BedokFunland JC's A Level H2 Chemistry Qns (Part 2)

last SG · 2010-07-26T23:40:35+00:00

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blacktoast94 asked :

also, is there some kind of framework to follow for explanation questions?
like for example they ask u to derive some structures from the information given, and explain fully.

everytime got this kind of question i'll get the strucutres correct but the explanation 0 marks.

besides naming the reaction, what else? issit explain the mechanism?

thanks!

You need to write out (eg. in table format, or in normal writing format, this is optional and up to you), a list of structural deductions, based on the data given in the question.

For example :
"Since a positive iodoform reaction occurs when compound A is warmed with alkaline aqueous iodine to generate CHI3, either the -CH(OH)CH3 group or the -C=OCH3 group is present in compound A. However, since the 2,4-DNPH test for carbonyl compounds did not generate an orange precipitate, it can be deduced that the -CH(OH)CH3 group instead of the -C=OCH3 group, is present in compound A."

Mechanisms for such "Structural Deductive Elucidation" type are not required, unless specifically asked for in the question. Besides, how many JC students can draw the full iodoform or 2,4-DNPH mechanism? (I do teach these mechanisms to my students for their own deeper understanding, but Cambridge won't require you to memorize or draw them out.)

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'A' level students take note :

Cambridge requires the use of comparative adjectives in your answers to such comparative-type questions.

Hi I just want to clarify whether the explanation is correct.

Explain why graphite has a lower melting point than diamond. (1 mark).

The answer is "Graphite has fewer covalent bonds than Diamond."

What if the student wrote this:

"Graphite has weak van der wals forces between the layers. Lesser energy is required to overcome such forces. As a result, it has a lower melting point than diamond".

Is this answer acceptable?

From my own understanding, diamond has 4 covalent bonds between the carbon atoms while graphite has 3 covalent bonds between the carbon atoms and weak van der wals forces. More energy is required to break the extra covalent bond in diamond than the weak van der wals forces between the layers of atoms in graphite.

If the question were to ask, "why is diamond harder than graphite?"

To order to mechanically break apart a piece of diamond, a lot more energy is required to overcome the much stronger covalent bonds in diamond. In contrast, in order to mechanically slice away layers of graphite (which we do when we write using pencils), less energy is required to overcome the weaker van der Waals attraction between the graphene layers of graphite.

A comparison (note the use of comparative adjectives more / less / higher / lower / stronger / weaker in my answer above) must be made to contrast the requirements for mechanically breaking apart pieces of both allotropes, rather than just describing the bonding in only either allotrope.

Turns out that to completely melt graphite, you need to overcome both the intramolecular covalent bonds within the graphene layers, as well as the van der Waals forces between the graphene layers.

Still, this question is rather unfair, because this is still not the full picture. Each C atom in graphite is sp2 hybridized and the covalent bonds within the graphene layers is actually stronger (due to partial double bond character) than the covalent bonds within diamond.

So you have greater number of covalent bonds within diamond, versus stronger covalent bonds within graphite. So who wins? Turns out the melting points of diamond and graphite are actually pretty close (despite graphite being soft and diamond behing ultra-hard).

So, this is still not a fair question (so Cambridge is also arguably at fault), because students (especially at 'O' levels) can't be expected to be aware of, and compare the magnitude of energy required, to overcome the greater number of covalent bonds in diamond, versus the stronger covalent bonds in graphite.

Okay. Thanks for the update.

This is indeed tricky. So if such qn will to appear, which answer would be better? Or both answers are acceptable? But at this level (I mean O level), should we assume the the covalent bonds are similar in both diamond and graphite?

Or I should explain the ambuigity of the qn to the student?

1. If you feel your student is up to it, then show him the bigger picture, afterwhich you proceed to...

2. Instruct him to remember and regurgitate the simplified answer (but still incomplete picture) that "diamond has more covalent bonds than graphite" in the exams.

3. If your student is not up to it, then skip straight to step 2.

Thank you :)

Anyway I re-checked the Mark Scheme and the answer given by the mark scheme is:

"Graphite has fewer strong bonds to break."

And according to the examiner reports, it states:

"To compare the two, mention of both diamond and graphite was required. That graphite ‘has weak bonds’ is not true, as the diagram shows. The Examiners required the answer that graphite has some weak bonds but diamond has only strong bonds."

Yes, the examiner report confirms what I said in my previous posts, about usage of comparative adjectives, difference in number of covalent bonds, and strength of covalent bonds (stronger in graphite, weaker in diamond).

The so-called 'weak bonds' (in the examiner's report) do not refer to covalent bonds, but van der Waals forces. This should be specified more clearly in the examiner's report, which are to be read by teachers, not students, so Cambridge shouldn't use oversimplified (and technically erroneous) terms like "weak bonds" when referring to van der Waals forces, lest it may confuse some people reading it.

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blacktoast asked :

usually for SN1 mechanism, the nucleophile can attack the carbocation from both sides, hence forming 2 isomers cuz of the chiral centre. but what if there is a tertiary halogenoalkane, and all the groups bonded to it are -CH3? then can the nucleohiple still attack from both sides? how to draw the mechanism and end product? .

Two enantiomers are obtained only if the product contains a chiral carbon. If the carbocation intermediate involved is the trimethylmethyl carbocation, the product would not contain a chiral carbon, and hence does not have optial isomers. Therefore, there would only be one possible product, regardless of which side the nucleophile attacked the trigonal planar carbocation.

blacktoast asked again :

so should i still draw the intermediate step whereby the nucleophile attacks from both sides?

No need to draw two mechanism pathways (ie. one from either side of the trigonal planar carbocation attacked), since there is only one product. Just draw the trigonal planar carbocation intermediate being attacked by the nucleophile from either direction.

Students take note :

For SN2 mechanism, you need to draw the reactant, transition state, product.

For SN1 mechanism, although there is reactant, transition state, intermediate, transition state, product; but you only need to draw only the reactant, intermediate, product (ie. transition states not required for SN1).

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KickMe asked :

Is excess acidified KCr2O7 implying strong or mild oxidation? No heat is mentioned.

If temperature is not mentioned, then the question is ambiguous, and then the onus would be on you, as the exam-smart student, to specify both possible answers, but with qualification or explanation, "if heated under reflux, the product is..." and "if temperature is kept cold, then the product is..."

Note that only primary alcohols can be oxidized by K2Cr2O7 to give different products. Secondary alcohols will only produce ketones.

Alkyl benzenes and alkenes can only be oxidized by KMnO4, and not K2Cr2O7.

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Kickme asked :

Give the reagents and condition to prepare butan-2-ol from butan-1-ol. What I have in mind is dehydration of butan-1-ol followed by hydrating the alkene formed. But this will mean that butan-1-ol will still be present. Are there other way to synthesize butan-2-ol?

No problem, since the major product of hydration of but-1-ene, will be butan-2-ol.

To understand why (Cambridge may ask you to explain this), draw the full mechanism, and discuss the relative stabilities of the two possible carbocation intermediates.

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kickme asked :

My notes only show the oxidation of primary, secondary and tertiary alcohols but does not mention methanol. If methanol can be oxidised, how can i distinguish methanol from propanol?

Using K2Cr2O7 won't make a difference. Using hot KMnO4 however, will oxidize methanol to methanal to methanoic acid to carbonic(IV) acid, which exists in equilibrium with, and hence can decompose into, CO2 and H2O. Efferversence of CO2 can be thus observed for methanol, but not for propanol.

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reaction ralative rate

RCH3 to RCH2Cl 1

R2CH2 to R2CHCl 7

R3CH to R3CCl 21

Using this information, and considering the number of hydrogen atoms of each type
(primary, secondary or tertiary) within the molecule, predict the relative ratio of the
two possible products J and K from the chlorination of 2-methylpropane. Explain your answer.

CH3 CH3 CH3

CH3 CH CH3 yielding CH3 C Cl CH3 CH

CH3 CH3 CH2Cl

2-methylpropane J K

SOLUTION:

ratio J / K = 21 / 1

EXPALNATION: According to the information given tertiary hydrogen atom will be repalced in 2-methylpropane to produce J in higher quantity where as there is only ONE primary replaceable hydrogen in the minor product K. Hence the ratio would be

J:K = 21 : 1

Is this the right expalnation or am I missing something UltimaOnline?

This question requires the student to consider both factors in his/her calculations : number of H atoms substitutable and stability of alkyl radical intermediates.

To get J :

(number of H atoms substitutable) x (stability of alkyl radical intermediates)

= 1 x 21 = 21

To get K :

(number of H atoms substitutable) x (stability of alkyl radical intermediates)

= 9 x 1 = 9

Hence ratio of products J to K = 21 / 9 or 7J : 3K

If Cambridge asks "what factors affect product distribution in free radical substitution?" the student should qualitatively explain both factors.

But if Cambridge asks "what is the relative percentages of the expected product distribution?", then :

If the Cambridge exam question does not provide data for the relative stabilities of alkyl radical intermediates (which can be phrased as "relative rates of reaction to generate primary, secondary and tertiary alkyl halides" as seen in the original question above), then the student need only give a quantitative answer based on calculations that only consider "number of H atoms substitutable".

If the Cambridge exam question does provide data for the relative stabilities of alkyl radical intermediates (such as in the original question above), then the student needs to consider both factors "number of H atoms substitutable" and "stability of alkyl radical intermediates" by multiplying the values associated with both factors, to obtain the true product distribution.

Most schools only teach the simpler factor of "number of H atoms substitutable", but this is misleading because in practice, both factors (which you would notice tend to contradict each other : tertiary alkyl halides have the least no. of H atoms to substitute away, but yet tertiary alkyl radicals are the most stabilized by hyperconjugation from the alkyl groups) are equally important and teaches the student that in Chemistry, just as in Real Life, it is myopic and does oneself injustice, to simply consider one side of the story or one particular viewpoint only; one must consider all sides of the story or all relevant viewpoints to understand and appreciate the truer, bigger, more beautiful picture of the Universe and of Life.

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Compound A has the molecular formula C9H12O. Heating A under reflux with acidified KMnO4 produced benzoic acid. Treatment of A with acidified potassium dichromate (VI) produced a ketone B which has a positive result with iodoform test.

Compound A reacts with I2 and aqueous sodium hydroxide to give a yellow solid D and another product which on acidification gave a compound C.

When A was dehydrated, 2 structural isomers E and F were produced, both of which decolourised bromine water. Oxidation of E (but not F) gave compound C as one of the products.

Make deductions to elucidate the structures of compounds A to F.

Answers :

A is C6H5CH2CH(OH)CH3
B is C6H5CH2COCH3
C is C6H5CH2COOH
D is CHI3
E is C6H5CH2CHCH2
F is C6H5CHCHCH3

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Just one simple question... so simple that it bugs me when I can't determine which is the preferred answer.

Alright, so we were asked how we can distinguish between I2 and Br2.

There were 2 options I find quite Ok

First is reaction with sodium thoiosulfate. This seems reasonable because we know that I- ions will become I2 and Br- becomes NaBr. If I am not wrong, NaBr should be white (as is all group I ionic compounds?) and I2 gives a yellow solution

Second option is using aq Fe2+ then add aq NaOH.

we know I2 does not react witih Fe2+... so Br2 WILL, giving Fe3+ and Br -. NaOH and react with Fe3+ to give Fe(OH)3 ppt?

I chose the first answer anyway... was the answer that just "came" to me at that time. Any other take? Thanks

Method 1 :
Assuming standard molarities, thiosulfate will indeed reduce both iodine and bromine, to iodide and bromide respectively, and is itself oxidized to tetrathionate and sulfate(VI) respectively. All of which are (equally) colourless.

Method 2 :
Assuming standard molarities, calculating cell potential (ie. reduction potential at cathode + oxidation potential at anode) will show that only one halogen is a sufficiently strong oxidizing agent to oxidize Fe2+ to Fe3+. Thereafter, adding OH- will precipitate Fe(OH)2(s) and Fe(OH)3(s) in the two separate test solutions, which have distinct colours of dark green and reddish brown respectively.

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Q1. The species C6H5CH2CH(CH3)Cl undergoes an intramolecular reaction (with a Lewis acid catalyst). What is the name for this reaction?

Answer :

In regards to the benzene ring nucleophile, the reaction is an electrophilic aromatic substitution.

In regards to the alkyl chloride electrophile, the reaction is a SN1 nucleophilic substitution.

Considering both the nucleophile and electrophile, the reaction is known as a Friedel-Crafts alkylation of the benzene ring.

Although the electrophile is a secondary alkyl halide, and can theoretically undergo either SN1 or SN2 reaction, however because the benzene ring is a weak nucleophile (since its pi electrons are delocalized by resonance (through the sideways overlapping of p orbitals), and are thus less available for nucleophilic attack), an SN2 mechanism which requires a strong nucleophile, is therefore not possible.

And since a Lewis acid catalyst is used, a (secondary) carbocation electrophile intermediate is generated (that is readily attacked by the benzene ring nucleophile), and hence this electrophilic aromatic substitution reaction proceeds via an SN1 nucleophilic substitution mechanism, and the overall reaction is known as a Friedel-Crafts alkylation of the benzene ring.

Q2. The species C6H5CH2CH2COCl undergoes an intramolecular reaction (with or without a Lewis acid catalyst). What is the name for this reaction?

Answer :

In regards to the benzene ring nucleophile, the reaction is an electrophilic aromatic substitution.

In regards to the acyl chloride electrophile, the reaction is a nucleophilic acyl substitution.

If a Lewis acid catalyst is not used, the nucleophilic substitution on the acyl halide electrophile proceeds via an addition-elimination mechanism. (In industrial practice, a Lewis acid catalyst is always used to lower activation energy and speed up the reaction, because time = money.)

If a Lewis acid catalyst is used, a (resonance stabilized) acylium cation electrophile intermediate is generated (that is readily attacked by the benzene ring nucleophile), and hence this Lewis acid catalyzed electrophilic aromatic substitution - cum - nucleophilic acyl substitution reaction is known overall as a Friedel-Crafts acylation of the benzene ring.

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1. 20.0cm3 of 0.10 mol dm-3 aq Na2CO3 was titrated witih 0.20 mol dm-3 aqueous HCl.

The 2 base hydrolysis constants of the carbonate ion are Kb1 = 2.08 X 10^-3 mol dm-3

Kb2 = 2.32 X 10^-8 mol dm-3

Calculate pH of solution when 0cm3, 5cm3, 15cm3,20cm3 and 25 cm3 of HCl was added.

Sketch the titration curve labeling the points and indicating buffer regions.

Do ICF (moles) tables for the different volumes of NaOH(aq) added. You'll find that 0cm3 and 5cm3 are before the 1st equivalence point, while 15cm3 is between 1st and 2nd equivalence points, and 20cm3 is for 2nd equivalence point, and 25cm3 is past the 2nd equivalence point.

To find pH at these various points :

0cm3 : Use Kb of CO3 2-

5cm3 : Use Henderson-Hasselbalch with [CO3 2-] / [HCO3 -]

15cm3 : Use Henderson-Hasselbalch with [HCO3 -] / [H2CO3]

20cm3 : Use Ka of H2CO3. Be mindful of the volume change.

25cm3 : Calculate [H+] contributed from HCl only, and thus calculate pH. Write a statement to explain that because HCl is a strong acid and H2CO3 is a weak acid, the protons dissociated from HCl suppresses the proton dissociation from the weak acid H2CO3 (as predicted by Le Chatelier's principle), and thus we only need to consider [H+] from HCl. Be mindful of the volume change.

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A buffer solution is titrated with NaOH(aq). Upon addition of 3 volumes of NaOH(aq), the buffer solution becomes most effective. Equivalence point is attained when 8 volumes of NaOH(aq) is added.

a) Calculate the ratio of the molarities of both members of the conjugate acid-base pair of the buffer solution, before the NaOH(aq) is added.

b) Given that the original pH of the buffer solution (ie. before NaOH(aq) is added) is 4.16, calculate the proton dissociation constant of the acidic component of the buffer.

Solution :

a) At maximum buffer capacity :

HA + OH- ---> A- + H2O

Initial (mol)         8y            3y              x-8y           n.a.
Change (mol)   -3y           -3y              +3y           n.a
Final (mol)         5y             0               x-5y           n.a.

At equivalence point :

HA + OH- ---> A- + H2O

Initial (mol)         8y            8y              x-8y           n.a.
Change (mol)   -8y           -8y              +8y           n.a.
Final (mol)          0              0                x               n.a.

First, complete the ICF table for equivalence point. Next, complete the ICF table for maximum buffer capacity, bearing in mind that whether for maximum buffer capacity or equivalence point, the initial moles of HA and A- are the same.

Since at maximum buffer capacity, [HA] = [A-], this implies 5y = x-5y and hence x = 10y.

Substituting x = 10y into the initial moles of A-, we have moles of A- = 2y.

Therefore, the ratio of the amounts of both members of the conjugate acid-base pair present in the buffer (before NaOH(aq) is added), is 8y HA : 2y A-, ie. 4 HA : 1 A-

b) Using the Henderson-Hasselbalch equation, we have

pH = pKa + log ( [base] / [acid] )

4.16 = pKa + log (1/4)

pKa = 4.762

Ka = 1.73 x 10^-5

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kickme asked :

Give the organic products of the reaction of (i) ethanal and (ii) propanone with,

a)LiAlH4 followed by dilute H2SO4
I understand that LiAlH4 implies reduction but what does H2SO4 do?

b) H2N-OH
Is the nucleophile OH-? It is possible for NH2+ to be the nucleophile instead?

Just a question: is it true that any compound with NH2+ can be used as brady's reagent?

Visualize the mechanism :
The H- guy (nucleophile) from LiAlH4 shoots out its balls (lone pair) to attack the chio (delta +ve) girl (electrophilic), the pi bond shifts up to O (to avoid violating C's octet), and you get the alkoxide (ie. alkyl oxide) intermediate. The H2SO4 protonates the alkoxide ion intermediate to generate the final product : the alcohol.

-----------------------------------

In hydroxylamine, both the N and the O atoms are nucleophilic centers, and for some reactions you'll get a mixture of products. Generally though, the N atom is more nucleophilic, because of two reasons :

N is less elecronegative than O, and thus more willing to 'shoot out its balls' (ie. lone pair) to nucleophilically attack the 'girl' (ie. electrophile). Furthermore, the N atom is bonded to two (less electronegative and thus electron donating) H atoms, in contrast to only one H atom for the O atom.

-----------------------------------

Visualize the mechanism : the N atom attacks the C atom, the pi bond shifts up, proton transfer (from N to O), 2nd protonation of O to generate good leaving group H2O+, N gets its balls back becomes horny again, attacks the C atom 2nd time, H2O+ eliminated as H2O, N loses proton to lost its +ve formal charge and regains its balls (lone pair) again.

If the OH group wasn't present, the product would be an imine (C=N functional group). Since the OH group remains bonded to the N atom, the final product with the (RRC=N-OH) functional group is known as an oxime :
http://en.wikipedia.org/wiki/Oxime

-----------------------------------

No, Brady's reagent refers specifically to 2,4-Dinitrophenylhydrazine. The two nitro groups are electron-withdrawing by both induction and resonance, and hence weakens the balls (electron density) of the benzene and amine nucleophile. Being a highly weakened, sterically hindered (benzene ring is large and bulky) nucleophile is important, because recal that 2,4-DNPH reacts only with very chio or feminine (electrophilic) electrophiles like aldehydes and ketones, but not with less chio or feminine (electrophilic) electrophiles like carboxylic acids, esters and amides, even though these compounds also contain the carbonyl C=O) group.

The reason for the reduced electrophilicity of the carboxylic acids, esters and amides, relates to resonance. In aldehydes and ketones, when the C=O pi bond shifts up, the alpha C atoms lack the balls (lone pair) and thus cannot donate electrons by resonance to stabilize the carbocation resonance contributor. In contrast, in carboxylic acids, esters and amides, the adjacent O or N atom can donate a lone pair by resonance to form a pi bond with the carbonyl carbon, preventing a +ve formal charge. Hence in the resonance hybrid, the carbonyl C atom of carboxylic acids, esters and amides are less chio, feminine and electrophilic, and won't be attacked by the weak balls guy (weakened nucleophile) the 2,4-DNPH.

Generally, the more stabilized (eg. by resonance, by induction, by hydrogen bonding, etc) you are, the less reactive you are. Carboxylic acids, esters and amides are stabilized by resonance, aldehydes and ketones are not.

--------------------------------------

kickme replied :

No, it must be LiAlH4 (in dry ether) followed by protonation (eg. via hydrolysis).

Yes, the mechanism is nucleophilic addition for the 1st step only. Google or Wikipedia out the mechanism to form imines and enamines.

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Is this ___________________ in the syllabus?

If you wanna boost your probability of scoring a distinction grade and stand victorious atop all the dead bodies on the bell-curve, then you should have the will and courage to boldly go where no H2 student hath gone before...

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blacktoast94 asked :

Hi all! does anyone know how to do this?

partial hydrolysis of a polypeptide with 2 different enzymes, A and B, produced the following peptides. deduce the order in which the amino acids are bonded together in the original polypeptide.

enzyme A:

try-lys-gly
leu-ala
arg-try-his-meth
arg-gly

enzyme B:

gly-arg-try
try-lys
ala-arg-gly
his-meth-leu

can explain how to go about doing such questions too? my lecture notes don't explain :/ . thank you! :)

There is no chemistry formula for such questions. Treat it as a jigsaw logic puzzle. Just copy-and-paste the smaller bits (ie. the oligopeptides) to form the larger picture (ie. the polypeptide). Once done, confirm your answer by checking that all the oligopeptides are indeed all present in the correct sequence in your hypothesized polypeptide.

Based on the work of both enzymes, the polypeptide (ie. the answer) is :
try-lys-gly-arg-try-his-meth-leu-ala-arg-gly

enzyme A:
try-lys-gly (confirmed)
leu-ala (confirmed)
arg-try-his-meth (confirmed)
arg-gly (confirmed)

enzyme B:
gly-arg-try (confirmed)
try-lys (confirmed)
ala-arg-gly (confirmed)
his-meth-leu (confirmed)

Note that the required assumption (which may or may not be stated or asked by the question) is that the convention for writing peptide sequences (for all oligopeptides and the polypeptide) is to put the N-terminus on the left and write the sequence from N- to C-terminus.

http://en.wikipedia.org/wiki/N-terminus
http://en.wikipedia.org/wiki/C-terminus

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blacktoast94 wrote :

Hi ultima thank you for your reply! :D i'm really grateful that you are able to help me all the time ^_^ .

i also have another question: i'm quite confused about alpihatic alcohols. are they neutral or basic? carboxylic acid and alcohol can undergo acid-base reaction, yet, according to what i've learnt for organic the aliphatic alcohols are neutral or even v. v. slightly acidic
so which one should i use if the alchohol is an aqueous solution? :qmarkblackface:

You're totally welcome blacktoast :xizao::qmarkstickout:, and also welcome to continue asking your questions here... countdown : only 8 months left to 2012 'A' levels! :biggrins:

As Einstein said, everything is relative. Alcohols are relatively neutral, meaning a solution of an aliphatic alcohol has a pH of 7 at rtp. But everything (including alcohols and water) can function as a base (if placed in a strongly acidic environment) or as an acid (if placed in a strongly basic environment).

Say you wanted to protonate your alcohol. Simply add concentrated strong acids eg. H2SO4. Once protonated, the OH group becomes OH2+ which is a good leaving group, allowing for nucleophilic substitution to occur readily.

Say you wanted to deprotonate your alcohol. Simply add sodium metal. This is both a redox reaction (since Na is oxidized to Na+, and H+ from alcohol is reduced to H2 gas) and a Bronsted-Lowry acid-base proton transfer reaction (since the alcohol is deprotonated to generate its conjugate base, the alkoxide ion).

You're expected to know, as part of the H2 syllabus, all three points described above : that alcohols are relatively neutral (compared to more acidic organic species such as phenol, and compared to more basic species such as amines; and explain all of these by electronegativity argument, inductive effects, resonance delocalization, etc), but you must also know how to use alcohol as an acid (ie. deprotonate it) and how to use alcohol as a base (ie. protonate it).

Countdown to 2012 'A' Levels! :biggrins:

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Metformin is an anti-diabetic, anti-cancer drug, that has proven effective in the treament and even prevention, of both diseases. It is available here in Singapore, but requires a doctor's prescription (so if you intend to take Metformin to prevent diabetes or cancer, you still need to ask your doctor for a prescription first).

Draw the mechanism for the synthesis of Metformin. (Hint : it's a simple two-step mechanism : nucleophilic addition, followed by (intramolecular) proton transfer reaction).

http://en.wikipedia.org/wiki/Metformin

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