CHEM 236_Spring 2010_Exam 4_April 28

CHEM 236_Spring - First Three Letters of Last Name NAME Network ID Section CHEMISTRY 236 SPRING 2010 EXAM IV Note The last pages of this exam are a

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: First Three Letters of Last Name NAME ________________________ Network ID ______________________ Section __________________ CHEMISTRY 236 SPRING 2010 EXAM IV APRIL 28, 2010 Note: The last pages of this exam are a periodic table, IR and NMR values. 1 2 3 4 5 6 7 8 (6) (4) (24) (12) (20) (12) (12) (10) 100 CHEM 236 – Spring 2010 2 Exam IV 1. [6 points] a. Draw a correct structure for: 3‐ethyl‐1,6‐heptadiyne b. Provide a correct name for: 2. [4 points] Draw the resonance structures of the carbocation that would be formed in the first step of the SN1 reaction of: I CHEM 236 – Spring 2010 3 Exam IV 3. [24 points] Fill in the blanks with a correct product or reactant for the following reactions: a. b. HgSO4 H2SO4 O c. d. CHEM 236 – Spring 2010 4 Exam IV 4. Answer the following: a. [8 points] Assign the 1H NMR spectrum of 4‐m ethylpent‐1‐en‐3‐one. Explain your reasoning for each assignment. b. [4 points] The base peak in the mass spectrum of 4‐methylpent‐1‐en‐3‐one occurs at m/z = 55. Draw a mechanism demonstrating how this peak arises from fragmentation of the molecular ion (m/z = 98). CHEM 236 – Spring 2010 5 Exam IV 5. [20 points] Provide additional reactants and/or reagents that would be needed to carry out the following conversions. Show the structures for any isolable intermediates. CO2CH3 CH3O2CC to CCO2CH3 CO2CH3 CHEM 236 – Spring 2010 6 Exam IV 6. Provide structures for A, B, C, and D and answer the questions. a. [6 points] The reduction of 4‐methylpent‐1‐en‐3‐one with hydrogen leads to two different products, A and B, both of formula C6H12O. Given the following spectroscopic data, identify A and B. Explain how the given data supports your proposed structures. A -1 IR (cm ) 13 C NMR (δ) B 3400 (broad), 1650 (moderate) 1710 (strong) 137, 115, 85, 36, and 17 213, 38, 31, 17, 8 CHEM 236 – Spring 2010 7 Exam IV b. [6 points] The reaction of cyclohexadiene and p‐quinnone gives two isomeric products C and D, C12H12O2. In protic solvents C and D undergo carbonyl to enol conversions to give E. Draw the structures of C and D and a mechanism for their formation. Suggest a reason why E is more stable than C or D. CHEM 236 – Spring 2010 8 Exam IV 7. [12 points] a. Provide a structural definition of an aromatic or heteroaromatic compound. b. Under each of the following structures indicate whether the compound is aromatic (use A) or non‐aromatic (use N). Note the two structures in the third row have ionic charges. CHEM 236 – Spring 2010 9 Exam IV 8. [10 points] Provide mechanisms for the reactions shown: CHEM 236 – Spring 2010 10 Exam IV CHEM 236 – Spring 2010 11 Exam IV CHEM 236 – Spring 2010 12 Exam IV CHEM 236 – Spring 2010 13 Exam IV ...
View Full Document

This note was uploaded on 12/30/2011 for the course CHEM 236 taught by Professor Baranger during the Spring '08 term at University of Illinois at Urbana–Champaign.

Ask a homework question - tutors are online