A compound A with a molecular formula C3H8O is oxidised with acidified potassium dichromate(VI) to form a liquid B. B reacts with hydrogen cyanide to form a compound C that contains 3 carbon atoms. What are the structures of A B and C?

Well, this is what I have... If #"A"# is #"H"_3"C"-("CH"_2)_2-"OH"#, then #"B"# is #"H"_3"C"-"CH"_2-"COOH"# and #"C"# #=# #"B"# because #"HCN"# doesn't effectively react with carboxylic acids.

Likely the intention is the secondary alcohol for #"A"#, acetone for #"B"#, and the cyanohydrin for #"C"#, even though #"C"# has 4 carbons.

There are three isomers of #"C"_3"H"_8"O"# (#"A"#):

1. #"H"_3"C"-"O"-"CH"_2"CH"_3#
2. #"H"_3"C"-("CH"_2)_2-"OH"#
3. #"H"_3"C"-("C"-"OH")-"CH"_3#

The ether does not react with #"H"_2"Cr"_2"O"_7#, while the alcohols get oxidized. The #1^@# alcohol should go all the way to propanoic acid (#"B"#), and the #2^@# alcohol would stop at acetone (#"B"#); #"H"_2"Cr"_2"O"_7# is a very good oxidizing agent.

In reacting with #"HCN"#, I would expect the carbonyl oxygen to get protonated, followed by nucleophilic attack of the now-electrophilic carbonyl carbon.

That, however, should not happen with carboxylic acids. I showed the intermediate to hopefully make it clear.

It looks like #"B"# must be the carboxylic acid in order to leave #"C"# with only #3# carbons. Otherwise there must be one carbon added from #"CN"^(-)#, necessarily giving #4#.

Considering that the final product #C# contains #3# carbon atoms in its longest carbon chain, I present here a possible answer.

The starting material #A# has got molecular formula #C_3H_8O#.
It follows general molecular formula #C_nH_(2n+2)O# of saturated ether or alcohol. But compound #A# is oxidized by acidified #K_2Cr_2O_7# to a liquid #B# which further reacts with #HCN# to form #C#. So the compound #A# must be an alcohol as ether is not oxidisable under given condition,

The possible structural formula of saturated alcohols having molecular formula #C_3H_8O# are.

(1) #CH_2CH_2CH_2OH#,prpoan-1-ol
and
(2) #CH_2CH(OH)CH_3#,prpoan-2-ol

For compound (1)

(1) #CH_2CH_2CH_2OH#,prpoan-1-ol which being a primary alcohol is first oxidized to propanal #CH_3CH_2CHO# and finally to propanoic acid #CH_3CH_2COOH# which does not react with #HCN# used here as final reactant. Hence #A# should not be prpoan-1-ol

#CH_2CH_2CH_2OHstackrel(color(red)([O]))->CH_3CH_2CHOstackrel(color(red)([O]))->CH_3CH_2COOH#

For compound (2)

(2) #CH_2CH(OH)CH_3#,prpoan-2-ol which being a secondary alcohol is oxidized to propanone #CH_3COCH_3# and its further oxidation is difficult . This reacts with #HCN# used here as final reactant to form 2-hydroxy-2-methylpropanenitrile. Hence #A# should be prpoan-2-ol

#CH_2CH(OH)CH_3stackrel(color(red)([O]))->CH_3COCH_3stackrel(color(red)([HCN]))->CH_3CH(OH)(CN)CH_3#