교과서 질문
Show how you might synthesize the following compounds starting with bromobenzene, and alkyl or alkenyl halides of four carbon atoms or fewer.
a. 3-phenylprop-1-ene
1164
views
Ch. 15 - Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy
Wade9th EditionOrganic ChemistryISBN: 9780135213728
당신이 사용하는 게 아니라요?교과서 변경다른 교과서를 선택하세요
TOC.tableOfContents
- Ch.1 - Structure and Bonding107
- Ch. 2 - Acids and Bases; Functional Groups115
- Ch.3 - Structure and Stereochemistry of Alkanes100
- Ch.4 - The Study of Chemical Reactions99
- Ch.5 - Stereochemistry93
- Ch.6 - Alkyl Halides; Nucleophilic Substitution118
- Ch. 7 - Structure and Synthesis of Alkenes; Elimination158
- Ch.8 - Reactions of Alkenes171
- Ch.9 - Alkynes91
- Ch.10 - Structure and Synthesis of Alcohols143
- Ch.11 - Reactions of Alcohols104
- Ch. 12 - Infrared Spectroscopy and Mass Spectrometry22
- Ch. 13 - Nuclear Magnetic Resonance Spectroscopy45
- Ch. 14 - Ethers, Epoxides, and Thioethers72
- Ch. 15 - Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy89
- Ch. 16 - Aromatic Compounds74
- Ch. 17 - Reactions of Aromatic Compounds126
- Ch. 18 - Ketones and Aldehydes193
- Ch. 19 - Amines129
- Ch. 20 - Carboxylic Acids77
- Ch. 21 - Carboxylic Acid Derivatives141
- Ch. 22 - Condensations and Alpha Substitutions of Carbonyl Compounds175
- Ch. 23 - Carbohydrates and Nucleic Acids93
- Ch. 24 - Amino Acids, Peptides, and Proteins54
모든 교과서
Wade 9th EditionCh. 15 - Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy문제 10
Wade 9th EditionCh. 15 - Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy문제 1015장, 문제 10
When N-bromosuccinimide is added to hex-1-ene in CCl4 and a sunlamp is shone on the mixture, three products result.
(a) Give the structures of these three products.
(b) Propose a mechanism that accounts for the formation of these three products
검증된 단계별 안내1
Understand the role of N-bromosuccinimide (NBS): NBS is a reagent commonly used for allylic bromination in the presence of light or a radical initiator. It selectively brominates the allylic position of alkenes.
Analyze the structure of hex-1-ene: Hex-1-ene is a six-carbon chain with a double bond between the first and second carbons. The allylic positions are the carbons adjacent to the double bond (C3 and C5).
Step (a): Identify the three products: (1) The major product is the allylic bromination product where bromine is added to the allylic position (C3 or C5). (2) A minor product results from bromination at the other allylic position. (3) A third product forms due to addition of bromine across the double bond, resulting in a dibromoalkane.
Step (b): Propose the mechanism: (1) Initiation: Light (hv) generates bromine radicals (Br•) from NBS. (2) Propagation: A bromine radical abstracts an allylic hydrogen from hex-1-ene, forming an allylic radical. (3) The allylic radical reacts with Br2 (generated from NBS) to form the allylic bromination product. (4) Competing reaction: Bromine radicals can also add across the double bond, leading to the dibromoalkane.
Conclude the mechanism: The formation of the three products is explained by the selectivity of NBS for allylic bromination and the competing addition reaction across the double bond. The major product is the allylic bromination product, while the other two products are minor due to competing pathways.
비슷한 문제에 대한 검증된 영상 답변:
이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
15m주요 개념
질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.
Electrophilic Addition
Electrophilic addition is a fundamental reaction mechanism in organic chemistry where an electrophile reacts with a nucleophile, typically involving alkenes. In this case, hex-1-ene acts as a nucleophile, reacting with N-bromosuccinimide, which serves as the electrophile. This process leads to the formation of more stable products through the addition of bromine across the double bond.
추천 영상:
09:23
1,2 vs 1,4 Addition
Radical Mechanism
A radical mechanism involves the formation of free radicals, which are highly reactive species with unpaired electrons. In the presence of light, such as from a sunlamp, N-bromosuccinimide can undergo homolytic cleavage to generate bromine radicals. These radicals can then add to the alkene, leading to the formation of various products through radical intermediates.
추천 영상:
03:28The mechanism of Radical Polymerization.
Product Distribution
Product distribution refers to the different products formed in a chemical reaction and their relative amounts. In the reaction of hex-1-ene with N-bromosuccinimide, the conditions (like light and solvent) can influence which products are favored. Understanding the stability of intermediates and the pathways available helps predict the distribution of the three products formed in this reaction.
추천 영상:
11:07Bridged-Products
관련 실천
교과서 질문
Propose a mechanism for each reaction, showing explicitly how the observed mixtures of products are formed.
(e) 3-chlorobut-1-ene + AgNO3, H2O → but-2-en-1-ol + but-3-en-2-ol
892
views
교과서 질문
When Br2 is added to buta-1,3-diene at –15 °C, the product mixture contains 60% of product A and 40% of product B. When the same reaction takes place at 60 °C, the product ratio is 10% A and 90% B.
a. Propose structures for products A and B. (Hint: In many cases, an allylic carbocation is more stable than a bromonium ion.)
b. Propose a mechanism to account for formation of both A and B.
c. Show why A predominates at –15 °C and B predominates at 60 °C.
1058
views
교과서 질문
When Br2 is added to buta-1,3-diene at –15 °C, the product mixture contains 60% of product A and 40% of product B. When the same reaction takes place at 60 °C, the product ratio is 10% A and 90% B.
d. If you had a solution of pure A, and its temperature were raised to 60 °C, what would you expect to happen? Propose a mechanism to support your prediction.
758
views
교과서 질문
Show how you might synthesize the following compounds starting with bromobenzene, and alkyl or alkenyl halides of four carbon atoms or fewer.
b. 5-methylhex-2-ene
1034
views
교과서 질문
Addition of 1-bromobut-2-ene to magnesium metal in dry ether results in formation of a Grignard reagent. Addition of water to this Grignard reagent gives a mixture of but-1-ene and but-2-ene (cis and trans). When the Grignard reagent is made using 3-bromobut-1-ene, addition of water produces exactly the same mixture of products in the same ratios. Explain this curious result
1009
views