Originally posted by Theplayfulgirl brainy:
Methyl alcohols and methyl ketones undergo iodoform reaction. Can methyl carboxylic acids undergo a similar reaction to generate iodoform?
The key to understanding any organic chemistry reaction, is its mechanism :
Mechanism to oxidize methyl alcohols to methyl ketones (occuring spontaneously under the "alkaline aqueous iodine" conditions of the iodoform test) :
No iodoform generated with methyl carboxylic acids (or esters or amides).
Under the alkaline conditions of the iodoform test, carboxylic acids are deprotonated, generating carboxylate ions, which are thus anionic, rendering the alpha C's protons non-acidic.
The first step of the iodoform mechanism (for methyl ketones; for methyl alcohols there is an additonal prior step which oxidizes the methyl alcohol to methyl ketone) involves alpha deprotonation of the methyl ketone's alpha protons (by the OH- base) to generate the enolate ion nucleophile, which subsequently attacks the diiodine electrophile.
This substitution of the alpha protons with iodine atoms repeats itself until the CH3 group becomes the CI3 group, which is a far better leaving group because CI3- anion has a much lower charge density compared to the CH3- anion.
The penultimate step involves the OH- nucleophile attacking the methyl ketone's acyl/carbonyl C atom, eliminating CI3 as CI3-. The final step involves a proton transfer from the (newly generated) carboxylic acid, to the CI3-, generating the yellow precipitate of triiodomethane (aka iodoform).
However (considering the first step of deprotonation), if you were to use a methyl carboxylic acid (instead of a methyl alcohol or a methyl ketone), a dinegative anionic charge on the conjugate base is simply too destabilizing. Hence methyl carboxylate ions cannot be further deprotonated, and therefore the iodoform reaction cannot proceed with methyl carboxylic acids.
Methyl amides and esters, also do not undergo the iodoform reaction, because the carbonyl O atom already bears additional electron density by resonance (as the O atom of the ester, or N atom of the amide, already donates its lone pair by resonance to the carbonyl group). Hence methyl esters and amides also (like carboxylate ions), are unable to be further deprotonated, and therefore unable to undergo the iodoform reaction.
Originally posted by hoay:
Nov 2006 P2 Q.3(d)
CH2=CHCH2Br is more electrophilic compared to CH2=CHBr.
Suggest an explanation for this.
There are two reasons for this observation (which applies to both aryl halides and vinyl halides).
The main reason is :
Due to the sideways overlap of the p orbital of the Br atom and the pi orbital of the alkene functional group in CH2=CHBr, a lone pair on the Br atom is delocalized by resonance to form a pi bond with the sp2 hybridized C atom of the alkene, resulting in the C-Br bond having partial double bond character (ie. more endothermic bond dissociation enthalpy) in the resonance hybrid, which thus significantly increases the Ea required for nucleophilic substitutions.
The other reason is :
In CH2=CHBr, the incoming nucleophile (eg. OH-) will be repelled by the nearby (ie. close proximity) electron-rich nucleophilic alkene double bond functional group, which thus significantly increases the Ea required for nucleophilic substitutions.
Originally posted by hoay:
Generally speaking increasing the concentration of a reactant incraesing the yield of a product. But if the yield of the product increases will the value of the Kc increases because the reaction moves to formward direction?? If not why?? What can change the value of Kc??
Only changing temperature will change the Kc value. Changing any other variable (eg. moles or molarity or partial pressure of a reactant or product) will change the Qc value.
So even if we say position of equilibrium has shifted, it doesn't necessarily mean the Kc value changed. It could be that Kc value has changed (if we changed temperature), or it could be that Qc value has changed (if we changed the moles or molarity or partial pressure of a reactant or a product).
Either way, when changing either the Qc value or the Kc value, we say the position of equilibrium (ie. Kc) will have shifted, relative to Qc.
If Qc is now less than Kc, the position of equilibrium now lies to the right, and the rate of the forward reaction will exceed the rate of the backward reaction, until Qc = Kc and equilibrium is re-estabilished.
If Qc is now more than Kc, the position of equilibrium now lies to the left, and the rate of the backward reaction will exceed the the rate of the forward reaction, until Qc = Kc and equilibrium is re-estabilished.
Jasmine ngjiamin asked :
Why does diamond have a higher melting point than silicon?
Both species in their standard sates have giant covalent lattice structures.
Hence, melting involves breaking the multitude of covalent bonds in their lattice structures.
In diamond (carbon), the C-C sigma bonds are formed by the more effective overlap using the less diffuse 2nd electron shell's sp3 hybridized orbitals, as opposed to the case of silicon in which the Si-Si sigma bonds are formed by the less effective overlap using the more diffuse 3rd electron shell's sp3 hybridized orbitals.
Consequently, the bond dissociation enthalpy/energy and thus the activation enthalpy/energy involved in melting diamond, is significantly more endothermic compared to that involved in melting silicon.
Originally posted by POP222:
If steam and water has the same volume at 100 degree celsius and atmospheric pressure, will they have equal number of molecules. Are the molecules able to move from water into the steam and from steam into water? It's in a closed setup.
At boiling point (for a given pressure) of a species, Gibbs free energy for the reaction in either direction is zero, and the system is at equilibrium between the liquid and gaseous states. Hence, the rate of the forward reaction (eg. vaporization of water) is equal to the rate of the backward reaction (eg. condensation of water).
Hence, the final answer to your final question is, yes of course.
Originally posted by hoay:
Standard electrode potential is the emf measured between a test electrode and S.H.E.
under S.T.P. Standard redox potential is same as standard electrode potential??
How will we define standard cell potential in words??
There is no standard (pun intended) word definition for cell potential.
For A level purposes, Cambridge only requires students to define electrode potential in words (as you've done), and cell potential by formula, ie.
Cell Potential = Reduction Potential @ Cathode + Oxidation Potential @ Anode.
Cell Potential = Reduction Potential @ Cathode - Reduction Potential @ Anode.
Regarding RTP vs STP, the Data Booklet values are for standard conditions (of 1 bar for all gases, and 1 mol/dm3 for all solutions), and for temperature at 25 deg C.
Originally posted by hoay:
Both Standard electrode potential and standard cell potential are same thing since the equation that you gave above regarding Cell potential will still give the standard electrode potential of any cell in consideration. Is that so?
Technically, the term Standard Electrode Potential refers to either the Oxidation Potential or (more usually, by convention) the Reduction Potential, for a particular Electrode (which could be either the Cathode or the Anode, relative to the SHE), under Standard conditions.
In contrast, the term Standard Cell Potential compulsorily takes into consideration both Redox Potentials, again under Standard conditions, of both Electrodes (ie. both the Cathode and the Anode) that constitute the ElectroChemical (usually Galvanic aka Voltaic) Cell.
Perhaps the source of many students' (and some school teachers' as well) confusion, is that the Standard Electrode Potential = Standard Cell Potential, if and only if, one of the Electrodes in the Cell, happens to be (by deliberate choice) the Standard Hydrogen Electrode (SHE).
Because the SHE's Electrode Potential (either/both Reduction and/or Oxidation Potentials) is designated by humans to be taken as the default zero Volts (ie. 0 V), hence the Standard Cell Potential for such a setup (and only such a setup) = the Electrode Potential for the Electrode being investigated.
See Wikipedia :
Originally posted by CaiHongRainx:
Hii, I have a few qns here.
Miscellanous qn: Do we still need to know what’s a mass spectrometer, NMR spectrum for H2 Chem?
Stoichiometry: A giant molecule contains a large amount of carbon, mainly of isotopes 12C and 13C. It was found that the relative atomic mass of carbon in the molecule is 12.20. What is the ratio of 12C to 13C ?
It happens that this is a MCQ qn. So the approach is just (12×0.8) + (13×0.2) = 12.2.
I was wondering, what if this is not a MCQ qn? I know we can let them be x and y.
So it becomes 12x + 13y = 12.2 . However there is only one eqn.
Gasesous State: Under extereme conditions, gallium chloride Ga2Cl6 can also be formed via the endothermic dimerisation of GaCl3. The two species can exist in dynamic equilibrium under suitable conditions.
2GaCl3(g) two way arrow Ga2Cl6(g)
(1) In an experiment, a total pressure of 7.75×10-3 N m-2 was measured when a 0.661g sample of GaCl3 was allowed to dimerise in a 1.80dm3 vessel at 200 degrees.
Calculate x, the fraction of GaCl3 that dimerised at 200 degrees, assuming ideal gas behaviour.
I apply pV=nRT, where turns n is 0.00354904
Initial mass is 0.661g of GaCl3, n = 0.00375141
Fraction that dimerised, x = (0.00375141 – 0.00354904) / 0.00375141 = 0.053945
But the answer is 0.108! (No working provided)
Thanks for answering!!!
For the 1st qn, let X be the % of C12. Then (X/100)(12)+((100-X)/100)(13) = 12.2 and solve for X.
For the 2nd qn, the number of moles when using PV=nRT is the total number of moles of both gaseous species (ie. both the monomer and the dimer).
2GaCl3 <---> Ga2Cl6
Initial : 0.00375141 and 0
Change : -2X and +X
Equilibrium : (0.00375141 - 2X) and X
Total moles of gaseous species = (0.00375141 - 2X) + X = 0.00354904 (from PV=nRT)
Solve for X.
Originally posted by Cre8ion:
Hi, would appreciate if someone could answer the following question :)
When we calculate Kc of reactions that place in aqueous medium, do we take into account the concentration of water, if it is a reactant or product?
ie. CH3CO2H (l) + C2H5OH (l) <=> CH3CO2C2H5 (l) + H20 (l)
Is the Kc = [CH3CO2C2H5][H20] / [C2H5OH][CH3CO2H] ?
Thanks ! :)
It depends on whether water was already initial present as a solvent (ie. in large excess), or if it functions solely as either a reactant or product.
Notice that in your particular question, it is implied that no water was initially present. Hence water here is a product, and must therefore be included in your Kc expression and formula.
Originally posted by Cre8ion:
Hi, it would be great if I could have some help with the following questions :)
1. Given the reasoning that the standard enthalpy change of neutralisation between weak acids/bases and strong acids/bases is lower as compared to those between strong acids and bases because energy is absorbed to ionize the undissociated weak acid/base, why is the standard enthalpy change of the reaction:
CH3COOH + NH3 -> H2O + CH3COONH4
greater than that of neutralisation between weak acids/bases and strong acids/bases since there should be more energy absorbed to ionize the greater number of undissociated acid/base?
2. Since the units of enthalpy change of reaction is kJ/mol, why does changing the stochiometry coefficients still change the enthalpy change of a reaction? Should this not be analogous to the concept of heat capacity and specific heat capacity, where the value of the specific heat capacity is constant for a particular material, regardless of mass, since the units of specific heat capacity already “accounts” for the mass?
Q1. There is a fundamental misconception with your question. Neutralization enthalpy is defined as per mole of water generated. No water is generated in the reaction between weak acids and weak bases.
Q2. kJ/mol is not per mole of any particular reactant or product, but per mole of the equation.
Convert 3-methylhex-2-ene-1-ol to 2-hydroxy-4-methyl-hepta-1-amine in only and exactly 3 steps.
Answer (mouseover to highlight) :
1. Oxidize to aldehyde using pyridinium chlorochromate (PCC).
2. Nucleophilic addition of CN- (unto the electrophilic acyl carbon), followed by protonation. Reagent required is HCN (in the cold, 10-20 deg C to minimize vaporization of toxic HCN) with trace amount of NaCN or KCN or NaOH or KOH).
3. Simultaneous hydrogenation of both the alkene and nitrile functional groups, using gaseous hydrogen and transition metal catalyst (eg. platinum, palladium or nickel).
Another BedokFunland JC original H2 Chemistry question.
Qn1. Explain why, despite cis-alkenedioic acids (eg. cis-butenedioic acid) being more polar compared to its trans isomer, the cis isomer may in some cases have a lower melting and/or boiling point.
In the cis-alkenedioic acid, some of the molecule's capacity for hydrogen bonding has been used up to form intra-molecular hydrogen bonds, and consequently the cis isomer has less extensive inter-molecular hydrogen bonds, which accordingly require a lower amount of energy to overcome, and hence the cis isomer (despite being more polar) may have a lower melting and/or boiling point compared to the trans isomer. For the same reason, the cis-isomer may have lower solubility (ie. less extensive hydrogen bonding with water) compared to the trans isomer.
Qn2. Explain why, despite cis-alkenes being more polar compared to its trans isomer, and hence understandably having the higher boiling point, may in some cases have the lower melting point.
As a trans-alkene is more symmetrical compared to its cis isomer, the trans isomer packs more efficiently and closely into its crystal lattice solid structure, which consequenty (ie. as a result of being able to achieve greater intermolecular proximities) allows for more extensive and stronger intermolecular van der Waals attractions. Hence, a trans-alkene (being more symmetrical) may have the higher melting point compared to its cis isomer, even as the cis-alkene (being more polar) may have the higher boiling point compared to its trans isomer.
Originally posted by hoay:
Energy is the capacity of doing work. Enthalpy is the total heat content of a system. What is the difference between energy and enthalpy??
For 'A' level Chemistry purposes, there is no difference, and Cambridge will accept either term.
"bond dissociation enthalpy" = "bond dissociation energy",
"bond formation enthalpy" = "bond formation energy",
"lattice dissociation enthalpy" = "lattice dissociation energy",
"lattice formation enthalpy" = "lattice formation energy",
FYI, at higher levels, although there are several different ways to define the technical difference between enthalpy and energy, but for 'A' level Chemistry purposes, it'll more than suffice for the interested student to take it that :
Energy change is heat change under constant volume.
Enthalpy change is heat change under constant pressure.
For most chemical reactions, to 3 significant figures, there is negligible difference between the enthalpy change and energy change values, and thus Cambridge accepts either term used interchangeably.
Ideally when applying Hess Law, it's still good practice (though not compulsory) to standardize all processes in the cycle, in either "enthalpy" or "energy" values, rather than a mixture of both.
(Similarly, "Oxidation Number" versus "Oxidation State"; there is a technical difference at higher levels, but not for 'A' level Chemistry purposes).
Originally posted by hoay:
But what do you eman by the statement that "enathalpy is the total heat content of a system."Does this mean all type of energies contained by a system viz. Potentilalenergy, Kinetic energy, vibrational energy, rotational energy?
All the energies you've mentioned, are included in both "enthalpy" and "energy", but "enthalpy" also measures the system's volume and pressure.
If you need to discuss this further, please visit a dedicated physics or chemistry forum, because anything beyond what's required for A level Chemistry (specifically the Singapore H2 syllabus) is not within my personal interest to discuss.
Originally posted by hoay:
Q.1 The eletrolysis of impure copper is carried out using CuSO4 as electrolyte, copper being pure Cu, while anode as Impure Cu.
The three observations are :
A) Anode dissolves
B) The mass of cathode increases
C) the blue color of solution fades.
Is the third observation correct??
Q.2 What impurities are present in impure Copper?
Q.3 What if the electrolyte is Copper nitrate?
Q.4 What will be the products left after this electrolysis is carried out? I mean the salts which are left behind.
Q1. Incorrect. Any changes to [Cu2+] are negligible, as far as 'O' and 'A' levels are concerned.
Q2. Other metal ions, both more electropositive and less electropositive, than copper.
Q3. No difference, as sulfate(VI) and nitrate(V) anions both have their heteroatoms already in their most positive OS possible, and cannot be oxidized further at the anode (to which they are electrostatically attracted).
Q4. Depends on the specific impurities present. Less electropositive metal ions are reduced at the cathode, while more electropositive metal ions remained in the aqueous oxidized state.
Impurities in this context, are essentially :
other metals which are less electropositive than Cu
other metals which are more electropositive than Cu
The former are mostly already in their reduced state (eg. Ag, Au, etc), and simply drop to the bottom as anodic sludge.
The latter are mostly already in their oxidized state (eg. Fe2+, Al3+, etc), and simply dissolve into the electrolyte solution.
Notice that both groups do not take part in any redox reaction at the anode, and therefore do not deprive the Cu (at the impure anode) from being oxidized at the anode.
Hence, at the anode, Cu is being oxidized to Cu2+, for every 2 moles of electrons transferred.
At the cathode, only Cu2+ is being reduced to Cu (as more electropositive metals, eg. Fe2+, Al3+, etc) prefer to remain in their oxidized, aqueous state.
Hence, at the cathode, Cu2+ is being reduced to Cu, for every 2 moles of electrons transferred.
Therefore, the [Cu2+] in the electrolyte solution remains almost unchanged (barring negligible factors such as evaporation of solvent, negligible oxidation of impurities closely similar to Cu in electropositivity, etc; all of which are not the concern of the 'O' or 'A' levels).
Another BedokFunland JC H2 Chemistry Qn
In organic chemistry reactions, nucleophilic substitution competes with elimination. If the elimination products are desired over the nucleophilic substitution products, what two factors can we use to promote the reaction towards elimination, and away from nucleophilic substitution?
Solvent and Temperature.
Using an ethanol solvation shell promotes the reaction towards elimination, while using an aqueous solvation (ie. hydration) shell, promotes the reaction towards nucleophilic substitution.
This is because the ethanol solvation shell poses greater steric hinderance (compared to a hydration shell), consequently forcing the OH- ion to function as a Bronsted-Lowry base, rather than as a nucleophile.
Because elimination reactions involve a positive entropy change, hence as predicted by the formula for Gibbs free energy, utilizing a higher reaction temperature, promotes elimination (ie. the Gibbs free energy change or delta G value, becomes more negative and the reaction more thermodynamically feasible) to a greater degree, than it does nucleophilic sustitution (which involve relatively negligible entropy change, and whose enthalpy change could be either endothermic or exothermic).
Note that nucleophilic substitution still requires a high temperature (as the breaking of covalent bonds involve a high activation energy barrier), and hence heating (usually under reflux) in both nucleophilic substitution and elimination is still necessary; it's just that the higher the temperature, the greater the tendency towards elimination because of the positive entropy change.
Another BedokFunland JC Question
What is the difference between "peptide" and "amide"? When asked to label a -CONH- group, which word should be written?
What does the term "secondary structure" of a protein refer to?
Both words refer to the -CONH- group, but peptide refers to this group within proteins, while amide refers to this group (everywhere else) outside proteins.
To be more specific, peptide groups are -CONH- generated from the nucleophilic acyl substitution / addition-elimination / condensation reaction, between the alpha carboxylic acid COOH group of one (alpha) amino acid, with the alpha amine NH2 group of another (alpha) amino acid, within a protein.
Amide groups refer to all other types of -CONH- groups, most usually those in synthetic molecules, eg. drugs and polymers.
Special mention : when (in the theoretical event of) a -CONH- group is generated from the reaction between one amino acid's R group's COOH group, and another amino acid's R group's NH2 group, (or one alpha group reacting with one R group), the consequent -CONH- group is still regarded as an amide group, rather than a peptide group (peptide groups are generated strictly from alpha COOH and alpha NH2 groups).
On a related note, when asked what the "secondary structure" of proteins refers to, Cambridge requires the H2 Chem student to specify, "A protein's secondary structure refers to the alpha-helix or beta-pleated sheet structure of a polypeptide, generated as a consequece of the hydrogen bonding between peptide groups (and not the R groups) of amino acids in a protein."