Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Organic Chemistry Exam: Radical-Chain Bromination and Reaction Mechanisms, Exams of Organic Chemistry

An organic chemistry exam focused on radical-chain bromination and reaction mechanisms. It includes multiple-choice questions covering topics such as termination steps, rate-determining steps, polyhalogenation, and selectivity in radical chain substitution reactions. The exam also features free-response questions requiring students to complete reaction mechanisms and propose multi-step syntheses. This exam is designed to assess students' understanding of reaction intermediates, major products, and stereochemistry in organic reactions. It provides a comprehensive review of key concepts in organic chemistry, suitable for university-level study and exam preparation. Useful for students to test their knowledge and understanding of organic chemistry principles, reaction mechanisms, and synthesis strategies. It also helps in identifying areas where further study may be needed, making it a valuable resource for exam preparation and self-assessment.

Typology: Exams

2023/2024

Available from 06/01/2025

Usvah-Hasan
Usvah-Hasan 🇺🇸

5

(1)

101 documents

1 / 6

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
1
CHEM 3361/02 EXAM V Fall 2024 TOTAL POINTS 105
TURN IN ONLY THE FREE RESPONSE ANSWER SHEET AND THE BUBBLE SHEET. Keep this exam.
Record the following Version on the bubble sheet: A
Part I. Multiple Choice. Record your answers on the bubble sheet. (85.5pts, 4.5pts/ea.)
For Problems 1-9, consider the following mechanism steps in the radical-chain bromination of
propane. Some bond energies are given to you below in kcal/mole.
CH3CH2CH2--H
(CH3)2CH--H
CH3CH2CH2--Br
(CH3)2CH--Br
H--Br
Br--Br
98
94
69
68
87
46
1. Which of the above represents a termination step?
A. I B. II C. III D. IV E. V
2. Which of the above represents the rate-determining step leading to the major product of
monobromination of propane?
A. I B. II C. III D. IV E. V
3. Which of the above represents a step leading to polyhalogenation?
A. I B. II C. III D. IV E. V
pf3
pf4
pf5

Partial preview of the text

Download Organic Chemistry Exam: Radical-Chain Bromination and Reaction Mechanisms and more Exams Organic Chemistry in PDF only on Docsity!

CHEM 3361/02 EXAM V Fall 2024 TOTAL POINTS 105

TURN IN ONLY THE FREE RESPONSE ANSWER SHEET AND THE BUBBLE SHEET. Keep this exam. Record the following Version on the bubble sheet: A Part I. Multiple Choice. Record your answers on the bubble sheet. (85.5pts, 4.5pts/ea.) For Problems 1- 9 , consider the following mechanism steps in the radical-chain bromination of propane. Some bond energies are given to you below in kcal/mole. CH 3 CH 2 CH 2 --H (CH 3 ) 2 CH--H CH 3 CH 2 CH 2 --Br (CH 3 ) 2 CH--Br H--Br Br--Br 98 94 69 68 87 46

  1. Which of the above represents a termination step? A. I B. II C. III D. IV E. V
  2. Which of the above represents the rate-determining step leading to the major product of monobromination of propane? A. I B. II C. III D. IV E. V
  3. Which of the above represents a step leading to polyhalogenation? A. I B. II C. III D. IV E. V
  1. What would be the ratio of 1 - bromopropane to 2-bromopropane if the monohalogenation of were entirely unselective? A. 1:1 B. 3 :1 C. 2 :1 D. 6:1 E. 3: 2
  2. What is H in kcal/mol for the RATE DETERMINING STEP for the formation of isopropyl bromide (2-bromopropane) from Br 2 and propane? a. +7 b. - 7 c. +22 d. - 22 e. 11
  3. What is H in kcal/mol for the RATE DETERMINING STEP for the formation of n-propyl bromide (1-bromopropane) from Br 2 and propane? a. +7 b. - 7 c. +22 d. - 22 e. 11
  4. Which of the following reaction diagrams best represents a comparison of the rate determining steps above?
  5. The greater selectivity of bromine over chlorine in radical chain substitution reactions is because the A. H-Br bond is weaker than an H-Cl bond, making the first propagation endothermic. B. H-Br bond is weaker than an H-Cl bond, making the first propagation step exothermic. C. H-Br bond is stronger than an H-Cl bond, making the first propagation step endothermic. D. H-Br bond is stronger than an H-Cl bond, making the first propagation step exothermic.
  1. What is the product of the following reaction sequence?
  2. What is the product of the following reaction sequence? For Problems 1 6 - 18 , choose the most accurate description for the given product from the reaction below. A. One chiral center, 100% of one configuration. B. One chiral center gotten as a racemic mixture. C. Two chiral centers, both racemic. D. Two chiral centers, one with 100% of one configuration, the other racemic. E. Two chiral centers, one with 100% of one configuration, the other with an excess of one configuration over the other.
  3. I (D) 17. II (A) 18. III (B)

CHEM 3361/02 EXAM IV Fall 2024 NAME _____________________________________________ Free Response Answer Sheet. (24 pts, 12 pts/ea) Part II. Mechanism. Complete both initiation steps and both propagation steps filling in all non-bonding electrons showing curly-arrow mechanisms leading to the major product of the following reaction. (Continued on next page.)