Mr yeong I still don’t understand the liquid + liquid = liquid + liquid. The notes say the density and concentration of pure liquid is constant. why do we include inside the Kc?

In summary, a phenyl group is a specific type of arene group, but not all arenes are phenyl groups.

Here’s why:

1. Clockwise Path: When following the arrow in the clockwise direction from the 4th carbon, you encounter a specific set of atoms/groups in a particular order.

2. Counterclockwise Path: When following the arrow in the counterclockwise direction, you encounter a different set of atoms/groups in a different order.

   
   

These different paths indicate that the two sides of the molecule differ in terms of their connectivity or substituent groups, which is what makes the 4th carbon a chiral center.

Yes, if you swap the two substituents on the line and the wedge/dash instead of the ones on the dash and wedge, you will also reverse the configuration of the chiral center. Any swap of two groups at a chiral center, regardless of which specific pairs of bonds are involved (line, wedge, or dash), will invert the configuration from R to S or S to R.

FYI: not in syllabus

How to determine R and S configuration (the 2 enantiomers)

https://www.chemistrysteps.com/how-to-determine-r-and-s-configuration/

Change of atomisation of an element : the element is in its standard state

Then we do not have to mention ‘giant ionic lattice breaking down and ions becoming free to move, leading to increase in entropy’?

In contrast, when going from P to S, the added electron enters a 3p orbital that already contains one electron. This means pairing occurs for the first time in the 3p subshell, introducing intra-orbital repulsion.

The configuration becomes less stable — it loses the half-filled symmetry of P (3p³), becoming neither half-filled nor fully filled (like 3p⁶). This increased repulsion and loss of symmetry lead to a dip in ionisation energy.

And is isoelectric point = equivalence point?

But we did not learn oxidation of amides so do we need to know this?

is it because if the 1.77 and 0.68 equations are chosen, it would be a disproportionation reaction since Ecell would still be positive? Or is it because in this qn Fe2+ must be oxidised

And is pKa + pKb = pKw = 14 all the time?

And Ka x Kb = Kw?

Objective 9 -> Ka is a better measure of acid strength, compared to a and pH

For carboxylic acid i can use Na2CO3 and effervescence of CO2 is observed

For acyl chloride i can use H2O and white fumes of HCl will be evolved

But i cannot think of a test to identify ester

why does having low ionisation energy and low electronegativity result in the suitabililty of reducing agent?

will there be a case where there is low ionisation energy but high electronegativity vice versa? if yes, how do we know the suitability of reducing agent then?

(b) Why LiAlH4 is a stronger reducing agent than NaBH4?

That two groups on the benzene ring have to be in the 1,2–positions, in order for the reaction to take place to form the cyclic compound E. Also, do I not have to factor in how the substituents direct the incoming bromine when forming the compound

• pyridine and pyrrole --> compare the availability of the lone pair in the p orbital of N atom

• H-C≡C-H vs H2C=CH2 vs H3C-CH3 --> compare the s character of the hybridised orbital and hence the distance of the lone pair from the nucleus

This is beyond our usual methods of comparing acidity and basicity of compounds via resonance or inductive effect

• EWG/EDG, and relative strength due to electronegativity (e.g. Cl vs F)

May I know when we are supposed to use which method? How do we identify what type of comparison we should be using, and for which type of questions?

1) Why is it that the pink is correct and blue is wrong for ii)

2) would I be marked down if I wrote blue instead in the Alevels

3) my confusion is since relative Mr is also the sum of relative Ar of all atoms in the compound, and in this case as n is u know shouldn’t I leave my answer in terms of n

eg conc or volume of reagents

also isnt I2 having a large atomic radius it has the least effective orbital overlap and thus least energy required to overcome the bond? how come I2 has a largest boiling point?

And if this is true, can it be used to answer the question of why bromine reacts less vigorously than chlorine with alkanes or should we stick to talking about about enthalpy change.

Or is the carbon in -CH2^- still sp3 hybridised and does not have an unhybridised p orbital that overlaps with the adjacent -COOH group.

then, comparing amides and acrylic chlorides, why is N able to delocalise its lone pair of electrons into the pi bond of c=o but Cl in cocl can’t?

Explain what isoelectric point is a nd able to locate where is it on the titration curve?

Able to sketch the titration curve when NaOH is added to an amino acid

With the aid of equations to show how buffer for an amino acid (e.g. glutamic acid above) works at buffer region at pH = 4.1

Statement 2 is incorrect as there is only 1 planar intermediate formed. Can you explain this statement in greater detail