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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,...

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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 a
ow) 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 equili
ium. (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-
omo-1-chloro-1-fluoroprop-1-ene (1 point):



(d) (1R,3R)-3-methylcyclohexan-1-ol (1 point):


(e) 3-amino-5-
omo-4-cyclopropyl-2-fluoroheptan-1-al (1 point):

3. (a) Identify the type of hy
idization (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 hy
idization (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 After Jul 25, 2021

Solution

Swapnil answered on Jul 26 2021
158 Votes
1
    
    2A
    The leaving groups whose conjugate bases are weak bases and the negative charge is stabilized 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 ca
ocation intermediate after the leaving group...
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