bestrobber97 asked :
Hello Ultima,
I'm taking my A's this year and need some help with a question. In
Paper 3, Q3(d), why can't P be CH3CH(Cl)CH(OH)CH2COCH3 ? And on a
side note, what are the 3 isomeric products that dehydration of P
would give? I know the 2 obvious ones, but can't seem to think of
the third :(
Thanks in advance :)
Compound P, C6H11O2Cl, reacts with 2,4-dinitrophenylhydrazine to
form an orange precipitate and decolourises hot, acidified
potassium manganate(VII). When compound P is heated with excess
aqueous sodium hydroxide, compound Q, C6H12O3, is formed. Aqueous
iodine is subsequently added to the hot mixture of this reaction
and compound R, C4H4O5Na2, is formed. P reacts with excess
concentrated sulfuric acid at 170 °C to give three possible
isomeric products S, T and U. S, but not T and U, reacts with
excess ammonia to form an amine. Suggest the structures of P, Q and
R, explaining your reasoning.
P cannot be CH3CH(Cl)CH(OH)CH2COCH3 because the qn stated that from
Q to R (when reacted with alkaline aqueous iodine), 2 O atoms are
added, and 2 C atoms are removed, meaning that 2 groups separately
underwent the iodoform reaction. Based on degree of unsaturation (1
for Q and 2 for R), you can deduce that 1 group must have been
CH3CH(OH)R and the other group must have been CH3COR.
Hence, P must have been CH3CH(OH)CH(Cl)CH2COCH3.
S is CH2=CHCH(NH2)CH2COCH3, being a terminal alkene has no
geometrical isomerism.
T and U are cis & trans (or E & Z) CH3CH=CClCH2COCH3, with
the vinyl halide C-Cl bond of the resonance hybrid having partial
double bond character, which is thus strong and does not cleave
readily, and hence T & U are resistant to nucleophilic
substitution by NH3.