Compound B has a molecular formula of C,H,0. What is the Unsaturation Number (UN) or Double-Bond Equivalent (DBE)? UN/DBE: Based on the UN/DBE you calculated, could there be a double bond in compound B? yes no Based on the UN/DBE you calculated, could there be a ring in compound B? yes no

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Given all the following information, please fill out everything i didn't I am trying to use this to study and I am very confused, everything I have filled out is correct I confirmed with my professor, but I am so confused on the rest of it

Compound B has a molecular formula of C,H0. What is the Unsaturation Number (UN) or Double-Bond
Equivalent (DBE)?
UN/DBE:
Based on the UN/DBE you calculated, could there be a double bond in compound B?
yes
no
Based on the UN/DBE you calculated, could there be a ring in compound B?
yes
no
Based on the molecular formula and the UN/DBE you calculated, check all the functional groups that could be present in
compound B.
O ketone
O benzene ring
|| ester
O alkyne
O alcohol
O aldehyde
O carboxylic acid
O alkene
O nitrile
O amine
О ерохide
O ether
O ring
Examine the IR spectrum provided for compound B.
The peaks below 1500 cannot be used to make any structural assignments because they are in the fingerprint region.
There are the two important peaks in this IR spectrum: peak 1 which is 3400 cm-1, and peak 2 which is 2950 cm-1.
What is peak 1 most likely due to?
What is peak 2 most likely due to?
О-Hor N-H
OC-O or N-0
С-Н (sp)
C=0
C-H (sp?)
O-H or N-H
OC-H (sp³)
O C-H (sp²)
O C=0
C-H (sp')
O C-0 or N-0
OC-H (sp)
Earlier, using the molecular formula and the UN/DBE, you came up with a list of functional groups that could be present in
compound B. Now, using the IR data you should be able to narrow this list.
Check all the functional groups that could be present in compound B.
О ерохide
alkyne
O alkene
O amine
O ring
aldehyde
O nitrile
ester
O ketone
I alcohol
O ether
O benzene ring
O carboxylic acid
Examine the 'H spectrum of compound B.
There is a table summarizing the spectral data. Using that data, assign each peak to the type of alkyl group.
ppm
Integration
Splitting
A
3.60
1 H
singlet
B
3.20
1 H
triplet
C
1.90
2H
multiplet
D
0.90
12 H
doublet
Based on the IR spectrum you determined that there will be an alcohol functional group. Which signal in the 'H spectrum
is due to that OH?
O A
B
OD
Based on the carbon spectrum you know that the molecule has a high degree of symmetry. Yet compound B has only one
O. Therefore the OH group and the carbon that it is attached to have to belong in the plane of symmetry. Which 'H signal
is due to the hydrogen that is on the same carbon as the OH group?
O A
B
Examine signal D. It has an integration of 12. How many equivalent methyl (CH,) groups are there in compound B?
number of equivalent methyl groups:
Now, you need to arrange the methyl groups in such a way that each of them is bound to a carbon that only has one proton
on it.
Hint: Calculate how many carbons and hydrogens you have not used yet.
What is the name of this alkyl fragment?
O ethyl group
O tert-butyl group
isopropyl group
O n-propyl group
Propose a structure for compound B.
Once you have a structure, it is always helpful to go back through the spectral data and make sure that the structure you
came up is consistent with all of it.
O O O O
Transcribed Image Text:Compound B has a molecular formula of C,H0. What is the Unsaturation Number (UN) or Double-Bond Equivalent (DBE)? UN/DBE: Based on the UN/DBE you calculated, could there be a double bond in compound B? yes no Based on the UN/DBE you calculated, could there be a ring in compound B? yes no Based on the molecular formula and the UN/DBE you calculated, check all the functional groups that could be present in compound B. O ketone O benzene ring || ester O alkyne O alcohol O aldehyde O carboxylic acid O alkene O nitrile O amine О ерохide O ether O ring Examine the IR spectrum provided for compound B. The peaks below 1500 cannot be used to make any structural assignments because they are in the fingerprint region. There are the two important peaks in this IR spectrum: peak 1 which is 3400 cm-1, and peak 2 which is 2950 cm-1. What is peak 1 most likely due to? What is peak 2 most likely due to? О-Hor N-H OC-O or N-0 С-Н (sp) C=0 C-H (sp?) O-H or N-H OC-H (sp³) O C-H (sp²) O C=0 C-H (sp') O C-0 or N-0 OC-H (sp) Earlier, using the molecular formula and the UN/DBE, you came up with a list of functional groups that could be present in compound B. Now, using the IR data you should be able to narrow this list. Check all the functional groups that could be present in compound B. О ерохide alkyne O alkene O amine O ring aldehyde O nitrile ester O ketone I alcohol O ether O benzene ring O carboxylic acid Examine the 'H spectrum of compound B. There is a table summarizing the spectral data. Using that data, assign each peak to the type of alkyl group. ppm Integration Splitting A 3.60 1 H singlet B 3.20 1 H triplet C 1.90 2H multiplet D 0.90 12 H doublet Based on the IR spectrum you determined that there will be an alcohol functional group. Which signal in the 'H spectrum is due to that OH? O A B OD Based on the carbon spectrum you know that the molecule has a high degree of symmetry. Yet compound B has only one O. Therefore the OH group and the carbon that it is attached to have to belong in the plane of symmetry. Which 'H signal is due to the hydrogen that is on the same carbon as the OH group? O A B Examine signal D. It has an integration of 12. How many equivalent methyl (CH,) groups are there in compound B? number of equivalent methyl groups: Now, you need to arrange the methyl groups in such a way that each of them is bound to a carbon that only has one proton on it. Hint: Calculate how many carbons and hydrogens you have not used yet. What is the name of this alkyl fragment? O ethyl group O tert-butyl group isopropyl group O n-propyl group Propose a structure for compound B. Once you have a structure, it is always helpful to go back through the spectral data and make sure that the structure you came up is consistent with all of it. O O O O
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