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help!
ok i know i always post here and it must seem like i'm lazy for not doing my hwk but this is hard! ok...
A compound A has molecular formula C4H6O2. It reacts with hydrogen cyanide to form compound B C6H18O2N2. A is readily oxidised by acidified potassium dichromate (VI) to an acidic compound C, C4H6O4. When 1g of C is dissolved in water and titrated with 1.0 mol dm-3 sodium hydroxide, NaOH, 16.9cm3 of NaOH are required for neutralisation. Suggest structural formulae for A,B and C and explain the above reactions.
Ok. So i know compound A is an aldehyde, B is a hydroxynitrile and C is a carboxylic acid. And i can explain the reactions. but can anyone suggest the structural formulae of A, B and C? Please help!
A compound A has molecular formula C4H6O2. It reacts with hydrogen cyanide to form compound B C6H18O2N2. A is readily oxidised by acidified potassium dichromate (VI) to an acidic compound C, C4H6O4. When 1g of C is dissolved in water and titrated with 1.0 mol dm-3 sodium hydroxide, NaOH, 16.9cm3 of NaOH are required for neutralisation. Suggest structural formulae for A,B and C and explain the above reactions.
Ok. So i know compound A is an aldehyde, B is a hydroxynitrile and C is a carboxylic acid. And i can explain the reactions. but can anyone suggest the structural formulae of A, B and C? Please help!
go for the neutralisation data first:
moes NaOH required = 0.0169 x 1 moles = 0.0169 moles
molecular formula of C is C4H6O4 so its Mr = 118
so 1g of C is 1/118 moles = 0.00847
by inspection 1 mole acid (C) is neutralised by 2 moles of NaOH so it contains two acidic hydrogens - in this case two COOH groups
reaction of A with HCN adds two CN groups so the compound A must be a dicarbonyl (nucleophilic addition).
you are told that it is easily oxidised to C (a dicarboxylic acid) so A must be a dialdehyde. The aldehyde group only occurs at the ends of a molecule (CHO group has only one bond remaining) so the compound A is OHC-CH2CH2-CHO butan-di -al.
Compound B is the addition product HOCH(CN)-CH2CH2-CH(CN)OH (you have a misprint it should be eight hydrogens not eighteen!)
moes NaOH required = 0.0169 x 1 moles = 0.0169 moles
molecular formula of C is C4H6O4 so its Mr = 118
so 1g of C is 1/118 moles = 0.00847
by inspection 1 mole acid (C) is neutralised by 2 moles of NaOH so it contains two acidic hydrogens - in this case two COOH groups
reaction of A with HCN adds two CN groups so the compound A must be a dicarbonyl (nucleophilic addition).
you are told that it is easily oxidised to C (a dicarboxylic acid) so A must be a dialdehyde. The aldehyde group only occurs at the ends of a molecule (CHO group has only one bond remaining) so the compound A is OHC-CH2CH2-CHO butan-di -al.
Compound B is the addition product HOCH(CN)-CH2CH2-CH(CN)OH (you have a misprint it should be eight hydrogens not eighteen!)
compound B C6H18O2N2
charco
go for the neutralisation data first:
moes NaOH required = 0.0169 x 1 moles = 0.0169 moles
molecular formula of C is C4H6O4 so its Mr = 118
so 1g of C is 1/118 moles = 0.00847
by inspection 1 mole acid (C) is neutralised by 2 moles of NaOH so it contains two acidic hydrogens - in this case two COOH groups
reaction of A with HCN adds two CN groups so the compound A must be a dicarbonyl (nucleophilic addition).
you are told that it is easily oxidised to C (a dicarboxylic acid) so A must be a dialdehyde. The aldehyde group only occurs at the ends of a molecule (CHO group has only one bond remaining) so the compound A is OHC-CH2CH2-CHO butan-di -al.
Compound B is the addition product HOCH(CN)-CH2CH2-CH(CN)OH (you have a misprint it should be eight hydrogens not eighteen!)
moes NaOH required = 0.0169 x 1 moles = 0.0169 moles
molecular formula of C is C4H6O4 so its Mr = 118
so 1g of C is 1/118 moles = 0.00847
by inspection 1 mole acid (C) is neutralised by 2 moles of NaOH so it contains two acidic hydrogens - in this case two COOH groups
reaction of A with HCN adds two CN groups so the compound A must be a dicarbonyl (nucleophilic addition).
you are told that it is easily oxidised to C (a dicarboxylic acid) so A must be a dialdehyde. The aldehyde group only occurs at the ends of a molecule (CHO group has only one bond remaining) so the compound A is OHC-CH2CH2-CHO butan-di -al.
Compound B is the addition product HOCH(CN)-CH2CH2-CH(CN)OH (you have a misprint it should be eight hydrogens not eighteen!)
charco, i bow down and WORSHIP you! Thank you soo much
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