Finding the repeating unit of polymerisation given two constituent molecules
$begingroup$
The compounds $ce{H2NCH2CH2NH2}$ and $ce{HOOCCH2COOH}$ react to form a polymer. What is the structure of the repeating unit of the polymer?
A. $require{enclose}ce{enclose{horizontalstrike}{(}HNCH2CONHCH2CH2NHCOenclose{horizontalstrike}{)}}$
B. $ce{enclose{horizontalstrike}{(}HNCH2CH2NHCOCH2COenclose{horizontalstrike}{)}}$
C. $ce{enclose{horizontalstrike}{(}OCCH2CONHCH2NHCOenclose{horizontalstrike}{)}}$
D. $ce{enclose{horizontalstrike}{(}HNCH2CH2NHCOCH2NHenclose{horizontalstrike}{)}}$
I came across this question which requires me to find the structure of the repeating unit of the polymer made by the reaction of two different compounds. However, I am not sure as to which repeating unit would suffice for this particular polymer.
As far as I know, there should be double bonds present at the ends of the repeating unit, so that the pi bond can break and allow the formation of additional bonds, and hence, a chain of molecules. However, using that line of thought I cannot seem to figure out the answer to the question.
I just need a direction, a hint or a clue as to how I should approach problems like such.
polymers
edited 16 hours ago
Karsten Theis
4,092542
asked 18 hours ago
Selena CarlosSelena Carlos
617
$endgroup$
add a comment |
$begingroup$
The compounds $ce{H2NCH2CH2NH2}$ and $ce{HOOCCH2COOH}$ react to form a polymer. What is the structure of the repeating unit of the polymer?
A. $require{enclose}ce{enclose{horizontalstrike}{(}HNCH2CONHCH2CH2NHCOenclose{horizontalstrike}{)}}$
B. $ce{enclose{horizontalstrike}{(}HNCH2CH2NHCOCH2COenclose{horizontalstrike}{)}}$
C. $ce{enclose{horizontalstrike}{(}OCCH2CONHCH2NHCOenclose{horizontalstrike}{)}}$
D. $ce{enclose{horizontalstrike}{(}HNCH2CH2NHCOCH2NHenclose{horizontalstrike}{)}}$
I came across this question which requires me to find the structure of the repeating unit of the polymer made by the reaction of two different compounds. However, I am not sure as to which repeating unit would suffice for this particular polymer.
As far as I know, there should be double bonds present at the ends of the repeating unit, so that the pi bond can break and allow the formation of additional bonds, and hence, a chain of molecules. However, using that line of thought I cannot seem to figure out the answer to the question.
I just need a direction, a hint or a clue as to how I should approach problems like such.
polymers
edited 16 hours ago
Karsten Theis
4,092542
asked 18 hours ago
Selena CarlosSelena Carlos
617
$endgroup$
4
$begingroup$
What you are mentioning is addition polymerization which malonic acid and ethylenediamine wouldn't dream of doing. Condensation occurs, rather.
$endgroup$
– William R. Ebenezer
18 hours ago
$begingroup$
Draw out the skeletal structures with different colors so you can see how the building blocks combine. If you have monomers A and B, and A can't link with A and B can't link with B, the only polymer you can get is A-B-A-B-A...
$endgroup$
– Karsten Theis
16 hours ago
1
$begingroup$
"As far as I know, there should be double bonds present at the ends of the repeating unit"; This is not correct.
$endgroup$
– A.K.
11 hours ago
add a comment |
$begingroup$
The compounds $ce{H2NCH2CH2NH2}$ and $ce{HOOCCH2COOH}$ react to form a polymer. What is the structure of the repeating unit of the polymer?
A. $require{enclose}ce{enclose{horizontalstrike}{(}HNCH2CONHCH2CH2NHCOenclose{horizontalstrike}{)}}$
B. $ce{enclose{horizontalstrike}{(}HNCH2CH2NHCOCH2COenclose{horizontalstrike}{)}}$
C. $ce{enclose{horizontalstrike}{(}OCCH2CONHCH2NHCOenclose{horizontalstrike}{)}}$
D. $ce{enclose{horizontalstrike}{(}HNCH2CH2NHCOCH2NHenclose{horizontalstrike}{)}}$
I came across this question which requires me to find the structure of the repeating unit of the polymer made by the reaction of two different compounds. However, I am not sure as to which repeating unit would suffice for this particular polymer.
As far as I know, there should be double bonds present at the ends of the repeating unit, so that the pi bond can break and allow the formation of additional bonds, and hence, a chain of molecules. However, using that line of thought I cannot seem to figure out the answer to the question.
I just need a direction, a hint or a clue as to how I should approach problems like such.
polymers
edited 16 hours ago
Karsten Theis
4,092542
asked 18 hours ago
Selena CarlosSelena Carlos
617
$endgroup$
The compounds $ce{H2NCH2CH2NH2}$ and $ce{HOOCCH2COOH}$ react to form a polymer. What is the structure of the repeating unit of the polymer?
A. $require{enclose}ce{enclose{horizontalstrike}{(}HNCH2CONHCH2CH2NHCOenclose{horizontalstrike}{)}}$
B. $ce{enclose{horizontalstrike}{(}HNCH2CH2NHCOCH2COenclose{horizontalstrike}{)}}$
C. $ce{enclose{horizontalstrike}{(}OCCH2CONHCH2NHCOenclose{horizontalstrike}{)}}$
D. $ce{enclose{horizontalstrike}{(}HNCH2CH2NHCOCH2NHenclose{horizontalstrike}{)}}$
I came across this question which requires me to find the structure of the repeating unit of the polymer made by the reaction of two different compounds. However, I am not sure as to which repeating unit would suffice for this particular polymer.
As far as I know, there should be double bonds present at the ends of the repeating unit, so that the pi bond can break and allow the formation of additional bonds, and hence, a chain of molecules. However, using that line of thought I cannot seem to figure out the answer to the question.
I just need a direction, a hint or a clue as to how I should approach problems like such.
polymers
polymers
edited 16 hours ago
Karsten Theis
4,092542
asked 18 hours ago
Selena CarlosSelena Carlos
617
edited 16 hours ago
Karsten Theis
4,092542
asked 18 hours ago
Selena CarlosSelena Carlos
617
edited 16 hours ago
Karsten Theis
4,092542
edited 16 hours ago
Karsten Theis
4,092542
edited 16 hours ago
Karsten Theis
4,092542
4,092542
asked 18 hours ago
Selena CarlosSelena Carlos
617
asked 18 hours ago
Selena CarlosSelena Carlos
617
asked 18 hours ago
Selena CarlosSelena Carlos
617
617
4
$begingroup$
What you are mentioning is addition polymerization which malonic acid and ethylenediamine wouldn't dream of doing. Condensation occurs, rather.
$endgroup$
– William R. Ebenezer
18 hours ago
$begingroup$
Draw out the skeletal structures with different colors so you can see how the building blocks combine. If you have monomers A and B, and A can't link with A and B can't link with B, the only polymer you can get is A-B-A-B-A...
$endgroup$
– Karsten Theis
16 hours ago
1
$begingroup$
"As far as I know, there should be double bonds present at the ends of the repeating unit"; This is not correct.
$endgroup$
– A.K.
11 hours ago
add a comment |
4
$begingroup$
What you are mentioning is addition polymerization which malonic acid and ethylenediamine wouldn't dream of doing. Condensation occurs, rather.
$endgroup$
– William R. Ebenezer
18 hours ago
$begingroup$
Draw out the skeletal structures with different colors so you can see how the building blocks combine. If you have monomers A and B, and A can't link with A and B can't link with B, the only polymer you can get is A-B-A-B-A...
$endgroup$
– Karsten Theis
16 hours ago
1
$begingroup$
"As far as I know, there should be double bonds present at the ends of the repeating unit"; This is not correct.
$endgroup$
– A.K.
11 hours ago
4
4
$begingroup$
What you are mentioning is addition polymerization which malonic acid and ethylenediamine wouldn't dream of doing. Condensation occurs, rather.
$endgroup$
– William R. Ebenezer
18 hours ago
$begingroup$
What you are mentioning is addition polymerization which malonic acid and ethylenediamine wouldn't dream of doing. Condensation occurs, rather.
$endgroup$
– William R. Ebenezer
18 hours ago
$begingroup$
Draw out the skeletal structures with different colors so you can see how the building blocks combine. If you have monomers A and B, and A can't link with A and B can't link with B, the only polymer you can get is A-B-A-B-A...
$endgroup$
– Karsten Theis
16 hours ago
$begingroup$
Draw out the skeletal structures with different colors so you can see how the building blocks combine. If you have monomers A and B, and A can't link with A and B can't link with B, the only polymer you can get is A-B-A-B-A...
$endgroup$
– Karsten Theis
16 hours ago
1
1
$begingroup$
"As far as I know, there should be double bonds present at the ends of the repeating unit"; This is not correct.
$endgroup$
– A.K.
11 hours ago
$begingroup$
"As far as I know, there should be double bonds present at the ends of the repeating unit"; This is not correct.
$endgroup$
– A.K.
11 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You are starting out with amino groups on one monomer and carboxylate groups on the other monomer. As William R. Ebenezer states in the comment, these combine via condensation (to form an amide, which is written as -NHCO- or -CONH- in condensed formulas). 
The linkages in the answer choices between repeating units are not visible until you write another repeating unit next to the ones shown. A. and B. show amide linkages between repeating units, while C. and D. show something else.
Here is the exercise rewritten to emphasize the amide bonds in the repeating units:
The compounds $ce{H2NCH2CH2NH2}$ and $ce{HOOCCH2COOH}$ react to form a polymer. What is the structure of the repeating unit of the polymer?
A. $require{enclose}ce{enclose{horizontalstrike}{(}HNCH2CO-NHCH2CH2NH-COenclose{horizontalstrike}{)}}$
B. $ce{enclose{horizontalstrike}{(}HNCH2CH2NH-COCH2COenclose{horizontalstrike}{)}}$
C. $ce{enclose{horizontalstrike}{(}OCCH2CO-NHCH2NH-COenclose{horizontalstrike}{)}}$
D. $ce{enclose{horizontalstrike}{(}HNCH2CH2NH-COCH2NHenclose{horizontalstrike}{)}}$
Now that we emphasized the amide linkages, you can see that the first monomer in A, the second monomer in C and the second monomer in D don't match your starting materials. We can also exclude C. and D. because repeating units are not linked via amides. The answer has to be B.
answered 15 hours ago
Karsten TheisKarsten Theis
4,092542
$endgroup$
add a comment |
$begingroup$
You got the answer you have looking for in Karsten Theis' answer. However, your statement of "as far as I know, there should be double bonds present at the ends of the repeating unit" is not entirely correct for all polymers in general (see A.K.'s comment). Yes, some have double bonds within the chain and at the end. But, most polymers forms without assistance from double bonds and, hence do not contain double bonds (unsaturations). Here is two best known polymers we used in our day to day life:
Both ate fabrics and none has end unsaturation. The reaction shows you that none need unsaturation to condense.
answered 3 hours ago
Mathew MahindaratneMathew Mahindaratne
5,952724
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You are starting out with amino groups on one monomer and carboxylate groups on the other monomer. As William R. Ebenezer states in the comment, these combine via condensation (to form an amide, which is written as -NHCO- or -CONH- in condensed formulas).
The linkages in the answer choices between repeating units are not visible until you write another repeating unit next to the ones shown. A. and B. show amide linkages between repeating units, while C. and D. show something else.
Here is the exercise rewritten to emphasize the amide bonds in the repeating units:
The compounds $ce{H2NCH2CH2NH2}$ and $ce{HOOCCH2COOH}$ react to form a polymer. What is the structure of the repeating unit of the polymer?
A. $require{enclose}ce{enclose{horizontalstrike}{(}HNCH2CO-NHCH2CH2NH-COenclose{horizontalstrike}{)}}$
B. $ce{enclose{horizontalstrike}{(}HNCH2CH2NH-COCH2COenclose{horizontalstrike}{)}}$
C. $ce{enclose{horizontalstrike}{(}OCCH2CO-NHCH2NH-COenclose{horizontalstrike}{)}}$
D. $ce{enclose{horizontalstrike}{(}HNCH2CH2NH-COCH2NHenclose{horizontalstrike}{)}}$
Now that we emphasized the amide linkages, you can see that the first monomer in A, the second monomer in C and the second monomer in D don't match your starting materials. We can also exclude C. and D. because repeating units are not linked via amides. The answer has to be B.
answered 15 hours ago
Karsten TheisKarsten Theis
4,092542
$endgroup$
add a comment |
$begingroup$
You are starting out with amino groups on one monomer and carboxylate groups on the other monomer. As William R. Ebenezer states in the comment, these combine via condensation (to form an amide, which is written as -NHCO- or -CONH- in condensed formulas).
The linkages in the answer choices between repeating units are not visible until you write another repeating unit next to the ones shown. A. and B. show amide linkages between repeating units, while C. and D. show something else.
Here is the exercise rewritten to emphasize the amide bonds in the repeating units:
The compounds $ce{H2NCH2CH2NH2}$ and $ce{HOOCCH2COOH}$ react to form a polymer. What is the structure of the repeating unit of the polymer?
A. $require{enclose}ce{enclose{horizontalstrike}{(}HNCH2CO-NHCH2CH2NH-COenclose{horizontalstrike}{)}}$
B. $ce{enclose{horizontalstrike}{(}HNCH2CH2NH-COCH2COenclose{horizontalstrike}{)}}$
C. $ce{enclose{horizontalstrike}{(}OCCH2CO-NHCH2NH-COenclose{horizontalstrike}{)}}$
D. $ce{enclose{horizontalstrike}{(}HNCH2CH2NH-COCH2NHenclose{horizontalstrike}{)}}$
Now that we emphasized the amide linkages, you can see that the first monomer in A, the second monomer in C and the second monomer in D don't match your starting materials. We can also exclude C. and D. because repeating units are not linked via amides. The answer has to be B.
answered 15 hours ago
Karsten TheisKarsten Theis
4,092542
$endgroup$
add a comment |
$begingroup$
You are starting out with amino groups on one monomer and carboxylate groups on the other monomer. As William R. Ebenezer states in the comment, these combine via condensation (to form an amide, which is written as -NHCO- or -CONH- in condensed formulas).
The linkages in the answer choices between repeating units are not visible until you write another repeating unit next to the ones shown. A. and B. show amide linkages between repeating units, while C. and D. show something else.
Here is the exercise rewritten to emphasize the amide bonds in the repeating units:
The compounds $ce{H2NCH2CH2NH2}$ and $ce{HOOCCH2COOH}$ react to form a polymer. What is the structure of the repeating unit of the polymer?
A. $require{enclose}ce{enclose{horizontalstrike}{(}HNCH2CO-NHCH2CH2NH-COenclose{horizontalstrike}{)}}$
B. $ce{enclose{horizontalstrike}{(}HNCH2CH2NH-COCH2COenclose{horizontalstrike}{)}}$
C. $ce{enclose{horizontalstrike}{(}OCCH2CO-NHCH2NH-COenclose{horizontalstrike}{)}}$
D. $ce{enclose{horizontalstrike}{(}HNCH2CH2NH-COCH2NHenclose{horizontalstrike}{)}}$
Now that we emphasized the amide linkages, you can see that the first monomer in A, the second monomer in C and the second monomer in D don't match your starting materials. We can also exclude C. and D. because repeating units are not linked via amides. The answer has to be B.
answered 15 hours ago
Karsten TheisKarsten Theis
4,092542
$endgroup$
You are starting out with amino groups on one monomer and carboxylate groups on the other monomer. As William R. Ebenezer states in the comment, these combine via condensation (to form an amide, which is written as -NHCO- or -CONH- in condensed formulas).
The linkages in the answer choices between repeating units are not visible until you write another repeating unit next to the ones shown. A. and B. show amide linkages between repeating units, while C. and D. show something else.
Here is the exercise rewritten to emphasize the amide bonds in the repeating units:
The compounds $ce{H2NCH2CH2NH2}$ and $ce{HOOCCH2COOH}$ react to form a polymer. What is the structure of the repeating unit of the polymer?
A. $require{enclose}ce{enclose{horizontalstrike}{(}HNCH2CO-NHCH2CH2NH-COenclose{horizontalstrike}{)}}$
B. $ce{enclose{horizontalstrike}{(}HNCH2CH2NH-COCH2COenclose{horizontalstrike}{)}}$
C. $ce{enclose{horizontalstrike}{(}OCCH2CO-NHCH2NH-COenclose{horizontalstrike}{)}}$
D. $ce{enclose{horizontalstrike}{(}HNCH2CH2NH-COCH2NHenclose{horizontalstrike}{)}}$
Now that we emphasized the amide linkages, you can see that the first monomer in A, the second monomer in C and the second monomer in D don't match your starting materials. We can also exclude C. and D. because repeating units are not linked via amides. The answer has to be B.
answered 15 hours ago
Karsten TheisKarsten Theis
4,092542
edited 14 hours ago
answered 15 hours ago
Karsten TheisKarsten Theis
4,092542
answered 15 hours ago
Karsten TheisKarsten Theis
4,092542
answered 15 hours ago
Karsten TheisKarsten Theis
4,092542
4,092542
add a comment |
add a comment |
$begingroup$
You got the answer you have looking for in Karsten Theis' answer. However, your statement of "as far as I know, there should be double bonds present at the ends of the repeating unit" is not entirely correct for all polymers in general (see A.K.'s comment). Yes, some have double bonds within the chain and at the end. But, most polymers forms without assistance from double bonds and, hence do not contain double bonds (unsaturations). Here is two best known polymers we used in our day to day life:
Both ate fabrics and none has end unsaturation. The reaction shows you that none need unsaturation to condense.
answered 3 hours ago
Mathew MahindaratneMathew Mahindaratne
5,952724
$endgroup$
add a comment |
$begingroup$
You got the answer you have looking for in Karsten Theis' answer. However, your statement of "as far as I know, there should be double bonds present at the ends of the repeating unit" is not entirely correct for all polymers in general (see A.K.'s comment). Yes, some have double bonds within the chain and at the end. But, most polymers forms without assistance from double bonds and, hence do not contain double bonds (unsaturations). Here is two best known polymers we used in our day to day life:
Both ate fabrics and none has end unsaturation. The reaction shows you that none need unsaturation to condense.
answered 3 hours ago
Mathew MahindaratneMathew Mahindaratne
5,952724
$endgroup$
add a comment |
$begingroup$
You got the answer you have looking for in Karsten Theis' answer. However, your statement of "as far as I know, there should be double bonds present at the ends of the repeating unit" is not entirely correct for all polymers in general (see A.K.'s comment). Yes, some have double bonds within the chain and at the end. But, most polymers forms without assistance from double bonds and, hence do not contain double bonds (unsaturations). Here is two best known polymers we used in our day to day life:
Both ate fabrics and none has end unsaturation. The reaction shows you that none need unsaturation to condense.
answered 3 hours ago
Mathew MahindaratneMathew Mahindaratne
5,952724
$endgroup$
You got the answer you have looking for in Karsten Theis' answer. However, your statement of "as far as I know, there should be double bonds present at the ends of the repeating unit" is not entirely correct for all polymers in general (see A.K.'s comment). Yes, some have double bonds within the chain and at the end. But, most polymers forms without assistance from double bonds and, hence do not contain double bonds (unsaturations). Here is two best known polymers we used in our day to day life:
Both ate fabrics and none has end unsaturation. The reaction shows you that none need unsaturation to condense.
answered 3 hours ago
Mathew MahindaratneMathew Mahindaratne
5,952724
answered 3 hours ago
Mathew MahindaratneMathew Mahindaratne
5,952724
answered 3 hours ago
Mathew MahindaratneMathew Mahindaratne
5,952724
answered 3 hours ago
Mathew MahindaratneMathew Mahindaratne
5,952724
5,952724
add a comment |
add a comment |
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4
$begingroup$
What you are mentioning is addition polymerization which malonic acid and ethylenediamine wouldn't dream of doing. Condensation occurs, rather.
$endgroup$
– William R. Ebenezer
18 hours ago
$begingroup$
Draw out the skeletal structures with different colors so you can see how the building blocks combine. If you have monomers A and B, and A can't link with A and B can't link with B, the only polymer you can get is A-B-A-B-A...
$endgroup$
– Karsten Theis
16 hours ago
1
$begingroup$
"As far as I know, there should be double bonds present at the ends of the repeating unit"; This is not correct.
$endgroup$
– A.K.
11 hours ago