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Schaum's Outline of Organic Chemistry

George Hademenos, George Hademenos

Chapter 8

ALKYNES AND DIENES - all with Video Answers

Educators

Chapter Questions

Problem 1

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Problem 2

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03:34

Problem 2

Apply the MO theory to the $\pi$ system of ethene.
Each of the doubly-bonded C's, $\mathrm{C}=\mathrm{C}$, has a $p$ AO. These two $p$ AO's provide two molecular orbitals, a lowerenergy bonding $\mathrm{MO}$ and a higher-energy antibonding $\mathrm{MO}^*$. Each $p \mathrm{AO}$ has one electron, giving two electrons for placement in the $\pi$ molecular orbitals. Molecular orbitals receive electrons in the order of their increasing energies, with no more than two of opposite spins in any given molecular orbital. For ethene, the two $p$ electrons, shown as $\uparrow$ and $\downarrow$, are placed in the bonding $\mathrm{MO}(\pi)$; the antibonding $\mathrm{MO}^*\left(\pi^*\right)$ is devoid of electrons. Note the simplification of showing only the signs of the upper lobes of the interacting $p$ orbitals. The stationary waves, with any nodes, are shown superimposed on the energy levels.

Susan Hallstrom
Susan Hallstrom
Numerade Educator
05:22

Problem 3

Draw models of $(a) s p$ hybridized $\mathrm{C}$ and $(b) \mathrm{C}_2 \mathrm{H}_2$ to show bonds formed by orbital overlap.
(a) See Fig. 8-1(a). Only one of three $p$ orbitals of $\mathrm{C}$ is hybridized. The two unhybridized $p$ orbitals $\left(p_z\right.$ and $\left.p_y\right)$ are at right angles to each other and also to the axis of the $s p$ hybrid orbitals.
(b) See Fig. 8-1(b). Sidewise overlap of the $p_y$ and $p_z$ orbitals on each C forms the $\pi_y$ and $\pi_z$ bonds, respectively.

VS
Vivek Singh
Numerade Educator
03:42

Problem 4

Why is the $\mathrm{C} \equiv \mathrm{C}$ distance $(0.120 \mathrm{~nm})$ shorter than the $\mathrm{C}=\mathrm{C}(0.133 \mathrm{~nm})$ and $\mathrm{C}-\mathrm{C}(0.154 \mathrm{~nm})$.
The carbon nuclei in $\mathrm{C} \equiv \mathrm{C}$ are shielded by six electrons (from three bonds) rather than by four or two electrons as in $\mathrm{C}=\mathrm{C}$ or $\mathrm{C}-\mathrm{C}$, respectively. With more shielding electrons present, the $\mathrm{C}$ 's of $-\mathrm{C}=\mathrm{C}-$ can get closer, thereby affording more orbital overlap and stronger bonds.

LJ
Lena Jake
Numerade Educator
03:54

Problem 5

Explain how the orbital picture of $-\mathrm{C}=\mathrm{C}-$ accounts for $(a)$ the absence of geometric isomers in $\mathrm{CH}_3 \mathrm{C}=\mathrm{CC}_2 \mathrm{H}_5 ;($ b) the acidity of an acetylenic $\mathrm{H}$, e.g.
$$
\mathrm{HC}=\mathrm{CH}+\mathrm{NH}_2^{-} \longrightarrow \mathrm{HC} \equiv \mathrm{C}^{-}+\mathrm{NH}_3 \quad\left(\mathrm{p} K_a=25\right)
$$
(a) The $s p$ hybridized bonds are linear, ruling out cis-trans isomers in which substituents must be on different sides of the multiple bond.
(b) We apply the principle: "The more $s$ character in the orbital used by the $\mathrm{C}$ of the $\mathrm{C}-\mathrm{H}$ bond, the more acidic is the H." Therefore the order of acidity of hydrocarbons is
<smiles>C#CCC(C)C</smiles>

Anupa Sharad Medhekar
Anupa Sharad Medhekar
Numerade Educator
01:43

Problem 6

(a) Relate the observed $\mathrm{C}-\mathrm{H}$ and $\mathrm{C}-\mathrm{C}$ bond lengths and bond energies given in Table $8-1$ in terms of the hybrid orbitals used by the $\mathrm{C}$ 's involved. (b) Predict the relative $\mathrm{C}-\mathrm{C}$ bond lengths in $\mathrm{CH}_3 \mathrm{CH}_3$, $\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_2$, and $\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}$.
$$
\begin{array}{|l|c|c|c|}
\hline \text { Compound } & \text { Bond } & \text { Bond Length, nm } & \text { Bond Energy, kJ/mol } \\
\hline \text { (1) } \mathrm{CH}_3-\mathrm{CH}_3 & -\mathrm{C}-\mathrm{H} & 0.110 & 410 \\
(2) \quad \mathrm{CH}_2=\mathrm{CH}_2 & =\mathrm{C}-\mathrm{H} & 0.108 & 423 \\
(3) \mathrm{H}-\mathrm{C}=\mathrm{C}-\mathrm{H} & =\mathrm{C}-\mathrm{H} & 0.106 & 460 \\
\text { (4) } \mathrm{CH}_3-\mathrm{CH}_3 & \mathrm{C}-\mathrm{C}- & 0.154 & 356 \\
\text { (5) } \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2 & \mathrm{C}-\mathrm{C}= & 0.151 & 377 \\
\text { (6) } \mathrm{CH}_3-\mathrm{C} \equiv \mathrm{C}-\mathrm{H} & \mathrm{C}-\mathrm{C} \equiv & 0.146 & 423 \\
\hline
\end{array}
$$
Bond energy increases as bond length decreases; the shorter bond length makes for greater orbital overlap and a stronger bond.
(a) The hybrid nature of $\mathrm{C}$ is: (1) $\mathrm{C}_{s p^3}-\mathrm{H}_s$, (2) $\mathrm{C}_{s p^2}-\mathrm{H}_s$, (3) $\mathrm{C}_{s p}-\mathrm{H}_s$, (4) $\mathrm{C}_{s p^3}-\mathrm{C}_{s p^3}$, (5) $\mathrm{C}_{s p^3}-\mathrm{C}_{s p^2}$ and (6) $\mathrm{C}_{s p} 3-\mathrm{C}_{s p}$. In going from (1) to (3) the $\mathrm{C}-\mathrm{H}$ bond length decreases as the $s$ character of the hybrid orbital used by $\mathrm{C}$ increases. The same situation prevails for the $\mathrm{C}-\mathrm{C}$ bond in going from (4) to (6). Bonds to $\mathrm{C}$ therefore become shorter as the $s$ character of the hybridized orbital used by $\mathrm{C}$ increases.
(b) The hybrid character of the $\mathrm{C}$ 's in the $\mathrm{C}-\mathrm{C}$ bond is: for $\mathrm{CH}_3-\mathrm{CH}_3, \mathrm{C}_{s p}{ }^3-\mathrm{C}_{s p} 3 ; \mathrm{H}_2 \mathrm{C}=\mathrm{CH}-\mathrm{CH}_3=\mathrm{CH}_2$, $\mathrm{C}_{s p^2}-\mathrm{C}_{s p^2}$; and $\mathrm{H}-\mathrm{C}=\mathrm{C}-\mathrm{C}=\mathrm{C}-\mathrm{H}, \mathrm{C}_{s p}-\mathrm{C}_{s p}$. Bond length becomes shorter as $s$ character increases and hence relative $\mathrm{C}-\mathrm{C}$ bond lengths should decrease in the order

Hitendra Singh
Hitendra Singh
Numerade Educator
05:36

Problem 7

Explain why $\mathrm{CH}_3 \mathrm{CHBrCH}_2 \mathrm{Br}$ does not react with $\mathrm{KOH}$ to give $\mathrm{CH}_2=\mathrm{CHCH}_2 \mathrm{Br}$.
In E2 eliminations the more acidic $\mathrm{H}$ is removed preferably. The inductive effect of the Br's increases the acidities of the H's on the C's to which the Br's are bonded. To get this product the less acidic $\mathrm{H}$ (one of the $\mathrm{CH}_3$ group) must be removed.

Nima Gharibi
Nima Gharibi
Numerade Educator
02:42

Problem 8

Outline a synthesis of propyne from isopropyl or propyl bromide.
The needed vic-dihalide is formed from propene, which is prepared from either of the alkyl halides.

Ian Kaigh
Ian Kaigh
Numerade Educator
02:08

Problem 9

Synthesize the following compounds from $\mathrm{HC} \equiv \mathrm{CH}$ and any other organic and inorganic reagents (do not repeat steps): (a) 1-pentyne, (b) 2-hexyne.
(a) $\mathrm{H}-\mathrm{C}=\mathrm{C}-\mathrm{H} \stackrel{\mathrm{NaNH}_2}{\longrightarrow} \mathrm{H}-\mathrm{C} \equiv \overline{\mathrm{C}}: \stackrel{+}{\mathrm{N} a} \stackrel{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 1}{\longrightarrow} \mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$
(b) $\stackrel{+}{\mathrm{N}} \mathrm{a}: \overrightarrow{\mathrm{C}} \equiv \mathrm{C}-\mathrm{H} \stackrel{\mathrm{CH}_3 \mathrm{I}}{\longrightarrow} \mathrm{CH}_3-\mathrm{C} \equiv \mathrm{C}-\mathrm{H} \stackrel{\mathrm{NaNH}_2}{\longrightarrow} \mathrm{CH}_3-\mathrm{C} \equiv \stackrel{-\mathrm{C}}{\longrightarrow} \stackrel{+}{\mathrm{Na}} \stackrel{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{I}}{\longrightarrow} \mathrm{CH}_3-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$

Anand Jangid
Anand Jangid
Numerade Educator
02:33

Problem 10

Industrially, acetylene is made from calcium carbide, $\mathrm{CaC}_2+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{HC} \equiv \mathrm{CH}+\mathrm{Ca}(\mathrm{OH})_2$. Formulate the reaction as a Brönsted acid-base reaction.
The carbide anion $\mathrm{C}_2^{2-}$ is the base formed when $\mathrm{HC}=\mathrm{CH}$ loses two $\mathrm{H}^{+}$s.

Ronald Prasad
Ronald Prasad
Numerade Educator
02:21

Problem 11

In terms of the mechanism, explain why alkynes are less reactive than alkenes towards electrophilic addition of, e.g., $\mathrm{HX}$ or $\mathrm{BR}_2$.

The mechanism of electrophilic addition is similar for alkenes and alkynes. When HX adds to a triple bond the intermediate is a carbocation having a positive charge on an $s p$-hybridized $\mathrm{C}$ atom,
<smiles>CC=CC</smiles>

This vinyl-type carbocation is less stable than its analog formed from an alkene, which has the positive charge on an $s p^2$-hybridized $\mathrm{C}$ atom,
<smiles>C[C](C)C(C)C</smiles>
An addendum such as $\mathrm{Br}_2$ forms an intermediate bromonium-type ion
<smiles></smiles>
In this ion some positive charge is dispersed to the C's, which, because of their $s p$-like hybrid character, are less able to bear the positive charge than the $s p^2 \mathrm{C}$ 's in the alkene's bromonium ion. Such situations cause alkynes to be less reactive than alkenes toward $\mathrm{Br}_2$.

Dr. Satish Ingale
Dr. Satish Ingale
Numerade Educator
02:21

Problem 12

Alkynes differ from alkenes in adding nucleophiles such as $\mathrm{CN}^{-}$. Explain.
The intermediate carbanion from addition of $\mathrm{CN}^{-}$to an alkyne has the unshared electron pair on an $s p^2$ hybridized $\mathrm{C}$. It is more stable and is formed more readily than the $s p^3$-hybridized carbanion formed from a nucleophile and an alkene.

Dr. Satish Ingale
Dr. Satish Ingale
Numerade Educator
03:12

Problem 13

Dehydrohalogenation of 3-bromohexane gives a mixture of cis-2-hexene and trans-2-hexene. How can this mixture be converted to pure $(a)$ cis-2-hexene? $(b)$ trans-2-hexene?
Relatively pure alkene geometric isomers are prepared by stereoselective reduction of alkynes.
(a) Hydrogenation of 2-hexyne with Lindlar's catalyst gives $98 \%$ cis-2-hexene.
$$
\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_2 \mathrm{CH}_2 \mathrm{CH}_3
$$
cis- and trans-2-Hexene
<smiles>CCCC(Br)Br</smiles>
(b) Reduction of 2-hexyne with $\mathrm{Na}$ in liquid $\mathrm{NH}_3$ gives the trans product.

Lottie Adams
Lottie Adams
Numerade Educator
03:25

Problem 14

Outline steps for the conversion of $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}$ to (a) $\mathrm{CH}_3 \mathrm{CBr}=\mathrm{CH}_2$, (b) $\mathrm{CH}_3 \mathrm{CCl}_2 \mathrm{CH}_3$, (c) $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHBr}$.
As usual, we think backward (the retrosynthetic approach). Each product is made from $\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CH}$, which in turn is synthesized from $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2$.

Lottie Adams
Lottie Adams
Numerade Educator
03:55

Problem 15

Will the following compounds react? Give any products and the reason for their formation.
(a) $\mathrm{CH}_3-\mathrm{C}=\mathrm{C}-\mathrm{H}+$ aq. $\mathrm{Na}^{+} \mathrm{OH}^{-} \longrightarrow$
(b) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{C}-\mathrm{MgI}+\mathrm{CH}_3 \mathrm{OH} \longrightarrow$
(c) $\mathrm{CH}_3 \mathrm{C}=\mathrm{C}^{-} \mathrm{Na}^{+}+\mathrm{NH}_4^{+} \longrightarrow$
(a) No. The products would be the stronger acid $\mathrm{H}_2 \mathrm{O}$ and the stronger base $\mathrm{CH}_3 \mathrm{C}=\mathrm{C}$ :-
(b) Yes. The products are the weaker acid $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{C}: \mathrm{H}$ and the weaker base $\mathrm{Mgl}\left(\mathrm{OCH}_3\right)$.
(c) Yes. The products are the weaker acid propyne and the weaker base $\mathrm{NH}_3$.

Lottie Adams
Lottie Adams
Numerade Educator
02:51

Problem 16

Deduce the structure of a $\mathrm{C}_5 \mathrm{H}_8$ compound which forms a precipitate with $\mathrm{Ag}^{+}$and is reduced to 2-methylbutane.

The precipitate shows an acetylene bond at the end of a chain with an acidic $\mathrm{H}$. With $-\mathrm{C} \equiv \mathrm{CH}$ the other three carbons must be present, as a $\left(\mathrm{CH}_3\right)_2 \mathrm{CH}-$ group, because of reduction of $\left(\mathrm{CH}_3\right)_2 \mathrm{CH}-\mathrm{C} \equiv \mathrm{CH}$ to $\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}_2 \mathrm{CH}_3$.

Cotton Starr
Cotton Starr
Numerade Educator
01:20

Problem 17

Name by the IUPAC method and classify as cumulated, conjugated, or isolated:
(a) $\mathrm{H}_2 \mathrm{C}=\mathrm{CH}-\mathrm{CH}=\mathrm{CHCH}_3$
(b) (b)
<smiles>C=CCC(=CCC)CC</smiles>
(c) $\mathrm{H}_2 \mathrm{C}=\mathrm{C}=\mathrm{CH}_2$
(d) $\mathrm{H}_2 \mathrm{C}=\mathrm{CH}-\mathrm{CH}=\mathrm{CHCH}=\mathrm{CH}_2$
(a) 1,3-Pentadiene. Conjugated diene since it has alternating double and single bonds, i.e., $-\mathrm{C}=\mathrm{C}-\mathrm{C}=\mathrm{C}-$. (b) 4Ethyl-1,4-heptadiene. Isolated diene since the double bonds are separated by at least one $s p^3$-hybridized $\mathrm{C}$, i.e., $-\mathrm{C}=\mathrm{C}-\left(\mathrm{CH}_2\right)_n-\mathrm{C}=\mathrm{C}-$. (c) 1,2-Propadiene (allene). Cumulated diene since 2 double bonds are on the same $\mathrm{C}$, i.e., $-\mathrm{C}=\mathrm{C}=\mathrm{C}-$. (d) 1,3,5-Hexatriene. Conjugated since it has alternating single and double bonds.

Lottie Adams
Lottie Adams
Numerade Educator
01:15

Problem 18

Compare the stabilities of the three types of dienes from the following heats of hydrogenation, $\Delta H_h$ (in $\mathrm{kJ} / \mathrm{mol}$ ). (For comparison, $\Delta H_h$ for 1-pentene is -126 .)
Conjugated
<smiles>C=CC=CC</smiles>
Isolated
$$
\mathrm{H}_2 \mathrm{C}=\mathrm{CH}-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}_2-252
$$
1,4-Pentadiene
Cumulated
$$
\mathrm{H}_2 \mathrm{C}=\mathrm{C}=\mathrm{CH}-\mathrm{CH}_2 \mathrm{CH}_3
$$
$-297$
1,2-Pentadiene
The calculated $\Delta H_h$, assuming no interaction between the double bonds, is $2(-126)=-252$. The more negative the observed value of $\Delta H_h$ compared to -252 , the less stable the diene; the less negative the observed value, the more stable the diene. Conjugated dienes are most stable and cumulated dienes are least stable; under the proper conditions allenes tend to rearrange to conjugated dienes.

Nikhil Choudhary
Nikhil Choudhary
Numerade Educator
01:01

Problem 19

Give steps for the conversion $\mathrm{HC} \equiv \mathrm{CCH}_2 \mathrm{CH}_2 \mathrm{CH}_3 \longrightarrow \mathrm{H}_2 \mathrm{C}=\mathrm{CH}-\mathrm{CH}=\mathrm{CHCH}_3$.
$$
\begin{aligned}
\mathrm{HC}=\mathrm{CCH}_2 \mathrm{CH}_2 \mathrm{CH}_3 \stackrel{\mathrm{H}_2 / \mathrm{Pt}(\mathrm{Pb})}{\longrightarrow} \mathrm{H}_2 \mathrm{C}=\mathrm{CHCH}_2 \mathrm{CH}_2 \mathrm{CH}_3 \underset{\text { (allylic substitution) }}{\stackrel{\mathrm{Cl}_2}{\longrightarrow}} \\
\mathrm{H}_2 \mathrm{C}=\mathrm{CHCHClCH}_2 \mathrm{CH}_3 \stackrel{\text { alc, } \mathrm{KOH}}{\longrightarrow} \mathrm{H}_2 \mathrm{C}=\mathrm{CHCH}_2=\mathrm{CHCH}_3
\end{aligned}
$$

Narayan Hari
Narayan Hari
Numerade Educator
01:13

Problem 20

Account for the stability of conjugated dienes by $(a)$ extended $\pi$ bonding. $(b)$ resonance theory.
(a) The four $p$ orbitals of conjugated dienes are adjacent and parallel (Fig. 8-2) and overlap to form an extended $\pi$ system involving all four $\mathrm{C}$ s. This results in greater stability and decreased energy.
(b) A conjugated diene is a resonance hybrid:
Structure (i) has 11 bonds and makes a more significant contribution than the other two structures, which have only 10 bonds. Since the contributing structures are not equivalent, the resonance energy is small.

Anand Jangid
Anand Jangid
Numerade Educator
06:27

Problem 22

Apply the MO theory to 1,3-butadiene and compare the relative energies of its molecular orbitals with those of ethene (Problem 8.21).

Four $p$ AO's (see Fig. 8-2) give four molecular orbitals, as shown in Fig. 8-3. Wherever there is a switch from + to -, there is a node, as indicated by a heavy dot. Note that $\pi_1$ of the diene has a lower energy than $\pi$ of ethene.
In a linear $\pi$ system the relative energies of the molecular orbitals are determined by the pairwise overlaps of adjacent $p$ orbitals along the chain. An excess of bonding interactions, + with + or - with - , denotes a bonding MO; an excess of antibonding interactions, + with - , denotes an antibonding $\mathrm{MO}^*$.

Caleb Prus
Caleb Prus
Numerade Educator
03:11

Problem 23

Explain how the energies shown in Fig. 8-3 are consistent with the fact that a conjugated diene is more stable than an isolated diene.

The energy of $\pi_1+\pi_2$ of the conjugated diene is less than twice the energy of an ethene $\pi$ bond. Two ethene $\pi$ bonds correspond to an isolated diene.

Matthew Lueckheide
Matthew Lueckheide
Numerade Educator
06:38

Problem 24

(a) Apply the MO theory to the allyl system. Indicate the relative energies of the molecular orbitals and state if they are bonding, nonbonding, or antibonding. (b) Insert the electrons for the carbocation $\mathrm{C}_3 \mathrm{H}_5^{+}$, the free radical $\mathrm{C}_3 \mathrm{H}_5$. and the carbanion $\mathrm{C}_3 \mathrm{H}_5^{-}$, and compare the relative energies of these three species.
(a) Three $p$ AO's give three molecular orbitals, as indicated in Fig. 8-4. Since there are an odd number of $p$ AO's in this linear system, the middle-energy molecular orbital is nonbonding $\left(\pi_2^n\right)$. Note that the node of this $\mathrm{MO}^n$ is at a $\mathrm{C}$, indicated by a $0 . \mathrm{An} \mathrm{MO}^{\mathrm{n}}$ can be recognized if the number of bonding pairs equals the number of antibonding pairs or if there is no overlap.
(b) $R^{+} \quad \pi_1 \uparrow \downarrow \quad$ R. $\begin{array}{ll}\pi_2^n \uparrow & \pi_1 \uparrow \downarrow\end{array} \quad R^{-} \begin{array}{r}\pi_2^n \uparrow \downarrow \\ \pi_1 \uparrow \downarrow\end{array}$
The electrons in the $\pi_2^n$ orbital do not appreciably affect the stability of the species. Therefore all three species are more stable than the corresponding alkyl systems $\mathrm{C}_3 \mathrm{H}_7^{+}, \mathrm{C}_3 \mathrm{H}_7^{-}$, and $\mathrm{C}_3 \mathrm{H}_7^{-}:$. The extra electrons do increase the repulsive forces between electrons slightly, so the order of stability is $\mathrm{C}_3 \mathrm{H}_5^{+}>\mathrm{C}_3 \mathrm{H}_5^{-}>\mathrm{C}_3 \mathrm{H}_5^{-}$.

Susan Hallstrom
Susan Hallstrom
Numerade Educator
05:16

Problem 25

Explain 1,4-addition in terms of the mechanism of electrophilic addition.
The electrophile $\left(\mathrm{H}^{+}\right)$adds to form an allylic carbocation with positive charge delocalized at $\mathrm{C}^2$ and $\mathrm{C}^4$ (resonance forms II and III). This cation adds the nucleophile at $\mathrm{C}^2$ to form the 1,2 -addition product or at $\mathrm{C}^4$ to form the 1,4-addition product.

Ian Kaigh
Ian Kaigh
Numerade Educator
03:22

Problem 26

Use an enthalpy-reaction diagram to explain the following observations. Start from the allylic carbocation, the common intermediate.
The different products arise from enthalpy differences in the second step, the reaction of $\mathrm{Br}^{-}$and the allyl $\mathrm{R}^{+}$. See Fig. 8-5. At $-80^{\circ} \mathrm{C}$ the 1,2 -adduct, the rate-controlled product, is favored because its formation has the lower $\Delta H^{\ddagger}$. 1,2-Adduct formation can reverse to refurnish the intermediate allylic carbocation, $\mathrm{R}^{+}$. At $40^{\circ} \mathrm{C}, \mathrm{R}^{+} \mathrm{goes}$ through the higher-energy transition state for formation of the more stable 1,4-adduct, the thermodynamic-controlled product. The 1,4-adduct accumulates because the addition, having a greater $\Delta H^{\ddagger}$, is more difficult to reserve than that for the 1,2 -adduct. The 1,4 -adduct has a lower enthalpy because it has more $\mathrm{R}$ groups on the $\mathrm{C}=\mathrm{C}$.

Madeline Currie
Madeline Currie
Numerade Educator
10:50

Problem 27

Explain why the conjugated 1,3-pentadiene reacts with one mole of $\mathrm{Br}_2$ at a faster rate than does the isolated 1,4-pentadiene.
The reaction products are shown:

The intermediate carbocation formed from the conjugated diene is allylic and is more stable than the isolated carbocation from the isolated diene. Since the transition state for the rate-controlling first step leading to the lowerenthalpy allylic $\mathrm{R}^{+}$also has a lower enthalpy, $\Delta H^{\ddagger}$ for this reaction is smaller and the reaction is faster. It is noteworthy that although the conjugated diene is more stable, it nevertheless reacts faster.

Dr. Satish Ingale
Dr. Satish Ingale
Numerade Educator
00:58

Problem 28

(a) Which additional C's in the following $\mathrm{R}^{+}$bear some + charge?
(b) With which of these C's will : $\mathrm{Nu}^{-}$react to give the thermodynamic-controlled product?
(a) $\mathrm{C}^3, \mathrm{C}^5$, and $\mathrm{C}^7$. These are alternating sites.

Sachin Rao
Sachin Rao
Numerade Educator
01:09

Problem 29

Write structural formulas for major and minor products from acid-catalyzed dehydration of $\mathrm{H}_2 \mathrm{C}=\mathrm{CHCH}_2 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_3$.
Dehydration can occur by removal of $\mathrm{H}$ from either $\mathrm{C}^3$ or $\mathrm{C}^5$.

Lottie Adams
Lottie Adams
Numerade Educator
05:35

Problem 30

Write the structures of the intermediate $\mathrm{R}^{+} \mathrm{s}$ and the two products obtained from the reaction of $\mathrm{H}_2 \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_3\right) \mathrm{CH}=\mathrm{CH}_2$ with $(a) \mathrm{HBr},($ b $) \mathrm{Cl}_2$.
(a) $\mathrm{H}^{+}$adds to $\mathrm{C}^l$ to form the more stable allylic $3^{\circ} \mathrm{R}^{+}$, rather than to $\mathrm{C}^2$ or $\mathrm{C}^3$ to form the nonallylic $1^{\circ} \mathrm{R}^{+}$'s $\mathrm{H}_2 \mathrm{C}-\mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{CH}=\mathrm{CH}_2$ and $\mathrm{H}_2 \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_3\right) \mathrm{CH}_2-\mathrm{CH}_2$, respectively; or to $\mathrm{C}^4$ to yield the $2^{\circ}$ allylic $\mathrm{R}^{+} \mathrm{H}_2 \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_3\right) \mathrm{C} \mathrm{HCH}_3$.

(b) $\mathrm{Cl}^{+}$also adds to $\mathrm{C}^{\prime}$ to form a hybrid allylic $\mathrm{R}^{+}$.

Anupa Sharad Medhekar
Anupa Sharad Medhekar
Numerade Educator
05:43

Problem 31

Write initiation and propagation steps in radical-catalyzed addition of $\mathrm{BrCCl}_3$ to 1,3-butadiene and show how the structure of the intermediate accounts for: $(a)$ greater reactivity of conjugated dienes than alkenes, $(b)$ orientation in addition.

(a) The allyl radical formed in the first propagation step is more stable and requires a lower $\Delta H^{\star}$ than the alkyl free radical from alkenes. The order of free-radical stability is allyl $>3^{\circ}>2^{\circ}>1^{\circ}$.
(b) The 1,4-orientation is similar to ionic addition because of the relative stabilities of the two products.

LP
Layhna Plagmann
Numerade Educator
02:16

Problem 32

For the conjugated and isolated dienes of molecular formula $\mathrm{C}_6 \mathrm{H}_{10}$ tabulate $(a)$ structural formula and IUPAC name, $(b)$ possible geometric isomers, $(c)$ ozonolysis products.
In Table $8-2$ a box is placed about the $\mathrm{C}=\mathrm{C}$ associated with geometric isomers.

Lottie Adams
Lottie Adams
Numerade Educator

Problem 33

Show reagents and reactions needed to prepare the following compounds from the indicated starting compounds. (a) Acetylene to ethylidene iodide (1,1-diiodoethane). (b) Propyne to isopropyl bromide. (c) 2-Butyne to racemic 2,3-dibromobutane. (d) 2-Bromobutane to trans-2-butene. (e) $n$-Propyl bromide to 2-hexyne. $(f)$ 1-Pentene to 2-pentyne.
(a)
(b)
$$
\begin{aligned}
& \mathrm{H}-\mathrm{C}=\mathrm{C}-\mathrm{H} \stackrel{\mathrm{HI}}{\longrightarrow} \mathrm{H}_2 \mathrm{C}=\mathrm{CHI} \stackrel{\mathrm{HI}}{\longrightarrow} \mathrm{CH}_3 \mathrm{CHI}_2 \\
& \mathrm{CH}_3 \mathrm{C}=\mathrm{C}-\mathrm{H} \stackrel{\mathrm{H}_2 / \mathrm{Pt}}{\longrightarrow} \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2 \stackrel{\mathrm{HBr}}{\longrightarrow} \mathrm{CH}_3 \mathrm{CHBrCH}_3
\end{aligned}
$$

Add $\mathrm{H}_2$ first; the reaction can be stopped after $1 \mathrm{~mol}$ is added.
(c) $\mathrm{CH}_3 \mathrm{C}_2 \mathrm{CCH}_3 \stackrel{\mathrm{H}_2 / \mathrm{Pt}(\mathrm{Pb})}{\longrightarrow}$ cis- $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_3 \stackrel{\mathrm{Br}_3}{\longrightarrow}$ rac-( \pm$)-\mathrm{CH}_3 \mathrm{CHBrCHBrCH}_3$ (trans addition)
(d) $* \mathrm{CH}_3 \mathrm{CHBrCH}_2 \mathrm{CH}_3 \stackrel{\text { alc. } \mathrm{KoH}}{\longrightarrow}$ cis + trans $-\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_3 \stackrel{\mathrm{Br}_2}{\longrightarrow}$
$$
\begin{aligned}
& \mathrm{CH}_3 \mathrm{CHBrCHBrCH}_3 \\
& \text { meso } \text { and rac }
\end{aligned} \stackrel{\mathrm{KNH}_2}{\longrightarrow} \mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CCH}_3 \stackrel{\mathrm{Na}, \mathrm{NH}_3}{\longrightarrow} \text { trans }-\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_3
$$
meso and rac
(e) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br} \stackrel{\text { alc. } \mathrm{KOH}}{\longrightarrow} \mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2 \stackrel{\mathrm{Br}_2}{\longrightarrow} \mathrm{CH}_3 \mathrm{CHBrCH}_2 \mathrm{Br} \stackrel{\mathrm{KNH}_2}{\longrightarrow}$
$$
\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CH} \stackrel{\mathrm{Na}}{\longrightarrow} \mathrm{CH}_3 \mathrm{C} \equiv \mathrm{C}^{-} \mathrm{Na}^{+} \stackrel{n-\mathrm{C}_3 \mathrm{H}_3 \mathrm{Br}}{\longrightarrow} \mathrm{CH}_3 \mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3
$$
(f) $\mathrm{H}_2 \mathrm{C}=\mathrm{CHCH}_2 \mathrm{CH}_2 \mathrm{CH}_3 \stackrel{\mathrm{HBr}}{\longrightarrow} \mathrm{CH}_3 \mathrm{CHBrCH}_2 \mathrm{CH}_2 \mathrm{CH}_3 \stackrel{\text { alc. } \mathrm{KOH}}{\longrightarrow}$
$$
\begin{aligned}
& \mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_2 \mathrm{CH}_3 \stackrel{\mathrm{Br}_2}{\longrightarrow} \\
& \text { (Saytzeff product) }
\end{aligned}
$$
(Saytzeff product)
$\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CCH}_2 \mathrm{CH}_3$ (not the less stable allene, $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}=\mathrm{CHCH}_3$ )

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04:18

Problem 34

Write a structural formula for organic compounds (A) through (N):
(a) $\mathrm{HC} \equiv \mathrm{CCH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ (B)
(b) $\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CH} \stackrel{\mathrm{CH}_3 \mathrm{MgBr}}{\longrightarrow}$ $(\mathrm{C}$, a gas $)+(\mathrm{D})$ (E)
(c) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C}=\mathrm{CH}+\mathrm{Na}^{+} \mathrm{NH}_2^{-} \longrightarrow$ (F) $\stackrel{\mathrm{C}_2 \mathrm{H}_3 1}{\longrightarrow}$ (G) $\stackrel{\mathrm{H}_3 \mathrm{O}^{+}, \mathrm{Hg}^{4+}}{\longrightarrow}$ (H)
(d) $\mathrm{CH}_3 \mathrm{C}=\mathrm{CH}+\mathrm{BH}_3 \longrightarrow$ (I) $\stackrel{\mathrm{CH}_3 \mathrm{COOH}}{\longrightarrow}$
(J) $\frac{\text { dil. aq. }}{\mathrm{KMnO}_4}(\mathrm{~K})$
(e)
<smiles>CC(C)C(C)Cl</smiles>
$\mathrm{KOH}$
(L) $\frac{\mathrm{BrCCl}_3}{\text { peroxide }}$
$(\mathrm{M})+(\mathrm{N})$

Anish Wadhwa
Anish Wadhwa
Numerade Educator
01:01

Problem 35

Assign numbers from 1 for LEAST to 5 for MOST to indicate the relative reactivity on $\mathrm{HBr}$ addition to the following compounds:
(a) $\mathrm{H}_2 \mathrm{C}=\mathrm{CH}-\mathrm{CH}_2 \mathrm{CH}_3$
(b) $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$
(c) $\mathrm{H}_2 \mathrm{C}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_2$
(d) $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_2$
(e)
<smiles>C=C(C)C(=C)C</smiles>
Conjugated dienes form the more stable allyl $\mathrm{R}^{+}$'s and therefore are more reactive than alkenes. Alkyl groups on the unsaturated C's increase reactivity. Relative reactivities are: (a) $1,($ b) $2,(c) 3,(d) 4,(e) 5$.

Narayan Hari
Narayan Hari
Numerade Educator
02:28

Problem 36

For the reaction of propyne with $(a) \mathrm{HOBr},(b) \mathrm{Br}_2+\mathrm{NaOH}$, give the structures of the products and the mechanisms of their formation.
(a)
<smiles>OO[In]Br</smiles>
tautomerize
<smiles>CC(=O)C=O</smiles>
Bromoaceto
(b) Propyne reacts with strong bases to form a nucleophilic carbanion which displaces $: \mathrm{Br}^{--}$from $\mathrm{Br}_2$ by attacking $\mathrm{Br}$ to form 1-bromopropyne.
I-Bromopropyne

Madeline Currie
Madeline Currie
Numerade Educator
03:51

Problem 37

${ }^{14} \mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2$ is subjected to allylic free-radical bromination. Will the reaction product be exclusively labeled $\mathrm{H}_2 \mathrm{C}=\mathrm{CH}^{14} \mathrm{CH}_2 \mathrm{Br}$ ? Explain.

No. The product consists of an equal number of $\mathrm{H}_2 \mathrm{C}=\mathrm{CH}^{14} \mathrm{CH}_2 \mathrm{Br}$ and ${ }^{14} \mathrm{CH}_2=\mathrm{CHCH}_2 \mathrm{Br}$ molecules. $\mathrm{H}$ abstraction produces a resonance hybrid of two contributing structures having both ${ }^{12} \mathrm{C}$ and ${ }^{14} \mathrm{C}$ as equally reactive, free-radical sites that attack $\mathrm{Br}_2$.

Jorge Villanueva
Jorge Villanueva
Numerade Educator
03:46

Problem 38

(a) Write a schematic structure for the mer of the polymer from head-to-tail reaction of 2-methyl1,3-butadiene. (b) Account for this orientation in polymerization. (c) Show how the structure is deduced from the product
obtained from ozonolysis of the polymer.
(a) 1,4-Addition with regular head-to-tail orientation produces a polymer with the following repeating unit (mer):
<smiles>CCC=C(C)CC</smiles>
(b) This orientation results from more rapid formation of the more stable intermediate free radical.
The $1^{\circ}$ allylic site is more reactive than the $3^{\circ}$ allylic site. Attack at the other terminal $=\mathrm{CH}_2$ gives the less stable free radical.
(c) Write the ozonolysis products with the O's pointing at each other. Now erase the O's and join the C's by a double bond.

Susan Hallstrom
Susan Hallstrom
Numerade Educator
01:33

Problem 39

(a) Calculate the heat of hydrogenation, $\Delta H_h$, of acetylene to ethylene if the $\Delta H_h$ 's to ethane are $-137 \mathrm{~kJ} / \mathrm{mol}$ for ethylene and $-314 \mathrm{~kJ} / \mathrm{mol}$ for acetylene. $(b)$ Use these data to compare the ease of hydrogenation of acetylene to ethylene with that of ethylene to ethane.
(a) Write the reaction as the algebraic sum of two other reactions whose terms cancel out to give wanted reactants, products and enthalpy. These are the hydrogenation of acetylene to ethane and the dehydrogenation of ethane to ethylene (reverse of hydrogenation of $\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2$ ).
(Eq. 1)
$$
\begin{array}{lll}
\mathrm{H}-\mathrm{C}=\mathrm{C}-\mathrm{H}+2 \mathrm{H}_2 & \longrightarrow \mathrm{CH}_3-\mathrm{CH}_3 & -314 \mathrm{~kJ} / \mathrm{mol} \\
\mathrm{CH}_3-\mathrm{CH}_3 & \longrightarrow \mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2+\mathrm{H}_2 & +137 \mathrm{~kJ} / \mathrm{mol} \\
\mathrm{H}-\mathrm{C}=\mathrm{C}-\mathrm{H}+\mathrm{H}_2 \longrightarrow \mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2 & -177 \mathrm{~kJ} / \mathrm{mol}
\end{array}
$$
(Eq. 2)

Equation 2 (dehydrogenation) is the reverse of hydrogenation $\left(\Delta H_h=-137 \mathrm{~kJ} / \mathrm{mol}\right)$. Hence the $\Delta H_h$ of Eq. 2 has a + value.
(b) Acetylene is less stable thermodynamically relative to ethylene than ethylene is to ethane because $\Delta H_h$ for acetylene $\rightarrow$ ethylene is $-177 \mathrm{~kJ} / \mathrm{mol}$, while for ethylene $\rightarrow$ ethane it is $-137 \mathrm{~kJ} / \mathrm{mol}$. Therefore acetylene is more easily hydrogenated and the process can be stopped at the ethylene stage. In general, hydrogenation of alkynes can be stopped at the alkene stage.

Dominador Tan
Dominador Tan
Numerade Educator

Problem 40

Deduce the structural formula of a compound of molecular formula $\mathrm{C}_6 \mathrm{H}_{10}$ which adds $2 \mathrm{~mol}_{\text {of }} \mathrm{H}_2$ to form 2-methylpentane, forms a carbonyl compound in aqueous $\mathrm{H}_2 \mathrm{SO}_4-\mathrm{HgSO}_4$ solution and does not react with ammoniacal $\mathrm{AgNO}_3$ solution, $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+} \mathrm{NO}_3^{-}$.

There are two degrees of unsaturation since the compound $\mathrm{C}_6 \mathrm{H}_{10}$ lacks four $\mathrm{H}$ 's from being an alkane. The addition of $2 \mathrm{~mol}$ of $\mathrm{H}_2$ excludes a cyclic compound. It may be either a diene or an alkyne, and the latter functional group is established by hydration to a carbonyl compound. The skeleton must be
<smiles>CCCCC(C)C</smiles>
as established by the reduction product. The two possible alkynes with this skeleton are
$$
\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}_2 \mathrm{C}=\mathrm{CH} \text { and }\left(\mathrm{CH}_3\right)_2 \mathrm{CH}-\mathrm{C}=\mathrm{C}-\mathrm{CH}_3
$$

The negative test for a l-alkyne with $\mathrm{Ag}^{+}$establishes the second structure, 4-methyl-2-pentyne.

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11:06

Problem 41

The allene 2,3-pentadiene $\left(\stackrel{L}{\mathrm{C}} \mathrm{H}_3 \stackrel{2}{\mathrm{C}} \mathrm{H}=\stackrel{3}{\mathrm{C}}=\stackrel{\dot{\mathrm{C}}}{\mathrm{C}} \stackrel{s}{\mathrm{C}} \mathrm{H}_3\right)$ does not have a chiral $\mathrm{C}$ but is resolved into enantiomers. (a) Draw an orbital picture that accounts for the chirality [see Problem 5.23(d)]. (b) What structural features must a chiral allene have?
(a) $\mathrm{C}^3$ is $s p$ hybridized and forms two $\sigma$ bonds by $s p-s p^2$ overlap with the orbitals of $\mathrm{C}^2$ and $\mathrm{C}^4$. The two remaining $p$ orbitals of $\mathrm{C}^3$ form two $\pi$ bonds, one with $\mathrm{C}^2$ and one with $\mathrm{C}^4$. These $\pi$ bonds are at right angles to each other. The $\mathrm{H}$ and $\mathrm{CH}_3$ on $\mathrm{C}^2$ are in a plane at right angles to the plane of the $\mathrm{H}$ and $\mathrm{CH}_3$ on $\mathrm{C}^4$. See Fig. 8-6.
Because there is no free rotation about the two $\pi$ bonds, the two H's and two $\mathrm{CH}_3$ 's have a fixed spatial relationship. Whenever the two substituents on $\mathrm{C}^2$ are different and the two substituents on $\mathrm{C}^4$ are different, the molecule lacks symmetry and is chiral.
(b) Individually, the terminal $\mathrm{C}$ 's of the allenic system must have two different attached groups; e.g. $\mathrm{RHC}=\mathrm{C}=\mathrm{CHR}^{\prime}(\mathrm{R})$. The groups could be other than $\mathrm{H}$ 's. $\mathrm{H}_2 \mathrm{C}=\mathrm{C}=\mathrm{CHR}$ is not chiral.

Anish Wadhwa
Anish Wadhwa
Numerade Educator
01:16

Problem 42

Heating $\mathrm{C}_4 \mathrm{H}_9 \mathrm{Br}(\mathrm{A})$ with alcoholic $\mathrm{KOH}$ forms an alkene, $\mathrm{C}_4 \mathrm{H}_8$ (B), which reacts with bromine to give $\mathrm{C}_4 \mathrm{H}_8 \mathrm{Br}_2$ (C). (C) is transformed by $\mathrm{KNH}_2$ to a gas, $\mathrm{C}_4 \mathrm{H}_6$ (D), which forms a precipitate when passed through ammoniacal $\mathrm{CuCl}$. Give the structures of compounds (A) through (D).

The precipitate with ammoniacal $\mathrm{CuCl}$ indicates that (D) is a 1 -alkyne, which can only be 1-butyne. The reactions and compounds are:
(D)
(C)
(B)
(A)
(A) cannot be $\mathrm{CH}_3 \mathrm{CHBrCH}_2 \mathrm{CH}_3$, which would give mainly $\mathrm{H}_3 \mathrm{CCH}=\mathrm{CHCH}_3$ and finally $\mathrm{CH}_3 \mathrm{C}_2 \mathrm{CCH}_3$.

Raghvendra Singh
Raghvendra Singh
Numerade Educator
01:21

Problem 43

Is the fact that conjugated dienes are more stable and more reactive than isolated dienes an incongruity?

No. Reactivity depends on the relative $\Delta H^*$ values. Although the ground-state enthalpy for the conjugated diene is lower than that of the isolated diene, the transition-state enthalpy for the conjugated system is lower by a greater amount (see Fig. 8-7).

Grigoriy Sereda
Grigoriy Sereda
Numerade Educator
01:01

Problem 44

Explain why 1,3-butadiene and $\mathrm{O}_2$ do not react unless irradiated by uv light to give the 1,4-adduct.
<smiles>C1=CCOOC1</smiles>
Ordinary ground-state $\mathrm{O}_2$ is a diradical,
<smiles>COO</smiles>
One bond could form, but the intermediate has 2 electrons with the same spin and a second bond cannot form.
<smiles>CC=C[CH]CPOOCC=CC</smiles>
When irradiated, $\mathrm{O}_2$ is excited to the singlet spin-paired state
<smiles>[O]O[O-]</smiles>
Singlet $\mathrm{O}_2$ reacts by a concerted mechanism to give the product.

Narayan Hari
Narayan Hari
Numerade Educator
01:38

Problem 45

Is there any inconsistency between the facts that the $\mathrm{C}-\mathrm{H}$ bond in acetylene has the greatest bond energy of all $\mathrm{C}-\mathrm{H}$ bonds and that it is also the most acidic?

No. Bond energy is a measure of homolytic cleavage, $\equiv \mathrm{C}: \mathrm{H} \longrightarrow \equiv \mathrm{C} \cdot+\cdot \mathrm{H}$. Acidity is due to a heterolytic cleavage, $\mathbf{m} \mathrm{C}: \mathrm{H}+$ Base $\longrightarrow$ æ­£: $:^{-}+\mathrm{H}^{+}($Base $)$.

Hitendra Singh
Hitendra Singh
Numerade Educator