Schaum's Outline of Organic Chemistry

George Hademenos, George Hademenos

Chapter 9 CYCLIC HYDROCARBONS - all with Video Answers

Educators

Chapter Questions

Problem 1

Draw structural formulas for (a) bromocycloheptane, $(b)$ 1-ethylcyclopentene, (c) bicyclo[3.1.0]hexane, $(d)$ cyclobutylcyclohexane.

Sima Sarker

Numerade Educator

04:01

In $(f) \mathrm{C}^1, \mathrm{C}^2, \mathrm{C}^3, \mathrm{C}^4, \mathrm{C}^5, \mathrm{C}^6$ constitute one ring; $\mathrm{C}^1, \mathrm{C}^7, \mathrm{C}^4, \mathrm{C}^5, \mathrm{C}^6$, another; $\mathrm{C}^1, \mathrm{C}^2, \mathrm{C}^4, \mathrm{C}^4, \mathrm{C}^7$, a third.
(a) This fused-ring bicyclic compound has a four-carbon bridge made up of $\mathrm{C}^2, \mathrm{C}^3, \mathrm{C}^4$, and $\mathrm{C}^5$, and a two-carbon bridge of $\mathrm{C}^7$ and $\mathrm{C}^8$; the bridge between the bridgehead $\mathrm{C}$ 's has 0 carbon atoms. There are eight $\mathrm{C}$ 's in the compound. The name is bicyclo[4.2.0]octane. (b) Consider the cyclopropane ring to be a substituent on the longest carbon chain. The name is 1-cyclopropyl-3-methyl-1-pentene. (c) The substituents, written alphabetically, are numbered so that the $C$ 's have the lowest possible numbers. The name is 3-bromo-1,1-dimethylcyclohexane. (d) 1,1,3-Trimethylcyclopentane. (e) 3-Nitrocyclohexene (not 6-nitrocyclohexene). The doubly bonded C's are numbered $I$ and 2 , to give the smaller number to the substituent. ( $f$ ) The numbering of C's of bicyclic compounds starts with the bridgehead $C$ closest to a substituent. Substituents on the largest bridge get the smallest numbers. The name is $2,2,7,7-$ tetramethylbicyclo[2.2.1]heptane.

Vasu Makani

Numerade Educator

02:17

Problem 3

Write structural formulas for the organic compounds designated by a ?. Indicate the stereochemistry where necessary and account for the products.

Raghvendra Singh

Numerade Educator

01:46

Problem 3

Give the names, structural formulas and stereochemical designations of the isomers of (a) bromo chloro cyclobutane, (b) dichlorocyclobutane, (c) bromo chloro cyclopentane, (d) iodocyclopentane, (e) dimethylcyclohexane. Indicate chiral C's.
(a) There is only one structure for 1-bromo-1-chlorocyclobutane:
<smiles>ClC1(Br)CCC1</smiles>
With 1-bromo-2-chlorocyclobutane there are cis and trans isomers and both substituted C's are chiral. Both geometric isomers form racemic mixtures.
In 1-bromo-3-chlorocyclobutane there are cis and trans isomers, but no enantiomers; $\mathrm{C}^{\prime}$ and $\mathrm{C}^3$ are not chiral, because a plane perpendicular to the ring bisects them and their four substituents. The sequence of atoms is identical going around the ring clockwise or counterclockwise from $\mathrm{C}^l$ to $\mathrm{C}^3$.
In these structural formulas, the other atoms on $\mathrm{C}^{\prime}$ and $\mathrm{C}^3$ are directly in back of those shown and are bisected by the indicated plane.
(b) Same as (a) except that the cis-1,2-dichlorocyclobutane has a plane of symmetry (dashed line below) and is meso.
(c) There are nine isomers because both 1,2- and 1,3-isomers have cis and trans geometric isomers, and these have enantiomers.
(d) The diiodo cyclopentanes are similar to the bromochloro derivative, except that both the cis-1,2- and the cis-1,3diiodo derivatives are meso. They both have planes of symmetry.
(e) There are nine isomeric methylcyclohexanols.

Ahmed Ali

Numerade Educator

Problem 4

(a) Calculate $\Delta H$ of combustion per $\mathrm{CH}_2$ unit for the first four cycloalkanes, given the following $\Delta H$ 's of combustion, in $\mathrm{kJ} / \mathrm{mol}$ : cyclopropane, -2091 ; cyclobutane, -2744 ; cyclopentane, -3320 ; cyclohexane, -3952. (b) Write (i) the thermochemical equation for the combustion of cyclopropane and (ii) the theoretical equation for the combustion of $\mathrm{CH}_2$ unit of any given ring. (c) How do ring stability and ring size correlate for the first four cycloalkanes?
(a) Divide the given $\Delta H$ values by the number of $\mathrm{CH}_2$ units in the ring (3, 4, 5, and 6 , respectively), to obtain: cyclopropane, -697 ; cyclobutane, -686 ; cyclopentane, -664 ; cyclohexane, -659 . Observe that these per-unit $\Delta H$ 's are in the reverse order of the total $\Delta H$ 's.
(b) (i) $\mathrm{C}_3 \mathrm{H}_6+\frac{9}{2} \mathrm{O}_2 \longrightarrow 3 \mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O} \quad \Delta \mathrm{H}=-2091 \mathrm{~kJ} / \mathrm{mol}$
(ii) $-\mathrm{CH}_2-+\frac{3}{2} \mathrm{O}_2 \longrightarrow \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} \quad \Delta \mathrm{H}<0$
(c) In (ii) of (b), $\Delta H \equiv H\left(\mathrm{CO}_2\right)+H\left(\mathrm{H}_2 \mathrm{O}\right)-H\left(\frac{3}{2} \mathrm{O}_2\right)-H$ (unit). Thus, for the four different ring-memberships under consideration,
$$
H(\text { unit })=\text { constant }-\Delta H
$$

Now, from (a), $\Delta H$ increases (becomes less negative) with increasing size of the ring. Thus, $H$ (unit) decreases with increasing size, which implies that $H$ (ring) also decreases with increasing size. But a decreasing $H$ (ring) means an increasing ring stability. In short, stability increases with ring size.

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01:43

Problem 5

Account for the ring strain in cyclopropane in terms of geometry and orbital overlap.
The C's of cyclopropane form an equilateral triangle with $\mathrm{C}-\mathrm{C}-\mathrm{C}$ bond angles of $60^{\circ}-\mathrm{a}$ significant deviation from the tetrahedral bond angle of $109.5^{\circ}$. This deviation from the "normal" bond angle constitutes angle strain, a major component of the ring strain of cyclopropane. In terms of orbital overlap, the strongest chemical bonds are formed by the greatest overlap of atomic orbitals. For sigma bonding, maximum overlap is achieved when the orbitals meet head-to-head along the bond axis, as in Fig. 9-1(a). This type of overlap in cyclopropane could not lead to ring closure for $s p^3$-hybridized C's because it would demand bond angles of $109.5^{\circ}$. Hence the overlap must be off the bond axis to give a bent bond, as shown in Fig. 9-1 $(b)$.

In order to minimize the angle strain the C's assume more $p$ character in the orbitals forming the ring and more $s$ character in the external bonds, in this case the $\mathrm{C}-\mathrm{H}$ bonds. Additional $p$ character narrows the expected angle, while more $s$ character expands the angle. The observed $\mathrm{H}-\mathrm{C}-\mathrm{H}$ bond angle of $114^{\circ}$ confirms this suggestion. Clearly, there are deviations from pure $p, s p, s p^3$, and $s p^2$ hybridizations.

Hitendra Singh

Numerade Educator

02:24

Problem 6

What factor besides angle strain contributes to the ring strain of cyclopropane?
Because the cyclopropane molecule is a planar ring, the three pairs of H's eclipse each other as in $n$-butane (see Fig. 4-4) to introduce an eclipsing (torsional) strain.

Ian Kaigh

Numerade Educator

00:58

Problem 7

(a) Why is the $\Delta H$ of cis-1,2-dimethylcyclopropane greater than that of its trans isomer? (b) Which isomer is more stable?
(a) There is more eclipsing in the cis isomer because the methyl groups are closer. $(b)$ The trans is more stable.

Grigoriy Sereda

Numerade Educator

02:24

Problem 8

Explain why the ring strain of cyclobutane is only slightly less than that of cyclopropane.
If the C's of the cyclobutane ring were coplanar, they would form a rigid square with internal bond angles of $90^{\circ}$. The deviation from $109.5^{\circ}$ would not be as great as that for cyclopropane, and there would be less angle strain in cyclopropane. However, this is somewhat offset by the fact that the eclipsing strain involves four pairs of H's, one pair more than in cyclopropane.

Actually, eclipsing strain is reduced because cyclobutane is not a rigid flat molecule. Rather, there is an equilibrium mixture of two flexible puckered conformations that rapidly flip back and forth (Fig. 9-2), thereby relieving eclipsing strain. Puckering more than offsets the slight increase in angle strain (angle is now $88^{\circ}$ ). The boxed H's in Fig. 9-2 alternate between up-and-down and outward-projecting positions relative to the ring.

Ian Kaigh

Numerade Educator

01:30

Problem 9

Depict the flexible puckered conformation of cyclobutane in a Newman projection.
See Fig. 9-3. The circle represents the $\mathrm{C}$ 's of any given ring $\mathrm{C}-\mathrm{C}$ bond. The other $\mathrm{C}$ 's of the ring bridge these two C's, one to the front $\mathrm{C}$ (heavy line) and the other to the back C (ordinary line). This Newman projection formula reveals that the H's on adjacent C's are skewed in the puckered conformation.

Aadit Sharma

Numerade Educator

Problem 10

Are the following compounds stable?
(a) No. A trans-cyclohexene is too strained. The trans unit $\mathrm{C}-\mathrm{C}=\mathrm{C}-\mathrm{C}$ cannot be bridged by two more C's. trans-Cycloalkenes are stable for eight-membered, (b), and larger rings. (c) No. Cycloalkanes of fewer than eight C's are too strained. The triple bond imposes linearity on four $\mathrm{C}$ 's, $\mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{C}$, which cannot be bridged by two more C's but can be bridged by four C's, as in $(d)$. (e) No. A bridgehead bicyclic cannot have a double bond at a bridgehead position (unless one of the rings has at least eight $\mathrm{C}$ 's). Such a bridgehead $\mathrm{C}$ and the three atoms bonded to it cannot assume the flat, planar structure required of an $s p^2$-hybridized C. This is known as Bredt's rule. $(f)$ Yes. This exists because one of the bridges has no $\mathrm{C}$, and these bridgehead $\mathrm{C}$ 's can easily use $s p^2$-hybridized orbitals to form triangular, planar sigma bonds. $(\mathrm{g})$ Yes. Compounds (spiranes) having a single $\mathrm{C}$ which is a junction for two separate rings are known for all size rings. However, the rings must be at right angles.
(h) No. Three- and six-membered rings cannot be fused trans, since there is too much strain.

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01:10

Problem 11

Among the cycloalkanes with up to 13 ring C's, cyclohexane has the least ring strain. Would one expect this if cyclohexane had a flat hexagonal ring?

No. A flat hexagon would have considerable ring strain. It would have six pairs of eclipsed $\mathrm{H}$ 's and $\mathrm{C}-\mathrm{C}-\mathrm{C}$ bond angles of $120^{\circ}$, which deviate by $10.5^{\circ}$ from the tetrahedral angle, 109.5 .

Cyclohexane minimizes its ring strain by being puckered rather than flat. The two extreme conformations are the more stable chair and the less stable boat. The twist-boat conformer is less stable than the chair by about $23 \mathrm{~kJ} / \mathrm{mol}$, but is more stable than the boat. It is formed from the boat by moving one "flagpole" to the left and the other to the right. See Fig. 9-4.

John Nicolle

Numerade Educator

12:59

Problem 12

In terms of eclipsing interactions explain why $(a)$ the chair conformation is more stable than the boat conformation and $(b)$ the twist-boat conformation is more stable than the boat conformation.
(a) In the boat form, Fig. 9-4(b), the following pairs of $\mathrm{C}-\mathrm{C}$ bonds are eclipsed: $\mathrm{C}^{\prime}-\mathrm{C}^2$ with $\mathrm{C}^3-\mathrm{C}^4$, and $\mathrm{C}^{\prime}-\mathrm{C}^6$ with $\mathrm{C}^4-\mathrm{C}^5$. Furthermore, the $\mathrm{H}^{\prime}$ s on $\mathrm{C}^2-\mathrm{C}^3$ and $\mathrm{C}^5-\mathrm{C}^6$ are also eclipsed. Additional strain arises from the crowding of the "flagpole" $\mathrm{H}$ 's on $\mathrm{C}^{\prime}$ and $\mathrm{C}^4$, which point toward each other. This is called steric strain because the $\mathrm{H}$ 's tend to occupy the same space. In the chair form, Fig. 9-4(a), all C-C bonds are skew and one pair of H's on adjacent C's is anti and the other pair is gauche (see the Newman projection).
(b) Twisting $\mathrm{C}^l$ and $\mathrm{C}^4$ of the boat form away from each other gives the flexible twist-boat conformation in which the steric and all the eclipsed interactions are reduced [Fig. 9-4(c)].

Susan Hallstrom

Numerade Educator

11:17

Problem 13

Give the conformational designations of the boxed H's in Fig. 9-2.
Axial on the left; equatorial on the right.
2. Monosubstituted Cyclohexanes

Replacing $\mathrm{H}$ by $\mathrm{CH}_3$ gives two different chair conformations; in Fig. 9-5(e) $\mathrm{CH}_3$ is axial, in Fig. 9-5 $f(f)$ $\mathrm{CH}_3$ is equatorial. For methylcyclohexane, the conformer with the axial $\mathrm{CH}_3$ is less stable and has $7.5 \mathrm{~kJ} / \mathrm{mol}$ more energy. This difference in energy can be analyzed in either of two ways:
1,3-Diaxial interactions (transannular effect). In Fig. 9-5(e) the axial $\mathrm{CH}_3$ is closer to the two axial H's than is the equatorial $\mathrm{CH}_3$ to the adjacent equatorial H's in Fig. 9-5(f). The steric strain for each $\mathrm{CH}_3-\mathrm{H} \mathrm{1,3-diaxial} \mathrm{interaction} \mathrm{is} 3.75 \mathrm{~kJ} / \mathrm{mol}$, and the total is $7.5 \mathrm{~kJ} / \mathrm{mol}$ for both.
Gauche interaction (Fig. 9-6). An axial $\mathrm{CH}_3$ on $\mathrm{C}^{\prime}$ has a gauche interaction with the $\mathrm{C}^2-\mathrm{C}^3$ bond of the ring. One gauche interaction is also $3.75 \mathrm{~kJ} / \mathrm{mol}$; for the two the difference in energy is $7.5 \mathrm{~kJ} / \mathrm{mol}$. The equatorial $\mathrm{CH}_3$ indicated as $\left(\mathrm{CH}_3\right)$ is anti to the $\mathrm{C}^2-\mathrm{C}^3$ ring bond.
In general, a given substituent prefers the less crowded equatorial position to the more crowded axial position.

Ronald Prasad

Numerade Educator

08:56

Problem 14

(a) Draw the possible chair conformational structures for the following pairs of dimethylcyclohexanes: (i) cis- and trans-1,2-; (ii) cis- and trans-1,3-; (iii) cis- and trans-1,4-. (b) Compare the stabilities of the more stable conformers for each pair of geometric isomers. (c) Determine which of the isomers of dimethylcyclohexane are chiral.

When using chair conformers, the better way to determine whether substituents are cis or trans is to look at the axial rather than the equatorial groups. If one axial bond is up and the other is down, the isomer is trans; if both axial bonds are up (or down), the geometric isomer is cis.
(a) (i) In the 1,2-isomer, since one axial bond is up and one is down, they are trans (Fig. 9-7). The equatorial bonds are also trans although this is not obvious from the structure. In the cis-1,2-isomer an $\mathrm{H}$ and $\mathrm{CH}_3$ are trans to each other (Fig. 9-8).
(ii) In the 1,3-isomer both axial bonds are up (or down) and cis (Fig. 9-9). In the more stable conformer [Fig. 9-9(a)] both $\mathrm{CH}_3$ 's are equatorial. In the trans isomer, one $\mathrm{CH}_3$ is axial and one equatorial (Fig. 9-10).
(iii) In the 1,4-isomer the axial bonds are in opposite directions and are trans (Fig. 9-11).
(b) Since an (e) substituent is more stable than an (a) substituent, in each case the $\left(\mathrm{CH}_3\right.$ 's ee) isomer is more stable than the $\left(\mathrm{CH}_3\right.$ 's ea) isomer.
(i) trans $>$ cis
(ii) cis > trans
(iii) trans $>$ cis
(c) The best way to detect chirality in cyclic compounds is to examine the flat structures as in Problem 9.3(e); trans1,2- and trans-1,3- are the chiral isomers.

George Bennett

Numerade Educator

03:30

Problem 15

Give your reasons for selecting the isomers of dimethylcyclohexane shown in Figs. 9-7 to 9-11 that exist as: (a) a pair of configurational enantiomers, each of which exists in one conformation; (b) a pair of conformational diastereomers; (c) a pair of configurational enantiomers, each of which exists as a pair of conformational diastereomers; $(d)$ a single conformation; $(e)$ a pair of conformational enantiomers.

(a) trans-1,3-Dimethylcyclohexane is chiral and exists as two enantiomers. Each enantiomer is (ae) and has only one conformer.
(b) Both cis-1,3- and trans-1,4-dimethylcyclohexane have conformational diastereomers, the stable (ee) and unstable (aa). Neither has configurational isomers.
(c) trans-1,2-Dimethylcyclohexane is a racemic form of a pair of configurational enantiomers. Each enantiomer has (ee) and (aa) conformational diastereomers.
(d) cis-1,4-Dimethylcyclohexane has no chiral C's and has only a single (ae) conformation.
(e) cis-1,2-Dimethylcyclohexane has two (ae) conformers that are nonsuperimposable mirror images.

Grigoriy Sereda

Numerade Educator

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Problem 16

Use 1,3-interactions and gauche interactions, when needed, to find the difference in energy between (a) cis- and trans-1,3-dimethylcyclohexane; (b) (ee) trans-1,2- and (aa) trans-1,2-dimethylcyclohexane.

Each $\mathrm{CH}_3 / \mathrm{H}$ 1,3-interaction and each $\mathrm{CH}_3 / \mathrm{CH}_3$ gauche interaction imparts $3.75 \mathrm{~kJ} / \mathrm{mol}$ of instability to the molecule.
(a) In the cis-1,3-isomer (Fig. 9-9) the more stable conformer has (ee) $\mathrm{CH}_3$ 's and thus has no 1,3-interactions. The trans isomer has (ea) $\mathrm{CH}_3$ 's. The axial $\mathrm{CH}_3$ has two $\mathrm{CH}_3 / \mathrm{H}$ 1,3-interactions, accounting for $2(3.75)=7.5 \mathrm{~kJ} / \mathrm{mol}$ of instability. The cis isomer is more stable than the trans isomer by $7.5 \mathrm{~kJ} / \mathrm{mol}$.
(b) See Fig. $9-12$; (ee) is more stable than (aa) by $15.0-3.75=11.25 \mathrm{~kJ} / \mathrm{mol}$.

Susan Hallstrom

Numerade Educator

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Problem 17

Write the structure of the preferred conformation of (a) trans-1-ethyl-3-isopropylcyclohexane, $(b)$ cis-2-chloro-cis-4-chlorocyclohexyl chloride.

(a) The trans-1,3-isomer is (ea); the bulkier group, in this case $i$-propyl, is equatorial, and the smaller group, in this case ethyl, is axial. See Fig. 9-13(a).
(b) See Fig. 9-13(b).

Vipin Singh

Numerade Educator

02:09

Problem 18

You wish to determine the relative rates of reaction of an axial and an equatorial $\mathrm{Br}$ in an $\mathrm{S}_{\mathrm{N}} 2$ displacement. Can you compare (a) cis- and trans-1-methyl-4-bromocyclohexane? (b) cis- and trans-1-t-butyl-4 bromocyclohexane? (c) cis-3,5-dimethyl-cis-1-bromocyclohexane and cis-3,5-dimethyl-trans-1-bromocyclohexane?
(a) The trans substituents are (ee). The cis substituents are (ea). Although $\mathrm{CH}_3$ is bulkier and has a greater (e) preference than has $\mathrm{Br}$, the difference in preference is small and an appreciable number of molecules exist with the $\mathrm{Br}$ (e) and the $\mathrm{CH}_3$ (a). At no time are there conformers with $\mathrm{Br}$ only in an (a) position. These isomers, therefore, cannot be used for this purpose.
(b) The bulky $t$-butyl group can only be (e). In practically all molecules of the $c$ is isomer, $\mathrm{Br}$ is forced to be (a). All molecules of the trans isomer have an (e) Br. Because $t$-butyl "freezes" the conformation and prevents interconversion, these isomers can be used.
(c) The cis-3,5-dimethyl groups are almost exclusively (ee) to avoid severe $\mathrm{CH}_3 / \mathrm{CH}_3$ 1,3-interactions were they to be (aa). These cis- $\mathrm{CH}_3$ 's freeze the conformation. When $\mathrm{Br}$ at $\mathrm{C}^l$ is cis, it has an (e) position; when it is trans, it has an (a) position. These isomers can be used.

Alkendra Singh

Numerade Educator

01:43

Problem 19

The carbene : $\mathrm{CCl}_2$ generated from chloroform, $\mathrm{CHCl}_3$, and $\mathrm{KOH}$ in the presence of alkenes gives substituted cyclopropanes. Write the equation for the reaction of : $\mathrm{CCl}_2$ and propene.
Another class of intermolecular cyclizations are the

Lottie Adams

Numerade Educator

05:16

Problem 20

Although cyclopropanes are less reactive than alkenes, they undergo similar addition reactions. (a) Account for this by geometry and orbital overlap. (b) How does $\mathrm{HBr}$ addition to 1,1-dimethylcyclopropane resemble Markovnikov addition?
(a) Because of the ring strain in cyclopropane (Problem 9.5), there is less orbital overlap (Fig. 9-1) and the sigma electrons are accessible to attack by electrophiles.
(b) The proton of $\mathrm{HBr}$ is attacked by an electron pair of a bent cyclopropane sigma bond to form a carbocation that adds $\mathrm{Br}^{-}$to give a 1,3-Markovnikov-addition product.

Ian Kaigh

Numerade Educator

02:09

Problem 21

Which conformation of 1,3-butadiene participates in the Diels-Alder reaction with, e.g., ethene?
The two conformations of 1,3-butadiene are s-cis (cisoid) and s-trans (transoid):
<smiles>C=CC=C</smiles>
<smiles>C=CC=C</smiles>
$s$-cis
s-trans

Although s-trans is the more favorable conformer, reaction occurs with $s$-cis because this conformation has its double bonds on the same side of the single bond connecting them; hence, the stable form of cyclohexene with a cis double bond is formed. Reaction of the $s$-trans conformer with ethene would give the impossibly strained trans-cyclohexene [Problem $9.10(a)$ ]. As the $s$-cis conformer reacts, the equilibrium between the two conformers shifts toward the $s$-cis side, and in this way all the unreactive $s$-trans reverts to the reactive $s$-cis conformer.

Alkendra Singh

Numerade Educator

15:58

Problem 22

Outline a synthesis of the following alicyclic compounds from acyclic compounds.

(a)CAN'T COPY
(b)CAN'T COPY
(c)CAN'T COPY
(d)CAN'T COPY

Zubair Abdulla

Numerade Educator

07:08

Problem 23

Starting with cyclopentanol, show the reactions and reagents needed to prepare $(a)$ cyclopentene, (b) 3-bromocyclopentane, (c) 1,3-cyclopentadiene, (d) trans-1,2-dibromocyclopentane, (e) cyclopentane.

Tom Rutherford

Numerade Educator

06:27

Problem 24

Complete the following reactions:
(c) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3$. Under these conditions the strained three-membered ring opens. (d) $\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}$. Again, the three-membered ring opens. (e) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$. The strained four-membered ring opens, but a higher temperature is needed than in part $(c)$. $(f)$ No reaction. The five-membered ring has no ring strain. $(g)$ No reaction. Even the strained rings are stable toward oxidation. $(h) \nabla^{\mathrm{COOH}}$, cyclopropanecarboxylic acid.

Rajesh Singh

Numerade Educator

02:26

Problem 25

Cycloalkanes with more than six C's are difficult to synthesize by intramolecular ring closures, yet they are stable. On the other hand, cyclopropanes are synthesized this way, yet they are the least stable cycloalkanes. Are these facts incompatible? Explain.

No. The relative ease of synthesis of cycloalkanes by intramolecular cyclization depends on both ring stability and the probability of bringing the two ends of the chain together to form a $\mathrm{C}$-to-C bond, thereby closing the ring. This probability is greatest for smallest rings and decreases with increasing ring size. The interplay of ring stability and this probability factor are summarized below (numbers represent ring sizes).

Probability of ring closure Thermal stability
$$
\begin{aligned}
& 3>4>5>6>7>8>9 \\
& 6>7,5>8,9>4>3 \\
& 5>3,6>4,7,8,9
\end{aligned}
$$

Ease of synthesis
The high yield of cyclopropane indicates that a favorable probability factor outweighs the ring instability. For rings with more than six C's the ring stability effect is outweighed by the highly unfavorable probability factor.

Crystal Wang

Numerade Educator

05:30

Problem 26

Account for the fact that intramolecular cyclizations to rings with more than six C's are effected at extremely low concentrations (Ziegler method).

Chains can also react intermolecularly to form longer chains. Although intramolecular reactions are ordinarily faster than intermolecular reactions, the opposite is true in the reaction of chains leading to rings with more than six C's. This side reaction from collisions between different chains is minimized by carrying out the reaction in extremely dilute solutions.

Prashant Bana

Numerade Educator

02:07

Problem 27

When applying the Woodward-Hoffmann rules to the Diels-Alder reaction, $(a)$ would the same conclusion be drawn if the LUMO of the dienophile interacts with the HOMO of the diene? (b) Would the reaction be light-catalyzed?
(a) Yes; see Fig. 9-17(a). (b) No; see Fig. 9-17(b).

Dr. Satish Ingale

Numerade Educator

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Problem 28

Use Woodward-Hoffmann rules to predict whether the following reaction would be expected to occur thermally or photochemically.
The MO energy levels of the allyl carbanion $\pi$ system showing the distribution of the four $\pi$ electrons (two from $\pi$ double bond and two unshared) are indicated in Fig. 9-18. The 0 is used whenever a node point is at an atom. The allowed reaction occurs thermally as shown in Fig. 9-19(a). The photoreaction is forbidden [Fig. 9-19(b)].

Dakota Smith

Numerade Educator

Problem 29

Pick out the isoprene units in the terpenes limonene, myrcene and $\alpha$-phellandrene, and in vitamin A, shown below.
In the structures below, dashed lines separate the isoprene units.

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02:07

Problem 30

Draw formulas for (a) isopropylcyclopentane, (b) cis-1,3-dimethylcyclooctane, (c) bicyclo[4.4.1] undecane, (d) trans-1-propyl-4-butylcyclohexane.

Lottie Adams

Numerade Educator

02:08

Problem 31

Name each of the following compounds and indicate which, if any, is chiral.

(a) The rings are numbered as shown, starting at a bridgehead $\mathrm{C}$ and going around the larger ring first, in such a way that the first doubly-bonded $C$ attains the lowest possible number. The name is bicyclo[4.3.0]non-7-ene. The molecule is chiral; $\mathrm{C}^i$ and $\mathrm{C}^6$ are chiral centers.
(b) 2,3-Diethylcyclopentene. The molecule is chiral; $\mathrm{C}^3$ is a chiral center.
(c) Bicyclo[4.4.2]dodecane. The molecule is achiral.

David Collins

Numerade Educator

02:01

Problem 32

Draw the structural formulas and give the stereochemical designation (meso, rac, cis, trans, achiral) of all the isomers of trichlorcyclobutane.

Nicole Smina

Numerade Educator

03:44

Problem 33

Show steps in the synthesis of cyclohexane from phenol, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}$.

Shazia Naz

Numerade Educator

02:09

Problem 35

Explain why (a) a carbene is formed by dehydrohalogenation of $\mathrm{CHCl}_3$ but not from methyl, ethyl, or n-propyl chlorides; (b) cis-1,3- and trans-1,4-di-tert-butylcyclohexane exist in chair conformations, but their geometric isomers, trans-1,3- and cis-1,4-, do not.
(a) Carbene is formed from $\mathrm{CHCl}_3$ because the three strongly electronegative $\mathrm{Cl}$ 's make this compound sufficiently acidic to have its proton abstracted by a base. $\mathrm{CH}_3 \mathrm{Cl}$ has only one $\mathrm{Cl}$ and is considerably less acidic. Carbene formation is an $\alpha$-elimination of $\mathrm{HCl}$ from the same $\mathrm{C}$; it does not occur with ethyl or propyl chlorides because protons are more readily eliminated from the $\beta$ ' 's to form alkenes.
(b) Both cis-1,3 and trans-1,4 compounds exist in the chair form because of the stability of their (ee) conformers. Trans-1,3- and cis-1,4- are (ea). An axial $t$-butyl group is very unstable, so that a twist-boat with a quasi-(ee) conformation (Fig. 9-20) is more stable than the chair.

Alkendra Singh

Numerade Educator

Problem 36

Assign structures or configurations for A through D. (a) Two isomers, A and B, with formula $\mathrm{C}_8 \mathrm{H}_{14}$ differ in that one adds $1 \mathrm{~mol}$ and the other $2 \mathrm{~mol}$ of $\mathrm{H}_2$. Ozonolysis of A gives only one product, $\mathrm{O}=\mathrm{CH}\left(\mathrm{CH}_2\right)_6 \mathrm{CH}=\mathrm{O}$, while the same reaction with $1 \mathrm{~mol}$ of $\mathrm{B}$ produces $2 \mathrm{~mol}$ of $\mathrm{CH}_2=\mathrm{O}$ and $1 \mathrm{~mol}$ of $\mathrm{O}=\mathrm{CH}\left(\mathrm{CH}_2\right)_6 \mathrm{CH}=\mathrm{O}$. (b) Two stereoisomers, $\mathrm{C}$ and D, of 3,4-dibromocyclopentane-1,1-dicarboxylic acid undergo decarboxylation as shown:
<smiles>O=C(O)C(CBr)(CBr)C(=O)O</smiles>
<smiles>CC(=O)OC(Br)CBr</smiles>
a gem-dicarboxylic acid
a monocarboxylic acid

C gives one, while D yields two, monocarboxylic acids.
(a) Both compounds have two degrees of unsaturation (see Problem 6.34). B absorbs two moles of $\mathrm{H}_2$ and has two multiple bonds. A absorbs one mole of $\mathrm{H}_2$ and has a ring and a double bond; it is a cycloalkene. As a cycloalkene, A can form only a single product, a dicarbonyl compound, on ozonolysis.
Since one molecule of B gives three carbonyl molecules, it must be a diene and not an alkyne.
$$
\begin{array}{cl}
\mathrm{H}_2 \mathrm{C}=\mathrm{O}+\mathrm{O}=\underset{\text { 1.6-Hexanedial }}{\mathrm{HC}\left(\mathrm{CH}_2\right)_4 \mathrm{CH}=\mathrm{O}}+\mathrm{O}=\mathrm{CH}_2 \stackrel{\mathrm{O}_3}{\stackrel{\mathrm{H}_2 \mathrm{C}}{\mathrm{C}}=\underset{1.7-\text { Octadiene }(\mathrm{B})}{\mathrm{CH}\left(\mathrm{CH}_2\right)_4 \mathrm{CH}}=\mathrm{CH}_2} \\
\stackrel{2 \mathrm{H}_2}{\longrightarrow} \underset{\text { Octane }}{\mathrm{CH}_3 \mathrm{CH}_2\left(\mathrm{CH}_2\right)_4 \mathrm{CH}_2 \mathrm{CH}_3}
\end{array}
$$

Octane
(b) The Br's of the dicarboxylic acid may be cis or trans. Decarboxylation of the cis isomer yields two isomeric products in which both Br's are cis (E) or trans (F) with respect to $\mathrm{COOH}$. The cis isomer is D. In the monocarboxylic acid $\mathrm{G}$ formed from $\mathrm{C}$ (trans isomer), one $\mathrm{Br}$ is cis and the other trans with respect to $\mathrm{COOH}$ and there is only one isomer.

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05:16

Problem 37

Outline the reactions and reagents needed to synthesize the following from any acyclic compounds having up to four C's and any needed inorganic reagents: $(a)$ cis-1-methyl-2-ethylcyclopropane; $(b)$ trans-1,1dichloro-2-ethyl-3-n-propylcyclopropane; $(c)$ 4-cyanocyclohexene; $(d)$ bromocyclobutane.
(a) The cis-disubstituted cyclopropane is prepared by stereospecific additions of singlet carbene to cis-2-pentene.

The alkene, having five C's, is best formed from 1-butyne, a compound with four C's.
(b) Add dichlorocarbene to trans-3-heptene, which is formed from 1-butyne.

Lottie Adams

Numerade Educator

Problem 38

Write planar structures for the cyclic derivatives formed in the following reactions, and give their stereochemical labels.
(a) 3-Cyclohexenol + dil. aq. $\mathrm{KMnO}_4 \longrightarrow$
(b) 3-Cyclohexenol $+\mathrm{HCO}_3 \mathrm{H}$ and then $\mathrm{H}_2 \mathrm{O} \longrightarrow$
(c) (+)-trans-1,2-Dibromocyclopropane $+\mathrm{Br}_2 \stackrel{\text { ltght }}{\longrightarrow}$
(d) meso-cis-1,2-Dibromocyclopropane $+\mathrm{Br}_2 \stackrel{\text { light }}{\longrightarrow}$
(e) 1-Methylcyclohexene $+\mathrm{HBr}$ (peroxide) $\longrightarrow$ (an anti addition)

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12:32

Problem 39

Use cyclohexanol and any inorganic reagents to synthesize $(a)$ trans-1,2-dibromocyclohexane, $(b)$ cis-1,2-dibromocyclohexane, (c) trans-1,2-cyclohexanediol.

Nima Gharibi

Numerade Educator

01:28

Problem 40

Decalin, $\mathrm{C}_{10} \mathrm{H}_{18}$, has cis and trans isomers that differ in the configurations about the two shared $\mathrm{C}$ 's as shown below. Draw their conformational structural formulas.

Vasu Makani

Numerade Educator

13:43

Problem 41

Explain the following facts in terms of the structure of cyclopropane. (a) The H's of cyclopropane are more acidic than those of propane. (b) The $\mathrm{Cl}$ of chlorocyclopropane is less reactive toward $\mathrm{S}_{\mathrm{N}} 2$ and $\mathrm{S}_{\mathrm{N}} 1$ displacements than the $\mathrm{Cl}$ in $\mathrm{CH}_3 \mathrm{CHClCH}_3$.
(a) The external $\mathrm{C}-\mathrm{H}$ bonds of cyclopropane have more $s$ character than those of an alkane (Fig. 9-1). The more $s$ character in the $\mathrm{C}-\mathrm{H}$ bond, the more acidic the $\mathrm{H}$.
(b) The $\mathrm{C}-\mathrm{Cl}$ bond of chlorocyclopropane also has more $s$ character, which diminishes the reactivity of the $\mathrm{Cl}$. Remember that vinyl chlorides are inert in $\mathrm{S}_{\mathrm{N}} 2$ and $\mathrm{S}_{\mathrm{N}} 1$ reactions. The $\mathrm{R}^{+}$formed during the $\mathrm{S}_{\mathrm{N}} 1$ reaction would have very high energy, since the $\mathrm{C}$ would have to use $s p^2$ hybrid orbitals needing a bond angle of $120^{\circ}$. The angle strain of the $\mathrm{R}^{+}$is much more severe $\left(120^{\circ}-60^{\circ}\right)$ than in cyclopropane itself $\left(109^{\circ}-60^{\circ}\right)$.

Dr. Satish Ingale

Numerade Educator

02:52

Problem 42

Use quantitative and qualitative tests to distinguish between (a) cyclohexane, cyclohexene and 1,3cyclohexadiene; (b) cyclopropane and propene.
(a) Cyclohexane does not decolorize $\mathrm{Br}_2$ in $\mathrm{CCl}_4$. The uptake of $\mathrm{H}_2$, measured quantitatively, is $2 \mathrm{~mol}$ for $1 \mathrm{~mol}$ of the diene, but $1 \mathrm{~mol}$ for $1 \mathrm{~mol}$ of the cycloalkene.
(b) Cyclopropane resembles alkenes and alkynes, and differs from other cycloalkanes in decolorizing $\mathrm{Br}_2$ slowly, adding $\mathrm{H}_2$, and reacting readily with $\mathrm{H}_2 \mathrm{SO}_4$. However, it is like other cycloalkanes and differs from multiplebonded compounds in not decolorizing aqueous $\mathrm{KMnO}_4$.

Bhumika Jayee

Numerade Educator

03:03

Problem 43

(a) Give the structure of the major product, A, whose formula is $\mathrm{C}_5 \mathrm{H}_8$, resulting from the dehydration of cyclobutylmethanol. On hydrogenation, A yields cyclopentane. (b) Give a mechanism for this reaction.
(a) Compound $\mathrm{A}$ is cyclopentene, which gives cyclopentane on hydrogenation.

Nicholas Sacco

Numerade Educator