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Concept explainers
(a)
Interpretation:
The structure of the alkene that could have been used to produce deuterated bromoalkene when treated with deuterium bromide (
Concept introduction:
The electrophilic addition of a Bronsted acid to the carbon-carbon double bond in
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Answer to Problem 11.8P
The structure of the corresponding alkene that has been used to produce the given deuterated bromoalkane is
Explanation of Solution
The structure of the given bromoalkane is
In the structure above, the carbon atom to which the bromine atom is attached must be the carbon bearing the positive charge. Thus, a stable tertiary carbocation must have been formed during the mechanism. Deuterium is an isotope of hydrogen having similar chemical properties. When an alkene is treated with deuterium bromide, the first step of addition is that the deuterium gets attached to the doubly bonded carbon atom which has got a higher number of hydrogens attached (or least substituted). Thus, the carbon atom to which deuterium is attached must be having a double bond in the original alkene. The resulting carbocation that would form will be secondary and can rearrange to a more stable, tertiary carbocation via 1, 2-hydride shift. Thus, a less stable carbocation undergoes carbocation rearrangement reaction to form the most stable carbocation. In the next step, the bromine ion attacks this tertiary carbocation to form the given product. Thus, the structure of the corresponding alkene that could have been used to produce the given deuterated bromoalkane must be
The complete mechanism is shown below:
In an electrophilic addition reaction, the carbocation rearrangement reaction takes place to form a rearranged product.
(b)
Interpretation:
The structure of the alkene that could have been used to produce deuterated bromoalkene when treated with deuterium bromide (
Concept introduction:
The electrophilic addition of a Bronsted acid to the carbon-carbon double bond in alkenes is susceptible to carbocation rearrangements due to the stability of the carbocation. The carbocation rearrangement occurs via either 1, 2 hydride shift or 1, 2 methyl shift depending on the stability of the carbocation formed. The stability order for carbocation is benzylic > tertiary > secondary > primary > methyl etc. The normal electrophilic addition gives 1, 2-addition product, but due to the rearrangement reaction, the rearranged product may be formed instead of 1, 2-addition product. In the reaction of an alkene and hydrogen halide or deuterium halide, the carbon to which the deuterium is attached is the carbon bearing the positive charge. Further, the carbon atom to which the halide atom is attached must be the carbon bearing the positive charge, that is, carbocation intermediate. Rearrangements of the carbocation intermediate may be observed if a more stable carbocation is possible.
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Answer to Problem 11.8P
The structure of the corresponding alkene that has been used to produce the given deuterated bromoalkane is
Explanation of Solution
The structure of the given bromoalkane is
In the structure above, the carbon atom to which the bromine atom is attached must be the carbon bearing the positive charge. Thus, a stable tertiary carbocation must have been formed during the mechanism. Deuterium is an isotope of hydrogen having similar chemical properties. When an alkene is treated with deuterium bromide, the first step of addition is that the deuterium gets attached to the doubly bonded carbon atom which has got a higher number of hydrogens attached (or least substituted). Thus, the carbon atom to which deuterium is attached must be having a double bond in the original alkene. The resulting carbocation that would form will be secondary and can rearrange to a more stable, tertiary carbocation via 1, 2-hydride shift. Thus, a less stable carbocation undergoes carbocation rearrangement reaction to form the most stable carbocation. In the next step, the bromine ion attacks this tertiary carbocation to form the given product.
Thus, the structure of the corresponding alkene that could have been used to produce the given deuterated bromoalkane must be
The complete mechanism is shown below:
In an electrophilic addition reaction, the carbocation rearrangement reaction takes place to form a rearranged product.
(c)
Interpretation:
The structure of the alkene that could have been used to produce deuterated bromoalkene when treated with deuterium bromide (
Concept introduction:
The electrophilic addition of a Bronsted acid to the carbon-carbon double bond in alkenes is susceptible to carbocation rearrangements due to the stability of the carbocation. The carbocation rearrangement occurs via either 1, 2 hydride shift or 1, 2 methyl shift depending on the stability of the carbocation formed. The stability order for carbocation is benzylic > tertiary > secondary > primary > methyl etc. The normal electrophilic addition gives 1, 2-addition product, but due to the rearrangement reaction, the rearranged product may be formed instead of 1, 2-addition product. In the reaction of an alkene and hydrogen halide or deuterium halide, the carbon to which the deuterium is attached is the carbon bearing the positive charge. Further, the carbon atom to which the halide atom is attached must be the carbon bearing the positive charge, that is, carbocation intermediate. Rearrangements of the carbocation intermediate may be observed if a more stable carbocation is possible.
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Answer to Problem 11.8P
The structure of the corresponding alkene that has been used to produce the given deuterated bromoalkane is
Explanation of Solution
The structure of the given bromoalkane is
In the structure above, the carbon atom to which the bromine atom is attached must be the carbon bearing the positive charge. Thus, a stable tertiary carbocation must have been formed during the mechanism. Deuterium is an isotope of hydrogen having similar chemical properties. When an alkene is treated with deuterium bromide, the first step of addition is that the deuterium gets attached to the doubly bonded carbon atom, which has got a higher number of hydrogens attached (or least substituted). Thus, the carbon atom to which deuterium is attached must be having a double bond in the original alkene. The resulting carbocation that would form will be secondary and can rearrange to a more stable, tertiary carbocation via 1, 2-hydride shift. Thus, a less stable carbocation undergoes carbocation rearrangement reaction to form the most stable carbocation. In the next step, the bromine ion attacks this tertiary carbocation to form the given product.
Thus, the structure of the corresponding alkene that has been used to produce the given deuterated bromoalkane is
The complete mechanism is shown below:
In an electrophilic addition reaction, the carbocation rearrangement reaction takes place to form a rearranged product.
(d)
Interpretation:
The structure of the alkene that could have been used to produce deuterated bromoalkene when treated with deuterium bromide (
Concept introduction:
The electrophilic addition of a Bronsted acid to the carbon-carbon double bond in alkenes is susceptible to carbocation rearrangements due to the stability of the carbocation. The carbocation rearrangement occurs via either 1, 2 hydride shift or 1, 2 methyl shift depending on the stability of the carbocation formed. The stability order for carbocation is benzylic > tertiary > secondary > primary > methyl etc. The normal electrophilic addition gives 1, 2-addition product, but due to the rearrangement reaction, the rearranged product may be formed instead of 1, 2-addition product. In the reaction of an alkene and hydrogen halide or deuterium halide, the carbon to which the deuterium is attached is the carbon bearing the positive charge. Further, the carbon atom to which the halide atom is attached must be the carbon bearing the positive charge, that is, carbocation intermediate. Rearrangements of the carbocation intermediate may be observed if a more stable carbocation is possible.
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Answer to Problem 11.8P
The structure of the corresponding alkene that has been used to produce the given deuterated bromoalkane is
Explanation of Solution
The structure of the given bromoalkane is
In the structure above, the carbon atom to which the bromine atom is attached must be the carbon bearing the positive charge. Thus, a stable tertiary carbocation must have been formed during the mechanism. Deuterium is an isotope of hydrogen having similar chemical properties. When an alkene is treated with deuterium bromide, the first step of addition is that the deuterium gets attached to the doubly bonded carbon atom, which has got a higher number of hydrogens attached (or least substituted). Thus, the carbon atom to which deuterium is attached must be having a double bond in the original alkene. The resulting carbocation that would form will be secondary and can rearrange to a more stable, tertiary carbocation via 1, 2-hydride shift. Thus, a less stable carbocation undergoes carbocation rearrangement reaction to form the most stable carbocation. In the next step, the bromine ion attacks this tertiary carbocation to form the given product.
Thus, the structure of the corresponding alkene that has been used to produce the given deuterated bromoalkane is
The mechanism for the reaction is shown below:
In an electrophilic addition reaction, the carbocation rearrangement reaction takes place to form a rearranged product.
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Chapter 11 Solutions
Organic Chemistry: Principles and Mechanisms (Second Edition)
- 3. Predict the major product and give a mechanism for the following reactions: (CH3)3COH/H₂SO4 a) b) NC CH₂O c) LOCH, (CH3)3COH/H2SO4 H,SO -OHarrow_forwardIndicate if the aldehyde shown reacts with the provided nucleophiles in acid or base conditions. a NaBH4 be Li eli -NH2 P(Ph3) f KCN g OH excess h CH3OH i NaCHCCH3arrow_forwardPredict the major products of the following organic reaction: + A ? Some important notes: • Draw the major product, or products, of the reaction in the drawing area below. • If there aren't any products, because no reaction will take place, check the box below the drawing area instead. • Be sure to use wedge and dash bonds when necessary, for example to distinguish between major products that are enantiomers. Explanation Check Click and drag to start drawing a structure. C © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Centearrow_forward
- Polar solutes are most likely to dissolve into _____, and _____ are most likely to dissolve into nonpolar solvents. A. nonpolar solutes; polar solvents B. nonpolar solvents; polar solvents C. polar solvents; nonpolar solutes D. polar solutes; nonpolar solventsarrow_forwardDeducing the Peactants Can the molecule on the right-hand side of this organic reaction be made in good yield from no more than two reactants, in one step, by moderately heating the reactants? ? Δ If your answer is yes, then draw the reactant or reactants in the drawing area below. You can draw the reactants in any arrangement you like. If your answer is no, check the box under the drawing area instead. Explanation Check Click and drag to start drawing a structure. © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center Xarrow_forwardDraw all 8 stereoisomers, circling each pair of enantiomer(s)/ mirror image compound(s)arrow_forward
- Bookmarks Profiles Tab Window Help Chemical Formula - Aktiv Che X + → C 11 a app.aktiv.com Google Chrome isn't your default browser Set as default Question 12 of 16 Q Fri Feb 2 Verify it's you New Chrome availabl- Write the balanced molecular chemical equation for the reaction in aqueous solution for mercury(I) nitrate and chromium(VI) sulfate. If no reaction occurs, simply write only NR. Be sure to include the proper phases for all species within the reaction. 3 Hg(NO3)2(aq) + Cг2(SO4)3(aq) → 3 Hg₂SO (s) + 2 Cr(NO3), (aq) ean Ui mate co ence an climate bility inc ulnerabili women, main critic CLIMATE-INI ernational + 10 O 2 W FEB 1 + 4- 3- 2- 2 2 ( 3 4 NS 28 2 ty 56 + 2+ 3+ 4+ 7 8 9 0 5 (s) (1) Ch O 8 9 (g) (aq) Hg NR CI Cr x H₂O A 80 Q A DII A F2 F3 FA F5 F6 F7 F8 F9 #3 EA $ do 50 % 6 CO & 7 E R T Y U 8 ( 9 0 F10 34 F11 川 F12 Subr + delete 0 { P }arrow_forwardDeducing the reactants of a Diels-Alder reaction n the molecule on the right-hand side of this organic reaction be made in good yield from no more than two reactants, in one step, by moderately heating the reactants? ? Δ • If your answer is yes, then draw the reactant or reactants in the drawing area below. You can draw the reactants in any arrangement you like. • If your answer is no, check the box under the drawing area instead. Explanation Check Click and drag to start drawing a structure. >arrow_forwardPredict the major products of the following organic reaction: + Some important notes: A ? • Draw the major product, or products, of the reaction in the drawing area below. • If there aren't any products, because no reaction will take place, check the box below the drawing area instead. • Be sure to use wedge and dash bonds when necessary, for example to distinguish between major products that are enantiomers. Explanation Check Click and drag to start drawing a structure.arrow_forward
- if the answer is no reaction than state that and please hand draw!arrow_forward"I have written solutions in text form, but I need experts to rewrite them in handwriting from A to Z, exactly as I have written, without any changes."arrow_forwardDon't used hand raiting and don't used Ai solutionarrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning
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