Hi, the following qns are asked with reference to
VJC_P3_2014
1. 1(a)(iii) not very sure how to calculate the ratio
2. 2(d) how do we know how to construct the equation between P
and NaOH?
3. 3(i) why KMnO4 cannot be used?
4. 4(b) I know that J contains COOH. However when it is reduced
won't it be converted to primary alcohol and alcohol will react
with conc sulfuric acid so I am a bit confused.
Thanks
Q1. At equilibrium, by definition, no electric current flows
through the circuit, Gibbs free energy change = Cell Potential = 0
volts.
Kc = [Products] / [Reactants]
Hence Kc = [Mg2+] / [Cu2+]
Since Mg2+ is the RHS product and Cu2+ is the LHS reactant.
Q2. This is a disproportionation reaction in which highly reactive
and unstable P4, undergoes (base-promoted) hydrolysis, and is
simultaneously oxidized to oxo-acidic hypophosphorous acid
(immediately deprotonated into its conjugate base salt), and
reduced to hydridic-basic phosphine.
Note the electronegativities of O > P > H, which will help
you to better understand the oxidation-to-acidic and
reduction-to-basic nature of the hydrolysis reaction.
The chemistry and mechanism for this reaction is, of course, beyond
the A level syllabus. But sufficient clues are given in the
question, specifically data that enables you to calculate the
empirical formula of species C, and of course, the textual
description of the reaction of A + NaOH + H2O* --> B + C.
* Whenever there is (aq), eg. NaOH(aq), the astute or distinction
student will be open to the possibility of hydrolysis (ie. reaction
with H2O), especially in acidic or alkaline conditions. Eg. In
Organic Chemistry, whenever KMnO4 / H+ or KMnO4 / OH- is used,
don't just think oxidation, but open yourself to the possibility of
acidic or alkaline hydrolysis of any ester / amide / nitrile / etc
groups simultaneously occurring as well, in challenging deductive
elucidation questions.
Q3. Heating under reflux with acidified KMnO4, the alkyl side
chains of the benzene rings (there are 2) would be oxidized,
yielding ortho-phthalic acid (from benzene ring on the left),
benzoic acid (from benzene ring on the right), 2-methylpropandioic
acid (centre fragment), and (from the methyl group beta to the
benzene ring on the left) carbonic(IV) acid (which decomposes into
CO2(g) and H2O(l) with thermodynamically favourable positive
entropy change, and as predicted by Le Chatelier's
principle).
Q4. While J is a primary alcohol, its only beta C atom happens to
be a tertiary C atom, ie. no beta H atom (required for elimination
together with the alpha OH group during dehydration of alcohols to
alkenes) is present. This clues you in on the branched structure of
fragment J, and hence of parent G as well.