Hi UltimaOnline,
Here are some DHS MCQs on which I would like seek your
help:
DHS14/P1/12 [Ans: D]
Remarks: I do not understand the titration curve. My understanding
is as follows - the first unlabelled equivalence point between A
and B signifies the end of neutralisation between maleic acid and
sodium hydroxide, and at the point, only -O2CHCHCO2- is present.
However, why is the second equivalence point is at C, (when an
additional 10cm³ of maleic acid is added from the first equivalence
point), instead of an addition 30cm³ of maleic acid (given that
there is 30cm³ of -O2CHCHCO2- at first equivalence pt)? Can I
confirm that at D, the only compound present is HO2CHCHCO2-? So
where is “mixture” to resist the pH change? If there were an option
further down the graph from D, I would have picked that, but D
doesn’t seem correct to me.
DHS14/P1/20
Remarks: I understand that elimination is out because it wouldn’t
yield a product which rotates plane-polarised light. I also
understand that both SN1 (since an optically pure enantiomer of
2-bromobutane was used since an optically pure enantiomer of
2-bromobutane was used) and SN2 mechanisms will yield an optically
active product. Am I supposed to draw any inference from the
reduction of the angle of rotation or is that just a red
herring?
DHS14/P1/25
Remarks: I can understand that C is not the answer because it
cannot be deduced from the information above. However, how do I
know whether C-16O bond is stronger than C-18O bond?
DHS14/P1/28 [Ans: C]
Remarks: I’m rather confused here. Do the arrows in the choices
signify a successive addition of the compounds without any
discarding of the aqueous layer? Why do I add water in the first
and last steps and how do I go about ascertaining the sequence of
the addition?
Once again, thank you very much for your help! :)
Hi Gohby,
DHS14/P1/12. Since you're adding a weak acid to a strong acid,
hence maximum buffer capacity is double equivalence point instead
of half equivalence point. Being a diprotic acid, there are of
course, 2 maximum buffer capacities. The 1st being equal molarities
of HA- and A2- (which occurs at [10+20]/2=15 cm3), and the 2nd
being equal molarities of H2A and HA- (which occurs at 40 cm3).
Clearly, the more effective buffer occurs at equal molarities of
H2A and HA- (ie. weak acid and amphiprotic species) rather than
equal molarities of HA- and A2- (ie. amphiprotic and strong base),
as illustrated by the larger horizontal region of the titration
curve. Besides, 15cm3 isn't even an option in this MCQ.
DHS14/P1/20. If it were SN1 only, and the reaction was complete
(as implied), then the product would be optically inactive. If it
were SN2 only, and the reaction was complete (as implied), then the
product would rotate plane polarized light by 13.5 degrees in
either direction. Hence a mixture of both SN1 and SN2 must have
occurred.
DHS14/P1/25. Option D is correct because the reactant is cyclic,
hence there is only 1 hydrolysis product, not 2. Addressing your
specific query : heavier isotopes form stronger bonds. This is Uni
level Chemistry, and for A level purposes, Cambridge can still
tempt the students with beyond-syllabus MCQ choices, to test if the
student is sufficiently proficient to recognize the within-syllabus
correct MCQ answer, and disregard the other options.
DHS14/P1/28. Problem is alcohol, being part non-polar and part
protic polar, is miscible with both water and alkyl halide. Addding
[water and excess conc HCl] serves to protonate the alcohol, which
then migrates into the aqueous layer. Discard the aqueous layer by
decantation. Add excess Na2CO3(aq) to remove excess HCl(aq) in the
form of CO2(g) and NaCl(aq). Add S2O3 2-(aq) to reduce all the
I2(s/aq/alc) to I-(aq), forming S4O6 2-(aq). Add water to remove
all ionic impurities, then carry out decantation to leave behind
the alkyl halide.
No prob, Ghoby! :)
