Module 6 Problem Set_ Principles of Organic Chemistry with Lab-2021-Gallaher
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Module 6 Problem Set: Principles of Organic Chemistry with Lab-2021-Gallaher
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Module 6 Problem Set
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14
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Question 1
Name the following amines using IUPAC systematic nomenclature
methodology:
a.
2-aminopentane
Correct!
Correct!
b.
N-ethyl-N-methylpropy
ou Answered
ou Answered
N-
methylpropylamine
Correct Answer
Correct Answer
12/14/23, 8:43 PM
Module 6 Problem Set: Principles of Organic Chemistry with Lab-2021-Gallaher
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c.
N-methylcyclohexylam
Correct!
Correct!
d.
N-methylpropylamine
ou Answered
ou Answered
N-ethyl-N-
methylpropylamine
Correct Answer
Correct Answer
0 / 0 pts
Question 2
Your Answer:
Classify each of the following amines as primary, secondary, or tertiary:
A. secondary
B. secondary
C. primary
D. tertiary
Answer:
a. Primary
b. Secondary
c. Secondary
d. Tertiary
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Question 3
Your Answer:
Assuming no significant difference in molecular weight, arrange the
following in order of increasing boiling point (lowest BP to highest BP),
and provide a brief explanation for your ranking.
Amine
Alcohol
Alkane
Alkane > Amine > Alcohol
Answer:
Alkane < Amine < Alcohol
The alkane has no H-bonding attractions. While both the
amine and alcohol are capable of forming H-bonds, the O-H is
more polar than the N-H, and as a result, the hydrogen
bonding in alcohols is stronger than that of amines.
0 / 0 pts
Question 4
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Your Answer:
Name the three major synthetic routes to the synthesis of amines as
presented in the module.
alkylation, reduction or nitrogen compounds, reductive amination
Answer:
1.
Alkylation of ammonia and other amines (reaction with
alkyl halides)
2.
Reduction of nitrogen compounds (aromatic nitro, amides,
nitriles)
3.
Reductive amination
0 / 0 pts
Question 5
Your Answer:
Briefly explain why amines are stronger bases than amides.
The chemistry of all amines is dominated by the lone pair of electrons on
the nitrogen atom. The lone pair makes amines both basic and
nucleophilic.
Answer:
Amines and amides can both act as bases by using the lone
pair on the nitrogen atom to accept a proton. The lone pair of
an amide is less available to accept a proton because it is tied
up in resonance with the adjacent carbonyl. Since no
resonance is available in amines, the lone pair is more
available and hence more basic.
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Question 6
Your Answer:
Provide the IUPAC name and subclassification (1°,2°,3°) for each of the
following amides:
a) n-mthyl-n-cyclo-pentide, 3°
b) n-methylpropamide, 2°
c) pentanamide, 1°
Answer:
a) N,N-diethylcyclohexanecarboxamide; Tertiary
b) N-methylpropanamide; Secondary
c) Pentanamide; Primary
0 / 0 pts
Question 7
Arrange the following list of carboxylic acid derivatives according to their
reactivity, from least reactive to most reactive:
Acid Anhydride
12/14/23, 8:43 PM
Module 6 Problem Set: Principles of Organic Chemistry with Lab-2021-Gallaher
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Your Answer:
Ester
Acid Halide
Amide
amide > ester > acid anhydride > acid halide
Answer:
Amide < Ester < Acid Anhydride <
Acid Halide
(Least Reactive) -------------------->
(Most Reactive)
0 / 0 pts
Question 8
What subclassification of amine product would form in each of the
following alkylation reactions? You do not have to generate the IUPAC
name of any product, just state the degree of structural substitution as 1°,
2°,3°, or quaternary.
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Your Answer:
A. 3°
B. 2°
C. quaternary
D. 1°
Answer:
a. Tertiary
b. Secondary
c. Quaternary
d. Primary
0 / 0 pts
Question 9
Your Answer:
Explain why most amides have a planar geometry around the nitrogen
atom.
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The rotation around the bond is greatly restricted due to the resonance
contribution.
Answer:
Even though the C-N bond is usually represented in
structures as a single covalent bond, rotation around this
bond is greatly restricted due to the resonance contribution of
the lone pair of electrons from the nitrogen to the carbonyl
carbon. The resonance hybrid has sufficient C=N character
such that it behaves as if it was a true double bond. The
geometric consequences of this resonance in amides are that
the carbonyl carbon and the nitrogen, as well as the two
atoms or groups attached to them, are lying in the same plane
(trigonal planar) molecular geometry.
0 / 0 pts
Question 10
Your Answer:
Explain why the
diazonium group is such a good leaving group when it is
replaced by a nucleophile in a synthetic reaction.
When the diazonium group leaves, it leaves very quickly as a gas and is
replaced quickly.
Answer:
Upon substitution by a nucleophile, the diazonium group
leaves as
nitrogen gas
, which bubbles from the reaction
mixture in a very fast and efficient reaction.
12/14/23, 8:43 PM
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Question 11
Your Answer:
What does it mean to say that the nitrogen of an amine is in “a reduced
state”?
meaning all bonds to the nitrogen are to either carbon or hydrogen
Answer:
This means that all bonds to the nitrogen are either to carbon
or hydrogen atoms.
0 / 0 pts
Question 12
Your Answer:
Explain why
reductive amination
is
the preferred method for the synthesis
of amines, as compared to direct alkylation.
Reductive amination avoids the potential for polyalkylation and production
of mixed products.
Answer:
In laboratory practice, reductive amination is the preferred method
for the synthesis of amines (as compared to direct alkylation), as it
avoids the potential for polyalkylation and the production of a
mixture of products.
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0 / 0 pts
Question 13
Your Answer:
Name the following nitriles using IUPAC systematic methodology.
pentanenitrile
Answer:
a. pentanenitrile
b. 3-methylbutanenitrile
c. cyclopentanecarbonitrile
10 / 10 pts
Question 14
Your Answer:
As a reminder, the questions in this review quiz are a requirement of the
course and the best way to prepare for the module exam. Did you
complete all questions in their entirety?
yes
Quiz Score:
10
out of 10
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Maps MTH 119, section...
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Using standard absolute entropies at 298 K, calculate the entropy change for the system when 1.57 moles of Fe3O4(s) react at standard conditions.
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- 4,6,7,8) Question 19 of 20 (5 points) Question Attempt: 1 of 1 Time Remaining: 30:47 Hannah 10 = 11 = 12 =13 =14 =15 = 16 17 18 19 20 Predict the reactants of this chemical reaction. That is, fill in the left side of the chemical equation. Be sure the equation you submit is balanced. (You can edit both sides of the equation to balance it, if you need to.) do Note: you are writing the molecular, and not the net ionic equation. - CaBr, (aq) + H,O(1) Submit Assignment Continue Accessibility Privacy Center 2021 McGraw Hill LLC All Rights Reserved. Terms of Use IIarrow_forwardBb Take Test: QUI A Ims.uhb.edu.sa/webapps/assessment/take/take.jsp?cours... لا تعيد السمكات إلى ا. . . الاثاد M Gmail O YouTube Human Biology Ch... A Apps Remaining Time: 15 minutes, 04 seconds. * Question Completion Status: 0.5 points Saved Question 6 Find the major product of following dehydration. OH CH3 CH,-CH-CH-CH2-CH3 + H20 CH3 OA CH2=CH-ĊH-CH2-CH3 OCH3 O B. CH3-C=c-CH2-CH3 OC. CH3-C-CH,- -CH2-CH2-CH3 O CH3 OD. CH3-C-CH-CH2-CH3 CH3 CH3-CH=C-CH2-CH3 » A Moving to another question will save this response. Question 6 of 10> .w. MacEarrow_forwardView History Bookmarks People Tab Window Help S, Amaya K - Outlo x B Chapter 5- CHEM-1110-941- x Question 20 - Chapter 5- Con x education.com/ext/map/index.html?_con%3con&external_browser%3D0&launchUrl=https%253A%252F%252Flms Saved 3 attempts left Check my work Enter your answer in the provided box. It takes 155 s for an unknown gas to effuse through a porous wall and 97 s for the same volume of N2 gas to effuse at the same temperature and pressure. What is the molar mass of the unknown gas? g/mol ... 115 MAR 1664 18 ottv Aarrow_forward
- We are stuck on 9. We get how Q is 16 and rxn goes left. at eq 2+ x. 2 + x. -2 x We get x= 1. Therefore k = 4. Please help us. Questin 9. See conditions from 8 This is not a graded question as it is a practice question . I am 60 years old and helping my son prepare for the AP exam in a few months. We do questions at the back of the textbook by Zumdahl and Zumdahlarrow_forwardESF X empt/quiz_start_frame_auto.d21?ou=9739790&isprv=&qi=11007766&cfql=0&dnb=0&fromQB=0&inProgress=1 ystems Question 5 (1 point) Ben starts walking along a path at 3 mi hr¹. Amanda starts jogging along the same path 1 hour later at 5 mi hr. After how much time jogging does Amanda catch up to Ben? 3.5 hr 2.5 hr 2 hr 16arrow_forwardB Quizzes - SCH3UE-15-Chemistry- X + A tdsb.elearningontario.ca/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d2l?ou=19297865&isprv=&drc=1&qi=5956344&cfql=0&dnb=D0... 2 E TDSB Bookmarks P Summary - PayPal Steam Guard Mobil.. M Booter Panel ul Asylum Stresser - H.. A Free Stresser - The. 6 Str3ssed Booter - L. O Home | IP Stresser 3 - Packages >> What can be added to a solution containing Sr2*(aq) ions to precipitate the ions from the solution? potassium sulfate barium hydroxide O lithium nitrate Two of the above are correct. (a). (b), and (c) are all correct. 围 i 11:15arrow_forward
- B Quizzes - SCH3UE-15-Chemistry- X + A tdsb.elearningontario.ca/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d2l?ou=19297865&isprv=&drc=1&qi=5956344&cfql=0&dnb=D0... 2 E TDSB Bookmarks P Summary - PayPal Steam Guard Mobil. Booter Panel l Asylum Stresser - H. A Free Stresser - The. O Str3ssed Booter - . O Home | IP Stresser 3 - Packages >> In testing for the presence of lead(II) ions in solution, Jamaal adds a solution of sodium nitrate. What will be the conclusion of the test if no precipitate forms? Lead(II) ions are not present in solution. Lead(II) ions are present in the solution. Lead(II) ions are present, but in a very small concentration. More sodium nitrate solution needs to be added before a conclusion can be made. No conclusion can be made in this test. 围 1 11:14arrow_forwardPlease helparrow_forwardDrowser Time Limit: 0:20:00 Time Left:0:15:19 Glenton Morris: Attempt 1 160 40 3.0 Page 1: 140 - 35 3.0 120 10 25 100 2.0 80 60 Z 10 Page 2: 05- 20 0 2. Time is 10 Timetsk 10 Page 3: Of all the graphs shown above, what plot type is correctly represented? O second order Page 4: O no plot shown is valid 4. 1st order 3rd order OL X (W) FONarrow_forward
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