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

last SG · 2010-07-26T23:40:35+00:00

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Question :

3-methylpentane can undergo reaction with chlorine to form monosubstituted compounds that are optically active. How many possible isomers, including stereoisomers, can be formed in the reaction?

Answer :

8 mono-halogenated isomers may be generated, which are :

R-1-chloro-3-methylpentane
S-1-chloro-3-methylpentane
2R,3R-2-chloro-3-methylpentane
2R,3S-2-chloro-3-methylpentane
2S,3R-2-chloro-3-methylpentane
2S,3S-2-chloro-3-methylpentane
3-chloro-3-methylpentane
3-chloromethylpentane

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Solution :

This is a bit of an unfair question by Cambridge, because of the subjective "weakly" adjective. Both B and C are arguably correct, but when faced with two possible answers, you gotta choose the better answer.

If option B was simply ethanoic acid, the question would be more problematic. But here, due to the strong electron-withdrawing by induction effect of the 3 Cl atoms, the strength of trichloroethanoic acid is increased by a tremendous magnitude.

Compare the pKa values :
http://en.wikipedia.org/wiki/Phenol vs http://en.wikipedia.org/wiki/Trichloroacetic_acid

When the pKa value of an acid is very low, the acid is very strong, which means that the conjugate base does not undergo hydrolysis to any significant extent.

If you've understood everything I said so far, the best answer to this MCQ should now be obvious to you.

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A 100cm^3 solution contains 0.2 mol/dm^3 MgCl2 and 0.1 mol/dm^3 CuCl2. A solution of sodium hydroxide is added to the mixture. Mg(OH)2 starts precipitating when 40 cm^3 of sodium hydroxide has been added.

Ksp of Mg(OH)2 is 6.3 x 10^-10, Ksp of Cu(OH)2 is 2.2 x 10^-20

What is the concentration, in mold/dm^3, of Cu^2+ in the solution when Mg(OH)2 just precipitates?

Solution :

This question has several red herrings.

From Ksp of Mg(OH)2, and new molarity of Mg2+, find molarity of OH- at that point. Plug this molarity into the Ksp of Cu(OH)2, find molarity of Cu2+.

Answer is 4.99 x 10^-12 mol/dm^3

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1. Which of the following statements contains one mole of the stated particle?
A Molecules in 19.0 g of fluorine gas.
B Electrons in 24.0 dm3 of hydrogen gas at room temperature and pressure.
C Neutrons in 1.00 g of helium gas.
D Protons in 2.02 g of neon gas.

2. An experiment is conducted to investigate the kinetics of reaction between bromopropane and 0.1 mol dm-3 sodium hydroxide.

The rate equation is as follows:

Rate = k [bromopropane] [OH-]

The half-life of bromopropane in one of the experiments is t minutes.

What is the new half-life (in minutes) of bromopropane when the concentration of bromopropane is doubled and concentration of sodium hydroxide is reduced to 0.01 mol dm-3?

A 0.05t
B 0.1t
C 5t
D 10t

3.Consider the following equilibrium system:

H2 (g) + I2 (g) 2HI (g) H = +53 kJ mol-1

Which of the following change is incorrect?

A Numerical value of Kp is not equal to Kc at 25 C.

B Increasing the mass of H2 will not cause the equilibrium constant to increase.

C Increasing temperature increases the rate constant and equilibrium constant.

D Rate of forward reaction is equal to rate of backward reaction when equilibrium is reached.

4. The solubility product of iron(II) carbonate is 2.1  10–11 while that of silver carbonate is 8.1  10–12 at 25°C.

Which of the following statements is true?

A Addition of silver nitrate increases the solubility of silver carbonate.

B Addition of sulfuric acid to a solution containing iron(II) carbonate increases the solubility product of iron(II) carbonate.

C Iron(II) carbonate precipitates first when sodium carbonate is added to a solution containing equal concentrations of iron(II) and silver ions.

D The solubility of iron(II) carbonate is higher than the solubility of silver carbonate.

5. Liquid E has an H of vapourisation of 10.0 kJ mol1 and a boiling point of
266 K.

What is the S of condensation of vapour E?

A –26.6 J mol1 K1

B –37.6 J mol1 K1

C +26.6 J mol1 K1

D +37.6 J mol1 K1

6. 2-methylpropane can react with bromine in the presence of sunlight to give two monosubstituted halogenoalkanes, 1-bromo-2-methylpropane and 2-bromo-2-methylpropane.
Given the relative rates of abstracting H atoms are:
Type of H atom primary secondary tertiary
Relative rate of abstraction 1 4 6

What is the expected ratio of 1-bromo-2-methylpropane to 2-bromo-2-methylpropane formed?

A 9 : 1 C 6 : 1

B 3 : 2 D 1 : 1

7. In which sequences are the molecules quoted in order of decreasing boiling points?

1 CH3(CH2)3CH3, (CH3)2CHCH2CH3, CH3C(CH3)2CH3

2 AlBr3, AlCl3, AlF3

3 SO2, SiO2, CO2

8. A cell consisting of a V2+ (aq), V3+ (aq) | Pt (s) half-cell and a Au3+ (aq) | Au (s) half-cell is shown below using conventional notation.
Pt (s) | V2+ (aq), V3+ (aq) || Au3+ (aq) | Au (s) Ecell = +1.76 V
Which of the following statements is true?
1 The mass of the Au electrode increase.
2 The negative electrode is the Pt electrode.
3 The standard electrode potential for Au3+ (aq) | Au (s) is +2.02 V.

9. Oxytetracycline is a class of broad-spectrum antibiotics used to treat a variety of infections.

Which of the following statements about oxytetracycline is correct?
1 One mole of oxytetracycline reacts with three moles of thionyl chloride.
2 One mole of oxytetracycline reacts with two moles of hot sodium hydroxide to liberate one mole of ammonia gas.
3 One mole of oxytetracycline reacts with six moles of ethanoyl chloride.

10. The mass percentage of magnesium in a mixture of magnesium chloride and
magnesium nitrate was found to be 21.25%. What mass of magnesium chloride
is present in 100 g of the mixture?

A 47 g
B 51 g
C 53 g
D 56 g

11. Methane was burned in an incorrectly adjusted burner. The methane was converted into a
mixture of carbon dioxide and carbon monoxide in the ratio of 98:2, together with water
vapour.
What will be the volume of oxygen consumed when y dm3
of methane is burned?

A 1.99y dm3
C 0.995y dm3
B 1.995y dm3
D 0.99y dm3

12. To identify an oxide of nitrogen, 0.10 mol of the oxide was mixed with 10 dm3
of hydrogen gas and passed over a heated catalyst. At the end of the reaction, 0.4 dm3
of hydrogen gas remained. The ammonia produced required 125 cm3 of 1.6 mol dm-3
HCl for neutralisation. All gasoues volumes were measured at room temperature and pressure.

What is the formula of the oxide of nitrogen?

A NO C NO2
B N2O D N2O4

13. Which atom has the highest ratio of unpaired electrons to paired electrons in its ground
state?

A boron C nitrogen

B carbon D oxygen

14. Use of the Data Booklet is relevant to this question.

Based on bond energies listed in the Data Booklet, what are the possible products of the
following reaction?

•CH3 + CH3CH2Cl ?

A CH4 + CH3CHCl C CH3CH2CH3 + Cl•
B CH3CH2CH2• + HCl D CH3CH2CH2Cl + H•

15. Which of the following quantities is equal to the Avogadro constant?

1 The number of oxygen atoms in 49.9 g of allactite, Mn7(AsO4)2(OH)8, of molar mas
798 g mol^-1

2 The number of aqueous chloride ions in a solution containing 0.5 mol of the comple
[Cr(H2O)5Cl]Cl2

3 The number of ions in 168 g of Reinecke’s salt, NH4[Cr(NH3)2(SCN)4], of molar mas
336 g mol^-1

16. Which of the following is hydrogen bonded in the liquid state?

1 CH3NH2

2 CH3CHO

3 CH2F2

17. Which of the following mixture produce ND3 gas upon heating?
[D = an isotope of hydrogen]

1 CaO (s) and ND4Cl (s)

2 CH3CN and NaOD in D2O

3 CH3CONH2 and NaOD in D2O

(Partial) Solutions :

Q1. From sample mass of neon gas, find moles of neon gas. Then find moles of protons (1mol of Ne contains 10mol of protons).

Q2. Changing the molarity of a reactant does not alter its half-life. But changing the molarity of other reactants, will.

Q3. Kp and Kc have the same mathematical value, under standard conditions. Alternatively, this MCQ can be solved by elimination.

Q4. From Ksp, find molar solubility. Fe(OH)2 will have a smaller molar solubility.
Note : usually for such questions, you can find the actual value of [OH-] at the point when the less soluble compound precipitates out. Plugging this molarity into the Ksp for the other (more soluble) compound, you can obtain the required molarity of the cation, which will be found to be mathematically larger than the existing molarity at this point.

Q5. At boiling point (or melting point, for that matter), the system is at equilibrium and delta G = 0.

Q6. Combine the two factors mathematically (by multiplying them together)
- number of H atoms subtitutable
- stability of alkyl radical intermediate

Q7. For the first case, branching decreases the surface area available for van der Waals attractions.
For the second case, the greater the number of electrons present, and the greater the molecular size, hence the more polarizable the electron charge clouds, hence the greater the magnitude of dipoles and partial charges, hence the stronger the electrostatic attractions, therefore the stronger the van der Waals.

Q8. From the cell notation, Au is the cathode, Pt is the anode. Cell potential = Reduction potential @ Cathode + Oxidation potential @ Anode.

Q9. The amide N atom and the enolic OH groups do not react with SOCl2.
1 mole of OH- is required for (base-promoted) hydrolysis of the terminal / primary amide, generating NH3. Another 1 mole of OH- is required to substitute away the -NR2 group, generating NR2H (where R = CH3).

Q10. Use algebra (only 1 algebraic variable required, since it's out of 100%) to solve.

Q11. Write two separate equations, one for complete combustion, one for incomplete combustion. Add up total O2 required.

Q12. From data, you can find the ratio of N to O, and also the actual moles of the oxide. Hence solve for x and y (in NxOy).

Q13. For N, each p orbital is only singly filled and hence unpaired.

Q14. Weaker bonds are more readily broken than stronger bonds, and stronger bonds are more readily formed than weaker bonds. This is why you cannot iodinate an alkane via free radical substitution.

Q15. Avogadro constant = 1 mole.

Q16. CH3NH2 can both accept and donate H bonds, while CH3CHO and CH2F2 can only accept H bonds.

Q17. It's all about the mechanism.

1 - CaO (s) and ND4Cl (s). D+ proton transferred from ND4+ to O2-, generating ND3 and OD-.

2 CH3CN and NaOD in D2O. OD- attacks C, pi bond shifts to N, N grabs D+ proton; repeat until you obtain a geminal triol (which dehydrates into a carboxylic acid) and ND2-, which grabs a D+ proton from the carboxylic acid to generate ND3.

3 CH3CONH2 and NaOD in D2O. OD- attacks C, pi bond shifts up, reforms carbonyl group, forming RCOOD while NH2 is eliminated as NH2-, which grabs a D+ proton from RCOOD to generate RCOO- and NH2D.

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KickMe asked :

Why is it that C2H6 has a greater entropy standard molar entropy that C2H4? Why does the stronger id-id interactions in C2H6 not result in a smaller entropy than C2H4?

Answer :

Regarding the van der Waals attractions, interestingly ethane does indeed have a slightly higher boiling point, but ethene has a slightly higher melting point (because melting point has to do with symmetry / stackability as well).

As long as the state (eg. gaseous versus gaseous) is the same, for entropy we do not look at the strength of the intermolecular attractions, but rather we consider the total number of permutations and combinations possible.

The more atoms are present, the more ways to permutate and combine, in terms of atom rotations, bond rotations, bond stretchings, bond bendings, etc. Hence entropy is greater for the molecule with greater number of atoms.

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NaCl is not hydrolyzed by water instaed water just hydrates it. AlCl3 is hydrolyzed by water producing acidic solution. CH3Cl in hydrolzsed by NaOH (aq) and by water to form alcohol... Hydrolysis is the reaction with water or NaOH(aq)?? Please calrify and define hydrolysis.

Technically, hydrolysis refers to the chemical reaction with water (don't worry about the "lysis" part, as either water or the other reactant will inevitably be broken up or "lysed" when the reaction occurs).

However, because water is a weaker nucleophile, therefore to save time and money, we use a stronger nucleophile OH- in nucleophilic substitution reactions involving water, as the richer the electron-rich species, the stronger its nucleophilic strength, the lower the Ea required, the faster the rate of reaction.

Furthermore, the final intended product is the same. For instance :

R-X reacts with water to generate R-OH2+, which loses a proton to X-, hence generating R-OH and H-X as the final products. Higher Ea, slower reaction.

R-X reacts with Na+OH- to generate R-OH, with Na+ and X- as counter ions, generating R-OH and Na+X- as the final products. Lower Ea, faster reaction.

On a related note, notice that physical & inorganic chemists are lazier compared to organic chemists in some ways (eg. physical/inorganic chemists' quick dative bond arrowhead versus organic chemists' full curved-arrow mechanism), but are stricter in other ways (eg. physical/inorganic chemists say "hydration refers to physical interaction with water, while hydrolysis refers to chemical reaction with water"; but organic chemists say, "since dehydration means removing water from alcohol to alkene, then let's just call adding water to alkene as hydration; no need to specify hydrolysis rather than hydration").

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EmpireCity asked :

AlCl3 reacts with AlH4 and (CH3)3N to give (CH3)3NAlH3.
Which statement about (CH3)3NAlH3 is correct?
A It contains hydrogen bonding
B It is dimeric
C The Al atom is electron deficient
D The bonds around Al atom are tetrahedrally arranged.

Why is B untrue when the product has dative bonding?

Mechanism :

Hydride ion (H-) transfers from AlH4- to AlCl3, generating AlH3 (a Lewis acid or 'electrophile'), which is then available for trimethylamine (CH3)3N (a Lewis base or 'nucleophile') to attack, forming the Lewis acid-base product of (CH3)3NAlH3 with AlCl3H- as a byproduct.

This (arguably) isn't considered 'dimerization' because an H- ion is lost from the two 'monomers' when forming the 'dimer' product. (It's a bit of an ambiguous area because you can also argue that polymerization can be either addition polymerization or condensation polymerization). So it's still a rather lousy option to include in this weak question.

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The rate of metabolic removal of paracetamol from the human body is a first order reaction with k = 0.26 / hour. How much time will have elapsed, between taking a paracetamol pill, until only 25% remains in the body?

Solution :

[Final] / [Initial] = ( 1 / 2 ) ^ n where n = number of half lives
1 / 4 = ( 1 / 2 ) ^ n
n = 2

Since half-life = ln2 / k = 2.666 hours

Hence time elapsed = (number of half lives) x (duration of half life) = 2 x 2.666 = 5.332 hours

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The mass percentage of magnesium in a mixture of magnesium chloride and
magnesium nitrate was found to be 21.25%. What mass of magnesium chloride
is present in 100 g of the mixture?

Solution :

% by mass of Mg in MgCl2 = 2.5498 x 10^-1
% by mass of Mg in Mg(NO3)2 = 1.6386 x 10^-1

Since % by mass of Mg in mixture = 21.25%, hence :
21.25 / 100 = X (2.5498x10^-1) + (1-X) (1.6386x10^-1)
X = 53.38 g

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Rate = k [Br2] [NO]^2

With 3 mol/dm3 of NO, time taken for the molarity of Br2 to decrease to 1/8 its original molarity, is 15 hours. If 6 mol/dm3 of NO is used, what's the time taken for the molarity of Br2 to decrease to 1/8 its original molarity?

Solution :

Since 15 hours elapsed to reduce molarity from 1k to 1/8k,

[Final] / [Initial] = (1/2) ^ n
1/8 = (1/2) ^ n
n = 3

Since 3 (old) half-lives = 15 hours, hence 1 (old) half-life = 5 hours.

Since molarity of the second-order reactant is doubled, the rate of reaction increases by (change in molarity) ^ (order of reactant) = 2 ^ 2 = 4 times.

Hence (new) half-life of the other reactant is the (old) half-life decreased by 4 times, ie. 5 / 4 = 1.25 hours.

Since molarity of Br2 is to be reduced from 1k to 1/8k,

[Final] / [Initial] = (1/2) ^ n
1/8 = (1/2) ^ n
n = 3

Since (new) half-live = 1.25 hours, hence time elapsed to decrease molarity of Br2 from 1k to 1/8k = 3 x 1.25 = 3.75 hours.

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2012 H2 Chemistry Paper 1 (Answers)

Q1. C
Q2. C
Q3. D
Q4. B
Q5. C
Q6. A
Q7. D
Q8. C
Q9. A
Q10. D
Q11. D
Q12. C
Q13. B
Q14. A
Q15. A
Q16. D
Q17. D
Q18. C
Q19. C
Q20. C
Q21. D
Q22. B
Q23. B
Q24. D
Q25. C
Q26. A
Q27. C
Q28. B
Q29. B
Q30. A
Q31. B
Q32. D
Q33. A
Q34. B
Q35. B
Q36. B
Q37. D
Q38. A
Q39. B
Q40. B

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Snowmine asked :

Give a reason for each of the following statement:

When testing for sulfate ions, the test solution should be acidified with dilute nitric acid before aqueous barium nitrate is added.

Suggested answer:
The hydrogen ions from the acid will react with carbonate ion to form carbon dioxide.

2H+ + CO2−3 -> CO2 + H2O

This helps to remove carbonate ions and hence prevents the formation of barium carbonate, which is also a white precipitate.

I took this from an assessment book but I feel that the answer is weird.
In the first place, the question didn't mention anything about carbonate ion hence why did the suggested answer talks about carbonate ion instead of sulfate ion?

Can someone please enlighten me! THANKS!

SkyedAngel added :

my chem teacher went through before but I didn't really listen. Something about if you don't remove the carbonate ions then your precipitate of whatever could actually be the carbonate compound not sulfate, end up you get wrong results.

I intervened to assist :

Because carbonate and hydrogen carbonate anions may be generated from the hydrolysis of atmospheric carbon dioxide, and because most carbonates (other than Na, K, NH4+, etc) are insoluble and can result in a false positive white precipitate, which can mislead the student into erroneously thinking sulfate anions are present.

If after acidification, a white precipitate is still observed when barium nitrate is added, you can safely deduce it's due to sulfate ions, and not due to carbonate ions (since acidification will protonate the carbonate ion into carbonic acid, which exists in equilibrium with, and hence can decompose into carbon dioxide and water).

Bonus (for JC students) :
Because carbon dioxide is relatively insoluble in water, due to limited hydrogen bonding and weak van der Waals interactions, and also because being electrophilic it undergoes hydrolysis (ie. attack by water nucleophile) rather than hydration, it readily leaves the reaction mixture, pulling the position of equilibrium over to the right as predicted by Le Chatelier's principle, resulting in dissolving of any carbonate precipitate under acidic conditions. Furthermore, the generation of gaseous carbon dioxide is thermodynamically favoured due to a significantly positive entropy change.

JC H2 Chem students should attempt drawing the mechanisms for both the forward and reverse reactions :

1) hydrolysis of CO2 to generate H2CO3 intermediate which exists in equilibrium with, and partially dissociates into H+ and HCO3- ions (ie. small Ka values, weak monoprotic acid, even weaker diprotic acid).

2) acidification / protonation of a metal carbonate to generate H2CO3 intermediate, which exists in equilibrium with, and hence decomposes into CO2 and H2O (you memorized this equation at Ordinary levels, now prove yourself a worthy Advanced level student by drawing out this mechanism to demonstrate true understanding rather than blind memorizing).

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Both water and ice have H-bonding between their molecules. Ice is less denser than the liquid water because of the longer H-bonds in ice as compared to liquid water and secondly ice has expnaded structure.

What do you mean by the longer H-bonds in ice? Have they been meausred?

Open structure in ice...well the liquid water has also open structure. I did not get this. Please explain.

The H bonds are 'forced' to be longer in ice, due to the rigid lattice structural arrangement in solid ice. No such rigid lattice arrangement occurs in liquid water, which means closer proximity, and therefore shorter (hence stronger) H bonds, are possible for liquid water.

The difference in lengths of the H bonds in solid vs liquid water, is unlikely to be part of the Cambridge mark scheme (at least for the Singapore syllabus). The exact H bond length in solid ice, also depends on the exact cystalline form of ice, for which over a dozen different forms have been identified.

Interestingly, for Cambridge A level Chemistry, students need not be familiar with the most common crystalline form of ice, which is ice 1-hexagonal (because this would be more difficult for A level students), but instead Cambridge allows A level Chem students to draw out or describe the less common, but more familiar (due to the similarity to the lattice structure of diamond, which all 'O' and 'A' level Chem students are already familiar with) crystalline form of ice 1-cubic (ie. the diamond structure).

The so-called 'open structure' of ice, refers to the fact that there are lots of empty (or 'open') spaces between the H2O molecules in solid ice. In contrast, there is very little empty space between H2O molecules in liquid water.

Why? Because H2O molecules are forced to spread apart with some distance between them, in the fixed lattice (ie. regular repeating pattern) crystalline structures in solid ice (analogy : 100 disciplined soldiers doing a weapons precision drill parade, would have to keep some distance apart from each other and occupy a huge space in an open field), while in the liquid state, the H2O molecules are milling in close contact with each other, with no fixed lattice structure (analogy : 100 frenzied women now squeezing chaotically at some super discount sale in some small clothes/shoes departmental store).

This is the reason why solid ice is less dense than liquid water.

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Many thanks for unceasing support and excellent explanation in the field of chemistry in general and A-level chemistry in particular.

Another question from nov 2012:

Q. Rat poison needs to be insoluble in rain water but soluble at the low pH of stomach contents. What is a suitable barium compound to use for rat poison?

A barium carbonate
B barium chloride
C barium hydroxide
D barium sulfate

D is the answer.

And if i am not mistaken I used the salt hydrolysis approach for this question.

A will have ph > 7

B will have ph > 7

C will have ph > 7.

Is it correct?

Totally welcome.

The answer should be BaCO3.

Although both BaSO4 and BaCO3 are insoluble (ie. sparingly soluble) at neutral pH, in acidic pH only BaCO3(s) undergoes a reaction to generate the toxic Ba2+(aq) ions.

The solubility of BaSO4 remains largely unchanged with pH, since Ba(OH)2 and H2SO4 are both considered strong base and strong acid respectively, hence there is little common ion effect or shifting of positions of equilibrium, which are the underlying principles of salt hydrolysis (to generate a significant amount of either the hydroxonium ion or the hydroxide ion, depending on whether the salt is acidic or basic).

The reaction of BaCO3(s) is as follows :

CO3 2- carbonate(IV) ions function as Bronsted-Lowry bases, and are protonated in acidic pH to generate H2CO3 carbonic(IV) acid, which exist in equilibrium with, and hence can decompose into, CO2(g) and H2O(l). Because CO2(g) leaves the reaction mixture, it pulls the position of equilibrium over to the right, as predicted by Le Chatelier's principle. Furthermore, non-gaseous reactants generating a gaseous product implies a positive entropy change of reaction, which is thus thermodynamically favourable and thermodynamically driven.

The toxic Ba2+(aq) is thus freed from the solid BaCO3(s) lattice structure, and available to do its merciless, murderous, lethal work coursing through the arteries and veins of the helpless rodent, who will soon experience multiple organ failure and die a terrible, excruciating, shrieking, convulsive, woeful death. The counter anion for the Ba2+(aq) ion, would be the Cl-(aq) ions of the HCl(aq) acid in the rat's stomach.

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Ah. I had some questions:

1. pH of ethers : acidic, basic or neutral

2. The addition reaction into benzen ring (=>xycloankalines): Is it nucleophllic or electrophillic

Ethers are neutral (ie. neither acidic nor basic). Their conjugate bases are not resonance stabilized (unlike carboxylic acids, phenols or enols). Oxygen atoms are a little too electronegative to be willing to donative a dative bond to be protonated readily (unlike say, N atoms of amines). But between being an acid and being a base, ethers would rather behave as a Bronsted-Lowry base. This behaviour underscores the underlying mechanisms for both generating vs cleavage of ethers :

To cleave ethers using a strong hydrohalic acid :

R-O-R + H-X ---> [R-OH-R]+ + -X ---> R-X + R-O-H

and to generate ethers, with a strong acid catalyst :

R-O-H + R-O-H + H+ ---> [R-OH2]+ + R-O-H ---> [R-OH-R]+ + H2O ---> R-O-R + H2O + H+

Mechanism :

Guys like Girls, hence Nucleophiles like Electrophiles. Only Nucleophiles have the balls (ie. lone pair, or pi bond pair) to attack (ie. donate a dative bond to) Electrophiles.

Benzene is obviously the guy (although he's a weak guy coz his balls are delocalized by resonance), who attacks the girl (though to force chemical intercourse/reaction to occur, you gotta either give the guy more balls = substitute electron-donating by resonance groups onto the benzene ring, or you gotta make the girl more chio = using a Lewis acid catalyst to induce a formal positive charge on the girl), hence all such reactions that directly involve the benzene ring guy shooting out his balls (pi electrons) to any girl electrophile, is called Electrophilic Aromatic Substitution.

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