Organic chemistry help (1 Viewer)

[csi]

Hi guys,

I need a hand on these questions:

1. Draw the structural formula of the organic product formed between:
a) butanoic acid and ethanamide
b) butanoic acid, 2-methylpropan-1-ol and an acid catalyst
2. Write a balance chemical equation for:
a) the dissociation of butanamide in water.
b) 1-chlorobutane and ammonia
c) 1-bromoethane and sodium hydroxide
3. What are the products formed if 2-methylbutylethanoate and potassium hydroxide?
4. What are the reagents needed to make propenamide (hint: a carboxylic acid and another reagent)?

Thanks guys!

 

[TheShy]

a) Condensation reaction, similar to esterification, but instead of a ester link, there will be a amide link
b) esterification reaction, the name of the ester will have the alcohol with suffix -yl first followed by the acid with a suffix of -oate. In this case, you name the alcohol first, 2-methylpropyl butanoate.
2. a) Butanamide is a base, so by bronsted-lowry, ittl be a proton accepter, so the dissociation will be . just gotta add state symbols
gotta ask jazz for b and c, but i doubt that those reactions will come up in hsc, since hsc course only has addition and substitution reactions, which none of those fall into.
3. this one is hydrolysis of ester, pre much reverse of esterification reaction. The products should be sodium ethanoate and 2-methylbutan-1-ol, but i might be wrong here
4. Once again, not quite sure of this but probably an amide? gotta ask jazz or someone else, cos i dont think this is in hsc course, thought might be wrong

 

[csi]

a) Condensation reaction, similar to esterification, but instead of a ester link, there will be a amide link
b) esterification reaction, the name of the ester will have the alcohol with suffix -yl first followed by the acid with a suffix of -oate. In this case, you name the alcohol first, 2-methylpropyl butanoate.
2. a) Butanamide is a base, so by bronsted-lowry, ittl be a proton accepter, so the dissociation will be . just gotta add state symbols
gotta ask jazz for b and c, but i doubt that those reactions will come up in hsc, since hsc course only has addition and substitution reactions, which none of those fall into.
3. this one is hydrolysis of ester, pre much reverse of esterification reaction. The products should be sodium ethanoate and 2-methylbutan-1-ol, but i might be wrong here
4. Once again, not quite sure of this but probably an amide? gotta ask jazz or someone else, cos i dont think this is in hsc course, thought might be wrong

Thanks

 

[Velocifire]

a) Condensation reaction, similar to esterification, but instead of a ester link, there will be a amide link
b) esterification reaction, the name of the ester will have the alcohol with suffix -yl first followed by the acid with a suffix of -oate. In this case, you name the alcohol first, 2-methylpropyl butanoate.
2. a) Butanamide is a base, so by bronsted-lowry, ittl be a proton accepter, so the dissociation will be . just gotta add state symbols
gotta ask jazz for b and c, but i doubt that those reactions will come up in hsc, since hsc course only has addition and substitution reactions, which none of those fall into.
3. this one is hydrolysis of ester, pre much reverse of esterification reaction. The products should be sodium ethanoate and 2-methylbutan-1-ol, but i might be wrong here
4. Once again, not quite sure of this but probably an amide? gotta ask jazz or someone else, cos i dont think this is in hsc course, thought might be wrong

wow, glad to see you back!

 

[CM_Tutor]

@TheShy's answers are mostly correct, but have a few gaps... and there is a problem with several questions:

1(a) is a problem, see below.

1(b) gives an ester product, CH₃CH₂CH₂C(=O)OCH₂CH(CH₃)₂, named 2-methyl-1-propyl butanoate

2(a) is a problem. TheShy's answer is the one that the examiner is seeking, but it is problematic, firstly because butanamide will be a much much weaker base than 1-aminobutane, and secondly because amides can also act as acids. See below.

2(b) is a substitution reaction to make the amine 1-aminobutane, most likely as the salt butylammonium chloride, CH₃CH₂CH₂CH₂NH₃⁺Cl^-

2(c) is a substitution reaction or an elimination reaction, depending on conditions. Aqueous NaOH / KOH converts haloalkanes to alcohols. NaOH / KOH in alcohol instead promotes elimination to form alkenes.

CH₃CH₂Br + NaOH(aq) -----> CH₃CH₂OH + NaBr

CH₃CH₂Br + NaOH ---in ethanol---> CH₂CH₂ + H₂O + NaBr

3. TheShy is correct, potassium hydroxide will hydrolyse an ester. Note the difference in products from acid and base hydrolysis (which TheShy has handled correctly here):

2-methyl-1-butyl ethanoate + water ----acid catalysis----> 2-methyl-1-butanol + ethanoic acid

2-methyl-1-butyl ethanoate + KOH ----base catalysis----> 2-methyl-1-butanol + potassium ethanoate

4. Another problem question

-----

Ok, so what is the problem?

• Firstly, amides are distinctly different from amines. In amines, there is free rotation about the C-N bond (which is a single bond) and the N lone pair is available to be protonated (ie, it acts like a L-B base). Amides have neither of these properties. The C-N bond of an amide is much stronger than would be predicted from amines, it is shorter, and it does not allow free rotation. In fact, it has significant double bond character because it exists as a hybrid of the structure we draw... R-C(=O)-NHR' ... and a structure with two separated charges... R-C(-O^-)=NHR'⁺. This means the N atom "lone pair" is actually involved in bonding and is unavailable to be used to form a coordinate covalent bond and act as a L-B base. In fact, the conjugate acid form of an amide has the extra H atom on the O and a c to N double bond, wit the + charge on the N.
• The page at https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book:_Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/24:_Organonitrogen_Compounds_II-_Amides_Nitriles__Nitro_Compounds/24.02:_Amides_as_Acids_and_Bases ilustrates the basicity of amides, which are generally treated as neutral compounds.
• The chemistry given for forming amides, by condensation of a carboxylic acid and ammonia or a suitable amine, is flawed. Putting a carboxylic acid with a weak base like ammonia or an amine will result in a L-B acid-bas equilibrium forming and almost no condensation. In reality, an amide is prepared by converting the carboxylic acid first to an acid chloride (RCOOH ---> RCOCl) using thionyl chloride (SOCl₂), then reacting the acid chloride with ammonia or the amine. The direct reaction can be promoted by heating to well over 100 degC or adding a coupling reagent like DCC
• On this basis, the answer sought for 4 is propenoic acid plus ammonia, but it won't actually work.
• If 1(a) said butanoic acid plus ethanamine, then the same reaction would be intended with the product being N-ethylbutanamide.
• If 1(a) really means ethanamide, then I wonder if the question is suggesting a product like CH₃CH₂CH₂C(=O)-NH-C(=O)CH₃, but I am very doubtful that this would form. It's not a functional group that is in the syllabus, or that I am familiar with, and the reaction seems unlikely given the lack of basicity of ethanamide. My guess is there is a typo and the question intends ethanamine.

 

[csi]

@TheShy's answers are mostly correct, but have a few gaps... and there is a problem with several questions:

1(a) is a problem, see below.

1(b) gives an ester product, CH₃CH₂CH₂C(=O)OCH₂CH₂CH(CH₃)₂, named 2-methyl-1-propyl butanoate

2(a) is a problem. TheShy's answer is the one that the examiner is seeking, but it is problematic, firstly because butanamide will be a much much weaker base than 1-aminobutane, and secondly because amides can also act as acids. See below.

2(b) is a substitution reaction to make the amine 1-aminobutane, most likely as the salt butylammonium chloride, CH₃CH₂CH₂CH₂NH₃⁺Cl^-

2(c) is a substitution reaction or an elimination reaction, depending on conditions. Aqueous NaOH / KOH converts haloalkanes to alcohols. NaOH / KOH in alcohol instead promotes elimination to form alkenes.

CH₃CH₂Br + NaOH(aq) -----> CH₃CH₂OH + NaBr

CH₃CH₂Br + NaOH ---in ethanol---> CH₂CH₂ + H₂O + NaBr

3. TheShy is correct, potassium hydroxide will hydrolyse an ester. Note the difference in products from acid and base hydrolysis (which TheShy has handled correctly here):

2-methyl-1-butyl ethanoate + water ----acid catalysis----> 2-methyl-1-butanol + ethanoic acid

2-methyl-1-butyl ethanoate + KOH ----base catalysis----> 2-methyl-1-butanol + potassium ethanoate

4. Another problem question

-----

Ok, so what is the problem?

• Firstly, amides are distinctly different from amines. In amines, there is free rotation about the C-N bond (which is a single bond) and the N lone pair is available to be protonated (ie, it acts like a L-B base). Amides have neither of these properties. The C-N bond of an amide is much stronger than would be predicted from amines, it is shorter, and it does not allow free rotation. In fact, it has significant double bond character because it exists as a hybrid of the structure we draw... R-C(=O)-NHR' ... and a structure with two separated charges... R-C(-O^-)=NHR'⁺. This means the N atom "lone pair" is actually involved in bonding and is unavailable to be used to form a coordinate covalent bond and act as a L-B base. In fact, the conjugate acid form of an amide has the extra H atom on the O and a c to N double bond, wit the + charge on the N.
• The page at https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book:_Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/24:_Organonitrogen_Compounds_II-_Amides_Nitriles__Nitro_Compounds/24.02:_Amides_as_Acids_and_Bases ilustrates the basicity of amides, which are generally treated as neutral compounds.
• The chemistry given for forming amides, by condensation of a carboxylic acid and ammonia or a suitable amine, is flawed. Putting a carboxylic acid with a weak base like ammonia or an amine will result in a L-B acid-bas equilibrium forming and almost no condensation. In reality, an amide is prepared by converting the carboxylic acid first to an acid chloride (RCOOH ---> RCOCl) using thionyl chloride (SOCl₂), then reacting the acid chloride with ammonia or the amine. The direct reaction can be promoted by heating to well over 100 degC or adding a coupling reagent like DCC
• On this basis, the answer sought for 4 is propenoic acid plus ammonia, but it won't actually work.
• If 1(a) said butanoic acid plus ethanamine, then the same reaction would be intended with the product being N-ethylbutanamide.
• If 1(a) really means ethanamide, then I wonder if the question is suggesting a product like CH₃CH₂CH₂C(=O)-NH-C(=O)CH₃, but I am very doubtful that this would form. It's not a functional group that is in the syllabus, or that I am familiar with, and the reaction seems unlikely given the lack of basicity of ethanamide. My guess is there is a typo and the question intends ethanamine.

Thanks for such a detailed explanation!