alcohol reactions qn (1 Viewer)

Masaken

Unknown Member
Joined
May 8, 2021
Messages
1,686
Location
in your walls
Gender
Female
HSC
2023
'Compound A reacts with dilute H2SO4 to produce compounds B and C. Compounds A and B both react with HBr to produce compound D. No colour change is observed when compound B is treated with acidified potassium permanganate solution. Compound A has a molar mass of 56 g/mol. Give the structures and IUPAC names of compounds A to D. Justify your answer with reference to the information provided.'

alright so i've established that A is an alkene because it's a hydration reaction in the presence of sulfuric acid as a catalyst and that B and C are alcohols (presumably structural isomers of one another). however, i'm not sure with what is meant to be indicated with B?? i think they're trying to say no reaction occurs, but i'm not sure what conclusion you're meant to take away from that , could someone please help? thanks in advance

i'm aware that the reaction with B and C with the potassium permanganate is an oxidation reaction as the permanganate is an oxidising agent but that's about it
 

Eagle Mum

Well-Known Member
Joined
Nov 9, 2020
Messages
532
Gender
Female
HSC
N/A
o i've established that A is an alkene because it's a hydration reaction in the presence of sulfuric acid as a catalyst and that B and C are alcohols (presumably structural isomers of one another). however, i'm not sure with what is meant to be indicated with B?? i think they're trying to say no reaction occurs, but i'm not sure what conclusion you're meant to take away from that, could someone please help? thanks in advance
As you’ve said, they are structural isomers.
Potassium permanganate doesn’t react with tertiary alcohols, therefore B must be the tertiary form of butanol, whereas C is likely to be primary butanol.

Note: Given alkenes have the common molecular structure of CnH2n with one carbon and two hydrogen atoms having molecular mass of 14, compound A must be butene (56/14= 4), which is the smallest simple alkene that can produce a tertiary alcohol since there are not enough carbon groups with ethanol & propanol.

To obtain the tertiary form of butanol, compound A has to the branched chain form of butene, but only the straight chain form of butene can produce secondary butanol, hence C is primary butanol.
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top