assume that $e=+1$.)
(a) Divide by 12 to get the number of $\mathrm{C}$ 's; the remainder of the weight is due to $\mathrm{H}$ 's. (i) $\mathrm{C}_7 \mathrm{H}_7^{+}$, (ii) $\mathrm{C}_5 \mathrm{H}_5^{+}$, (iii) $\mathrm{C}_7 \mathrm{H}_7^{+}$.
(b) (i) If one $\mathrm{N}$ is present, subtracting 14 leaves a mass of 29 , which means $2 \mathrm{C}$ 's (mass of 24 ) are present. Therefore the formula is $\mathrm{C}_2 \mathrm{H}_5 \mathrm{~N}^{+}$. If $2 \mathrm{~N}$ 's are present, it is $\mathrm{CH}_3 \mathrm{~N}_2^{+}$. (ii) $\mathrm{CH}_3 \mathrm{~N}_3^{+}, \mathrm{C}_2 \mathrm{H}_5 \mathrm{~N}_2^{+}$or $\mathrm{C}_3 \mathrm{H}_7 \mathrm{~N}^{+}$.

Ian Kaigh
Ian Kaigh
Numerade Educator
09:13

Problem 32

(a) Do parent (molecular) ions, $\mathrm{RS}^{+}$, of hydrocarbons ever had odd $m / e$ values? $(b)$ If an $\mathrm{RS}^{+}$ contains only $\mathrm{C}, \mathrm{H}$ and $\mathrm{O}$, may its $m / e$ value be either odd or even? (c) If an $\mathrm{RS}^{+}$contains only $\mathrm{C}, \mathrm{H}$ and $\mathrm{N}$, may its $m / e$ value be either odd or even? ( $d$ ) Why cannot an ion, $m / e=3 \mathrm{I}$, be $\mathrm{C}_2 \mathrm{H}_7^{+}$? What might it be?
(a) No. Hydrocarbons, and their parent ions, must have an even number of H's: $\mathrm{C}_n \mathrm{H}_{2 n+2}, \mathrm{C}_n \mathrm{H}_{2 n}, \mathrm{C}_n \mathrm{H}_{2 n-2}$, $\mathrm{C}_n \mathrm{H}_{2 n-6}$, etc. Since the atomic weight of $\mathrm{C}$ is even (12), the $m / e$ values must be even.
(b) The presence of $\mathrm{O}$ in a formula does not change the ratio of $\mathrm{C}$ to $\mathrm{H}$. Since the mass of $\mathrm{O}$ is even (16), the mass of $\mathrm{RS}^{+}$with $\mathrm{C}, \mathrm{H}$ and $\mathrm{O}$ must be even.
(c) The presence of each $\mathrm{N}(m=14)$ requires an additional $\mathrm{H}\left(\mathrm{C}_n \mathrm{H}_{2 n+3} \mathrm{~N}, \mathrm{C}_n \mathrm{H}_{2 n+1} \mathrm{~N}, \mathrm{C}_n \mathrm{H}_{2 n-1} \mathrm{~N}\right)$. Therefore, if the number of N's is odd, an odd number of $\mathrm{H}$ 's and an odd $m / e$ value result. An even number of N's requires an even number of $\mathrm{H}$ 's and an even $m / e$ value. These statements apply only to parent ions, not to fragment ions.
(d) The largest number of H's for two C's is six $\left(\mathrm{C}_2 \mathrm{H}_6\right)$. Some possibilities are $\mathrm{CH}_3 \mathrm{O}^{+}$and $\mathrm{CH}_5 \mathrm{~N}^{+}$.

David Collins
David Collins
Numerade Educator

Problem 33

Write equations involving the electron-dot formulas for each fragmentation used to explain the following. (a) Isobutane, a typical branched-chain alkane, has a lower-intensity $\mathrm{RS}^{+}$peak than does $n$-butane, a typical unbranched alkane. (b) All $1^{\circ}$ alcohols, $\mathrm{RCH}_2 \mathrm{CH}_2 \mathrm{OH}$, have a prominent fragment cation at $m / e=31$. (c) All $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{R}$-type hydrocarbons have a prominent fragment cation at $m / e=91$. (d) Alkenes of the type $\mathrm{H}_2 \mathrm{C}=\mathrm{CHCH}_2 \mathrm{R}$ have a prominent fragment cation at $m / e=41$. (e) Aldehydes,
<smiles>[R]C=O</smiles>
show intense peaks at $(m / e)_{\mathrm{RS}}-1$ and $m / e=29$.
(a) Cleavage of $\mathrm{C}-\mathrm{C}$ is more likely than cleavage of (the stronger) $\mathrm{C}-\mathrm{H}$. Fragmentation of $\mathrm{RS}^{+}$for isobutane,
<smiles>C[C+](C)CCC(C)C(C)C</smiles>
gives a $2^{\circ} \mathrm{R}^{+}$, which is more stable than the $1^{\circ} \mathrm{R}^{+}$from $n$-butane,

Hence $\mathrm{RS}^{+}$of isobutane undergoes fragmentation more readily than does $\mathrm{RS}^{+}$of $n$-butane, and fewer $\mathrm{RS}^{+}$ fragments of isobutane survive. Consequently, isobutane, typical of branched-chain alkanes, has a low-intensity $\mathrm{RS}^{+}$peak compared with $n$-butane.

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04:09

Problem 34

Why is less than $1 \mathrm{mg}$ of parent compound used for mass spectral analysis?
A relatively small number of molecules are taken to prevent collision and reaction between fragments. Combination of fragments might lead to ions with larger masses than $\mathrm{RS}^{+}$, making it impossible to determine the molecular weight. The fragmentation pattern would also become confusing.

Dr. Satish Ingale
Dr. Satish Ingale
Numerade Educator
05:45

Problem 35

Give the structure of a compound, $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}$, whose mass spectrum shows $m / e$ values of 15,43 , $57,91,105$, and 148 .

The value 15 suggests a ${ }^{+} \mathrm{CH}_3$. Because $43-15=28$, the mass of a $\mathrm{C}=\mathrm{O}$ group, the valule of 43 could mean an acetyl, $\mathrm{CH}_3 \mathrm{CO}$, group in the compound. The highest value, 148 , gives the molecular weight. Cleaving an acetyl

Nima Gharibi
Nima Gharibi
Numerade Educator
01:18

Problem 36

How could mass spectroscopy distinguish among the three deuterated forms of ethyl methyl ketone?
(1) $\mathrm{DCH}_2 \mathrm{CH}_2 \mathrm{COCH}_3$
(2) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COCH}_2 \mathrm{D}$
(3) $\mathrm{CH}_3 \mathrm{CHDCOCH}_3$
The expected pcaks for each compound are shown in Table 12-6; each has a different combination of peaks.

Lottie Adams
Lottie Adams
Numerade Educator
00:34

Problem 37

Match the type of spectrometer with the kind of information which it can provide the chemist.
1. Mass A. functional groups
2. Infrared B. molecular weights
3. Ultraviolet C. proton environment
4. Nuclear magnetic resonance D. conjugation

Amy Jiang
Amy Jiang
Numerade Educator
02:00

Problem 38

Why do colored organic compounds, such as $\beta$-carotene (the orange pigment isolated from carrots), have extended conjugation?

The more effective electron delocalization in molecules with extended conjugation narrows the energy gap between the HOMO and the LUMO. Thus, visible radiation of lower frequency (longer wavelength) is absorbed in the HOMO $\rightarrow$ LUMO electron transition.

Joseph Fritchman
Joseph Fritchman
Numerade Educator
04:25

Problem 39

(a) Account for the fact that benzene absorbs at $254 \mathrm{~nm}$ in the uv, and phenol, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}$, absorbs at $280 \mathrm{~nm}$. (b) Where would one expect 1,3,5-hexatriene to absorb, relative to benzene?
(a) The $p$ orbital on $\mathrm{O}$, housing a pair of $n$ electrons, overlaps with the cyclic $\pi$ system of benzene, thereby extending the electron delocalization and decreasing the energy for the HOMO $\rightarrow$ LUMO transition. Consequently, phenol absorbs at longer wavelengths.
(b) The triene absorbs at a longer $\lambda_{\max }(275 \mathrm{~nm})$. Since the cyclic $\pi$ system of benzene has a lower energy than the linear $\pi$ system of the triene, benzene absorbs radiation of shorter wavelength.

Eileen Sullivan
Eileen Sullivan
Numerade Educator
00:48

Problem 40

Which peaks in the ir spectra distinguish cyclohexane from cyclohexene?
One of the $\mathrm{C}-\mathrm{H}$ stretches in cyclohexene is above $3000 \mathrm{~cm}^{-1}\left(\mathrm{C}_{s p^2}-\mathrm{H}\right)$; in cyclohexane the $\mathrm{C}-\mathrm{H}$ stretches are below $3000 \mathrm{~cm}^{-1}\left(\mathrm{C}_{s p},-\mathrm{H}\right)$. Cyclohexene has a $\mathrm{C}=\mathrm{C}$ stretch at about $1650 \mathrm{~cm}^{-1}$.

Lottie Adams
Lottie Adams
Numerade Educator
08:44

Problem 41

The mass spectrum of a compound containing $\mathrm{C}, \mathrm{H}, \mathrm{O}$, and $\mathrm{N}$ gives a maximum $m / e$ of 121 . Its ir spectrum shows peaks at $700,750,1520,1685$, and $3100 \mathrm{~cm}^{-1}$, and a twin peak at $3440 \mathrm{~cm}^{-1}$. What is a reasonable structure for the compound?

The molecular weight is 121 . Since the mass is odd, there must be an odd number of N's [Problem 12.32(c)]. The ir data indicate the following groups to be present:
$$
\begin{array}{ll}
1520 \mathrm{~cm}^{-1}: & \text { aromatic ring (1450-1600 } \mathrm{cm}^{-1} \text { range) } \\
1685 \mathrm{~cm}^{-1}: & \mathrm{C}=\mathrm{O} \text { stretch of amide structure }-\mathrm{CO}-\mathrm{N}\left(1630-1690 \mathrm{~cm}^{-1} \text { range }\right) \\
3100 \mathrm{~cm}^{-1}: & \text { aromatic } \mathrm{C}-\mathrm{H} \text { bond }\left(3000-3100 \mathrm{~cm}^{-1} \text { range }\right) \\
3440 \mathrm{~cm}^{-1}: & -\mathrm{N}-\mathrm{H} \text { in amine or amide }\left(3300-3500 \mathrm{~cm}^{-1} \text { range }\right) \\
700,750 \mathrm{~cm}^{-1}: & \text { monosubstituted phenyl }
\end{array}
$$

A twin peak due to symmetric and antisymmetric $\mathrm{N}-\mathrm{H}$ stretches means an $\mathrm{NH}_2$ group. By putting the pieces together we find that the compound is benzamide,
<smiles>NC(=O)c1ccccc1</smiles>

Zubair Abdulla
Zubair Abdulla
Numerade Educator

Problem 42

The ir spectrum of methyl salicylate, $o-\mathrm{HOC}_6 \mathrm{H}_4 \mathrm{COOCH}_3$, has peaks at 3300, 1700, 3050, 1540, 1590 , and $2990 \mathrm{~cm}^{-1}$. Correlate these peaks with the following structures: $(a) \mathrm{CH}_3,(b) \mathrm{C}=\mathrm{O},(c) \mathrm{OH}$ group on the ring, and $(d)$ aromatic ring.
(a) $2990 \mathrm{~cm}^{-1}$;
(b) $1700 \mathrm{~cm}^{-1}$;
(c) $3300 \mathrm{~cm}^{-1}$;
(d) $3050,1540,1590 \mathrm{~cm}^{-1}$.

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03:17

Problem 43

Calculate $\varepsilon_{\max }$ for a compound whose maximum absorbance is $A_{\max }=1.2$. The cell length $(l)$ is $1.0 \mathrm{~cm}$ and the concentration is $0.076 \mathrm{~g} / \mathrm{L}$. The mass spectrum of the compound has the largest $m / e$ value at 100.
The molecular weight is $100 \mathrm{~g} / \mathrm{mol}$; therefore, $c=7.6 \times 10^{-4} \mathrm{~mol} / \mathrm{L}$
and
$$
\varepsilon_{\max }=\frac{A_{\max }}{C l}=\frac{1.2}{\left(7.6 \times 10^{-4}\right)(1.0)}=1600
$$

Tianyu Li
Tianyu Li
Numerade Educator

Problem 45

A compound, $\mathrm{C}_3 \mathrm{H}_6 \mathrm{O}$, contains a $\mathrm{C}=\mathrm{O}$ group. How could nmr establish whether this compound is an aldehyde or a ketone?

If an aldehyde, the compound is $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}$, with three multiplet peaks and a downfield signal for
<smiles>CC=O</smiles>
$(\delta=9-10 \mathrm{ppm})$. If a ketone, it is $\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{O}$, with one singlet.

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02:24

Problem 46

The nmr spectrum of a dichloropropane shows a quintuplet and, downfield, a triplet of about twice the intensity. Is the isomer 1,1-, 1,2-, 1,3- or 2,2-dichloropropane?
We would expect the following signals:
1,1-Dichloropropane, $\mathrm{Cl}_2 \mathrm{CH}^a \mathrm{CH}_2^b \mathrm{CH}_3^c$ : a triplet $\left(\mathrm{H}^c\right)$, a complex multiplet more downfield $\left(\mathrm{H}^b\right)$, and a triplet still more downfield $\left(\mathrm{H}^a\right)$.
1.2-Dichloropropane, $\mathrm{ClCH}_2^a \mathrm{CH}^b \mathrm{CH}_3^c$ : a doublet $\left(\mathrm{H}^c\right)$, another doublet more downfield $\left(\mathrm{H}^a\right)$, a complex multiplet most downfield $\left(\mathrm{H}^b\right)$.
1.3-Dichloropropane, $\mathrm{ClCH}_2^a \mathrm{CH}_2^b \mathrm{CH}_2^a \mathrm{Cl}$ : a quintuplet $\left(\mathrm{H}^b\right)$, and, downfield, a triplet $\left(\mathrm{H}^a\right)$.
2,2-Dichloropropane,
<smiles>CC(C)(Cl)Cl</smiles>
a singlet $\left(\mathrm{H}^a\right)$.

The compound is 1,3-dichloropropane.

Lottie Adams
Lottie Adams
Numerade Educator
09:19

Problem 47

Consider the coupled ${ }^{13} \mathrm{C}$ nmr spectra of $(a)$ 1,3,5-trimethylbenzene, (b) 1,2,3-trimethylbenzene, (c) n-propylbenzene. (i) Write each structure and label the different kinds of C's by numerals $1,2, \ldots$. (ii) Show the splitting for each peak by the letters $s$ for singlet, $\mathrm{d}$ for doublet, $\mathrm{t}$ for triplet, and $\mathrm{q}$ for quartet.

Zubair Abdulla
Zubair Abdulla
Numerade Educator
08:53

Problem 48

Indicate whether the following statements are true or false and give a reason in each case. (a) The ir spectra are identical for the enantiomers
<smiles>CCC(C)C(C)Br</smiles>
(b) The nmr spectra of the compounds in $(a)$ are also identical. (c) The ir spectrum of I-hexene has more peaks than the uv spectrum. (d) Compared to $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}$, the $\mathrm{n} \rightarrow \pi^*$ for $\mathrm{H}_2 \mathrm{C}=\mathrm{CHCHO}$ has shifted to a shorter wavelength (blue shift).
$(a)$ and $(b)$ True. The compounds are enantiomers which have identical vibrational modes and proton resonances.
(c) True. The ir spectrum has peaks for stretching and bending of all bonds, while the uv spectrum has only one peak for excitation of a $\pi$ electron.
(d) False. The shift is to longer wavelength (red shift), since
<smiles>C=CC=O</smiles>
is a conjugated system.

Kim Trang Nguyen
Kim Trang Nguyen
Numerade Educator
02:24

Problem 49

Assign the nmr spectra shown in Fig. 12-10 to the appropriate monochlorination products of 2,4dimethylpentane $\left(\mathrm{C}_7 \mathrm{H}_{15} \mathrm{Cl}\right)$ and justify your assignment. Note the integration assignments drawn in the spectra.
The three possible structures are:
<smiles>CC(C)CC(C)CCl</smiles>
<smiles>CC(C)CC(C)(C)Cl</smiles>
<smiles>CC(C)C(Cl)C(C)C</smiles>
1-Chloro-2,4-
2-Chloro-2, 4-
3-Chloro-2,4-
dimethylpentane (I)
dimethylpentane (II)
dimethylpentane (III)

The best clue is the most downfield signal arising from the $\mathrm{H}$ 's closest to $\mathrm{Cl}$. In spectrum $(a)$ the signal with the highest $\delta$ value is a doublet, integrating for two $\mathrm{H}$ 's, that corresponds only to structure I $\left(\mathrm{ClCH}_2-\right)$. This is confirmed by the nine H's of the $3 \mathrm{CH}_3$ 's that are most upfield and the four $2^{\circ}$ and $3^{\circ} \mathrm{H}$ 's with signals between these.
In spectrum $(b)$ the most downfield signal is a triplet, for one $\mathrm{H}$, which arises from the
<smiles>CCCC(Cl)CC</smiles>
grouping in III. In addition, the most upfield signal is a doublet, integrating for $12 \mathrm{H}$ 's, which is produced by the H's of the $4 \mathrm{CH}_3$ 's split by the $3^{\circ} \mathrm{H}$.

This leaves Il for spectrum $(c)$. The most downfield group of irregular signals, integrating for three H's, comes from the two $2^{\circ}$ and one $3^{\circ} \mathrm{H}$ on $\mathrm{C}^3$ and $\mathrm{C}^4$, respectively. The most upfield doublet, integrating for six $\mathrm{H}$ 's, arises from the two equivalent $\mathrm{CH}_3$ 's on $\mathrm{C}^4$ split by the $\mathrm{C}^4 3^{\circ} \mathrm{H}$. The two $\mathrm{CH}_3$ 's on $\mathrm{C}^2$ give rise (six H's) to the singlet of median $\delta$ value.

Lottie Adams
Lottie Adams
Numerade Educator
06:24

Problem 50

Deduce structures for the compound whose spectral data are presented in Fig. 12-11, Table 12-7, and Fig. 12-12. Assume an $\mathrm{O}$ is present in the molecule. There was no uv absorption above $180 \mathrm{~nm}$.

$$
\begin{array}{|l|r|r|r|r|r|r|r|r|r|r|r|r|r|}
\hline m / e & 26 & 27 & 29 & 31 & 39 & 41 & 42 & 43 & 44 & 45 & 59 & 87 & 102 \\
\hline \begin{array}{l}
\text { Relative intensity, } \\
\% \text { of base peak }
\end{array} & 3 & 18 & 6 & 4 & 11 & 17 & 6 & 61 & 4 & 100 & 11 & 21 & 0.63 \\
\hline
\end{array}
$$

$$
\begin{array}{|l|r|r|r|r|r|r|r|r|r|r|r|r|r|}
\hline m / e & 26 & 27 & 29 & 31 & 39 & 41 & 42 & 43 & 44 & 45 & 59 & 87 & 102 \\
\hline \begin{array}{l}
\text { Relative intensity, } \\
\% \text { of base peak }
\end{array} & 3 & 18 & 6 & 4 & 11 & 17 & 6 & 61 & 4 & 100 & 11 & 21 & 0.63 \\
\hline
\end{array}
$$

Zubair Abdulla
Zubair Abdulla
Numerade Educator