Test 5 (Lecture 5 / Module 5): 5 points 1. For each reaction below, look at the reagents and conditions (above and below the arrow) and draw the final product. Determine if it proceed via: SN1, SN2,...

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Test 5 (Lecture 5 / Module 5): 5 points 1. For each reaction below, look at the reagents and conditions (above and below the arrow) and draw the final product. Determine if it proceed via: SN1, SN2, or neither (no reaction) (2 points): (a) (0.4 points): Answer: (b) (0.4 points): Answer: (c) (0.4 points): Answer: (d) (0.4 points): Answer: (e) (0.4 points): Answer: 2. (a) Circle ONLY GOOD LEAVING GROUPS which can be replaced in SN1 or SN2 reactions (0.5 points): (b) Which one is THE BEST LEAVING GROUP (0.5 points): Hint: use the chart on page 151 (Chapter 9.4): Answer: 3. Identify if the following (SOLVENT, SUBSTRATE, NUCLEOPHILE, LEAVING GROUP) are best for SN1 or SN2 reactions or BOTH (2 points): SN1 SN2 SECONDARY (SUBSTRATE) ‒ (example) YES (example) TOSYLATE (OTs) GOOD (LEAVING GROUP) YES (example) YES (example) (NUCLEOPHILE) (SUBSTRATE) (NUCLEOPHILE) (NUCLEOPHILE) (SUBSTRATE) TRIFLATE (OTf) (LEAVING GROUP) (SUBSTRATE) WATER (H2O) (SOLVENT) DMSO (SOLVENT) (SOLVENT) MIDTERM (Lecture 1 – 5 / Module 1 – 5): 20 points 1. Show a mechanism for the acid-base reaction and determine the position of equilibrium. (5 points): (a) (2.5 points) Answer: (b) (2.5 points): Answer: 2. Draw the structures of the following compounds. (5 points): (a) 2-chloro-3-ethyl-2,4-difluoro-4-methylnonane (1 point): (b) 4-ethyl-2,4-dimethylcyclohexa-2,5-dien-1-one (1 point): (c) (Z)-2-bromo-1-chloro-1-fluoroprop-1-ene (1 point): (d) (1R,3R)-3-methylcyclohexan-1-ol (1 point): (e) 3-amino-5-bromo-4-cyclopropyl-2-fluoroheptan-1-al (1 point): 3. (a) Identify the type of hybridization (sp, sp2, sp3) of each atom (C atoms O atoms and N atoms) of the following compound (2 points): sp = BLUE circle or sphere sp2 = GREEN circle or sphere sp3 = RED circle or sphere (b) Identify the type of hybridization (sp, sp2, sp3) ONLY of N (NITROGEN) atoms of the following compound (2 points): 4. Draw a chair conformation, a flipped chair conformation and identify the most stable chair conformation and explain why (6 points): (a) (b) (c) (d) (e) (f) BONUS QUESTIONS: 1. For the compounds below (a) draw both Conjugate Base 1 and Conjugate Base 2. Determine which of the two protons is more acidic using the ARIO rule and explain your reasoning - if the reason is R (number of Resonance Structures) please draw all the resonance structures. Make sure you keep track of all formal charges and lone pairs (1 point): Conjugate Base 1: Resonance: Conjugate Base 2: Resonance: Answer: 2. Answer the questions and fill each box below (total 2 points = 0.1 points x 20 boxes): Find all of CHIRAL centers “Circle or label them if they exist” Is it a MESO form? YES/NO Name each CHIRAL center (R or S if they exist) Draw the ENANTIOMER “if it exists” Remember: meso forms do not have enantiomers Draw a DIASTEREOMER 3. Draw Newman projection for the compound below (2 points): Hint: You have to draw two connected Newman projections. Name: (1S,2R,5S)-2-isopropyl-5-methylcyclohexan-1-ol (A) Newman 1 (1 points): (B) Newman 2 (1 points):
Answered 1 days AfterJul 25, 2021

Answer To: Test 5 (Lecture 5 / Module 5): 5 points 1. For each reaction below, look at the reagents and...

Swapnil answered on Jul 26 2021
145 Votes
1
    
    2A
    The leaving groups whose conjugate bases are weak bases and the negative charge is stabi
lized are good leaving groups. Weak bases have low pKa value. Hence, the leaving groups which have low pKa value as conjugate base, will act as good leaving groups in SN1 Or SN2 reactions.
The list of the approximate pKa values of the conjugate bases of the substituents in the given compound is given in the table:
    Base
    pka values
    -OMe
    17
    F-
    3.1
    Br-
    -9
    I-
    -10
    Cl-
    -7
    -OTs
    -3
It can be seen that, the -OTs, Cl-, Br-, I- will act as leaving group and the -OMe and F- are non-leaving groups.
    2B
    The -OTs base having lower pKa value will act as excellent leaving group in this case according to me because of the better localization of the negative charge on the three oxygen atoms present in the group. (However, it will prefer SN1 mechanism due to the presence of tertiary carbocation intermediate after the leaving group...
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