yazyj
9E1AD864-CBA4-4A0C-8538-A1A0A3A0243A.jpeg  6D497A85-62FF-48C6-9B3F-08F904D0AA05.jpeg  332E3C6A-4A01-4FF9-8BE9-34EA9B69CC40.jpeg

My thought process when doing this question:

-I treated all the elements as if they belonged to period 3.

-I referred to data booklet’s 2nd IE to obtain Al’s 2nd IE for rough guage (1820) 

-Looked at the graph and thought H was Ar and worked backwards to get C which I thought was Al.

could you tell me where I went wrong? Or which steps
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yazyj
Q2: 

When is it possible for molecules to have 2 diff types of bonds? (Other than the familiar ones like inter and intramolecular hydrogen bonds)

And also, what does the water molecule’s ion dipole have to do with oil removal? Why isn’t id-id with R’ not enough to remove oil?

4F22BC75-BB42-4EE3-B87B-B23293952B53.jpeg
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yazyj
Q3: What are the kinds of molecules that undergo dimerisation? And the conditions/ criteria to undergo dimerisation. So far I’m only aware of 2 (AlCl3 and BeCl2)
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yazyj
Q4: 

How do I go about solving such problems? 

88178BB7-F9BD-48BA-B497-0F745EEA00B4.jpeg  86DBACAA-211F-4449-8D8B-DE4E4968455D.jpeg 
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yazyj
Q5: Are all steps in free radical substitution exothermic?
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yazyj
Q6: How else do I know induction is occurring (other than presence of electronegative atom and sigma bonds)? 

And also, when talking about electronegative atom in induction, we rarely talking about any other atoms except halogens. Could other elements have inductive effect too?

And how else do I identify resonance structures?

When discussing about resonance structures, is it a must to talk about delocalisation of electrons? And are partial double bonds always formed during the process?
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yazyj
Q7: Is there a rough rule of thumb to know when to add water as a product to equations?
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BedokFunlandJC

yazyj wrote:
9E1AD864-CBA4-4A0C-8538-A1A0A3A0243A.jpeg  6D497A85-62FF-48C6-9B3F-08F904D0AA05.jpeg  332E3C6A-4A01-4FF9-8BE9-34EA9B69CC40.jpeg

My thought process when doing this question:

-I treated all the elements as if they belonged to period 3.

-I referred to data booklet’s 2nd IE to obtain Al’s 2nd IE for rough guage (1820) 

-Looked at the graph and thought H was Ar and worked backwards to get C which I thought was Al.

could you tell me where I went wrong? Or which steps

 

For such questions, you're supposed to figure out the answer by pattern recognition, rather than using actual IE values from Data Booklet (which should be done only as a last resort, if you've no idea how to proceed without such).

If the graph was for 1st IE, then G would be Group 18 (period 3). But since the graph is for 2nd IE, then G must be Group 1 (period 4), hence working backwards, A is Group 13, ie. Al.

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BedokFunlandJC
yazyj wrote:
Q2: 

When is it possible for molecules to have 2 diff types of bonds? (Other than the familiar ones like inter and intramolecular hydrogen bonds)

And also, what does the water molecule’s ion dipole have to do with oil removal? Why isn’t id-id with R’ not enough to remove oil?

4F22BC75-BB42-4EE3-B87B-B23293952B53.jpeg


Case by case basis, by looking at the functional groups present in a molecule, then consider all possible interactions. Eg. a carboxylate group can either accept a H bond (if a H bond donor is present), or have ionic bond interactions (if a cation is present).

Imagine your hand is oily, and you rub your hands with soap. Without washing away with water, can you get rid of the oil + soap mixture on your hands? This is why soap molecules must no only have favourable interactions with oil molecules, but must also have favourable interactions with water molecules. Hence soap molecules must have both an ionic end (for ion-permanent dipole interactions with water) and a hydrophobic non-polar end (for van der Waals interactions with fats and oils).
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BedokFunlandJC
yazyj wrote:
Q4: 

How do I go about solving such problems? 

88178BB7-F9BD-48BA-B497-0F745EEA00B4.jpeg  86DBACAA-211F-4449-8D8B-DE4E4968455D.jpeg
 



By mathematically combining (ie. multiplying) both factors : No of H atoms on the molecule substitutable to obtain that particular product, as well as kinetic reactivity of pri vs sec vs tert H atoms being substituted.

1st product : any one of the 12 pri H atoms substitutable, hence 12 x 1 = 12
2nd product : any one of the 4 sec H atoms substitutable, hence 4 x 7 = 28
3nd product : any one of the 2 tert H atoms substitutable, hence 2 x 21 = 42
Reduce to lowest ratio = 6 : 14 : 21
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BedokFunlandJC
yazyj wrote:
Q5: Are all steps in free radical substitution exothermic?


Initiation is always endothermic, termination is always exothermic.

Propagation steps are either endothermic (ie. bromination, since the H-Br bond formed is weak) or exothermic (ie. chlorination, since the H-Cl bond formed is strong).

Among the many various possible propagation steps, the ones which are most exothermic and hence most enthalpically favourable, are the ones which are most thermodynamically feasible / spontaneous (ie. most -ve Gibbs free energy change), and hence those propagation steps are the ones which will occur.

Since this Prelim Qn specified only one propagation step, choose the single most feasible (ie. most exothermic or enthalpically favorable) propagation step, generating the most stable secondary alkyl radical (there are 2 little errors in this Prelim paper's Mark Scheme for this Qn, as a less stable primary alkyl radical is chosen instead of the more stable secondary alkyl radical; furthermore the unpaired electron is shown on the wrong atom). 

Free radical iodination cannot occur because the reaction is too endothermic thus too enthapically unfavourable, ie. Gibbs free energy change is +ve hence thermodynamically unfeasible / non-spontaneous.
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