Schaum's Outline of Organic Chemistry

George Hademenos, George Hademenos

Chapter 6 ALKENES - all with Video Answers

Educators

Chapter Questions

(a) Suggest a mechanism for the dehydration of $\mathrm{CH}_3 \mathrm{CHOHCH}_3$ that proceeds through a carbocation intermediate. Assign a catalytic role to the acid and keep in mind that the $\mathrm{O}$ in $\mathrm{ROH}$ is a basic site like the $\mathrm{O}$ in $\mathrm{H}_2 \mathrm{O}$. (b) Select the slow rate-determining step and justify your choice. (c) Use transition states to explain the order of reactivity of $\mathrm{ROH}: 3^{\circ}>2^{\circ}>1^{\circ}$.
(a)CAN'T COPY
Instead of $\mathrm{HSO}_4^{-}$, a molecule of alcohol could act as the base in Step 3 to give $\mathrm{ROH}_2^{+}$.
(b) Carbocation formation, Step 2, is the slow step, because it is a heterolysis leading to a very high-energy carbocation possessing an electron-deficient $\mathrm{C}$.
(c) The order of reactivity of the alcohols reflects the order of stability of the incipient carbocation $\left(3^{\circ}>2^{\circ}>1^{\circ}\right)$ in the TS of Step 2, the rate-determining step. See Fig. 6-2.

Nikhil Choudhary

Numerade Educator

01:53

Problem 2

Supply the structural formula and IUPAC name for $(a)$ trichloroethylene, $(b)$ sec-butylethylene, (c) sym-divinylethylene.
Alkenes are also named as derivatives of ethylene. The ethylene unit is shown in a box.
(a) CAN'T COPY
(b)CAN'T COPY
(c) CAN'T COPY

Ronald Prasad

Numerade Educator

02:07

Problem 2

(a) Describe the stereochemistry of glycol formation with peroxyformic acid $\left(\mathrm{HCO}_3 \mathrm{H}\right)$ if cis-2butene gives a racemic glycol and trans-2-butene gives the meso form. (b) Give a mechanism for cis.

Ian Kaigh

Numerade Educator

04:35

Problem 3

Write the structural formula for $(a)$ the major trimeric alkene formed from $\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CH}_2$, labeling the mer; $(b)$ the dimeric alkene from $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2$. [Indicate the dimeric $\mathrm{R}^{+}$.]

Vasu Makani

Numerade Educator

03:16

Problem 3

Predict (a) the geometry of ethylene, $\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2 ;(b)$ the relative $\mathrm{C}$-to- $\mathrm{C}$ bond lengths in ethylene and ethane; $(c)$ the relative $\mathrm{C}-\mathrm{H}$ bond lengths and bond strengths in ethylene and ethane; $(d)$ the relative bond strengths of $\mathrm{C}-\mathrm{C}$ and $\mathrm{C}=\mathrm{C}$.
(a) Each $\mathrm{C}$ in ethylene (ethene) uses $s p^2 \mathrm{HO}$ 's (Fig. 2-8) to form three trigonal $\sigma$ bonds. All five $\sigma$ bonds (four $\mathrm{C}-\mathrm{H}$ and one $\mathrm{C}-\mathrm{C}$ ) must lie in the same plane: ethylene is a planar molecule. All bond angles are approximately $120^{\circ}$.
(b) The $\mathrm{C}=\mathrm{C}$ atoms, having four electrons between them, are closer to each other than the $\mathrm{C}-\mathrm{C}$ atoms, which are separated by only two electrons. Hence, the $\mathrm{C}=\mathrm{C}$ length $(0.134 \mathrm{~nm})$ is less than the $\mathrm{C}-\mathrm{C}$ length $(0.154 \mathrm{~nm})$.
(c) The more $s$ character in the hybrid orbital used by $\mathrm{C}$ to form a $\sigma$ bond, the closer the electrons are to the nucleus and the shorter is the $\sigma$ bond. Thus, the $\mathrm{C}-\mathrm{H}$ bond length in ethylene $(0.108 \mathrm{~nm})$ is less than the length in ethane $(0.110 \mathrm{~nm})$. The shorter bond is also the stronger bond.
(d) Since it takes more energy to break two bonds than one bond, the bond energy of $\mathrm{C}=\mathrm{C}$ in ethylene $(611 \mathrm{~kJ} / \mathrm{mol})$ is greater than that of $\mathrm{C}-\mathrm{C}$ in ethane $(348 \mathrm{~kJ} / \mathrm{mol})$. However, note that the bond energy of the double bond is less than twice that of the single bond. This is so because it is easier to break a $\pi$ bond than a $\sigma$ bond.

Freddie Montague

Numerade Educator

02:28

Problem 4

Supply the structural formulas of the alkenes and the reagents which react to form: $(a)\left(\mathrm{CH}_3\right)_3 \mathrm{CI}$, (b) $\mathrm{CH}_3 \mathrm{CHBr}_2$, (c) $\mathrm{BrCH}_2 \mathrm{CHClCH}_3$, (d) $\mathrm{BrCH}_2 \mathrm{CHOlICH}_2 \mathrm{Cl}$.
(a) $\mathrm{CH}_3-\stackrel{\mathrm{C}}{\mathrm{C}}=\mathrm{CH}_2+\mathrm{Hl}$
(b) $\mathrm{H}_2 \mathrm{C}=\mathrm{CHBr}+\mathrm{HBr}$
(c) $\mathrm{H}_2 \mathrm{C}=\mathrm{CCl}-\mathrm{CH}_3+\mathrm{HBr}+$ peroxide
(d) $\mathrm{BrCH}_2-\mathrm{CH}=\mathrm{CH}_2+\mathrm{HOCl}$ or $\mathrm{H}_2 \mathrm{C}=\mathrm{CHCH}_3+\mathrm{BrCl}$ or $\mathrm{H}_2 \mathrm{C}=\mathrm{CH}-\mathrm{CH}_2 \mathrm{Cl}+\mathrm{HOBr}$

Eugene Schneider

University of Minnesota - Twin Cities

05:30

Problem 4

Which of the following alkenes exhibit geometric isomerism? Supply structural formulas and names for the isomers.
(a)CAN'T COPY
(b)CAN'T COPY
(c)CAN'T COPY
(d)CAN'T COPY
(e)CAN'T COPY
(f)CAN'T COPY
(a) No geometric isomers because one double-bonded $\mathrm{C}$ has two $\mathrm{C}_2 \mathrm{H}_5$ 's.
(b) No geometric isomers; one double-bonded $\mathrm{C}$ has two H's.
(c) Has geometric isomers because each double-bonded $\mathrm{C}$ has two different substituents:
CAN'T COPY
(d) There are two geometric isomers because one of the double bonds has two different substituents.
CAN'T COPY
(e) Both double bonds meet the conditions for geometric isomers and there are four diastereomers of 2,4-heptadiene.
CAN'T COPY
Note that $c i s$ and trans and $E$ and $Z$ are listed in the same order as the bonds are numbered.
(f) There are now only three isomers because cis-trans and trans-cis geometries are identical.
CAN'T COPY

Vasu Makani

Numerade Educator

10:06

Problem 5

Write structural formulas for $(a)(E)$-2-methyl-3-hexene (trans). $(b)$ (S)-3-chloro-1-pentene, (c) (R), (Z)-2-chloro-3-heptene (cis).
CAN'T COPY

Kaitlynn Wade

Numerade Educator

03:05

Problem 5

Give 4 simple chemical tests to distinguish an alkene from an alkane.
A positive simple chemical test is indicated by one or more detectable events. such as a change in color. formation of a precipitate, evolution of a gas, uptake of a gas, evolution of heat.

Alkanes give none of these tests.

Aadit Sharma

Numerade Educator

02:55

Problem 6

How do boiling points and solubilities of alkanes compare with those of corresponding alkanes?

Alkanes and alkenes are nonpolar compounds whose corresponding structures have almost identical molecular weights. Boiling points of alkenes are close to those of alkanes and similarly have $20^{\circ}$ increments per $\mathrm{C}$ atom. Both are soluble in nonpolar solvents and insoluble in water, except that lower-molecular-weight alkenes are slightly more water-soluble because of attraction between the $\pi$ bond and $\mathrm{H}_2 \mathrm{O}$.

Ayushi Sambyal

Numerade Educator

01:27

Problem 7

Show the directions of individual bond dipoles and net dipole of the molecule for $(a)$ 1,1dichloroethylene, (b) cis- and trans-1,2-dichloroethylene.

The individual dipoles are shown by the arrows on the bonds between $\mathrm{C}$ and $\mathrm{CI}$. The net dipole for the molecule is represented by an arrow that bisects the angle between the two $\mathrm{Cl}$ s. $\mathrm{C}-\mathrm{H}$ dipoles are insignificant and are disregarded.

Lottie Adams

Numerade Educator

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Problem 8

How can heats of combustion be used to compare the differences in stability of the geometric isomers of alkenes?

The thermodynamic stability of isomeric hydrocarbons is determined by burning them to $\mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$ and comparing the heat evolved per mole $(-\Delta H$ combustion). The more stable isomer has the smaller $(-\Delta H)$ value. Trans alkenes have the smaller values and hence are more stable than the cis isomers. This is supported by the exothermic ( $\Delta H$ negative) conversion of cis to trans isomers by ultraviolet light and some chemical reagents.

The cis isomer has higher energy because there is greater repulsion between its alkyl groups on the same side of the double bond than between an alkyl group and an $\mathrm{H}$ in the trans isomer. These repulsions are greater with larger alkyl groups, which produce larger energy differences between geometric isomers.

Susan Hallstrom

Numerade Educator

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Problem 9

(a) How does the greater enthalpy of cis-vis a vis trans-2-butene affect the ratio of isomers formed during the dehydrohalogenation of 2-chlorobutane? (b) How does replacing the $\mathrm{CH}_3$ groups of 2-chlorobutane by tbutyl groups to give $\mathrm{CH}_3 \mathrm{C}\left(\mathrm{CH}_3\right)_2 \mathrm{CH}_2 \mathrm{CHClC}\left(\mathrm{CH}_3\right)_2 \mathrm{CH}_3$ alter the distribution of the alkene geometric isomers?
(a) The transition states for the formation of the geometric isomers reflect the relative stabilities of the isomers. The greater repulsion between the nearby $\mathrm{CH}_3$ 's in the cis-like transition state (TS), causes this TS to have a higher enthalpy of activation $\left(\Delta H^{\ddagger}\right)$ than the trans-like TS. Consequently, the trans isomer predominates.
(b) The repulsion of the bulkier $t$-butyl groups causes a substantial increase in the $\Delta H^*$ of the cis-like TS, and the trans isomer is practically the only product.

Susan Hallstrom

Numerade Educator

Problem 10

Give the structural formulas for the alkenes formed on dehydrobromination of the following alkyl bromides and underline the principal product in each reaction: (a) 1-bromobutane, (b) 2-bromobutane, (c) 3-bromopentane, (d) 2-bromo-2-methylpentane, (e) 3-bromo-2-methylpentane, (f) 3-bromo-2,3-dimethylpentane.
The $\mathrm{Br}$ is removed with an atom from an adjacent $\mathrm{C}$.
(a)CAN'T COPY
(b)CAN'T COPY
(c)CAN'T COPY
(d)CAN'T COPY
(e)CAN'T COPY
(f)CAN'T COPY

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00:40

Problem 12

Account for the fact that dehydration of: (a) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}$ yields mainly $\mathrm{CH}_3 \mathrm{CH}^2 \mathrm{CHCH}_3$ rather than $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}=\mathrm{CH}_2,(b)\left(\mathrm{CH}_3\right)_3 \mathrm{CCHOHCH}$ yields mainly $\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_3\right)_2$.
(a) The carbocation $\left(\mathrm{R}^{+}\right)$formed in a reaction like Step 2 of Problem $6.1 \mathrm{I}(\mathrm{a})$ is $\mathrm{I}^{\circ}$ and rearranges to a more stable $2^{\circ}$ $\mathrm{R}_2 \mathrm{CH}^{+}$by a hydride shift (indicates as $\sim \mathrm{H} ;$; the $\mathrm{H}$ migrates with its bonding pair of electrons).
(b) The $2^{\circ} \mathrm{R}_2 \mathrm{CH}^{+}$formed undergoes a methide shift $\left(\sim: \mathrm{CH}_3\right)$ to the more stable $3^{\circ} \mathrm{R}_3 \mathrm{C}^{+}$.
Carbocations are always prone to rearrangement, especially when rearrangement leads to a more stable carbocation. The alkyl group may actually begin to migrate as the leaving group (e.g., $\mathrm{H}_2 \mathrm{O}$ ) is departing-even before the carbocation is fully formed.

Nikhil Choudhary

Numerade Educator

01:00

Problem 13

Assign numbers from 1 for LEAST to 3 for MOST to indicate the relative ease of dehydration and justify your choices.
(a) 1,(b) 3, (c) 2. The ease of dehydration depends on the relative ease of forming an $\mathrm{R}^{+}$, which depends in turn on its relative stability. This is greatest for the $3^{\circ}$ alcohol $(b)$ and least for the $1^{\circ}$ alcohol $(a)$.

Narayan Hari

Numerade Educator

04:18

Problem 14

Give structural formulas for the reactants that form 2-butene when treated with the following reagents: (a) heating with conc. $\mathrm{H}_2 \mathrm{SO}_4,(b)$ alcoholic $\mathrm{KOH}$, (c) zinc dust and alcohol, (d) hydrogen and a catalyst.
(a) $\mathrm{CH}_3 \mathrm{CHOHCH}_2 \mathrm{CH}_3$
(b) $\mathrm{CH}_3 \mathrm{CHBrCH}_2 \mathrm{CH}_3$
(c) $\mathrm{CH}_3 \mathrm{CHBrCHBrCH}_3$
(d) $\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CCH}_3$.

Anish Wadhwa

Numerade Educator

02:20

Problem 15

Write the structural formula and name of the principal organic compound formed in the following reactions:

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

Marissa Turner

Numerade Educator

01:15

Problem 16

Given the following heats of hydrogenation, $-\Delta H_h$, in $\mathrm{kJ} / \mathrm{mol}$ : I-pentene, 125.9; cis-2-pentene, 119.7; trans-2-pentene, 115.5. (a) Use an enthalpy diagram to derive two generalizations about the relative stabilities of alkenes. (b) Would the $\Delta H_h$ of 2-methyl-2-butene be helpful in making your generalizations? (c) The corresponding heats of combustion, $-\Delta H_c$, are: 3376,3369 , and $3365 \mathrm{~kJ} / \mathrm{mol}$. Are these values consistent with your generalizations in part (a)? (d) Would the $\Delta H_c$ of 2-methyl-2-butene be helpful in your comparison? (e) Suggest a relative value for the $\Delta H_c$ of 2-methyl-2-butene.
(a) See Fig. 6-3. The lower $\Delta H_h$, the more stable the alkene. (1) The alkene with more alkyl groups on the double bond is more stable; 2-pentene $>1$-pentene. (2) The trans isomer is usually more stable than the cis. Bulky alkyl groups are anti-like in the trans isomer and eclipsed-like in the cis isomer.
(b) No. The alkenes being compared must give the same product on hydrogenation.
(c) Yes. Again the highest value indicates the least stable isomer.
(d) Yes. On combustion all four isomers give the same products, $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{CO}_2$.
(e) Less than $3365 \mathrm{~kJ} / \mathrm{mol}$, since this isomer is a trisubstituted alkene and the 2-pentenes are disubstituted.

Nikhil Choudhary

Numerade Educator

01:02

Problem 17

What is the stereochemistry of the catalytic addition of $\mathrm{H}_2$ if trans- $\mathrm{CH}_3 \mathrm{CBr}=\mathrm{CBrCH}_3$ gives rac- $\mathrm{CH}_3 \mathrm{CHBrCHBrCH}_3$ and its cis isomer gives the meso product?
In hydrogenation reactions, two $\mathrm{H}$ atoms add stereoselectively syn (cis) to the $\pi$ bond of the alkene.

David Collins

Numerade Educator

03:44

Problem 18

Unsymmetrical reagents like $\mathrm{HX}$ add to unsymmetrical alkenes such as propene according to Markovnikov's rule: the positive portion, e.g., $\mathrm{H}$ of $\mathrm{HX}$, adds to the $\mathrm{C}$ that has more H's ("the rich get richer"). Explain by stability of the intermediate cation.

Dr. Satish Ingale

Numerade Educator

05:35

Problem 19

Give the structural formula of the major organic product formed from the reaction of $\mathrm{CH}_3 \mathrm{CH}^2=\mathrm{CH}_2$ with: (a) $\mathrm{Br}_2$, (b) $\mathrm{HI},(c) \mathrm{BrOH},($ d $) \mathrm{H}_2 \mathrm{O}$ in acid, (e) cold $\mathrm{H}_2 \mathrm{SO}_4,(f) \mathrm{BH}_3$ from $\mathrm{B}_2 \mathrm{H}_6$. (g) peroxyformic acid $\left(\mathrm{H}_2 \mathrm{O}_2\right.$ and $\mathrm{HCOOH}$ ).

The positive $(\delta+)$ part of the addendum is an electrophile $\left(\mathrm{E}^{+}\right)$which forms $\mathrm{CH}_3 \stackrel{+}{\mathrm{C}} \mathrm{HCH}_2 \mathrm{E}$ rather than $\mathrm{CH}_3 \mathrm{CHECH}_2$. The Nu: part then forms a bond with the carbocation.
The $\mathrm{E}^{+}$is in a box; the Nu:- is encircled.

Anupa Sharad Medhekar

Numerade Educator

04:03

Problem 20

Account for the anti-Markovnikov orientation in Problem $6.19(f)$.
The electron-deficient $\mathrm{B}$ of $\mathrm{BH}_3$, as an electrophilic site, racts with the $\pi$ electrons of $\mathrm{C}=\mathrm{C}$, as the nucleophilic site. In typical fashion, the bond is formed with the $\mathrm{C}$ having the greater number of $\mathrm{H}$ 's, in this case, the terminal $\mathrm{C}$. As this bond forms, one of the $\mathrm{Hs}$ of $\mathrm{BH}_3$ begins to break away from the $\mathrm{B}$ as it forms a bond to the other doubly bonded $\mathrm{C}$ atom giving a four-center transition state shown in the equation. The product from this step, $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{BH}_2(n-$ propyl borane), reacts stepwise in a similar fashion with two more molecules of propene, eventually to give $\left(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2\right)_3 \mathrm{~B}$.

Lottie Adams

Numerade Educator

08:33

Problem 21

(a) What principle is used to relate the mechanisms for dehydration of alcohols and hydration of alkenes? $(b)$ What conditions favor dehydration rather than hydration reactions?
(a) The principle of microscopic reversibility states that every reaction is reversible, even if only to a microscopic extent. Furthermore, the reverse process proceds through the same intermediates and transition states, but in the opposite order.
$$
\mathrm{RCH}_2 \mathrm{CH}_2 \mathrm{OH} \stackrel{\mathrm{H}}{=} \mathrm{RCH}=\mathrm{CH}_2+\mathrm{H}_2 \mathrm{O}
$$
(b) Low $\mathrm{H}_2 \mathrm{O}$ concentration and high temperature favor alkene formation by dehydration, because the volatile alkene distills out of the reaction mixture and shifts the equilibrium. Hydration of alkenes occurs at low temperature and with dilute acid which provides a high concentration of $\mathrm{H}_2 \mathrm{O}$ as reactant.

Timothy Gailey

Numerade Educator

01:11

Problem 22

Why are dry gaseous hydrogen halides $(\mathrm{HX})$ acids and not their aqueous solutions used to prepare alkyl halides from alkenes?

Dry hydrogen halides are stronger acids and better electrophiles than the $\mathrm{H}_3 \mathrm{O}^{+}$formed in their water solutions. Furthermore, $\mathrm{H}_2 \mathrm{O}$ is a nucleophile that can react with $\mathrm{R}^{\prime}$ to give an alcohol.

Nicole Smina

Numerade Educator

04:09

Problem 23

Arrange the following alkenes in order of increasing reactivity on addition of hydrohalogen acids: (a) $\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2$, (b) $\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CH}_2$, (c) $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_3$.

The relative reactivities are directly related to the stabilities of the intermediate $\mathrm{R}^{+} \mathrm{s}$. Isobutylene, $(b)$, is most reactive because it forms the $3 \cdots\left(\mathrm{CH}_3\right)_2 \mathrm{CCH}_3$. The next-most reactive compound is 2-butene, $(c)$, which forms the $2=\mathrm{CH}_3 \overrightarrow{\mathrm{C}} \mathrm{HCH}_2 \mathrm{CH}_3$. Ethylene forms the $1 \mathrm{CH}_3 \dot{C H}_2$ and is least reactive. The order of increasing reactivity is: $(a)<(c)<(b)$.

Anupa Sharad Medhekar

Numerade Educator

01:13

Problem 24

The addition of $\mathrm{HBr}$ to some alkenes gives a mixture of the expected alkyl bromide and an isomer formed by rearrangement. Outline the mechanism of formation and structures of products from the reaction of $\mathrm{HBr}$ with (a) 3-methyl-1-butene, (b) 3,3-dimethyl-1-butene.
No matter how formed, an $\mathrm{R}^{+}$can undergo $\mathrm{H}$ : or : $\mathrm{CH}_3$ (or other alkyl) shifts to form a more stable $\mathrm{R}^{\prime+}$.

Lottie Adams

Numerade Educator

02:25

Problem 25

Compare and explain the relative rates of addition to alkenes (reactivities) of $\mathrm{HCl}, \mathrm{HBr}$ and HI.

The relative reactivity depends on the ability of $\mathrm{HX}$ to donate an $\mathrm{H}^{+}$(acidity) to form an $\mathrm{R}^{+}$in the ratecontrolling first step. The acidity and reactivity order is $\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}^{\prime}$.

Aadit Sharma

Numerade Educator

02:09

Problem 26

(a) What does each of the following observations tell you about the mechanism of the addition of $\mathrm{Br}_2$ to an alkene? (i) In the presence of a $\mathrm{Cl}^{-}$salt, in addition to the vic-dibromide, some vic-bromochloroalkane is isolated but no dichloride is obtained. (ii) With cis-2-butene only rac-2,3-dibromobutane is formed. (iii) With trans-2butene only meso-2,3-dibromobutane is produced. (b) Give a mechanism compatible with these observations.
(a) (i) $\mathrm{Br}_2$ adds in two steps. If $\mathrm{Br}_2$ added in one step, no bromo chloroalkane would be formed. Furthermore, the first step must be the addition of an electrophile (the $\mathrm{Br}^{+}$part of $\mathrm{Br}_2$ ) followed by addition of a nucleophile, which could now be $\mathrm{Br}^{-}$or $\mathrm{Cl}^{-}$. This explains why the products must contain at least one $\mathrm{Br}$. (ii) One $\mathrm{Br}$ adds from above the plane of the double bond, the second $\mathrm{Br}$ adds from below. This is an anti (trans) addition. Since a $\mathrm{Br}^{+}$can add from above to either $\mathrm{C}$, a racemic form results.
\text { (iii) This substantiates the anti addition. }
The reaction is also stereospecific because different stereoisomers give stereochemically different products, e.g. cis $\rightarrow$ racemic and trans $\rightarrow$ meso. Because of this stereospecificity, the intermediate cannot be the free carbocation $\mathrm{CH}_3 \mathrm{CHBr}_{\mathrm{C}}^{\mathrm{C}} \mathrm{HCH}_3$. The same carbocation would arise from either cis-or trans-2-butene, and the product distribution from both reactants would be identical.
(b) The open carbocation is replaced by a cyclic bridged ion having $\mathrm{Br}^{+}$partially bonded to each $\mathrm{C}$ (bromonium ion). In this way the stereochemical differences of the starting materials are retained in the intermediate. In the second step, the nucleophile attacks the side opposite the bridging group to yield the anti addition product.
$\mathrm{Br}_2$ does not break up into $\mathrm{Br}^{+}$and $\mathrm{Br}^{-}$. More likely, the $\pi$ electrons attack one of the $\mathrm{Br}$ 's, displacing the other as an anion (Fig. 6-5).

Alkendra Singh

Numerade Educator

04:45

Problem 27

Alkenes react with aqueous $\mathrm{Cl}_2$ or $\mathrm{Br}_2$ to yield vic-halohydrins, $-\mathrm{CXCOH}$. Give a mechanism for this reaction that also explains how $\mathrm{Br}_2$ and $\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CH}_2$ give $\left(\mathrm{CH}_3\right)_2 \mathrm{C}(\mathrm{OH}) \mathrm{CH}_2 \mathrm{Br}$.

The reaction proceeds through a bromonium ion [Problem $6.26(b)$ ] which reacts with the nucleophilic $\mathrm{H}_2 \mathrm{O}$ to give
<smiles>CC(C)[Ge](C)(C)C(C)(C)O</smiles>
This protonated halohydrin then loses $\mathrm{H}^{+}$to the solvent, giving the halohydrin. The partial bonds between the C's and $\mathrm{Br}$ engender $\delta+$ charges on the C's. Since the bromonium ion of 2-methylpropene has more partial positive charge on the $3^{\circ}$ carbon that on the $I^{\circ}$ carbon, $\mathrm{H}_2 \mathrm{O}$ binds to the $3^{\circ} \mathrm{C}$ to give the observed product. In general, $\mathrm{X}$ appears on the $\mathrm{C}$ with the greater number of $\mathrm{H}$ 's. The addition, like that of $\mathrm{Br}_2$ is anti because $\mathrm{H}_2 \mathrm{O}$ binds to the $\mathrm{C}$ from the side away from the side where the $\mathrm{Br}$ is positioned.

Ian Kaigh

Numerade Educator

05:12

Problem 29

(a) Suggest a mechanism for the dimerization of isobutylene, $\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CH}_2$. (b) Why does $\left(\mathrm{CH}_3\right)_3 \mathrm{C}^{+}$add to the "tail" carbon rather than to the "head" carbon? (c) Why are the Brönsted acids $\mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{HF}$ typically used as catalysts, rather than $\mathrm{HCl}, \mathrm{HBr}$, or $\mathrm{HI}$ ?

Ian Kaigh

Numerade Educator

Problem 31

Suggest a mechanism for alkane addition where the key step is an intermolecular hydride (H:) transfer.
See Steps 1 and 2 in Problem 6.29 for formation of the dimeric $\mathrm{R}^{+}$.

This intermolecular $\mathrm{H}$ : transfer forms the $\left(\mathrm{CH}_3\right)_3 \mathrm{C}^{+}$ion which adds to another molecule of $\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CH}_2$ to continue the chain. A $3^{\circ} \mathrm{H}$ usually transfers to leave a $3^2 \mathbf{R}^{+}$.

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03:33

Problem 32

Suggest a chain-propagating free-radical mechanism for addition of $\mathrm{HBr}$ in which $\mathrm{Br}$. attacks the alkene to form the more stable carbon radical.

Initiation Steps
$$
\begin{aligned}
& \mathrm{R}-\mathrm{O}-\mathrm{O}-\mathrm{R} \stackrel{\text { heat }}{\longrightarrow} 2 \mathrm{R}-\mathrm{O} \cdot(-\ddot{\mathrm{O}}-\ddot{\mathrm{O}}-\text { bond is weak }) \\
& \mathrm{RO}++\mathrm{HBr} \longrightarrow \mathrm{Br} \cdot+\mathrm{R}-\mathrm{O}-\mathrm{H}
\end{aligned}
$$

Propagation Steps For Chain Reaction
$$
\begin{aligned}
& \mathrm{CH}_3 \mathrm{CHBrC}_2 \leftrightarrow \times \mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2+\mathrm{Br} \cdot \longrightarrow \mathrm{CH}_3 \dot{\mathrm{C}} \mathrm{HCH}_2 \mathrm{Br} \\
& \left(1^{\circ}\right. \text { radical) } \\
& \left(2^{\circ}\right. \text { radical) } \\
& \mathrm{CH}_3 \dot{\mathrm{C}} \mathrm{HCH}_2 \mathrm{Br}+\mathrm{HBr} \longrightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}+\mathrm{Br} \text {. } \\
&
\end{aligned}
$$
The $\mathrm{Br}$ - generated in the second propagation step continues the chain.

Ian Kaigh

Numerade Educator

11:54

Problem 33

Give the products formed on ozonolysis of $($ a $) \mathrm{H}_2 \mathrm{C}=\mathrm{CHCH}_2 \mathrm{CH}_3$, (b) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHCH}_3$, (c) $\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CHCH}_2 \mathrm{CH}_3$, (d $)$ cyclobutene, (e) $\mathrm{H}_2 \mathrm{C}=\mathrm{CHCH}_2 \mathrm{CH}=\mathrm{CHCH}_3$.

To get the correct answers, erase the double bond and attach $\mathrm{a}=\mathrm{O}$ to each of the formerly double-bonded C's. The total numbers of $\mathrm{C}$ 's in the carbonyl products and in the alkene reactant must be equal.
(a) $\mathrm{H}_2 \mathrm{C}=\mathrm{O}+\mathrm{O}=\mathrm{CHCH}_2 \mathrm{CH}_3$.
(b) $\mathrm{CH}_3 \mathrm{CH}=\mathrm{O}$; the alkene is symmetrical and only one carbonyl compound is formed.
(c) $\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{O}+\mathrm{O}=\mathrm{CHCH}_2 \mathrm{CH}_3$.
(d) $\mathrm{O}=\mathrm{CHCH}_2 \mathrm{CH}_2 \mathrm{CH}=\mathrm{O}$; a cycloalkene gives only a dicarbonyl compound.
(e) $\mathrm{H}_2 \mathrm{C}=\mathrm{O}+\mathrm{O}=\mathrm{CHCH}_2 \mathrm{CH}=\mathrm{O}+\mathrm{O}=\mathrm{CHCH}_3$. Noncyclic polyenes give a mixture of monocarbonyl compounds formed from the terminal $\mathrm{C}$ 's and dicarbonyl compounds from the internal doubly bonded C's.

Ian Kaigh

Numerade Educator

04:02

Problem 34

Deduce the structures of the following alkenes.
(a) An alkene $\mathrm{C}_{10} \mathrm{H}_{20}$ on ozonolysis yields only
<smiles>CCCC(C)=O</smiles>
(b) An alkene $\mathrm{C}_9 \mathrm{H}_{18}$ on ozonolysis gives
<smiles>CC(C)(C)C=O</smiles>
and
<smiles>CCC(C)=O</smiles>
(c) A compound $\mathrm{C}_8 \mathrm{H}_{14}$ adds one mole of $\mathrm{H}_2$ and forms an ozonolysis the dialdehyde
<smiles>CC(C=O)CCC(C)C=O</smiles>
(d) A compound $\mathrm{C}_{\mathrm{R}} \mathrm{H}_{12}$ adds two moles of $\mathrm{H}_2$ and undergoes ozonolysis to give two moles of the dialdehyde $\mathrm{O}=\mathrm{CHCH}_2 \mathrm{CH}_2 \mathrm{CH}=\mathrm{O}$.
(a) The formation of only one carbonyl compound indicates that the alkene is symmetrical about the double bond. Write the structure of the ketone twice so that the $\mathrm{C}=\mathrm{O}$ groups face each other. Replacement of the two O's by a double bond gives the alkene structure.

Dalton Hilovsky

Numerade Educator

02:03

Problem 35

Use the concepts of $(a)$ resonance and $(b)$ extended $\pi$ orbital overlap (delocalization) to account for the extraordinary stability of the allyl-type radical.
<smiles>CC(C)=C(C)C(C)(C)C</smiles>
(a) Two equivalent resonance structures can be written:
therefore the allyl-type radical has considerable resonance energy (Section 2.7) and is relatively stable.
(b) The three C's in the allyl unit are $s p^2$-hybridized and each has a $p$ orbital lying in a common plane (Fig. 6-7). These three $p$ orbitals overlap forming an extended $\pi$ system, thereby delocalizing the odd electron. Such delocalization stabilizes the allyl-type free radical.

Adriano Chikande

Numerade Educator

01:10

Problem 36

Designate the type of cach set of $\mathrm{H}$ 's in $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_2 \mathrm{CH}_2-\mathrm{CH}\left(\mathrm{CH}_3\right)_2$ (e.g. $3^n$, allylic, etc.) and show their relative reactivity toward a $\mathrm{Br}$ - atom. using (1) for the most reactive, (2) for the next, etc.
Labeling the $\mathrm{H}$ 's as
$$
\stackrel{(a)}{\mathrm{CH}_3} \mathrm{CH}^{(b)}=\stackrel{(b)}{\mathrm{C}} \stackrel{(c)}{\mathrm{HCH}_2} \mathrm{CH}_2 \stackrel{(d)}{ } \mathrm{CH}^{(e)}\left(\mathrm{CH}_3\right)_2
$$
we have: (a) $1^{\circ}$, allylic (2); (b) vinylic (6); (c) 2 , allylic (1); (d) $2^{\prime}(4) ;(e) 3^{\circ}(3) ;(f) 1^{\circ}(5)$.

Narayan Hari

Numerade Educator

01:56

Problem 37

Write structures for the following:
(a) 2,3-dimethyl-2-pentene
(b) 4-chloro-2,4-dimethyl-2-pentene
(c) allyl bromide
(d) 2,3-dimethylcyclohexene
(e) 3-isopropyl-1-hexene
(f) 3-isopropyl-2,6-dimethyl-3-heptene

Adriano Chikande

Numerade Educator

Problem 38

(a) Give structural formulas and systematic names for all alkenes with the molecular formula $\mathrm{C}_6 \mathrm{H}_{12}$ that exist as stereomers. (b) Which stereomer has the lowest heat of combustion (is the most stable), and which is the least stable?
(a) For a molecule to possess stereoisomers, it must be chiral, exhibit geometric isomerism, or both. There are five such constitutional isomers of $\mathrm{C}_6 \mathrm{H}_{12}$. Since double-bonded C's cannot be chiral, one of the remaining four $\mathrm{C}$ 's must be chiral. This means that a sec-butyl group must be attached to the $\mathrm{C}=\mathrm{C}$ group, and the enantiomers are:
A molecule with a terminal $=\mathrm{CH}_2$ cannot have geometric isomers. The double bond must be internal and is first placed between $\mathrm{C}^2$ and $\mathrm{C}^3$ to give three more sets of stereoisomers:

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02:20

Problem 39

Write structural formulas for the organic compounds designated by a ? and show the stereochemistry where requested.

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

Marissa Turner

Numerade Educator

01:36

Problem 40

Draw an enthalpy-reaction progress diagram for addition of $\mathrm{Br}_2$ to an alkene. See Fig. 6-8.

Benjamin Angeles

Numerade Educator

02:23

Problem 41

Write the initiation and the propagation steps for a free-radical-catalyzed (RO-) addition of $\mathrm{CH}_3 \mathrm{CH}=\mathrm{O}$ to 1 -hexene to form methyl $n$-hexyl ketone.

Madeline Currie

Numerade Educator

02:08

Problem 42

Suggest a radical mechanism to account for the interconversion of cis and trans isomers by heating with $\mathrm{I}_2$.
$\mathrm{I}_2$ has a low bond dissociation energy $(151 \mathrm{~kJ} / \mathrm{mol})$ and forms $2 \mathrm{I} \cdot$ on heating. $\mathrm{I} \cdot$ adds to the $\mathrm{C}=\mathrm{C}$ to form a carbon radical which rotates about its sigma bond and assumes a different conformation. However, the $\mathrm{C}-\mathrm{I}$ bond is also weak $(235 \mathrm{~kJ} / \mathrm{mol})$ and the radical loses $\mathrm{I}$. under these conditions. The double bond is reformed and the two conformations produce a mixture of cis and trans isomers.

Natalie Johns

Numerade Educator

03:09

Problem 43

Write structures for the products of the following polar addition reactions:
(a) $\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CHCH}_3+\mathrm{I}-\mathrm{Cl} \longrightarrow$ ?
(b) $\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CH}_2+\mathrm{HSCH}_3 \longrightarrow$ ?
(c) $\left(\mathrm{CH}_3\right)_3 \stackrel{+}{\mathrm{N}}-\mathrm{CH}=\mathrm{CH}_2+\mathrm{HI} \longrightarrow$ ?
(d) $\mathrm{H}_2 \mathrm{C}=\mathrm{CHCF}_3+\mathrm{HCl} \longrightarrow$ ?

Benjamin Angeles

Numerade Educator

Problem 44

Explain the following observations: (a) $\mathrm{Br}_2$ and propene in $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$ gives not only $\mathrm{BrCH}_2 \mathrm{CHBrCH}_3$ but also $\mathrm{BrCH}_2 \mathrm{CH}\left(\mathrm{OC}_2 \mathrm{H}_5\right) \mathrm{CH}_3$. (b) lsobutylene is more reactive than 1 -butene towards peroxide-catalyzed addition of $\mathrm{CCl}_4$. (c) The presence of $\mathrm{Ag}^{\perp}$ salts enhance the solubility of alkenes in $\mathrm{H}_2 \mathrm{O}$.
(a) The intermediate bromonium ion reacts with both $\mathrm{Br}^{-}$and $\mathrm{C}_2 \mathrm{H}_5 \dddot{\mathrm{O}} \mathrm{H}$ as nucleophiles to give the two products. (b) The more stable the intermediate free radical, the more reactive the alkene. $\mathrm{H}_2 \mathrm{C}=\mathrm{CHCH}_2 \mathrm{CH}_3$ adds $\cdot \mathrm{CCl}_3$ to give the less stable 2 radical $\mathrm{Cl}_3 \mathrm{CCH}_2 \mathrm{CHCH}_2 \mathrm{CH}_3$. whereas $\mathrm{H}_2 \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_3\right)_2$ reacts to give the more stable $3^2$ radical $\mathrm{Cl}_3 \mathrm{CCH}_2 \dot{\mathrm{C}}\left(\mathrm{CH}_3\right)_2$. (c) Ag coordinates with the alkene by $p-d \pi$ bonding to give an ion similar to bromonium ion, but more stable:

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Problem 46

Outline the steps needed for the following syntheses in reasonable yield. Inorganic reagents and solvents may also be used. (a) 1-Chloropentane to 1.2-dichloropentane. (b) 1-Chloropentane to 2-chloro-pentane. (c) 1-ChLoropentane to 1-bromopentane. (d) 1-Bromobutane to 1.2-dihydroxybutane. (e) Isobutyl chloride to
<smiles>CC(C)(C)CC(C)(C)I</smiles>
Syntheses are best done by working backwards, keeping in mind your starting material.
(a) The desired product is a vic-dichloride made by adding $\mathrm{Cl}_2$ to the appropriate alkene, which in turn is made by dehydrochlorination the starting material.
$$
\mathrm{ClCH}_2 \mathrm{CH}_2 \mathrm{Cl}_2 \mathrm{CH}_2 \mathrm{CH}_3 \underset{\mathrm{KOH}}{\stackrel{\mathrm{al}}{\longrightarrow}} \mathrm{H}_2 \mathrm{C}=\mathrm{CHCl}_2 \mathrm{CH}_2 \mathrm{CH}_3 \stackrel{\stackrel{\circ}{\longrightarrow}}{\longrightarrow} \mathrm{ClCH}_2 \mathrm{CHClCH}_2 \mathrm{CH}_2 \mathrm{CH}_3
$$
(b) To get a pure product add $\mathrm{HCl}$ to $\mathrm{I}$-pentene as made in part (a).
$$
\mathrm{H}_2 \mathrm{C}=\mathrm{CHCH}_2 \mathrm{CH}_2 \mathrm{CH}_3+\mathrm{HCl} \longrightarrow \mathrm{H}_3 \mathrm{CCHClCH}_2 \mathrm{CH}_2 \mathrm{CH}_3
$$
(c) An anti-Markovnikov addition of $\mathrm{HBr}$ to 1-pentene [part $(a)]$.
$$
\mathrm{H}_2 \mathrm{C}=\mathrm{CHCH}_2 \mathrm{CH}_2 \mathrm{CH}_3+\mathrm{HBr} \stackrel{\text { peroxide }}{\longrightarrow} \mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3
$$
(d) Glycols are made by mild oxidation of alkenes.
$$
\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3 \underset{\mathrm{KOll}}{\stackrel{\text { alc }}{\longrightarrow}} \mathrm{H}_2 \mathrm{C}=\mathrm{CHCH}_2 \mathrm{CH}_3 \stackrel{\mathrm{KMnO}_4}{\mathrm{RT}}-\mathrm{HOCH}_2 \mathrm{CHOHCH}_2 \mathrm{CH}_3
$$
(e) The product has twice as many C's as does the starting material. The skeleton of C's in the product corresponds to that of the dimer of $\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CH}_2$.

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07:50

Problem 47

Show how propene can be converted to $(a)$ 1,5-hexadiene, (b) 1-bromopropene, (c) 4-methyl-1pentene.
(a) $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2 \stackrel{\mathrm{Cl}_2, 500^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{ClCH}_2 \mathrm{CH}=\mathrm{CH}_2 \frac{\text { 1. } \mathrm{Li} \text {. 2 } \text {. } \mathrm{Cul}}{3 . \mathrm{ClCH}_2 \mathrm{CH}=\mathrm{CH}_2} \mathrm{H}_2 \mathrm{C}=\mathrm{CHCH}_2 \mathrm{CH}_2 \mathrm{CH}=\mathrm{CH}_2$
(b) $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2 \stackrel{\mathrm{Br}_2\left(\mathrm{CCl}_4\right)}{-} \mathrm{CH}_3 \mathrm{CHBrCH}_2 \mathrm{Br} \stackrel{\text { alc. }}{\mathrm{KOH}}-\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHBr}$
(Little $\mathrm{CH}_3 \mathrm{CBr}=\mathrm{CH}_2$ is formed because the $2^{\circ} \mathrm{H}$ of $-\mathrm{CH}_2 \mathrm{Br}$ is more acidic than the $3^{\circ} \mathrm{H}$ of $-\stackrel{1}{\mathrm{C}} \mathrm{HBr}$.)
(a)

Colton K

Numerade Educator

04:18

Problem 48

(a) $\mathrm{Br}_2$ is added to $(S)-\mathrm{H}_2 \mathrm{C}=\dot{\mathrm{C}} \mathrm{HC} H \mathrm{BrCH}_3$. Give Fischer projections and $R . S$ designations for the products. Are the products optically active? (b) Repeat $(a)$ with $\mathrm{HBr}$.
(a) $\mathrm{C}^2$ becomes chiral and the configuration of $\mathrm{C}^3$ is unchanged. There are two optically active diastereomers of 1,2,3-tribromobutane. It is best to draw formulas with $\mathrm{H}$ 's on vertical lines.
(b) There are two diastereomers of 2,3-dibromobutane:

David Taylor

Numerade Educator

00:29

Problem 49

Polypropylene can be synthesized by the acid-catalyzed polymerization of propylene. $(a)$ Show the first three steps. (b) Indicate the repeating unit (mer).

Mishal Gul

Numerade Educator

02:53

Problem 50

List the five kinds of reactions of carbocations and give an example of each.
(a) They combine with nucleophiles.
(b) As strong acids they lose a vicinal $\mathrm{H}$ [deprotonate], to give an alkene.
(c) They rearrange to give a more stable carbocation.
(d) They add to an alkene to give a carbotation with higher molecular weight.
electrophile nucleophile
(e) They may remove : $\mathrm{H}$ (a hybride transfer) from a tertiary position in an alkane.

David Collins

Numerade Educator

03:09

Problem 51

From propene, prepare (a) $\mathrm{CH}_3 \mathrm{CHDCH}_3$, (b) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{D}$.

(a) $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2+\mathrm{HCl} \longrightarrow \mathrm{CH}_3 \mathrm{CHClCH}_3 \stackrel{\mathrm{Mg}}{\longrightarrow} \mathrm{CH}_3 \mathrm{CHMgClCH}_3 \stackrel{\mathrm{D} 2 \mathrm{O}}{\longrightarrow} \mathrm{CH}_3 \mathrm{CHDCH}_3$ or propene $+\mathrm{B}_2 \mathrm{D}_6 \longrightarrow\left(\mathrm{CH}_3 \mathrm{CHDCH}_2\right)_3 \mathrm{~B} \stackrel{\left(\mathrm{CH}_3 \mathrm{COC}\right) ! I}{\longrightarrow}$ product
(b) $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2 \stackrel{\mathrm{B}_2 \mathrm{H}_6}{\longrightarrow}\left(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2\right)_3 \mathrm{~B} \stackrel{\mathrm{CH}_3 \mathrm{COOD}}{\longrightarrow} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{D}$

Benjamin Angeles

Numerade Educator

04:18

Problem 53

Give the configuration, stereochemical designation and $R, S$ specification for the indicated tetrahydroxy products.

(a) syn Addition of encircled OH's:
(b) anti Addition of encircled $\mathrm{OH}$ 's:
(c) syn Addition; same products as part (b).
(d) syn Addition; same products as part (a).
(e) syn Addition:
One optically active stereoisomer is formed.
(f) anti Addition:
Two optically active diastereomers.

David Taylor

Numerade Educator

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Problem 54

Describe $(a)$ radical-induced, and $(b)$ anion-induced, polymerization of alkenes. (c) What kind of alkenes undergo anion-induced polymerization?
(a) See Problem 6.32 for formation of a free-radical initiator, $\mathrm{RO}$, which adds according to the Markovnikov rule.

The polymerization can terminate when the free-radical terminal $\mathrm{C}$ of a long chain forms a bond with the terminal $\mathrm{C}$ of another long chain (combination), $\mathrm{RC} \cdot+\cdot \mathrm{CR}^{\prime} \longrightarrow \mathrm{RC}-\mathrm{CR}^{\prime}$. Termination may also occur when the terminal free-radical C's of two long chains disproportionate, in a sort of auto-redox reaction. One $\mathrm{C}$ picks off an $\mathrm{H}$ from the $\mathrm{C}$ of the other chain, to give an alkane at one chain end and an alkene group at the other chain end:
(b) Typical anions are carbanions, R:-, generated from lithium or Grignard organometallics.
These types of polymerizations also have stereochemical consequences.
(c) Since alkenes do not readily undergo anionic additions, the alkene must have a functional group $\mathrm{X}$ (such as $-\mathrm{CN}$ or $\mathrm{O}=\mathrm{COR}$ ) on the $\mathrm{C}=\mathrm{C}$ that can stabilize the negative charge:

Lainey Roebuck

Numerade Educator

11:54

Problem 55

Alkenes undergo oxidative cleavage with acidic $\mathrm{KMnO}_4$ and as a result each $\mathrm{C}$ of the $\mathrm{C}=\mathrm{C}$ ends up in a molecule in its highest oxidation state. Give the products resulting from the oxidative cleavage of $(a)$ $\mathrm{H}_2 \mathrm{C}=\mathrm{CHCH}_2 \mathrm{CH}_3$, (b) $(E)$ - or $(Z)-\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_3,($ c $)\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CHCH}_2 \mathrm{CH}_3$.
(a) $\mathrm{H}_2 \mathrm{C}=\mathrm{CHCH}_2 \mathrm{CH}_3 \stackrel{\mathrm{KMnO}_4}{\longrightarrow} \mathrm{CO}_2+\mathrm{HOOCCH}_2 \mathrm{CH}_3 \quad\left(\mathrm{H}_2 \mathrm{C}=\right.$ gives $\left.\mathrm{CO}_2\right)$
(b) $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_3 \stackrel{\mathrm{KMnO}_4}{\longrightarrow} \mathrm{CH}_3 \mathrm{COOH}+\mathrm{HOOCCH}(\mathrm{RCH}=$ gives $\mathrm{RCOOH})$
(c)

Ian Kaigh

Numerade Educator