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Schaum's Outline of Organic Chemistry

George Hademenos, George Hademenos

Chapter 7

ALKYL HALIDES - all with Video Answers

Educators

Chapter Questions

01:03

Problem 1

CAN'T COPY

Ankur S
Ankur S
Numerade Educator
01:40

Problem 2

Compare and account for differences in the (a) dipole moment, $(b)$ boiling point, $(c)$ density and $(d)$ solubility in water of an alkyl halide RX and its parent alkane RH.
(a) $\mathrm{RX}$ has a larger dipole moment because the $\mathrm{C}-\mathrm{X}$ bond is polar. (b) $\mathrm{RX}$ has a higher boiling point since it has a larger molecular weight and also is more polar. (c) $\mathrm{RX}$ is more dense since it has a heavy $\mathrm{X}$ atom; the order of decreasing density is $\mathrm{RI}>\mathrm{RBr}>\mathrm{RCl}>\mathrm{RF}$. (d) $\mathrm{RX}$, like $\mathrm{RH}$, is insoluble in $\mathrm{H}_2 \mathrm{O}$, but $\mathrm{RX}$ is somewhat more soluble because some $\mathrm{H}$-bonding can occur:
$$
\stackrel{\circ}{\mathrm{R}}-\stackrel{\circ-}{\mathrm{X}}:-\cdot \stackrel{\delta}{\mathrm{H}}-\stackrel{\delta}{\mathrm{O}} \mathrm{H}
$$

This effect is greatest for RF.

Lottie Adams
Lottie Adams
Numerade Educator
03:21

Problem 3

Account for the following observations: (a) In a polar solvent such as water the $\mathrm{S}_{\mathrm{N}} 1$ and $\mathrm{E} 1$ reactions of a $3^{\circ}$ RX have the same rate. (b) $\left(\mathrm{CH}_3\right)_3 \mathrm{Cl}+\mathrm{H}_2 \mathrm{O} \rightarrow\left(\mathrm{CH}_3\right)_3 \mathrm{COH}+\mathrm{HI}$ but $\left(\mathrm{CH}_3\right)_3 \mathrm{Cl}+$ $\mathrm{OH}^{-} \rightarrow\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CH}_2+\mathrm{H}_2 \mathrm{O}+1^{-}$.

(a) The rate-controlling step for both $\mathrm{El}$ and $\mathrm{S}_{\mathrm{N}} 1$ reactions is the same:
$$
\ddot{\mathrm{R}}-\stackrel{\stackrel{-}{-} \mathrm{X}}{\stackrel{\text { slow }}{\longrightarrow}} \mathrm{R}^{+}+\mathrm{X}^{-}
$$
and therefore the rates are the same. (b) In a nucleophilic solvent in the absence of a strong base, a $3^{\circ} \mathrm{RX}$ undergoes an $\mathrm{S}_{\mathrm{N}} 1$ solvolysis. In the presence of a strong base $\left(\mathrm{OH}^{-}\right)$a $3^{\circ} \mathrm{RX}$ undergoes mainly $\mathrm{E} 2$ elimination.

Raghvendra Singh
Raghvendra Singh
Numerade Educator
02:01

Problem 3

Give the products of the following reactions:
(a) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}+\mathrm{HI} \longrightarrow$
(b) $n-\mathrm{C}_4 \mathrm{H}_9 \mathrm{OH}+\mathrm{NaBr}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow$
(c) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}+\mathrm{PI}_3\left(\mathrm{P}+\mathrm{I}_2\right) \longrightarrow$
(d) $\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}_2 \mathrm{OH}+\mathrm{SOCl}_2 \longrightarrow$
(e) $\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2+\mathrm{Br}_2 \longrightarrow$
(f) $\mathrm{CH}_3 \mathrm{CH}=\mathrm{C}\left(\mathrm{CH}_3\right)_2+\mathrm{HI} \longrightarrow$
(g) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}+\mathrm{I}^{-} \longrightarrow$

Sima Sarker
Sima Sarker
Numerade Educator
00:58

Problem 4

How is conformational analysis used to explain the $6: 1$ ratio of trans-to cis-2-butene formed on dehydrochlorination of 2 -chlorobutane?
For either enantiomer there are two conformers in which the $\mathrm{H}$ and $\mathrm{Cl}$ eliminated are anti to each other. Conformer I has a less crowded, lower-enthalpy transition state than conformer II, Its $\Delta H^*$ is less and reaction rate greater; this accounts for the greater amount of trans isomer obtained from conformer I and the smaller amount of cis isomer from conformer II.

Grigoriy Sereda
Grigoriy Sereda
Numerade Educator
01:16

Problem 4

Which of the following chlorides can be made in good yield by light-catalyzed monochlorination of the corresponding hydrocarbon?
(a) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Cl}$
(c) $\left(\mathrm{CH}_3\right)_3 \mathrm{CCH}_2 \mathrm{Cl}$
(b) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl}$
(d) $\left(\mathrm{CH}_3\right)_3 \mathrm{CCl}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:35

Problem 5

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

Raghvendra Singh
Raghvendra Singh
Numerade Educator
00:34

Problem 5

Give reactions for tests that can be carried out rapidly in a test tube to differentiate the following compounds: hexane, $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCl}, \mathrm{H}_2 \mathrm{C}=\mathrm{CHCH}_2 \mathrm{Cl}$ and $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl}$.

Hexane is readily distinguished from the other three compounds because there is a negative test for $\mathrm{Cl}^{-}$after $\mathrm{Na}^{-}$ fusion and treatment with acidic $\mathrm{AgNO}_3$. The remaining three compounds are differentiated by their reactivity with alcoholic $\mathrm{AgNO}_3$ solution. $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCl}$ is a vinylic chloride and does not react even on heating. $\mathrm{H}_2 \mathrm{C}=\mathrm{CHCH} \mathrm{Cl}_2$ is most reactive (allylic) and precipitates $\mathrm{AgCl}$ in the cold, while $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl}$ gives a precipitate of $\mathrm{AgCl}$ on warming with the reagent.

Aadit Sharma
Aadit Sharma
Numerade Educator
03:12

Problem 6

What generalizations about the relationship of basicity and nucleophilicity can be made from the following relative rates of nucleophilic displacements:
(a) $\mathrm{OH}^{-} \gg \mathrm{H}_2 \mathrm{O}$ and $\mathrm{NH}_2^{-} \gg \mathrm{NH}_3$
(b) $\mathrm{H}_3 \mathrm{C}:^{-}>: \stackrel{\mathrm{O}_{\mathrm{H}}}{ }{ }^{-}>::_{\mathrm{F}}^{-}$
(c) ${: \mathrm{I}^{-}}^{-}$: $\ddot{\mathrm{Br}}:^{-}>:_{\mathrm{C}}^{\mathrm{l}}:^{-}>: \ddot{\mathrm{F}}$ :
(d) $\mathrm{CH}_3 \mathrm{O}^{-}>\mathrm{OH}^{-}>\mathrm{CH}_3 \mathrm{COO}^{-}$
(a) Bases are better nucleophiles than their conjugate acids.
(b) In going from left to right in the Periodic Table, basicity and nucleophilicity are directly related - they both decrease.
(c) In going down a Group in the Periodic Table they are inversely related, in that nucleophilicity increases and basicity decreases.
(d) When the nucleophilic and basic sites are the same atom (here an $\mathrm{O}$ ), nucleophilicity parallels basicity.

The order in Problem 7.6(c) may occur because the valence electrons of a larger atom could be more available for bonding with the $\mathrm{C}$, being further away from the nucleus and less firmly held. Alternatively, the greater ease of distortion of the valence shell (induced polarity) makes easier the approach of the larger atom to the $\mathrm{C}$ atom. This property is called polarizability. The larger, more polarizable species (e.g. I, Br, $\mathrm{S}$, and $\mathrm{P}$ ) exhibit enhanced nucleophilicity; they are called soft bases. The smaller, more weakly polarizable bases (e.g. N, O, and F) have diminished nucleophilicity; they are called hard bases.

Himanshu Kushwaha
Himanshu Kushwaha
Numerade Educator
02:59

Problem 7

Explain why the order of reactivity of Problem 7.6(c) is observed in nonpolar, weakly polar aprotic, and polar protic solvents, but is reversed in polar aprotic solvents.

In nonpolar and weakly polar aprotic solvents, the salts of : $\mathrm{Nu}^{-}$are present as ion-pairs (or ion-clusters) in which the nearby cations diminish the reactivity of the anion. Since, with a given cation, ion-pairing is strongest with the smallest ion, $\mathrm{F}^{-}$, and weakest with the largest ion, $\mathrm{I}^{-}$, the reactivity of $\mathrm{X}^{-}$decreases as the size of the anion decreases. In polar protic solvents, hydrogen-bonding, which also lessens the reactivity of $\mathrm{X}^{-}$, is weakest with the largest ion,
again making the largest ion more reactive. Polar aprotic solvents solvate only the cations, leaving free, unencumbered anions. The reactivities of all anions are enhanced, but the effect is more pronounced the smaller the anion. Hence, the order of Problem 7.6(c) is reversed.

Molly Cary
Molly Cary
Numerade Educator

Problem 8

Write equations for the reaction of $\mathrm{RCH}_2 \mathrm{X}$ with
(a) : $\mathrm{I}^{-}$
(c) : $\mathrm{O}^{\prime}{ }^{\prime-}$
(d) $\mathrm{R}^{\prime}:-$
(e)
<smiles>[R][C-]=[O+]</smiles>
(f) $\mathrm{H}_3 \mathrm{~N}$ :
(g) $: \mathrm{CN}^{-}$
and classify the functional group in each product.
(a) $:_{\mathrm{I}}^{-}+\mathrm{RCH}_2 \mathrm{X} \longrightarrow \mathrm{RCH}_2 \mathrm{I}+: \ddot{\mathrm{X}}^{-} \quad$ Iodide
(c) $-: \mathrm{OR}^{\prime}+\mathrm{RCH}_2 \mathrm{X} \longrightarrow \mathrm{RCH}_2 \mathrm{OR}^{\prime}+: \ddot{\mathrm{X}}^{-} \quad$ Ether
(d) ${ }^{-}: \mathrm{R}^{\prime}+\mathrm{RCH}_2 \mathrm{X} \longrightarrow \mathrm{RCH}_2 \mathrm{R}^{\prime}+: \ddot{\mathrm{X}}^{-} \quad$ Alkane (coupling)
(e) $-: \mathrm{OOOCR}^{\prime}+\mathrm{RCH}_2 \mathrm{X} \longrightarrow \mathrm{RCH}_2 \mathrm{OOCR}^{\prime}+: \ddot{\mathrm{X}}_{-}^{-}$Ester
(f) $: \mathrm{NH}_3+\mathrm{RCH}_2 \mathrm{X} \longrightarrow \mathrm{RCH}_2 \mathrm{NH}_3^{+}+: \ddot{X}^{-} \quad$ Ammonium salt
(g) ${ }^{-} \mathrm{CN}+\mathrm{RCH}_2 \mathrm{X} \longrightarrow \mathrm{RCH}_2 \mathrm{CN}+: \ddot{\mathrm{X}}^{-} \quad$ Nitrile (or Cyanide)

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02:33

Problem 9

Compare the effectiveness of acetate $\left(\mathrm{CH}_3 \mathrm{COO}^{-}\right)$, phenoxide $\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{O}^{-}\right)$and benzenesulfonate $\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{SO}_3^{-}\right)$anions as leaving groups if the acid strengths of their conjugate acids are given by the $\mathrm{p} K_a$ values 4.5 , 10.0 , and 2.6 , respectively.
The best leaving group is the weakest base, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{SO}_3^{-}$; the poorest is $\mathrm{C}_6 \mathrm{H}_5 \mathrm{O}^{-}$, which is the strongest base.
Sulfonates are excellent leaving groups-much better than the halides. One of the best leaving groups $\left(10^8\right.$ times better than $\left.\mathrm{Br}^{-}\right)$is $\mathrm{CF}_3 \mathrm{SO}_3^{-}$, called triflate.

Ryder Mora
Ryder Mora
Numerade Educator
01:03

Problem 10

Give the 3 steps for the mechanism of the $\mathrm{S}_{\mathrm{N}} 1$ hydrolysis of $3^{\circ} \mathrm{Rx}, \mathrm{Me}_3 \mathrm{CBr}$.
(I) $\mathrm{Me}_3 \mathrm{C}: \mathrm{Br}: \longrightarrow \mathrm{Me}_3 \mathrm{C}^{+}+\mathrm{Br}^{--}$
(2) $\mathrm{Me}_3 \mathrm{C}^{+}+: \mathrm{OH}_2 \leftarrow \mathrm{Me}_3 \mathrm{C}: \mathrm{OH}_2^{+}$
(3) $\mathrm{Me}_3 \mathrm{CO}_2^{+}+\mathrm{H}_2 \mathrm{O} \leftarrow \mathrm{Me}_3 \mathrm{CO} \ddot{H} \mathrm{H}+\mathrm{H}_3 \mathrm{O}^{+}$

Narayan Hari
Narayan Hari
Numerade Educator
01:33

Problem 11

Give examples of the four charge-types of $S_N 2$ reactions, as shown in the first line of Table 7-1.
$$
\begin{aligned}
& \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}+: \mathrm{O}_{\mathrm{H}}-\longrightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{O} \mathrm{H} \\
& \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}+: \mathrm{BH}_3 \longrightarrow
\end{aligned}
$$
a sulfonium cation
Dimethyl sulfide
(table cant copy)

Nicole Smina
Nicole Smina
Numerade Educator
03:21

Problem 12

(a) Give an orbital representation for an $\mathrm{S}_{\mathrm{N}} 2$ reaction with ( $S$ )-RCHDX and : $\mathrm{Nu}^{-}$, if in the transition state the $\mathrm{C}$ on which displacement occurs uses $s p^2$ hybrid orbitals. (b) How does this representation explain (i) inversion, (ii) the order of reactivity $3^n>2^a>1^{-n}$ ?
(a) See Fig. 7-1.
(b) (i) The reaction is initiated by the nucleophile beginning to overlap with the tail of the $s p^3$ hybrid orbital holding $\mathrm{X}$. In order for the tail to become the head, the configuration must change; inversion occurs. (ii) As H's on the attacked C are replaced by R's, the TS becomes more crowded and has a higher enthalpy. With a $3^{\circ} \mathrm{RX}$, there is a higher $\Delta H^{\ddagger}$ and a lower rate.

Raghvendra Singh
Raghvendra Singh
Numerade Educator
04:11

Problem 13

(a) Give a representation of an $\mathrm{S}_{\mathrm{N}} 1$ TS which assigns a role to the nucleophilic protic solvent molecules (HS:) needed to solvate the ion. (b) In view of this representation, explain why (i) the reaction is first-order;
(ii) $\mathrm{R}^{+}$reacts with solvent rather than with stronger nucleophiles that may be present; (iii) catalysis by $\mathrm{Ag}^{+}$takes place; (iv) the more stable the $\mathrm{R}^{+}$, the less inversion and the more racemization occurs.
(a)
Solvent-assisted $\mathrm{S}_{\mathrm{N}} 1 \mathrm{TS}$
(b) (i) Although the solvent HS: appears in the TS, solvents do not appear in the rate expression. (ii) HS: is already partially bonded, via solvation, with the incipient $\mathrm{R}^{+}$. (iii) $\mathrm{Ag}^{+}$has a stronger affinity for $\mathrm{X}^{-}$than has a solvent molecule; the dissociation of $\mathrm{X}^{-}$is accelerated. (iv) The HS: molecule solvating an unstable $\mathrm{R}^{+}$is more apt to form a bond, causing inversion. When $\mathrm{R}^{+}$is stable, the TS gives an intermediate that reacts with another HS: molecule to give a symmetrically solvated cation,
$$
\left[\frac{\mathrm{HS}-\mathrm{C}--\mathrm{SH}}{\mathrm{U}}\right]^{+}
$$
which collapses to a racemic product:
<smiles>C[C@@H](C(C)(C)S)C(C)(C)[SH2]</smiles>
The more stable is $\mathrm{R}^{+}$, the more selective it is and the more it can react with the nucleophilic anion $\mathrm{Nu}^{-}$.

Marissa Turner
Marissa Turner
Numerade Educator
12:23

Problem 14

Give differences between $\mathrm{S}_{\mathrm{N}} 1$ and $\mathrm{S}_{\mathrm{N}} 2$ transition states.
1. In the $\mathrm{S}_{\mathrm{N}} 1$ TS there is considerable positive charge on $\mathrm{C}$; there is much weaker bonding between the attacking group and leaving groups with $\mathrm{C}$. There is little or no charge on $\mathrm{C}$ in the $\mathrm{S}_{\mathrm{N}} 2 \mathrm{TS}$.
2. The $\mathrm{S}_{\mathrm{N}} 1$ TS is approached by separation of the leaving group; the $\mathrm{S}_{\mathrm{N}} 2$ TS by attack of :Nu${ }^{-}$or :Nu.
3. The $\Delta H^*$ of the $\mathrm{S}_{\mathrm{N}} \mathrm{I}$ TS (and the rate of the reaction) depends on the stability of the incipient $\mathrm{R}^{+}$. When $\mathrm{R}^{+}$ is more stable, $\Delta H^{\ddagger}$ is lower and the rate is greater. The $\Delta H^{\ddagger}$ of the $\mathrm{S}_{\mathrm{N}} 2$ TS depends on the steric effects. When there are more R's on the attacked $\mathrm{C}$ or when the attacking : $\mathrm{Nu}^{-}$is bulkier, $\Delta H^{\ddagger}$ is greater and the rate is less.

Susan Hallstrom
Susan Hallstrom
Numerade Educator
08:09

Problem 15

How can the stability of an intermediate $\mathrm{R}^{+}$in an $\mathrm{S}_{\mathrm{N}} 1$ reaction be assessed from its enthalpyreaction diagram?

The intermediate $\mathrm{R}^{+}$is a trough between two transition-state peaks. More stable $\mathrm{R}^{+}$'s have deeper troughs and differ less in energy from the reactants and products.

Julia G.
Julia G.
Numerade Educator
01:01

Problem 16

(a) Formulate $\left(\mathrm{CH}_3\right)_3 \mathrm{COH}+\mathrm{HCl} \rightarrow\left(\mathrm{CH}_3\right)_3 \mathrm{CCl}+\mathrm{H}_2 \mathrm{O}$ as an $\mathrm{S}_{\mathrm{N}}$ l reaction. (b) Formulate the reaction $\mathrm{CH}_3 \mathrm{OH}+\mathrm{HI} \rightarrow \mathrm{CH}_3 1+\mathrm{H}_2 \mathrm{O}$ as an $\mathrm{S}_{\mathrm{N}} 2$ reaction.
(a)
Step $2 \quad\left(\mathrm{CH}_3\right)_3 \mathrm{CO}_2 \stackrel{+}{\stackrel{\text { slow }}{=}}\left(\mathrm{CH}_3\right)_3 \mathrm{C}^{+}+\mathrm{H}_2 \mathrm{O}$
Step $3 \quad\left(\mathrm{CH}_3\right)_3 \mathrm{C}^{+}+\mathrm{Cl}^{-} \stackrel{\text { tast }}{\longrightarrow}\left(\mathrm{CH}_3\right)_3 \mathrm{CCl}$
(b) Step $1 \quad \mathrm{CH}_3 \mathrm{OH}+\mathrm{HI} \stackrel{\text { fast }}{=} \mathrm{CH}_3 \stackrel{+}{\mathrm{O}} \mathrm{H}_2+1^{-}$
Step 2

Narayan Hari
Narayan Hari
Numerade Educator
01:08

Problem 17

ROH does not react with $\mathrm{NaBr}$, but adding $\mathrm{H}_2 \mathrm{SO}_4$ forms $\mathrm{RBr}$. Explain. $\mathrm{Br}^{-}$, an extremely weak Brönsted base. cannot displace the strong base $\mathrm{OH}^{-}$. In acid, $\mathrm{ROH}_2$ is first formed. Now, $\mathrm{Br}^{-}$displaces $\mathrm{H}_2 \mathrm{O}$, which is a very weak base and a good leaving group.

Mishal Gul
Mishal Gul
Numerade Educator
11:40

Problem 18

Optically pure $(S)-(+)-\mathrm{CH}_3 \mathrm{CHBr}-n-\mathrm{C}_6 \mathrm{H}_{13}$ has $[\alpha]_{\mathrm{D}}^{25}=+36.0^{\circ}$. A partially racemized sample having a specific rotation of $+30^{\circ}$ is reacted with dilute $\mathrm{NaOH}$ to form $(R)-(-)-\mathrm{CH}_3 \mathrm{CH}(\mathrm{OH})-n-\mathrm{C}_6 \mathrm{H}_{13}$ $\left([x]_{\mathrm{D}}^{25}=-5.97^{\circ}\right)$, whose specific rotation is $-10.3^{\circ}$ when optically pure. (a) Write an equation for the reaction using projection formulas. (b) Calculate the percent optical purity of reactant and product. (c) Calculate percentages of racemization and inversion. (d) Calculate percentages of frontside and backside attack. (e) Draw a conclusion concerning the reactions of $2^{\circ}$ alkyl halides. $(f)$ What change in conditions would increase inversion?
(a)
(b) The percentage of optically active enantiomer (optical purity) is calculated by dividing the observed specific rotation by that of pure enantiomer and multiplying the quotient by $100 \%$. The optical purities are:
$$
\text { Bromide }=\frac{+30^{\circ}}{+36^{\circ}}(100 \%)=83 \% \quad \text { Alcohol }=\frac{-5.97^{\circ}}{-10.3^{\circ}}(100 \%)=58 \%
$$
(c) The percentage of inversion is calculated by dividing the percentage of optically active alcohol of opposite configuration by that of reacting bromide. The percentage of racemization is the difference between this percentage and $100 \%$.
$$
\begin{aligned}
\text { Percentage inversion } & =\frac{58 \%}{83 \%}(100 \%)=70 \% \\
\text { Percentage racemization } & =100 \%-70 \%=30 \%
\end{aligned}
$$
(d) Inversion involves only backside attack, while racemization results from equal backside and frontside attack. The percentage of backside reaction is the sum of the inversion and one-half of the racemization; the percentage of frontside attack is the remaining half of the percentage of racemization.
$$
\begin{aligned}
& \text { Percentage backside reaction }=70 \%+\frac{1}{2}(30 \%)=85 \% \\
& \text { Percentage frontside reaction }=\frac{1}{2}(30 \%)=15 \%
\end{aligned}
$$
(e) The large percentage of inversion indicates chiefly $\mathrm{S}_{\mathrm{N}} 2$ reaction, while the smaller percentage of racemization indicates some $\mathrm{S}_{\mathrm{N}} 1$ pathway. This duality of reaction mechanism is typical of $2^{\circ}$ alkyl halides.
(f) The $\mathrm{S}_{\mathrm{N}} 2$ rate is increased by raising the concentration of the nucleophile-in this case, $\mathrm{OH}^{-}$.

Nicholas Sacco
Nicholas Sacco
Numerade Educator
01:01

Problem 19

Account for the following stereochemical results:
$\mathrm{H}_2 \mathrm{O}$ is more nucleophilic and polar than $\mathrm{CH}_3 \mathrm{OH}$. It is better able to react to give HS:-- ${ }_{\mathrm{R}}-\cdots \cdot \mathrm{SH}$ (see Problem 7.13 ), leading to racemization.

Narayan Hari
Narayan Hari
Numerade Educator
02:05

Problem 20

$\mathrm{NH}_3$ reacts with $\mathrm{RCH}_2 \mathrm{X}$ to form an ammonium salt, $\mathrm{RCH}_2 \mathrm{NH}_3^{+} \mathrm{X}^{-}$. Show the transition state, indicating the partial charges.
<smiles>[R]C([Y])[C@@H]([Y])[NH3+]</smiles>
$\mathrm{N}$ gains $\delta+$ as it begins to form a bond.

Susan Hallstrom
Susan Hallstrom
Numerade Educator
01:22

Problem 21

$\mathrm{H}_2 \mathrm{C}=\mathrm{CHCH}_2 \mathrm{Cl}$ is solvolyzed faster than $\left(\mathrm{CH}_3\right)_2 \mathrm{CHCl}$. Explain.
Solvolyses go by an $\mathrm{S}_{\mathrm{N}} 1$ mechanism. Relative rates of different reactants in $\mathrm{S}_{\mathrm{N}} 1$ reactions depend on the stabilities of intermediate carbocations. $\mathrm{H}_2 \mathrm{C}=\mathrm{CHCH}_2 \mathrm{Cl}$ is more reactive because
$$
\left[\mathrm{H}_2 \mathrm{C}=\mathrm{CH}=\mathrm{CH}_2\right]^{+}
$$
is more stable than $\left(\mathrm{CH}_3\right)_2 \mathrm{CH}^{+}$. See Problem 6.35 for a corresponding explanation of stability of an allyl radical.)

Catherine Lemar
Catherine Lemar
Numerade Educator

Problem 22

In terms of $(a)$ the inductive effect and $(b)$ steric factors, account for the decreasing stability of $\mathrm{R}^{+}$:

(a) Compared to $\mathrm{H}, \mathrm{R}$ has an electron-releasing inductive effect. Replacing $\mathrm{H}$ 's on the positive $\mathrm{C}$ by $\mathrm{CH}_3$ 's disperses the positive charge and thereby stabilizes $\mathrm{R}^{+}$.
(b) Steric acceleration also contributes to this order of $\mathrm{R}^{+}$stability. Some steric strain of the three $\mathrm{Me}$ 's in $\mathrm{Me}_3 \mathrm{C}-\mathrm{Br}$ separated by a $109^{\circ}$ angle $\left(s p^3\right)$ is relieved upon going to a $120^{\circ}$ separation in $\mathrm{R}^{+}$with a $\mathrm{C}$ using $s p^2$ hybrid orbitals.
Carbocations have been prepared as long-lived species by the reaction $\mathrm{RF}+\mathrm{SbF}_5 \rightarrow \mathrm{R}^{+}+\mathrm{SbF}_6^{-}, \mathrm{SbF}_5, \mathrm{a}$ covalent liquid, is called a superacid because it is a stronger Lewis acid than $\mathrm{R}^{+}$.

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01:31

Problem 23

How does the $\mathrm{S}_{\mathrm{N}} 1$ mechanism for the hydrolysis of 2-bromo-3-methylbutane, a $2^{\circ}$ alkyl bromide, to give exclusively the $3^{\circ}$ alcohol 2-methyl-2-butanol, establish a carbocation intermediate?

The initial slow step is dissociation to the $2^{\circ}$ 1,2-dimethylpropyl cation. A hydride shift yields the more stable $3^{\circ}$ carbocation, which reacts with $\mathrm{H}_2 \mathrm{O}$ to form the $3^{\circ}$ alcohol.

Henry R
Henry R
Numerade Educator
02:26

Problem 24

$\mathrm{RBr}$ reacts with $\mathrm{AgNO}_2$ to give $\mathrm{RNO}_2$ and $\mathrm{RONO}$. Explain. The nitrite ion, has two different nucleophilic sites: the $\mathrm{N}$ and either $\mathrm{O}$. Reaction with the unshared pair on $\mathrm{N}$ gives $\mathrm{RNO}_2$, while RONO is formed by reaction at $O$. [Anions with two nucleophilic sites are called ambident anions.]

Lottie Adams
Lottie Adams
Numerade Educator
01:14

Problem 25

In terms of transition-state theory, account for the following solvent effects: $(a)$ The rate of solvolysis of a $3^{\circ} \mathrm{RX}$ increases as the polarity of the protic nucleophilic solvent $(\mathrm{SH})$ increases, e.g.
$$
\mathrm{H}_2 \mathrm{O}>\mathrm{HCOOH}>\mathrm{CH}_3 \mathrm{OH}>\mathrm{CH}_3 \mathrm{COOH}
$$
(b) The rate of the $\mathrm{S}_{\mathrm{N}^2}$ reaction $: \mathrm{Nu}^{-}+\mathrm{RX} \rightarrow \mathrm{RNu}+: \mathrm{X}^{-}$decreases slightly as the polarity of protic solvent increases. (c) The rate of the $\mathrm{S}_{\mathrm{N}^2}$ reaction : $\mathrm{Nu}+\mathrm{RX} \rightarrow \mathrm{RNu}^{+}+: \mathrm{X}^{-}$increases as the polarity of the solvent increases. $(d)$ The rate of reaction in $(b)$ is greatly increased in a polar aprotic solvent. $(e)$ The rate of the reaction in $(b)$ is less in nonpolar solvents than in aprotic polar solvents.
See Table 7-2.

(e) In nonpolar solvents, $\mathrm{Nu}^{-}$is less reactive because it is ion-paired with its countercation, $\mathrm{M}^{+}$.
$\mathrm{S}_{\mathrm{N}} 2$ reactions with $\mathrm{Nu}^{-}$'s are typical of reactions between water-soluble salts and organic substrates that are soluble only in nonpolar solvents. These incompatible reactants can be made to mix by adding small amounts of phase-transfer catalysts, such as quaternary ammonium salts, $\mathrm{Q}^{+} \mathrm{A}^{-} . \mathrm{Q}^{+}$has a watersoluble ionic part and nonpolar $\mathrm{R}$ groups that tend to be soluble in the nonpolar solvent. Hence, $\mathrm{Q}^{+}$ shuttles between the two immiscible solvents while transporting $\mathrm{Nu}^{-}$, as $\mathrm{Q}^{+} \mathrm{Nu}^{-}$, into the nonpolar solvent; there $\mathrm{Nu}^{-}$quickly reacts with the organic substrate. $\mathrm{Q}^{+}$then moves back to the water while it transports the leaving group, $\mathrm{X}^{-}$, as $\mathrm{Q}^{+} \mathrm{X}^{-}$. Since the positive charge on the $\mathrm{N}$ of $\mathrm{Q}^{+}$is surrounded by the $\mathrm{R}$ groups, ion-pairing between $\mathrm{Nu}^{-}$and $\mathrm{Q}^{+}$is loose, and $\mathrm{Nu}^{-}$is quite free and very reactive.

Sana Riaz
Sana Riaz
Numerade Educator
08:23

Problem 26

Write equations for and explain the use of tetrabutyl ammonium chloride, $\mathrm{Bu}_4 \mathrm{~N}^{+} \mathrm{Cl}^{-}$, to facilitate the reaction between 1-heptyl chloride and cyanide ion.
$$
\begin{aligned}
\mathrm{Bu}_4 \mathrm{~N}^{+} \mathrm{Cl}^{-}+\mathrm{Na}^{+} \mathrm{CN}^{-} & \longrightarrow \mathrm{Bu}_4 \mathrm{~N}^{+} \mathrm{CN}^{-}+\mathrm{Na}^{+} \mathrm{Cl}^{-} \quad \text { (in water) } \\
n-\mathrm{C}_7 \mathrm{H}_{15} \mathrm{Cl}+\mathrm{Bu}_4 \mathrm{~N}^{+} \mathrm{CN}^{-} & \longrightarrow n-\mathrm{C}_7 \mathrm{H}_{15} \mathrm{CN}+\mathrm{Bu}_4 \mathrm{~N}^{+} \mathrm{Cl}^{-} \quad \text { (in nonpolar phase) }
\end{aligned}
$$

The phase-transfer catalyst, $\mathrm{Bu}_4 \mathrm{~N}^{+} \mathrm{Cl}^{-}$, reacts with $\mathrm{CN}^{-}$to form a quaternary cyanide salt that is slightly soluble in the organic phase because of the bulky, nonpolar, butyl groups. Reaction with $\mathrm{CN}^{-}$to form the nitrile is rapid because it is not solvated or ion-paired in the organic phase; it is a free, strong nucleophile. The phase-transfer catalyst, regenerated in the organic phase, returns to the aqueous phase, and the chain process is propagated.

Marissa Turner
Marissa Turner
Numerade Educator
02:53

Problem 27

(a) Why do alkyl halides rarely undergo the El reaction? $(b)$ How can the El mechanism be promoted?
(a) RX reacts by the El mechanism only when the base is weak and has a very low concentration; as the base gets stronger and more concentrated, the E2 mechanism begins to prevail. On the other hand, if the base is too weak or too dilute, either the $\mathrm{R}^{+}$reacts with the nucleophilic solvent to give the $\mathrm{S}_{\mathrm{N}} 1$ product or, in nonpolar solvents, RX fails to react.
(b) By the use of catalysts, such as $\mathrm{Ag}^{+}$, which help pull away the leaving group $\mathrm{X}^{-}$.

Madeline Currie
Madeline Currie
Numerade Educator
03:55

Problem 28

Why does $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{I}$ undergo loss of $\mathrm{HI}$ with strong base faster than $\mathrm{CD}_3 \mathrm{CH}_2 \mathrm{I}$ loses $\mathrm{Dl}$ ?
These are both $\mathrm{E} 2$ reactions in which the $\mathrm{C}-\mathrm{H}$ or $\mathrm{C}-\mathrm{D}$ bonds are broken in the rate-controlling step. Therefore, the H-D isotope effect accounts for the faster rate of reaction of $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{I}$.

Nima Gharibi
Nima Gharibi
Numerade Educator
05:42

Problem 29

Explain the fact that whereas 2-bromopentane undergoes dehydrohalogenation with $\mathrm{C}_2 \mathrm{H}_5 \mathrm{O}^{-} \mathrm{K}^{+}$to give mainly 2-pentene (the Sayzteff product), with $\mathrm{Me}_3 \mathrm{CO}^{-} \mathrm{K}^{+}$it gives mainly 1-pentene (the anti-Sayzteff, Hofmann, product).
Since $\mathrm{Me}_3 \mathrm{CO}^{-}$is a bulky base, its attack is more sterically hindered at the $2^{\circ} \mathrm{H}$ than at the $1^{\circ} \mathrm{H}$. With $\mathrm{Me}_3 \mathrm{CO}^{-}$,

Dan Ni
Dan Ni
Numerade Educator
00:43

Problem 30

Assuming that anti elimination is favored, illustrate the stereospecificity of the E2 dehydrohalogenation by predicting the products formed from $(a)$ meso- and $(b)$ either of the enantiomers of 2,3-dibromobutane. Use the wedge-sawhorse and Newman projections.

Aadit Sharma
Aadit Sharma
Numerade Educator
03:55

Problem 31

Account for the percentages of the products, (i) $\left(\mathrm{CH}_3\right)_2 \mathrm{CHOC}_2 \mathrm{H}_5$ and (ii) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2$, of the reaction of $\mathrm{CH}_3 \mathrm{CHBrCH}_3$ with
(a) $\mathrm{C}_2 \mathrm{H}_5 \mathrm{ONa} / \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \rightarrow 79 \%$
(ii) $+21 \%$ (i)
(b) $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \rightarrow 3 \%$
(ii) $+97 \%$ (i)
(a) $\mathrm{C}_2 \mathrm{H}_5 \mathrm{O}^{-}$is a strong base and $\mathrm{E} 2$ predominates. (b) $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$ is weakly basic but nucleophilic, and $\mathrm{S}_{\mathrm{N}} 1$ is favored.

Lottie Adams
Lottie Adams
Numerade Educator
05:21

Problem 33

Write the structure of the only tertiary halide having the formula $\mathrm{C}_5 \mathrm{H}_{11} \mathrm{Br}$.
In order for the halide to be tertiary, $\mathrm{Br}$ must be attached to a $\mathrm{C}$ that is attached to three other $\mathrm{C}$ 's; i.e., to no $\mathrm{H}$ 's. This gives the skeletal arrangement
involving four C's. A fifth $\mathrm{C}$ must be added, which must be attached to one of the C's on the central C, giving:
and with the H's.

Carina Carlos
Carina Carlos
Numerade Educator
02:59

Problem 34

On substitution of one $\mathrm{H}$ by a $\mathrm{Cl}$ in the isomers of $\mathrm{C}_5 \mathrm{H}_{12},(a)$ which isomer gives only a primary halide? (b) Which isomers give secondary halides? (c) Which isomer gives a tertiary halide?
(a) 2,2-Dimethylpropane. (b) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ gives 2-chloropentane and 3-chloropentane. 2-Methylbutane gives 2-chloro-3-methylbutane. (c) 2-Methylbutane gives 2-chloro-2-methylbutane.

Alkendra Singh
Alkendra Singh
Numerade Educator
00:56

Problem 35

Complete the following table:
CANT COPY

John Connell
John Connell
Numerade Educator

Problem 36

Give the organic product in the following substitution reactions. The solvent is given above the arrow.

(a) $\mathrm{HS}^{-}+\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHBrCH}_3 \stackrel{\mathrm{CH}_3 \mathrm{OH}}{\longrightarrow}$
(b) $\mathrm{I}^{-}+\left(\mathrm{CH}_3\right)_3 \mathrm{CBr} \stackrel{\mathrm{HCOOH}}{\longrightarrow}$
(c) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}+\mathrm{AgCN} \longrightarrow 2$ products
(d) $\mathrm{CH}_3 \mathrm{CHBrCH}_3+\mathrm{CH}_3 \mathrm{NH}_2 \longrightarrow$
(e) $\mathrm{CH}_3 \mathrm{CHBrCH}_3+\left(\mathrm{CH}_3\right)_2 \ddot{\mathrm{S}}: \longrightarrow$
(f) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}+: \mathrm{P}\left(\mathrm{C}_6 \mathrm{H}_5\right)_3 \longrightarrow$
(g) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}+$
<smiles>[3H][C@H]1CC[Si](O)(O)S1</smiles>
(thiosulfate ion)

(a) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHSHCH}_3$ (a mercaptan).
(b) $\left(\mathrm{CH}_3\right)_3 \mathrm{COCH} ; 3^{\circ} \mathrm{RX}$ undergoes $\mathrm{S}_{\mathrm{N}} 1$ solvolysis.
(c) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CN}+\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NC} ;: \mathrm{C}=\mathrm{N}$ :- $^{-}$is an ambident anion (Problem 7.24).
(d)
<smiles>CCCNC(C)C</smiles>
$\mathrm{Br}^{-}$; an ammonium salt.
(e) $\left[\left(\mathrm{CH}_3\right)_2 \mathrm{CHS}\left(\mathrm{CH}_3\right)_2\right]^{+} \mathrm{Br}^{-}$; a sulfonium salt.
(f) $\left[\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{P}\left(\mathrm{C}_6 \mathrm{H}_5\right)_3\right]^{+} \mathrm{Br}^{-}$; a phosphonium salt.
$(g)$
<smiles>CCSS(C)([O-])[O-]</smiles>
$\mathrm{S}$ is a more nucleophilic site than is $\mathrm{O}$.

Check back soon!
00:29

Problem 37

Account for the observation that catalytic amounts of $\mathrm{KI}$ enhance the rate of reaction of $\mathrm{RCH}_2 \mathrm{Cl}$ with $\mathrm{OH}^{-}$to give the alcohol $\mathrm{RCH}_2 \mathrm{OH}$.

Since $\mathrm{I}^{-}$is a better nucleophile than $\mathrm{OH}^{-}$, it reacts rapidly with $\mathrm{RCH}_2 \mathrm{Cl}$ to give $\mathrm{RCH}_2 \mathrm{I}$. But since $\mathrm{I}^{-}$is a much better leaving group than $\mathrm{Cl}^{-}, \mathrm{RCH}_2 \mathrm{I}$ reacts faster with $\mathrm{OH}^{-}$than does $\mathrm{RCH}_2 \mathrm{Cl}$. Only a catalytic amount of $\mathrm{I}^{-}$is needed, because the regenerated $\mathrm{I}^{-}$is recycled in the reaction.

Mishal Gul
Mishal Gul
Numerade Educator
02:28

Problem 38

Account for the following products from reaction of $\mathrm{CH}_3 \mathrm{CHOHC}\left(\mathrm{CH}_3\right)_3$ with $\mathrm{HBr}$ : (a)
<smiles>CC(C)C(C)C(C)(C)C</smiles>
(b) $\mathrm{H}_2 \mathrm{C} \equiv \mathrm{CHC}\left(\mathrm{CH}_3\right)_3$,
(c) $\left(\mathrm{CH}_3\right)_2 \mathrm{CHCHBr}\left(\mathrm{CH}_3\right)_2$,
(d) $\left(\mathrm{CH}_3\right)_2 \mathrm{CH}=\mathrm{CH}\left(\mathrm{CH}_3\right)_2$, (e)

For this $2^{\circ}$ alcohol, oxonium-ion formation is followed by loss of $\mathrm{H}_2 \mathrm{O}$ to give a $2^{\circ}$ carbocation that rearranges to a $3^{\circ}$ carbocation. Both carbocations react by two pathways: they form bonds to $\mathrm{Br}^{-}$to give alkyl bromides or they lose $\mathrm{H}^{+}$to yield alkenes.

ES
Eugene Schneider
University of Minnesota - Twin Cities
01:28

Problem 39

Show steps for the following conversions:
(a) $\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3 \longrightarrow \mathrm{CH}_3 \mathrm{CHBrCH}_2 \mathrm{CH}_3$
(c) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3 \longrightarrow \mathrm{CH}_2 \mathrm{ClCHClCH}$ Cl
(b) $\mathrm{CH}_3 \mathrm{CHBrCH}_2 \mathrm{CH}_3 \longrightarrow \mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$
To do syntheses, it is best to work backward while keeping in mind your starting material. As you do this, keep asking what is needed to make what you want. Always try to use the fewest steps.

(a) The precursors of the possible product are the corresponding alcohol, 1-butene, and 2-butene. The alcohol is a poor choice, because it would have to be made from either of the alkenes and an extra step would be needed. 2Butene cannot be made directly from the $1^{\circ}$ halide, but 1-butene can.
$$
\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3 \stackrel{\text { alc, } \mathrm{KOH}}{\longrightarrow} \mathrm{CH}_2=\mathrm{CHCH}_2 \mathrm{CH}_3 \stackrel{\mathrm{HBr}}{\longrightarrow} \mathrm{CH}_3 \mathrm{CHBrCH}_2 \mathrm{CH}_3
$$
(b) To ensure getting 1-butene, the needed precursor, use a bulky base for the dehydrohalogenation of the starting material.
$$
\mathrm{CH}_3 \mathrm{CHBrCH}_2 \mathrm{CH}_3 \stackrel{\mathrm{Me}_3 \mathrm{CO}^{-} \mathrm{K}^{+}}{\longrightarrow} \mathrm{CH}_2=\mathrm{CHCH}_2 \mathrm{CH}_3 \underset{\text { peroxides }}{\stackrel{\text { HBr }}{\longrightarrow}}-\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3
$$
(c) The precursor for vic dichlorides is the corresponding alkene; in this case, $\mathrm{H}_2 \mathrm{C}-\mathrm{CHCH}_2 \mathrm{Cl}$, which is made by allylic chlorination of propene. Although free-radical chlorination of propane gives a mixture of isomeric propyl chlorides, the mixture can be dehydrohalogenated to the same alkene, making this particular initial chlorination a useful reaction.
$$
\begin{aligned}
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3 \underset{\mathrm{Uv}}{\stackrel{\mathrm{Cl}_2}{\longrightarrow}} \mathrm{ClCH}_2 \mathrm{CH}_2 \mathrm{CH}_3+\mathrm{CH}_3 \mathrm{CHClCH}_3 \underset{\mathrm{KOH}}{\stackrel{\text { alc. }}{\longrightarrow}} \mathrm{CH}_2=\mathrm{CHCH}_3 \underset{\mathrm{uv}}{\stackrel{\mathrm{Cl}_2}{\longrightarrow}} \mathrm{CH}_2= & \mathrm{CHCH}_2 \mathrm{Cl} \\
& \downarrow \mathrm{Cl}_2 \\
& \mathrm{ClCH}_2 \mathrm{CHClCH}_2 \mathrm{Cl}
\end{aligned}
$$

Grigoriy Sereda
Grigoriy Sereda
Numerade Educator
01:38

Problem 40

Indicate the products of the following reactions and point out the mechanism as $\mathrm{S}_{\mathrm{N}} 1, \mathrm{~S}_{\mathrm{N}} 2$, E1 or E2.
(a) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}+\mathrm{LiAlH}_4$ (source of : $\mathrm{H}^{-}$)
(d) $\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{Br}+\mathrm{Mg}$ (ether)
(b) $\left(\mathrm{CH}_3\right)_3 \mathrm{CBr}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$, heat at $60^{\circ} \mathrm{C}$
(e) $\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}+\mathrm{Mg}$ (ether)
(c) $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCl}+\mathrm{NaNH}_2$
(f) $\mathrm{CH}_3 \mathrm{CHBrCH}_3+\mathrm{NaOCH}_3$ in $\mathrm{CH}_3 \mathrm{OH}$
(a) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3$; an $\mathrm{S}_{\mathrm{N}} 2$ reaction, : $\mathrm{H}^{-}$of $\mathrm{AlH}_4^{-}$replaces $\mathrm{Br}^{-}$.
(b)
(c) $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCl}+\mathrm{NaNH}_2 \longrightarrow \mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CH}+\mathrm{NH}_3+\mathrm{NaCl}$
Vinyl halides are quite inert toward $\mathrm{S}_{\mathrm{N}} 2$ reactions.
(d) $\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{Br}+\mathrm{Mg} \longrightarrow \mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2+\mathrm{MgBr}_2$
This is an $\mathrm{E} 2$ type of $\beta$-elimination via an alkyl magnesium iodide.
$$
\mathrm{Mg}+\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{Br} \longrightarrow \mathrm{BrMg}^{-} \overparen{\mathrm{CH}_2}-\mathrm{CH}_2-\bigodot_{\mathrm{Br}} \longrightarrow \mathrm{MgBr}_2+\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2
$$
(e) This reaction resembles that in $(d)$ and is an internal $\mathrm{S}_{\mathrm{N}} 2$ reaction.
(f) This $2^{\circ} \mathrm{RBr}$ undergoes both $\mathrm{E} 2$ and $\mathrm{S}_{\mathrm{N}} 2$ reactions to form propylene and isopropyl methyl ether.
$$
\mathrm{CH}_3 \mathrm{CHBrCH}_3+\stackrel{+}{\mathrm{NaO}}-\overline{C H} \mathrm{CH}_3\left(\mathrm{CH}_3 \mathrm{OH}\right) \longrightarrow \underset{\mathrm{OCH}_3}{\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2}+\underset{\mathrm{CH}_3 \mathrm{CHCH}_3}{\mathrm{CH}_3}
$$

Crystal Wang
Crystal Wang
Numerade Educator

Problem 41

Give structures of the organic products of the following reactions and account for their formation:
(a) $\mathrm{ClCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}+\mathrm{NaCN} \stackrel{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}{\longrightarrow}$
(b) $\mathrm{CH}_3 \mathrm{CHBrCH}_3+\mathrm{NaI}$ acetone
(c) $\mathrm{ClCH}_2 \mathrm{CH}=\mathrm{CH}_2+\mathrm{NaI} \stackrel{\text { acetone. }}{\longrightarrow}$

(a) $\mathrm{ClCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CN}$. $\mathrm{Br}^{-}$is a better leaving group than $\mathrm{Cl}^{-}$.
(b) $\mathrm{CH}_3 \mathrm{CHICH}_3$. Equilibrium is shifted to the right because $\mathrm{Nal}$ is soluble in acetone, while $\mathrm{NaBr}$ is not and precipitates.
(c) $\mathrm{ICH}_2 \mathrm{CH}=\mathrm{CH}_2 . \mathrm{Nal}$ is soluble in acetone and $\mathrm{NaCl}$ is insoluble.

Check back soon!
05:33

Problem 42

Account for the following observations when (S)- $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CHID}$ is heated in acetone solution with NaI: $(a)$ The enantiomer is racemized. (b) If radioactive ${ }^* 1^{-}$is present in excess, the rate of racemization is twice the rate at which the radioactive ${ }^* 1^{-}$is incorporated into the compound.
(a) Since enantiomers have identical energy, reaction proceeds in both directions until a racemic equilibrium mixture is formed.
(b) Each radioactive $* \mathrm{I}^{-}$incorporated into the compound forms one molecule of enantiomer. Now one unreacted molecule and one molecule of its enantiomer, resulting from reaction with ${ }^* \mathrm{I}^{-}$, form a racemic modification. Since two molecules are racemized when one ${ }^* 1^{-}$reacts, the rate of racemization will be twice that at which ${ }^* \mathrm{I}^{-}$ reacts.

David Collins
David Collins
Numerade Educator
07:01

Problem 43

Indicate the effect on the rate of $\mathrm{S}_{\mathrm{N}} 1$ and $\mathrm{S}_{\mathrm{N}} 2$ reactions of the following: (a) Doubling the concentration of substrate (RL) or $\mathrm{Nu}^{-}$. (b) Using a mixture of ethanol and $\mathrm{H}_2 \mathrm{O}$ or only acetone as solvent. (c) Increasing the number of $\mathrm{R}$ groups on the $\mathrm{C}$ bonded to the leaving group, $\mathrm{L}$. (d) Using a strong $\mathrm{Nu}^{-}$.
(a) Doubling either $[\mathrm{RL}]$ or $\left[\mathrm{Nu}^{-}\right]$doubles the rate of the $\mathrm{S}_{\mathrm{N}} 2$ reaction. For $\mathrm{S}_{\mathrm{N}} 1$ reactions the rate is doubled only by doubling $[R L]$ and is not affected by any change in $\left[\mathrm{Nu}^{-}\right]$.
(b) A mixture of ethanol and $\mathrm{H}_2 \mathrm{O}$ has a high dielectric constant and therefore enhances the rate of $\mathrm{S}_{\mathrm{N}} 1$ reactions. This usually has little effect on $\mathrm{S}_{\mathrm{N}} 2$ reactions. Acetone has a low dielectric constant and is aprotic and favors $\mathrm{S}_{\mathrm{N}} 2$ reactions.
(c) Increasing the number of $\mathrm{R}$ 's on the reaction site enhances $\mathrm{S}_{\mathrm{N}} 1$ reactivity through electron release and stabilization of $\mathrm{R}^{+}$. The effect is opposite in $\mathrm{S}_{\mathrm{N}} 2$ reactions because bulky $\mathrm{R}$ 's sterically hinder formation of, and raise $\Delta H^*$ for, the transition state.
(d) Strong nucleophiles favor $\mathrm{S}_{\mathrm{N}} 2$ reactions and do not affect $\mathrm{S}_{\mathrm{N}} 1$ reactions.

Catherine Smith
Catherine Smith
Missouri State University
01:01

Problem 44

List the following alkyl bromides in order of decreasing reactivity in the indicated reactions.
(I)
<smiles>CCC(C)(C)Br</smiles>
(II)
$$
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}
$$
(a) $\mathrm{S}_{\mathrm{N}} 1$ reactivity, (b) $\mathrm{S}_{\mathrm{N}} 2$ reactivity, (c) reactivity with alcoholic $\mathrm{AgNO}_3$.
(a) Reactivity for the $\mathrm{S}_{\mathrm{N}} 1$ mechanism is $3^{\circ}$ (I) $>2^{\circ}$ (III) $>1^{\circ}$ (II).
(b) The reverse reactivity for $\mathrm{S}_{\mathrm{N}} 2$ reactions gives $\mathrm{I}^{\prime}$ (II) $>2^{\circ}$ (III) $>3^{\circ}$ (I).
(c) $\mathrm{Ag}^{+}$catalyzes $\mathrm{S}_{\mathrm{N}} 1$ reactions and the reactivities are $3^{\circ}(\mathrm{I})>2^{\circ}$ (III) $>1^{\circ}$ (II).

Narayan Hari
Narayan Hari
Numerade Educator
07:30

Problem 45

Potassium tert-butoxide, $\mathrm{K}^{+} \overline{\mathrm{OCMe}} \mathrm{C}_3$, is used as a base in E2 reactions. (a) How does it compare in effectiveness with ethylamine, $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2$ ? (b) Compare its effectiveness in the solvents tert-butyl alcohol and dimethylsulfoxide (DMSO). (c) Give the major alkene product when it reacts with $\left(\mathrm{CH}_3\right)_2 \mathrm{CClCH}_2 \mathrm{CH}_3$.
(a) $\mathrm{K}^{+} \mathrm{OCMe}_3$ is more effective because it is more basic. Its larger size also precludes $\mathrm{S}_{\mathrm{N}} 2$ reactions.
(b) Its reactivity is greater in aprotic DMSO because its basic anion is not solvated. $\mathrm{Me}_3 \mathrm{COH}$ reduces the effectiveness of $\mathrm{Me}_3 \mathrm{CO}^{-}$by $\mathrm{H}$-bonding.
(c) $\mathrm{Me}_3 \mathrm{CO}^{-}$is a bulky base and gives the anti-Saytzeff (Hofmann) product $\mathrm{CH}_2=\mathrm{C}\left(\mathrm{CH}_3\right) \mathrm{CH}_2 \mathrm{CH}_3$.

Zubair Abdulla
Zubair Abdulla
Numerade Educator
02:28

Problem 46

Give structures of all alkenes formed and underline the major product expected from E2 elimination of: (a) 1-chloropentane, (b) 2-chloropentane.
(a) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl} \longrightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}=\mathrm{CH}_2$
A $1^{\circ}$ alkyl halide, therefore one alkene.
(b)
A $2^{\circ}$ alkyl halide flanked by two R's; therefore two alkenes are formed. The more substituted alkene is the major product because of its greater stability.

ES
Eugene Schneider
University of Minnesota - Twin Cities

Problem 48

State whether each of the following $\mathrm{R}^{+}$'s is stabilized or destabilized by the attached atom or group:
(a)
<smiles>C[C](C)C(C)(C)F</smiles>
(b) $: \ddot{\mathrm{F}}_3 \mathrm{C}^{+}$
(c)
<smiles>C[C](C)N</smiles>
(d)
<smiles>C[C](C)N</smiles>

If an electron-withdrawing group is adjacent to the positive $\mathrm{C}$, it will tend to destabilize the carbocation. Electron Donating groups, on the other hand, delocalize the + charge and serve to stabilize the carbocation.
(a) Destabilized. The strongly electron-withdrawing F's place a $\delta+$ on the atom adjacent to $\mathrm{C}^{+}$:
<smiles></smiles>
(Arrows indicate withdrawn electron density.)
(b) Stabilized. Each $\mathrm{F}$ has an unshared pair of clectrons in a $p$ orbital which can be shifted to $-\stackrel{+}{\mathrm{C}}-$ via $p-p$ orbital overlap.
<smiles>CC(F)=C(C)[I+]</smiles>
(c) Stabilized. The unshared pair of electrons on $\mathrm{N}$ can be contributed to $\mathrm{C}^{+}$.
<smiles>C[CH+]=C(C)C(C)=N</smiles>
(d) Destabilized. The adjacent $\mathrm{N}$ has a + charge.

Check back soon!
01:21

Problem 49

Account for the formation of
$$
\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_2 \mathrm{CN} \text { and }
$$
<smiles>C=CC(C)C#N</smiles>
from the reaction with $\mathrm{CN}^{-}$of 1 -chloro-2-butene, $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_2 \mathrm{Cl}$.
Formation of 1-cyano-2-butene results from $\mathrm{S}_{\mathrm{N}} 2$ reaction at the terminal $\mathrm{C}$.
$$
\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}^{-} \mathrm{CH}_2^{-}-\mathrm{Cl}^{-\mathrm{CN}^{-}} \longrightarrow \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_2-\mathrm{CN}+\mathrm{Cl}^{-}
$$

Attack by $\mathrm{CN}^{-}$can also occur at $\mathrm{C}^3$ with the $\pi$ electrons of the double bond acting as nucleophile to displace $\mathrm{Cl}^{-}$in an allylic rearrangement:

Hitendra Singh
Hitendra Singh
Numerade Educator
01:25

Problem 50

Calculate the rate for the $\mathrm{S}_{\mathrm{N}} 2$ reaction of 0.1-M $\mathrm{C}_2 \mathrm{H}_5 \mathrm{I}$ with 0.1-M $\mathrm{CN}^{-}$if the reaction rate for $0.01 \mathrm{M}$ concentration is $5.44 \times 10^{-9} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}$.
The rates are proportional to the products of the concentrations,
$$
\begin{aligned}
\frac{\text { Rate }}{5.44 \times 10^{-9} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}} & =\frac{[0.1][0.1]}{[0.01][0.01]} \\
\text { Rate } & =100 \times 5.44 \times 10^{-9} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}=5.44 \times 10^{-7} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}
\end{aligned}
$$

Narayan Hari
Narayan Hari
Numerade Educator
03:21

Problem 52

Will the following reactions be primarily displacement or elimination?
(a) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl}+\mathrm{I}^{-} \longrightarrow$
(b) $\left(\mathrm{CH}_3\right)_3 \mathrm{CBr}+\mathrm{CN}^{-}$(ethanol) $\longrightarrow$
(c) $\mathrm{CH}_3 \mathrm{CHBrCH}_3+\mathrm{OH}^{-}\left(\mathrm{H}_2 \mathrm{O}\right) \longrightarrow$
(d) $\mathrm{CH}_3 \mathrm{CHBrCH}_3+\mathrm{OH}^{-}$(ethanol) $\longrightarrow$
(e) $\left(\mathrm{CH}_3\right)_3 \mathrm{CBr}+\mathrm{H}_2 \mathrm{O} \longrightarrow$
(a) $\mathrm{S}_{\mathrm{N}} 2$ displacement. $\mathrm{I}^{-}$is a good nucleophile, and a poor base.
(b) E2 elimination. A $3^{\circ}$ halide and a fairly strong base.
(c) Mainly $\mathrm{S}_{\mathrm{N}} 2$ displacement.
(d) Mainly E2 elimination. A less polar solvent than that in (c) favors E2.
(e) $\mathrm{S}_{\mathrm{N}} \mathrm{I}$ displacement. $\mathrm{H}_2 \mathrm{O}$ is not basic enough to remove a proton to give elimination.

Raghvendra Singh
Raghvendra Singh
Numerade Educator
03:21

Problem 53

Depending on the solvent, $\mathrm{ROH}$ reacts with $\mathrm{SOCl}_2$ to give $\mathrm{RCl}$ by two pathways, each of which involves formation of a chlorosulfite ester,
<smiles>O=[Se](=O)Cl</smiles>
along with $\mathrm{HCl}$. Use the following stereochemical results to suggest mechanisms for the two pathways: In pyridine, a $3^{\circ}$ amine base,
$$
\text { (R)- } \mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_2 \mathrm{CH}_3 \longrightarrow(S)-\mathrm{CH}_3 \mathrm{CHClCH}_2 \mathrm{CH}_3
$$
and, in ether, the same $(R)-\mathrm{ROH} \longrightarrow(R)-\mathrm{RCl}$.
Pyridine (Py) reacts with the initially formed $\mathrm{HCl}$ to give $\mathrm{PyH}^{+} \mathrm{Cl}^{-}$and the free nucleophilic $\mathrm{Cl}^{-}$attacks the chiral $\mathrm{C}$ with inversion, displacing the $\mathrm{OSOCl}^{-}$as $\mathrm{SO}_2$ and $\mathrm{Cl}^{-}$. With no change in priority, inversion gives the $(S)$ $\mathrm{RCl}$. Ether is too weakly basic to cause enough dissociation of $\mathrm{HCl}$. In the absence of $\mathrm{Cl}^{-}$, the $\mathrm{Cl}$ of the $-\mathrm{OSOCl}$ attacks the chiral $\mathrm{C}$ from the side to which the group is attached. This internal nucleophilic substitution $\left(\mathbf{S}_{\mathrm{N}} \mathbf{i}\right)$ reaction proceeds through an ion-pair and leads to retention of configuration.

Raghvendra Singh
Raghvendra Singh
Numerade Educator